[
    {
        "image_filename": "designv10_0_0001975_tro.2013.2281564-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001975_tro.2013.2281564-Figure4-1.png",
        "caption": "Fig. 4. (a) Cut-away view of cable-disk contact. (b) Assumed belt friction model for cable-disk contact.",
        "texts": [
            " This enables the model to accurately represent the dynamic sliding friction when the cable sliding velocity is not near zero, while ensuring a continuous force profile when the sliding direction changes. This approach does not allow for a greater static friction than dynamic sliding frictional force, as is normally observed in mechanical systems, but significantly simplifies the dynamic analysis in this initial study. Due to the \u201cwrap\u201d of the cabling around the cable routing holes, as illustrated in Fig. 4(a), a belt friction model is used in this analysis. Fig. 4(b) shows an illustration of the key model parameters, and (31), shown below, provides the numerical expression of the model, where \u03bc is the coefficient of saturated viscous friction and \u03b7i,j is the contact angle defined in (32), shown below. Based on (31), (33), shown below, may be formulated to determine the magnitude of the frictional force F i,j,f r at the jth hole of the ith disk. This magnitude is the difference in the left and right tensions Ti,j and T(i+1),j . A key benefit of this approach is that it allows the solver to estimate the frictional force magnitude with estimates for the left and right tensions"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure3.54-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure3.54-1.png",
        "caption": "FIGURE 3.54. Lateral force ratio Fy/Fz as a function of slip ratio s for different sideslip \u03b1.",
        "texts": [
            " However, as long as the angles and slips are within the linear range of tire behavior, a superposition can be utilized to estimate the output forces. Driving and braking forces change the lateral force Fy generated at any sideslip angle \u03b1. This is because the longitudinal force pulls the tireprint in the direction of the driving or braking force and hence, the length of lateral displacement of the tireprint will also change. Figure 3.53 illustrates how a sideslip \u03b1 affects the longitudinal force ratio Fx/Fz as a function of slip ratio s. Figure 3.54 illustrates the effect of sideslip \u03b1 on the lateral force ratio Fy/Fz as a function of slip ratio s. Figure 3.55 and 3.56 illustrate the same force ratios as Figures 3.53 and 3.54 when the slip ratio s is a parameter. Proof. Consider a turning tire under a sideslip angle \u03b1. The tire develops a lateral force Fy = \u2212C\u03b1 \u03b1. Applying a driving or braking force on this tire will reduce the lateral force while developing a longitudinal force Fx = \u03bcx (s) Fz. Experimental data shows that the reduction in lateral force in presence of a slip ratio s is similar to Figure 3.54. Now assume the sideslip \u03b1 is reduced to zero. Reduction \u03b1 will increase the longitudinal force while decreasing the lateral force. Increasing the longitudinal force is experimentally similar to Figure 3.55. A turning tire under a slip ratio s develops a longitudinal force Fx = \u03bcx (s) Fz. Applying a sideslip angle \u03b1 will reduce the longitudinal force while developing a lateral force. Experimental data shows that the reduction in longitudinal force in presence of a sideslip \u03b1 is similar to Figure 3.53. Now assume the slip ratio s and hence, the driving or breaking force is reduced to zero. Reduction s will increase the lateral force while decreasing the longitudinal force. Increasing the lateral force is similar to Figure 3.54. Example 116 Pacejka model. An approximate equation is presented to describe force Equations (3.175) 3. Tire Dynamics 153 154 3. Tire Dynamics 3. Tire Dynamics 155 or (3.176). This equation is called the Pacejka model. F = A sin \u00a9 B tan\u22121 \u00a3 Cx\u2212D \u00a1 Cx\u2212 tan\u22121 (Cx) \u00a2\u00a4\u00aa (3.183) A = \u03bcFz (3.184) C = C\u03b1 AB (3.185) B,D = shape factors (3.186) The Pacejka model is substantially empirical. However, when the parameters A, B, C, D, C1, and C2 are determined for a tire, the equation expresses the tire behavior well enough",
            " Assume Fz = 4000N and we need a lateral force Fy = \u22123000N. If \u03b1 = 4deg, what would be the required camber angle \u03b3? Estimate the coefficients C\u03b1 and C\u03b3 . 8. High camber angle. Consider a tire with C\u03b3 = 300N/deg and C\u03b1 = 700N/deg. If the camber angle is \u03b3 = 18deg how much lateral force will develop for a zero sideslip angle? How much sideslip angle is needed to reduce the value of the lateral force to Fy = \u22123000N? 3. Tire Dynamics 163 9. Sideslip and longitudinal slip. Consider the tire for which we have estimated the behavior shown in Figure 3.54. Assume a vehicle with that tire is turning with a constant speed on a circle such that \u03b1 = 4deg. What should be the sideslip angle \u03b1 if we accelerate the vehicle such that s = 0.05, or decelerate the vehicle such that s = \u22120.05? 10. F Motion of the air in tire. What do you think about the motion of the pressurized air within the tires, when the vehicle moves with constant velocity or constant acceleration? 4 The maximum achievable acceleration of a vehicle is limited by two factors: maximum torque at driving wheels, and maximum traction force at tireprints"
        ],
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    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure2.8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure2.8-1.png",
        "caption": "Fig. 2.8. Single-layer winding of an AFPM machine with m1 = 3, 2p = 6, s1 = 36, y1 = Q1 = 6 and q1 = 2.",
        "texts": [
            "7) The distribution factor of a polyphase winding for the fundamental space harmonic \u03bd = 1 is defined as the ratio of the phasor sum\u2014to\u2014arithmetic sum of EMFs induced in each coil and expressed by the following equation: kd1 = sin(\u03c0/2m1) q1 sin[\u03c0/(2m1q1)] (2.8) The pitch factor for the fundamental space harmonic \u03bd = 1 is defined as the ratio of the phasor sum\u2014to\u2014arithmetic sum of the EMFs per coil side and expressed as: kp1 = sin ( \u03b2 \u03c0 2 ) (2.9) The winding factor for fundamental is the product of the distribution factor (eqn (2.8)) times pitch factor (eqn (2.9)), i.e. kw1 = kd1kp1 (2.10) The angle in electrical degrees between neighbouring slots is \u03b3 = 360o s1 p (2.11) Fig. 2.8 shows a single layer winding distributed in s1 = 36 slots of a threephase, 2p = 6 AFPM machine. Toroidal stator windings are used in twin-rotor double-sided AFPM machines (Fig. 2.4). The toroidal stator winding of a three-phase, six-pole AFPM machine with twin external rotor is shown in Fig. 2.9. Each phase of the winding has an equal number of coils connected in opposition so as to cancel the possible flux circulation in the stator core. Those coils are evenly distributed along the stator core diametrically opposing each other so the only possible number of poles are 2, 6, 10, ",
            " General equations given in Chapter 2 for the performance calculations will be developed further and adjusted to the construction of AFPM machines with ferromagnetic cores. Application of the FEM analysis to performance calculations is also emphasized. Single-sided AFPM machines with stator ferromagnetic cores have a single PM rotor disc opposite to a single stator unit consisting of a polyphase winding and ferromagnetic core (Fig. 2.1). The stator ferromagnetic cores can be slotted or slotless. The stator winding is always made of flat wound coils (Fig. 2.8). The PMs can be mounted on the surface of the rotor or embedded (buried) in the rotor disc. In the case of a slotless stator the magnets are almost always surface mounted, while in the case of a slotted stator with a small air gap between the rotor and stator core, the magnets can be either surface mounted on the disc (Fig. 2.1) or buried in the rotor disc (Fig. 2.6). Large axial magnetic forces on bearings are the main drawback of single-sided AFPM machines with ferromagnetic stator cores. In double-sided AFPM machines with ideal mechanical and magnetic symmetry, the axial magnetic forces are balanced",
            "3) or outer PM rotor discs with iron-cored stator fixed in the middle (Figs 2.4, 2.5 and 2.6). As with single-sided AFPM machines the stator ferromagnetic cores can be slotted or slotless, and the rotor magnets can be surface mounted, embedded or buried 124 4 AFPM Machines With Iron Cores (Fig. 2.6). Again, in the case of a slotless stator with a large air gap between the rotor and stator core the magnets are almost always surface mounted. The stator windings of double-sided AFPM machines can be flat wound (slotted or slotless) as shown in Fig. 2.8 or toroidally wound (normally slotless) as shown in Fig. 2.9. An example of a commercial double-sided AFPM servo motor with ferromagnetic core is shown in Figs. 4.1 and 4.2. External stators have slotted ring-shaped cores made of nonoriented electrotechnical steel ribbon. The inner rotor does not have any ferromagnetic material. PMs are mounted on a nonmagnetic rotating disc. Fig. 4.3 shows a double-sided AFPM synchronous generator with the stator core wound of amorphous alloy ribbon manufactured by LE Incorporated , Indianapolis, IN, U"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure13.13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure13.13-1.png",
        "caption": "FIGURE 13.13. A full car vibrating model of a vehicle.",
        "texts": [
            "269) Comparing these results with the results in example 490, shows the antiroll bar affects only the roll mode of vibration. A half car model needs a proper antiroll bar to increase the roll natural frequency. It is recommended to have the roll mode as close as possible to the body bounce natural frequency to have narrow resonance zone around the body bounce. Avoiding a narrow resonance zone would be simpler. 13.6 Full Car Vibrating Model A general vibrating model of a vehicle is called the full car model. Such a model, that is shown in Figure 13.13, includes the body bounce x, body roll \u03d5, body pitch \u03b8, wheels hop x1, x2, x3, and x4 and independent road excitations y1, y2, y3, and y4. A full car vibrating model has seven DOF with the following equations 13. Vehicle Vibrations 865 of motion. mx\u0308+ cf \u00b3 x\u0307\u2212 x\u03071 + b1\u03d5\u0307\u2212 a1\u03b8\u0307 \u00b4 + cf \u00b3 x\u0307\u2212 x\u03072 \u2212 b2\u03d5\u0307\u2212 a1\u03b8\u0307 \u00b4 +cr \u00b3 x\u0307\u2212 x\u03073 \u2212 b1\u03d5\u0307+ a2\u03b8\u0307 \u00b4 + cr \u00b3 x\u0307\u2212 x\u03074 + b2\u03d5\u0307+ a2\u03b8\u0307 \u00b4 +kf (x\u2212 x1 + b1\u03d5\u2212 a1\u03b8) + kf (x\u2212 x2 \u2212 b2\u03d5\u2212 a1\u03b8) +kr (x\u2212 x3 \u2212 b1\u03d5+ a2\u03b8) + kr (x\u2212 x4 + b2\u03d5+ a2\u03b8) = 0 (13.270) Ix\u03d5\u0308+ b1cf \u00b3 x\u0307\u2212 x\u03071 + b1\u03d5\u0307\u2212 a1\u03b8\u0307 \u00b4 \u2212 b2cf \u00b3 x\u0307\u2212 x\u03072 \u2212 b2\u03d5\u0307\u2212 a1\u03b8\u0307 \u00b4 \u2212b1cr \u00b3 x\u0307\u2212 x\u03073 \u2212 b1\u03d5\u0307+ a2\u03b8\u0307 \u00b4 + b2cr \u00b3 x\u0307\u2212 x\u03074 + b2\u03d5\u0307+ a2\u03b8\u0307 \u00b4 +b1kf (x\u2212 x1 + b1\u03d5\u2212 a1\u03b8)\u2212 b2kf (x\u2212 x2 \u2212 b2\u03d5\u2212 a1\u03b8) \u2212b1kr (x\u2212 x3 \u2212 b1\u03d5+ a2\u03b8) + b2kr (x\u2212 x4 + b2\u03d5+ a2\u03b8) +kR \u00b5 \u03d5\u2212 x1 \u2212 x2 w \u00b6 = 0 (13"
        ],
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    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure6.25-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure6.25-1.png",
        "caption": "Figure 6.25 (a) The total roof load is 2.8 kN/m2; (b) this generates a uniformly distributed load on the beam equal to 5.6 kN/m.",
        "texts": [
            "24b shows one of the frames. Each frame consists of an 8- metre beam that is simply supported at both ends by a column. The 5-metre columns are rigidly joined to a square footing, located at a certain depth below ground level. The dead weight of the roof slabs is 2 kN/m2. The weight of the waterproof roof covering and insulation is set at 0.3 kN/m2. In addition, a live load of 0.5 kN/m2 is taken into account. The total load on the roof slabs is therefore p = (2 kN/m2) + 0.3 kN/m2) + (0.5 kN/m2) = 2.8 kN/m2. Figure 6.25a shows a load of 2.8 kN acting on a square metre. If one takes an arbitrary strip of the roof of 1-metre width, the total load on the strip would be (4 m)(1 m)(2.8 kN/m2) = 11.2 kN. Each beam carries half of this, or in other words, 5.6 kN over a 1-metre length, (see Figure 6.25b). Over the full length, the beam is therefore loaded by a uniformly distributed load of 5.6 kN/m. We also have to include the dead weight of the beam. If we assume a dead weight of 6 kN/m, the total 232 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM load on the beam is (see Figure 6.26) q = (5.6 kN/m) + (6 kN/m) = 11.6 kN/m. The beam is simply supported. The support reactions, which have to be provided by the columns, amount to 1 2 \u00d7 (11.6 kN/m)(8 m) = 46.4 kN. Equal and opposite forces are acting on the columns"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001456_j.commatsci.2017.03.053-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001456_j.commatsci.2017.03.053-Figure3-1.png",
        "caption": "Fig. 3. Schematic of a molten zone and HAZ scanning through a 1-layer thick 3D domain, two rasters in the layer have fully traversed the domain and the third is nearing completion.",
        "texts": [
            " Remaining modifications required to simulate the multi-pass and multi-layer nature of AM processes include scan pattern details such as cross-hatching, hatch spacing, number of scans per layer, and layer height. These are incorporated into the model as adjustable user inputs. The model incorporates layer remelting and epitaxial growth between layers through adjustment of the molten zone width W, length L, depth D, and HAZ. A snapshot of a molten zone and HAZ traveling through a simulation domain is shown in Fig. 3. Finally, while the AM process is by definition, \u2018\u2018additive\u201d, simulation domains typically do not grow with time. To address this inconsistency, inactive regions are \u2018\u2018masked\u201d and evolution of these sites is not allowed prior to initial interaction with the molten zone. This masking does not affect time-evolution of the simulation, which is advanced by a fixed interval each time step. Sites become \u2018\u2018unmasked\u201d and evolve when they fall within or below the active build layer. As simulation time progresses, initial spins within each new layer are completely disordered, and nonperiodic boundary conditions are used to prevent unphysical interactions at simulation boundaries"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure8.15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure8.15-1.png",
        "caption": "Fig. 8.15. Adept-One robot with direct drives Fig. 8.16. Vane actuator (ADEPT - USA)",
        "texts": [
            " Such angular electromagnets, or motors, are applied (in 1986) only in the case of a smaller assembly robot as in the case of the Adept Company (USA). This demonstrates the only basic deficiency in the form of a very complex control system, only able to control such a system in stable manner with hig4torque and low inertia characteristics. They also still possess a rather high mass, so generally only the degrees of freedom near the robot pedestal can be driven directly by that type of actuator. This is also the case with the Adept-one and Adept-two robot of the above-mentioned firm, illustrated in Fig. 8.15. 240 8 Elements, Structures and Application of Industrial Robots In the case of the hydraulic robot drives for powering the gripper degrees of freedom the so-called Vane-type actuators are often used (Fig. 8.16) which in the one-vane version work through an angle of up to 3000 and in the two-vane version up to 150\u00b0. These actuators are basically of very simple design but demand high technology for the internal surface treatment and manufacture of the linear seals along the vane's edges. With smaller assembly robots (reaching 0"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure6.19-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure6.19-1.png",
        "caption": "FIGURE 6.19. Limit position for a slider-crank mechanism.",
        "texts": [
            " Applied Mechanisms The acceleration of point B with respect to point A is GaB/A = G\u03b13 \u00d7 Gr3 + G\u03c93 \u00d7 \u00a1 G\u03c93 \u00d7 Gr3 \u00a2 = \u23a1\u23a3 \u2212b\u03b13 sin \u03b83 \u2212 b\u03c923 cos \u03b83 b\u03b13 cos \u03b83 \u2212 b\u03c923 sin \u03b83 0 \u23a4\u23a6 . (6.160) Example 236 Limit positions for a slider-crank mechanism. When the output slider of a slider-crank mechanism stops while the input link can turn, we say the slider is at a limit position. It happens when the angle between the input and coupler links is either 180 deg or 360 deg. Limit positions of a slider-crank mechanism are usually dictated by the design requirements. A limit position for a slider-crank mechanism is shown in Figure 6.19. We show the limit angle of the input link by \u03b82L1 , \u03b82L2 , and the corresponding horizontal distance of the slider by sMax, smin. They can be calculated by the following equations: \u03b82L1 = sin\u22121 \u2219 e b+ a \u00b8 (6.161) sMax = q (b+ a) 2 \u2212 e2 (6.162) \u03b82L2 = sin\u22121 \u2219 e b\u2212 a \u00b8 (6.163) smin = q (b\u2212 a) 2 \u2212 e2 (6.164) The length of stroke that the slider travels repeatedly would be s = sMax \u2212 smin = q (b+ a) 2 \u2212 e2 \u2212 q (b\u2212 a) 2 \u2212 e2. (6.165) 6. Applied Mechanisms 339 Example 237 F Quick return slider-crank mechanism"
        ],
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    },
    {
        "image_filename": "designv10_0_0001967_tcsi.2018.2797257-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001967_tcsi.2018.2797257-Figure2-1.png",
        "caption": "Fig. 2. DC motor.",
        "texts": [
            " Due to the abrupt change of actual operating parameters and the time-varying characteristics of TR, the DC motor model can be described by S-MJSs. In addition, SMC is an influential and robust tool in the stabilization and tracking control of the nonlinear dynamical systems and disturbances, which has some advantages in the robustness to the uncertainties, insensitivity to the external disturbances, and fast convergence. In the control process of DC motor model, we will apply the SMC approach to obtain better performance. Consider the DC motor model in Fig. 2, taken from [14], described by x\u03071(t) = \u2212bi ji x1(t) + Kt ji x2(t), u(t) = Kv x1(t) + Rx2(t), y(t) = x2(t), (46) where the state vectors are given as x1(t) = v(t) and x2(t) = i(t) standing for speed of shaft and electric current; Kv , Kt and R denote electromotive force, torque constant, and electric resistor, respectively. Further, ji and bi are defined as ji = Jm + jci n2 and bi = bm + bci n2 . Also, the DC motor system follows a SMP {rt , t \u2265 0} taking value in S = {1, 2} with the TR matrices given as \u03bc = [\u22120"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure5.25-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure5.25-1.png",
        "caption": "Figure 5.25 The hinge forces at S follow from the force equilibrium of AS or BS.",
        "texts": [
            " The magnitude of the horizontal support reactions follow from the moment equilibrium about S of one of the parts AS or BS. 172 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM If one selects the left-hand part AS, this gives \u2211 T (AS) z |S = Ah \u00d7 4 \u2212 Av \u00d7 4 = 0 (d) or, if one assumes the right-hand part BS \u2211 T (BS) z |S = \u221260 \u00d7 2 \u2212 Bh \u00d7 4 + Bv \u00d7 4 = 0. (e) Both equations are equivalent. The solution is Ah = Bh = 15 kN. All the support reactions are shown in Figure 5.24c. b. The interaction forces in hinge S follow from the force equilibrium of AS or BS (see Figure 5.25). The equilibrium of the left-hand part AS gives Sh = Sv = 15 kN. Check: For these forces, the left-hand part BS is also in equilibrium. c. To find the forces acting on joint D, the joint is isolated (see Figure 5.26a). There are three interaction forces acting between joint D and member SD. The magnitude of these forces is found from the equilibrium of member SD. In the same way, one can use the equilibrium of BD to find the magnitude of the three interaction forces between joint D and member BD"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.35-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.35-1.png",
        "caption": "FIGURE 7.35. Illustration of a positive four-wheel steering vehicle in a left turn.",
        "texts": [
            "81) Similarly, we may eliminate R1 R1 = 1 2 wr + c2 tan \u03b4ir (7.82) = \u22121 2 wr + c2 tan \u03b4or (7.83) between (7.77) and (7.78) to provide the kinematic condition between the rear steering angles \u03b4ir and \u03b4or. cot \u03b4or \u2212 cot \u03b4ir = wr c2 (7.84) Using the following constraint c1 \u2212 c2 = l (7.85) 7. Steering Dynamics 413 we may combine Equations (7.81) and (7.84) wf cot \u03b4of \u2212 cot \u03b4if \u2212 wr cot \u03b4or \u2212 cot \u03b4ir = l (7.86) to find the kinematic condition (7.73) between the steer angles of the front and rear wheels for a positive 4WS vehicle. Figure 7.35 illustrates a negative 4WS vehicle in a left turn. The turning center O is on the left, and the inner wheels are the left wheels that are closer to the turning center. The front inner and outer steer angles \u03b4if , \u03b4of may be calculated from the triangles 4OAE and 4OBF , while the rear inner and outer steer angles \u03b4ir, \u03b4or may be calculated from the triangles 4ODG and 4OCH as follows. tan \u03b4if = c1 R1 \u2212 wf 2 (7.87) tan \u03b4of = c1 R1 + wf 2 (7.88) 414 7. Steering Dynamics \u2212 tan \u03b4ir = \u2212c2 R1 \u2212 wr 2 (7"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003774_j.1469-185x.1974.tb01299.x-Figure13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003774_j.1469-185x.1974.tb01299.x-Figure13-1.png",
        "caption": "Fig. 13. Illustrations of the envelope model for a spherical micro-organism. (Q) The arrow represents the direction of the metachronal wave. --, The instantaneous surface over the organism. -.-, The mean position of the envelope is represented by the surface I = Q; - ,The actual surface of the organism. Only the upper hemisphere has been included as the flow is axisymmetric.",
        "texts": [
            " These lines are the crests of the metachronal waves, and they show up most clearly under darkground illumination. The wave crests are composed of many cilia moving together in their effective stroke (see Tamm & Horridge, 1970). The cilia that comprise these waves show a beat pattern that can be represented approximately by Fig. 4b, if one neglects the (minor) lateral movements in the three-dimensional beat, and they lie Mechanics of ciliary motion 91 close together throughout their beat, so that the wave has a continuous outline similar to that shown later in Fig. 13b. There is a strong suggestion of mechanical metachronal coordination by least mutual interference because adjacent cilia are closely packed and tend to be nearly parallel with one another. Because of the form of the waves the cilia are unlikely to exert much individual effect on the fluid, but the movement of the whole wave over the body as a nearly solid ridge must move the water quite effectively (Sleigh, 1962; Parducz, 1967). Between one and four waves pass a given point in each second (i.e. the frequency of beat is 1-4 Hz), and these waves move at roughly equal intervals and even speeds of 1oo-400pm/s, although sometimes waves are seen to coalesce posteriorly",
            " This is because the cilium anterior to the beating one has just completed its stroke, while the posterior cilium is beginning its effective beat, thus giving the appearance that the central cilium (beating one) is bisecting the angle made by the other two (i.e. the cilia converge). I t is apparent from Fig. 3 (a) that the cilium has very little freedom of movement in the effective stroke in symplectic metachronism. In the recovery stroke, most of the length of the cilium is spread out along the organisms\u2019 surface, so again the cilia will be close together. Following these observations, we have used an extensible envelope (Section V) over the surface of the numerous oscillating cilia (see Fig. 13) to model the exterior flow field. In other metachronal wave patterns, such as the antiplectic or diaplectic types, the cilia in the effective stroke tend to beat more independently, so for a correct analysis of their movement the influence of the individual cilia should be incorporated in the model. A model more suitable to these metachronal wave patterns is discussed in detail in Section VI. The fluid-mechanical principles of flow around an oscillating elongated body such as a cilium or flagellum are determined in part by the low numerical values of the cilium Reynolds number Rec",
            " Fluid motion around micro-organisms can be easily observed, so we can compare our theoretical models with experimental observations, but there is little information on fluid flow in the tubes of the reproductive and respiratory tracts. If the theory correlates reasonably well with the observations, we can extend our models to flow in tubes to help understanding of ciliary flows. The two most important mathematical models used for the envelope approach were the spherical and planar ones. For the spherical model (see Fig. 13) we are able to take into account the influence that finiteness has on the organism's velocity, while in the case of the planar model we can say something about the influence of planeness on propulsive velocities. The correct theoretical model we may expect to be somewhere between these two extremes. Many organisms are, in fact, either planar or elongated (e.g. Opalina and Puvamecium) so that the cilia can beat more effectively in the direction of propulsion. The first model that we will discuss using the envelope approach to ciliary propulsion is the one with spherical geometry. The theoretical calculations can be found in the paper by Blake ( I ~ ~ I U ) , which follows the approach used by Lighthill (1952). I t was found that metachronal wave shapes that were peaked (see Fig. 13c) with a form similar to that in Opalina produced velocities between 50 and IOO ,um/s. The direction in which the organism was propelled was opposite to that of the metachronal wave, as is seen in organisms with symplectic metachronism. These velocities compare favourably with those observed for Opalina, which exhibits symplectic metachronism. It should be pointed out though that not all wave shapes produced such large velocities. Indeed, with some wave shapes the organism was propelled in the same direction as the metachronal wave so that the metachronism was antiplectic, but the predicted velocities in these cases bear no comparison to those observed for micro-organisms in nature"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure1.11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure1.11-1.png",
        "caption": "Figure 1.11 The Oosterschelde barrier; the subsoil has to be able to bear the structure.",
        "texts": [
            " The division between structural mechanics and the mechanics of materials is only effective for so-called statically-determinate structures,2 or structures in which the force flow can be determined directly from the equilibrium. For calculations relating to structures other than those that are statically-determinate (so-called statically-indeterminate structures) one has to use elements from both structural mechanics and the mechanics of materials. The behaviour of a structure must be investigated \u201cbeyond the base\u201d. For example, it is important that the slides in the Oosterschelde barrier in Figure 1.11 are sufficiently strong, but it is equally important that the structure can be properly carried by the subsoil. Since the behaviour of soil clearly 1 Stability is defined as the reliability of the equilibrium. Since the stability of the equilibrium depends on the stiffness of the structure, the stability demand can also be interpreted as a stiffness demand. 2 The concepts statically-determinate and statically-indeterminate are covered in more detail in Chapter 4. 1 Introduction 7 differs from that of regular solid material, the investigation into the forces and deformations in soil is part of a separate field of expertise known as soil mechanics"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure3.59-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure3.59-1.png",
        "caption": "Fig. 3.59: Kinematic design for: a) CaPaMan 2; b) CaPaMan 3.",
        "texts": [
            " Finally, the size of MP and FP are given by rp and rf, respectively, where H is the center point of MP, O is the center point of FP, Hk is the center point of the k-th BJ, and Ok is the middle point of the frame link ak. Platform MP is driven by the three leg mechanisms through the corresponding articulation points H1, H2, H3. A motor on the input crankshaft can actuate each leg mechanism so that the device is a 3-d.o.f.s spatial mechanism. In addition, the kinematic architecture has been useful to propose an alternative lowcost solution for the leg design, as shown in Fig. 3.59. The versions CaPaMan 2 in Fig. 3.59 a) and CaPaMan 3 in Fig. 3.59 b) have been obtained from the kinematic design of CaPaMan in Fig. 3.57 by substituting the prismatic joint with a suitable revolute joint and the four-bar linkage with a slider-crank mechanism, respectively. Chapter 3: Fundamentals of the Mechanics of Robots204 In order to describe the motion of MP with respect to FP, a world frame OXYZ has been assumed as fixed to FP and a moving frame HXPYPZP has been fixed to MP, as shown in Fig. 3.57. Particularly, OXYZ has been fixed with Z axis orthogonal to the FP plane, X axis as coincident with the line joining O to O1 , and Y axis to give a Cartesian frame"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.19-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.19-1.png",
        "caption": "FIGURE 8.19. A McPherson suspension.",
        "texts": [
            " De Dion design is a modification of a beam axle that may be used as a dead axle or to attach the differential to the chassis and transfer the driving power to the drive wheels by employing universal joints and split shafts. Figure 8.17 illustrates a De Dion suspension. 8.2 Independent Suspension Independent suspensions is introduced to let a wheel to move up and down without affecting the opposite wheel. There are many forms and designs of independent suspensions. However, double A-arm and McPherson strut suspensions are the simplest and the most common designs. Figure 8.18 illustrates a sample of a double A-arm and Figure 8.19 shows a McPherson suspension. Kinematically, a double A-arm suspension mechanism is a four-bar linkage with the chassis as the ground link, and coupler as the wheel carrying link. A McPherson suspension is an inverted slider mechanism that has the chassis as the ground link and the coupler as the wheel carrying link. A double A-arm and a McPherson suspension mechanism on the left and right wheels are schematically shown in Figures 8.20 and 8.21 respectively. Double A-arm, is also called double wishbone, or short/long arm suspension"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure9.25-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure9.25-1.png",
        "caption": "FIGURE 9.25. A simplified models of a trebuchet.",
        "texts": [],
        "surrounding_texts": [
            "Dynamics of a rigid vehicle may be considered as the motion of a rigid body with respect to a fixed global coordinate frame. The principles of Newton and Euler equations of motion that describe the translational and rotational motion of the rigid body are reviewed in this chapter. 9.1 Force and Moment In Newtonian dynamics, the forces acting on a system of connected rigid bodied can be divided into internal and external forces. Internal forces are acting between connected bodies, and external forces are acting from outside of the system. An external force can be a contact force, such as the traction force at the tireprint of a driving wheel, or a body force, such as the gravitational force on the vehicle\u2019s body. External forces and moments are called load, and a set of forces and moments acting on a rigid body, such as forces and moments on the vehicle shown in Figure 9.1, is called a force system. The resultant or total force F is the sum of all the external forces acting on a body, and the resultant or 522 9. Applied Dynamics total moment M is the sum of all the moments of the external forces. F = X i Fi (9.1) M = X i Mi (9.2) Consider a force F acting on a point P at rP . The moment of the force about a directional line l passing through the origin is Ml = lu\u0302 \u00b7 (rP \u00d7F) (9.3) where u\u0302 is a unit vector on l. The moment of the force F, about a point Q at rQ is MQ = (rP \u2212 rQ)\u00d7F (9.4) so, the moment of F about the origin is M = rP \u00d7F. (9.5) The moment of a force may also be called torque or moment. The effect of a force system is equivalent to the effect of the resultant force and resultant moment of the force system. Any two force systems are equivalent if their resultant forces and resultant moments are equal. If the resultant force of a force system is zero, the resultant moment of the force system is independent of the origin of the coordinate frame. Such a resultant moment is called couple. When a force system is reduced to a resultant FP andMP with respect to a reference point P , we may change the reference point to another point Q and find the new resultants as FQ = FP (9.6) MQ = MP + (rP \u2212 rQ)\u00d7FP = MP + QrP \u00d7FP . (9.7) The momentum of a moving rigid body is a vector quantity equal to the total mass of the body times the translational velocity of the mass center of the body. p = mv (9.8) The momentum p is also called translational momentum or linear momentum. Consider a rigid body with momentum p. The moment of momentum, L, about a directional line l passing through the origin is Ll = lu\u0302 \u00b7 (rC \u00d7 p) (9.9) 9. Applied Dynamics 523 where u\u0302 is a unit vector indicating the direction of the line, and rC is the position vector of the mass center C. The moment of momentum about the origin is L = rC \u00d7 p. (9.10) The moment of momentum L is also called angular momentum. A bounded vector is a vector fixed at a point in space. A sliding or line vector is a vector free to slide on its line of action. A free vector is a vector that may move to any point as long as it keeps its direction. Force is a sliding vector and couple is a free vector. However, the moment of a force is dependent on the distance between the origin of the coordinate frame and the line of action. The application of a force system is emphasized by Newton\u2019s second and third laws of motion. The second law of motion, also called the Newton\u2019s equation of motion, states that the global rate of change of linear momentum is proportional to the global applied force. GF = Gd dt Gp = Gd dt \u00a1 mGv \u00a2 (9.11) The third law of motion states that the action and reaction forces acting between two bodies are equal and opposite. The second law of motion can be expanded to include rotational motions. Hence, the second law of motion also states that the global rate of change of angular momentum is proportional to the global applied moment. GM = Gd dt GL (9.12) Proof. Differentiating from angular momentum (9.10) shows that Gd dt GL = Gd dt (rC \u00d7 p) = \u00b5 GdrC dt \u00d7 p+ rC \u00d7 Gdp dt \u00b6 = GrC \u00d7 Gdp dt = GrC \u00d7 GF = GM. (9.13) Kinetic energy K of a moving body point P with mass m at a position GrP , and having a velocity GvP , is K = 1 2 mGv2P = 1 2 m \u00b3 Gd\u0307B + BvP + B G\u03c9B \u00d7 BrP \u00b42 . (9.14) 524 9. Applied Dynamics The work done by the applied force GF on m in moving from point 1 to point 2 on a path, indicated by a vector Gr, is 1W2 = Z 2 1 GF \u00b7 dGr. (9.15) However, Z 2 1 GF \u00b7 dGr = m Z 2 1 Gd dt Gv \u00b7 Gvdt = 1 2 m Z 2 1 d dt v2dt = 1 2 m \u00a1 v22 \u2212 v21 \u00a2 = K2 \u2212K1 (9.16) that shows 1W2 is equal to the difference of the kinetic energy between terminal and initial points. 1W2 = K2 \u2212K1 (9.17) Equation (9.17) is called principle of work and energy. Example 342 Position of center of mass. The position of the mass center of a rigid body in a coordinate frame is indicated by BrC and is usually measured in the body coordinate frame. BrC = 1 m Z B Br dm (9.18) \u23a1\u23a3 xC yC zC \u23a4\u23a6 = \u23a1\u23a2\u23a2\u23a3 1 m R B x dm 1 m R B y dm 1 m R B z dm \u23a4\u23a5\u23a5\u23a6 (9.19) Applying the mass center integral on the symmetric and uniform L-section rigid body with \u03c1 = 1 shown in Figure 9.2 provides the position of mass center C of the section. The x position of C is xC = 1 m Z B xdm = 1 A Z B x dA = \u2212b 2 + ab\u2212 a2 4ab+ 2a2 (9.20) and because of symmetry, we have yC = \u2212xC = b2 + ab\u2212 a2 4ab+ 2a2 . (9.21) 9. Applied Dynamics 525 When a = b, the position of C reduces to yC = \u2212xC = 1 2 b. (9.22) Example 343 F Every force system is equivalent to a wrench. The Poinsot theorem states: Every force system is equivalent to a single force, plus a moment parallel to the force. Let F andM be the resultant force and moment of a force system. We decompose the moment into parallel and perpendicular components,Mk andM\u22a5, to the force axis. The force F and the perpendicular moment M\u22a5 can be replaced by a single force F0 parallel to F. Therefore, the force system is reduced to a force F0 and a moment Mk parallel to each other. A force and a moment about the force axis is called a wrench. The Poinsot theorem is similar to the Chasles theorem that states: Every rigid body motion is equivalent to a screw, which is a translation plus a rotation about the axis of translation. Example 344 F Motion of a moving point in a moving body frame. The velocity and acceleration of a moving point P as shown in Figure 5.12 are found in Example 200. GvP = Gd\u0307B + GRB \u00a1 BvP + B G\u03c9B \u00d7 BrP \u00a2 (9.23) GaP = Gd\u0308B + GRB \u00a1 BaP + 2 B G\u03c9B \u00d7 BvP + B G\u03c9\u0307B \u00d7 BrP \u00a2 +GRB \u00a1 B G\u03c9B \u00d7 \u00a1 B G\u03c9B \u00d7 BrP \u00a2\u00a2 (9.24) Therefore, the equation of motion for the point mass P is GF = mGaP = m \u00b3 Gd\u0308B + GRB \u00a1 BaP + 2 B G\u03c9B \u00d7 BvP + B G\u03c9\u0307B \u00d7 BrP \u00a2\u00b4 +m GRB \u00a1 B G\u03c9B \u00d7 \u00a1 B G\u03c9B \u00d7 BrP \u00a2\u00a2 . (9.25) 526 9. Applied Dynamics Example 345 Newton\u2019s equation in a rotating frame. Consider a spherical rigid body, such as Earth, with a fixed point that is rotating with a constant angular velocity. The equation of motion for a moving point vehicle P on the rigid body is found by setting Gd\u0308B = B G\u03c9\u0307B = 0 in the equation of motion of a moving point in a moving body frame (9.25) BF = mBaP +mB G\u03c9B \u00d7 \u00a1 B G\u03c9B \u00d7 BrP \u00a2 + 2mB G\u03c9B \u00d7 B r\u0307P (9.26) 6= mBaP which shows that the Newton\u2019s equation of motion F = ma must be modified for rotating frames. Example 346 Coriolis force. The equation of motion of a moving vehicle point on the surface of the Earth is BF = mBaP +mB G\u03c9B \u00d7 \u00a1 B G\u03c9B \u00d7 BrP \u00a2 + 2mB G\u03c9B \u00d7 BvP (9.27) which can be rearranged to BF\u2212mB G\u03c9B \u00d7 \u00a1 B G\u03c9B \u00d7 BrP \u00a2 \u2212 2mB G\u03c9B \u00d7 BvP = mBaP . (9.28) Equation (9.28) is the equation of motion for an observer in the rotating frame, which in this case is an observer on the Earth. The left-hand side of this equation is called the effective force Feff , Feff = BF\u2212mB G\u03c9B \u00d7 \u00a1 B G\u03c9B \u00d7 BrP \u00a2 \u2212 2mB G\u03c9B \u00d7 BvP (9.29) because it seems that the particle is moving under the influence of this force. The second term is negative of the centrifugal force and pointing outward. The maximum value of this force on the Earth is on the equator r\u03c92 = 6378.388\u00d7 103 \u00d7 \u00b5 2\u03c0 24\u00d7 3600 366.25 365.25 \u00b62 = 3.3917\u00d7 10\u22122m/ s2 (9.30) which is about 0.3% of the acceleration of gravity. If we add the variation of the gravitational acceleration because of a change of radius from R = 6356912m at the pole to R = 6378388m on the equator, then the variation of the acceleration of gravity becomes 0.53%. So, generally speaking, a sportsman such as a pole-vaulter who has practiced in the north pole can show a better record in a competition held on the equator. The third term is called the Coriolis force or Coriolis effect, FC, which is perpendicular to both \u03c9 and BvP . For a mass m moving on the north hemisphere at a latitude \u03b8 towards the equator, we should provide a 9. Applied Dynamics 527 lateral eastward force equal to the Coriolis effect to force the mass, keeping its direction relative to the ground. FC = 2mB G\u03c9B \u00d7 Bvm = 1.4584\u00d7 10\u22124 Bpm cos \u03b8 kgm/ s2 (9.31) The Coriolis effect is the reason why the west side of railways, roads, and rivers wears. The lack of providing the Coriolis force is the reason for turning the direction of winds, projectiles, flood, and falling objects westward. Example 347 Work, force, and kinetic energy in a unidirectional motion. A vehicle with mass m = 1200 kg has an initial kinetic energy K = 6000 J. The mass is under a constant force F = F I\u0302 = 4000I\u0302 and moves from X(0) = 0 to X(tf ) = 1000m at a terminal time tf . The work done by the force during this motion is W = Z r(tf ) r(0) F \u00b7 dr = Z 1000 0 4000 dX = 4\u00d7 106Nm = 4MJ (9.32) The kinetic energy at the terminal time is K(tf ) =W +K(0) = 4006000 J (9.33) which shows that the terminal speed of the mass is v2 = r 2K(tf ) m \u2248 81.7m/ s. (9.34) Example 348 Direct dynamics. When the applied force is time varying and is a known function, then, F(t) = m r\u0308. (9.35) The general solution for the equation of motion can be found by integration. r\u0307(t) = r\u0307(t0) + 1 m Z t t0 F(t)dt (9.36) r(t) = r(t0) + r\u0307(t0)(t\u2212 t0) + 1 m Z t t0 Z t t0 F(t)dt dt (9.37) This kind of problem is called direct or forward dynamics. 528 9. Applied Dynamics 9.2 Rigid Body Translational Dynamics Figure 9.3 depicts a moving body B in a global coordinate frame G. Assume that the body frame is attached at the mass center of the body. Point P indicates an infinitesimal sphere of the body, which has a very small mass dm. The point mass dm is acted on by an infinitesimal force df and has a global velocity GvP . According to Newton\u2019s law of motion we have df = GaP dm. (9.38) However, the equation of motion for the whole body in a global coordinate frame is GF = mGaB (9.39) which can be expressed in the body coordinate frame as BF = mB GaB +m B G\u03c9B \u00d7 BvB (9.40)\u23a1\u23a3 Fx Fy Fz \u23a4\u23a6 = \u23a1\u23a3 max +m (\u03c9yvz \u2212 \u03c9zvy) may \u2212m (\u03c9xvz \u2212 \u03c9zvx) maz +m (\u03c9xvy \u2212 \u03c9yvx) \u23a4\u23a6 . (9.41) In these equations, GaB is the acceleration vector of the body mass center C in the global frame,m is the total mass of the body, and F is the resultant of the external forces acted on the body at C. 9. Applied Dynamics 529 Proof. A body coordinate frame at the mass center is called a central frame. If frame B is a central frame, then the center of mass, C, is defined such that Z B Brdm dm = 0. (9.42) The global position vector of dm is related to its local position vector by Grdm = GdB + GRB Brdm (9.43) where GdB is the global position vector of the central body frame, and therefore, Z B Grdm dm = Z B GdB dm+ GRB Z m Brdm dm = Z B GdB dm = GdB Z B dm = mGdB. (9.44) A time derivative of both sides shows that mGd\u0307B = mGvB = Z B Gr\u0307dm dm = Z B Gvdm dm (9.45) and another derivative is mGv\u0307B = mGaB = Z B Gv\u0307dm dm. (9.46) However, we have df = Gv\u0307P dm and therefore, mGaB = Z B df . (9.47) The integral on the right-hand side accounts for all the forces acting on the body. The internal forces cancel one another out, so the net result is the vector sum of all the externally applied forces, F, and therefore, GF = m GaB = m Gv\u0307B . (9.48) In the body coordinate frame we have BF = BRG GF = m BRG GaB = m B GaB = m BaB +m B G\u03c9B \u00d7 BvB. (9.49) 530 9. Applied Dynamics The expanded form of the Newton\u2019s equation in the body coordinate frame is then equal to BF = m BaB +m B G\u03c9B \u00d7 BvB\u23a1\u23a3 Fx Fy Fz \u23a4\u23a6 = m \u23a1\u23a3 ax ay az \u23a4\u23a6+m \u23a1\u23a3 \u03c9x \u03c9y \u03c9z \u23a4\u23a6\u00d7 \u23a1\u23a3 vx vy vz \u23a4\u23a6 = \u23a1\u23a3 max +m (\u03c9yvz \u2212 \u03c9zvy) may \u2212m (\u03c9xvz \u2212 \u03c9zvx) maz +m (\u03c9xvy \u2212 \u03c9yvx) \u23a4\u23a6 . (9.50) 9.3 Rigid Body Rotational Dynamics The rigid body rotational equation of motion is the Euler equation BM = Gd dt BL = BL\u0307+B G\u03c9B \u00d7 BL = BI B G\u03c9\u0307B + B G\u03c9B \u00d7 \u00a1 BI B G\u03c9B \u00a2 (9.51) where L is the angular momentum BL = BI B G\u03c9B (9.52) and I is the moment of inertia of the rigid body. I = \u23a1\u23a3 Ixx Ixy Ixz Iyx Iyy Iyz Izx Izy Izz \u23a4\u23a6 (9.53) The elements of I are functions of the mass distribution of the rigid body and may be defined by Iij = Z B \u00a1 r2i \u03b4mn \u2212 ximxjn \u00a2 dm , i, j = 1, 2, 3 (9.54) where \u03b4ij is Kronecker\u2019s delta. \u03b4mn = \u00bd 1 if m = n 0 if m 6= n (9.55) The expanded form of the Euler equation (9.51) is Mx = Ixx\u03c9\u0307x + Ixy\u03c9\u0307y + Ixz\u03c9\u0307z \u2212 (Iyy \u2212 Izz)\u03c9y\u03c9z \u2212Iyz \u00a1 \u03c92z \u2212 \u03c92y \u00a2 \u2212 \u03c9x (\u03c9zIxy \u2212 \u03c9yIxz) (9.56) 9. Applied Dynamics 531 My = Iyx\u03c9\u0307x + Iyy\u03c9\u0307y + Iyz\u03c9\u0307z \u2212 (Izz \u2212 Ixx)\u03c9z\u03c9x \u2212Ixz \u00a1 \u03c92x \u2212 \u03c92z \u00a2 \u2212 \u03c9y (\u03c9xIyz \u2212 \u03c9zIxy) (9.57) Mz = Izx\u03c9\u0307x + Izy\u03c9\u0307y + Izz\u03c9\u0307z \u2212 (Ixx \u2212 Iyy)\u03c9x\u03c9y \u2212Ixy \u00a1 \u03c92y \u2212 \u03c92x \u00a2 \u2212 \u03c9z (\u03c9yIxz \u2212 \u03c9xIyz) . (9.58) which can be reduced to M1 = I1\u03c9\u03071 \u2212 (I2 \u2212 I2)\u03c92\u03c93 M2 = I2\u03c9\u03072 \u2212 (I3 \u2212 I1)\u03c93\u03c91 (9.59) M3 = I3\u03c9\u03073 \u2212 (I1 \u2212 I2)\u03c91\u03c92 in a special Cartesian coordinate frame called the principal coordinate frame. The principal coordinate frame is denoted by numbers 123 to indicate the first, second, and third principal axes. The parameters Iij , i 6= j are zero in the principal frame. The body and principal coordinate frame sit at the mass center C. Kinetic energy of a rotating rigid body is K = 1 2 \u00a1 Ixx\u03c9 2 x + Iyy\u03c9 2 y + Izz\u03c9 2 z \u00a2 \u2212Ixy\u03c9x\u03c9y \u2212 Iyz\u03c9y\u03c9z \u2212 Izx\u03c9z\u03c9x (9.60) = 1 2 \u03c9 \u00b7 L (9.61) = 1 2 \u03c9T I \u03c9 (9.62) that in the principal coordinate frame reduces to K = 1 2 \u00a1 I1\u03c9 2 1 + I2\u03c9 2 2 + I3\u03c9 2 3 \u00a2 . (9.63) Proof. Let mi be the mass of the ith particle of a rigid body B, which is made of n particles and let ri = Bri = \u00a3 xi yi zi \u00a4T (9.64) be the Cartesian position vector of mi in a central body fixed coordinate frame Oxyz. Assume that \u03c9 = B G\u03c9B = \u00a3 \u03c9x \u03c9y \u03c9z \u00a4T (9.65) is the angular velocity of the rigid body with respect to the ground, expressed in the body coordinate frame. 532 9. Applied Dynamics The angular momentum of mi is Li = ri \u00d7mir\u0307i = mi [ri \u00d7 (\u03c9 \u00d7 ri)] = mi [(ri \u00b7 ri)\u03c9 \u2212 (ri \u00b7 \u03c9) ri] = mir 2 i\u03c9 \u2212mi (ri \u00b7 \u03c9) ri. (9.66) Hence, the angular momentum of the rigid body would be L = \u03c9 nX i=1 mir 2 i \u2212 nX i=1 mi (ri \u00b7 \u03c9) ri. (9.67) Substitution for ri and \u03c9 gives us L = \u00b3 \u03c9x \u0131\u0302+ \u03c9y j\u0302+ \u03c9z k\u0302 \u00b4 nX i=1 mi \u00a1 x2i + y2i + z2i \u00a2 \u2212 nX i=1 mi (xi\u03c9x + yi\u03c9y + zi\u03c9z) \u00b7 \u00b3 xi\u0131\u0302+ yij\u0302+ zik\u0302 \u00b4 (9.68) and therefore, L = nX i=1 mi \u00a1 x2i + y2i + z2i \u00a2 \u03c9x \u0131\u0302 + nX i=1 mi \u00a1 x2i + y2i + z2i \u00a2 \u03c9y j\u0302 + nX i=1 mi \u00a1 x2i + y2i + z2i \u00a2 \u03c9z k\u0302 \u2212 nX i=1 mi (xi\u03c9x + yi\u03c9y + zi\u03c9z)xi\u0131\u0302 \u2212 nX i=1 mi (xi\u03c9x + yi\u03c9y + zi\u03c9z) yij\u0302 \u2212 nX i=1 mi (xi\u03c9x + yi\u03c9y + zi\u03c9z) zik\u0302 (9.69) or L = nX i=1 mi \u00a3\u00a1 x2i + y2i + z2i \u00a2 \u03c9x \u2212 (xi\u03c9x + yi\u03c9y + zi\u03c9z)xi \u00a4 \u0131\u0302 + nX i=1 mi \u00a3\u00a1 x2i + y2i + z2i \u00a2 \u03c9y \u2212 (xi\u03c9x + yi\u03c9y + zi\u03c9z) yi \u00a4 j\u0302 + nX i=1 mi \u00a3\u00a1 x2i + y2i + z2i \u00a2 \u03c9z \u2212 (xi\u03c9x + yi\u03c9y + zi\u03c9z) zi \u00a4 k\u0302 (9.70) 9. Applied Dynamics 533 which can be rearranged as L = nX i=1 \u00a3 mi \u00a1 y2i + z2i \u00a2\u00a4 \u03c9x\u0131\u0302 + nX i=1 \u00a3 mi \u00a1 z2i + x2i \u00a2\u00a4 \u03c9y j\u0302 + nX i=1 \u00a3 mi \u00a1 x2i + y2i \u00a2\u00a4 \u03c9zk\u0302 \u2212 \u00c3 nX i=1 (mixiyi)\u03c9y + nX i=1 (mixizi)\u03c9z ! \u0131\u0302 \u2212 \u00c3 nX i=1 (miyizi)\u03c9z + nX i=1 (miyixi)\u03c9x ! j\u0302 \u2212 \u00c3 nX i=1 (mizixi)\u03c9x + nX i=1 (miziyi)\u03c9y ! k\u0302. (9.71) By introducing the moment of inertia matrix I with the following elements, Ixx = nX i=1 \u00a3 mi \u00a1 y2i + z2i \u00a2\u00a4 (9.72) Iyy = nX i=1 \u00a3 mi \u00a1 z2i + x2i \u00a2\u00a4 (9.73) Izz = nX i=1 \u00a3 mi \u00a1 x2i + y2i \u00a2\u00a4 (9.74) Ixy = Iyx = \u2212 nX i=1 (mixiyi) (9.75) Iyz = Izy = \u2212 nX i=1 (miyizi) (9.76) Izx = Ixz = \u2212 nX i=1 (mizixi) . (9.77) we may write the angular momentum L in a concise form Lx = Ixx\u03c9x + Ixy\u03c9y + Ixz\u03c9z (9.78) Ly = Iyx\u03c9x + Iyy\u03c9y + Iyz\u03c9z (9.79) Lz = Izx\u03c9x + Izy\u03c9y + Izz\u03c9z (9.80) 534 9. Applied Dynamics or in a matrix form L = I \u00b7 \u03c9 (9.81)\u23a1\u23a3 Lx Ly Lz \u23a4\u23a6 = \u23a1\u23a3 Ixx Ixy Ixz Iyx Iyy Iyz Izx Izy Izz \u23a4\u23a6\u23a1\u23a3 \u03c9x \u03c9y \u03c9z \u23a4\u23a6 . (9.82) For a rigid body that is a continuous solid, the summations must be replaced by integrations over the volume of the body as in Equation (9.54). The Euler equation of motion for a rigid body is BM = Gd dt BL (9.83) where BM is the resultant of the external moments applied on the rigid body. The angular momentum BL is a vector defined in the body coordinate frame. Hence, its time derivative in the global coordinate frame is GdBL dt = BL\u0307+B G\u03c9B \u00d7 BL. (9.84) Therefore, BM = dL dt = L\u0307+ \u03c9 \u00d7 L = I\u03c9\u0307 + \u03c9\u00d7 (I\u03c9) (9.85) or in expanded form BM = (Ixx\u03c9\u0307x + Ixy\u03c9\u0307y + Ixz\u03c9\u0307z) \u0131\u0302 +(Iyx\u03c9\u0307x + Iyy\u03c9\u0307y + Iyz\u03c9\u0307z) j\u0302 +(Izx\u03c9\u0307x + Izy\u03c9\u0307y + Izz\u03c9\u0307z) k\u0302 +\u03c9y (Ixz\u03c9x + Iyz\u03c9y + Izz\u03c9z) \u0131\u0302 \u2212\u03c9z (Ixy\u03c9x + Iyy\u03c9y + Iyz\u03c9z) \u0131\u0302 +\u03c9z (Ixx\u03c9x + Ixy\u03c9y + Ixz\u03c9z) j\u0302 \u2212\u03c9x (Ixz\u03c9x + Iyz\u03c9y + Izz\u03c9z) j\u0302 +\u03c9x (Ixy\u03c9x + Iyy\u03c9y + Iyz\u03c9z) k\u0302 \u2212\u03c9y (Ixx\u03c9x + Ixy\u03c9y + Ixz\u03c9z) k\u0302 (9.86) and therefore, the most general form of the Euler equations of motion for a rigid body in a body frame attached to C are Mx = Ixx\u03c9\u0307x + Ixy\u03c9\u0307y + Ixz\u03c9\u0307z \u2212 (Iyy \u2212 Izz)\u03c9y\u03c9z \u2212Iyz \u00a1 \u03c92z \u2212 \u03c92y \u00a2 \u2212 \u03c9x (\u03c9zIxy \u2212 \u03c9yIxz) (9.87) My = Iyx\u03c9\u0307x + Iyy\u03c9\u0307y + Iyz\u03c9\u0307z \u2212 (Izz \u2212 Ixx)\u03c9z\u03c9x \u2212Ixz \u00a1 \u03c92x \u2212 \u03c92z \u00a2 \u2212 \u03c9y (\u03c9xIyz \u2212 \u03c9zIxy) (9.88) Mz = Izx\u03c9\u0307x + Izy\u03c9\u0307y + Izz\u03c9\u0307z \u2212 (Ixx \u2212 Iyy)\u03c9x\u03c9y \u2212Ixy \u00a1 \u03c92y \u2212 \u03c92x \u00a2 \u2212 \u03c9z (\u03c9yIxz \u2212 \u03c9xIyz) . (9.89) 9. Applied Dynamics 535 Assume that we are able to rotate the body frame about its origin to find an orientation that makes Iij = 0, for i 6= j. In such a coordinate frame, which is called a principal frame, the Euler equations reduce to M1 = I1\u03c9\u03071 \u2212 (I2 \u2212 I2)\u03c92\u03c93 (9.90) M2 = I2\u03c9\u03072 \u2212 (I3 \u2212 I1)\u03c93\u03c91 (9.91) M3 = I3\u03c9\u03073 \u2212 (I1 \u2212 I2)\u03c91\u03c92. (9.92) The kinetic energy of a rigid body may be found by the integral of the kinetic energy of a mass element dm, over the whole body. K = 1 2 Z B v\u03072dm = 1 2 Z B (\u03c9 \u00d7 r) \u00b7 (\u03c9 \u00d7 r) dm = \u03c92x 2 Z B \u00a1 y2 + z2 \u00a2 dm+ \u03c92y 2 Z B \u00a1 z2 + x2 \u00a2 dm+ \u03c92z 2 Z B \u00a1 x2 + y2 \u00a2 dm \u2212\u03c9x\u03c9y Z B xy dm\u2212 \u03c9y\u03c9z Z B yz dm\u2212 \u03c9z\u03c9x Z B zx dm = 1 2 \u00a1 Ixx\u03c9 2 x + Iyy\u03c9 2 y + Izz\u03c9 2 z \u00a2 \u2212Ixy\u03c9x\u03c9y \u2212 Iyz\u03c9y\u03c9z \u2212 Izx\u03c9z\u03c9x (9.93) The kinetic energy can be rearranged to a matrix multiplication form K = 1 2 \u03c9T I \u03c9 (9.94) = 1 2 \u03c9 \u00b7 L. (9.95) When the body frame is principal, the kinetic energy will simplify to K = 1 2 \u00a1 I1\u03c9 2 1 + I2\u03c9 2 2 + I3\u03c9 2 3 \u00a2 . (9.96) Example 349 A tilted disc on a massless shaft. Figure 9.4 illustrates a disc with mass m and radius r, mounted on a massless shaft. The shaft is turning with a constant angular speed \u03c9. The disc is attached to the shaft at an angle \u03b8. Because of \u03b8, the bearings at A and B must support a rotating force. We attach a principal body coordinate frame at the disc center as shown in the figure. The angular velocity vector in the body frame is B G\u03c9B = \u03c9 cos \u03b8 \u0131\u0302+ \u03c9 sin \u03b8 j\u0302 (9.97) 536 9. Applied Dynamics and the mass moment of inertia matrix is BI = \u23a1\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3 mr2 2 0 0 0 mr2 4 0 0 0 mr2 4 \u23a4\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6 . (9.98) Substituting (9.97) and (9.98) in (9.90)-(9.92), with 1 \u2261 x, 2 \u2261 y, 3 \u2261 z, provides that Mx = 0 (9.99) My = 0 (9.100) Mz = mr2 4 \u03c9 cos \u03b8 sin \u03b8. (9.101) Therefore, the bearing reaction forces FA and FB are FA = \u2212FB = \u2212Mz l = \u2212mr2 4l \u03c9 cos \u03b8 sin \u03b8. (9.102) Example 350 Steady rotation of a freely rotating rigid body. The Newton-Euler equations of motion for a rigid body are GF = mGv\u0307 (9.103) BM = I B G\u03c9\u0307B + B G\u03c9B \u00d7 BL. (9.104) 9. Applied Dynamics 537 Consider a situation in which the resultant applied force and moment on the body are zero. GF = BF = 0 (9.105) GM = BM = 0 (9.106) Based on the Newton\u2019s equation, the velocity of the mass center will be constant in the global coordinate frame. However, the Euler equation reduces to \u03c9\u03071 = I2 \u2212 I3 I1 \u03c92\u03c93 (9.107) \u03c9\u03072 = I3 \u2212 I1 I22 \u03c93\u03c91 (9.108) \u03c9\u03073 = I1 \u2212 I2 I3 \u03c91\u03c92 (9.109) that show the angular velocity can be constant if I1 = I2 = I3 (9.110) or if two principal moments of inertia, say I1 and I2, are zero and the third angular velocity, in this case \u03c93, is initially zero, or if the angular velocity vector is initially parallel to a principal axis. Example 351 Angular momentum of a two-link manipulator. A two-link manipulator is shown in Figure 9.5. Link A rotates with angular velocity \u03d5\u0307 about the z-axis of its local coordinate frame. Link B is attached to link A and has angular velocity \u03c8\u0307 with respect to A about the xA-axis. We assume that A and G were coincident at \u03d5 = 0, therefore, the rotation matrix between A and G is GRA = \u23a1\u23a3 cos\u03d5(t) \u2212 sin\u03d5(t) 0 sin\u03d5(t) cos\u03d5(t) 0 0 0 1 \u23a4\u23a6 . (9.111) Frame B is related to frame A by Euler angles \u03d5 = 90deg, \u03b8 = 90deg, and \u03c8, hence, ARB = \u23a1\u23a3 c\u03c0c\u03c8 \u2212 c\u03c0s\u03c0s\u03c8 \u2212c\u03c0s\u03c8 \u2212 c\u03c0c\u03c8s\u03c0 s\u03c0s\u03c0 c\u03c8s\u03c0 + c\u03c0c\u03c0s\u03c8 \u2212s\u03c0s\u03c8 + c\u03c0c\u03c0c\u03c8 \u2212c\u03c0s\u03c0 s\u03c0s\u03c8 s\u03c0c\u03c8 c\u03c0 \u23a4\u23a6 \u23a1\u23a3 \u2212 cos\u03c8 sin\u03c8 0 sin\u03c8 cos\u03c8 0 0 0 \u22121 \u23a4\u23a6 (9.112) 538 9. Applied Dynamics and therefore, GRB = GRA ARB (9.113) = \u23a1\u23a3 \u2212 cos\u03d5 cos\u03c8 \u2212 sin\u03d5 sin\u03c8 cos\u03d5 sin\u03c8 \u2212 cos\u03c8 sin\u03d5 0 cos\u03d5 sin\u03c8 \u2212 cos\u03c8 sin\u03d5 cos\u03d5 cos\u03c8 + sin\u03d5 sin\u03c8 0 0 0 \u22121 \u23a4\u23a6 . The angular velocity of A in G, and B in A are G\u03c9A = \u03d5\u0307K\u0302 (9.114) A\u03c9B = \u03c8\u0307\u0131\u0302A. (9.115) Moment of inertia matrices for the arms A and B can be defined as AIA = \u23a1\u23a3 IA1 0 0 0 IA2 0 0 0 IA3 \u23a4\u23a6 (9.116) BIB = \u23a1\u23a3 IB1 0 0 0 IB2 0 0 0 IB3 \u23a4\u23a6 . (9.117) These moments of inertia must be transformed to the global frame GIA = GRB AIA GRT A (9.118) GIB = GRB BIB GRT B . (9.119) 9. Applied Dynamics 539 The total angular momentum of the manipulator is GL = GLA + GLB (9.120) where GLA = GIA G\u03c9A (9.121) GLB = GIB G\u03c9B = GIB \u00a1 G A\u03c9B + G\u03c9A \u00a2 . (9.122) Example 352 Poinsot\u2019s construction. Consider a freely rotating rigid body with an attached principal coordinate frame. HavingM = 0 provides a motion under constant angular momentum and constant kinetic energy L = I \u03c9 = cte (9.123) K = 1 2 \u03c9T I \u03c9 = cte. (9.124) Because the length of the angular momentum L is constant, the equation L2 = L \u00b7 L = L2x + L2y + L2z = I21\u03c9 2 1 + I22\u03c9 2 2 + I23\u03c9 2 3 (9.125) introduces an ellipsoid in the (\u03c91, \u03c92, \u03c93) coordinate frame, called the momentum ellipsoid. The tip of all possible angular velocity vectors must lie on the surface of the momentum ellipsoid. The kinetic energy also defines an energy ellipsoid in the same coordinate frame so that the tip of the angular velocity vectors must also lie on its surface. K = 1 2 \u00a1 I1\u03c9 2 1 + I2\u03c9 2 2 + I3\u03c9 2 3 \u00a2 (9.126) In other words, the dynamics of moment-free motion of a rigid body requires that the corresponding angular velocity \u03c9(t) satisfy both Equations (9.125) and (9.126) and therefore lie on the intersection of the momentum and energy ellipsoids. For clarity, we may define the ellipsoids in the (Lx, Ly, Lz) coordinate system as L2x + L2y + L2z = L2 (9.127) L2x 2I1K + L2y 2I2K + L2z 2I3K = 1. (9.128) Equation (9.127) is a sphere and Equation (9.128) defines an ellipsoid with\u221a 2IiK as semi-axes. To have a meaningful motion, these two shapes must intersect. The intersection may form a trajectory, as shown in Figure 9.6. 540 9. Applied Dynamics It can be deduced that for a certain value of angular momentum there are maximum and minimum limit values for acceptable kinetic energy. Assuming I1 > I3 > I3 (9.129) the limits of possible kinetic energy are Kmin = L2 2I1 (9.130) Kmax = L2 2I3 (9.131) and the corresponding motions are turning about the axes I1 and I3 respectively. Example 353 F Alternative derivation of Euler equations of motion. Assume that the moment of the small force df is shown by dm and a mass element is shown by dm, then, dm = Grdm \u00d7 df = Grdm \u00d7 Gv\u0307dm dm. (9.132) The global angular momentum dl of dm is equal to dl = Grdm \u00d7 Gvdm dm (9.133) and according to (9.12) we have dm = Gd dt dl (9.134) Grdm \u00d7 df = Gd dt \u00a1 Grdm \u00d7 Gvdm dm \u00a2 . (9.135) 9. Applied Dynamics 541 Integrating over the body results inZ B Grdm \u00d7 df = Z B Gd dt \u00a1 Grdm \u00d7 Gvdm dm \u00a2 = Gd dt Z B \u00a1 Grdm \u00d7 Gvdm dm \u00a2 . (9.136) However, utilizing Grdm = GdB + GRB Brdm (9.137) where GdB is the global position vector of the central body frame, can simplify the left-hand side of the integral toZ B Grdm \u00d7 df = Z B \u00a1 GdB + GRB Brdm \u00a2 \u00d7 df = Z B GdB \u00d7 df + Z B G Brdm \u00d7 df = GdB \u00d7 GF+ GMC (9.138) where MC is the resultant external moment about the body mass center C. The right-hand side of Equation (9.136) is Gd dt Z B \u00a1 Grdm \u00d7 Gvdm dm \u00a2 = Gd dt Z B \u00a1\u00a1 GdB + GRB Brdm \u00a2 \u00d7 Gvdm dm \u00a2 = Gd dt Z B \u00a1 GdB \u00d7 Gvdm \u00a2 dm+ Gd dt Z B \u00a1 G Brdm \u00d7 Gvdm \u00a2 dm = Gd dt \u00b5 GdB \u00d7 Z B Gvdmdm \u00b6 + Gd dt LC = Gd\u0307B \u00d7 Z B Gvdmdm+ GdB \u00d7 Z B Gv\u0307dmdm+ d dt LC . (9.139) We use LC for angular momentum about the body mass center. Because the body frame is at the mass center, we haveZ B Grdm dm = mGdB = mGrC (9.140)Z B Gvdmdm = mGd\u0307B = mGvC (9.141)Z B Gv\u0307dmdm = mGd\u0308B = mGaC (9.142) and therefore, Gd dt Z B \u00a1 Grdm \u00d7 Gvdm dm \u00a2 = GdB \u00d7 GF+ Gd dt GLC . (9.143) 542 9. Applied Dynamics Substituting (9.138) and (9.143) in (9.136) provides the Euler equation of motion in the global frame, indicating that the resultant of externally applied moments about C is equal to the global derivative of angular momentum about C. GMC = Gd dt GLC . (9.144) The Euler equation in the body coordinate can be found by transforming (9.144) BMC = GRT B GMC = GRT B Gd dt LC = Gd dt GRT B LC = Gd dt BLC = BL\u0307C + B G\u03c9B \u00d7 BLC . (9.145) 9.4 Mass Moment of Inertia Matrix In analyzing the motion of rigid bodies, two types of integrals arise that belong to the geometry of the body. The first type defines the center of mass and is important when the translation motion of the body is considered. The second is the moment of inertia that appears when the rotational motion of the body is considered. The moment of inertia is also called centrifugal moments, or deviation moments. Every rigid body has a 3 \u00d7 3 moment of inertia matrix I, which is denoted by I = \u23a1\u23a3 Ixx Ixy Ixz Iyx Iyy Iyz Izx Izy Izz \u23a4\u23a6 . (9.146) The diagonal elements Iij , i = j are called polar moments of inertia Ixx = Ix = Z B \u00a1 y2 + z2 \u00a2 dm (9.147) Iyy = Iy = Z B \u00a1 z2 + x2 \u00a2 dm (9.148) Izz = Iz = Z B \u00a1 x2 + y2 \u00a2 dm (9.149) and the off-diagonal elements Iij , i 6= j are called products of inertia Ixy = Iyx = \u2212 Z B xy dm (9.150) 9. Applied Dynamics 543 Iyz = Izy = \u2212 Z B yz dm (9.151) Izx = Ixz = \u2212 Z B zx dm. (9.152) The elements of I for a rigid body, made of discrete point masses, are defined in Equation (9.54). The elements of I are calculated about a body coordinate frame attached to the mass center C of the body. Therefore, I is a frame-dependent quantity and must be written like BI to show the frame it is computed in. BI = Z B \u23a1\u23a3 y2 + z2 \u2212xy \u2212zx \u2212xy z2 + x2 \u2212yz \u2212zx \u2212yz x2 + y2 \u23a4\u23a6 dm (9.153) = Z B \u00a1 r2I\u2212 r rT \u00a2 dm (9.154) = Z B \u2212r\u0303 r\u0303 dm. (9.155) Moments of inertia can be transformed from a coordinate frame B1 to another coordinate frame B2, both installed at the mass center of the body, according to the rule of the rotated-axes theorem B2I = B2RB1 B1I B2RT B1 . (9.156) Transformation of the moment of inertia from a central frame B1 located at B2rC to another frame B2, which is parallel to B1, is, according to the rule of parallel-axes theorem, B2I = B1I +mr\u0303C r\u0303TC . (9.157) If the local coordinate frame Oxyz is located such that the products of inertia vanish, the local coordinate frame is called the principal coordinate frame and the associated moments of inertia are called principal moments of inertia. Principal axes and principal moments of inertia can be found by solving the following equation for I:\u00af\u0304\u0304\u0304 \u00af\u0304 Ixx \u2212 I Ixy Ixz Iyx Iyy \u2212 I Iyz Izx Izy Izz \u2212 I \u00af\u0304\u0304\u0304 \u00af\u0304 = 0 (9.158) det ([Iij ]\u2212 I [\u03b4ij ]) = 0. (9.159) Since Equation (9.159) is a cubic equation in I, we obtain three eigenvalues I1 = Ix I2 = Iy I3 = Iz (9.160) 544 9. Applied Dynamics that are the principal moments of inertia. Proof. Two coordinate frames with a common origin at the mass center of a rigid body are shown in Figure 9.7. The angular velocity and angular momentum of a rigid body transform from the frame B1 to the frame B2 by vector transformation rule B2\u03c9 = B2RB1 B1\u03c9 (9.161) B2L = B2RB1 B1L. (9.162) However, L and \u03c9 are related according to Equation (9.52) B1L = B1I B1\u03c9 (9.163) and therefore, B2L = B2RB1 B1I B2RT B1 B2\u03c9 = B2I B2\u03c9 (9.164) which shows how to transfer the moment of inertia from the coordinate frame B1 to a rotated frame B2 B2I = B2RB1 B1I B2RT B1 . (9.165) Now consider a central frame B1, shown in Figure 9.8, at B2rC , which rotates about the origin of a fixed frame B2 such that their axes remain parallel. The angular velocity and angular momentum of the rigid body transform from frame B1 to frame B2 by B2\u03c9 = B1\u03c9 (9.166) B2L = B1L+ (rC \u00d7mvC) . (9.167) 9. Applied Dynamics 545 Therefore, B2L = B1L+mB2rC \u00d7 \u00a1 B2\u03c9\u00d7B2rC \u00a2 = B1L+ \u00a1 m B2 r\u0303C B2 r\u0303TC \u00a2 B2\u03c9 = \u00a1 B1I +m B2 r\u0303C B2 r\u0303TC \u00a2 B2\u03c9 (9.168) which shows how to transfer the moment of inertia from frame B1 to a parallel frame B2 B2I = B1I +mr\u0303C r\u0303TC . (9.169) The parallel-axes theorem is also called the Huygens-Steiner theorem. Referring to Equation (9.165) for transformation of the moment of inertia to a rotated frame, we can always find a frame in which B2I is diagonal. In such a frame, we have B2RB1 B1I = B2I B2RB1 (9.170) or \u23a1\u23a3 r11 r12 r13 r21 r22 r23 r31 r32 r33 \u23a4\u23a6\u23a1\u23a3 Ixx Ixy Ixz Iyx Iyy Iyz Izx Izy Izz \u23a4\u23a6 = \u23a1\u23a3 I1 0 0 0 I2 0 0 0 I3 \u23a4\u23a6\u23a1\u23a3 r11 r12 r13 r21 r22 r23 r31 r32 r33 \u23a4\u23a6 (9.171) which shows that I1, I2, and I3 are eigenvalues of B1I. These eigenvalues can be found by solving the following equation for \u03bb:\u00af\u0304\u0304\u0304 \u00af\u0304 Ixx \u2212 \u03bb Ixy Ixz Iyx Iyy \u2212 \u03bb Iyz Izx Izy Izz \u2212 \u03bb \u00af\u0304\u0304\u0304 \u00af\u0304 = 0. (9.172) 546 9. Applied Dynamics The eigenvalues I1, I2, and I3 are principal moments of inertia, and their associated eigenvectors are called principal directions. The coordinate frame made by the eigenvectors is the principal body coordinate frame. In the principal coordinate frame, the rigid body angular momentum is\u23a1\u23a3 L1 L2 L3 \u23a4\u23a6 = \u23a1\u23a3 I1 0 0 0 I2 0 0 0 I3 \u23a4\u23a6\u23a1\u23a3 \u03c91 \u03c92 \u03c93 \u23a4\u23a6 . (9.173) Example 354 Principal moments of inertia. Consider the inertia matrix I I = \u23a1\u23a3 20 \u22122 0 \u22122 30 0 0 0 40 \u23a4\u23a6 . (9.174) We set up the determinant (9.159)\u00af\u0304\u0304\u0304 \u00af\u0304 20\u2212 \u03bb \u22122 0 \u22122 30\u2212 \u03bb 0 0 0 40\u2212 \u03bb \u00af\u0304\u0304\u0304 \u00af\u0304 = 0 (9.175) which leads to the following characteristic equation. (20\u2212 \u03bb) (30\u2212 \u03bb) (40\u2212 \u03bb)\u2212 4 (40\u2212 \u03bb) = 0 (9.176) Three roots of Equation (9.176) are I1 = 30.385, I2 = 19.615, I3 = 40 (9.177) and therefore, the principal moment of inertia matrix is I = \u23a1\u23a3 30.385 0 0 0 19.615 0 0 0 40 \u23a4\u23a6 . (9.178) Example 355 Principal coordinate frame. Consider the inertia matrix I I = \u23a1\u23a3 20 \u22122 0 \u22122 30 0 0 0 40 \u23a4\u23a6 (9.179) the direction of a principal axis xi is established by solving\u23a1\u23a3 Ixx \u2212 Ii Ixy Ixz Iyx Iyy \u2212 Ii Iyz Izx Izy Izz \u2212 Ii \u23a4\u23a6\u23a1\u23a3 cos\u03b1i cos\u03b2i cos \u03b3i \u23a4\u23a6 = \u23a1\u23a3 0 0 0 \u23a4\u23a6 (9.180) 9. Applied Dynamics 547 for direction cosines, which must also satisfy cos2 \u03b1i + cos 2 \u03b2i + cos 2 \u03b3i = 1. (9.181) For the first principal moment of inertia I1 = 30.385 we have\u23a1\u23a3 20\u2212 30.385 \u22122 0 \u22122 30\u2212 30.385 0 0 0 40\u2212 30.385 \u23a4\u23a6\u23a1\u23a3 cos\u03b11 cos\u03b21 cos \u03b31 \u23a4\u23a6 = \u23a1\u23a3 0 0 0 \u23a4\u23a6 (9.182) or \u221210.385 cos\u03b11 \u2212 2 cos\u03b21 + 0 = 0 (9.183) \u22122 cos\u03b11 \u2212 0.385 cos\u03b21 + 0 = 0 (9.184) 0 + 0 + 9.615 cos \u03b31 = 0 (9.185) and we obtain \u03b11 = 79.1 deg (9.186) \u03b21 = 169.1 deg (9.187) \u03b31 = 90.0 deg . (9.188) Using I2 = 19.615 for the second principal axis\u23a1\u23a3 20\u2212 19.62 \u22122 0 \u22122 30\u2212 19.62 0 0 0 40\u2212 19.62 \u23a4\u23a6\u23a1\u23a3 cos\u03b12 cos\u03b22 cos \u03b32 \u23a4\u23a6 = \u23a1\u23a3 0 0 0 \u23a4\u23a6 (9.189) we obtain \u03b12 = 10.9 deg (9.190) \u03b22 = 79.1 deg (9.191) \u03b32 = 90.0 deg . (9.192) The third principal axis is for I3 = 40\u23a1\u23a3 20\u2212 40 \u22122 0 \u22122 30\u2212 40 0 0 0 40\u2212 40 \u23a4\u23a6\u23a1\u23a3 cos\u03b13 cos\u03b23 cos \u03b33 \u23a4\u23a6 = \u23a1\u23a3 0 0 0 \u23a4\u23a6 (9.193) which leads to \u03b13 = 90.0 deg (9.194) \u03b23 = 90.0 deg (9.195) \u03b33 = 0.0 deg . (9.196) 548 9. Applied Dynamics Example 356 Moment of inertia of a rigid rectangular bar. Consider a homogeneous rectangular link with mass m, length l, width w, and height h, as shown in Figure 9.9. The local central coordinate frame is attached to the link at its mass center. The moments of inertia matrix of the link can be found by the integral method. We begin with calculating Ixx Ixx = Z B \u00a1 y2 + z2 \u00a2 dm = Z v \u00a1 y2 + z2 \u00a2 \u03c1dv = m lwh Z v \u00a1 y2 + z2 \u00a2 dv = m lwh Z h/2 \u2212h/2 Z w/2 \u2212w/2 Z l/2 \u2212l/2 \u00a1 y2 + z2 \u00a2 dx dy dz = m 12 \u00a1 w2 + h2 \u00a2 (9.197) which shows Iyy and Izz can be calculated similarly Iyy = m 12 \u00a1 h2 + l2 \u00a2 (9.198) Izz = m 12 \u00a1 l2 + w2 \u00a2 . (9.199) Since the coordinate frame is central, the products of inertia must be zero. To show this, we examine Ixy. Ixy = Iyx = \u2212 Z B xy dm = Z v xy\u03c1dv = m lwh Z h/2 \u2212h/2 Z w/2 \u2212w/2 Z l/2 \u2212l/2 xy dxdy dz = 0 (9.200) 9. Applied Dynamics 549 Therefore, the moment of inertia matrix for the rigid rectangular bar in its central frame is I = \u23a1\u23a3 m 12 \u00a1 w2 + h2 \u00a2 0 0 0 m 12 \u00a1 h2 + l2 \u00a2 0 0 0 m 12 \u00a1 l2 + w2 \u00a2 \u23a4\u23a6 . (9.201) Example 357 Translation of the inertia matrix. The moment of inertia matrix of the rigid body shown in Figure 9.10, in the principal frame B(oxyz) is given in Equation (9.201). The moment of inertia matrix in the non-principal frame B0(ox0y0z0) can be found by applying the parallel-axes transformation formula (9.169). B0 I = BI +m B0 r\u0303C B0 r\u0303TC (9.202) The mass center is at B0 rC = 1 2 \u23a1\u23a3 l w h \u23a4\u23a6 (9.203) and therefore, B0 r\u0303C = 1 2 \u23a1\u23a3 0 \u2212h w h 0 \u2212l \u2212w l 0 \u23a4\u23a6 (9.204) that provides B0 I = \u23a1\u23a3 1 3h 2m+ 1 3mw2 \u221214 lmw \u221214hlm \u221214 lmw 1 3h 2m+ 1 3 l 2m \u221214hmw \u221214hlm \u2212 14hmw 1 3 l 2m+ 1 3mw2 \u23a4\u23a6 . (9.205) 550 9. Applied Dynamics Example 358 Principal rotation matrix. Consider a body inertia matrix as I = \u23a1\u23a3 2/3 \u22121/2 \u22121/2 \u22121/2 5/3 \u22121/4 \u22121/2 \u22121/4 5/3 \u23a4\u23a6 . (9.206) The eigenvalues and eigenvectors of I are I1 = 0.2413 , \u23a1\u23a3 2.351 1 1 \u23a4\u23a6 (9.207) I2 = 1.8421 , \u23a1\u23a3 \u22120.8511 1 \u23a4\u23a6 (9.208) I3 = 1.9167 , \u23a1\u23a3 0 \u22121 1 \u23a4\u23a6 . (9.209) The normalized eigenvector matrix W is equal to the transpose of the required transformation matrix to make the inertia matrix diagonal W = \u23a1\u23a3 | | | w1 w 2 w 3 | | | \u23a4\u23a6 = 2RT 1 = \u23a1\u23a3 0.856 9 \u22120.515 6 0.0 0.364 48 0.605 88 \u22120.707 11 0.364 48 0.605 88 0.707 11 \u23a4\u23a6 . (9.210) We may verify that 2I \u2248 2R1 1I 2RT 1 =WT 1I W = \u23a1\u23a3 0.2413 \u22121\u00d7 10\u22124 0.0 \u22121\u00d7 10\u22124 1.842 1 \u22121\u00d7 10\u221219 0.0 0.0 1.916 7 \u23a4\u23a6 . (9.211) Example 359 F Relative diagonal moments of inertia. Using the definitions for moments of inertia (9.147), (9.148), and (9.149) it is seen that the inertia matrix is symmetric, andZ B \u00a1 x2 + y2 + z2 \u00a2 dm = 1 2 (Ixx + Iyy + Izz) (9.212) and also Ixx + Iyy \u2265 Izz (9.213) Iyy + Izz \u2265 Ixx (9.214) Izz + Ixx \u2265 Iyy. (9.215) 9. Applied Dynamics 551 Noting that (y \u2212 z) 2 \u2265 0 it is evident that \u00a1 y2 + z2 \u00a2 \u2265 2yz and therefore Ixx \u2265 2Iyz (9.216) and similarly Iyy \u2265 2Izx (9.217) Izz \u2265 2Ixy. (9.218) Example 360 F Coefficients of the characteristic equation. The determinant (9.172)\u00af\u0304\u0304\u0304 \u00af\u0304 Ixx \u2212 \u03bb Ixy Ixz Iyx Iyy \u2212 \u03bb Iyz Izx Izy Izz \u2212 \u03bb \u00af\u0304\u0304\u0304 \u00af\u0304 = 0 (9.219) for calculating the principal moments of inertia, leads to a third-degree equation for \u03bb, called the characteristic equation. \u03bb3 \u2212 a1\u03bb 2 + a2\u03bb\u2212 a3 = 0 (9.220) The coefficients of the characteristic equation are called the principal invariants of [I]. The coefficients of the characteristic equation can directly be found from the following equations: a1 = Ixx + Iyy + Izz = tr [I] (9.221) a2 = IxxIyy + IyyIzz + IzzIxx \u2212 I2xy \u2212 I2yz \u2212 I2zx = \u00af\u0304\u0304\u0304 Ixx Ixy Iyx Iyy \u00af\u0304\u0304\u0304 + \u00af\u0304\u0304\u0304 Iyy Iyz Izy Izz \u00af\u0304\u0304\u0304 + \u00af\u0304\u0304\u0304 Ixx Ixz Izx Izz \u00af\u0304\u0304\u0304 = 1 2 \u00a1 a21 \u2212 tr \u00a3 I2 \u00a4\u00a2 (9.222) a3 = IxxIyyIzz + IxyIyzIzx + IzyIyxIxz \u2212 (IxxIyzIzy + IyyIzxIxz + IzzIxyIyx) = IxxIyyIzz + 2IxyIyzIzx \u2212 \u00a1 IxxI 2 yz + IyyI 2 zx + IzzI 2 xy \u00a2 = det [I] (9.223) 552 9. Applied Dynamics Example 361 F The principal moments of inertia are coordinate invariants. The roots of the inertia characteristic equation are the principal moments of inertia. They are all real but not necessarily different. The principal moments of inertia are extreme. That is, the principal moments of inertia determine the smallest and the largest values of Iii. Since the smallest and largest values of Iii do not depend on the choice of the body coordinate frame, the solution of the characteristic equation is not dependent of the coordinate frame. In other words, if I1, I2, and I3 are the principal moments of inertia for B1I, the principal moments of inertia for B2I are also I1, I2, and I3 when B2I = B2RB1 B1I B2RT B1 . We conclude that I1, I2, and I3 are coordinate invariants of the matrix [I], and therefore any quantity that depends on I1, I2, and I3 is also coordinate invariant. The matrix [I] has only three independent invariants and every other invariant can be expressed in terms of I1, I2, and I3. Since I1, I2, and I3 are the solutions of the characteristic equation of [I] given in (9.220), we may write the determinant (9.172) in the form (\u03bb\u2212 I1) (\u03bb\u2212 I2) (\u03bb\u2212 I3) = 0. (9.224) The expanded form of this equation is \u03bb3 \u2212 (I1 + I2 + I3)\u03bb 2 + (I1I2 + I2I3 + I3I1) a2\u03bb\u2212 I1I2I3 = 0. (9.225) By comparing (9.225) and (9.220) we conclude that a1 = Ixx + Iyy + Izz = I1 + I2 + I3 (9.226) a2 = IxxIyy + IyyIzz + IzzIxx \u2212 I2xy \u2212 I2yz \u2212 I2zx = I1I2 + I2I3 + I3I1 (9.227) a3 = IxxIyyIzz + 2IxyIyzIzx \u2212 \u00a1 IxxI 2 yz + IyyI 2 zx + IzzI 2 xy \u00a2 = I1I2I3. (9.228) Being able to express the coefficients a1, a2, and a3 as functions of I1, I2, and I3 determines that the coefficients of the characteristic equation are coordinate-invariant. Example 362 F Short notation for the elements of inertia matrix. Taking advantage of the Kronecker\u2019s delta (5.138) we may write the el- 9. Applied Dynamics 553 ements of the moment of inertia matrix Iij in short notation forms Iij = Z B \u00a1\u00a1 x21 + x22 + x23 \u00a2 \u03b4ij \u2212 xixj \u00a2 dm (9.229) Iij = Z B \u00a1 r2\u03b4ij \u2212 xixj \u00a2 dm (9.230) Iij = Z B \u00c3 3X k=1 xkxk\u03b4ij \u2212 xixj ! dm (9.231) where we utilized the following notations: x1 = x x2 = y x3 = z. (9.232) Example 363 F Moment of inertia with respect to a plane, a line, and a point. The moment of inertia of a system of particles may be defined with respect to a plane, a line, or a point as the sum of the products of the mass of the particles into the square of the perpendicular distance from the particle to the plane, line, or point. For a continuous body, the sum would be definite integral over the volume of the body. The moments of inertia with respect to the xy, yz, and zx-plane are Iz2 = Z B z2dm (9.233) Iy2 = Z B y2dm (9.234) Ix2 = Z B x2dm. (9.235) The moments of inertia with respect to the x, y, and z axes are Ix = Z B \u00a1 y2 + z2 \u00a2 dm (9.236) Iy = Z B \u00a1 z2 + x2 \u00a2 dm (9.237) Iz = Z B \u00a1 x2 + y2 \u00a2 dm (9.238) and therefore, Ix = Iy2 + Iz2 (9.239) Iy = Iz2 + Ix2 (9.240) Iz = Ix2 + Iy2 . (9.241) 554 9. Applied Dynamics The moment of inertia with respect to the origin is Io = Z B \u00a1 x2 + y2 + z2 \u00a2 dm = Ix2 + Iy2 + Iz2 = 1 2 (Ix + Iy + Iz) . (9.242) Because the choice of the coordinate frame is arbitrary, we can say that the moment of inertia with respect to a line is the sum of the moments of inertia with respect to any two mutually orthogonal planes that pass through the line. The moment of inertia with respect to a point has similar meaning for three mutually orthogonal planes intersecting at the point. 9.5 Lagrange\u2019s Form of Newton\u2019s Equations of Motion Newton\u2019s equation of motion can be transformed to d dt \u00b5 \u2202K \u2202q\u0307r \u00b6 \u2212 \u2202K \u2202qr = Fr r = 1, 2, \u00b7 \u00b7 \u00b7n (9.243) where Fr = nX i=1 \u00b5 Fix \u2202fi \u2202q1 + Fiy \u2202gi \u2202q2 + Fiz \u2202hi \u2202qn \u00b6 . (9.244) Equation (9.243) is called the Lagrange equation of motion, where K is the kinetic energy of the n degree-of-freedom (DOF ) system, qr, r = 1, 2, \u00b7 \u00b7 \u00b7 , n are the generalized coordinates of the system, F = \u00a3 Fix Fiy Fiz \u00a4T is the external force acting on the ith particle of the system, and Fr is the generalized force associated to qr. Proof. Let mi be the mass of one of the particles of a system and let (xi, yi, zi) be its Cartesian coordinates in a globally fixed coordinate frame. Assume that the coordinates of every individual particle are functions of another set of coordinates q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, and possibly time t. xi = fi(q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, t) (9.245) yi = gi(q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, t) (9.246) zi = hi(q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, t) (9.247) If Fxi, Fyi, Fzi are components of the total force acting on the particle mi, then the Newton equations of motion for the particle would be Fxi = mix\u0308i (9.248) Fyi = miy\u0308i (9.249) Fzi = miz\u0308i. (9.250) 9. Applied Dynamics 555 We multiply both sides of these equations by \u2202fi \u2202qr \u2202gi \u2202qr \u2202hi \u2202qr respectively, and add them up for all the particles to have nX i=1 mi \u00b5 x\u0308i \u2202fi \u2202qr + y\u0308i \u2202gi \u2202qr + z\u0308i \u2202hi \u2202qr \u00b6 = nX i=1 \u00b5 Fxi \u2202fi \u2202qr + Fyi \u2202gi \u2202qr + Fzi \u2202hi \u2202qr \u00b6 (9.251) where n is the total number of particles. Taking a time derivative of Equation (9.245), x\u0307i = \u2202fi \u2202q1 q\u03071 + \u2202fi \u2202q2 q\u03072 + \u2202fi \u2202q3 q\u03073 + \u00b7 \u00b7 \u00b7+ \u2202fi \u2202qn q\u0307n + \u2202fi \u2202t (9.252) we find \u2202x\u0307i \u2202q\u0307r = \u2202 \u2202q\u0307r \u00b5 \u2202fi \u2202q1 q\u03071 + \u2202fi \u2202q2 q\u03072 + \u00b7 \u00b7 \u00b7+ \u2202fi \u2202qn q\u0307n + \u2202fi \u2202t \u00b6 = \u2202fi \u2202qr . (9.253) and therefore, x\u0308i \u2202fi \u2202qr = x\u0308i \u2202x\u0307i \u2202q\u0307r = d dt \u00b5 x\u0307i \u2202x\u0307i \u2202q\u0307r \u00b6 \u2212 x\u0307i d dt \u00b5 \u2202x\u0307i \u2202q\u0307r \u00b6 . (9.254) However, x\u0307i d dt \u00b5 \u2202x\u0307i \u2202q\u0307r \u00b6 = x\u0307i d dt \u00b5 \u2202fi \u2202qr \u00b6 = x\u0307i \u00b5 \u22022fi \u2202q1\u2202qr q\u03071 + \u00b7 \u00b7 \u00b7+ \u22022fi \u2202qn\u2202qr q\u0307n + \u22022fi \u2202t\u2202qr \u00b6 = x\u0307i \u2202 \u2202qr \u00b5 \u2202fi \u2202q1 q\u03071 + \u2202fi \u2202q2 q\u03072 + \u00b7 \u00b7 \u00b7+ \u2202fi \u2202qn q\u0307n + \u2202fi \u2202t \u00b6 = x\u0307i \u2202x\u0307i \u2202qr (9.255) and we have x\u0308i \u2202x\u0307i \u2202q\u0307r = d dt \u00b5 x\u0307i \u2202x\u0307i \u2202q\u0307r \u00b6 \u2212 x\u0307i \u2202x\u0307i \u2202qr (9.256) 556 9. Applied Dynamics which is equal to x\u0308i x\u0307i q\u0307r = d dt \u2219 \u2202 \u2202q\u0307r \u00b5 1 2 x\u03072i \u00b6\u00b8 \u2212 \u2202 \u2202qr \u00b5 1 2 x\u03072i \u00b6 . (9.257) Now substituting (9.254) and (9.257) in the left-hand side of (9.251) leads to nX i=1 mi \u00b5 x\u0308i \u2202fi \u2202qr + y\u0308i \u2202gi \u2202qr + z\u0308i \u2202hi \u2202qr \u00b6 = nX i=1 mi d dt \u2219 \u2202 \u2202q\u0307r \u00b5 1 2 x\u03072i + 1 2 y\u03072i + 1 2 z\u03072i \u00b6\u00b8 \u2212 nX i=1 mi \u2202 \u2202qr \u00b5 1 2 x\u03072i + 1 2 y\u03072i + 1 2 z\u03072i \u00b6 = 1 2 nX i=1 mi d dt \u2219 \u2202 \u2202q\u0307r \u00a1 x\u03072i + y\u03072i + z\u03072i \u00a2\u00b8 \u22121 2 nX i=1 mi \u2202 \u2202qr \u00a1 x\u03072i + y\u03072i + z\u03072i \u00a2 . (9.258) where 1 2 nX i=1 mi \u00a1 x\u03072i + y\u03072i + z\u03072i \u00a2 = K (9.259) is the kinetic energy of the system. Therefore, the Newton equations of motion (9.248), (9.249), and (9.250) are converted to d dt \u00b5 \u2202K \u2202q\u0307r \u00b6 \u2212 \u2202K \u2202qr = nX i=1 \u00b5 Fxi \u2202fi \u2202qr + Fyi \u2202gi \u2202qr + Fzi \u2202hi \u2202qr \u00b6 . (9.260) Because of (9.245), (9.246), and (9.247), the kinetic energy is a function of q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn and time t. The left-hand side of Equation (9.260) includes the kinetic energy of the whole system and the right-hand side is a generalized force and shows the effect of changing coordinates from xi to qj on the external forces. Let us assume that the coordinate qr alters to qr+ \u03b4qr while the other coordinates q1, q2, q3, \u00b7 \u00b7 \u00b7 , qr\u22121, qr+1, \u00b7 \u00b7 \u00b7 , qn and time t are unaltered. So, the coordinates of mi are changed to xi + \u2202fi \u2202qr \u03b4qr (9.261) yi + \u2202gi \u2202qr \u03b4qr (9.262) zi + \u2202hi \u2202qr \u03b4qr (9.263) 9. Applied Dynamics 557 Such a displacement is called virtual displacement. The work done in this virtual displacement by all forces acting on the particles of the system is \u03b4W = nX i=1 \u00b5 Fxi \u2202fi \u2202qr + Fyi \u2202gi \u2202qr + Fzi \u2202hi \u2202qr \u00b6 \u03b4qr. (9.264) Because the work done by internal forces appears in opposite pairs, only the work done by external forces remains in Equation (9.264). Let\u2019s denote the virtual work by \u03b4W = Fr (q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, t) \u03b4qr. (9.265) Then we have d dt \u00b5 \u2202K \u2202q\u0307r \u00b6 \u2212 \u2202K \u2202qr = Fr (9.266) where Fr = nX i=1 \u00b5 Fxi \u2202fi \u2202qr + Fyi \u2202gi \u2202qr + Fzi \u2202hi \u2202qr \u00b6 . (9.267) Equation (9.266) is the Lagrange form of equations of motion. This equation is true for all values of r from 1 to n. We thus have n second-order ordinary differential equations in which q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn are the dependent variables and t is the independent variable. The coordinates q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn are called generalized coordinates and can be any measurable parameters to provide the configuration of the system. The number of equations and the number of dependent variables are equal, therefore, the equations are theoretically sufficient to determine the motion of all mi. Example 364 Equation of motion for a simple pendulum. A pendulum is shown in Figure 9.11. Using x and y for the Cartesian position of m, and using \u03b8 = q as the generalized coordinate, we have x = f(\u03b8) = l sin \u03b8 (9.268) y = g(\u03b8) = l cos \u03b8 (9.269) K = 1 2 m \u00a1 x\u03072 + y\u03072 \u00a2 = 1 2 ml2\u03b8\u0307 2 (9.270) and therefore, d dt \u00b5 \u2202K \u2202\u03b8\u0307 \u00b6 \u2212 \u2202K \u2202\u03b8 = d dt (ml2\u03b8\u0307) = ml2\u03b8\u0308. (9.271) The external force components, acting on m, are Fx = 0 (9.272) Fy = mg (9.273) 558 9. Applied Dynamics and therefore, F\u03b8 = Fx \u2202f \u2202\u03b8 + Fy \u2202g \u2202\u03b8 = \u2212mgl sin \u03b8. (9.274) Hence, the equation of motion for the pendulum is ml2\u03b8\u0308 = \u2212mgl sin \u03b8. (9.275) Example 365 A pendulum attached to an oscillating mass. Figure 9.12 illustrates a vibrating mass with a hanging pendulum. The pendulum can act as a vibration absorber if designed properly. Starting with coordinate relationships xM = fM = x (9.276) yM = gM = 0 (9.277) xm = fm = x+ l sin \u03b8 (9.278) ym = gm = l cos \u03b8 (9.279) we may find the kinetic energy in terms of the generalized coordinates x and \u03b8. K = 1 2 M \u00a1 x\u03072M + y\u03072M \u00a2 + 1 2 m \u00a1 x\u03072m + y\u03072m \u00a2 = 1 2 Mx\u03072 + 1 2 m \u00b3 x\u03072 + l2\u03b8\u0307 2 + 2lx\u0307\u03b8\u0307 cos \u03b8 \u00b4 (9.280) Then, the left-hand side of the Lagrange equations are d dt \u00b5 \u2202K \u2202x\u0307 \u00b6 \u2212 \u2202K \u2202x = (M +m)x\u0308+ml\u03b8\u0308 cos \u03b8 \u2212ml\u03b8\u0307 2 sin \u03b8 (9.281) d dt \u00b5 \u2202K \u2202\u03b8\u0307 \u00b6 \u2212 \u2202K \u2202\u03b8 = ml2\u03b8\u0308 +mlx\u0308 cos \u03b8. (9.282) 9. Applied Dynamics 559 The external forces acting on M and m are FxM = \u2212kx (9.283) FyM = 0 (9.284) Fxm = 0 (9.285) Fym = mg. (9.286) Therefore, the generalized forces are Fx = FxM \u2202fM \u2202x + FyM \u2202gM \u2202x + Fxm \u2202fm \u2202x + Fym \u2202gm \u2202x = \u2212kx (9.287) F\u03b8 = FxM \u2202fM \u2202\u03b8 + FyM \u2202gM \u2202\u03b8 + Fxm \u2202fm \u2202\u03b8 + Fym \u2202gm \u2202\u03b8 = \u2212mgl sin \u03b8 (9.288) and finally the Lagrange equations of motion are (M +m)x\u0308+ml\u03b8\u0308 cos \u03b8 \u2212ml\u03b8\u0307 2 sin \u03b8 = \u2212kx (9.289) ml2\u03b8\u0308 +mlx\u0308 cos \u03b8 = \u2212mgl sin \u03b8. (9.290) Example 366 Kinetic energy of the Earth. Earth is approximately a rotating rigid body about a fixed axis. The two motions of the Earth are called revolution about the sun, and rotation about an axis approximately fixed in the Earth. The kinetic energy of the Earth due to its rotation is K1 = 1 2 I\u03c921 = 1 2 2 5 \u00a1 5.9742\u00d7 1024 \u00a2\u00b56356912 + 6378388 2 \u00b62\u00b5 2\u03c0 24\u00d7 3600 366.25 365.25 \u00b62 = 2.5762\u00d7 1029 J 560 9. Applied Dynamics and the kinetic energy of the Earth due to its revolution is K2 = 1 2 Mr2\u03c922 = 1 2 \u00a1 5.9742\u00d7 1024 \u00a2 \u00a1 1.49475\u00d7 1011 \u00a22\u00b5 2\u03c0 24\u00d7 3600 1 365.25 \u00b62 = 2.6457\u00d7 1033 J where r is the distance from the sun and \u03c92 is the angular speed about the sun. The total kinetic energy of the Earth is K = K1 +K2. However, the ratio of the revolutionary to rotational kinetic energies is K2 K1 = 2.6457\u00d7 1033 2.5762\u00d7 1029 \u2248 10000. Example 367 F Explicit form of Lagrange equations. Assume the coordinates of every particle are functions of the coordinates q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn but not the time t. The kinetic energy of the system made of n massive particles can be written as K = 1 2 nX i=1 mi \u00a1 x\u03072i + y\u03072i + z\u03072i \u00a2 = 1 2 nX j=1 nX k=1 ajkq\u0307j q\u0307k (9.291) where the coefficients ajk are functions of q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn and ajk = akj . (9.292) The Lagrange equations of motion d dt \u00b5 \u2202K \u2202q\u0307r \u00b6 \u2212 \u2202K \u2202qr = Fr r = 1, 2, \u00b7 \u00b7 \u00b7n (9.293) are then equal to d dt nX m=1 amrq\u0307m \u2212 1 2 nX j=1 nX k=1 ajk \u2202qr q\u0307j q\u0307k = Fr (9.294) or nX m=1 amrq\u0308m + nX k=1 nX n=1 \u0393rk,nq\u0307kq\u0307n = Fr (9.295) where \u0393ij,k is called the Christoffel operator \u0393ij,k = 1 2 \u00b5 \u2202aij \u2202qk + \u2202aik \u2202qj \u2212 \u2202akj \u2202qi \u00b6 . (9.296) 9. Applied Dynamics 561 9.6 Lagrangian Mechanics Assume for some forces F = \u00a3 Fix Fiy Fiz \u00a4T there is a function V , called potential energy, such that the force is derivable from V F = \u2212\u2207V. (9.297) Such a force is called potential or conservative force. Then, the Lagrange equation of motion can be written as d dt \u00b5 \u2202L \u2202q\u0307r \u00b6 \u2212 \u2202L \u2202qr = Qr r = 1, 2, \u00b7 \u00b7 \u00b7n (9.298) where L = K \u2212 V (9.299) is the Lagrangean of the system and Qr is the nonpotential generalized force. Proof. Assume the external forces F = \u00a3 Fxi Fyi Fzi \u00a4T acting on the system are conservative. F = \u2212\u2207V (9.300) The work done by these forces in an arbitrary virtual displacement \u03b4q1, \u03b4q2, \u03b4q3, \u00b7 \u00b7 \u00b7 , \u03b4qn is \u2202W = \u2212\u2202V \u2202q1 \u03b4q1 \u2212 \u2202V \u2202q2 \u03b4q2 \u2212 \u00b7 \u00b7 \u00b7 \u2202V \u2202qn \u03b4qn (9.301) then the Lagrange equation becomes d dt \u00b5 \u2202K \u2202q\u0307r \u00b6 \u2212 \u2202K \u2202qr = \u2212 \u2202V \u2202q1 r = 1, 2, \u00b7 \u00b7 \u00b7n. (9.302) Introducing the Lagrangean function L = K \u2212 V converts the Lagrange equation to d dt \u00b5 \u2202L \u2202q\u0307r \u00b6 \u2212 \u2202L \u2202qr = 0 r = 1, 2, \u00b7 \u00b7 \u00b7n (9.303) for a conservative system. The Lagrangean is also called kinetic potential. If a force is not conservative, then the virtual work done by the force is \u03b4W = nX i=1 \u00b5 Fxi \u2202fi \u2202qr + Fyi \u2202gi \u2202qr + Fzi \u2202hi \u2202qr \u00b6 \u03b4qr = Qr \u03b4qr (9.304) and the equation of motion would be d dt \u00b5 \u2202L \u2202q\u0307r \u00b6 \u2212 \u2202L \u2202qr = Qr r = 1, 2, \u00b7 \u00b7 \u00b7n (9.305) where Qr is the nonpotential generalized force doing work in a virtual displacement of the rth generalized coordinate qr. 562 9. Applied Dynamics Example 368 Spherical pendulum. A pendulum analogy is utilized in modeling of many dynamical problems. Figure 9.13 illustrates a spherical pendulum with mass m and length l. The angles \u03d5 and \u03b8 may be used as describing coordinates of the system. The Cartesian coordinates of the mass as a function of the generalized coordinates are \u23a1\u23a3 X Y Z \u23a4\u23a6 = \u23a1\u23a3 r cos\u03d5 sin \u03b8 r sin \u03b8 sin\u03d5 \u2212r cos \u03b8 \u23a4\u23a6 (9.306) and therefore, the kinetic and potential energies of the pendulum are K = 1 2 m \u00b3 l2\u03b8\u0307 2 + l2\u03d5\u03072 sin2 \u03b8 \u00b4 (9.307) V = \u2212mgl cos \u03b8. (9.308) The kinetic potential function of this system is then equal to L = 1 2 m \u00b3 l2\u03b8\u0307 2 + l2\u03d5\u03072 sin2 \u03b8 \u00b4 +mgl cos \u03b8 (9.309) which leads to the following equations of motion: \u03b8\u0308 \u2212 \u03d5\u03072 sin \u03b8 cos \u03b8 + g l sin \u03b8 = 0 (9.310) \u03d5\u0308 sin2 \u03b8 + 2\u03d5\u0307\u03b8\u0307 sin \u03b8 cos \u03b8 = 0. (9.311) Example 369 Controlled compound pendulum. A massive arm is attached to a ceiling at a pin joint O as illustrated in Figure 9.14. Assume that there is viscous friction in the joint where an ideal motor can apply a torque Q to move the arm. The rotor of an ideal motor has no moment of inertia by assumption. 9. Applied Dynamics 563 The kinetic and potential energies of the manipulator are K = 1 2 I\u03b8\u0307 2 = 1 2 \u00a1 IC +ml2 \u00a2 \u03b8\u0307 2 (9.312) V = \u2212mg cos \u03b8 (9.313) where m is the mass and I is the moment of inertia of the pendulum about O. The Lagrangean of the manipulator is L = K \u2212 V = 1 2 I\u03b8\u0307 2 +mg cos \u03b8 (9.314) and therefore, the equation of motion of the pendulum is M = d dt \u00b5 \u2202L \u2202\u03b8\u0307 \u00b6 \u2212 \u2202L \u2202\u03b8 = I \u03b8\u0308 +mgl sin \u03b8. (9.315) The generalized force M is the contribution of the motor torque Q and the viscous friction torque \u2212c\u03b8\u0307. Hence, the equation of motion of the manipulator is Q = I \u03b8\u0308 + c\u03b8\u0307 +mgl sin \u03b8. (9.316) Example 370 An ideal 2R planar manipulator dynamics. An ideal model of a 2R planar manipulator is illustrated in Figure 9.15. It is called ideal because we assume the links are massless and there is no friction. The masses m1 and m2 are the mass of the second motor to run 564 9. Applied Dynamics the second link and the load at the endpoint. We take the absolute angle \u03b81 and the relative angle \u03b82 as the generalized coordinates to express the configuration of the manipulator. The global positions of m1 and m2 are\u2219 X1 Y2 \u00b8 = \u2219 l1 cos \u03b81 l1 sin \u03b81 \u00b8 (9.317)\u2219 X2 Y2 \u00b8 = \u2219 l1 cos \u03b81 + l2 cos (\u03b81 + \u03b82) l1 sin \u03b81 + l2 sin (\u03b81 + \u03b82) \u00b8 (9.318) and therefore, the global velocity of the masses are\u2219 X\u03071 Y\u03071 \u00b8 = \u2219 \u2212l1\u03b8\u03071 sin \u03b81 l1\u03b8\u03071 cos \u03b81 \u00b8 (9.319) \u2219 X\u03072 Y\u03072 \u00b8 = \u23a1\u23a3 \u2212l1\u03b8\u03071 sin \u03b81 \u2212 l2 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 sin (\u03b81 + \u03b82) l1\u03b8\u03071 cos \u03b81 + l2 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 cos (\u03b81 + \u03b82) \u23a4\u23a6 . (9.320) The kinetic energy of this manipulator is made of kinetic energy of the masses and is equal to K = K1 +K2 = 1 2 m1 \u00b3 X\u03072 1 + Y\u0307 2 1 \u00b4 + 1 2 m2 \u00b3 X\u03072 2 + Y\u0307 2 2 \u00b4 = 1 2 m1l 2 1\u03b8\u0307 2 1 + 1 2 m2 \u00b5 l21\u03b8\u0307 2 1 + l22 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b42 + 2l1l2\u03b8\u03071 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 cos \u03b82 \u00b6 . (9.321) 9. Applied Dynamics 565 The potential energy of the manipulator is V = V1 + V2 = m1gY1 +m2gY2 = m1gl1 sin \u03b81 +m2g (l1 sin \u03b81 + l2 sin (\u03b81 + \u03b82)) . (9.322) The Lagrangean is then obtained from Equations (9.321) and (9.322) L = K \u2212 V (9.323) = 1 2 m1l 2 1\u03b8\u0307 2 1 + 1 2 m2 \u00b5 l21\u03b8\u0307 2 1 + l22 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b42 + 2l1l2\u03b8\u03071 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 cos \u03b82 \u00b6 \u2212 (m1gl1 sin \u03b81 +m2g (l1 sin \u03b81 + l2 sin (\u03b81 + \u03b82))) . which provides the required partial derivatives as follows: \u2202L \u2202\u03b81 = \u2212 (m1 +m2) gl1 cos \u03b81 \u2212m2gl2 cos (\u03b81 + \u03b82) (9.324) \u2202L \u2202\u03b8\u03071 = (m1 +m2) l 2 1\u03b8\u03071 +m2l 2 2 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 +m2l1l2 \u00b3 2\u03b8\u03071 + \u03b8\u03072 \u00b4 cos \u03b82 (9.325) d dt \u00b5 \u2202L \u2202\u03b8\u03071 \u00b6 = (m1 +m2) l 2 1\u03b8\u03081 +m2l 2 2 \u00b3 \u03b8\u03081 + \u03b8\u03082 \u00b4 +m2l1l2 \u00b3 2 \u03b8\u03081 + \u03b8\u03082 \u00b4 cos \u03b82 \u2212m2l1l2\u03b8\u03072 \u00b3 2\u03b8\u03071 + \u03b8\u03072 \u00b4 sin \u03b82 (9.326) \u2202L \u2202\u03b82 = \u2212m2l1l2\u03b8\u03071 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 sin \u03b82 \u2212m2gl2 cos (\u03b81 + \u03b82) (9.327) \u2202L \u2202\u03b8\u03072 = m2l 2 2 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 +m2l1l2\u03b8\u03071 cos \u03b82 (9.328) d dt \u00b5 \u2202L \u2202\u03b8\u03072 \u00b6 = m2l 2 2 \u00b3 \u03b8\u03081 + \u03b8\u03082 \u00b4 +m2l1l2\u03b8\u03081 cos \u03b82 \u2212m2l1l2\u03b8\u03071\u03b8\u03072 sin \u03b82 (9.329) Therefore, the equations of motion for the 2R manipulator are Q1 = d dt \u00b5 \u2202L \u2202\u03b8\u03071 \u00b6 \u2212 \u2202L \u2202\u03b81 = (m1 +m2) l 2 1\u03b8\u03081 +m2l 2 2 \u00b3 \u03b8\u03081 + \u03b8\u03082 \u00b4 +m2l1l2 \u00b3 2 \u03b8\u03081 + \u03b8\u03082 \u00b4 cos \u03b82 \u2212m2l1l2\u03b8\u03072 \u00b3 2\u03b8\u03071 + \u03b8\u03072 \u00b4 sin \u03b82 +(m1 +m2) gl1 cos \u03b81 +m2gl2 cos (\u03b81 + \u03b82) (9.330) 566 9. Applied Dynamics Q2 = d dt \u00b5 \u2202L \u2202\u03b8\u03072 \u00b6 \u2212 \u2202L \u2202\u03b82 = m2l 2 2 \u00b3 \u03b8\u03081 + \u03b8\u03082 \u00b4 +m2l1l2\u03b8\u03081 cos \u03b82 \u2212m2l1l2\u03b8\u03071\u03b8\u03072 sin \u03b82 +m2l1l2\u03b8\u03071 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 sin \u03b82 +m2gl2 cos (\u03b81 + \u03b82) . (9.331) The generalized forces Q1 and Q2 are the required forces to drive the generalized coordinates. In this case, Q1 is the torque at the base motor and Q2 is the torque of the motor at m1. The equations of motion can be rearranged to have a more systematic form Q1 = \u00a1 (m1 +m2) l 2 1 +m2l2 (l2 + 2l1 cos \u03b82) \u00a2 \u03b8\u03081 +m2l2 (l2 + l1 cos \u03b82) \u03b8\u03082 \u22122m2l1l2 sin \u03b82 \u03b8\u03071\u03b8\u03072 \u2212m2l1l2 sin \u03b82 \u03b8\u0307 2 2 +(m1 +m2) gl1 cos \u03b81 +m2gl2 cos (\u03b81 + \u03b82) (9.332) Q2 = m2l2 (l2 + l1 cos \u03b82) \u03b8\u03081 +m2l 2 2\u03b8\u03082 +m2l1l2 sin \u03b82 \u03b8\u0307 2 1 +m2gl2 cos (\u03b81 + \u03b82) . (9.333) Example 371 Mechanical energy. If a system of masses mi are moving in a potential force field Fmi = \u2212\u2207iV (9.334) their Newton equations of motion will be mir\u0308i = \u2212\u2207iV i = 1, 2, \u00b7 \u00b7 \u00b7n. (9.335) The inner product of equations of motion with r\u0307i and adding the equations nX i=1 mir\u0307i \u00b7 r\u0308i = \u2212 nX i=1 r\u0307i \u00b7\u2207iV (9.336) and then, integrating over time 1 2 nX i=1 mir\u0307i \u00b7 r\u0307i = \u2212 Z nX i=1 ri \u00b7\u2207iV (9.337) shows that K = \u2212 Z nX i=1 \u00b5 \u2202V \u2202xi xi + \u2202V \u2202yi yi + \u2202V \u2202zi zi \u00b6 = \u2212V +E (9.338) 9. Applied Dynamics 567 where E is the constant of integration. E is called the mechanical energy of the system and is equal to kinetic plus potential energies. E = K + V (9.339) Example 372 Falling wheel. Figure 9.16 illustrates a wheel turning, without slip, over a cylindrical hill. We may use the conservation of mechanical energy to find the angle at which the wheel leaves the hill. At the initial instant of time, the wheel is at point A. We assume the initial kinetic and potential, and hence, the mechanical energies are zero. When the wheel is turning over the hill, its angular velocity, \u03c9, is \u03c9 = v r (9.340) where v is the speed at the center of the wheel. At any other point B, the wheel achieves some kinetic energy and loses some potential energy. At a certain angle, where the normal component of the weight cannot provide more centripetal force, mg cos \u03b8 = mv2 R+ r . (9.341) the wheel separates from the surface. Employing the conservation of energy, we have EA = EB (9.342) KA + VA = KB + VB. (9.343) The kinetic and potential energy at the separation point B are KB = 1 2 mv2 + 1 2 IC\u03c9 2 (9.344) VB = \u2212mg (R+ r) (1\u2212 cos \u03b8) (9.345) 568 9. Applied Dynamics where IC is the mass moment of inertia for the wheel about its center. Therefore, 1 2 mv2 + 1 2 IC\u03c9 2 = mg (R+ r) (1\u2212 cos \u03b8) (9.346) and substituting (9.340) and (9.341) provides\u00b5 1 + IC mr2 \u00b6 (R+ r) g cos \u03b8 = 2g (R+ r) (1\u2212 cos \u03b8) (9.347) and therefore, the separation angle is \u03b8 = cos\u22121 2mr2 IC + 3mr2 . (9.348) Let\u2019s examine the equation for a disc wheel with IC = 1 2 mr2. (9.349) and find the separation angle. \u03b8 = cos\u22121 4 7 (9.350) \u2248 0.96 rad \u2248 55.15 deg Example 373 Turning wheel over a step. Figure 9.17 illustrates a wheel of radius R turning with speed v to go over a step with height H < R. We may use the principle of energy conservation and find the speed of the wheel after getting across the step. Employing the conservation of energy, 9. Applied Dynamics 569 we have EA = EB (9.351) KA + VA = KB + VB (9.352) 1 2 mv21 + 1 2 IC\u03c9 2 1 + 0 = 1 2 mv22 + 1 2 IC\u03c9 2 2 +mgH (9.353)\u00b5 m+ IC R2 \u00b6 v21 = \u00b5 m+ IC R2 \u00b6 v22 + 2mgH (9.354) and therefore, v2 = vuutv21 \u2212 2gH 1 + IC mR2 . (9.355) The condition for having a real v2 is v1 > vuut 2gH 1 + IC mR2 . (9.356) The second speed (9.355) and the condition (9.356) for a solid disc are v2 = r v21 \u2212 4 3 Hg (9.357) v1 > r 4 3 Hg (9.358) because we assumed that IC = 1 2 mR2. (9.359) Example 374 Trebuchet. A trebuchet, shown schematically in Figure 9.18, is a shooting weapon of war powered by a falling massive counterweight m1. A beam AB is pivoted to the chassis with two unequal sections a and b. The figure shows a trebuchet at its initial configuration. The origin of a global coordinate frame is set at the pivot point. The counterweight m1 is at (x1, y1) and is hinged at the shorter arm of the beam at a distance c from the end B. The mass of the projectile is m2 and it is at the end of a massless sling with a length l attached to the end of the longer arm of the beam. The three independent variable angles \u03b1, \u03b8, and \u03b3 describe the motion of the device. We consider the parameters a, b, c, d, l, m1, and m2 constant, and determine the equations of motion by the Lagrange method. Figure 9.19 illustrates the trebuchet when it is in motion. The position coordinates of masses m1 and m2 are x1 = b sin \u03b8 \u2212 c sin (\u03b8 + \u03b3) (9.360) y1 = \u2212b cos \u03b8 + c cos (\u03b8 + \u03b3) (9.361) 570 9. Applied Dynamics and x2 = \u2212a sin \u03b8 \u2212 l sin (\u2212\u03b8 + \u03b1) (9.362) y2 = \u2212a cos \u03b8 \u2212 l cos (\u2212\u03b8 + \u03b1) . (9.363) Taking a time derivative provides the velocity components x\u03071 = b\u03b8\u0307 cos \u03b8 \u2212 c \u00b3 \u03b8\u0307 + \u03b3\u0307 \u00b4 cos (\u03b8 + \u03b3) (9.364) y\u03071 = b\u03b8\u0307 sin \u03b8 \u2212 c \u00b3 \u03b8\u0307 + \u03b3\u0307 \u00b4 sin (\u03b8 + \u03b3) (9.365) x\u03072 = l (c\u2212 \u03b1\u0307) cos (\u03b1\u2212 \u03b8)\u2212 a\u03b8\u0307 cos (\u03b8) (9.366) y\u03072 = a\u03b8\u0307 sin \u03b8 \u2212 l \u00b3 \u03b8\u0307 \u2212 \u03b1\u0307 \u00b4 sin (\u03b1\u2212 \u03b8) . (9.367) 9. Applied Dynamics 571 which shows that the kinetic energy of the system is K = 1 2 m1v 2 1 + 1 2 m2v 2 2 = 1 2 m1 \u00a1 x\u030721 + y\u030721 \u00a2 + 1 2 m2 \u00a1 x\u030722 + y\u030722 \u00a2 = 1 2 m1 \u00b3\u00a1 b2 + c2 \u00a2 \u03b8\u0307 2 + c2\u03b3\u03072 + 2c2\u03b8\u0307\u03b3\u0307 \u00b4 \u2212m1bc\u03b8\u0307 \u00b3 \u03b8\u0307 + \u03b3\u0307 \u00b4 cos \u03b3 + 1 2 m2 \u00b3\u00a1 a2 + l2 \u00a2 \u03b8\u0307 2 + l2\u03b1\u03072 \u2212 2l2\u03b8\u0307\u03b1\u0307 \u00b4 \u2212m2al\u03b8\u0307 \u00b3 \u03b8\u0307 \u2212 \u03b1\u0307 \u00b4 cos (2\u03b8 \u2212 \u03b1) . (9.368) The potential energy of the system can be calculated by y-position of the masses. V = m1gy1 +m2gy2 = m1g (\u2212b cos \u03b8 + c cos (\u03b8 + \u03b3)) +m2g (\u2212a cos \u03b8 \u2212 l cos (\u2212\u03b8 + \u03b1)) (9.369) Having the energies K and V , we can set up the Lagrangean L. L = K \u2212 V (9.370) Using the Lagrangean, we are able to find the three equations of motion. d dt \u00b5 \u2202L \u2202\u03b8\u0307 \u00b6 \u2212 \u2202L \u2202\u03b8 = 0 (9.371) d dt \u00b5 \u2202L \u2202\u03b1\u0307 \u00b6 \u2212 \u2202L \u2202\u03b1 = 0 (9.372) d dt \u00b5 \u2202L \u2202\u03b3\u0307 \u00b6 \u2212 \u2202L \u2202\u03b3 = 0. (9.373) The trebuchet appeared in 500 to 400 B.C. China and was developed by Persian armies around 300 B.C. It was used by the Arabs against the Romans during 600 to 1200 A.D. The trebuchet is also called the manjaniq, catapults, or onager. The \"Manjaniq\" is the root of the words \"machine\" and \"mechanic\". 9.7 Summary The translational and rotational equations of motion for a rigid body, expressed in the global coordinate frame, are GF = Gd dt Gp (9.374) GM = Gd dt GL (9.375) 572 9. Applied Dynamics where GF and GM indicate the resultant of the external forces and moments applied on the rigid body, measured at the mass center C. The vector Gp is the momentum and GL is the moment of momentum for the rigid body at C p = mv (9.376) L = rC \u00d7 p. (9.377) The expression of the equations of motion in the body coordinate frame are BF = Gp\u0307+ B G\u03c9B \u00d7 Bp = m BaB +m B G\u03c9B \u00d7 BvB (9.378) BM = BL\u0307+B G\u03c9B \u00d7 BL = BI B G\u03c9\u0307B + B G\u03c9B \u00d7 \u00a1 BI B G\u03c9B \u00a2 (9.379) where I is the moment of inertia for the rigid body. I = \u23a1\u23a3 Ixx Ixy Ixz Iyx Iyy Iyz Izx Izy Izz \u23a4\u23a6 . (9.380) 9. Applied Dynamics 573 The elements of I are functions of the mass distribution of the rigid body and are defined by Iij = Z B \u00a1 r2i \u03b4mn \u2212 ximxjn \u00a2 dm , i, j = 1, 2, 3 (9.381) where \u03b4ij is Kronecker\u2019s delta. Every rigid body has a principal body coordinate frame in which the moment of inertia is in the form BI = \u23a1\u23a3 I1 0 0 0 I2 0 0 0 I3 \u23a4\u23a6 . (9.382) The rotational equation of motion in the principal coordinate frame simplifies to M1 = I1\u03c9\u03071 \u2212 (I2 \u2212 I2)\u03c92\u03c93 M2 = I2\u03c9\u03072 \u2212 (I3 \u2212 I1)\u03c93\u03c91 (9.383) M3 = I3\u03c9\u03073 \u2212 (I1 \u2212 I2)\u03c91\u03c92. The equations of motion for a mechanical system having n DOF can also be found by the Lagrange equation d dt \u00b5 \u2202L \u2202q\u0307r \u00b6 \u2212 \u2202L \u2202qr = Qr r = 1, 2, \u00b7 \u00b7 \u00b7n (9.384) L = K \u2212 V (9.385) where L is the Lagrangean of the system, K is the kinetic energy, V is the potential energy, and Qr is the nonpotential generalized force. Qr = nX i=1 \u00b5 Qix \u2202fi \u2202q1 +Qiy \u2202gi \u2202q2 +Qiz \u2202hi \u2202qn \u00b6 (9.386) The parameters qr, r = 1, 2, \u00b7 \u00b7 \u00b7 , n are the generalized coordinates of the system, Q = \u00a3 Qix Qiy Qiz \u00a4T is the external force acting on the ith particle of the system, and Qr is the generalized force associated to qr. When (xi, yi, zi) are Cartesian coordinates in a globally fixed coordinate frame for the particle mi, then its coordinates may be functions of another set of generalized coordinates q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn and possibly time t. xi = fi(q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, t) (9.387) yi = gi(q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, t) (9.388) zi = hi(q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, t) (9.389) 574 9. Applied Dynamics 9.8 Key Symbols a, b, w, h length a acceleration C mass center d position vector of the body coordinate frame df infinitesimal force dm infinitesimal mass dm infinitesimal moment E mechanical energy F force FC Coriolis force g gravitational acceleration H height I moment of inertia matrix I1, I2, I3 principal moment of inertia K kinetic energy l directional line L moment of momentum L = K \u2212 V Lagrangean m mass M moment p momentum P,Q points in rigid body r radius of disc r position vector R radius R rotation matrix t time u\u0302 unit vector to show the directional line v \u2261 x\u0307, v velocity V potential energy w eigenvector W work W eigenvector matrix x, y, z, x displacement \u03b4ij Kronecker\u2019s delta \u0393ij,k Christoffel operator \u03bb eigenvalue \u03d5, \u03b8, \u03c8 Euler angles \u03c9,\u03c9 angular velocity k parallel \u22a5 orthogonal 9. Applied Dynamics 575 Exercises 1. Kinetic energy of a rigid link. Consider a straight and uniform bar as a rigid bar. The bar has a mass m. Show that the kinetic energy of the bar can be expressed as K = 1 6 m (v1 \u00b7 v1 + v1 \u00b7 v2 + v2 \u00b7 v2) where v1 and v2 are the velocity vectors of the endpoints of the bar. 2. Discrete particles. There are three particles m1 = 10kg, m2 = 20 kg, m3 = 30 kg, at r1 = \u23a1\u23a3 1 \u22121 1 \u23a4\u23a6 r1 = \u23a1\u23a3 \u22121\u22123 2 \u23a4\u23a6 r1 = \u23a1\u23a3 2 \u22121 \u22123 \u23a4\u23a6 . Their velocities are v1 = \u23a1\u23a3 2 1 1 \u23a4\u23a6 v1 = \u23a1\u23a3 \u221210 2 \u23a4\u23a6 v1 = \u23a1\u23a3 3 \u22122 \u22121 \u23a4\u23a6 . Find the position and velocity of the system at C. Calculate the system\u2019s momentum and moment of momentum. Calculate the system\u2019s kinetic energy and determine the rotational and translational parts of the kinetic energy. 3. Newton\u2019s equation of motion in the body frame. Show that Newton\u2019s equation of motion in the body frame is\u23a1\u23a3 Fx Fy Fz \u23a4\u23a6 = m \u23a1\u23a3 ax ay az \u23a4\u23a6+ \u23a1\u23a3 0 \u2212\u03c9z \u03c9y \u03c9z 0 \u2212\u03c9x \u2212\u03c9y \u03c9x 0 \u23a4\u23a6\u23a1\u23a3 vx vy vz \u23a4\u23a6 . 4. Work on a curved path. A particle of mass m is moving on a circular path given by GrP = cos \u03b8 I\u0302 + sin \u03b8 J\u0302 + 4 K\u0302. Calculate the work done by a force GF when the particle moves from \u03b8 = 0 to \u03b8 = \u03c0 2 . (a) GF = z2 \u2212 y2 (x+ y) 2 I\u0302 + y2 \u2212 x2 (x+ y) 2 J\u0302 + x2 \u2212 y2 (x+ z) 2 K\u0302 576 9. Applied Dynamics (b) GF = z2 \u2212 y2 (x+ y) 2 I\u0302 + 2y x+ y J\u0302 + x2 \u2212 y2 (x+ z) 2 K\u0302 5. Principal moments of inertia. Find the principal moments of inertia and directions for the following inertia matrices: (a) [I] = \u23a1\u23a3 3 2 2 2 2 0 2 0 4 \u23a4\u23a6 (b) [I] = \u23a1\u23a3 3 2 4 2 0 2 4 2 3 \u23a4\u23a6 (c) [I] = \u23a1\u23a3 100 20 \u221a 3 0 20 \u221a 3 60 0 0 0 10 \u23a4\u23a6 6. Rotated moment of inertia matrix. A principal moment of inertia matrix B2I is given as [I] = \u23a1\u23a3 3 0 0 0 5 0 0 0 4 \u23a4\u23a6 . The principal frame was achieved by rotating the initial body coordinate frame 30 deg about the x-axis, followed by 45 deg about the z-axis. Find the initial moment of inertia matrix B1I. 7. Rotation of moment of inertia matrix. Find the required rotation matrix that transforms the moment of inertia matrix [I] to an diagonal matrix. [I] = \u23a1\u23a3 3 2 2 2 2 0.1 2 0.1 4 \u23a4\u23a6 8. F Cubic equations. The solution of a cubic equation ax3 + bx2 + cx+ d = 0 9. Applied Dynamics 577 where a 6= 0, can be found in a systematic way. Transform the equation to a new form with discriminant 4p3 + q2, y3 + 3py + q = 0 using the transformation x = y \u2212 b 3a , where, p = 3ac\u2212 b2 9a2 q = 2b3 \u2212 9abc+ 27a2d 27a3 . The solutions are then y1 = 3 \u221a \u03b1\u2212 3 p \u03b2 y2 = e 2\u03c0i 3 3 \u221a \u03b1\u2212 e 4\u03c0i 3 3 p \u03b2 y3 = e 4\u03c0i 3 3 \u221a \u03b1\u2212 e 2\u03c0i 3 3 p \u03b2 where, \u03b1 = \u2212q + p q2 + 4p3 2 \u03b2 = \u2212q + p q2 + 4p3 2 . For real values of p and q, if the discriminant is positive, then one root is real, and two roots are complex conjugates. If the discriminant is zero, then there are three real roots, of which at least two are equal. If the discriminant is negative, then there are three unequal real roots. Apply this theory for the characteristic equation of the matrix [I] and show that the principal moments of inertia are real. 9. Kinematics of a moving car on the Earth. The location of a vehicle on the Earth is described by its longitude \u03d5 from a fixed meridian, say, the Greenwich meridian, and its latitude \u03b8 from the equator, as shown in Figure 9.20. We attach a coordinate frame B at the center of the Earth with the x-axis on the equator\u2019s plane and the y-axis pointing to the vehicle. There are also two coordinate frames E and G where E is attached to the Earth and G is the global coordinate frame. Show that the angular velocity of B and the velocity of the vehicle are B G\u03c9B = \u03b8\u0307 \u0131\u0302B + (\u03c9E + \u03d5\u0307) sin \u03b8 j\u0302B + (\u03c9E + \u03d5\u0307) cos \u03b8 k\u0302 B GvP = \u2212r (\u03c9E + \u03d5\u0307) cos \u03b8 \u0131\u0302B + r\u03b8\u0307 k\u0302. Calculate the acceleration of the vehicle. 578 9. Applied Dynamics \u03b8 X Y Z x y z r \u03c9E Z (a) (b) G B Y z x y E \u03d5 \u03b8 y z \u03c9E P P 9. Applied Dynamics 579 (c) the pivot O has a uniform motion on a circle rO = R cos\u03c9t I\u0302 +R sin\u03c9t J\u0302. 13. Equations of motion from Lagrangean. Consider a physical system with a Lagrangean as L = 1 2 m (ax\u0307+ by\u0307) 2 \u2212 1 2 k (ax+ by) 2 . and find the equations of motion. The coefficients m, k, a, and b are constant. 14. Lagrangean from equation of motion. Find the Lagrangean associated to the following equations of motions: (a) mr2\u03b8\u0308 + k1l1\u03b8 + k2l2\u03b8 +mgl = 0 580 9. Applied Dynamics (b) r\u0308 \u2212 r \u03b8\u0307 2 = 0 r2 \u03b8\u0308 + 2r r\u0307 \u03b8\u0307 = 0 15. Trebuchet. Derive the equations of motion for the trebuchet shown in Figure 9.18. 16. Simplified trebuchet. Three simplified models of a trebuchet are shown in Figures 9.23 to 9.25. Derive and compare their equations of motion. 9. Applied Dynamics 581 10"
        ]
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        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure1.13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure1.13-1.png",
        "caption": "Figure 1.13 When detailing the column-beam joints, like here in reinforced concrete, the dimensions are no longer negligible.",
        "texts": [
            " We can then use the findings to make predictions relating to the behaviour of a structure. It is from this predictive capactity that the science of mechanics derives its practical use. results are exact; for this reason, mechanics is often called an exact science. Mechanics involves measurable, physical quantities. A quantity X is generally characterised by a numerical value {X} and a unit [X]. This can be symbolically described as ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM schematisation for the joint (see Figure 1.13). mulas, based on certain schematisations and modelling. Mathematics sub- Within a given schematisation, mathematical models are used, and the Mechanics uses a range of concepts that offer a schematisation of reality. schematisations. 1 Introduction 9 quantity = numerical value \u00d7 unit X = {X} \u00d7 [X]. The unit [X] is the degree to which the quantity X is measured. In mechanics, one uses the International Units System (Syst\u00e8me International d\u2019Unit\u00e9), abbreviated in all languages to SI. The SI includes \u2022 seven basic units (Table 1"
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    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.60-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.60-1.png",
        "caption": "FIGURE 8.60.",
        "texts": [],
        "surrounding_texts": [
            "where, a1 is the longitudinal distance between C and the front axle, b1 is the lateral distance between C and the tireprint of the tire 1, and h is the height of C from the ground level. If P is a point in the tire frame at T1rP T1rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 . (8.62) then its coordinates in the body frame are BrP = BRT1 T1rP + BdT1 = \u23a1\u23a3 a1 + xP cos \u03b41 \u2212 yP sin \u03b41 yP cos \u03b41 \u2212 b1 + xP sin \u03b41 zP \u2212 h \u23a4\u23a6 . (8.63) The rotation matrix BRT1 is a result of steering about the z1-axis. BRT1 = \u23a1\u23a3 cos \u03b41 \u2212 sin \u03b41 0 sin \u03b41 cos \u03b41 0 0 0 1 \u23a4\u23a6 (8.64) Employing Equation (8.28), we may examine a wheel point P at W rP W rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 (8.65) 8. Suspension Mechanisms 497 and find the body coordinates of the point BrP = BRT1 T1rP + BdT1 = BRT1 \u00a1 T1RW W rP + T1dW \u00a2 + BdT1 = BRT1 T1RW W rP + BRT1 T1dW + BdT1 = BRW W rP + BRT1 T1dW + BdT1 (8.66) BrP = \u23a1\u23a3 a1 + xP cos \u03b41 \u2212 yP cos \u03b3 sin \u03b41 + (Rw + zP ) sin \u03b3 sin \u03b41 xP sin \u03b41 \u2212 b1 + yP cos \u03b3 cos \u03b41 \u2212 (Rw + zP ) cos \u03b41 sin \u03b3 (Rw + zP ) cos \u03b3 + yP sin \u03b3 \u2212 h \u23a4\u23a6 (8.67) where, BRW = BRT1 T1RW = \u23a1\u23a3 cos \u03b41 \u2212 cos \u03b3 sin \u03b41 sin \u03b3 sin \u03b41 sin \u03b41 cos \u03b3 cos \u03b41 \u2212 cos \u03b41 sin \u03b3 0 sin \u03b3 cos \u03b3 \u23a4\u23a6 (8.68) T1dW = \u23a1\u23a3 0 \u2212Rw sin \u03b3 Rw cos \u03b3 \u23a4\u23a6 . (8.69) Example 334 F Wheel-body to vehicle transformation. The wheel-body coordinate frames are always parallel to the vehicle frame. The origin of the wheel-body coordinate frame of the wheel number 1 is at BdW1 = \u23a1\u23a3 a1 \u2212b1 \u2212h+Rw \u23a4\u23a6 . (8.70) Hence the transformation between the two frames is only a displacement. Br = BIW1 W1r+ BdW1 (8.71) 8.6 F Caster Theory The steer axis may have any angle and any location with respect to the wheel-body coordinate frame. The wheel-body frame C (xc, yc, zc) is a frame at the center of the wheel at its rest position, parallel to the vehicle coordinate frame. The frame C does not follow any motion of the wheel. The steer axis is the kingpin axis of rotation. Figure 8.51 illustrates the front and side views of a wheel and its steering axis. The steering axis has angle \u03d5 with (yc, zc) plane, and angle \u03b8 with (xc, zc) plane. The angles \u03d5 and \u03b8 are measured about the yc and xc axes, 498 8. Suspension Mechanisms respectively. The angle \u03d5 is the caster angle of the wheel, while the angle \u03b8 is the lean angle. The steering axis of the wheel, as shown in Figure 8.51, is at a positive caster and lean angles. The steering axis intersect the ground plane at a point that has coordinates (sa, sb,\u2212Rw) in the wheelbody coordinate frame. If we indicate the steering axis by the unit vector u\u0302, then the components of u\u0302 are functions of the caster and lean angles. C u\u0302 = \u23a1\u23a3 u1 u2 u3 \u23a4\u23a6 = 1p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 \u23a1\u23a3 cos \u03b8 sin\u03d5 \u2212 cos\u03d5 sin \u03b8 cos \u03b8 cos\u03d5 \u23a4\u23a6 (8.72) The position vector of the point that u\u0302 intersects the ground plane, is called the location vector s that in the wheel-body frame has the following coordinates: Cs = \u23a1\u23a3 sa sb \u2212Rw \u23a4\u23a6 (8.73) We express the rotation of the wheel about the steering axis u\u0302 by a zero pitch screw motion s\u030c. CTW = C s\u030cW (0, \u03b4, u\u0302, s) = \u2219 CRW Cs\u2212 CRW Cs 0 1 \u00b8 = \u2219 CRW CdW 0 1 \u00b8 (8.74) Proof. The steering axis is at the intersection of the caster plane \u03c0C and the lean plane \u03c0L, both expressed in the wheel-body coordinate frame. The 8. Suspension Mechanisms 499 two planes can be indicated by their normal unit vectors n\u03021 and n\u03022. C n\u03021 = \u23a1\u23a3 0 cos \u03b8 sin \u03b8 \u23a4\u23a6 (8.75) C n\u03022 = \u23a1\u23a3 \u2212 cos\u03d50 sin\u03d5 \u23a4\u23a6 (8.76) The unit vector u\u0302 on the intersection of the caster and lean planes can be found by u\u0302 = n\u03021 \u00d7 n\u03022 |n\u03021 \u00d7 n\u03022| (8.77) where, n\u03021 \u00d7 n\u03022 = \u23a1\u23a3 cos \u03b8 sin\u03d5 \u2212 cos\u03d5 sin \u03b8 cos \u03b8 cos\u03d5 \u23a4\u23a6 (8.78) |n\u03021 \u00d7 n\u03022| = q cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 (8.79) and therefore, C u\u0302 = \u23a1\u23a3 u1 u2 u3 \u23a4\u23a6 = \u23a1\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3 cos \u03b8 sin\u03d5p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 \u2212 cos\u03d5 sin \u03b8p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 cos \u03b8 cos\u03d5p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 \u23a4\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6 . (8.80) Steering axis does not follow any motion of the wheel except the wheel hop in the z-direction. We assume that the steering axis is a fixed line with respect to the vehicle, and the steer angle \u03b4 is the rotation angle about u\u0302. The intersection point of the steering axis and the ground plane defines the location vector s. Cs = \u23a1\u23a3 sa sb \u2212Rw \u23a4\u23a6 (8.81) The components sa and sb are called the forward and lateral locations respectively. Using the axis-angle rotation (u\u0302, \u03b4), and the location vector s, we can define the steering process as a screw motion s\u030c with zero pitch. Employing Equations (5.473)-(5.477), we find the transformation screw for wheel frame W to wheel-body frame C. CTW = C s\u030cW (0, \u03b4, u\u0302, s) (8.82) = \u2219 CRW Cs\u2212 CRW Cs 0 1 \u00b8 = \u2219 CRW Cd 0 1 \u00b8 500 8. Suspension Mechanisms CRW = I cos \u03b4 + u\u0302u\u0302T vers \u03b4 + u\u0303 sin \u03b4 (8.83) CdW = \u00a1\u00a1 I\u2212 u\u0302u\u0302T \u00a2 vers \u03b4 \u2212 u\u0303 sin \u03b4 \u00a2 Cs. (8.84) u\u0303 = \u23a1\u23a3 0 \u2212u3 u2 u3 0 \u2212u1 \u2212u2 u1 0 \u23a4\u23a6 (8.85) vers \u03b4 = 1\u2212 cos \u03b4 (8.86) Direct substitution shows that CRW and CdW are: CRW = \u23a1\u23a3 u21 vers \u03b4 + c\u03b4 u1u2 vers \u03b4 \u2212 u3s\u03b4 u1u3 vers \u03b4 + u2s\u03b4 u1u2 vers \u03b4 + u3s\u03b4 u22 vers \u03b4 + c\u03b4 u2u3 vers \u03b4 \u2212 u1s\u03b4 u1u3 vers \u03b4 \u2212 u2s\u03b4 u2u3 vers \u03b4 + u1s\u03b4 u23 vers \u03b4 + c\u03b4 \u23a4\u23a6 (8.87) CdW = \u23a1\u23a3 (s1 \u2212 u1 (s3u3 + s2u2 + s1u1)) vers \u03b4 + (s2u3 \u2212 s3u2) sin \u03b4 (s2 \u2212 u2 (s3u3 + s2u2 + s1u1)) vers \u03b4 + (s3u1 \u2212 s1u3) sin \u03b4 (s3 \u2212 u3 (s3u3 + s2u2 + s1u1)) vers \u03b4 + (s1u2 \u2212 s2u1) sin \u03b4 \u23a4\u23a6 (8.88) The vector CdW indicates the position of the wheel center with respect to the wheel-body frame. The matrix CTW is the homogeneous transformation from wheel frame W to wheel-body frame C, when the wheel is steered by the angle \u03b4 about the steering axis u\u0302. Example 335 F Zero steer angle. To examine the screw transformation, we check the zero steering. Substituting \u03b4 = 0 simplifies the rotation matrix CRW and the position vector CdW to I and 0 CRW = \u23a1\u23a3 1 0 0 0 1 0 0 0 1 \u23a4\u23a6 (8.89) CdW= \u23a1\u23a3 0 0 0 \u23a4\u23a6 (8.90) indicating that at zero steering, the wheel frameW and wheel-body frame C are coincident. Example 336 F Steer angle transformation for zero lean and caster. Consider a wheel with a steer axis coincident with zw. Such a wheel has no lean or caster angle. When the wheel is steered by the angle \u03b4, we can find the coordinates of a wheel point P in the wheel-body coordinate frame using transformation method. Figure 8.52 illustrates a 3D view, and Figure 8.53 a top view of such a wheel. 8. Suspension Mechanisms 501 502 8. Suspension Mechanisms Assume W rP = [xw, yw, zw] T is the position vector of a wheel point, then its position vector in the wheel-body coordinate frame C is CrP = CRW W rP = Rz,\u03b4 W rP = \u23a1\u23a3 cos \u03b4 \u2212 sin \u03b4 0 sin \u03b4 cos \u03b4 0 0 0 1 \u23a4\u23a6\u23a1\u23a3 xw yw zw \u23a4\u23a6 = \u23a1\u23a3 xw cos \u03b4 \u2212 yw sin \u03b4 yw cos \u03b4 + xw sin \u03b4 zw \u23a4\u23a6 . (8.91) We assumed that the wheel-body coordinate is installed at the center of the wheel and is parallel to the vehicle coordinate frame. Therefore, the transformation from the frame W to the frame C is a rotation \u03b4 about the wheel-body z-axis. There would be no camber angle when the lean and caster angles are zero and steer axis is on the zw-axis. Example 337 F Zero lean, zero lateral location. The case of zero lean, \u03b8 = 0, and zero lateral location, sb = 0, is important in caster dynamics of bicycle model. The screw transformation for this case will be simplified to C u\u0302 = \u23a1\u23a3 u1 u2 u3 \u23a4\u23a6 = \u23a1\u23a3 sin\u03d5 0 cos\u03d5 \u23a4\u23a6 (8.92) Cs = \u23a1\u23a3 sa 0 \u2212Rw \u23a4\u23a6 (8.93) CRW = \u23a1\u23a3 sin2 \u03d5 vers \u03b4 + cos \u03b4 \u2212 cos\u03d5 sin \u03b4 sin\u03d5 cos\u03d5 vers \u03b4 cos\u03d5 sin \u03b4 cos \u03b4 \u2212 sin\u03d5 sin \u03b4 sin\u03d5 cos\u03d5 vers \u03b4 sin\u03d5 sin \u03b4 cos2 \u03d5 vers \u03b4 + cos \u03b4 \u23a4\u23a6 (8.94) Cd = \u23a1\u23a2\u23a3 cos\u03d5 (sa cos\u03d5+Rw sin\u03d5) vers \u03b4 \u2212 (sa cos\u03d5+Rw sin\u03d5) sin \u03b4 \u22121 2 (Rw \u2212Rw cos 2\u03d5+ sa sin 2\u03d5) vers \u03b4 \u23a4\u23a5\u23a6 . (8.95) Example 338 F Position of the tireprint. The center of tireprint in the wheel coordinate frame is at rT W rT = \u23a1\u23a3 0 0 \u2212Rw \u23a4\u23a6 . (8.96) 8. Suspension Mechanisms 503 If we assume the width of the tire is zero and the wheel is steered, the center of tireprint would be at CrT = CTW W rT = \u23a1\u23a3 xT yT zT \u23a4\u23a6 (8.97) where xT = \u00a1 1\u2212 u21 \u00a2 (1\u2212 cos \u03b4) sa + (u3 sin \u03b4 \u2212 u1u2 (1\u2212 cos \u03b4)) sb (8.98) yT = \u2212 (u3 sin \u03b4 + u1u2 (1\u2212 cos \u03b4)) sa + \u00a1 1\u2212 u22 \u00a2 (1\u2212 cos \u03b4) sb (8.99) zT = (u2 sin \u03b4 \u2212 u1u3 (1\u2212 cos \u03b4)) sa \u2212 (u1 sin \u03b4 + u2u3 (1\u2212 cos \u03b4)) sb \u2212Rw (8.100) or xT = sb \u00c3 cos \u03b8 cos\u03d5 sin \u03b4p cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 + 1 4 sin 2\u03b8 sin 2\u03d5 (1\u2212 cos \u03b4) cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 ! +sa \u00b5 1\u2212 cos2 \u03b8 sin2 \u03d5 cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 \u00b6 (1\u2212 cos \u03b4) (8.101) yT = \u2212sa \u00c3 cos \u03b8 cos\u03d5 sin \u03b4p cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 \u2212 1 4 sin 2\u03b8 sin 2\u03d5 (1\u2212 cos \u03b4) cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 ! +sb \u00b5 1\u2212 cos2 \u03d5 sin2 \u03b8 cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 \u00b6 (1\u2212 cos \u03b4) (8.102) zT = \u2212Rw \u2212 sb cos \u03b8 sin\u03d5+ sa cos\u03d5 sin \u03b8p cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 sin \u03b4 + 1 2 sb cos 2 \u03d5 sin 2\u03b8 \u2212 sa cos 2 \u03b8 sin 2\u03d5 cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 (1\u2212 cos \u03b4) (8.103) Example 339 F Wheel center drop. The zT coordinate in (8.100) or (8.103) indicates the amount that the center of the tireprint will move in the vertical direction with respect to the wheel-body frame when the wheel is steering. If the steer angle is zero, \u03b4 = 0, then zT is at zT = \u2212Rw. (8.104) Because the center of tireprint must be on the ground, H = \u2212Rw \u2212 zT indicated the height that the center of the wheel will drop during steering. H = \u2212Rw \u2212 zT (8.105) = sb cos \u03b8 sin\u03d5+ sa cos\u03d5 sin \u03b8p cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 sin \u03b4 \u22121 2 sb cos 2 \u03d5 sin 2\u03b8 \u2212 sa cos 2 \u03b8 sin 2\u03d5 cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 (1\u2212 cos \u03b4) 504 8. Suspension Mechanisms The zT coordinate of the tireprint may be simplifie for different designs: 1\u2212 If the lean angle is zero, \u03b8 = 0, then zT is at zT = \u2212Rw \u2212 1 2 sa sin 2\u03d5 (1\u2212 cos \u03b4)\u2212 sb sin\u03d5 sin \u03b4. (8.106) 2\u2212 If the lean angle and lateral location are zero, \u03b8 = 0, sb = 0, then zT is at zT = \u2212Rw \u2212 1 2 sa sin 2\u03d5 (1\u2212 cos \u03b4) . (8.107) In this case, the wheel center drop may be expressed by a dimensionless equation. H sa = 1 2 sin 2\u03d5 (1\u2212 cos \u03b4) (8.108) Figure 8.54 illustrates H/sa for the caster angle \u03d5 = 5deg, 0 deg, \u22125 deg, \u221210 deg, \u221215 deg, \u221220 deg, and the steer angle \u03b4 in the range \u221210 deg < \u03b4 < 10 deg. In street cars, we set the steering axis with a positive longitudinal location sa > 0, and a few degrees negative caster angle \u03d5 < 0. In this case the wheel center drops as is shown in the figure. 3\u2212 If the caster angle is zero, \u03d5 = 0, then zT is at zT = \u2212Rw + 1 2 sb sin 2\u03b8 (1\u2212 cos \u03b4)\u2212 sa sin \u03b8 sin \u03b4. (8.109) 4\u2212 If the caster angle and lateral location are zero, \u03d5 = 0, sb = 0, then zT is at zT = \u2212Rw \u2212 sa sin \u03b8 sin \u03b4. (8.110) 8. Suspension Mechanisms 505 In this case, the wheel center drop may be expressed by a dimensionless equation. H sa = \u2212 sin \u03b8 sin \u03b4 (8.111) Figure 8.55 illustrates H/sa for the lean angle \u03b8 = 5deg, 0, \u22125 deg, \u221210 deg, \u221215 deg, \u221220 deg and the steer angle \u03b4 in the range \u221210 deg < \u03b4 < 10 deg. The steering axis of street cars is usually set with a positive longitudinal location sa > 0, and a few degrees positive lean angle \u03b8 > 0. In this case the wheel center lowers when the wheel number 1 turns to the right, and elevates when the wheel turns to the left. Comparison of Figures 8.54 and 8.55 shows that the lean angle has much more affect on the wheel center drop than the caster angle. 5\u2212 If the lateral location is zero, sb = 0, then zT is at zT = \u2212Rw \u2212 sa cos\u03d5 sin \u03b8p cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 sin \u03b4 \u22121 2 sa cos2 \u03b8 sin 2\u03d5 cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 (1\u2212 cos \u03b4) (8.112) and the wheel center drop,H, may be expressed by a dimensionless equation. H sa = \u22121 2 cos2 \u03b8 sin2 \u03d5 (1\u2212 cos \u03b4) cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 \u2212 cos\u03d5 sin \u03b8 sin \u03b4p cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 (8.113) Example 340 F Position of the wheel center. As given in Equation (8.88), the wheel center is at CdW with respect to 506 8. Suspension Mechanisms the wheel-body frame. CdW = \u23a1\u23a3 xW yW zW \u23a4\u23a6 (8.114) Substituting for u\u0302 and s from (8.72) and (8.73) in (8.88) provides the coordinates of the wheel center in the wheel-body frame as xW = (sa \u2212 u1 (\u2212Rwu3 + sbu2 + sau1)) (1\u2212 cos \u03b4) + (sbu3 +Rwu2) sin \u03b4 (8.115) yW = (sb \u2212 u2 (\u2212Rwu3 + sbu2 + sau1)) (1\u2212 cos \u03b4) \u2212 (Rwu1 + sau3) sin \u03b4 (8.116) zW = (\u2212Rw \u2212 u3 (\u2212Rwu3 + sbu2 + sau1)) (1\u2212 cos \u03b4) + (sau2 \u2212 sbu1) sin \u03b4 (8.117) or xW = sa (1\u2212 cos \u03b4) + \u00b5 1 2 Rw sin 2\u03d5\u2212 sa sin 2 \u03d5 \u00b6 cos2 \u03b8 + 1 4 sb sin 2\u03b8 sin 2\u03d5 cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 (1\u2212 cos \u03b4) + (sb cos \u03b8 \u2212Rw sin \u03b8)p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 cos\u03d5 sin \u03b4 (8.118) yW = sb (1\u2212 cos \u03b4) \u2212 1 2 \u00a1 Rw sin 2\u03b8 + sb sin 2 \u03b8 \u00a2 cos2 \u03d5\u2212 1 4 sa sin 2\u03b8 sin 2\u03d5 cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 (1\u2212 cos \u03b4) \u2212 Rw sin\u03d5+ sa cos\u03d5p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 cos \u03b8 sin \u03b4 (8.119) zW = \u2212Rw (1\u2212 cos \u03b4) + \u00b5 Rw cos 2 \u03b8 + 1 2 sb sin 2\u03b8 \u00b6 cos2 \u03d5\u2212 1 2 sa cos 2 \u03b8 sin 2\u03d5 cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 (1\u2212 cos \u03b4) \u2212sa cos\u03d5 sin \u03b8 + sb cos \u03b8 sin\u03d5p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 sin \u03b4 (8.120) The zW coordinate indicates how the center of the wheel will move in the vertical direction with respect to the wheel-body frame, when the wheel is steering. It shows that zW = 0, as long as \u03b4 = 0. 8. Suspension Mechanisms 507 The zW coordinate of the wheel center may be simplified for different designs: 1\u2212 If the lean angle is zero, \u03b8 = 0, then zW is at zW = \u2212Rw \u00a1 1\u2212 cos2 \u03d5 \u00a2 (1\u2212 cos \u03b4)\u2212 sb sin\u03d5 sin \u03b4 \u22121 2 sa sin 2\u03d5 (1\u2212 cos \u03b4) . (8.121) 2\u2212 If the lean angle and lateral location are zero, \u03b8 = 0, sb = 0, then zW is at zW = \u2212Rw \u00a1 1\u2212 cos2 \u03d5 \u00a2 (1\u2212 cos \u03b4)\u2212 1 2 sa sin 2\u03d5 (1\u2212 cos \u03b4) . (8.122) 3\u2212 If the caster angle is zero, \u03d5 = 0, then zW is at zW = \u2212Rw \u00a1 1\u2212 cos2 \u03b8 \u00a2 (1\u2212 cos \u03b4)\u2212 sa sin \u03b8 sin \u03b4 + 1 2 sb sin 2\u03b8 (1\u2212 cos \u03b4) . (8.123) 4\u2212 If the caster angle and lateral location are zero, \u03d5 = 0, sb = 0, then zW is at zW = \u2212Rw \u00a1 1\u2212 cos2 \u03b8 \u00a2 (1\u2212 cos \u03b4)\u2212 sa sin \u03b8 sin \u03b4. (8.124) 5\u2212 If the lateral location is zero, sb = 0, then zT is at zW = \u2212Rw (1\u2212 cos \u03b4)\u2212 sa cos\u03d5 sin \u03b8p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 sin \u03b4 + Rw cos 2 \u03b8 cos2 \u03d5\u2212 1 2 sa cos 2 \u03b8 sin 2\u03d5 cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 (1\u2212 cos \u03b4) (8.125) In each case of the above designs, the height of the wheel center with respect to the ground level can be found by adding H to zW . The equations for calculating H are found in Example 340. Example 341 F Camber theory. Having a non-zero lean and caster angles causes a camber angle \u03b3 for a steered wheel. To find the camber angle of an steered wheel, we may determine the angle between the camber line and the vertical direction zc. The camber line is the line connecting the wheel center and the center of tireprint. The coordinates of the center of tireprint (xT , yT , zT ) are given in Equations (8.101)-(8.103), and the coordinates of the wheel center (xW , yW , zW ) are given in Equations (8.118)-(8.120). The line connecting (xT , yT , zT ) to (xW , yW , zW ) may be indicated by the unit vector l\u0302c l\u0302c = (xW \u2212 xT ) I\u0302 + (yW \u2212 yT ) J\u0302 + (zW \u2212 zT ) K\u0302q (xW \u2212 xT ) 2 + (yW \u2212 yT ) 2 + (zW \u2212 zT ) 2 (8.126) 508 8. Suspension Mechanisms in which I\u0302 , J\u0302 , K\u0302, are the unit vectors of the wheel-body coordinate frame C. The camber angle is the angle between l\u0302c and K\u0302, which can be found by the inner vector product. \u03b3 = cos\u22121 \u00b3 l\u0302c \u00b7 K\u0302 \u00b4 = cos\u22121 (zW \u2212 zT )q (xW \u2212 xT ) 2 + (yW \u2212 yT ) 2 + (zW \u2212 zT ) 2 (8.127) As an special case, let us determine the camber angle when the lean angle and lateral location are zero, \u03b8 = 0, sb = 0. In this case, we have xT = sa \u00a1 1\u2212 sin2 \u03d5 \u00a2 (cos \u03b4 \u2212 1) (8.128) yT = \u2212sa cos\u03d5 sin \u03b4 (8.129) zT = zT = \u2212Rw \u2212 1 2 sa sin 2\u03d5 (1\u2212 cos \u03b4) (8.130) xW = \u00b5 sa + 1 2 Rw sin 2\u03d5\u2212 sa sin 2 \u03d5 \u00b6 (1\u2212 cos \u03b4) (8.131) yW = sb (1\u2212 cos \u03b4)\u2212Rw sin\u03d5+ sa cos\u03d5 sin \u03b4 (8.132) zW = \u00b5 Rw \u00a1 cos2 \u03d5\u2212 1 \u00a2 \u2212 1 2 sa sin 2\u03d5 \u00b6 (1\u2212 cos \u03b4) . (8.133) 8.7 Summary There are two general types of suspensions: dependent, in which the left and right wheels on an axle are rigidly connected, and independent, in which the left and right wheels are disconnected. Solid axle is the most common dependent suspension, while McPherson and double A-arm are the most common independent suspensions. The roll axis is the instantaneous line about which the body of a vehicle rolls. Roll axis is found by connecting the roll center of the front and rear suspensions of the vehicle. The instant center of rotation of a wheel with respect to the body is called suspension roll center. So, to find the roll center of the front or rear half of a car, we should determine the suspension roll centers, and find the intersection of the lines connecting the suspension roll centers to the center of their associated tireprints. Three coordinate frames are employed to express the orientation of a tire and wheel with respect to the vehicle: the wheel frame W , wheel-body frame C, and tire frame T . A wheel coordinate frame W (xw, yw, zw) is attached to the center of a wheel. It follows every translation and rotation of the wheel except the spin. Hence, the xw and zw axes are always in the tire-plane, while the yw-axis is always along the spin axis. When the wheel 8. Suspension Mechanisms 509 is straight and the W frame is parallel to the vehicle coordinate frame, we attach a wheel-body coordinate frame C (xc, yc, zc) at the center of the wheel parallel to the vehicle coordinate axes. The wheel-body frame C is motionless with respect to the vehicle coordinate and does not follow any motion of the wheel. The tire coordinate frame T (xt, yt, zt) is set at the center of the tireprint. The zt-axis is always perpendicular to the ground. The xt-axis is along the intersection line of the tire-plane and the ground. The tire frame does not follows the spin and camber rotations of the tire however, it follows the steer angle rotation about the zc-axis. We define the orientation and position of a steering axis by the caster angle \u03d5, lean angle \u03b8, and the intersection point of the axis with the ground surface at (sa, sb) with respect to the center of tireprint. Because of these parameters, a steered wheel will camber and generates a lateral force. This is called the caster theory. The camber angle \u03b3 of a steered wheel for \u03b8 = 0, and sb = 0 is: \u03b3 = cos\u22121 \u00b3 l\u0302c \u00b7 K\u0302 \u00b4 = cos\u22121 (zW \u2212 zT )q (xW \u2212 xT ) 2 + (yW \u2212 yT ) 2 + (zW \u2212 zT ) 2 (8.134) where xT = sa \u00a1 1\u2212 sin2 \u03d5 \u00a2 (cos \u03b4 \u2212 1) (8.135) yT = \u2212sa cos\u03d5 sin \u03b4 (8.136) zT = zT = \u2212Rw \u2212 1 2 sa sin 2\u03d5 (1\u2212 cos \u03b4) (8.137) xW = \u00b5 sa + 1 2 Rw sin 2\u03d5\u2212 sa sin 2 \u03d5 \u00b6 (1\u2212 cos \u03b4) (8.138) yW = sb (1\u2212 cos \u03b4)\u2212Rw sin\u03d5+ sa cos\u03d5 sin \u03b4 (8.139) zW = \u00b5 Rw \u00a1 cos2 \u03d5\u2212 1 \u00a2 \u2212 1 2 sa sin 2\u03d5 \u00b6 (1\u2212 cos \u03b4) . (8.140) 510 8. Suspension Mechanisms 8.8 Key Symbols a, b, c, d lengths of the links of a four-bar linkage ai distance of the axle number i from the mass center A,B, \u00b7 \u00b7 \u00b7 coefficients in equation for calculating \u03b83 b1 distance of left wheels from mass center B (x, y, z) vehicle coordinate frame C mass center C coupler point C (xc, yc, zc) wheel-body coordinate frame C T dW C expression of the position of W with respect to T e, \u03b1 polar coordinates of a coupler point g overhang h = z \u2212 z0 vertical displacement of the wheel center H wheel center drop Iij instant center of rotation between link i and link j IijImn a line connecting Iij and Imn I\u0302 , J\u0302 , K\u0302 unit vectors of the wheel-body frame C I identity matrix J1, J2, \u00b7 \u00b7 \u00b7 length function for calculating \u03b83 l\u0302c unit vector on the line (xT , yT , zT ) to (xW , yW , zW ) ms sprung mass mu unsprung mass n\u03021 normal unit vectors to \u03c0L n\u03022 normal unit vectors to \u03c0C P point q, p, f parameters for calculating couple point coordinate r position vector Rw tire radius TRW rotation matrix to go from W frame to T frame s position vector of the steer axis sa forward location of the steer axis sb lateral location of the steer axis s\u030cW (0, \u03b4, u\u0302, s) zero pitch screw about the steer axis T (xt, yt, zt) tire coordinate system TTW homogeneous transformation to go from W to T u\u0302 steer axis unit vector u\u0303 skew symmetrix matrix associated to u\u0302 uC position vector of the coupler point u\u0302z unit vector in the z-direction vx forward speed x, y suspension coordinate frame xC , yC coordinate of a couple point (xT , yT , zT ) wheel-body coordinates of the origin of T frame 8. Suspension Mechanisms 511 (xW , yW , zW ) wheel-body coordinates of the origin of W frame vers \u03b4 1\u2212 cos \u03b4 W (xwywzw) wheel coordinate system z vertical position of the wheel center z0 initial vertical position of the wheel center \u03b1 angle of a coupler point with upper A-arm \u03b3 camber angle \u03b4 steer angle \u03b5 = ms/mu sprung to unsprung mass ratio \u03b8 lean angle \u03b80 angle between the ground link and the z-direction \u03b8i angular position of link number i \u03b82 angular position of the upper A-arm \u03b83 angular position of the coupler link \u03b84 angular position of link lower A-arm \u03b8i0 initial angular position of \u03b8i \u03c0C caster plane \u03c0L lean plane \u03c5 trust angle \u03d5 caster angle \u03c9 angular velocity 512 8. Suspension Mechanisms Exercises 1. Roll center. Determine the roll ceneter of the kinematic models of vehicles shown in Figures 8.56 to 8.59. 2 1 Body 4 6 3 5 8 7 8. Suspension Mechanisms 513 514 8. Suspension Mechanisms 8. Suspension Mechanisms 515 4. F Position of the roll center and mass ceneter. Figure 8.66 illustrates the wheels and mass center C of a vehicle. Design a double A-arm suspension such that the roll center of the C 516 8. Suspension Mechanisms c d a 2\u03b8 3\u03b8 4\u03b8 A B M N x y 0\u03b8 \u03b1 e z Cb 8. Suspension Mechanisms 517 Determine CTW for \u03d5 = 8deg, \u03b8 = 12deg, and the location vector Cs Cs = \u23a1\u23a3 3.8 cm 1.8 cm \u2212Rw \u23a4\u23a6 . (a) The vehicle uses a tire 235/35ZR19. (b) The vehicle uses a tire P215/65R15 96H. 10. F Wheel drop. Find the coordinates of the tireprint for \u03d5 = 10deg \u03b8 = 10deg Cs = \u23a1\u23a3 3.8 cm 1.8 cm 38 cm \u23a4\u23a6 if \u03b4 = 18deg. How much is the wheel drop H. 11. F Wheel drop and steer angle. Draw a plot to show the wheel drop H at different steer angle \u03b4 for the given data in Exercise 10. 12. F Camber and steering. Draw a plot to show the camber angle \u03b3 at different steer angle \u03b4 for the following characteristics: \u03d5 = 10deg \u03b8 = 0deg Cs = \u23a1\u23a3 3.8 cm 0 cm 38 cm \u23a4\u23a6 Part III Vehicle Dynamics 9"
        ]
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure9.17-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure9.17-1.png",
        "caption": "Fig. 9.17. Exploded view of the axial flux PM motor with counter-rotating rotor: 1 \u2014 main propeller, 2 \u2014 counter-rotating propeller, 3 \u2014 radial bearing, 4 \u2014 outer shaft, 5 \u2014 PM rotor, 6 \u2014 motor bearing, 7 \u2014 assembly ring, 8 \u2014 stator, 9 \u2014 inner shaft.",
        "texts": [
            " In order to achieve the opposite motion of the two rotors, the stator winding 298 9 Applications coils have to be arranged in such a manner that counter-rotating magnetic fields are produced in the machine\u2019s annular air gaps. The stator is positioned between two rotors that consist of mild steel discs and axially magnetized NdFeB PMs. The magnets are mounted on the disc\u2019s surface from the stator sides. Each rotor has its own shaft that drives a propeller, i.e. the motor has two coaxial shafts that are separated by a radial bearing. The arrangement is shown in Fig. 9.17 [39]. 9.4 Electromagnetic Aircraft Launch System 299 Military aircrafts are catapulted from the decks of aircraft carriers with the aid of steam catapults. A steam catapult uses two parallel rows of slotted cylinders with pistons connected to the shuttle that tows the aircraft. The pistons under the steam pressure accelerate the shuttle until the aircraft will takeoff. Steam catapults have many drawbacks [79], i.e.: \u2022 operation without feedback control; \u2022 large volume (over 1100 m3) and mass (up to 500 t); \u2022 occupation of the \u201dprime\u201d real estate on the carrier and negative effect on the stability of the ship; \u2022 low efficiency (4 to 6%); \u2022 operational energy limit, approximately 95 MJ; \u2022 need for frequent maintenance"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000730_s0022112006002631-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000730_s0022112006002631-Figure5-1.png",
        "caption": "Figure 5. Diagram of the geometry for squirmer 1 and sphere 2. The squirming motion of squirmer 1 generates a translational velocity (with components Ur,2 and U\u03b8,2), (a) rotational velocity \u2126z,2 and a stresslet S2 on sphere 2; \u03b80 is the angle between the orientation vector e of squirmer 1 and r .",
        "texts": [
            " This simplification in the boundary condition for the BEM is used throughout this paper. 4.2. Comparison for far-field separation Let squirmer 1 be at the centre of a cylindrical coordinate system with orientation vector e and \u03b2 = 1 (B2 = B1). Let sphere 2, which has no squirming velocity, be at r with the same radius as squirmer 1. Let the z = 0 plane contain e and the centres of both squirmers; let \u03b80 be the angle between the orientation vector e of squirmer 1 and r . This coordinate system is illustrated in figure 5. The squirming motion of squirmer 1 generates a translational velocity (with components Ur,2 and U\u03b8,2), a rotational velocity \u2126z,2 and a stresslet S2 on sphere 2. These quantities are compared with the analytical results. If one wants to know the velocities and stresslet generated by two squirmers, one just needs to add up two cases, (i) particle 1 is a squirmer and particle 2 is an inert sphere and (ii) particle 1 is an inert sphere and particle 2 is a squirmer, because the flow problem is linear",
            " The difference is defined by the absolute value of the analytical velocity Uana r,2 minus the numerical velocity UBEM r,2 . It is found that this velocity difference is lower r,2 . than 10\u22122 if r > 3. Therefore the analytical equations (2.3) and (2.7) can be used for r 3 with an accuracy of approximately 1%. The stresslet can be given as (2.9) in the far field. Figures 8(a) and 8(b) show a comparison of the stresslet components Sxx,2 and Sxy,2 on sphere 2; for \u03b80 = \u03c0/4 and 3\u03c0/4. Here the x-axis is taken in the e-direction and the y-axis is taken in the \u03b80 = \u03c0/2 direction, as illustrated in figure 5. It is found that the analytical and numerical results again correspond well when r > 3, though not quite as well as for the velocities. At leading order, the stresslet decays as r\u22123, which is confirmed by the numerical results. The difference between the analysis and the numerical simulation for the stresslet was examined at various values of \u03b80, and the results are shown in figures 9(a) and 9(b). The strength of the stresslet for a solitary squirmer is of order 10, so the difference itself is rather larger than for the translational velocity (figure 7)",
            " Generally speaking, the pairwise additivity of forces is to be preferred, because it can express lubrication forces more precisely and can be adjusted to prevent particles overlapping. (iii) The stresslet for both farand near-field separation is known from (2.9) and (3.58), respectively. The near-field stresslet needs to be improved by using (4.15), which is possible if one has a database of forces. In constructing the database we exploited the linearity of the flow problem and assumed that particle 1 is a squirmer and particle 2 is an inert sphere, as illustrated in figure 5. If one wants to know the forces and torques generated by two squirmers, one just needs to add the case where particle 1 is an inert sphere and particle 2 is a squirmer, which can be obtained from the same database. The database covers a wide range of relative positions such that 2.01 r 1000 and 0 \u03b80 \u03c0 (see figure 5). The results for 1000 < r and r < 2.01 can be calculated from the analytical results obtained in \u00a7 2 and \u00a7 3. The results for \u03c0 <\u03b80 < 2\u03c0 can be obtained by exploiting the symmetry of the problem. The database covers the parameter range 0.1 \u03b2 10 as well. Some sample lines are shown in table 1 for the case with \u03b2 = 1; the total number of lines in this case is 54 481. The full database is available as a supplement to the online version of the paper. In this section, we will introduce some interesting features of the trajectories"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure2.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure2.6-1.png",
        "caption": "FIGURE 2.6. A four wheel brake car, parked uphill.",
        "texts": [
            "71) Hence, if a2 < a1 then \u03c6Mf < \u03c6Mr and therefore, a rear brake is more effective than a front brake on uphill parking as long as \u03c6Mr is less than the tilting angle, \u03c6Mr < tan\u22121 a1 h . At the tilting angle, the weight vector passes through the contact point of the rear wheel with the ground. Similarly we may conclude that when parked on a downhill road, the front brake is more effective than the rear brake. Example 49 Four-wheel braking. Consider a four-wheel brake car, parked uphill as shown in Figure 2.6. In these conditions, there will be two brake forces Fx1 on the front wheels and two brake forces Fx1 on the rear wheels. 2. Forward Vehicle Dynamics 49 The equilibrium equations for this car are 2Fx1 + 2Fx2 \u2212mg sin\u03c6 = 0 (2.72) 2Fz1 + 2Fz2 \u2212mg cos\u03c6 = 0 (2.73) \u22122Fz1a1 + 2Fz2a2 \u2212 (2Fx1 + 2Fx2)h = 0. (2.74) These equations provide the brake force and reaction forces under the front and rear tires. Fz1 = 1 2 mg a2 l cos\u03c6\u2212 1 2 mg h l sin\u03c6 (2.75) Fz2 = 1 2 mg a1 l cos\u03c6+ 1 2 mg h l sin\u03c6 (2.76) Fx1 + Fx2 = 1 2 mg sin\u03c6 (2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-FigureC.5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-FigureC.5-1.png",
        "caption": "FIGURE C.5 Single zero thickness conductor sheets.",
        "texts": [
            "21) PARTIAL INDUCTANCE FORMULAS FOR ORTHOGONAL GEOMETRIES 389 b1 = y2 \u2212 ys1, b2 = y2 \u2212 ye1 (C.22) Z = z2 \u2212 z1, rkm = \u221a a2 k + b2 m + Z2 . (C.23) This formula is useful for the case where the two conductors are of a different shape. Other conductor shapes can be taken into account by combining this formula with numerical integration for the cross-section dimension of one or both inductive cells. C.1.4 Lp11 for Rectangular Zero Thickness Current Sheet Fortunately, the analytical Lp for a zero thickness rectangular sheet shown in Fig. C.5 leads to a relatively simple formula. The result is Lp11 = \ud835\udf070\ud835\udcc1 6 \ud835\udf0b [ 3 log [u + (u2 + 1)1\u22152 ] + u2 + 1 u + 3 u log [ 1 u + ( 1 u2 + 1 )1\u22152 ] \u2212 [ u4\u22153 + (1 u )2\u22153 ]3\u22152 ] , (C.24) where \ud835\udcc1 = xe1 \u2212 xs1 and u = \ud835\udcc1\u2215(ye1 \u2212 ys1). We note that this relatively simple formula is very useful for semianalytical solutions in layered models. Importantly, it eliminates the singular behavior problem for the partial self-inductance. C.1.5 Lp12 for Two Parallel Zero Thickness Current Sheets Two zero thickness parallel current sheets are shown in Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003908_978-1-84882-730-1-Figure3.4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003908_978-1-84882-730-1-Figure3.4-1.png",
        "caption": "Figure 3.4 Definition of wind speed and direction",
        "texts": [
            " Wind-induced Forces and Moments For the case where the vessel is at rest (zero speed), the vector wi E of the windinduced forces and moments is given by wi E D 1 2 aV 2 w 2 6666664 CX . w/AFw CY . w/ALw CZ. w/AFw CK. w/ALwHLw CM . w/AFwHFw CN . w/ALwLoa 3 7777775 ; (3.26) where Vw is the wind speed, a is the air density, AFw is the frontal projected area, ALw is the lateral projected area,HFw is the centroid of AFw above the water line, HLw is the centroid of ALw above the water line, Loa is the over all length of the vessel, w is the angle of relative wind of the vessel bow, see Figure 3.4, and is given by w D \u02c7w ; (3.27) 3.3 Modeling of Ocean Vessels 51 with \u02c7w being the wind direction. All the wind coefficients (look-up tables)CX . w/, CY . w/, CZ. w/, CK. w/, CM . w/, and CN . w/ are computed numerically or by experiments in a wind tunnel, see [28]. For the case where the vessel is moving, the vector wi E is given by wi E D 1 2 aV 2 rw 2 6666664 CX . rw/AFw CY . rw/ALw CZ. rw/AFw CK. rw/ALwHLw CM . rw/AFwHFw CN . rw/ALwLoa 3 7777775 (3.28) where Vrw D q u2rw Cv2rw ; rw D arctan2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure3.18-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure3.18-1.png",
        "caption": "FIGURE 3.18. Effective radius Rw compared to tire radius Rg and loaded height Rh.",
        "texts": [
            " Tire Dynamics 109 \u03c4x is negative for x > 0 and is positive for x < 0, showing an inward longitudinal stress. Figure 3.16 illustrates the absolute value of a \u03c4x distribution for n = 1. The y-direction tangential stress \u03c4y may be modeled by the equation \u03c4y(x, y) = \u2212\u03c4yM \u00b5 x2n a2n \u2212 1 \u00b6 sin \u00b3y b \u03c0 \u00b4 n \u2208 N (3.33) where \u03c4y is positive for y > 0 and negative for y < 0, showing an outward lateral stress. Figure 3.17 illustrates the absolute value of a \u03c4y distribution for n = 1. 3.4 Effective Radius Consider a vertically loaded wheel that is turning on a flat surface as shown in Figure 3.18. The effective radius of the wheel Rw, which is also called a rolling radius, is defined by Rw = vx \u03c9w (3.34) where, vx is the forward velocity, and \u03c9w is the angular velocity of the wheel. The effective radius Rw is approximately equal to Rw \u2248 Rg \u2212 Rg \u2212Rh 3 (3.35) and is a number between the unloaded or geometric radius Rg and the loaded height Rh. Rh < Rw < Rg (3.36) 110 3. Tire Dynamics Proof. An effective radius Rw = vx/\u03c9w is defined by measuring a wheel\u2019s angular velocity \u03c9w and forward velocity vx"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003995_ac950655w-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003995_ac950655w-Figure2-1.png",
        "caption": "Figure 2. (a, bottom) Layout for a sensor array. Three IMPs were used for depositing sensor films, and the fourth was used for a reference film. (b, top) Enlarged view of a sensor array mounted on a ceramic base.",
        "texts": [
            " We have made use of the electrostatic interaction between the charged surface of a conducting polymer and the charges on biomolecules to target the adsorption of a specific enzyme to a specific microelectrode pair. Microelectrode Layout and Fabrication. Two different electrode layouts were used, as shown in Figures 1 and 2. Figure 1 shows the layout for a single-substrate microsensor. One interdigitated microelectrode pair (IMP) was used for the sensor device and the other for a reference device. Figure 2 shows the layout for a sensor array with three IMPs for three different sensors and one IMP for a reference device. The masks for these layouts were designed on a Sun-Sparc work station using EESOF Academy Tools. A hard copy was obtained on Flashscan Photoplotter. The starting substrates were (100) p-type Si of 5 \u2126-cm resistivity. They were cleaned using the RCA procedure and oxidized in dry oxygen. The oxidation temperature was 1000 \u00b0C, and the oxidation time was set to obtain an oxide thickness of 100 nm",
            " The results described so far clearly demonstrate that (i) the immobilization of glucose oxidase, urease, and lipase can be directed to the desired IMP, (ii) adsorption of the enzyme can be inhibited by holding the polymer film at a potential where it is either neutral or negatively charged, and (iii) the immobilization is sufficiently strong to permit multiple usage. Based on these results, we have fabricated a sensor array consisting of glucose, urea, and lipid sensors as well as a reference sensor. The electrode layout used was that shown in Figure 2. At first, all the four IMPs were coated with polyaniline. Glucose oxidase, urease, and lipase were separately immobilized from their respective solutions on one IMP each, while the fourth IMP was used as a reference sensor. The polymerization and immobilization procedures were as described earlier. The sensor response to each substrate was determined independently, and the resulting calibration plots were used to obtain concentrations of these substrates when present in a mixture. Each IMP was independently addressed, and the conductance of each patch of polymer film could be determined when all the four IMPs were exposed to a common pool of analyte solution"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure3.29-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure3.29-1.png",
        "caption": "Fig. 3.29 The robot begins to rotate about the heel since the ZMP exists at the heel due to the highly accelerated center of mass. Since the acceleration in the vertical direction is generated, the ZMP remains in the support polygon.",
        "texts": [
            "28, what will happen when robot modeled by a point mass moves horizontally with high acceleration? If there is enough friction between the sole and the ground, the horizontal acceleration will not be barred. As explained above, the ZMP exists on the line defined by the gravity and inertial forces, and their values can be obtained by substituting z\u0308 = 0 and pz = 0 into (3.84) px = x\u2212 zx\u0308 g . 100 3 ZMP and Dynamics The larger the acceleration of the robot becomes, the further from the support polygon the ZMP can be! Using Fig. 3.29, we will explain the mistake in this discussion. Since the center of mass is highly accelerated, the ZMP is shifted to the heel. Then the robot will begin to rotate about the heel. Since the acceleration in the vertical direction is generated, we have z\u0308 > 0. Taking this effect into consideration, the ZMP should be calculated by px = x\u2212 zx\u0308 z\u0308 + g . Since z\u0308 will increase according to the increment of x\u0308, the ZMP remains in the support polygon9. More concretely, to calculate the ZMP for given motion of the robot using (3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003794_robot.1990.126059-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003794_robot.1990.126059-Figure4-1.png",
        "caption": "Fig. 4: Definition of A' matrix.",
        "texts": [
            " With d i f f e ren t i a l d e f l e c t i o n and torsion a t the connecting mechanism we calculate t h e p o s i t i o n i n g e r r o r v e c t o r E d f o r t h e manipulator model shown in Fig. 3. The differential transformation vector for cel l i and the force/moment vector Fi are defined as: center of cell k of cel l k t o cel l i (T-matrix). where d i j : d i f f e ren t i a l t r a n s l a t i o n in j - aij : differential rotation about j-axis. f i j : force in j-direction. : moment about j-axis. We mkh rewrite equation (10) and (11) into direction. A i = Kil Fi (12) W e define the matrix A; which combines the (13) three matrices Bi. Ci and Di (see Fig. 4) as: A; = B i * C i * D i where Bi: transformation matrix from center point of c e l l (i-1) t o connecting s u r f a c e , both have t h e same orientation, Ci: differential transformation matrix from connecting surface of cell (i-1) t o next cel l i. Di: transformation matrix from connecting surface of cel l i t o center point of the same cell. The matrix Ci can be calculated using equation (10) and with sin e --> ei , cos Qi --> 1 follows: W e now calculate equation (15) and (16). where m is the to t a l number of cells"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000209_tsmc.1980.4308393-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000209_tsmc.1980.4308393-Figure2-1.png",
        "caption": "Fig. 2. Vector definitions between base origin and link origins and center of masses.",
        "texts": [
            " We suppose that Aj now represents a 3 x 3 rotation matrix relating the orientations of coordinate systemj- 1 andj. That is to say, if iv is a vector expressed in terms of the orientation of coordinate system j axes, then j-v= Ai-v. When a vector is presented without a left superscript, it is referred to the base coordinate orientation (v=\u00b0v). In the subsequent development, capital letters represent 3 x 3 matrices, lower case letters represent 3 x 1 vectors. iWk is defined as before, save that it is now the composition of 3 x 3 rotation matrices. Define the following vectors (see Fig. 2): pi vector from the base coordinate origin to the joint i coordinate origin, p* vector from coordinate origin i -1 to coordinate origin i, r, vector from the base coordinate origin to the link i center of mass, ri* vector from coordinate origin i to the link i center of mass, and ni = r,* /mi. The generalized forces for this system are derived in Appendix A: [ tr(m iiT+ a finj4 + aWSn pT T Jr (1 5)i JjW )mjgTaqr The Wi term has the same recursive expression as in (10), though presently referring to a 3 x 3 rotation matrix"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003719_j.1469-7998.1985.tb04944.x-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003719_j.1469-7998.1985.tb04944.x-Figure7-1.png",
        "caption": "FIG. 7. Outlines traced from photographs by Muybridge (1887) of a Fallow deer galloping. (a) and (b) show the most extended and most bent positions of the back. (c) and (d) are used in the calculation of internal kinetic energy. The solid circles (0) are the centres of mass of the segments defined by lines across the leg. The open circles (w) are the assumed centres for rotation of the legs.",
        "texts": [
            " The length of the aponeurosis, from pin 1 to the insertion on the ilium and sacrum, is about 360 mm, so a strain of 0.03 implies an extension of 11 mm. Vertebrate striated muscle fibres exerting isometric tension have a strain of about 0.013 (Huxley & Simmons, 1971). For 120 mm fibres, as in the deer longissimus, this implies an extension of less than 2 mm. Thus, if the force developed in galloping equalled the isometric tension, the aponeurosis and muscle fibres together would stretch by only about 13 mm. This is much less than the extension of about 35 mm that would be required for the movement shown in Fig. 7, if the muscle remained taut throughout. The function for the aponeurosis as a strain energy store, suggested in this paper, does not require it to be taut throughout the stride. It has to be taut only during stage (c) of the stride (Fig. l), while internal kinetic energy is being lost and regained. It seems likely that the longissimus muscle may contract almost isometrically during this stage of the stride, and that such changes in length of the muscle and aponeurosis as occur during this stage may be mainly elastic deformations"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure3.53-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure3.53-1.png",
        "caption": "Fig. 3.53: Parameters for the inverse kinematic problem of parallel manipulators.",
        "texts": [
            " Fundamental characteristics can be recognized in: - complex direct kinematics - limited complex-shaped workspace - high precision and stiff performances - volume occupancy - large payload - high speed and acceleration - singularities The inverse kinematics problem for parallel manipulators, which is essential for trajectory planning, is quite straightforward. The inverse kinematics describes the mapping from the generalized coordinates describing the position and orientation of the moving platform to the joint coordinates. Furthermore, there is a unique solution and each joint variable may be computed being given the desired pose of the robot. One can consider as generalized coordinates XYZ the position of the center point P of the moving platform with respect to a fixed frame (Xo Yo Zo), Fig. 3.53, and Euler angles defining the orientation of the moving platform with respect to a fixed frame. A matrix R defines the orthogonal 3 \u00d7 3 rotation matrix defined by the Euler angles, which describes the orientation of the frame attached to the moving platform with respect to the fixed frame, Fig. 3.53. Let Ai and Bi be the attachment points at the base and moving platform, respectively, and di the leg lengths. Let ai and bi be the position vectors of points Ai and Bi in the fixed and moving coordinate frames, respectively. Thus, the inverse kinematics problem can be solved by using iiii RBA abp \u2212+= (3.9.1) to extract the joint variables from leg lengths. The length of the i-th leg can be obtained by taking the dot product of the vector AiBi with itself, for i:1,\u2026,6 in the form [ ] [ ]ii t ii 2 i RR abpabpd \u2212+\u2212+= (3",
            "54 b) by substituting the spherical joints with revolute joints or by a combination of these. Indeed, even the general plate design of Fig. 3.54 a) can give a great variety of manipulator architectures when spherical joints are substituted by revolute and prismatic joints or a combination of these. Fundamentals of Mechanics of Robotic Manipulation 191 In the case of Fig. 3.54 the spherical joint connections in the plates are located to give a symmetrical design 3S with an equilateral triangle with the joints Aj and Bj (j = 1, 2, 3). Thus, referring to Fig. 3.53, the position vector of the center point P with respect to O X0Y0Z0 can be expressed as ( )321 3 1 OBOBOBOP ++= (3.9.3) with jjjjj +R OABAOB \u03b3= (j = 1, 2, 3) (3.9.4) where R\u03b3j expresses the coordinate transformation from frame AjXjYjZj to the fixed frame O X0Y0Z0. It can be written as a function of a structural rotation angle \u03b3j among X0 and Xj in the form 100 0cossin 0sincos R jj jj j \u03b3\u03b3 \u03b3\u2212\u03b3 =\u03b3 (3.9.5) Frame AjXjYjZj has been determined with Xj axis lying on the plane of the fixed plate pointing outward from O; Zj axis is orthogonal to the plate plane"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure3.37-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure3.37-1.png",
        "caption": "Figure 3.37 A body subject to three forces at three points: (a) moment equilibrium exists; (b) there is no moment equilibrium; (c) the closed force polygon shows that both bodies are in force equilibrium.",
        "texts": [
            " Solution: If two forces are exerted on a body, the body can only be in equilibrium if the two forces have a common line of action, an equal magnitude and an opposite direction. In vector notation: FA = \u2212 FB. From the moment equilibrium about A, it follows that the common line of action of FA and FB is along AB (see Figure 3.36b). In that case, the horizontal component of FB is (4/5)FB. From the horizontal force equilibrium, it follows that: \u2211 Fx = \u2212(28 kN) + 4 5FB = 0 \u21d2 FB = 35 kN. 78 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM 2. A body subject to forces at three points (see Figure 3.37). A body that is subject to three forces can be in equilibrium only if \u2022 the three forces are coplanar, \u2022 the forces form a closed force polygon (force equilibrium),1 and \u2022 their lines of action pass through a single point (moment equilibrium). The same closed force polygon (c) is applicable for both bodies (a) and (b) in Figure 3.37: there is therefore force equilibrium in both cases. In case (a) there is moment equilibrium. This is easily checked by determining the moment of the three forces about the intersection of the three lines of action: none of the forces contribute to the sum of the moments. There is no moment equilibrium in case (b). The system of forces forms a resultant couple. The magnitude of the couple is determined by deriving the sum of the moments about the intersection of two lines of action. Figure 3.38 shows the right-hand part BC of the three-hinged frame, mentioned earlier"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001604_c1sm06512b-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001604_c1sm06512b-Figure4-1.png",
        "caption": "Fig. 4 Colour coded steady-state distribution r(r) of reaction products around single (a) and double (b) free carriers. The insets show the corresponding distributions in the presence of one or two cargoes. (c) Misalignment of a free double carrier, driven by the large concentration gradient at the \u2018\u2018contact\u2019\u2019 point. For the same configurations as in (a)\u2013(c), the bottom panels show r(r) in the plane perpendicular to the wall and the symmetry axis and passing through the point of contact of the two active spheres (or through the centre of the free single carrier, respectively). The densities r(r) are represented in units of the total rate of solute production the characteristic diffusion time R2/D.",
        "texts": [
            " The translational and rotational velocities of the object are determined as integrals, over the surface of the particles, of the phoretic slip weighted by complex hydrodynamic resistance matrices (which depend only on the geometry of the system30,45). Consequently, the orientation of the motion and the magnitude of the phoretic velocity can be qualitatively understood from the distribution of the catalytic reaction product around the carrier\u2013cargo composites. 50 | Soft Matter, 2012, 8, 48\u201352 Because the axis of the Janus particle remains, on average, parallel to thewall (x\u2013y plane), the overall behaviour (translation, rotation, or both) can be inferred from the structure of the concentration distributions in this plane. Fig. 4 displays the calculated concentration distribution of the catalytic reaction products formed around the Janus spheres of radius R and various carrier\u2013cargo composites. For the numerical calculation it has been assumed that the particles are half-coated by the catalyst and they perform motion in close vicinity to the wall, without being in mechanical contact with it. If the symmetry axis of the carrier\u2013cargo composite coincides with that of the in plane concentration distribution, the motion involves only translational motion along that axis and no rotational component. This type of motion holds for perfectly aligned and rigidly \u2018\u2018glued\u2019\u2019 spheres, as shown in Fig. 4(a) and (b), and in their corresponding insets. However, if the spheres are not \u2018\u2018glued\u2019\u2019, as it is the case in the experiment, the misalignment of the catalytic particles with respect to the axis of the colloidal cluster is likely to happen. For instance, the concentration distribution in Fig. 4(b) implies that the catalytic spheres actually experience a tendency to turn such that the directions of the catalytic parts are facing away from each other. The larger concentration gradient near the contact point (red coloured area in Fig. 4(b)) imposes a misalignment of the particles, as shown in Fig. 4 (c). The same conclusion can be drawn for all the configurations in the insets of Fig. 4(b). Therefore, for any composite comprising more than one active particle or one cargo, the perfectly aligned configurations with a symmetry axis in the plane of motion are unstable so that purely translational motion cannot occur. The configuration in Fig. 4(c) shows that the overall distribution of the reaction product exhibits broken axial symmetry and will necessarily lead to a phoretic rotation of the whole composite. Such a rotation can emerge as clockwise- or anti-clockwise with equal probability, but once it emerges its direction is maintained. This explains the rotations observed experimentally (see the trajectories in Fig. 2), and in particular the systematic clockwise and anti-clockwise rotations exhibited by the double carrier configurations (Fig",
            " 2(a), right). Further geometrical misalignments, as shown in Fig. 1 (bottom right), additionally contribute to such rotations. These scenarios are qualitatively different from the commonly invoked diffusional rotation, which involves random changes in the instantaneous direction of motion due to thermal fluctuations.21,32 Additionally, one should note that the presence of the glass substrate (\u2018\u2018wall\u2019\u2019) modifies the structure of the steady state distributions of the reaction products around the particles. Fig. 4(d)\u2013(f) clearly demonstrate the \u2018\u2018deformations\u2019\u2019 of the concentration distributions in the y\u2013z plane with higher solute (O2 molecules) concentrations close to the wall. As already mentioned before, such an asymmetry in conjunction with the solute\u2013particle repulsive effective interactions in the system implies a tendency for the carrier tomove in the direction away from the wall. However, this competes with the stronger opposite tendency of sedimentation, which explains why the active motion remains confined near and parallel to the wall",
            "{ Furthermore, a comparison of the velocities of the single and double micro-motor(s) carrying two cargoes (the blue lines in Fig. 3(a) and (b)) reveals that at all fuel concentrations studied the single micro-motor performs as good, or even better, than a double carrier; for example, at 15% v/v of H2O2 the velocities are approximately 4.25 and 4.21 mm s 1, respectively. This can be explained qualitatively from visual inspection of the corresponding theoretical O2 concentration distributions r(r) shown in the insets of Fig. 4(a) and (b). The single carrier, loaded by one or two cargoes, exhibits carrier\u2013cargo and cargo\u2013cargo contact points (Fig. 4(a), inset). Due to the symmetry of the composite, the concentration of the reaction product is low in the vicinity of these contact points. Therefore, for the single carrier the concentration gradient along the surface of the composite runs mainly along theOx direction, with negligibly weak variations at the carrier\u2013cargo contacts. For the double carrier composite there is an additional type of contact points, between the two carriers. In contrast to the other points of contact, this one is associated with a large concentration of the reaction product (red area in Fig. 4(b), and insets). The resulting gradient of the concentration, oriented towards this point, induces a significant flow of the solvent along the surface of the particles and towards the carrier\u2013carrier contact point (along the Oy direction). This component of the flow does not contribute to the translational motion of composites (along Ox). Near the carrier\u2013carrier contact, this flow is eventually directed towards the contact point, and thus contributes in the negative Ox direction, which slows down the translational motion of the whole double carrier composite",
            " This results in an effectively larger \u2018\u2018driving force\u2019\u2019 for the single motor configuration, while the hydrodynamic resistances are expected to be similar (because in a first approximation they are determined by the cross-sections exposed in the Ox direction of motion). These effects of the carrier\u2013 carrier contact point on the solution flow are superposed to the already identified tendency of the double micro-motor to break the alignment of the catalytic caps (leading to configurations shown in Fig. 4(c)), which additionally decreases the efficiency of the double carrier composite with two cargoes. This journal is \u00aa The Royal Society of Chemistry 2012 A further direct comparison between the velocities attained in various configurations is difficult at the qualitative level because of the single-particle spinning and reorientation inside the composite as discussed above; a complete numerical analysis is required for elucidating the effects of the cargoes. In summary, we have investigated the motion (i"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure10.11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure10.11-1.png",
        "caption": "Figure 10.11 The stress vectors p(I) and p(II) in corresponding points on the positive and negative sectional plane are equal and opposite, so p(I) = \u2212 p(II).",
        "texts": [
            " 1 Note: it is wrong to say that a bending moment is positive when the couple acts on the positive sectional plane in accordance with the positive sense of rotation and on the negative sectional plane in accordance with the negative sense of rotation. This is shown in Figures 10.9a and 10.9d. 10 Section Forces 393 If we look at the same small area A on the negative sectional plane, there is an equal but opposite force, in accordance with the principle of action and reaction. The stress vectors p(I) and p(II) have the same magnitude at corresponding points ( A \u2192 0) on the positive and negative sectional plane, but have opposite directions (see Figure 10.11): p(I) = \u2212 p(II). The stress vector is defined in a particular point and for a particular sectional plane. If we want to indicate the force transfer (interaction) at a point of the cross-section, the stress vector p alone is not enough, as we also have to indicate the status of the sectional plane that is considered. This is done by means of the unit normal vector n on that plane. To describe the action of the forces that the matter to the right of the section exerts on the matter to the left, and vice versa, we introduce the following quantities, which are known as cross-sectional stresses (see Figure 10"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000206_978-0-387-74315-8-Figure6-3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000206_978-0-387-74315-8-Figure6-3-1.png",
        "caption": "Figure 6-3: Articulated body i",
        "texts": [
            " This process is limited to the calculation of inertias for articulated bodies free from kinematic loops. The presence of kinematic loops complicates matters considerably, and consideration of this problem is deferred until Chapter 9. We come now to the detailed description of the articulated-body method for forward dynamics, using the system model defined in Chapter 4. The algorithm consists of two steps: Step 1: Caleulating the artieulated-body inertias Define articulated body i as the articulated body comprising links i. . n of the robot, connected by joints i+l . . n (see Figure 6-3). Link i is always the handle of articulated body i. Let it and Pi be the articulated-body inertia and associated bias force of Iink i in articulated body i. Dur target is to find a pair of recurrence relations which will enable us to calculate \u00ce~ and , - -A - Pi from Ii+l and Pi+l\u00b7 99 Since velocity effects are to be included in p i' we must start by caIculating the velocities and velocity-product forces at each link in turn. Let ~i be the absolute velocity of link i and let p: be the bias force on link i due to velocity-product forces, then "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002737_j.addma.2021.102168-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002737_j.addma.2021.102168-Figure2-1.png",
        "caption": "Fig. 2. Relative density, microcrack density, and demonstrator: (a) effect of laser energy input on the relative density and the microcracking density, (b) CAD model of the combustion chamber, and (c) LPBF GH3536 combustion chamber.",
        "texts": [
            " With the further increment in laser energy density, the pores played a predominant role in the defects again in zone III. Different from the lack-of-fusion pores, these pores with spherical morphologies and small sizes were metallurgical pores. The formation of these metallurgical pores was mainly associated with the pore transformation from feedstock powder and entrapped gas [25]. In general, the microcracks in LPBF GH3536 samples could not be avoidable by merely optimizing the printing parameters. Fig. 2a summarizes the samples\u2019 relative densities and the microcrack densities in microcrack length/area as a function of energy input \u03c8 = vdh. The relative densities in zone II were higher than the other two zones, and the microcrack densities were also lower than zones I and III. Accordingly, the optimized printing parameters for GH3536 alloy were chosen from zone II: P = 240 W, v = 600 mm/s, d = 40 \u00b5m, and h = 110 \u00b5m. The large-scale combustion chamber with conformal cooling channels demonstrated no warpage, distortion, and visible macrocracks (Fig. 2b and c), which proved the industrial feasibility of the optimized printing parameters. The printed samples with dimension sizes of 10 \u00d7 10 \u00d7 10 mm3 and 60 \u00d7 10 \u00d7 10 mm3 were used for microstructure observation and mechanical evaluation, respectively. Fig. 3 shows the microstructure of the as-fabricated GH3536 sample along the build direction. The overlapped semi-elliptical melt pools with a depth of ~100 \u00b5m and a width of ~170 \u00b5m could be observed readily in the as-fabricated sample. A large number of microcracks with a length of 10\u2013100 \u00b5m were randomly distributed in the as-fabricated sample, but most of the microcracks were propagated along the build direction and preferentially located at the grain boundaries, which was consistent with LPBF-fabricated HastelloyX samples [18]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003443_0278364903022001005-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003443_0278364903022001005-Figure2-1.png",
        "caption": "Fig. 2. A variant of the Delta parallel robot.",
        "texts": [
            " Thus, the Gr\u00fcbler-Kutzbach criterion can be finally rewritten as (Huang, Kong and Fang 1997) M = d(n \u2212 g \u2212 1) + XXXg i=1fi + v, (4) where M denotes the mobility of the mechanism, d is the order of the mechanism, n is the number of links, g is the number of kinematic pairs, fi is the freedom of the ith pair and v is the number of redundant constraints. It is shown that a planar four-bar parallelogram in a kinematic chain can be treated as a generalized prismatic pair (Huang and Li 2002b). For the PM with closed loops in the limbs, such as Delta robot (Clavel 1988) and Tsai\u2019s robot (Stamper, Tsai and Walsh 1997), the concept of a generalized kinematic pair will aid the mobility analysis. Figure 2 shows a variant of the Delta robot (Clavel 1990), where a planar four-bar parallelogram with four universal joints in each vertex is in each limb. The 4U parallelogram may be treated as a generalized kinematic pair. The 4U parallelogram can be considered as a PM with two 2U limbs, as shown in Figure 2(b). We take the output link 3 as the moving platform and the fixed link 1 as the base platform. Considering the structural condition of the 2U limb, we set the central point of the first universal joint as the origin of the limb frame, and the x-axis coincident with the first revolute axis and the y-axis coincident with the second revolute axis. Thus, the limb twist system of limb 1 in the general configuration is given by $11 = (1 0 0 ; 0 0 0) $12 = (0 1 0 ; 0 0 0) $13 = (0 1 0 ; \u2212z1 0 x1) $14 = (1 0 0 ; 0 z1 0), (5) where (x1, 0, z1) is the coordinate of the central point of the second universal joint in limb 1",
            " The two forces are parallel and restrict a translational DoF along link 2 and a rotational DoF about the normal of the plane formed by links 2 and 4. Thus, the 4U parallelogram can be regarded as a generalized kinematic pair, denoted by a superscript g, with three DoFs: $g 1 = (0 1 0 ; 0 0 0) $g 2 = (0 0 0 ; 0 1 0) $g 3 = (0 0 0 ; \u2212z1 0 x1), (7) which express one rotational and two translational DoFs. When analyzing the mobility of such a robot, we replace the 4U parallelogram by the generalized kinematic pair. Thus, such a limb consists of three links and four kinematic pairs, as shown in Figure 2(c). The limb twist system of the robot is a 4-system, that is, $g i1 = (0 1 0 ; 0 0 0) $g i2 = (0 0 0 ; 0 1 0) $g i3 = (0 0 0 ; \u2212z1 0 x1) $g i4 = (0 1 0 ; a4 0 c4). (8) The limb constraint system is $r i1 = (0 0 0 ; 1 0 0) $r i2 = (0 0 0 ; 0 0 1), (9) which shows that each limb exerts two constraint couples on the moving platform, one being perpendicular to the base plane and the other being parallel to the base plane. The three limbs of the robot exert six constraint couples in total on the moving platform"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001877_0278364917694244-Figure12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001877_0278364917694244-Figure12-1.png",
        "caption": "Fig. 12. (a) MIT Cheetah\u2019s leg consisting of three links. The first and third links are kept in parallel to each other by the parallelogram mechanism. (b) Two link kinematic conversion of original link structure. Using the torques ush and ukn at the shoulder and knee joints, forces Fx\u2032 and Fz\u2032 are generated. (c) An additional actuator creates ab/adduction torque uab for each leg.",
        "texts": [
            " However, the experimental hardware is not constrained to the plane and has light legs which do require careful treatment in order to realize the foot-force centered controllers developed in the previous section. This section presents the implementation details to realize the control design on the real robot hardware. Figure 11 shows the overall control system for the Cheetah robot, with many of the control system blocks detailed further in this section. The MIT Cheetah 2 has 3D kinematics which extend beyond those captured in the planar simple model. As shown in Figure 12(a), each leg of the MIT Cheetah 2 consists of three links beginning from the shoulder joint. The motions of the first and last link from the shoulder are kinematically tied to be parallel to each other, resulting in two degrees of freedom in the leg swing plane. The first actuator torque ush, rotates the link represented by the thick solid black line in Figure 12(a), providing rotation of all three links relative to the body. The second actuator torque ukn rotates the link represented by the dashed red line, yielding rotation of the second link while the first and third links are kept in parallel. Because the first and third links are parallel, the original link structure can be kinematically converted to a mechanism with only two links shown in Figure 12(b). In addition to the actuators for the knee and shoulder angles, there is one more actuator to create ab/adduction torque uab for each leg (see Figure 12(c)). 6.1. Specification of leg forces and joint torques for 3D bounding To realize the force profiles from the simple model controller using available actuators, this subsection presents a solution to the static force-production problem. More specifically, we obtain a relationship based on kinematics only, which is used to realize horizontal, lateral, and vertical forces, as well as a net rolling torque (Fx, Fy, Fz, and \u03c4x in Figure 13) through available actuators. Solution of this problem is used to stabilize the posture during 3D bounding"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003608_1.2359478-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003608_1.2359478-Figure2-1.png",
        "caption": "Fig. 2 Geometry used to calculate curvatures and surface velocities",
        "texts": [
            " In the ransverse plane, the surface velocities at a contact point are given s Ui= iRi, where i is the angular velocity and i=1,2 for pinion nd gear surfaces, respectively. The radius of curvature Ri in the ransverse plane is a function of the roll angle of the correspondng point in contact and base radius, Ri=Rbi i. Due to the helix 0 / Vol. 129, JANUARY 2007 om: http://mechanicaldesign.asmedigitalcollection.asme.org/ on 06/05/201 angle, an inclination angle exists between the contact line and the tooth trace in the tangent plane as shown in Fig. 2. The angle is defined by the base helix angle b and the helix angle at the operating pitch cylinder op as =cos\u22121 cos op/cos b 38 . For a pair of spur gears in mesh, one contact line is equivalent to the contact between two virtual cylinders in contact. For a pair of helical gears, due to the inclination angle, modifications must be made to obtain this equivalence. Consequently, the contact line is discretized into many small segments and each segment is viewed as a contact line that is equivalent to a spur gear with an equivalent angular velocity \u0302= cos . For helical gears, principal directions are x and y, both defined in the common tangent plane as shown in Fig. 2. At any contact point q, the radii of curvature in the principal directions, Rix and Riy, and the radii of curvature in the lubricant entrainment direction e, Rie, can be related through Euler\u2019s equation 39 as Rie= cos2 /Rix + sin2 /Riy \u22121. For involute helical gears, the theoretical contact line is a straight line, Riy = , and the lubricant entrainment direction coincides with the direction that defines the transverse plane. Thus, the radii of curvatures and surface velocities in the direction x that is normal to the contact line are Rix=Rie cos2 =Ri cos2 and Uix= \u0302iRix = iRie cos3 = iRi cos3 , where i=1,2 for the pinion and gear surfaces, respectively"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002149_j.matdes.2015.06.063-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002149_j.matdes.2015.06.063-Figure1-1.png",
        "caption": "Fig. 1. Geometry of test specimen.",
        "texts": [
            " Stress and strain distributions at pores, and local stress concentration factors have been studied for their influence on mechanical performance of parts. AlSi12 specimens were manufactured using a commercially available system SLM 250 HL equipped with 400 W fiber laser using commercially available powder material containing 11.62 wt-% Si, 0.72 wt-% Fe, 0.03 wt-% Cu balanced by Al. The specimens were melted using laser power of 350 W, and a scan speed of 930 mm/s (batch I). The details of the process set up can be seen elsewhere [10]. The specimens (Fig. 1) conforming to ISO 1099 were manufactured in vertical orientation under argon atmosphere. Post-process stress relief (SR) was performed at 200 C followed by oven cooling to remove any residual stresses. Hot isostatic pressing (HIP) was carried out on half of the specimens for elimination of remnant porosity (batch II). The process was carried out at 497 C and 1000 bar for 2 h under argon atmosphere. Defect characterization was carried out in terms of two-dimensional pore area and in terms of three-dimensional defect volume"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure3.37-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure3.37-1.png",
        "caption": "FIGURE 3.37. Bottom view of a laterally deflected and turning tire.",
        "texts": [
            " The wheel will start sliding laterally when the lateral force reaches a maximum value FyM . At this point, the lateral force approximately remains constant and is proportional to the vertical load FyM = \u03bcy Fz (3.136) 3. Tire Dynamics 137 where, \u03bcy is the tire friction coefficient in the y-direction. A bottom view of the tireprint of a laterally deflected tire is shown in Figure 3.36. If the laterally deflected tire is turning forward on the road, the tireprint will also flex longitudinally. A bottom view of the tireprint for such a laterally deflected and turning tire is shown in Figure 3.37. Although the tire-plane remains perpendicular to the road, the path of the wheel makes an angle \u03b1 with tire-plane. As the wheel turns forward, undeflected treads enter the tireprint region and deflect laterally as well as longitudinally. When a tread moves toward the end of the tireprint, its lateral deflection increases until it approaches the tailing edge of the tireprint. The normal load decreases at the tail of the tireprint, so the friction force is lessened and the tread can slide back to its original position when leaving the tireprint region"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002362_j.ymssp.2020.106640-Figure12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002362_j.ymssp.2020.106640-Figure12-1.png",
        "caption": "Fig. 12. Contact region of various kinds of spalling pattern.",
        "texts": [
            " The contact stress of the spalled gear can also be acquired through the proposed method. The contact stress during the whole meshing process (contact pattern) is shown in Fig. 11a. The contact stress reaches zero in the spalling area, which means that the bottom of the spalling area is not in contact state. Stress concentration occurs at the edge of spalling area. This feature is captured by both two methods (see Fig. 11b), which verifies the computation accuracy of the proposed method. The spalling patterns can mainly be categorized as three types (see Fig. 12). The spalling pattern I is frequently adopted in traditional studies. This type of spalling will not be discussed because it rarely appears in the fatigue experiment. For the spalling pattern II, the bottom of the spalling area may be in contact when the spalling area is large and the spalling depth is small. For the spalling pattern III obtained by the fatigue experiment, the spalling depth is not a sensitive parameter (contact pattern, mesh stiffness and transmission error). In order to prove this viewpoint, the effects of spalling depth on meshing characteristics are discussed",
            " 14b). Typically, the spalling depth ranges from 30 lm to 300 lm. If the spalling depth is extremely large (hs > 500 lm), the global stiffness will be influenced [40]. In most cases, the physical spalling depth has little effect on the mesh stiffness and contact stress unless the spalling depth is too small or too large. Similar descriptions can also be found in Ref. [40]. For the spalling pattern obtained by the fatigue experiment, the spalling depth is about 20\u201350 lm (spalling pattern III in Fig. 12c). It is proved that the spalling depth is not a sensitive parameter for the cases in this manuscript. Therefore, only the two-dimensional spalling morphology is involved in the proposed model. Contact stress is widely used to evaluate the contact state of the gear pair under tooth modification. Similar descriptions can also be applied to elaborate the contact characteristics of the spalled gear pair. The contact stress is highly correlated to the geometric morphology of the pitting or spalling (see Figs",
            " Through the comparison of the simulated and experimental results, desirable agreement can be found for the prediction of the fault level. Tooth pairs D and E suffered from multiple teeth spalling, leading to larger vibration energy and more obvious impulse. However, there is still a large discrepancy between simulation and experiment. The main reason is that the level of vibration energy is different between simulation and experiment. The accelerometers are mounted at the bearing position (see Fig. 12) in the experiment. Energy decay is inevitable in the transfer path of the vibration signal. The interface between gear and shaft, inner race and outer race, outer race and housing will cause significant energy loss. Residual signal is widely used to highlight the weak fault characteristics. For the experiment signal, it is hard to determine the accurate angular position of the spalling defect. Therefore, only the residual signal obtained from simulation is analyzed. The residual signal of the spalled gear pair shows a strong impulse feature and the interval between each impulse is 0"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003941_physrevlett.101.268101-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003941_physrevlett.101.268101-Figure1-1.png",
        "caption": "FIG. 1. A collision of two self-propelled hard rods (the width of the rod is exaggerated for clarity). k\u0302 is a unit vector from rod 2 to rod 1 normal to the point of contact. Points on the side of the rods are identified by vectors i.",
        "texts": [
            " First, we show that the additional momentum transfer from self-propulsion lowers the density of the isotropic-nematic transition, thereby providing a microscopic identification for the physical mechanism responsible for the enhancement of orientational order observed in numerical simulations of motility assays [8]. Second, we demonstrate that self-propulsion greatly enhances the effect of confinement and the role of boundaries. The microscopic model.\u2014We consider quasi twodimensional hard rods of length \u2018 and thickness 2R confined to a plane, as shown in Fig. 1. The ith rod is characterized by the position ri of its center of mass and a unit vector u\u0302i \u00bc \u00f0cos i; sin i\u00de directed along its long axis. Each rod free-streams on the substrate, until it collides with another rod. The collision results in instantaneous linear and angular momentum transfer such that the total energy, linear and angular momenta of the two rods are conserved [20]. The microdynamics of the system is governed by coupled Langevin equations, PRL 101, 268101 (2008) P HY S I CA L R EV I EW LE T T E R S week ending 31 DECEMBER 2008 0031-9007=08=101(26)=268101(4) 268101-1 2008 The American Physical Society @vi @t \u00bc X j T\u00f0i; j\u00devi \u00fe Fu\u0302i i vi \u00fe i\u00f0t\u00de; (1) @",
            " Finally, the collision operator T\u00f0i; j\u00de generates the instantaneous momentum transfer between rods at contact and is given by T\u00f01; 2\u00de \u00bc Z 0 s1s2 Z k\u0302 jV12 k\u0302j \u00f0 V12 k\u0302\u00de\u00f0b12 1\u00de; (3) where k\u0302 is the unit normal at the point of contact directed from rod 2 to rod 1, si 2 \u00bd \u2018=2; \u2018=2 parametrize the distance of points along the axis of each rod from the center of mass, and R0 s1s2 . . . R\u2018=2 \u2018=2 ds1 R\u2018=2 \u2018=2 ds2 \u00f0R12\u00de . . . , where R12 \u00bc 0 corresponds to the condition that the two rods be in contact at one point, as for instance in Fig. 1, where R12 \u00bc r1 r2 \u00fe s1u\u03021 \u00f0\u2018=2\u00deu\u03022 2Rk\u0302. Also, V12 \u00bc v1 v2 \u00fe!1 1 !2 2, with i \u00bc siu\u0302i Rk\u0302, is the relative velocity of the two rods at the point of contact. The operator b12 replaces precollisional velocities with their postcollisional values, e.g., b\u00f01; 2\u00dev1;2 \u00bc v1;2 k\u0302A, where the momentum exchange A is obtained by requiring energy and momentum conservation. The explicit calculation of the T operator is given in [21]. Modified Smoluchowski equation.\u2014We are interested here in the overdamped limit, when inertial effects are negligible and the low density dynamics is described by a Smoluchowski equation for the the probability distribution c\u00f0x; t\u00de of rods at r oriented in the direction , with x \u00bc \u00f0r; \u00de, The derivation of the Smoluchowski equation is rather technical and will not be presented here"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure3.9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure3.9-1.png",
        "caption": "Figure 3.9 The resultant R of two parallel forces F1 and F2, in (a) the same and (b) opposite directions.",
        "texts": [
            " If R1 is the resultant of F1 and P1, and R2 is the resultant of F2 and P2, then the line of action of the resultant of all the forces, that is the resultant R of F1 and F2, passes through the intersection of the lines of action of R1 and R2. From the graphical construction, one can see that the line of action of the ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM 3 Statics of a Rigid Body 57 resultant R of two parallel forces F1 and F2, acting in the same direction, is between their lines of action, nearer the larger force, and such that the distances a and b to the lines of action of F1 and F2 respectively are reversed proportionally to the magnitudes of these forces (see Figure 3.9a): a b = F2 F1 . If the two parallel forces F1 and F2 have opposite directions, then the resultant R has the same direction as the larger of the two forces, and the line of action of R is outside the lines of action of F1 and F2 on the side of the larger force. Now too, the distances a and b from the line of action of R to the lines of action of F1 and F2 are reversed proportionally to the magnitude of these forces (see Figure 3.9b). In conclusion, for the resultant R of two parallel forces F1 and F2: \u2022 R is in the direction of the larger force; \u2022 the line of action of R is closer to the larger force; \u2022 R is between F1 and F2 if these forces are in the same direction; \u2022 R is outside F1 and F2 if these forces have opposite directions. Figure 3.10 shows the special case of two equal and opposite parallel forces F1 = F and F2 = F . If we want to graphically compound these forces in the way described above, using two equal and opposite additional forces P1 = P and P2 = P , with the common line of action AB, we again find two equal and opposite parallel forces R1 and R2 (see Figure 3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure13.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure13.1-1.png",
        "caption": "FIGURE 13.1. A full car vibrating model of a vehicle.",
        "texts": [
            " (a) Determine the best suspension for the secondary system with m2 = 1kg to act as a vibration absorber. (b) Determine the natural frequencies of the two system for the optimized vibration absorber. (c) Determine the nodal frequencies and amplitudes of the primary system. 826 12. Applied Vibrations 19. F Frequency response. Prove the following equations: G2 = FTB kY G2 = FTE e\u03c92nme G2 = FTR e\u03c92nme \u00b3 1 + mb m \u00b4 13 Vehicle Vibrations Vehicles are multiple-DOF systems as the one that is shown in Figure 13.1. The vibration behavior of a vehicle, which is called ride or ride comfort, is highly dependent on the natural frequencies and mode shapes of the vehicle. In this chapter, we review and examine the applied methods of determining the equations of motion, natural frequencies, and mode shapes of different models of vehicles. 13.1 Lagrange Method and Dissipation Function Lagrange equation, d dt \u00b5 \u2202K \u2202q\u0307r \u00b6 \u2212 \u2202K \u2202qr = Fr r = 1, 2, \u00b7 \u00b7 \u00b7n (13.1) or, d dt \u00b5 \u2202L \u2202q\u0307r \u00b6 \u2212 \u2202L \u2202qr = Qr r = 1, 2, \u00b7 \u00b7 \u00b7n (13"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure6.39-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure6.39-1.png",
        "caption": "FIGURE 6.39. An inverted slider-crank mechanism and a coupler point at C.",
        "texts": [
            " Point C is on the coupler link AB and is at a distance kb from A, where 0 < k < 1. The coordinates of point C are xC = a cos \u03b82 + ka cos \u03b82 = a (1 + k) cos \u03b82 (6.269) yC = a sin \u03b82 \u2212 ka sin \u03b82 = a (1\u2212 k) sin \u03b82 (6.270) and therefore, cos \u03b82 = xC a (1 + k) (6.271) sin \u03b82 = yC a (1\u2212 k) . (6.272) Using cos2 \u03b82 + sin2 \u03b82 = 1, we can show that the coupler point C will 362 6. Applied Mechanisms move on an ellipse. x2C a2 (1 + k) 2 + y2C a2 (1\u2212 k) 2 = 1 (6.273) 6.5.3 Coupler Point Curve for Inverted Slider-Crank Mechanism Figure 6.39 illustrates an inverted slider-crank mechanism and a coupler point at C. When the mechanism moves, the coupler point C will move on a coupler point curve with the following parametric equation: xC = a cos \u03b82 + c cos (\u03c0 \u2212 \u03b1\u2212 \u03b84) (6.274) yC = a sin \u03b82 + c sin (\u03c0 \u2212 \u03b1\u2212 \u03b84) (6.275) The angle \u03b82 is the input angle and acts as a parameter, and \u03b84 is the angle of the output link, given by Equation (6.173). \u03b84 = 2 tan\u22121 \u00c3 \u2212H \u00b1 \u221a H2 \u2212 4GI 2G ! (6.276) G = d\u2212 e\u2212 a cos \u03b82 (6.277) H = 2a sin \u03b82 (6.278) I = a cos \u03b82 \u2212 d\u2212 e. (6.279) Proof. We attach a planar Cartesian coordinate frame to the ground link at MN . Drawing a vertical line through C defines the variable angle \u03b3 = \u2220ACF as shown in Figure 6.39. We also define three angles \u03b21 = \u2220ANM , 6. Applied Mechanisms 363 \u03b22 = \u2220ANB, and \u03b23 = \u2220BEF , to simplify the calculations. From 4ACE, we find \u03b3 = \u03c0 \u2212 \u03b23 \u2212 \u03b1 (6.280) and from quadrilateral \u00a4EFNB, we find \u03b23 +\u2220EFN + \u03b22 + \u03b21 +\u2220EBN = 2\u03c0. (6.281) However, \u2220EBN = \u03c0 2 (6.282) \u2220EFN = \u03c0 2 (6.283) and therefore, \u03b23 + \u03b22 + \u03b21 = \u03c0. (6.284) The output angle \u03b84 is equal to \u03b84 = \u03c0 \u2212 (\u03b22 + \u03b21) , (6.285) and thus, \u03b84 = \u03b23. (6.286) Now the angle \u03b3 may be written as \u03b3 = \u03c0 \u2212 \u03b84 \u2212 \u03b1 (6.287) where \u03b84 is the output angle, found in Equation (6"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-FigureC.15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-FigureC.15-1.png",
        "caption": "FIGURE C.15 Arbitrary oriented rectangular bars.",
        "texts": [
            "80b) PARTIAL INDUCTANCE FORMULAS FOR NONORTHOGONAL GEOMETRIES 403 Again, this filament-to-filament partial mutual inductance is used in combination with numerical integration for different cross-sections, for example, [16]. Please note that a small number could be added to the coordinates for parallel wires as we did in the formula for the in plane case in the previous section. C.2.4 Lp for Two Cells or Bars with Same Current Direction In this example, we assume that all the current filaments are in the same direction, even if conductor cross sections are not parallel to the y\u2013z coordinates as shown in Figure C.15. In this case, the unit vectors are the same or, x\u0302 = a\u0302. Hence, the wire-to-wire formula (C.8) or the sheet-to-wire inductance in Section C.1.3 can be utilized. Then, we can represent the partial inductance in the form Lp12 = \ud835\udf070 4\ud835\udf0b \u222b xe1 xs1 \u222b ye1 ys1 \u222b ze1 zs1 \u222bx2 \u222bb2 \u222bc2 1 R1,2 dx1 dy1 dz1 dx2 db2 dc2, (C.81) where R1,2 = R[r1(x1, y1, z2), r2(x2, b2, c2)]. In this specific example, conductor 1 is in orthogonal coordinates and conductor 2 is in local coordinates. Of course, if we apply the partial inductance for a sheet and a wire in (C"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure9.42-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure9.42-1.png",
        "caption": "Figure 9.42 Once N14 is known we can find N13 from the equilibrium of joint E.",
        "texts": [
            " If determining the location of point B is complicated (not the case here) you could also first determine N10 from the moment equilibrium about D and then derive N9 from the moment equilibrium about A, for example: \u2211 Tz|D = \u2212(80 kN)(6 m) + N10 \u00d7 (4 m) = 0 \u21d2 N10 = +120 kN, 9 Trusses 345 \u2211 Tz|A = \u2212(80 kN)(3 m) \u2212 (120 kN)(3 m) + N10 \u00d7 (3 m) + N9 \u00d7 (3 m) = 0 \u21d2 N9 = +80 kN. When determining the force in member 13, we again encounter the problem that a section across member 13 cuts four members (see Figure 9.41). In this case, the problem cannot be easily solved from the equilibrium of joint E isolated in Figure 9.42. In order to find member force N13 from the force equilibrium of joint E, we first have to know one of the member forces N12 or N14. Here there is a special case, in which we can determine the force in member 14 by means of the section in Figure 9.41, even though it passes over four members. Since, in this section, three of the four unknown member forces intersect at point G, the fourth force, in this case N14, can be derived directly from the moment equilibrium about G. This gives (for the part shown to the left of the section, with force N14 moved to point E) \u2211 Tz|G = \u2212(80 kN)(9 m) + (120 kN)(3 m) \u2212 3 10N14 \u221a 10 \u00d7 (5 m) = 0 so that N14 = \u221224 \u221a 10 kN. The horizontal force equilibrium of joint E (Figure 9.42) now gives N12 = N14 = \u221224 \u221a 10 kN. The vertical force equilibrium gives N13 = \u2212 1 10 \u221a 10 \u00d7 (N12 + N14) = +48 kN. Member 13 is therefore a tension member. Table 9.2 provides a summary of all the member forces in the truss. 346 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Example 3 The third example relates to the somewhat more complicated truss in Figure 9.43, a so-called K-truss. This type of truss is sometimes used as wind bracing in bridges. Here, the K-truss has four fields and is loaded by two vertical forces of 120 kN and a horizontal force of 240 kN"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure12.14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure12.14-1.png",
        "caption": "Figure 12.14 (a) An isolated beam segment with uniformly distributed full load and (b) the associated parabolic bending moment diagram.",
        "texts": [
            "13c, we can recognise a cable AD hanging (as a parabola) under the influence of the uniformly distributed full load, and pushed upwards by the support reactions at B and C. The general validity of rule 16 is explained in Chapter 14. Due to a uniformly distributed load, the M diagram has the shape of a parabola. Since uniformly distributed loads occur frequently in practice we will discuss a number of the striking properties of parabolas below. They can be used to sketch a parabolic M diagram quickly. We will use the isolated beam segment in Figure 12.14a as starting point, with length and a uniformly distributed load q over the full length. In addition to shear forces, the section planes at A and B are subject to bending moments MA and MB, both causing tension at the underside. Figure 12.14b shows the associated M diagram. Properties of the parabolic M diagram include the following: \u2022 The tangents to the M diagram at A and B are found by drawing the M diagram for the resultant of the distributed load. The tangents at A and B intersect in the middle C of AB. \u2022 The vertically measured distance p between chord k and the parabola is p = 1 2qab (in which a + b = ). \u2022 In the middle C (a = b = 1 2 ) this distance is pC = 1 8q 2. \u2022 The intersection of the tangents at A and B is at a distance 2pC under Figure 12"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.54-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.54-1.png",
        "caption": "FIGURE 7.54. The initial position of a one-axle trailer pulled by a car moving forward in a straight line with a constant velocity is a circle about the hinge point.",
        "texts": [
            "209) Because \u03ba(t1) = 1/R(t) > 0, we conclude that f 0(t1) > 0, and it is not possible to have f 0(t1) \u2264 0. Example 296 F Straight motion of the car with constant velocity. Consider a car moving forward in a straight line with a constant velocity. We may use a normalization and set the speed of the car as 1 moving in positive x direction starting from x = 0. Using a two-dimensional vector expression we have r = \u2219 xc yc \u00b8 = \u2219 t 0 \u00b8 . (7.210) Because of (7.192), we get z(0) = r(0)\u2212 s(0) = \u2212s(0) (7.211) and therefore, the initial position of the trailer must lie on a unit circle as shown in Figure 7.54. Using two dimensional vectors, we may express z(0) as a function of \u03b8 z(0) = \u2212s(0) = \u2219 xt(0) yt(0) \u00b8 = \u2219 cos \u03b8 sin \u03b8 \u00b8 (7.212) and simplify Equation (7.202) as s\u0307 = \u2219 x\u0307t y\u0307t \u00b8 = \u2219 (t\u2212 xt) 2 \u2212yt (t\u2212 xt) \u00b8 . (7.213) 440 7. Steering Dynamics Equation (7.213) is a set of two coupled first-order ordinary differential equations with the solution s = \u2219 xt yt \u00b8 = \u23a1\u23a2\u23a2\u23a3 t+ e\u22122t \u2212 C1 e\u22122t + C1 C2e \u2212t e\u22122t + C1 \u23a4\u23a5\u23a5\u23a6 . (7.214) Applying the initial conditions (7.212) we find C1 = cos \u03b8 \u2212 1 cos \u03b8 + 1 (7.215) C2 = 2 sin \u03b8 cos \u03b8 + 1 "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002772_j.robot.2021.103785-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002772_j.robot.2021.103785-Figure6-1.png",
        "caption": "Fig. 6. (a) CAD structure of Modsnakejoint [93] (b) Module structure of AmphiBot-I [152]. Left: two connected joints. Right: internal structure of one module. (c) Actuated universal joint design of Snoopy snake robot [153]. (d) Module of ACM-R5 snake robot [154]. (e) Shammas and Choset\u2019s snake-like manipulator [101]. Left: assembled manipulator; right: schematic joint structure. (f) Joint structure of Perambulator-II snake robot [155]. (g) Mechanical design of Choset\u2019s angular bevel joint [156]. (h) Mechanical structure of Slim Slime Robot [95]. (i) Module structure of OmniTread-4 [146]. Left: the internal module structure; Middle: DC Motor and gear box [157]; right: illustration of pneumatic actuator driving joint.",
        "texts": [
            " Modular structures of snake robots Then, we will classify the modular structures according to the DOF and discuss module structures of snake robots. \u2022 1-DOF module: Due to the limited space in this review, we will only take two examples, Modsnake [93] and AmphiBot I [152], to explain the design concept of 1-DOF snake robot modules. There are many other examples such as S1 robot designed by Miller et al. [158], modular robotPolyBot designed by Yim et al. [45] and ACM III designed by Hirose et al. [125]. Fig. 6(a) shows the CAD drawing of one Modsnake robot module which was made with a modified Hitec HS-5955TG servo motor. The original servo bottom of HS-5955TG was replaced with a servo back in Fig. 6(a) so that it allowed extra space for a customized circuit and specially designed bearing system. Fig. 6(b) illustrates the module structure of AmphiBot-I snake robot [152]. An example of two connected modules is shown in Fig. 6. Each module\u2019s axis of rotation was perpendicular to the operating surface and a pair of passive wheels was equipped under each module. For 1-DOF modules, many planar snake robots were designed by connecting all modules with all rotation axes in parallel. Passive wheels were commonly equipped under modules to reduce the tangential friction. The advantage of such design is that dynamics modelling and motion control are well studied and easy to implement [159,160]. Autonomous planar snake robots [161,162] have been developed for obstacle avoidance and path planning",
            " It is shown that autonomous decentralized control mechanisms are the driving force for the development of snake robots with highly adaptive functions [32, 41]. \u2022 2-DOF module: There are mainly two types of mechanical design of 2-DOF snake robot module: (1) stacking two revolute motors orthogonally, and (2) parallel arranging two motors with the help of worm gears, angular swivel gears or bevel gears. Many other snake robots consist of 2-DOF modules. For example, Snoopy snake robot designed by Wolf et al. [98] applied two DC motors to build a universal actuated joint (see Fig. 6(c)). SHR robot [165] in the Jet Propulsion Laboratory (JPL) and anthropomorphic robot finger designed by Choi et al. [166] were designed based on Takanashi\u2019s angular swivel joints. Shammas has a good historical view on swivel joints in [153]. Recently, Liljeback et al. [164] proposed an interesting snake robot, Kulko, which was built by a number of spherical 2-DOF modules. Each module consisted of two intersecting half-circle rings which are driven by servo motors via worm gear and chain drive. The ACM-R4 [95] snake robot proposed by Yamada and Hirose was built by 2-DOF modules which used one motor for bending and another motor for driving a pair of wheels for propulsion. Fig. 6(d) shows the schematic of the modular structure of the ACM-R5 snake robot. The ACM-R5 has demonstrated impressive 3D snake motions including swimming underwater. The 2-DOF motor-based structure is the second most popular structure. Two DC motors are fixed into one module to provide two degrees of freedom. Two types of 2-DOF module were reviewed: (1) stacking two revolute motors orthogonally, and (2) parallel arranging two motors with the help of worm gears, angular swivel gears or bevel gears. The first solution has simple structure but bulky body while the second solution is compact but complex gear box and joint mechanism",
            " Regarding to modelling and terrain adaptability, these snake robots are like 3D snake robots with 1-DOF modules. \u2022 3-DOF module: Compared to 1-DOF modules, the 3-DOF structure is more compact so that the snake robot can be shorter [101\u2013103]. Due to the complexity of implementing three rotation axes in one compact module, there are only two of the most typical snake robots which have 3-DOF modules, including: (1) Shammas and Choset\u2019s snake-like hyper-redundant manipulator, i.e., Medusa [101] (see Fig. 6(e)) and (2) Ye et al. [103] reconfigurable snake robot, i.e., Perambulator-II (See Fig. 6(f)). The predecessor of the manipulator is a 2- DOF angular swivel joint designed by Choset et al. [156,167] (see Fig. 6(g)). In their latest design [101], one more motor was added to provide three DOFs, i.e., twisting, and all gears were redesigned to use standard off-the-shelf gears. Ma et al. [168] proposed a 3-DOF snake robot module. Ye et al. [103] replaced the DC motors with FutabaS3305 servo motors which were more powerful and accurate. The basic mechanical structures of two modules were similar. The module was called Modular Universal Unit (MUU) and was intended to create a reconfigurable snake robot. Moreover, a pneumatic module can have 2-DOF or 3-DOF motion and the assembled snake robot is able to move in 3D space",
            " The pneumatic actuator was the type of actuator used in early snake robots, i.e., Slim Slime robot (SSR) by Ohno et al. [58]. Recently, Liljeback et al. proposed a modular pneumatic snake robot [132]. Qi et al. [171] proposed a novel pneumatic soft snake robot that uses a travelling wave motion to move in a complex and narrow environment. Similar to the structure of pneumatic actuators, the hydrodynamic actuator has also been used for snake robot design. For example, Date et al. [172] proposed a snake-like propulsion mechanism driven and controlled by fluids. Fig. 6(h) shows the mechanical structure of one SSR module. Inside a module, three identical metal bellows were arranged parallel to each other with even spacing. The SSR (see Fig. 6(h)) is composed of six modules in series and was successfully able to perform locomotion in an inclined pipe. An upgrade of SSR, SSR2, SSR3, was proposed by Aoki et al. in 2004 [173], who designed a bridle bellows for each module in order to produce higher torque. Moreover, Zhang et al. [174] studied the adaptive stiffness control and motion optimization of a snake robot with a variable stiffness actuator, and changed its stiffness by controlling the magnetorheological fluid around the actuator",
            " In each module, a DC motor drives a number of tracks to provide forward propulsion and a pneumatic actuator connects two modules to control the connecting stiffness and bending angle between them. Two research groups designed hybrid actuators and implemented on snake robot. Borenstein et al. [104,146] developed OmniTread-8 and OmniTread-4. Osuka et al. [105] at Kyoto University developed MOIRA and MOIRA-II. OmniTread and MOIRA robots share a similar modular structure. We will take OmniTread as an example to discuss [106]. Fig. 6(i) shows the modular structure of OmniTread-4 (as shown in Fig. 5 for the assembled robot). The module was a rectangular box and four sides of the module were covered by tracks. OmniTread-8 and OmniTread-4 demonstrated operation over many terrain types such as gravel, dirt, ramp conditions and smooth surfaces. \u2022 Artificial muscle actuator: Designing multifunctional artificial muscle actuators with diversely complex deformation is crucial to the snake robots and their bioinspired applications. Hu et al"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000192_j.automatica.2004.11.029-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000192_j.automatica.2004.11.029-Figure1-1.png",
        "caption": "Fig. 1. Comparison of two controllers.",
        "texts": [
            " Here i can be chosen only once for each r, and the magnitude > 0 is adjusted with respect to C, Km, KM in order to stabilize (4) in finite time. In particular, the following controller is obtained with r =3: u = \u2212 sign(\u0308 + 2(|\u0307|3 + | |2)1/6 sign(\u0307 + | |2/3 sign )). (11) Let > 0. Then the 3-sliding homogeneity of (11) follows from the identity sign( \u0308+2(| 2\u0307|3+| 3 |2)1/6 sign( 2\u0307+| 3 |2/3sign 3 )) = sign(\u0308 + 2(|\u0307|3 + | |2)1/6 sign(\u0307 + | |2/3 sign )). The main drawback of these controllers is some trajectory chattering during the transient (see the simulation results in Fig. 1) caused by the complicated structure of the control discontinuity set. The output-feedback performance with noisy measurements is also problematic (coefficients i from Theorem 3 are relatively large). Recall that q is the least common multiple of 1, 2, . . . , r , and define the homogeneous norm and the saturation function Nr = Nr,r = (| |q/r + |\u0307|q/(r\u22121) + \u00b7 \u00b7 \u00b7 + | (r\u22121)|q)1/q , sat(z, \u03b5) = min[1,max(\u22121, z/\u03b5)]. Let i = 1, . . . , r \u2212 1. The new construction is as follows: 0,r = sign , i,r = sat([ (i) + iNi,r i\u22121,r ]/Nr\u2212i r , \u03b5i)",
            " The function g(x)=10 sin(0.05x)+5 was taken for the simulation. Define = y \u2212 g(x). The relative degree r equals 3 and (3) holds locally. The controller parameters and L are found by simulation. Apply the standard controller (11) with = 20, and the new output-feedback controller no. 3 from the list with =0.5 and the differentiator parameter L=400 (though already L=20 suffices, the performance with noisy measurements is worse). The control is taken 0 with t 0.5. The comparison of the controllers is shown in Fig. 1. The short initial chattering of u at t=0.5 in Fig. 1 is caused by the residual differentiator convergence (Fig. 2b). The chattering of the actual control is totally removed. The accuracies | |=|x\u2212xc| 3.2\u00b710\u22127, |x\u0307\u2212x\u0307c| 1.7\u00b710\u22124, |x\u0308\u2212x\u0308c| 1.5\u00b7 10\u22122 and |x \u2212 xc| 3.5 \u00b7 10\u22124, |x\u0307 \u2212 x\u0307c| 5.6 \u00b7 10\u22123, |x\u0308 \u2212 x\u0308c| 1.5 \u00b7 10\u22121 were obtained respectively with = 10\u22124 and = 10\u22123, which generally corresponds to Theorem 4. The performance of the new controller and of the internal second order differentiator in the presence of a highfrequency measurement noise with the magnitude 0"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003728_1.3099008-Figure21-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003728_1.3099008-Figure21-1.png",
        "caption": "Fig 21. Examples for systems with friction self-excited oscillations.",
        "texts": [
            "org/about-asme/terms-of-use Appl Mech Rev vol 51, no 5, May 1998 Among the nonlinear effects friction may have in dynamical systems are nonlinear damping properties, self-excited oscillations, multiple steady states, and chaotic behavior. One particular nonlinear effect which is most often associated with friction is stick-slip. This nonlinear effect shows up in many kinds of engineering systems and also in everyday life. Examples are the stick-slip oscillations of the string of a violin, squeaking chalks, creaking doors, squealing tramways in 'narrow curves, chattering machine tools or grating brakes (see Fig 21). lncidently, stick-slip was coined in 1939 by FP Bowden and LL Leben, who had built an apparatus at the University of Cambridge for studying the phenomenon (Rabinowicz, 1965). Systems with dry friction studied recently can be classified as follows: 1) Stick-slip oscillations of a single degree of freedom system with a) self excitation based on measurements (Antoniou etal , 1976; Feder and Feder, 1991; Gao and Kuhlmann-Wilsdorf, 1990; Grudzinski et al, 1992; Heslot et al, 1994; Ko and Brockley, 1970; Musiol, 1990; Sakamoto et al 1980; Sakamoto, 1985), based on simulations (Armstrong-Helouvry et al, 1994; Grudzinski et al, 1992; Hagedorn, 1978; Kauderer, 1958; Magnus, 1961; Rabinowicz, 1958; Sikora, 1991; Sikora, 1992), b) external excitation based on measurements (Wojewoda, 1992), based on simulations (Den Hartog, 1931; Marui and Kato, 1984; Moser, 1987; Schmieg and Vieisack, 1987; Shaw, 1986), c) simultaneous self and external excitation (Marui and Kato, 1984; Narayanan and Jayaraman, 1991; Oestreich et al, 1996; Popp and Stelter, 1989; Poppet al, 1996; Stelter, 1990; Stelter, 1992; Wiercigroch, 1994), 2) friction oscillators with displacement-dependent friction (Anderson and Ferri, 1990; Feeny and Moon, 1989; Feeny, 1995; Feeny and Liang, 1995), 3) systems modelling the squeal of brakes (Abdelhamid, 1995a; Abdelhamid, 1995b; Alifov and Frolov, 1977; Anderson and Ferri, 1990; Chan, 1995; Hulten, 1995; Lang, 1994; Nack and Joshi, 1995; Spurr, 1961; Tondl, 1970; Watson, 1991), 4) friction oscillators with multiple degrees of freedom (Harli, 1983; Oden, 1989; Popp and Stelter, 1990; Pfeiffor, 1992; Tworzydlo etal , 1992), 5) modelling the friction body with a tangential, normal and torsional degree of freedom (Aronov et al, 1984; Earles and Soar, 1972; Hess and Soom, 1992; Soom and Kim, 1981; Streator, 1992; Swayze and Akay, 1992; Tworzydlo etal , 1992), 6) sliders on wavy surfaces with contact oscillations (Anand and Soom, 1984; Hess and Soom, 1991a; Hess and Soom, 1991b; Soom and Chen, 1986; Soom and Kim, 1981; Wang and Soom, 1983), 7) friction body as continuum with travelling Schallamach waves (Dieterman, 1990), Feeny et al: Historical review of dry friction and stick-slip phenomena 331 Existence and uniqueness of solutions Including the nonlinearity associated with dry friction in the equations of motion of a dynamical system can have interesting effects on the dynamical response"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000790_j.engfailanal.2011.07.006-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000790_j.engfailanal.2011.07.006-Figure1-1.png",
        "caption": "Fig. 1. Model of the spur gear tooth as a non-uniform cantilever beam with a crack at tooth root.",
        "texts": [
            " Few work focus their attentions on the gear tooth propagating along tooth width, namely the crack depth varies along tooth width. A gear mesh stiffness calculation model with consideration of tooth crack propagating along tooth width is proposed in Section 1.2 which is based on the existing analytical model and method reviewed in Section 1.1 which assumes the tooth crack was though the tooth width with a constant depth. Deflections of a spur gear tooth can be determined by considering it as a non-uniform cantilever beam with an effective length d displayed in Fig. 1. Here, the crack is assumed to go through the whole tooth width [1,4,13] W with a constant depth q0 and a crack inclination angle ac. The bending, shear and axial compressive energy stored in a tooth can be represented by [12,13], Ub \u00bc F2 2Kb ; Us \u00bc F2 2Ks ; Ua \u00bc F2 2Ka \u00f01-a;b; c\u00de where Kb, Ks, Ka are the bending, shear and axial compressive stiffness in the same direction under the action of the force F. Based on the beam theory, the potential energy stored in a meshing gear tooth can be calculated by [1,12,13], Ub \u00bc Z d 0 M2 2EIx dx; Us \u00bc Z d 0 1:2F2 b 2GAx dx; Ua \u00bc Z d 0 F2 a 2EIx dx \u00f02-a;b; c\u00de where Ub, Us, Ua are the potential energy stored in the bending, shear and axial compressive deformations, respectively under the action of the mesh force F. And Fb, Fa and M are calculated by Fb \u00bc F cos a1; Fa \u00bc F sin a1; M \u00bc Fb x Fa h \u00f03-a;b; c\u00de Based on Eqs. (1)\u2013(3), the bending stiffness Kb can be obtained as, 1 Kb \u00bc Z d 0 \u00f0x cos a1 h sina1\u00de2 EIx dx \u00f04\u00de Shear stiffness Ks is calculated by, 1 Ks \u00bc Z d 0 1:2 cos2 a1 GAx dx \u00f05\u00de Axial compressive stiffness Ka is, 1 Ka \u00bc Z d 0 sin2 a1 EAx dx \u00f06\u00de In the formulas (2)\u2013(6), h, x, dx, a1; d are shown in Fig. 1. E is the Young modulus. G represents the shear modulus. Ix and Ax represent the area moment of inertia and area of the section where the distance between the section and the acting point of the applied force is x, and they can be obtained by Ix \u00bc 1 12 \u00f0hx \u00fe hx\u00de3W; hx 6 hq 1 12 \u00f0hx \u00fe hq\u00de3W; hx > hq ( \u00f07\u00de Ax \u00bc \u00f0hx \u00fe hx\u00deW; hx 6 hq \u00f0hx \u00fe hq\u00deW; hx > hq \u00f08\u00de G \u00bc E 2\u00f01\u00fe v\u00de \u00f09\u00de where v is the Poisson ratio. hq \u00bc hc q0 sin ac , q0 and ac are the depth and the inclination angle of the crack, respectively"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure16.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure16.6-1.png",
        "caption": "Figure 16.6 (a) Beam with (b) mechanism for determining the bending moment at D.",
        "texts": [
            " Conclusion: If the force sought performs negative work (rule 1) over a unit displacement (rule 3), then the influence line is equal to the deflection line of the mechanism. The influence line is positive where the displacement is in the direction of F (segment AB) and negative where the displacement is in the opposite direction of F (segment BC) (rule 2). Example 2 \u2013 Influence line for a bending moment The second example relates to the influence line for the bending moment at cross-section D of the beam in Figure 16.6a. Again we convert the beam into a mechanism by introducing a hinge at D. The action of the bending moment at D is replaced by the pair of moments MD that are applied to the mechanism at either side of the hinge (see Figure 16.6b). The direction of MD can be chosen arbitrarily. 750 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM We apply a virtual displacement to the mechanism by rotating beam segments AD and DBC at D through an angle \u03b4\u03b8 with respect to one another, but in such a way that MD performs negative work (rule 1) (see Figure 16.6c). Assume that load F is displaced by a distance \u03b4w, and that \u03b4w is positive in the direction of F (rule 2). For equilibrium, the virtual work is zero: \u2212MD\u03b4\u03b8 + F\u03b4w = 0 so that MD F = \u03b4w \u03b4\u03b8 . The bending moment MD is proportional to the displacement \u03b4w. The influence line for MD therefore has the same shape as the deflection line of the mechanism. The scale factor is the angle \u03b4\u03b8 . For a virtual displacement, \u03b4w and \u03b4\u03b8 are infinitesimally small, but their ratio is finite. The influence line can therefore be drawn on an enlarged scale. \u03b4\u03b8 can be defined as (see also Section 15.4.2): \u03b4\u03b8 = \u03b4a . If we select \u03b4a equal to , it is said that the angle \u03b4\u03b8 has the orthogonal unit value \u2013 this is also referred to as a unit rotation (rule 3). In that case the influence line is exactly the same as the deflection line of the mechanism (see Figure 16.6d). Conclusion: If we apply a unit rotation to the hinge, so that the bending moment performs negative work, then the influence line for that bending moment is the same as the deflection line of the mechanism. 16 Influence Lines 751 Figure 16.7 The construction of an angle \u03b4\u03b8 with orthogonal unit value. Figure 16.8 (a) Beam with (b) mechanism for determining the shear force at D, (c) virtual displacement for which VD performs negative work. Arcs can be used to construct an orthogonal unit angle as shown in Figure 16.7. In that figure, two arcs are drawn, but one arc is actually sufficient, as appears from the plot of the influence line in Figure 16.6d. Example 3 \u2013 Influence line for a shear force The third example relates to the influence line for the shear force in crosssection D of the beam in Figure 16.8a. The procedure is identical to that in the previous examples. The beam is transformed into a mechanism by introducing a slide joint or shear force hinge at D and replacing the action of the shear force at D by the pair of forces VD that are applied on the mechanism at either side of the slide joint (see Figure 16.8b). The direction of VD can be chosen arbitrarily"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure7.13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure7.13-1.png",
        "caption": "Figure 7.13 (a) Due to the horizontal component of the membrane force, the ring beam is pulled inwards on all sides. Here we can recognise the analogy of a closed ring with underpressure. (b) Compressive forces are generated in the ring. They can be determined using the boiler formula, or directly from the equilibrium of half a ring beam.",
        "texts": [
            " These are the distributed force n, which the pneu exerts on the ring beam, and the dead weight qdw of the ring beam (also a force per length). The vertical component nv tries to lift the ring beam. In order to prevent this, the dead weight qdw has to be larger than nv = 134 kN/m. If the ring beam is made of concrete, with a specific weight of 24 kN/m3, then the cross-section A of the beam has to obey qdw = A \u00d7 (24 kN/m3) \u2265 nv = 134 kN/m \u21d2 A \u2265 134 kN/m 24 kN/m3 = 5.6 m2. The cross-section of the ring beam has to be at least 5.6 m2. b. Due to the horizontal forces nh, the ring beam is pulled inwards from all sides (see Figure 7.13a). Here, you will recognise the loading case of the closed ring from Figure 7.8b, but now with an underpressure instead of an overpressure. 7 Gas Pressure and Hydrostatic Pressure 253 A compressive force N \u20321 is formed in the ring. This can be calculated using the boiler formula from the previous example, or directly from the equilibrium of the half ring beam in Figure 7.13b: N \u2032 = 2rbeamnh 2 = rbeamnh = (1000 m)(263 kN/m) = 263 MN. Comment: This force is relatively large for a concrete cross-section of 5.6 m2. The compressive force in the ring may therefore call for a larger cross-section. Example 3 A uniformly distributed load q is acting on the plane body in Figure 7.14. The load acts in the plane of the body along the entire outline and normal to the body. Questions: a. Show that the resultant of the distributed load on the body is zero, regardless of the shape of the body"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000512_027836499801701202-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000512_027836499801701202-Figure1-1.png",
        "caption": "Fig. 1. Sketch of a compass-like biped robot on a slope. The robot consists of two kneeless legs, each having a point mass, and a third point mass coincides with the hip joint. The prismatic-joint knee shown in the picture is an imaginary concoction meant to avoid the conceptual foot-clearance problem. See Section 2.1 for details.",
        "texts": [
            " We present a novel graphical technique based on the first return map that compactly captures the entire evolution of the gait, from symmetry to chaos. The fractal dimension of the attractor is computed to be 2.07. For a system with a four-dimensional state space, this indicates strong phasespace volume contraction. Additional passive dissipative elements in the robot joint result in a significant improvement in the stability and versatility of the gait, and provide a rich repertoire for simple control laws. 2. Modeling 2.1. Robot Model and Modeling Assumptions Figure 1 shows a sketch of a compass-like biped robot. The details of the model and the underlying assumptions are listed below: Mass: Concentrated at three points: ~ mass mH at the hip, and ~ masses m on each leg, located at distances a and b from the leg tip and the hip, respectively. The total mass of the robot mc = 2m + mh is constant and equal to 20 kg, whereas the mass ratio /-t = varies from 0.1 to 10 during the simulation trials. Leg: The legs are identical. The leg length I = a + b is constant and equal to 1 m, whereas the length ratio a b varies from 0.1 to 10 during the simulation trials. Actuation: The robot is unactuated. Ground: The robot walks down on a plane surface inclined at a constant angle 0 with the horizontal. Gait: The motion is constrained in the sagittal plane, and consists of the following two stages: ~ Swing: during this stage, the robot hip pivots around the point of support on the ground of its support leg. The other leg, called the nonsupport leg or the swing leg swings forward (the compass robot in Fig. 1 is in the swing stage). ~ Transition: this occurs instantaneously when the swing leg touches the ground and the previous support leg leaves the ground. Ground impact: The impact of the swing leg with the ground is assumed to be slipless plastic. This implies that during the instantaneous transition stage (see, for instance, the work of Hurmuzlu and Chang [1992]): ~ the robot configuration remains unchanged, and ~ the angular momentum of the robot about the impacting foot as well as the angular momentum of the pre-impact support leg about the hip are conserved"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure9.24-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure9.24-1.png",
        "caption": "Figure 9.24 More than three support reactions are needed for an immovable support of a compound truss, as the internal degrees of freedom also have to be eliminated. In this case, four support reactions are required, provided by two hinged supports.",
        "texts": [
            " For example by means of a hinged support together with a roller support or bar support, as shown in Figure 9.23. Compound trusses can be seen as systems of rigid bodies that have a certain degree of freedom with respect to one another. The possible movements with respect to one another are known as the internal degrees of freedom. The immovability of a compound truss always needs more than three support reactions, as the internal degrees of freedom also have to be eliminated. In this way, the truss in Figure 9.24 is not shape-retaining in itself, as the two constituent parts can rotate with respect to one another. The two hinged supports ensure the kinematic determinacy of the truss. In the examples, bar supports, roller supports, and hinged supports have been used. It should be clear that fixed supports are not used in trusses. 9 Trusses 333 Instead of looking at the degrees of freedom for simple or compound trusses, we can also determine how many support reactions are needed to keep the truss in equilibrium under every imaginable load"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure3.34-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure3.34-1.png",
        "caption": "Fig. 3.34: Free-body diagram of a manipulator link.",
        "texts": [
            " But it would not have permitted the computation of the angular velocity entry since there is no orientation vector whose derivative gives the angular velocity vector. One of the most important features of robotic manipulators is the positioning of objects or end-effector in static configurations. In this task a manipulator acts like a structure when the joints are locked in a suitable way. Thus, it is of practical interest to evaluate and simulate the static behavior of a manipulator. In addition, design considerations can be deduced by the computations for the static equilibrium. In Fig. 3.34 a mechanical model is presented in the form of a free-body diagram to determine conditions for static equilibrium of a manipulator link. The free-body diagram of a rigid body is obtained by analyzing forces and torques acting on the body when it is considered as disconnected from the neighboring bodies. Thus, actions on a link body of a manipulator can be considered as determined in Fig. 3.34: - fi is the force exerted on link i by the link i-1 through the joint connection; - ti is the torque exerted on link i by the link i-1 through the joint connection; - Fi is the external force exerted on link i by bodies that do not compose of the manipulator; it can include the weight of the body link; - Ti is the external torque exerted on link i by bodies that do not compose of the manipulator. All above-mentioned vectors are expressed with respect to the i-th frame on link i. The actions fi+1 and ti+1 are the actions exerted by link i on link i+1 and they are described in the i+1-th reference. The skew general architecture and points of application for forces and torques Chapter 3: Fundamentals of the Mechanics of Robots148 characterize the geometry of the link free-body diagram of Fig. 3.34. The general skew architecture refers also to the model for H\u2013D notation and represents a general model of a mechanical design from which any particular case can be deduced. The positions of the points of application for forces and torques can also be considered as structural characteristics for manipulator links. In fact, points Oi and Oi+1 are the origins of the i-th and (i+1)-th reference frames; Gi point is the center of mass of the link. It is of note that both forces and torques will be considered as resultant vectors of actions acting on the link, even in a distributed way",
            " Otherwise the corresponding point of application must be identified in the link geometry. It is of note that even the actions fi and ti of the neighborhood link through the joint can be considered as resultant vectors of distributed actions exerted on the joint kinematic surface. This means that a well-designed joint should have a symmetric mechanical design and operation so that Oi will be the point of application of the resultant vectors. Fundamentals of Mechanics of Robotic Manipulation 149 The mechanical model of Fig. 3.34 is useful also for deducing a formulation for the static equilibrium of a link. In fact, summing the forces and equating to zero, the equilibrium for translation is expressed as 0R i1i1iii =+\u2212 ++ Fff (3.5.1) in which vector fi+1 has been written with respect to the i-th frame by means of the rotation matrix iRi+1 describing the (i+1)-th frame relative to i-th frame. The rotation equilibrium can be expressed by summing torques about the origin point Oi in the form 0xRxR iii1i1ii1iii1i1iii =+\u2212+\u2212 +++++ FGOfOOTtt (3",
            " Fundamentals of Mechanics of Robotic Manipulation 153 in which g function vector indicates the algebraic nature of those dynamic equations. Both problems have practical interest when they are solved with efficient computational algorithms. Direct problem is fundamental for simulating the robot operation before building or operating them. Inverse problem is useful for regulating the robot operation during a given manipulation task. In addition, the solutions of both problems can be used and are used in control algorithms for high performance robot motion. The model of Fig.3.34 can be extended to the case of dynamics by adding the consideration of inertia forces and torques, as shown in Fig. 3.36. The inertia forces and torques are expressed as Gii in i m aF \u2212= (3.6.3) i Gi iii Gi i in i xIIT \u2212\u2212= (3.6.4) in which inertia characteristics are evaluated in terms of mass mi and inertia tensor Ii Gi of the i-th link. Indeed, inertia characteristics are determined when both mass distribution and kinematic status of a link are known. Chapter 3: Fundamentals of the Mechanics of Robots154 Mass distribution is characterized by the mass of a link and inertia tensor"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure14.53-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure14.53-1.png",
        "caption": "Figure 14.53 (a) Fixed bar type structure, loaded by two forces, with (b) the lines of force.",
        "texts": [
            " This is not surprising, for1 M = Ne. 1 Without taking into account the coordinate system and signs. 678 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM The scale factor is N . If N is a tensile force, the line of force is on the same side of the member axis as the M diagram. If N is a compressive force, the line of force and the M diagram are not on the same side. It is up to the reader to verify this, using the bending moment diagram in Figure 14.49b and the lines of force in Figure 14.52. Example 2 The structure ABCD in Figure 14.53a is fixed at A, and loaded by a horizontal force F at C and a vertical force F at D. Question: Determine the lines of force for AB, BC and BD. Solution: The lines of force are shown in Figure 14.53b. All cross-sections between C and B have to transfer the horizontal force F . The line of force for BC therefore coincides with the line of action of this horizontal force. All cross-sections between D and B have to transfer the vertical force F . The line of force for BD coincides with the line of action of this vertical force. All cross-sections between B and A have to transfer the resultant of the forces F at C and D. The line of force for AB coincides with the line of action of this resultant"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure10.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure10.1-1.png",
        "caption": "FIGURE 10.1. Vehicle body coordinate frame B(Cxyz).",
        "texts": [
            " Derive and compare their equations of motion. 9. Applied Dynamics 581 10 In this chapter we develop a dynamic model for a rigid vehicle in a planar motion. When the forward, lateral and yaw velocities are important and are enough to examine the behavior of a vehicle, the planar model is applicable. 10.1 Vehicle Coordinate Frame The equations of motion in vehicle dynamics are usually expressed in a set of vehicle coordinate frame B(Cxyz), attached to the vehicle at the mass center C, as shown in Figure 10.1. The x-axis is a longitudinal axis passing through C and directed forward. The y-axis goes laterally to the left from the driver\u2019s viewpoint. The z-axis makes the coordinate system a righthand triad. When the car is parked on a flat horizontal road, the z-axis is perpendicular to the ground, opposite to the gravitational acceleration g. To show the vehicle orientation, we use three angles: roll angle \u03d5 about the x-axis, pitch angle \u03b8 about the y-axis, and yaw angle \u03c8 about the zaxis. Because the rate of the orientation angles are important in vehicle 584 10"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001097_1.6271-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001097_1.6271-Figure3-1.png",
        "caption": "Fig. 3 Point mass model for outer-loop inversion.",
        "texts": [
            " An approximate model for the attitude dynamics of the helicopter was generated by linearizing the nonlinear model around hover and neglecting coupling between the attitude and translational dynamics as well as the stabilizer bar, \u03b1des = A\u03021   p q r   + A\u03022   u v w   + B\u0302      \u03b4lat \u03b4lon \u03b4ped   \ufe38 \ufe37\ufe37 \ufe38 des \u2212   \u03b4lat \u03b4lon \u03b4ped   \ufe38 \ufe37\ufe37 \ufe38 trim    (60) or \u03b1des = A\u03021\u03c9B + A\u03022vB + B\u0302(\u03b4mdes \u2212 \u03b4mtrim) (61) where A\u03021 and A\u03022 are the attitude and translational dynamics, respectively, \u03c9B is the angular velocity of the body with respect to the Earth expressed in the body frame, vB is the body velocity vector with respect to the Earth expressed in the body frame, and \u03b4mtrim is the trim control vector that is consistent with the linear model. Choosing the control matrix B\u0302 such that it is invertible, the moment controls may be evaluated as \u03b4mdes = B\u0302\u22121(\u03b1des \u2212 A\u03021\u03c9B \u2212 A\u03022vB) + \u03b4mtrim (62) The translational dynamics were modeled as a point mass with a thrust vector that may be oriented in a given direction as shown in Fig. 3. More involved inverses25 may be used, but the simple relationships between thrust, attitude, and accelerations suffice when used with adaptation, ades =   0 0 Z\u03b4coll  ( \u03b4colldes \u2212 \u03b4colltrim ) + Lbvg (63) where Z\u03b4coll is the control derivative for acceleration in the vertical axis. Lbv is the direction cosine matrix that transforms a vector from the vehicle (or local) frame to the body frame and g is an assumed gravity vector. The desired specific force along the body z axis may be evaluated as fsf = (ades \u2212 Lbvg)3 (64) The required collective input may be evaluated as \u03b4colldes = fsf / Z\u03b4coll + \u03b4colltrim (65) The attitude augmentation required to orient the thrust vector to attain the desired translational accelerations are given by the following small-angle corrections from the current reference body attitude and attitude command: 1 = ades2/ fsf, 2 = \u2212ades1 / fsf, 3 = 0 (66) For this simplified helicopter model, heading change has no effect on accelerations in the x, y plane, and hence, 3 = 0"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.32-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.32-1.png",
        "caption": "FIGURE 7.32. A negative four-wheel steering vehicle.",
        "texts": [
            " Steering Dynamics 409 reduces dramatically to an acceptable value when the vehicle is moving. Furthermore, an offset design sometimes makes more room to attach the other devices, and simplifies manufacturing. So, it may be used for small off-road vehicles, such as a mini Baja, and toy vehicles. 7.5 F Four wheel steering. At very low speeds, the kinematic steering condition that the perpendicular lines to each tire meet at one point, must be applied. The intersection point is the turning center of the vehicle. Figure 7.31 illustrates a positive four-wheel steering vehicle, and Figure 7.32 illustrates a negative 4WS vehicle. In a positive 4WS situation the front and rear wheels steer in the same direction, and in a negative 4WS situation the front and rear wheels steer opposite to each other. The kinematic condition between the steer angles of a 4WS vehicle is cot \u03b4of \u2212 cot \u03b4if = wf l \u2212 wr l cot \u03b4of \u2212 cot \u03b4if cot \u03b4or \u2212 cot \u03b4ir (7.73) where, wf and wr are the front and rear tracks, \u03b4if and \u03b4of are the steer angles of the front inner and outer wheels, \u03b4ir and \u03b4or are the steer angles of the rear inner and outer wheels, and l is the wheelbase of the vehicle"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001751_0278364913483183-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001751_0278364913483183-Figure5-1.png",
        "caption": "Fig. 5. Photograph of the electromagnetic coil setup. (a) All system components are shown with the coils in the operational position. (b) Two coils hinge open to allow for access to the workspace.",
        "texts": [
            " As a final feedback control element to better deal with wall effects when the wall position is known a priori, a feedforward term can be included which uses equations (26) and (27) to update the lag angles proportional to the increase in expected torque. However, the modest increase in drag near the wall precludes the use of such a wall-effect controller except when operating within 1\u20132 microrobot body-lengths of the wall. The magnetic microrobots are actuated by eight independent air or iron-core electromagnetic coils shown in Figure 5, which are aligned pointing to a common workspace center point with an approximate opening size of 12 cm. The currents in the electromagnetic coils are controlled using a PC with data acquisition system at a control bandwidth of 20 kHz, using linear electronic amplifiers (SyRen 25, Dimension Engineering Inc.) and at St Petersburg State University on January 25, 2014ijr.sagepub.comDownloaded from Hall-effect current sensors (ACS714, Allegro Microsystems Inc.). Imaging of the microrobots and workspace is accomplished by two CCD cameras (Foculus F0134SB) connected to variable magnification microscope lenses, providing up to a 26 \u00d7 20 mm field of view from the top and side perspectives"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure13.73-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure13.73-1.png",
        "caption": "Figure 13.73 The vertical support reaction at B follows from the moment equilibrium of ABEG about G.",
        "texts": [
            " At D, G and E draw the tangents to the M diagram. 13 Calculating M, V and N Diagrams 603 Solution (units kN and m): a. The structure BEGD is statically indeterminate to the second degree. The structure has five unknown support reactions while there are only three equilibrium equations. b. The horizontal support reaction at B follows from the shear force in BE. Bh = 5 kN (\u2192). Bar AE is a two-force member so that Av = 0. Introduce a cut at G, and investigate the moment equilibrium of ABEG about G (see Figure 13.73): \u2211 T|G = +12.5 + 23 \u00d7 1, 25 + 5 \u00d7 2.5 \u2212 Bv \u00d7 2.5 = 0 \u21d2 Bv = 21.5 kN (\u2191). The support reactions at A and C are found from the equilibrium of the structure as a whole: \u2211 Fvert = 0 \u21d2 Cv = 24.5 kN (\u2191),\u2211 T|A = 0 \u21d2 Ch = 32.5 kN (\u2192),\u2211 Fhor = 0 \u21d2 Ah = 37.5 kN (\u2190). In Figure 13.74, the support reactions are shown as they act in reality. c. By resolving the horizontal and vertical support reaction at C into components parallel to and normal to CD we find the normal force NCD and the 604 ENGINEERING MECHANICS"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure6.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure6.2-1.png",
        "caption": "Fig. 6.2. Segmental construction of a disc type coreless PM brushless machine: (a) single module (segment), (b) three-module assembly.",
        "texts": [
            " The whole winding is then embedded in a high mechanical integrity plastic or resin. Since the topology shown in Fig 6.1 does not use any ferromagnetic core with slots, the machine is free of cogging (detent) torque 194 6 AFPM Machines Without Stator and Rotor Cores 6.3 Air Gap Magnetic Flux Density 195 and core losses. The only eddy current losses are losses in the stator winding conductors and metallic parts (if they exist) that reinforce the stator coreless winding. The coreless machine designed as a segmental (modular) machine is shown in Fig. 6.2. The output shaft power can easily be adjusted to desired level by adding more modules. To obtain a high power (or high torque) density motor, the magnetic flux density in the air gap should be as high as possible. This can be achieved by using PMs arranged in \u201cHalbach array\u201d (Figs 3.15, 6.3 and Fig 6.4). The magnetic flux density distribution excited by Halbach arrays is described by eqns (3.42) to (3.46). In practice, the angle between magnetization vectors of adjacent magnets is 90o, 60o or 45o (Fig",
            " The coreles stator has a foil winding at both sides. The 8-pole PM rotor is designed as a twin external rotor. A low speed, three-phase, Y-connected AFPM brushless motor rated at 10 kW, 750 rpm, 28.5 A has been designed and investigated. The rotor does not have any ferromagnetic core and consists of trapezoidal coils embedded in a high mechanical integrity resin. Average quality sintered NdFeB PMs with Br = 1.2 T and Hc = 950 kA/m have been used. The double-disc construction is similar to that shown in Fig. 6.2b. The design data and calculated parameters are given in Table 6.1. All parameters except the air gap magnetic flux density Bmg and synchronous reactance Xsd = Xsq = Xs have been calculated analytically. The magnetic flux density and synchronous reactance has been calculated using the 2D FEM. The resistance R1 = 0.175 \u2126 and synchronous reactance Xs = 0.609 \u2126 per phase are for two stator modules connected electrically in series. The EMF constant kE = 5.013 Vs and torque constant kT = 2.394 Nm/A have been calculated for one stator segment"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001467_j.mechmachtheory.2010.05.001-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001467_j.mechmachtheory.2010.05.001-Figure4-1.png",
        "caption": "Fig. 4. An SPS or UPS leg.",
        "texts": [
            " (8), the input transmission index of the PSS leg can thus be obtained as \u03bb34 = jf34\u22c5v34j jf34\u22c5v34jmax = jf34\u22c5v34j = jcos\u03c8j \u00f015\u00de is also free from the coordinate system o\u2212xyz. which One may see from Eq. (15) that the input transmission index of the PSS leg is equal to the absolute value of the cosine of the angle between the translational direction of the prismatic joint and the unit vector along the coupler link. Moreover, this result also fits some similar legs, such as the PUS leg, the PRS leg, and the planar PRR leg. Notably, an SPS or UPS leg (shown in Fig. 4), where the prismatic joint between two passive joints is actuated, is usually employed in parallel manipulators. In such a leg, the position of the prismatic actuator is different from that in the PSS leg. However, the input transmission index of the SPS or UPS leg can still be obtained through the above result. Since the translational direction of the prismatic actuator is along the coupler link, the input transmission index of the SPS or UPS leg is constant and its value is equal to 1. As shown in Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002261_s00604-017-2609-1-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002261_s00604-017-2609-1-Figure2-1.png",
        "caption": "Fig. 2 Summary of enzymatic glucose oxidation mechanisms in the first, second and third generation biosensors",
        "texts": [
            " The drawbacks associated with these second-generation biosensors are that the mediator leakage affects the sensor performance. A further improvement in glucose sensing is achieved by eliminating the usage of mediator. The third-generation glucose sensors facilitate the direct electron transfer between the redox center of an enzyme and the electrode, leading to a high sensitivity and reproducibility [10\u201312]. The chemical-relay groups anchored GOD networks increases the electron transfer kinetics and offers high output current [13]. Figure 2 explains the mechanism of first, second and third generation glucose biosensors. The first glucose monitoring device (Ames Reflectance Meter) emerged in 1969. It was based on the catalytic activity of glucose oxidase enzyme and evaluated glucose levels in a 50 \u03bcL blood sample. Later in the 1970s, selfmonitoring of diabetes became accessible with the construction of the personal glucose monitor, which allowed multiple capillary blood glucose tests, insulin dose adaptation and, thus, better glucose control in terms of both hyper and hypoglycemia"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure9.79-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure9.79-1.png",
        "caption": "Figure 9.79 Zero-force members have a definite function in a truss. On the one hand, they ensure the truss retains its shape. On the other hand they can prevent buckling (in the plane of the structure) of (long) compressed members, such as the top chord in this truss.",
        "texts": [
            " When we talk about omitting zero-force members, this is done only to simplify the calculation. If the zero-force members are removed from the truss in reality, the truss becomes kinematically indeterminate. Zero-force members therefore have a genuine function in the truss. On the one hand they ensure the truss retains its shape, while on the other they can prevent buckling (in the plane of the structure) of (long) compressed members, such as the bottom chord in Figure 9.77, or the top chord in Figure 9.79. 9 Trusses 369 Example 3 You are given the truss in Figure 9.80. The diagonals are crossing members. Question: Determine all the zero-force members for the given load. Solution: In this truss, it is not possible to find a section across three members (that do not intersect in one point), nor is there a joint with less than two unknowns (member forces or support reactions). We therefore cannot determine the member forces with the method of sections, or with the method of joints, unless we first determine the support reactions"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000206_978-0-387-74315-8-Figure3-2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000206_978-0-387-74315-8-Figure3-2-1.png",
        "caption": "Figure 3-2: Articulated body with direction al apparent mass accelerate, so its acceleration is [OT Iim} f/f In fact the 3 X 3 matrix in the mass quadrant of the inertia for this articulated body, taking the sphere as the handle, is",
        "texts": [
            " Articulated-body inertias, when they exist, are symmetric, 6 posi\u0163ive definite matrices (with respect to spatial transpose); articulated-body inverse inertias are symmetric, positive definite or semi-definite; and inverse cross-inertias are generally non-symmetric, indefinite and singular. Articulated-body inertias are general symmetric matrices, not conforming to the special form of rigid-body inertias. Physically, this implies that there is no such thing as a centre of mass for an articulated body and that the apparent mass has direction al properties analogous to those of rotational inertia. To see this consider the simple articulated body shown in Figure 3-2. It consists of two bodies: a sphere of mass mI with a cylindrical shaft through the middle, and a cylinder of mass m2 which fits in the shaft. The shaft is parallel to the y axis and the two bodies are connected kinematically by a cylindric joint. Both bodies are initially at rest i'nd their centres of mass coincide at the centre of the sphere. A force fz applied to the sphere along a line parallel to the z axis and passing through the centre of the sphere causes both bodies to accelerate with acceleration [OT 1/(m1+m2) fzT]T, but a force fII applied to the sphere in the y direction through the centre of the sphere causes only the sphere to 6This is ccrtainly true for loop-free articlllatcd bodies with no kinematic contact with ground, but 1 havc not bccn ablc to prove this statement for thc general case"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000846_j.ymssp.2017.05.024-Figure14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000846_j.ymssp.2017.05.024-Figure14-1.png",
        "caption": "Fig. 14. Pitting growth modeling (slight to moderate to severe \u2013 from left to right) [58].",
        "texts": [
            " [142] presented a single rectangular tooth pit effect on the mesh stiffness of a wind turbine gearbox and analyzed vibration signal fault signatures in the time and frequency domain. Ma et al. [126] modeled a rectangular spalling and investigated the influence of spalling width, spalling length and spalling location on mesh stiffness, respectively. Saxena et al. [127] modeled a single spall in the rectangular, circular or V-shaped, respectively and investigated the effect of spall shape, size and location on gear mesh stiffness. Liang et al. [58] modeled tooth pitting propagation on a single tooth by increasing the number of circular pits as shown in Fig. 14. The mesh stiffness equations are derived using the potential energy method and expressed as a function of gear rotation angle, which is easy to use. Later, Liang et al. [39] modeled tooth pitting propagation to neighboring teeth and analyzed fault symptoms for pitting damage detection. In addition, statistical features that are insensitive to gear mesh damping and environmental noise are recommended. In Refs. [5,39,58,96,126,127,142], gear fault is reflected by gear mesh stiffness reduction which is evaluated using the potential energy method"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure7.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure7.3-1.png",
        "caption": "Fig. 7.3. Potentiometer",
        "texts": [
            " In the same ratio is the number of wires needed to 7.1 Positional Sensors 211 transmit the signals. Because of the more complex construction, absolute encoders are more expensive and less reliable than incremental ones. The advantage of the absolute encoder over the incremental one is that it does not accumulate the error induced by disturbances, and after reconnecting to power supplies it shows the absolute position at once, while the incremental encoder has to be brought into the referent position first. Potentiometers (Fig. 7.3) are the simplest positional transducers. They consist of a circular resistor (made of wire, carbon or some other resistive material) with the sliding contact on it connected to the rotating shaft. The resistance between the connections of the potentiometer is the function of the relative position of the shaft with the slider to the circular resistor. This signal is converted into digital information by means of the appropriate measuring electronics. The only moving parts of the resolver which might be the source of errors are the contact rings on the rotor"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000790_j.engfailanal.2011.07.006-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000790_j.engfailanal.2011.07.006-Figure3-1.png",
        "caption": "Fig. 3. Crack model at gear tooth root.",
        "texts": [
            " The total equivalent mesh stiffness of one tooth pair in mesh can be obtained by Ke \u00bc 1 1 Kb1 \u00fe 1 Ks1 \u00fe 1 Ka1 \u00fe 1 Kf 1 \u00fe 1 Kb2 \u00fe 1 Ks2 \u00fe 1 Ka2 \u00fe 1 Kf 2 \u00fe 1 Kh \u00f014\u00de Here, the subscripts 1, 2 mean the pinion and gear, respectively. At very early stage of crack propagation, it usually starts from some local positions where the stress concentrations are observed. In order to investigate the effect of a tooth root crack propagating along tooth width non-uniformly, a mesh stiffness model is developed by dividing a gear tooth into some independent thin pieces like that shown in Fig. 3b, so that the crack length along tooth width for each piece can be regarded as a constant which is reasonable when dx is small. Stiffness of each piece is denoted as Kt (x) and can be calculated with taking tooth bending, shear and axial compress into account based on Eqs. (4)\u2013(6). Kt (x) can be obtained by Kt\u00f0x\u00de \u00bc 1 1 Kb\u00f0x\u00de \u00fe 1 Ks\u00f0x\u00de \u00fe 1 Ka\u00f0x\u00de \u00f015\u00de Kb (x), Ks (x), Ka (x) are the stiffness under consideration of the effect of tooth bending, shear and axial compress for one piece of tooth. x is the distance between the piece of tooth and one end of the tooth which is shown in Fig. 3a and b. Then, the stiffness of the whole tooth can be obtained by integration of Kt (x) along tooth width, Kt \u00bc Z W 0 Kt\u00f0x\u00de \u00f016\u00de With the effect of fillet-foundation deflection and Hertzian contact, the equivalent mesh stiffness can be calculated by Ke \u00bc 1 1 Kt1 \u00fe 1 Kf 1 \u00fe 1 Kt2 \u00fe 1 Kf 2 \u00fe 1 Kh \u00f017\u00de Although the proposed mesh stiffness model can also be suitable to a spatial gear tooth crack, this study assumes that the crack propagation is in plane shown as A\u2013A in Fig. 3a and c with different crack depth along tooth width for simplicity. The crack depth along tooth width can be described as a function of x in the coordinate system XOY in Figs. 3c and 4. That means qx \u00bc f \u00f0x\u00de \u00f018\u00de Further, the crack depth is assumed to distribute along tooth width as a parabolic function as shown in Fig. 4. When the crack length Wc is less than tooth width W, the crack curve is denoted by the solid curve in Fig. 4. While the crack propagates through the whole tooth width, the crack curve is described as the dashed curve which propagates along crack depth q2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002772_j.robot.2021.103785-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002772_j.robot.2021.103785-Figure7-1.png",
        "caption": "Fig. 7. (a) Adaptive control architecture [43]. (b) Adaptive control architecture using policy gradient reinforcement learning [43]. (c) Illustration of friction adaptive lateral undulation gait [39,40]. (d) Illustration of backbone as a helical tread around a virtual core cylinder [193]. (e) Modsnake performs sidewinding on a slope [193]. (f) A sequence for raising a fifth segment (dashed plotting is initial position and solid plotting is the end position) [194]. (g) Backbone curve [195]. (h) Path following of a planar snake robot with obstacles [196]. Left: initial backbone guess, right: final backbone curve of snake robot to follow. (i) Reaction force and body deformation of a snake according to obstacles [70]. (Circles: pegs as obstacles, arrows with number indicate the direction and magnitude of the forces).",
        "texts": [
            " To generate adaptive motion, some learning algorithms, such as genetic algorithms or intensive learning, can be used on a large scale [220\u2013222]. They can help the robots to collect sensory information, analyse the internal state of the robot and the stimuli around it, and learn the appropriate response [96]. After the learning process, the robot will automatically adjust its movement mode or gait to the environment, while under automatic control. Therefore, the control architecture is closed and complete [43,223], as shown in Fig. 7(a) and (b). We will discuss how existing snake robots can adapt to uneven surfaces, change gaits regarding to dynamic friction of a surface, deal with slippery slopes, climb over obstacles, pass through narrow passages, avoid getting stuck, and recover from joint malfunction in subsequent sections. Moreover, due to its special body shape and application requirements, the snake robot also has a special movement mode, i.e., obstacle aided locomotion. In obstacle aided locomotion, snake robots can generate thrust to support their body movement by touching obstacles and using them as stationary push points [74,132,186,224]",
            " [39,40] proposed a local reflexive mechanism of autonomous decentralized control system, and took a two-dimensional snake robot as an example to show that the adaptability to the environment changes, as well as the robustness of body segment failure caused by local sensory reflexive control, which provides new ideas for the methodology of autonomous distributed control systems. To sense the friction force between the robot and ground, an elastic material was used to connect two adjacent modules. The experimental results showed successful amplitude and wavelength transition of lateral undulation gait between different friction floor surfaces (see Fig. 7(c)). It is worth mentioning that this decentralized control method has fault tolerance against local malfunction. Nansai et al. [226,227] designed a head control system capable of adapting to passive side-slipping for snake robots in order to move through the environments which have ground surface conditions with nonuniform frictional coefficients. Sloped terrain represents terrain with a certain incline rate, and the friction coefficient of its surface is determined by various factors such as the slope angle and the type of medium",
            " [231] proposed a control method that makes the head of a snake robot follow a l t [ f g s v c o t s s e s l s d t a t s a f 4 a s s e a w f t s o l s r r f o r c s n r b p a g T a m a j l f i t r o B c n arbitrary trajectory of two non-parallel planes. Apart from ateral undulation, sidewinding gait is also an efficient translaion gait especially on loose and slippery ground. Choset et al. 196] studied the stability of sidewinding gait on a slope and ind a solution for the minimum aspect ratio of the sidewinding ait to maintain stability [193]. They presented a virtual thread idewinding model in which the backbone of the snake robot was iewed as a helical tread moving around a virtual core elliptical ylinder (see Fig. 7(d)). Using such a model, the stability problem f snake robots was converted to look for the local extrema of he potential energy of the virtual ellipse as it rolls along the lope. They calculated the relationship between the maximum table gradient and the aspect ratio of two semi-major axes of the llipse. One of ModSnakes was used to evaluate their algorithm as hown in Fig. 7(e). Later, Marvi et al. [68] found that the desertived rattlesnake (Crotalus Cerastes) can effectively move through liding and pitching on sloping granular media such as sand unes. Experiments have shown that as the tilt angle increases, he rattlesnake increases the length of contact between the body nd the sand. In the physical model of the snake-shaped robot, his movement allows the body to ascend to the maximum sandy lope that is close to the slope stability. Bae et al. [232] proposed driving assistance mechanism that can prevent the snake robot rom sliding down on a slope and increase its speed",
            " Regarding aided climbing over an obstacle, a mall number of front modules of a snake robot are lifted to lean n the obstacle. Aided climbing allows the robot to use supports ike slopes or poles to climb. In the vertical cylinder climbing ituation, a snake robot can coil its body around the cylinder and oll its body up and down. Due to limited output torque of snake obot joints, the strategy of lifting other modules should be careully designed to avoid the joints being damaged by the weight f lifted modules. Nilsson et al. [194] also proposed a simple ecursive climbing algorithm (see Fig. 7(f) for an example) which an minimize the dynamic torque. Brown et al. [234] developed a nake robot named Millibot Train, which can climb vertical steps early half the robot length using manual configuration. Snake obots can move from one position to another on a large scale y applying adaptive motion behaviour [235]. Kon et al. [236] roposed a control method for semiautonomous step climbing by snake robot based on mixed integer quadratic programming to enerate the reference trajectory of the head of the snake robot",
            " [245] presented a navigation approach for snake robots to autonomously overcome unknown and challenging obstacles like stairs or large steps. In nature, snakes exploit obstacles such as rocks on the ground to obtain high efficiency locomotion. Snakes have been observed gliding forward by bracing the sides of its body against grass, stones or other obstacles on the surface of the ground: the reactions from these obstacles yield reaction force normal to the contacting surface between the body and obstacles [70] (as shown in Fig. 7(i)). In other words, snakes utilize the stationary obstacles to increase the friction between their bodies and environments [32]. In order to carry out such tasks in cluttered environments, snake robots must have a high degree of awareness and be capable of efficient obstacle exploitation to gain propulsion, this locomotion is defined as obstacle-aided locomotion [71\u201373]. Regardless of whether the object is stationary or moving, biological snakes or snake-like robots can push their body against the objects to propel themselves forward effectively",
            " They added Jacobian matrix of an bstacle avoidance point as a second constraint together with n end-effector Jacobian matrix to determine joint angles. The bstacle avoidance point was assigned to the closest point of the obot to obstacles. However, this method was hard to apply on snake robot with multiple DOFs especially operating in a 3D pace because of the large dimension of the Jacobian matrix and n infinite number of solutions of joint angles. To solve this problem, Chirikjian et al. [195,261,262] develped a backbone curve concept to describe the mechanism cenreline of a robot (see Fig. 7(g)). The geometric aspects of hyperedundant robot motion planning are reduced to the determinaion of the proper time varying behaviour of the backbone refernce set. To avoid obstacles, a virtual tunnel was built through an obstacle field from starting point to a goal. Choset et al. [196] improved Chirikjian\u2019s method by adapting a roadmap and proposed a generalized variance graph (GVG) to simplify the search of the virtual tunnel and backbone curve, and this GVG is an one-dimension map of an arbitrary dimension space. During this curve growing procedure, the head motion has been determined and the rest of the body just follows its predecessor to finish the whole-body path planning, so called follow-the-leader method [229]. In Choset\u2019s follow-the-leader approach [156,167], the robot shape is described by a parametric backbone curve (as shown in Fig. 7(g)). The principle of this method is that the inverse kinematic problem of hyper-redundant robot is reduced to finding a group of time varying modal participation factors that satisfy task requirements. The solid circle points in Fig. 7(h) are GVG vertices and the lines between them are GVG edges. The backbone curve is very close to the GVG edge, which is assured to be collision free, such as the left of Fig. 7(h). However, such backbone curves may have sharp corners and exceed physical joint angle limitations. Therefore, they [167] applied a cost function to minimize the path with respect to curvature so that the snake robot can easily follow the path, such as the backbone curve at the right of Fig. 7(h). The concept of follow-the-leader has been realized in CPG controllers as well [71,263]. Other obstacle avoidance algorithms include using mechanical intelligence [264,265] and behaviour based control [266]. Given an approximated robot path in a static obstacle environment, many path following algorithms can be used, such as B-spline path following algorithm proposed by Conkur\u2019s [267]. Hirose et al. [13] applied a neural network method based on tactile sensors to control the joint angles of ACM-III snake robot"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003874_ar9002015-Figure9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003874_ar9002015-Figure9-1.png",
        "caption": "FIGURE 9. Structure of a 1998 version of the continuous glucose monitoring subcutaneously implanted \u201cwired\u201d GOx electrodes: gray, \u201cwired\u201d enzyme sensing layer; pink, glucose transport controlling membrane; aqua, biocompatible poly(ethylene oxide) film. The three layers are deposited in the 90 \u00b5m deep recess formed by etching the gold of the tip of a polyimide insulated gold wire. The diameter of the wire is 250 \u00b5m; the thickness of the polyimide layer is 20 \u00b5m. Courtesy of Christopher Thomas, Abbott Diabetes Care.",
        "texts": [
            " For this reason, the reactive interferant elimination was replaced by down-shifting the redox potentials of the GOx-wiring redox hydrogels to potentials where the interferant-electrooxidation-associated currents were small,7,30,31 then negligible.28 Acceptably steady glucose sensitivity for >5 days was provided, along with a resolution of 0.2 mM across the 2-30 Vol. 43, No. 7 July 2010 963-973 ACCOUNTS OF CHEMICAL RESEARCH 969 mM range relevant to the management of diabetes, by a tailored glucose-flux-limiting membrane. When a membrane reduces the influx of glucose sufficiently, the current is determined by the glucose flux through the membrane, not by the less stable activity of the electrocatalyst. Figure 9 shows a schematic diagram of the implanted working electrodes that Ephraim and Adam wore and that Adam\u2019s UT team later implanted in rats. As counter-reference electrodes, commercially available EKG Ag/AgCl and NiOx/NiOy electrodes were used. The implanted electrodes, of 0.29 mm diameter, were formed in polyimide insulated gold wires, the tips of which were etched by Sten Eric Lindquist to form a 0.09 mm deep cavity. The gold tips were coated by Elisabeth Csoeregi with three successive layers of wired GOx, of glucose flux-reducing poly(ethylene glycol) diglycidyl ether-cross-linked poly(Nvinylimidazole), and of a bioinert, cross-linked, poly(ethylene glycol), such as its diacrylate",
            "25,27 The production versions of the FreeStyle Navigator system incorporated design changes to simplify manufacture and enhance performance. First, the sensor itself was changed from a hollow tube design to a coplanar printed circuit in which working, reference, and counter electrodes were incorporated onto a single polyester substrate, as shown in Figure 12. In this design, the three screen printed electrodes were stacked one atop each other, separated by intervening dielectric insulator, to minimize the size of the implanted portion of the sensor. In addition, the multilayer coating of Figure 9 was replaced by a single membrane made of a polymer cross-linked with a poly(glycidyl ether). It combines the dual functions of limiting glucose flux and providing a bioinert interface. This polymer is shown in Figure 13. It is applied by reproducible processes, yielding sensors with predictable sensitivity. Clinical trial results utilizing the membrane of Figure 13 and a sensor structure similar to that of Figure 12 were published in 2003.35 The FreeStyle Navigator system monitors and stores the subcutaneous glucose concentration at 1 min intervals"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000233_978-1-4614-0460-6-Figure2.37-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000233_978-1-4614-0460-6-Figure2.37-1.png",
        "caption": "Fig. 2.37 Free body diagram for inverted pendulum",
        "texts": [
            "9*clim; % Define simulation variables dt = tau/20; % sec/step steps = round(50*tau/dt); % Euler integration initial conditions x = 1e-3; % m dx = 0; % m/s % Initialize final position and time vectors y = zeros(1, steps); time = zeros(1, steps); for cnt = 1:steps ddx = (-(c - gamma)*dx - k*x)/m; % m/s^2 dx = dx + ddx*dt; % m/s x = x + dx*dt; % m % Write results to vectors y(cnt) = x; % m time(cnt) = cnt*dt; % s end figure(1) plot(time, y*1e3, 'k') xlim([0 max(time)]) ylim([-5 5]) set(gca,'FontSize', 14) xlabel('t (s)') ylabel('x (mm)') In order to model and simulate divergent instability, we will use the inverted pendulum, composed of a mass, m, supported by a massless rod of length, l, that rotates about the frictionless pivot, O. The rod is held in its vertical equilibrium position by springs and dampers as shown in Fig. 2.36. The free body diagram for 2.5 Unstable Behavior 69 a rotation, y, of the mass/rod counterclockwise from the equilibrium position is provided in Fig. 2.37. The forces that need to be considered include: \u2022 the gravity force, mg \u2022 the two spring forces, kd, where d \u00bc l 2 sin y\u00f0 \u00de is the horizontal deflection due to rotation of the rod as shown in Fig. 2.38 \u2022 the two viscous damping forces, c _d, where _d \u00bc dd dt \u00bc l 2 cos y\u00f0 \u00de _y is the horizontal velocity \u2022 the reaction forces at the pivot in the horizontal, Rx, and vertical, Ry, directions. In order to sum moments, M, about O, the forces can be redrawn to give the components perpendicular and parallel to the massless rod"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003443_0278364903022001005-Figure10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003443_0278364903022001005-Figure10-1.png",
        "caption": "Fig. 10. 3-DoF 3 \u2212 xRxRxRuRuR tranlational PM.",
        "texts": [
            " The standard base of the mechanism twist system is $m1 = (0 0 0 ; 1 0 0) $m2 = (0 0 0 ; 0 1 0) $m3 = (0 0 0 ; 0 0 1). (37) The standard base of the mechanism constraint system is $r m1 = (0 0 0 ; 1 0 0) $r m2 = (0 0 0 ; 0 1 0) $r m3 = (0 0 0 ; 0 0 1). (38) To form such a mechanism constraint system, the limb constraint system only contains couples and all the limb constraint couples must be non-coplanar. The limb constraint system can contain one constraint couple or two constraint couples or three constraint couples. Figure 10 shows a 3 \u2212 xRxRxRuRuR translational PM, which was presented by Frisoli et al (2000) and by Carricato and Parenti-Castelli (2001). The limb twist system is given by $i1 = (1 0 0 ; 0 0 0) $i2 = (1 0 0 ; 0 b2 c2) $i3 = (1 0 0 ; 0 b3 c3) $i4 = (0 m4 n4 ; a4 b4 c4) $i5 = (0 m4 n4 ; a5 b5 c5). (39) The limb constraint system is given by $r i1 = (0 0 0 ; 0 n4 \u2212 m4), (40) which is perpendicular to both $i3 and $i4. The three limb constraint couples $r 11, $r 21, and $r 31 are obviously non-coplanar in space and they form a 3-system given by eq (38)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure2.11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure2.11-1.png",
        "caption": "Fig. 2.11. Connection diagram of a three-phase, nine-coil winding of an AFPM brushless machine.",
        "texts": [
            " Consequently these windings are sometimes called tooth windings. In air-cored, double-layer non-overlap windings, the stator coils lie side-by-side against each other (see e.g. Fig. 1.5b). Note that the term slot can also be visualized when applied to air-cored windings. In single-layer, non-overlap slotted iron-cored windings, only one coil layer is in a slot and therefore every second tooth is wound (alternate teeth wound); 2.2 Windings 39 air-cored, single-layer non-overlap windings have their coils displaced from each other as shown e.g. in Fig. 2.11. In double-layer, non-overlap windings, the number of stator coils is always equal to the number of stator slots, i.e. Qc = ncm1 = s1, where nc is the number coils per phase. In single-layer non-overlap windings the number of stator coils is equal to half the number of stator slots i.e. Qc = s1/2. In concentrated, non-overlap windings, there is only one coil in a coil group or coil phase belt (z = 1), but in distributed non-overlap windings two or more coils are distributed and connected in series to form a coil group (z = 2, 3, ",
            " Electromagnetic analysis of AFPM machines with non-overlap stator windings are the same as those with overlap windings provided that the winding factors are calculated according to the following equations: the distribution factor for the fundamental space harmonic is calculated by [240] kd1 = sin(\u03c0/2m1) z sin[(\u03c0/(2m1z)] (2.14) 40 2 Principles of AFPM Machines For single and double layer slotted iron-cored non-overlap windings, the pitch factor for the fundamental space harmonic is calculated by using eqn (2.9) or else kp1 = sin( \u03c0 2 2p s1 ) = sin(\u03b8m/2) (2.15) where \u03b8m is the slot pitch angle (also corresponding to the coil span angle in the case of double layer windings shown in Fig. 1.5b or corresponding to the coil pitch angle in the case of single layer windings as shown in Fig. 2.11) given by \u03b8m = 2\u03c0p s1 (2.16) For air-cored, non-overlap windings, the pitch factor for the fundamental space harmonic is derived in [146]. For single layer non-overlap windings the pitch factor is given by kp1 = sin(\u03b8m/2) sin(\u03b8re/2) \u03b8re/2 (2.17) while for double layer, non-overlap windings, the pitch factor is given by kp1 = sin[(\u03b8m \u2212 \u03b8re)/2] sin(\u03b8re/2) \u03b8re/2 (2.18) where \u03b8re is the electrical angle corresponding to the coil-layer width. For example, for a double layer, air-cored non-overlap winding with m1 = 3, p = 12, s1 = Qc = 18 (thus nc = 6) and \u03b8re = 0"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003926_0167-6911(91)90050-o-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003926_0167-6911(91)90050-o-Figure1-1.png",
        "caption": "Fig. 1. Phase-plane behaviour of uncontrolled Duffing type system (2).",
        "texts": [],
        "surrounding_texts": [
            "(37, w, K ) ~ R u 1 3 7 = 0 , w = 0 } . []"
        ]
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure6.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure6.6-1.png",
        "caption": "FIGURE 6.6. Dead position for a four-bar linkage.",
        "texts": [
            "81) The sweep angle of the output link would be \u03c6 = \u03b84L2 \u2212 \u03b84L1 . (6.82) Example 225 Dead positions for a four-bar linkage. When the input link of a four-bar linkage locks, we say the linkage is at a dead position. It happens when the angle between the output and coupler links is either 180 deg or 360 deg. Limit positions of a four-bar linkage, if there are any, must be determined by the designer to make sure the linkage is never stuck in a dead position. A dead position for a four-bar linkage is shown in Figure 6.6. We show the dead angle of the output link by \u03b84D1 , \u03b84D2 , and the corresponding input angles by \u03b82D1 , \u03b82D2 . They can be calculated by the following 6. Applied Mechanisms 321 equations: \u03b82D1 = cos\u22121 \" a2 + d2 \u2212 (b+ c) 2 2ad # (6.83) \u03b84D1 = cos\u22121 \" a2 \u2212 d2 \u2212 (b+ c)2 2 (b+ c) d # (6.84) \u03b82D2 = cos\u22121 \" a2 + d2 \u2212 (b\u2212 c) 2 2ad # (6.85) \u03b84D2 = cos\u22121 \" a2 \u2212 d2 \u2212 (b\u2212 c)2 2ad # (6.86) Example 226 F Designing a four-bar linkage using Freudenstein\u2019s equation. Designing a mechanism can be thought of as determining the required lengths of the links to accomplish a specific task"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003685_9780470692059-Figure11.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003685_9780470692059-Figure11.1-1.png",
        "caption": "Figure 11.1 Photoselection of differently oriented molecules with plane-polarized (a) and unpolarized (b) light. Excited molecules are shaded. After Albrecht, A.C. (1961). Journal of Molecular Spectroscopy, 6, 84. Figure published by D\u00f6rr, F. (1971), Creation and Detection of the Excited States, pp. 53\u2013122, Dekker, New York.",
        "texts": [
            " Natural light is unpolarized; it has no preferential direction. Thus, in order to study molecules dynamics during the excited-state lifetime, molecular photoselection should be performed. This can be reached by exciting the fluorophores with a polarized light and by recording the emitted light in a polarized system. When excitation is performed with polarized light (definite orientation), absorption of the fluorophore will depend on the orientation of its dipole in the ground state compared to the polarized excitation light (Figure 11.1). Fluorophores with dipoles perpendicular to excitation light will not absorb. Fluorophores with dipoles parallel to excitation light will absorb the most. Thus, polarized excitation will induce photoselection in the fluorophore absorption. The electric vector of excitation light is oriented parallel to or in the same direction as the z-axis. Emitted light will be measured with a polarizer. When emission is parallel to the excitation, the measured intensity is called I\u2016. When the emission is perpendicular to the excitation light, the measured intensity is called I\u22a5"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001514_j.engfailanal.2015.06.004-Figure24-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001514_j.engfailanal.2015.06.004-Figure24-1.png",
        "caption": "Fig. 24. (a) 6 DOF model in Ref. [36], (b) 6 DOF model in Ref. [37].",
        "texts": [
            " For all analytical methods, the gear tooth is assumed as a variable cross-section cantilever beam. Fig. 22. (a) Single DOF model [65], (b) 4 DOF model [10], (c) 4 DOF model [31]. (see Fig. 23a). Taking the effects of torsional and lateral stiffness and damping of the shafts into account, Omar et al. [48] established a 9 DOF model for a gear pair (see Fig. 23b). On the basis of the TVMS model of a cracked gear pair, Chen and Shao [36] established a 6 DOF LMM of a spur gear system where the y axis is parallel to the action line of the gear pair (see Fig. 24a). Based on a similar model presented in Ref. [36] (see Fig. 24b), Mohammed et al. [37] investigated the influence of tooth root crack propagation on dynamic response of a gear system. Ma et al. [55] established a 6 DOF model of a perforated gear system, which is similar to the model in Ref. [36] except for ignoring the effect of tooth friction. Considering the torsional and lateral vibrations, Wu et al. [28] adopted a 6 DOF model of a gear system to analyze the effects of crack on the system vibration response [67] (see Fig. 25a). To study the influence of tooth friction, Bartelmus [67] presented an 8 DOF model (see Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure9.34-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure9.34-1.png",
        "caption": "Fig. 9.34. Exploded view of DuraHeartR centrifugal rotary pump. Courtesy of Terumo Corporation, Tokyo, Japan.",
        "texts": [],
        "surrounding_texts": [
            "When electricity was new, people had high hopes that it had curative powers. For example, electropathy (electrodes between patient\u2019s hands and ailing body part), very popular from 1850 to 1900, promised to cure most diseases and conditions, including mental illness. The 21st century biomedical engineering community has resurrected magnetic fields to treat depression, e.g. magnetic seizure therapy (high frequency, powerful electromagnets) or transcranial magnetic stimulation (strong pulse magnetic fields). At present, many medical devices use small permanent magnet (PM) electric motors as, for example, high-quality pumps, centrifuges, infusion pumps, hemodialysis machines, precision handpieces and implantable devices (ventricular assist devices, pacemakers, defibrillators, nerve stimulators, etc). Since the reliability of medical products is critical, the electric motor is considered a precision component rather than a commodity device [203]. This section focuses on very small PM brushless motors for implantable devices, in particular, motors for rotary blood pumps [102]. A left ventricular assist device (LVAD) is an electromechanical pump implanted inside the body and intended to assist a weak heart that cannot 314 9 Applications efficiently pump blood on its own. It is used by end-stage heart failure patients who are unable to receive a heart transplant due to donor availability, eligibility, or other factors. Motor driven pumps implanted in the human body must be free of shaft seals. This problem can be solved by embedding PMs in the pump rotor placed in a special enclosure and driven directly by the stator magnetic field. In this case the nonmagnetic air gap is large and high energy PMs are required. Electromagnetic pumps for LVADs can be classified into three categories: \u2022 1st generation (1G), i.e. electromagnetic pulse pumps; \u2022 2nd generation (2G), i.e. electromagnetic rotary pumps; \u2022 3rd generation (3G), i.e. electromagnetic pumps with magnetic or hydro- dynamic bearings. Electromagnetic 1G pumps were driven by electromagnets, linear oscillating motors or linear short-stroke actuators. The pump, integrated with a linear actuator, is heavy, large and noisy. DuraHeartR 3G LVAD developed by Terumo Corporation, Tokyo, Japan, combines centrifugal pump with magnetic levitation technologies (Figs 9.33 and 9.34) [102]. Magnetic levitation allows the impeller to be suspended within the blood chamber by electromagnets and position sensors. The three-phase, 8-pole, axial flux PM brushless motor with slotless stator resembles a floppy disc drive spindle motor (Fig. 9.33). NdFeB PMs are integrated with the impeller. The output power of the motor is 4.5 W, speed 2000 rpm and torque 0.0215 Nm [226]. 9.11 Ventricular Assist Devices 315 An axial flux slotless motor integrated with a centrifugal blood pump is shown in Figs. 9.35 and 9.36. The so-called VentrAssistTM manufactured now by Australian company Ventracor is a new cardiac LVAD, which has only one moving part - a hydrodynamically suspended impeller integrated with a PM rotor. The hydrodynamic forces act on tapered edges of the four blades. The stator of the brushless electric motor is of slotless type and has only upper and lower coils. The three coil winding and four pole rotor use the second harmonic of the magnetic field wave to produce the torque. To provide redundancy, body coils and cover coils are connected in parallel so that the motor still can run even if one coil is damaged. The housing and impeller shell are made of titanium alloy Ti-6Al-4V. Vacodym 510 HR NdFeB PMs are embedded into impeller. To reduce the reluctance for the magnetic flux, laminated silicon steel return paths (yokes) are designed. The 2D FEM simulation of the magnetic field distribution is shown in Figs 9.37 and 9.38 [102]. The measured performance characteristics are shown in Figs 9.39a and 9.39b [226]. For output power between 3 and 7 W and speed between 2000 and 2500 rpm, the efficiency is from 45 to 48% (Fig. 9.39a). At 3 W and 2250 rpm the winding losses are 1.7 W, eddy current loses in titanium 316 9 Applications 1.0 W and losses in laminated yokes are 0.7 W [226]. At load torque 0.03 Nm the fundamental phase current is 0.72 A (Fig. 9.39b). The device weighs 298 g and measures 60 mm in diameter, making it suitable for both children and adults. 9.12 Axial Flux Machines with Superconducting Field Excitation System 317"
        ]
    },
    {
        "image_filename": "designv10_0_0001109_pnas.1218869110-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001109_pnas.1218869110-Figure6-1.png",
        "caption": "Fig. 6. Each cilium is modeled by three semiflexible filaments consisting of chains of monomers that are connected by harmonic springs of length b (nearest neighbors) and c (next-nearest neighbors) to form a crane-like structure. Bond lengths are varied to induce a preferred curvature. The bond lengths of the \u201cred\u201d filament are varied to create the power and recovery strokes. In addition, the preferred bond lengths of the seven bonds at the base of the \u201cgreen\u201d filament are stretched by 10% to generate aplanarity during the recovery stroke. The power-stroke direction (PSD) is rotated by an angle \u0398 with respect to the main lattice direction.",
        "texts": [
            " We hope that our results will contribute to the design of new autonomous cilia-like rowers, which have this important property. A detailed description of the model, methods, and results is given in SI Materials and Methods. A brief summary is given below. Model.Weuseamechanistic, 3Dmodel of a cilium,which is designed to capture the active beat and the hydrodynamics of interacting cilia arrays. Each cilium is represented by a bundle of three parallel semiflexible filaments, each of which consists of a linear chain of beads and springs (Fig. 6); the filaments are interconnected by a second type of spring to keep them approximately at a fixed distance from each other (38). For activity, a dynamic spontaneous curvature is created by locally varying the lengths of springs of one of the three filaments. This mechanism is inspired by the connecting dynein motors moving along adjacent microtubules in the axoneme\u2014including the effect of a stall force. During the power stroke, the force distribution along the cilium is adjusted such that a nearly straight, extended conformation is achieved",
            " In the recovery stroke, the force distribution is constructed such that a strongly curved part travels from the anchoring part of the cilium to its free end. For MCWs to develop, a feedback between the hydrodynamic flow and the beat pattern is essential. Motivated by the \u201cgeometric clutch hypothesis\u201d (32), we assume that switching between power and recovery strokes is controlled by curvature thresholds of the individual cilium; i.e., no external clock is used to determine the beat pattern. The power-stroke direction (PSD) forms an angle \u0398 with the main lattice direction, as illustrated in Fig. 6. Correlation Function. To characterize MCWs, we define the phase of the beat of an individual cilium by B = cos(\u0398)\u0394x + sin(\u0398)\u0394y, where \u0394x = x(tip) \u2212 x(base) and \u0394y = y(tip) \u2212 y(base) are the projected displacements of the tip of the cilium from its base, and \u0398 is the power-stroke direction. This defines the phase field B(r, t), where r is Gc\u00f0r; t\u00de= \u00c6\u03b4B\u00f0r0; t0\u00de\u03b4B\u00f0r0 + r; t0 + t\u00de\u00e60=\u00c6\u03b4B\u00f0r0; t0\u00de2\u00e6; [5] where \u03b4B(r0, t0) = B(r0, t0) \u2212 B(r0, t0), and the average is taken over all lattice positions r0 and over a time interval of about 10 beat periods"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000554_978-3-540-85629-0-Figure7.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000554_978-3-540-85629-0-Figure7.3-1.png",
        "caption": "Fig. 7.3. One degree of freedom model with passive nonlinear parameters.",
        "texts": [
            " , z\u0307n]T the model in state space is finally obtained: x\u0307 = Ax + B u (7.4) where A = [ 0 I \u2212M\u22121K \u2212 M\u22121D ] , B = [ 0 \u2212M\u22121m ] , u = y\u0308. 7.1.3 Nonlinear model with one degree of freedom Usually, a mechanical system is characterized by nonlinear effects that may be considered by properly defining the \u2018passive\u2019 parameters of the model. Several nonlinear effects may be taken into account, such as Coulomb or viscous friction, nonlinear dumping (e.g. proportional to x\u03072), backslash, and so on. Let us consider the system of Fig. 7.3 where, in particular, the dumping element takes into consideration different frictional phenomena. The parameters d, \u03b1 and \u03b2 represent the viscous, the quadratic and the Coulomb damping respectively. The dynamics of the system is described now by 270 7 Dynamic Analysis of Trajectories mz\u0308 + dz\u0307 + kz = \u2212my\u0308 \u2212 \u03b1|z\u0307|z\u0307 \u2212 \u03b2 z\u0307 |z\u0307| . where z = x \u2212 y. This equation may also be written as z\u0308 + 2\u03b4\u03c9nz\u0307 + \u03c92 nz = \u2212\u03c92 n\u03b6 with \u03b6 = y\u0308 \u03c92 n + \u03b1 m\u03c92 n |z\u0307|z\u0307 + \u03b2 m\u03c92 n z\u0307 |z\u0307| . If a backslash effect has to be considered, it is necessary to define an additional nonlinear term, such as the one represented in Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure13.47-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure13.47-1.png",
        "caption": "Figure 13.47 Girder, posts and shores isolated from the threehinged shored frame, with joining forces and support reactions.",
        "texts": [
            " The dimensions are shown in the figure. Questions: a. Determine the support reactions at A and B. b. Determine the forces in shores DD\u2032 and EE\u2032, with the correct signs for tension and compression. c. Isolate CDSEG, and draw all the forces acting on it. d. For CDSEG, draw the N , V and M diagrams with the deformation symbols and the tangents at C, D, S, E and G to the M diagram. Include relevant values. Solution (units kN and m): a. Determining the support reactions is left to the reader (see Section 5.3, Example 1). b. In Figure 13.47, all parts of the frame have been isolated, and all the joining forces are shown. The support reactions at A and B are also shown. Both shores DD\u2032 and EE\u2032 are at 45\u25e6: 13 Calculating M, V and N Diagrams 587 Dh = Dv = 1 2NDD\u2032\u221a 2, Eh = Ev = 1 2NEE\u2032\u221a 2. Dh can be determined from the moment equilibrium of post AD\u2032C about C: \u2211 T|C = +32 \u00d7 5 + Dh \u00d7 2.5 = 0 \u21d2 Dh = Dv = \u221264 kN so that NDD\u2032 = \u221264 \u221a 2 kN (a compressive force). In the same way, Eh can be determined from the moment equilibrium of post BE\u2032G about G: \u2211 T|G = \u221232 \u00d7 7"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002204_tfuzz.2016.2594273-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002204_tfuzz.2016.2594273-Figure2-1.png",
        "caption": "Fig. 2: Two-link robotic manipulator.",
        "texts": [
            "5; saturation amplitudes v1,max = 15, v2,max = 4, v3,max = 10 and v1,min = \u221215, v2,min = \u22124, v3,min = \u221210. Then, simulation results are presented in Fig. 1. Fig. 1 displays the trajectories of states and saturated control input signals, from which it can be seen that although there exists saturation phenomenon in control input signals, all states are ultimately bounded. This verifies the effectiveness of the proposed control scheme. Next, to show the advantages and potential applicability of the proposed method, a practical example about control of two-link robot manipulator (see Fig. 2) is considered. Example 2 : The dynamic equation of two-link robot manipulator can be written as H(\u03b8)\u03b8\u0308 + C(\u03b8, \u03b8\u0307)\u03b8\u0307 +G+ F +D = \u03c5 (67) 1063-6706 (c) 2016 IEEE. Personal use is permitted, but republication/redistribution requires IEEE permission. See http://www.ieee.org/publications_standards/publications/rights/index.html for more information. with H(\u03b8) = [ H11 H1,2 H21 H22 ] , C(\u03b8, \u03b8\u0307) = [ C11 C12 C21 0 ] G = [g1, g2] T , F = [F1, F2] T D = [D1, D2] T , \u03c5 = [\u03c51, \u03c52] T where \u03b8 = [\u03b81, \u03b82] T , \u03b8\u0307 = [ \u03b8\u03071, \u03b8\u03072 ]T and \u03b8\u0308 = [ \u03b8\u03081, \u03b8\u03082 ]T denote the position of the joint angle, the joint angular velocity and the joint angular acceleration, respectively; \u03c5 is the actuator torque; the inertia matrix H(\u03b8) is symmetric and uniformly positive definite; C(\u03b8, \u03b8\u0307) is the Coriolis and centrifugal matrix forces; G is the gravitational torque; F and D denote uncertain friction term and delay term, respectively; \u03c5i (i = 1, 2) is the control saturation input satisfying \u03c5i(t) = sat(ui(t)) =  \u03c5i,max, ui(t) > \u03c5i,max ui(t), \u03c5i,min < ui(t) < \u03c5i,max \u03c5i,min, ui(t) \u2264 \u03c5i,min and H11 = I1 +m1l 2 c1 + I2 +m2l 2 2 +m2l 2 1 + 2m2l1l2 cos(\u03b4) cos(q2) + 2m2l1l2 sin(\u03b4) sin(q2) H12 = I2 +m2l 2 2 +m2l1l2 sin(\u03b4) cos(q2) +m2l1l2 sin(\u03b4) sin(q2) H21 = H12; H22 = I2 +m2l 2 2 C11 = (m2l1l2 sin(\u03b4) cos(q2)\u2212m2l1l2 cos(\u03b4) sin(q2))q\u03072 C12 = (m2l1l2 sin(\u03b4) cos(q2) \u2212m2l1l2 cos(\u03b4) sin(q2))(q\u03071 + q\u03072) C21 = (m2l1l2 cos(\u03b4) sin(q2)\u2212m2l1l2 sin(\u03b4) cos(q2))q\u03071 g1 = m1l1g cos(q1) +m2g(l2 cos(q1 + q2 \u2212 \u03b4) + l1 cos(q1)) g2 = m2l2g cos(q1 + q2 \u2212 \u03b4) Note that F1 = \u03be1q\u03071, F2 = \u03be2q\u03072 are unknown friction terms with \u03be1 and \u03be2 being friction coefficients"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure10.8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure10.8-1.png",
        "caption": "FIGURE 10.8. The force system at the tireprint of tire number 1.",
        "texts": [
            " Vehicle Planar Dynamics 597 Mz = Z m \u00a1 x \u00a1 v\u0307y + vx r + x r\u0307 \u2212 y r2 \u00a2 \u2212 y \u00a1 v\u0307x \u2212 vy r \u2212 y r\u0307 + x r2 \u00a2\u00a2 dm = r\u0307 Z m \u00a1 x2 + y2 \u00a2 dm+ (v\u0307y + vx r) Z m x dm \u2212 (v\u0307x \u2212 vy r) Z m y dm\u2212 2r2 Z m xy dm = Iz r\u0307 (10.80) because for a principal coordinate frame we haveZ m xdm = 0 (10.81)Z m y dm = 0 (10.82)Z m xy dm = 0. (10.83) 10.3 Force System Acting on a Rigid Vehicle To determine the force system on a rigid vehicle, first we define the force system at the tireprint of a wheel. The lateral force at the tireprint depends on the sideslip angle. Then, we transform and apply the tire force system on the body of the vehicle. 10.3.1 Tire Force and Body Force Systems Figure 10.8 depicts wheel number 1 of a vehicle. The components of the force system in the xy-plane applied on a rigid vehicle, because of the generated forces at the tireprint of the wheel number i, are Fxi = Fxwi cos \u03b4i \u2212 Fywi sin \u03b4i (10.84) Fyi = Fywi cos \u03b4i + Fxwi sin \u03b4i (10.85) Mzi = Mzwi + xiFyi \u2212 yiFxi . (10.86) Therefore, the total planar force system on the rigid vehicle in the body coordinate frame is BFx = X i Fxi = X i Fxw cos \u03b4i \u2212 X i Fyw sin \u03b4i (10.87) 598 10. Vehicle Planar Dynamics BFy = X i Fyi = X i Fyw cos \u03b4i + X i Fxw sin \u03b4i (10"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001963_j.bios.2013.11.007-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001963_j.bios.2013.11.007-Figure1-1.png",
        "caption": "Fig. 1. (a) Schematic diagram and (b) photograph of the cellulose-PPy battery cell before and after sealing it into a polymer-coated aluminum pouch (Nystrom et al., 2009).",
        "texts": [
            " The performance of this paper-based battery was lower than the counterpart fabricated on a polyethylene naphthalate substrate, indicating that the paper/electrolyte combination has a limited ability to take up anode oxidation products before suffering a reduction in ionic mobility. One obvious method to improve the performance is to evenly introduce the PPy through the cellulose fibers (Nystrom et al., 2009). The cellulose-PPy conductive paperlike battery was able to be charged with currents as high as 600 mA cm 2 with only 6% loss in capacity over 100 subsequent charge and discharge cycles (Fig. 1). In this study, they developed highly porous cellulose substrates with several homogenous nanometer-thick layers of PPy to obtain a high-surface area exhibiting an exceptionally high ion-exchange capacity. Ferreira et al. (2010, 2011) proposed Au/Cu-based paper batteries that could be interconnected in series and recharged using water (Fig. 2). The series of integrated batteries were able to supply a voltage of about 3 V and a current ranging from 0.7 mA to 25 mA and to successfully control the ON/OFF gate state of paper transistors (Ferreira et al"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003719_j.1469-7998.1985.tb04944.x-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003719_j.1469-7998.1985.tb04944.x-Figure3-1.png",
        "caption": "FIG. 3. Records of tensile tests on samples of the longissirnus aponeurosis of (a) the deer and (b) the dog. The broken line in (a) shows how tangents were drawn to determine the moduli given in Fig. 4. The stipple in (b) shows how the areas used to determine the strain energy were defined.",
        "texts": [
            " Mammalian striated muscle, contracting isometrically at its optimum length, exerts stresses of about 0.3 MPa (Wells, 1965). Table I11 shows the muscle forces corresponding to this stress for the deer and dog. The forces have been divided by the cross-sectional areas of the aponeuroses (taken from Table 11) to obtain stresses in the aponeuroses. Note that these stresses refer to isometric contraction of the muscle: larger stresses might act if the muscle were being forcibly extended (Flitney & Hirst, 1978). Figure 3 shows records of tensile tests on strips from the longissimus aponeurosis of deer and dog. They resemble closely our (unpublished) records of tests on tendons from the legs of various mammals, obtained by the same method. The energy dissipations (represented by the areas of the loops) are, however, large compared to those obtained by Ker (1981) in tests on plantaris tendons from sheep (Ovis aries). The difference is probably due to energy losses in the clamps, which Ker excluded from his results by using an extensometer. Gradients in Fig. 3, calculated from the scales of stress and strain, give values for Young\u2019s modulus. The modulus is low at low strains, and higher at high strains, as generally found in tests on tendon (for example, Ker, 1981). The moduli given in Fig. 4(a) are of tangents drawn to the (nearly straight) upper part of the rising phase of the loop, as shown by a broken line in Fig. 3(a). They are lower than moduli found for leg tendons in the same range of stresses. Ker (1981) obtained a mean modulus of 1.65 GPa for sheep plantaris tendons. Alexander & Dimery ( 1985) obtained moduli of about 1.3 GPa for Donkey (Equus usinus) leg tendons, at stresses above 30 MPa. We obtained moduli of 0.7 to 1-7 GPa for various leg tendons of the deer, at stresses above 20 MPa. The moduli given in Fig. 4(a) are probably too low, for two reasons. Firstly, it was difficult to remove all traces of muscle from the aponeurosis, so the cross-sectional areas calculated from the masses and lengths of the strips are too large",
            " Larger extensions were used in subsequent tests, until the specimen broke. The extensions were increased in quite large steps, so the highest stress shown in Fig. 4(a) for each specimen may be much smaller than the stress at which the specimen broke. Failure always occurred at one of the clamps, indicating that clamping had caused stress concentrations. Even the highest stresses in the graph may be well below the tensile strength of the aponeurosis. Areas, in graphs of force against extension, represent energy. The area that is stippled in Fig. 3(b) represents the strain energy stored in the specimen when the force was at its maximum value. It has been drawn half-way between the rising and falling arcs of the loop, on the assumption that equal energy losses occurred while the force was rising, and while it was falling. The strain energies given in Fig. 4(b) were measured in this way. They are expressed as energy per unit mass of strained aponeurosis. More strain energy would have been stored if higher stresses had been attained. It has already been suggested that the maximum stresses attained in the tests may have been well below the tensile strength of the aponeurosis",
            " Properties of the vertebral column Figures 5(a) and (b) are records of compressive tests on intervertebral discs of the deer. The specimens were made by cutting transversely through two adjacent centra, and included approximately half of each centrum as well as the intervening disc. Compressive stiffness (represented by the gradients of the records) increases from low values at low loads to almost constant values at loads above 2 kN. The stiffnesses given in Table IV are the gradients of tangents drawn as indicated in Fig. 3(a). part of the insertion of the muscle, or anterior to any part of the origin.) Thus, larger forces may act in life than were applied in the tests recorded in Table IV. Two epaxial muscles of the lumbar region, the sacrococcygeus and multifidus, were ignored in the preceding paragraph. They may contract simultaneously with longissimus, but their dimensions indicate that they exert much smaller forces (Alexander & Jayes, 1981). The joint between the last lumbar vertebra and the sacrum of the deer seems to allow a much wider range of flexion and extension than the other intervertebral joints",
            " Galloping Greyhounds bend their backs through about 90\" (Muybridge, 1887) but other breeds bend their backs less: for example, the film of Dalmatians analysed by Alexander, Jayes & Ker (1980) shows bending through only 50\". A Rough collie, like our specimen, would probably gallop more like a Dalmatian than like a Greyhound, in which case the length changes required of the longissimus muscle and aponeurosis (if they remained taut throughout) would be about 25 mm. The longest fascicles that we found in the dog aponeurosis were 170 mm long, and the mean length of the muscle fibres of the dog was 68 mm (Table 111). At the estimated stress of 17 MPa (Table 111), the strain in the aponeurosis would be about 0.03 (Fig. 3(b)). Assuming, as for the deer, that the strain in the muscle fibres would be 0.01 3, these strains would give a total length change of 6 mm. As in the case of the deer, the elastic deformations would be much less than would be required for the movements of galloping if the muscle remained taut throughout. The aim of this section is to obtain rough estimates of the internal kinetic energy fluctuations that occur in galloping. Only the distal parts of the legs (distal to the elbow and knee) will be considered"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002054_tie.2016.2593683-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002054_tie.2016.2593683-Figure4-1.png",
        "caption": "Fig. 4 Modal test of stator core and stator assembly",
        "texts": [
            " 3 that current harmonics do not induce extra spatial order but change the force amplitude and bring extra frequencies. The stator core is laminated by silicon steel sheet along the axial direction and the teeth are wound by coils. It is not practical to model these two parts according to their actual structure. Thus, an equivalent stator model is required for the accurate prediction of vibration and noise. In order to validate the equivalent model of stator core and coils, modal tests of stator core and stator assembly are respectively implemented, as shown in Fig. 4. In the test a vibration exciter is used to yield excitation force with a wide frequency band and frequency response function is measured to estimate modal parameters. Fig. 5 shows the corresponding FE structural model which is built in ANSYS. In this motor, a coil winds around a single tooth and the end part beyond the stator core is short. Therefore, only the coil part within the slot is modeled. The material anisotropy should be considered according to the actual stator structure. Because the electromagnetic force distributes almost uniformly along the axial direction, the circumferential stator modes contribute the most to the vibration"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000247_tsmc.1986.289285-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000247_tsmc.1986.289285-Figure1-1.png",
        "caption": "Fig. 1. Geometry of a manipulator link.",
        "texts": [
            " Only two types of single-degree-offreedom joints will be discussed: rotational, for which all relative motion between two links is restricted to rotation about a line fixed in both links, and translational, for which relative motion is restricted to translation along a line fixed in both links. In accordance with the notation developed by Denavit and Hartenburg [5], the links are numbered sequentially from 0 to n, starting at the base of the manipulator, and the geometry of each link is described as illustrated in Fig. 1. A right-handed set of orthogonal unit vectors, xi, yo, zi, is fixed in each link i (i = 0,.., n) with zi parallel to the line Li fixed in both links i and i-1; that is, parallel to joint axis i. The orientation of z0 is arbitrary. In addition, the common normal Ni between joint axes i and i + 1 has a length ai and its intersections with the joint axes labeled Oi and Qi, respectively. Unit vector xi is parallel to the common normal, and unit vector yi completes the orthogonal set. The kinematics of the link are completely characterized by its length ail offset bi, rotation angle Oi, and twist angle P3i"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure15.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure15.3-1.png",
        "caption": "Figure 15.3 (a) A beam subject to bending with (b) the associated load-displacement diagram.",
        "texts": [
            " In this case, the total amount of work A depends only on the position of the starting and end points and not on the shape of the route followed. In full A = Fxux + Fyuy + Fzyz = Fu cos \u03b1. Note that no work is performed if F and u are normal to one another (in that case \u03b1 = \u00b1\u03c0/2 and cos \u03b1 = 0). Forces that are constant in magnitude and direction include gravitational forces. The dimension of work is force multiplied by distance. The applicable SI unit is the joule, denoted as J: J = N \u00b7 m = kg \u00b7 m2/s2. 712 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Consider the simply supported beam in Figure 15.3a, with a point load F . Due to a load, the beam will bend. The sag at the concentrated load is u. The relationship between the load F and displacement u can be shown in a load-displacement diagram (see Figure 15.3b). The shape of the diagram depends on the properties of the material. The shape is not important at this stage. If with an increasing load the displacement u increases by an amount du, the force F performs work dA = F du. When the load and displacement have reached their final value, the total amount of work performed is: A = \u222b u 0 F du. The total amount of work performed is equal to the area under the loaddisplacement diagram. Performing work can be seen as a mechanical process of energy exchange between a body and its environment"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000206_978-0-387-74315-8-Figure4-2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000206_978-0-387-74315-8-Figure4-2-1.png",
        "caption": "Figure 4-2: Robot geometry and parameters",
        "texts": [
            " Normally the connectivity9 has to be given explicitly, but for the moment we will consider only robots consisting of a single un-branched open-Ioop kinematic chain, as shown in Figure 4-1, for which the connectivity is 9Connectivity is used here in the topological sense. 72 implicit in the nl1mbering scheme. More general robot mechanisms will be considered in Chapters 7 and 9. Let there be n movable links, numbered l..n, and n one-degree-of freedom joints, also numbered l..n, such that joint i connects link i to link i-l. Joint 1 connects link 1 to the immovable base member, which will be called link O for convenience. The next step is to define the geometry of the linkage, the types of joint and the inertia parameters (see Figure 4-2). To do this, a set of link coordinate systems is introduced, one attached to each link, where the coordinate system attached to link i is called i-coordinates. There is no restriction on where these coordinate systems are placed relative to their links, but certain placements allow the dynamics to be computed more efficiently than others. This topic is discussed in Chapter 8. Since the position of a link is frxed in its own coordinate system, we can express the inertia of link i and the axis of any joint embedded in link i as a constant in i-coordinates"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001803_icra.2015.7139759-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001803_icra.2015.7139759-Figure5-1.png",
        "caption": "Fig. 5: (a) and (b): Visualization of the possible reorientation of the propeller around XPi (case (a)) and YPi (case (b)). The angle of reorientation is denoted with \u03b1i in (a) and \u03b2i in (b)",
        "texts": [
            " In this section we present the design of a preliminary prototype obtained instantiating the general model introduced in Section II in a more particular case. A CAD of the prototype is shown in Fig. 4. First of all, to reduce the complexity and for the sake of symmetry, we have chosen \u03bbi = (i \u2212 1)\u03c03 and Lxi = 0.4 m \u2200 i = 1 . . . 6. With this choice, the origin OPi of each propeller frame is equally spaced with 60\u25e6 between each other from the center of the body frame OB to have a symmetric configuration in normal hovering position. Furthermore, as shown in Fig. 5 each propeller is mounted in an arc frame which is free to rotate in XPi and Y Pi , so that the tilt angle of \u03b1i and \u03b2i can be fixed as desired. The radius of the arc(Rarc) is designed equal to the length of the motor (with the propeller attached), so that OPi always stays at the same location in the XBY B plane with only its direction vector [XPi Y Pi ZPi ] T changing according to the \u03b1i and \u03b2i orientation. The arm in which each propeller set-up is suspended is designed to have a curved architecture with the radius of the curvature, more than the propeller radius (Rprop), so that independently from the value of \u03b1i and \u03b2i in a certain allowed interval, the propellers never come in contact with the arm during flight"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure9.4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure9.4-1.png",
        "caption": "Figure 9.4",
        "texts": [
            " Cross beams have been introduced 1 The proof for this cannot be given at this stage, but is based on the characteristic that the members in a truss are relatively weak with respect to bending, and relatively stiff with respect to extension (changing length). 9 Trusses 323 between the main beams, which are supported at the joints of the truss. Between the cross beams, stringers carry the deck (not shown). In this way, the traffic loading is directed via deck, stringers and cross beams as joint loads onto the trusses. Figure 9.4 (a) to (d) Four structures with rigid joints and loaded by forces at the joints. (e) to (h) The same structures, but now all the rigid joints are replaced by hinged joints. With hinged joints (a) ad (b) are kinematically determinate and can be considered to be trusses. For (c) and (d), the use of hinged joints generates a mechanism; they cannot be considered trusses. The force flow in (c) and (d) occurs mainly by bending. 324 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM One also often assumes that the dead weight of a truss applies at the joints"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002448_j.matdes.2017.06.040-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002448_j.matdes.2017.06.040-Figure5-1.png",
        "caption": "Fig. 5. Schematic representation of spatter creation during SLMprocess. (a) One laser beam is op spatters. (b) Two laser beams are working closely and greater number of spatters are creating",
        "texts": [
            " 4b, external light sources are utilized in the building chamber to provide enough light for capturing video at speeds of 6000 and 3000 frames per second (fps). . One laser scan track includes 42 image frames (each frame ~0.167ms). (a) The total area ach frame vs frame number. High energy density of the focused laser beam during the SLM process causes the melt pools to reach the evaporation point of the material at the center of the laser spot. Consequently, a recoil pressure is made above the melting pool as shown in Fig. 5a. While low recoil pressure facilitates the flattening of the molten pools in SLM, high recoil pressure causes the molten material to be removed by melt expulsion [20,21]. The ejected metal is cooled down very quickly and condensates, forming particles with different particle sizes depending on the duration of the condensation process [22]. These particles are called droplet spatters in this paper. Furthermore, the strong influences of the laser plume dynamics on non-melted powder around the melting pool leads to eject the powder near the melt pool from the powder bed. These particles are called powder spatter in this paper. Both types of spatters are deposited into the unmelted powder or the solidified layer. As shown in Fig. 5b, by increasing the number of the operating laser beams, much greater amount of spatters are induced. High-speed camera images during a laser track, while one laser is operating at spanof timeduring themanufacturing process (part 5 shown in Fig. 3a) are shown in Fig. 6. The laser track length and the operation time are 9.9 mm and 6 ms, respectively. It can be observed from Fig. 6a that there are several spatter particles left from the previous scan track. As the melting process started (Fig. 6a), the spatter particles are creating and dispersing around of the melting pool (Fig",
            " (c) Laser beam creates the next layer and the spatter C stays on the ng layer, Spatter C remains as an unmelted region in the final product. The shielding gas flow which depending on its local velocity profile, may avoid depositing spatter particles on powders which are exposed to the laser beam afterward or in the layers just consolidated by the laser beam(s). However, due to the size of spatter particles (see Fig. 11), remarkable amount of spatters will fall onto the powder bed or solidified layers (see Fig. 5). Fig. 12a and b show the SEM photography of the powders before SLM process (called clean powder), and the collectedmetallic powders that have not laser melted after SLMprocess (called dirty powder), respectively. The particles sizes range between 30 \u03bcm and 50 \u03bcm for clean powder, and 30 \u03bcm to 200 \u03bcm for dirty powders. Fig. 12d shows the morphology of one of the Al-Si10-Mg laser spatters. The spatter is much larger than the clean powder particle (see Fig. 12c) and is mainly spherical. The spherical shape of the spatter is due to the fact that molten metal is solidifying before impinging on the powder bed"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003881_mssp.2001.1414-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003881_mssp.2001.1414-Figure7-1.png",
        "caption": "Figure 7. Friction force directions.",
        "texts": [
            " 6, the relative velocity between the contacting teeth will be zero. This can be shown from the geometry where qh/of\"qp/op\"u p /u g (19) in which u p represents the angular velocity of the pinion and u g the angular velocity of the gear. The relative velocity of the contacting point of the pinion with respect to the contacting point of the gear is v p \"of]u p !qh]u g \"0. (20) The sliding friction force at the contacting point p of the pinion will thus be zero. When the contacting point is on the left-hand side of the pitch point p as shown in Fig. 7, for example point a, the relative velocity of the contacting point of the pinion with respect to the contacting point of the gear is given by v a \"(of!l 1 )]u p !(qh#l 1 )]u g (0. (21) The direction of this relative velocity v a will be up from the above formula, as shown in Fig. 7. The friction force on the contacting point of the pinion will be F\"F a and the direction of the resulting friction force will be down. When the contacting point is on the right-hand side of the pitch point p as shown in Fig. 7, for example point s, the relative velocity of the contacting point of the pinion to the contacting point of the gear is given by v s \"(of#l 2 )]u p !(qh!l 2 )]u g '0. (22) The direction of v s will then be down as shown in Fig. 7. The resulting friction force on the contacting point of the pinion will then be F\"F s and the direction of the resulting friction force will be up as shown. The friction force as applied to the gear will be in the opposite direction to that which is applied to the pinion, at the contacting positions discussed above. As the gear teeth move through the mesh cycle, the normal contact force between the teeth changes its value as well as the coe$cient of friction. Rebbechi et al. [8] has measured the dynamic coe$cient of friction, k, at values up to 0"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure3.34-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure3.34-1.png",
        "caption": "FIGURE 3.34. A tire turning on sand.",
        "texts": [
            " An alternative method for defining the slip ratio is s = \u23a7\u23aa\u23aa\u23a8\u23aa\u23a9 1\u2212 vx Rg\u03c9w Rg\u03c9w > vx driving Rg\u03c9w vx \u2212 1 Rg\u03c9w < vx braking (3.122) where vx is the speed of the wheel center, \u03c9w is the angular velocity of the wheel, and Rg is the tire radius. In another alternative definition, the following equation is used for longitudinal slip: s = 1\u2212 \u00b5 Rg\u03c9w vx \u00b6n where n = \u00bd +1 Rg\u03c9w \u2264 vx \u22121 Rg\u03c9w > vx (3.123) s \u2208 [0, 1] In this definition s is always between zero and one. When s = 1, then the tire is either locked while the car is sliding, or the tire is spinning while the car is not moving. Example 107 F Tire on soft sand. Figure 3.34 illustrates a tire turning on sand. The sand will be packed when the tire passes. The applied stresses from the sand on the tire are developed during the angle \u03b81 < \u03b8 < \u03b82 measured counterclockwise from vertical direction. It is possible to define a relationship between the normal stress \u03c3 and tangential stress \u03c4 under the tire \u03c4 = (c+ \u03c3 tan \u03b8) \u00b3 1\u2212 e r k [\u03b81\u2212\u03b8+(1\u2212s)(sin \u03b8)\u2212sin \u03b81] \u00b4 (3.124) 3. Tire Dynamics 135 where s is the slip ratio defined in Equation (3.122), and \u03c4M = c+ \u03c3 tan \u03b8 (3.125) is the maximum shear stress in the sand applied on the tire"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure4.33-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure4.33-1.png",
        "caption": "Fig. 4.33: A force control scheme for a pneumatic actuator of grippers by using a pressure proportional electrovalve.",
        "texts": [
            " In these cases designing a suitable pressure control system in order to vary the pressure in the push chamber of the cylinder can solve the force control problem. Several architectures can be used. Three practical solutions are reviewed in the form of the schemes shown in Figs 4.33, 4.34, and 4.35. Referring to Figs 4.33, 4.34, and 4.35, a closing/opening movement of the gripper fingers is obtained in correspondence to a pulling/pushing movement of the cylinder when the chamber 1 is in pressure and the other chamber 2 is discharging the flow. In Fig. 4.33 the pressure control in chamber 1 is obtained by directly using a pressure proportional electrovalve 3/2 (three ways/two positions), which can ensure the required pressure when a suitable electric analogue signal Vrif is imposed by the user or is sent by a control unit. Usually Vrif is computed as the V value in the Fig.4.32 in a preliminary step by considering the static equilibrium of the grasp, but even by suitable start-up calibration. Fundamentals of the Mechanics of Robotic Manipulation 289 In Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003540_s0006-3495(93)81129-9-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003540_s0006-3495(93)81129-9-Figure6-1.png",
        "caption": "FIGURE 6 Definition of the variables and axes associated with the modeling of swimming kinematics.",
        "texts": [
            " The latter is expressed as a function of the three Euler angles, which in turn are determined by the organism's linear and angular velocities at any instant. This procedure is discussed by Keller and Rubinow (60) and has since been applied by Ramia (37); hence only a brief outline is presented here. Let r(t) represent any position vector in the cell body fixed (x, y, z) frame (i.e. a frame fixed with respect to the cell body that hereafter will be referred to as the body frame) and R(t) be this vector referred to the globally fixed (X, Y, Z) frame (see Fig. 6) at a given time t. The transformation from r(t) to R(t) consists of a translation by Rc(t) and a rotation by A-\u0300 (t) (Keller and Rubinow (60)): R(t) = Rc(t) + A -(t)r(t), dRc(t) A-(t)Uq(t) (9) where U(t) = Uq is the linear velocity of the body frame (i.e., the previously defined instantaneous swimming velocity) and A- 1(t) is a 3 X 3 rotation matrix expressible in terms of the Euler angles +i1(t), +k2(t), and +P3(t) (Eq. 4-47 in Goldstein (61)). These angles in turn depend on the angular velocity of the body frame fQq according to the differential equations (Eq",
            " Next, the instanta- Mean swimming speed ui neous swimming linear velocity U and angular velocity Q are calculated using the BEM. Assuming these velocities to FIGURE 7 Flowchart outlining the steps involved in the approximate be constant throughout the small but finite time increment At, kinematic scheme. the position and orientation of the organism are then trans- Ramia et al. 765 Volume 65 August 1993 ORGANISM TRACKING SCIIEME formed according to the equation R(t + At) = UAt + W(fQ, At)R(t) (11) where R is the position vector of any given point on the cell body (as defined in Fig. 6); a similar equation exists for transforming any point Rf on the flagellum. This transformation is composed of a translation UAt and a rotation W(Ql, At)R(t). Here, W(Q, At) is a matrix representing a rotation through an angle QiAt about an axis the direction of which is defined by the direction cosines of the angular velocity vector (C,fly, fz)/fl and which passes through the cell body/flagellar joining pointRc (Eq. 1.70 ofPaul (62)). At the end of each time increment, the newly updated organism discretization defined by R(t + At) and R1{t + At) has assumed a new position, and the flagellar phase angle and orientation, along with the appropriate boundary conditions, are then used as new input for the BEM"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure4.39-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure4.39-1.png",
        "caption": "Figure 4.39 Examples of movable or kinematically indeterminate supports: the support permits (a) a rotation about RC and (b) a movement perpendicular to the supporting bars.",
        "texts": [
            " Three bar supports (at least) are required for an immovable or kinematically 1 The fact that the centre of rotation RC is a fixed point is true only if the rotation is still small. When considering Figure 4.42, one should not be confused by the fact that the displacements have been drawn to a large scale with respect to the dimensions of the structure. 132 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM determinate support of a rigid body. The bars may not all intersect in one point, or all be parallel, as is shown in Figure 4.39. In the support in Figure 4.39a, all bars intersect at the rotation centre RC, allowing the body to rotate. This support is movable or kinematically indeterminate. The support in Figure 4.39b, in which all bars are parallel to one another (intersect at a point at infinite distance), is also kinematically indeterminate, as the body is free to move in the direction perpendicular to the bar supports. The similarities between bar supports and roller supports were repeatedly pointed out in Section 4.2. Figure 4.38c also shows that two bar supports act as a hinged support at the rotation centre RC, the intersection of the two bars. An immovable support of a rigid body is therefore also possible with three roller supports, as in Figure 4"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure3.11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure3.11-1.png",
        "caption": "FIGURE 3.11. Normal stress \u03c3z applied on the round because of a stationary tire under a normal load Fz.",
        "texts": [
            " The normal component is the contact pressure \u03c3z, while the tangential component can be further decomposed in the x and y directions to make the longitudinal and lateral shear stresses \u03c4x and \u03c4y. For a stationary tire under normal load, the tireprint is symmetrical. Due to equilibrium conditions, the overall integral of the normal stress over the tireprint area AP must be equal to the normal load Fz, and the integral of shear stresses must be equal to zero. Z AP \u03c3z(x, y) dA = Fz (3.20)Z AP \u03c4x(x, y) dA = 0 (3.21)Z AP \u03c4y(x, y) dA = 0 (3.22) 3.3.1 Static Tire, Normal Stress Figure 3.11 illustrates a stationary tire under a normal load Fz along with the generated normal stress \u03c3z applied on the ground. The applied loads on the tire are illustrated in the side view shown in Figure 3.12. For a stationary tire, the shape of normal stress \u03c3z(x, y) over the tireprint area depends on tire and load conditions, however its distribution over the tireprint is generally in the shape shown in Figure 3.13. 3. Tire Dynamics 105 106 3. Tire Dynamics The normal stress \u03c3z(x, y) may be approximated by the function \u03c3z(x, y) = \u03c3zM \u00b5 1\u2212 x6 a6 \u2212 y6 b6 \u00b6 (3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003968_robot.1993.292011-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003968_robot.1993.292011-Figure1-1.png",
        "caption": "Figure 1: Model of a car with n trailers.",
        "texts": [
            " di is the distance from the wheels of trailer i to the wheels of trailer i - 1, where i E (2, . . . , n}. dl is then the distance from the wheels of trailer 1 to the wheels of the car. vo is the tangential velocity of the car and is an input to the system. The other input is the angular velocity of the car, w . We denote T \u201d = [Vo,Wl The tangential velocity of trailer i, v i , is given by i vi = cos(ei-l - ei)vi-l = n cos(ej-l -e j ) (2) j=l where i E (1, . . . , n}. An illustration of these definitions is presented in Fig. 1. A car with n trailers (1) is a nonholonomic system with rolling constraints for each trailer and the car. Denote by (xi, yi) the absolute position of trailer i where (xn,yn) = (x,y). The absolute position of the car is denoted by (20, yo). The nonholonomic constraints can then be expressed as, [3], sinOixi-cosOi+i=O, i~ {O,I,...,n} (3) From a geometrical consideration, Fig. 1, we easily see that n n xi = x + dj cost\u2019, , yi = y + dj sin 0, j=i+l j =i+ 1 where (x,y) = (xnlyn). Eq. (3) then implies n O = sinOjx-sinOi djsinOjej j=i+l n - cos eiy - cos ei dj cos ejej j =i+ 1 which gives the following nonholonomic constraints for i E {0,1, ..., n}: sin 8 i i - cos O i y - dj cos(& - Oj)e, = 0 2 j=i+l The system has, thus, (n + 3) - ( n + 1) = 2 degrees of freedom corresponding to the two independent velocity inputs. We will .nake a change of inputs under the assumption that the state p = [z, y, O n , "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure1.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure1.3-1.png",
        "caption": "Figure 1.3 A surfer balancing on his board. Source: Leidraad voor surfers, Vereniging Zeilscholen Nederland.",
        "texts": [
            " Above and below this disk, there are some 200 globular star clusters that revolve in ellipsoidal orbits 2 around the centre of the Galaxy. \u2022 A calculation of the forces exerted on a swimming shark. In Figure 1.2a, in which the shark is at rest, the shark is subject to forces resultant from its weight W and the upward force A caused by the water pressure. As a result, the animal tips over (see Figure 1.2b). If the shark\u2019s tail generates a thrust T , vertical forces are generated that keep the shark in vertical equilibrium (see Figure 1.2c). \u2022 Balancing on a surfboard (see Figure 1.3). \u2022 A calculation as to the deformation of an oil platform at sea subject to wave action. Figure 1.4 shows a concrete platform designed for the Norwegian Troll field with a water depth of 340 metres. The seabed consists of extremely weak clay. The sea conditions are extremely rough with waves over 10 metres in height. The mass of the deck is 60,000 tons (60 \u00d7 106 kg). \u2022 The description of water currents in a river, estuary, or sea. Figure 1.5 represents a current model for the North Sea. The arrows indicate the direction and strength of the current for certain areas"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure6.41-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure6.41-1.png",
        "caption": "FIGURE 6.41. A universal joint with four links: link 1 is the ground, link 2 is the input, link 4 is the output, and the cross-link 3 is a coupler link.",
        "texts": [
            "285) and thus, \u03b84 = \u03b23. (6.286) Now the angle \u03b3 may be written as \u03b3 = \u03c0 \u2212 \u03b84 \u2212 \u03b1 (6.287) where \u03b84 is the output angle, found in Equation (6.173). Therefore, the coordinates xC and yC can be calculated as two parametric functions of \u03b82 for a given set of a, d, c, e, and \u03b1. 6.6 F Universal Joint Dynamics The universal joint shown in Figure 6.40 is a mechanism used to connect rotating shafts that intersect in an angle \u03d5. The universal joint is also known as Hook\u2019s coupling, Hook joint, Cardan joint, or yoke joint. Figure 6.41 illustrates a universal joint. There are four links in a universal joint: link number 1 is the ground, which has a revolute joint with the input link 2 and the output link 4. The input and the output links are connected with a cross-link 3. The universal joint is a three-dimensional four-bar linkage for which the cross-link acts as a coupler link. The driver and driven shafts make a complete revolution at the same time, but the velocity ratio is not constant throughout the revolution. The angular velocity of the output shaft 4 relative to the input shaft 2 is called 364 6. Applied Mechanisms 6. Applied Mechanisms 365 speed ratio \u2126 and is a function of the angular position of the input shaft \u03b8, and the angle between the shafts \u03d5. \u2126 = \u03c94 \u03c92 = cos\u03d5 1\u2212 sin2 \u03d5 cos2 \u03b8 (6.288) Proof. A universal joint may appear in many shapes, however, regardless of how it is constructed, it has essentially the form shown in Figure 6.41. Each connecting shaft ends in a U -shaped yoke. The yokes are connected by a rigid cross-link. The ends of the cross-link are set in bearings in the yokes. When the driver yoke turns, the cross-link rotates relative to the yoke about its axis AB. Similarly, the cross-link rotates about the axis CD and relative to the driven yoke. Although the driver and driven shafts make a complete revolution at the same time, the velocity ratio is not constant throughout the revolution. A separate illustration of the input, output, and the cross links are shown in Figure 6",
            " Example 254 F Double universal joint. To eliminate the non-uniform speed ratio between the input and output shafts, connected by a universal joint, we can connect a second joint to make the intermediate shaft have a variable speed ratio with respect to both the input and the output shafts in such a way that the overall speed ratio between the input and output shafts remain equal to one. Example 255 F Alternative proof for universal joint equation. Consider a universal joint such as that shown in Figure 6.41. Looking along the axis of the input shaft, we see points A and B moving in a circle and points C and D moving in an ellipse as shown in Figure 6.47(a). This is because A and B trace a circle in a normal plane, and C and D trace a circle in a rotated plane by the angle \u03d5. Assume the universal joint starts rotating when the axis CD of the cross link is at the intersection of the planes of motion CD and AB, as shown in Figure 6.47(a). If the axis AB turns an angle \u03b8, then the projection of the axis CD will turn the same angle, as can be seen in Figure 6"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure9.11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure9.11-1.png",
        "caption": "Fig. 9.11. Single-sided AFPM brushless motor fitted to spoked wheel [215].",
        "texts": [],
        "surrounding_texts": [
            "Battery EVs are vehicles that use secondary batteries (rechargeable batteries or storage batteries) as their only source of energy. An EV power train can convert energy stored in a battery into vehicle motion and it can also reverse direction and convert the kinetic energy of the vehicle back into the storage battery through regenerative braking. AFPM brushless motors are used in EVs as in-wheel motors (Figs 6.8 and 9.11) [215, 225]. The pancake shape of an AFPM motor allows for designing a compact motorized wheel. With the aid of the AFPM brushless motor the differential mechanism can be replaced by an electronic differential system [101]. The configuration shown in Fig. 9.12a illustrates the use of a pair of electric motors mounted on the chassis to drive a pair of wheels through drive shafts, which incorporate constant velocity joints. In the configuration shown in Fig. 9.12b, the motors forming the electronic differential are mounted directly in the wheels of the vehicle. The electromechanical drive system is considerably simplified, when the motor is mounted in the wheel, because the drive shafts and constant velocity joints are now no longer needed. However, the resultant \u201cunsprung\u201d wheel mass of the vehicle is increased by the mass of the motor. Wheel motors of this direct drive configuration also suffer overloading because the speed of the rotor is lower than would be the case with a geared arrangement. This leads to an increase of the active materials volume required in the motor. The disadvantages of conventional wheel motors can be overcome by using the arrangement shown in Fig. 9.13. The two stators are directly attached to 292 9 Applications the vehicle body whilst the PM rotor is free to move in radial directions. It can be observed that in this case the wheel and disc rotor form the unsprung mass, whilst the stators of the motor become sprung mass supported on the chassis [101]. Given the advantages of using wheel motor in electric vehicle drive system such as more onboard space, better vehicle control and modular drive train topology, there is a growing interest in automotive industry to develop wheel motors for future electric vehicles. Figure 9.14 shows the AFPM wheel motor developed at General Motors Advanced Technology Center, Torrance, CA, U.S.A. The developed wheel hub motor has a single stator located between two PM rotor discs. This topology features higher torque density design as it utilizes copper from both surfaces of the stator for torque production [66, 199, 223]. To remove heat from inside stator winding, provision is made of an aluminum ring with internal liquid coolant passages along end winding of the outer stator winding [P149]. The design specification of the GM\u2019s wheel motor is summarised in Table 9.3. To demonstrate the performance and robustness of the AFPM wheel motor technology, two wheel motors were fitted on GM\u2019s S-10 show truck and underwent several thousand miles road tests. 9.2 Electric Vehicles 293 294 9 Applications"
        ]
    },
    {
        "image_filename": "designv10_0_0002351_s00170-017-0426-7-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002351_s00170-017-0426-7-Figure8-1.png",
        "caption": "Fig. 8 (i) Numerical prediction of temperature variance in the part for case study C1.1 (conformal insert) where temperatures ranged from 83 to 87 \u00b0C at ejection. (ii) Numerical prediction of temperature variance for the part for case study B1.1 (conventional insert) where temperatures ranged from 83 to 97 \u00b0C at ejection",
        "texts": [
            " The simulation was conducted with experimentally measured values of moulding machine parameters taken from physical sensor readings, including melt injection flow rate, melt temperature, as well as coolant flow rate, and inlet temperature (e.g. Table 9). Additionally, the specific heat and thermal conductivity of mould material parameters used during simulation were based on temperature-dependent material models available in the simulation software database [13]. As such, the potential modelling discrepancies with temperature-independent models reported in literature were minimised (for example [10]). Figure 8 presents the simulated numerical results for analysis cases C1.1 and B1.1, which show differences in temperature throughout the part at the end of the moulding cycle for the conformal and conventional inserts. Higher temperature variation in the part indicates cooling non-uniformity which can cause defects such as part warping. The results show that due to the more uniform distribution of cooling channels, the conformal cooling insert (Fig. 8(i)) exhibits higher uniformity in temperature compared to the standard insert (Fig. 8(ii)) with a difference in peak temperatures of 10 \u00b0C. The results highlight the performance benefits of conformal cooling. Similarly, Fig. 9 shows numerically predicted and experimentally measured temperatures at the mould thermocouple locations (Fig. 7) throughout the cycle for case studies C1.1 (conformal insert) and B1.1 (conventional insert). The results show several characteristics: & Inner thermocouples (1\u20134) showed the lowest increase in temperatures, attributable to close proximity of surrounding cooling channels and the associated high rate of cooling"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001391_bf02133524-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001391_bf02133524-Figure4-1.png",
        "caption": "Fig. 4.",
        "texts": [
            " Under this hypothesis: tbe/ap 9, solution to the bolonomic problem formulated in Sec. 3.1 for given F, 8, mimmizes: 1 e S e + 1 ~ ,HX+i~x__~f(3) (3 .14) (iv) ~,(f, x) ~ ~ - -~- subject to : e = B f - - N X - - 8 , X/> 0 (3.15) 1 1 (V) ~'_~(Q, X) -~ - ~ - (~CQ + ~ XHX + (~8 (3.16) subject to: I3Q = F , l ~ / q - - HX < K (L17) 1 \u00a3 (H - - I ~ Z N ) X - )~(I~Q e - - K) (vD (~) .o_,(x) ___ - T (3 18) (3) The only variables are f, X, since e can be eliminated by means of the compatibility equation included in the constraint set (3.15). (g Fig. 4 visualizes geometrically, for a case of strict convexity and of two variables, these dual optimization problems and their mutual connexions (feasible domains are shaded, lines of constant objective dashed, X \u00b0 indicates the optimal vector). subject to: fv][)(,) subject to: X~>0 ooCX) ~ + i ( H - I~ZN)X (H - - t~IZN)X /> l ~ Q S - - K . (3.19) (3.20) (3.21) Proof - - Only an outline is given below, since the underlying mathematical concepts are available in standard books, e. g. [26], the algebraic manipulations are elementary and demonstrations of this kind have been already developed in related papers, e"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002364_j.msea.2020.140168-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002364_j.msea.2020.140168-Figure1-1.png",
        "caption": "Fig. 1. Schematic diagram and the laboratory hammering equipment.",
        "texts": [
            " Therefore, thermal processing (or in situ heat treatment) is another approach to improve the microstructure and mechanical properties for AM alloys. In this paper, the interlayer deformation during the WAAM process of 2319 aluminum alloy was implemented by a self-developed pneumatic hammering device, which modifies internal microstructure and properties. The microstructural evolution during the WAAM process and the strengthening mechanisms were investigated. A laboratory hammering equipment was developed and used to investigate the WAAM process, as shown in Fig. 1. The WAAM setup comprises a Fronius Cold Metal Transfer (CMT) power source, a VR1550 wire feeder and a KUKA KR16 6-axis robot. The nominal chemical composition of 2219 plate and 2319 filler wire, which are quite close to each other, are listed in Table 1. Before WAAM, the 2219 aluminum substrate with dimensions of 500 \u00d7 500 \u00d7 30 mm was cleaned with an acetone and alkaline solution. According to our previous studies [14\u201316], wall-shaped components were deposited in alternate directions layer by layer by advanced CMT + Pulse (CMT + PA) mode on the substrate",
            " The hammering process was employed between each deposited layer with identical hammering parameters. Before hammering, the temperature of the deposited layer was cooled to 50 \u25e6C. The deposition parameters, including wire feeding speed, travel speed and shielding gas flow, are 7 m/min, 0.5 m/min and 25 L/min respectively. The hammering setup consisted of an air hammer and a 3-axis platform that can move along horizontal X, Y and vertical Z directions, as shown in X. Fang et al. Materials Science & Engineering A 800 (2021) 140168 Fig. 1. The hammering parameters are listed in Table 2. The deformation strain \u03b5 was defined as: \u03b5=(H \u2212 h)/H \u00d7 100% (1) where H is the initial bead height, h is the height after hammering. The samples were cut near the center of the fabricated samples to exclude the disturbing factors from the rough surface, using the sparkerosion wire cutting machine to prepare standard specimens for microstructure observation and tensile tests. An anodizing film was formed in a 25 g/L fluoroboric acid solution using an ElectroMet 4 electrolytic polishing machine (Buehler, Lake County, IL, USA) at a voltage of 26 V for 1 min"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003700_a:1008924521542-Figure13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003700_a:1008924521542-Figure13-1.png",
        "caption": "Figure 13. Spring mechanism at an ankle joint.",
        "texts": [],
        "surrounding_texts": [
            "In running in animals, the tendon acts as a shock absorber for landing and as an actuator for jumping. In the present study we use a spring mechanism (Raibert, 1986; Furusho et al., 1995) in place of a tendon. The mutual entrainment between oscillation of a N.O. network and oscillation of the spring-mass and environment system (S.M.E.S.) causes the quadruped to run. A bound gait appears when quadruped animals run fast. We use a bound gait for running because running in a bound gait does not need actuators around the roll axis and is suitable for the quadruped which has large powered actuators around the pitch axis at hip joints of hindlegs. It is difficult to transfer the robot\u2019s state from stationary standing to running in a bound gait, since the movement of the body is large in a bound gait, and the motions of the fore and hindlegs are much different from each other. We realized a bound gait by transition from a pronk gait."
        ]
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure6.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure6.2-1.png",
        "caption": "Fig. 6.2 Free rotating rigid body in zero-G: In general, free rotation in 3D does not occur around a fixed axis",
        "texts": [
            "9 [m3] is freely rotating in zero-G space around its center of mass. The angular velocity \u03c9 changes based on Euler\u2019s equation (6.4), whereas its posture R changes by the equation of rotation (2.33) from Chapter 2 \u03c9\u0307 = \u2212I\u22121(\u03c9 \u00d7 I\u03c9) (6.5) R\u0307 = \u03c9\u0302R. (6.6) 186 6 Dynamic Simulation We can integrate these differential equations numerically, assuming the angular acceleration and velocity \u03c9\u0307, \u03c9 are constant during the short period of \u0394t using { \u03c9(t+\u0394t) = \u03c9(t) + \u03c9\u0307\u0394t R(t+\u0394t) = e\u03c9\u0302\u0394tR(t). The simulated result is illustrated in Fig. 6.2. While no external forces exist, we see the object rotates in a complicated manner. The upper graph of Fig. 6.3 shows the change of the angular velocity start from its initial value \u03c9 = [1 1 1]T (rad/s). On the other hand, the angular momentum shown in the lower graph of Fig. 6.3 keeps constant. This shows conservation of the angular momentum which is expected, as well as the validity of the simulation. Like this simulation, generally, a floating rigid body in 3D space rotates in a complicated manner which does not have a fixed rotating axis1"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001516_s00170-012-4271-4-Figure14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001516_s00170-012-4271-4-Figure14-1.png",
        "caption": "Fig. 14 Non-assembly mechanisms was manufactured by SLM. a Manual-added supports designed in Magics 14.0 software; b before supports were removed; c after supports were removed",
        "texts": [
            " Considering the placing style (c), it not only needs smaller amount of manual-added supports, but also has smaller z-directional height than style (b). Another advantage of placing style (c) is that the supporting pattern is \u201cline,\u201d which is convenient to be removed; while the supporting patterns of placing styles (a) and (b) should be \u201cweb\u201d or \u201ccontour.\u201d which are much more difficult to be removed than \u201cline\u201d pattern. Based on above analysis, the fabrication of overhanging surface will commonly use placing style (c). Figure 14 shows the example of SLM manufactured nonassembly mechanisms using placing style (c), where the inclined angle of the key clearance part was adjusted to be a little bigger than the reliable building angle. It is found that the amount of manual-added supports were little (Fig. 14a, b), and all the metal supports could be easily removed. The non-assembly mechanisms can be swung directly and smoothly after supports were removed (Fig. 14c). Multi-layered part is built from multi-single layers, and a single layer is composed of multi-single tracks. During laser scanning the metal powder, the laser energy is transferred to heat absorbed by metal powder, which makes the powder melt and bind together. When the laser energy input is enough and the metal powder is absolutely melted, then regular and continuous track can be obtained [17], while the laser energy input is insufficient to melt the metal powder, then sintering status will be got. There exists a heat affect zone around the molten pool during laser scanning metal powder, making the metal powder around the molten pool not fully melted or in sintering status [18], as Fig. 15 shows, many stainless steel powder particles were adhered to the both sides of the tracks. The clearance was critical during SLM fabrication of non-assembly mechanisms [19] (Fig. 14). The smaller the clearance, the more stable the mechanisms will works. The adhered powder particles will affect much on the size of the clearance, leading to the movement of the mechanisms not so smooth, that\u2019s because the adhered powder particles were hard to escape from the key positions. In principle, the clearance of the non-assembly mechanisms should be designed as small as the size of the biggest powder particles (it was 30 \u03bcm in this experiment), but for the powder adhering phenomenon happened in heat affected zone, the clearance size must be designed much bigger than the theoretical size, only in this way that the freely moved nonassembly mechanisms can be directly manufactured by SLM"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure9.19-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure9.19-1.png",
        "caption": "Fig. 9.19. ECI drilling system: (a) general view, (b) power unit with AFPM synchronous motors, (c) PA-44 Kaman Aerospace AFPM motor. Courtesy of Tesco Corporation, Calgary, Alberta, Canada.",
        "texts": [
            " A rule of thumb for selecting the power source is that the power required to hoist the drill rods should be about three times the power that is required to turn the drill string. For high elevations, the power loss is about 3% for each 300 m. Large power AFPM brushless motors (Table 9.4) are especially suitable for portable drilling equipment because of their compact design, light weight, precise speed control, high efficiency and high reliability. In the ECI lightweight top drive drilling system (Fig. 9.19) manufactured by Tesco Corporation, Calgary, Alberta, Canada, the traditional induction or brush type d.c. motors have been replaced with high performance liquid cooled AFPM synchronous motors. This system is compatible with most 600 V a.c. rig power systems. The product highlights include: \u2022 Kaman PA44 AFPM synchronous motors that (Table 9.4) have been successfully tested to 60g triaxial loading, making them the ideal choice for rough drilling environments; \u2022 AFPM synchronous motors that allow a high level of precision in speed and torque control not available with other motors; \u2022 a modular design that allows drilling to continue at reduced power with one motor; \u2022 direct connection to the rigs a"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000233_978-1-4614-0460-6-Figure9.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000233_978-1-4614-0460-6-Figure9.3-1.png",
        "caption": "Fig. 9.3 Example showing rigid coupling of components I and II to form assembly III. The force F2 is applied to the assembly at coordinate X2 in order to determine H22 and H12",
        "texts": [
            " H11 \u00bc X1 F1 \u00bc x1 F1 \u00bc h11f1 \u00fe h12af2a F1 \u00bc h11f1 h12a h2a2a \u00fe h2b2b\u00f0 \u00de 1h2a1F1 F1 H11 \u00bc h11F1 h12a h2a2a \u00fe h2b2b\u00f0 \u00de 1h2a1F1 F1 \u00bc h11 h12a h2a2a \u00fe h2b2b\u00f0 \u00de 1h2a1 (9.8) We can also use f2a to determine the cross receptance H21. See Eq. 9.9. H21 \u00bc X2 F1 \u00bc x2a F1 \u00bc h2a1f1 \u00fe h2a2af2a F1 \u00bc h2a1f1 h2a2a h2a2a \u00fe h2b2b\u00f0 \u00de 1h2a1F1 F1 H21 \u00bc h2a1F1 h2a2a h2a2a \u00fe h2b2b\u00f0 \u00de 1h2a1F1 F1 \u00bc h2a1 h2a2a h2a2a \u00fe h2b2b\u00f0 \u00de 1h2a1 (9.9) We determine the direct and cross receptances, H22 and H12, respectively, by applying a force to the assembly coordinate X2. See Fig. 9.3. The component receptances are again h11 \u00bc x1 f1 , h2a2a \u00bc x2a f2a , h12a \u00bc x1 f2a , and h2a1 \u00bc x2a f1 for I and h2b2b \u00bc x2b f2b for II. The compatibility condition for the rigid coupling remains as x2b x2a \u00bc 0. However, the equilibrium condition is f2a \u00fe f2b \u00bc F2 because the force is applied to the coupling coordinate. To determine H22 \u00bc X2 F2 , we begin by writing the component displacements. For I, the displacements are: x1 \u00bc h12af2a and x2a \u00bc h2a2af2a: (9.10) f2b f2a x2b x2a II I x1 9.2 Two-Component Rigid Coupling 325 For II, we have x2b \u00bc h2b2bf2b"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002125_tfuzz.2016.2634162-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002125_tfuzz.2016.2634162-Figure1-1.png",
        "caption": "Fig. 1 Nonsymmetric dead-zone nonlinearity.",
        "texts": [
            ", n \u2212 1, x = [x1, x2, ..., xn] T \u2208 R are the state variables with x(0) = x0, xi = [x1, x2, ..., xi] T . x1 is the scalar output, and u is the scalar control input. fi(\u00b7) is smooth unknown nonlinear function, gi(\u00b7) is smooth known nonlinear function and y is the output signal. According to [44]\u2013[47], u is a nonsymmetric dead-zone input nonlinearity whose definition is u = D(w) \u2206 =  or (w \u2212 pr) , if w \u2265 pr 0 , if \u2212 pl < w < pr ol (w + pl) , if w \u2264 \u2212pl. (2) The nonsymmetric dead-zone nonlinearity is displayed in Fig. 1 in which w \u2208 \u211c is the dead-zone input signal. ol and or represent the left and the right slopes characteristic for the dead-zone. pl and pr stand for the breakpoints. The goal of this paper is to construct the control laws to regulate the output x1 to track the designed signal xd. To complete the control system design, the following assumptions and lemmas are introduced. Assumption 1: There is a large scalar Mn satisfying |w| \u2264 Mn. Assumption 2 [44]\u2013[47]: The coefficients or, ol, pr and pl are uncertainty, strictly positive constants with or \u0338= ol"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000492_02783649922066376-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000492_02783649922066376-Figure7-1.png",
        "caption": "Fig. 7. Two simple examples to compare and contrast the CoP (P ), GCoM (C), and FRI point (F ). At left, the foot is in static equilibrium since F is within the support line (although C is outside); P is coincident with F . At right, the foot is starting to rotate, since F is outside the support line (although C is inside); P is at the tip about which the foot rotates.",
        "texts": [
            " Next we consider an actuated system (Fig. 6b) with an ankle torque that precisely compensates for the gravitational moment but does not generate any shank motion; i.e., \u03b8\u03071 = 0 and \u03b8\u03081 = 0. To determine the position of the FRI point of this system, we use \u03c41 = \u2212O1G2 \u00d7 m2g and R1 = \u2212m2g in eq. (6). We get \u2211 FGi \u00d7 mig = 0. This means that F falls on the CG gravity line of the system. This property is valid not only for the foot/shank, but for any stationary mechanism (Shih et al. 1990). In the next example, shown in Figure 7 (left), the shank configuration corresponds to a GCoM position C outside the support polygon. The foot is, however, prevented from rotating by the ankle torque (ml2\u03b8\u0308\u2212mg cos \u03b8 ). This should be taken into consideration while planning the gait initiation of biped robots. It is noteworthy that to stop the robot from tipping over, some control laws accelerate the heavy robot body forward (Hirai et al. 1998). This generates a supplementary backward inertia force\u2014similar to this example\u2014which shifts the FRI point F backward, bringing it within the support polygon. Since the foot is stationary, F = P . Finally in Figure 7 (right), the shank is vertically upright with its GCoM well within the support line. Despite this, the foot starts to rotate due to the ankle torque ml2\u03b8\u0308 . The FRI point F is situated outside the support line at a horizontal distance OFy = l\u03b8\u0308 g (l + h) from O. The CoP is at the extreme frontal point of the support polygon. Although the focus of this work is the dynamics of biped robots and the introduction of the FRI point, it is the control of this point which is of importance to the robotics community"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002794_j.rcim.2021.102165-Figure10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002794_j.rcim.2021.102165-Figure10-1.png",
        "caption": "Fig. 10. Offline programming for trajectory planning based on CAD model.",
        "texts": [
            " Though the two DE and LM-DEH algorithms both can find optimal values, the number of iterations of the former is at least 2 times that of the latter, resulting in the poor efficiency. Those results demonstrate that the proposed kinematic parameter calibration method is very effective and greatly improves the absolute positioning accuracy of the robot. Since aero-engine blades are mostly composed of curved surfaces and curves, the trajectory of robotic grinding of blades is difficult to accurately obtain through teaching programming. For the path algorithms of robotic machining, most of the existing results are based on offline programming of a CAD model. Fig. 10 shows the trajectory generated based on the tool location data of the existing model and the CAD G. Luo et al. Robotics and Computer-Integrated Manufacturing 71 (2021) 102165 software through offline programming method. Robotic machining of blades requires that robot can follow the planned complex path and guarantee the fine path accuracy. The main factors affecting the path accuracy are errors related to the robot kinematic model and dynamic model. Robot kinematic calibration can ensure to make the planned path and the actual path as close as possible, thereby improving the absolute accuracy of path to a certain extent, as shown in Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002628_j.autcon.2020.103078-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002628_j.autcon.2020.103078-Figure3-1.png",
        "caption": "Fig. 3. hTetro shape transformation from O to L.",
        "texts": [
            " \u2211= \u2212 + \u2212C x x y y( ) ( )translation i j A B C D i j i j , \u03f5 , , , 2 2 (1) \u2211= \u00d7 \u2212 + \u2212C pi x x y y 2 ( ) ( )rotation i A C D i B i B \u03f5 , , 2 2 (2) = \u23a7 \u23a8 \u23aa \u23a9\u23aa C a C if a O Z L T J S I C if a clockwise anticlockwise C if a up down left right ( ) \u03f5 { , , , , , , } \u03f5 { , } \u03f5 { , , , } total t transformation t rotation t translation t (3) Transformation costs are dependent on the current shape and the desired shape. Table 1 shows the equations used to compute the transformation cost Ctransformation. Generally, a set of block rotations are required to complete a transformation. Since these rotations are done around a hinge, the cost is in terms of L, which is l/2, where l is the length of the block. An example of a transformation is shown in Fig. 3. For changing the shape from O to L, the robot must first change from O to I and then I to L. The cost for transitioning from O to I is +\u03c0 L( 5 2 ) , since both C and D blocks have to move an angle of \u03c0, and they are at a distance of L 2 and L 5 respectively. The cost for changing from I to L is L( 2 )pi 2 , since only D block moves this time an angle of \u03c0 2 at a distance of L 2 from the hinge. Hence, the overall cost turns out to be +\u03c0 L( 5 2 )3 2 . The overall cost of any action is given in Eq"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002411_j.powtec.2018.03.010-Figure16-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002411_j.powtec.2018.03.010-Figure16-1.png",
        "caption": "Fig. 16. Graphical illustration of adjusting scan line distance of Ti-6Al-4V specimens with porous structure: (a) 3D model, (b) strategies of scanning, and (c) layer loading to obtain the end dimensions [100].",
        "texts": [
            " Sheng Zhang et al. [100] investigated the effect of scan line spacing on the produced Ti-6Al-4V specimens. The Ti-6Al-4V specimens image fabricated by SLM is depicted in Fig. 14. The entire cross-section profile of the specimen can be seen from the side, while the pores are at the top surface of the figure, respectively. Fig. 15 shows the normal and rectangular-shaped pore distributions with thin wall struts of approximately 200, 300, 400, 500, 600, 700 \u03bcm in size for each specimen. The diagram in Fig. 16 shows the analysis and scanning strategies used to obtain the end dimension of Ti-6Al-4V specimens. It was clearly observed that the single laser scan line of approximately 200 \u03bcm spot size caused the thin wall struts. Also, through scan line distancemodification, the length of exceeding 200 \u03bcm as required by the pores was attained. Fig. 15(a) shows the resulted closed pores with irregular shapes of about 200 \u03bcm of scanned line distance. However, when the pore diameter is smaller than the largest powder size, it becomes an open pore",
            " [193,194] had successfully manufactured porous titanium alloy specimen in their studies by using SLM technique and investigated the geometric effect of the unit cell on the produced specimen mechanical properties. They demonstrated that porousmaterial with arbitrary porosity and good compressive strength can be achieved using the SLM process. Furthermore, Stamp et al. [195], Douglas T. et al. [196], and Warnke et al. [197] were all in agreement with Mullen et al. in manufacturing porous titanium alloy via SLM technology for biomedical applications. I. Yadroitsev et al. [198] alternatively introduced two different scanning strategies to fabricate Ti-6Al-4V via SLM as illustrated in Fig. 16. The \u201ctwo-zone technique\u201d (strategy A) [199] determines the highest density specimens, whereas, the second approach (strategy B) reduces the effect of the Heat-Affected Zone (HAZ) and remelting number of thefinalmicrostructure. Fig. 21(a) shows the smooth microstructure morphology of the as-made SLM specimens categorised as a \u03b1\u2019 martensitic [149], while Fig. 21(b) shows the observable melt pool boundaries of the tracks resulted from the different mechanical properties of the materials. The heating and cooling processes above the \u03b2-transit temperature (T\u03b1-\u03b2 = 1000 \u00b1 20 \u00b0C) guide the renucleation of the complete phases since the cooling rate is dependent on the alteration of \u03b2-phase"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001287_1.3109245-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001287_1.3109245-Figure1-1.png",
        "caption": "Fig. 1 Selective laser melting. \u201ea\u2026 Fabricat sition of a powder layer, \u201eII\u2026 scanning of the and \u201ec\u2026 Detailed view of the laser-powder in \u201e3\u2026 powder delivery piston, \u201e4\u2026 fabrication",
        "texts": [
            " Two factors destabilize the process with increasing the scanning velocity: increasing the length-to-width ratio of the melt pool and decreasing the width of its contact with the substrate. DOI: 10.1115/1.3109245 Keywords: powder bed, absorbing scattering medium, melt pool, capillary instability Introduction The process of selective laser melting SLM is a technique of irect manufacturing from metallic powder. Its principal differnce from conventional powder metallurgy processes is local eating and binding of powder by a scanning laser beam. Rapid anufacturing 1 of complex parts see examples in Fig. 1 a is ssured by this method. The SLM includes scattering and absorpion of laser radiation in the powder, heat conduction, melting and oalescence of powder particles, formation of the melt pool, and ts solidification. Figure 1 b shows a typical scheme of layer-byayer SLM fabrication. A SLM machine comprises of a fabrication late 1 with two holes to which two containers are attached. A oller 2 delivers powder from the left container and deposits a hin powder layer in the right container I . The thickness of the ayer about 50 m 2 is controlled by pistons moving up 3 nd down 4 . The deposited layer is scanned by a laser beam 5 , hich ensures local melting and binding to the previously fabri- ated layer II . A part of a complex shape can be obtained in the ight container after multiple deposition-scanning cycles III . The omplete remelting of the powder in the scanning zone is shown n Fig. 1 c , and its good adhesion to the substrate ensures obtainng functional parts with high mechanical properties 2,3 . The Contributed by the Heat Transfer Division of ASME for publication in the JOURAL OF HEAT TRANSFER. Manuscript received April 15, 2008; final manuscript reeived January 28, 2009; published online May 1, 2009. Review conducted by Ben . Li. ournal of Heat Transfer Copyright \u00a9 20 om: http://heattransfer.asmedigitalcollection.asme.org/ on 04/07/2014 Ter process of SLM is sensitive to a number of parameters such as the powder layer thickness, the power and the diameter of the laser beam, and the scanning speed",
            " nother reason to apply the model of slow consolidation, where he processes of radiation transfer and shrinkage are essentially eparated in space is the experimentally observed detachment of he remelted material from the loose powder 3 see also the iscussion in Sec. 4 . This work concerns modeling the laser-powder interaction zone ased on the assumption of slow consolidation and aims to clarify hysical processes at SLM. A single-line scan on a layer of unonsolidated powder is studied, as shown in Fig. 1 c . The subtrate is implied to be solid and thermally thick. The model inludes laser radiation transfer described in Sec. 2 and conductive eat transfer described in Sec. 3. Radiation Transfer The freshly deposited powder is not mechanically compressed n the SLM machine in order to not destroy the underlying conolidated part. Therefore, it has as high porosity as freely poured owder, which is in the range 40\u201360% for typical metallic pow- 72101-2 / Vol. 131, JULY 2009 om: http://heattransfer.asmedigitalcollection",
            " In the dimensionless form q = Q Q0 = f+ \u2212 f\u2212 2 + q+ \u2212 q\u2212 = a 4 \u2212 3 D 1 \u2212 2 e\u2212 1 \u2212 a e\u22122a + 1 + a e2a \u2212 3 + e\u22122 1 + a \u2212 1 \u2212 a e2a \u2212 + 1 \u2212 a \u2212 1 + a e2a \u2212 \u2212 3 1 \u2212 e\u2212 \u2212 e \u22122 4 \u2212 3 20 Absorptivity of the system powder-substrate defined as the fraction of the incident radiation passed the powder surface, is estimated as A = q 0 = a 4 \u2212 3 D 2 1 \u2212 2 e\u2212 \u2212 3 + e\u22122 1 + a \u2212 1 \u2212 a e2a + 1 \u2212 a \u2212 1 + a e\u22122a \u2212 3 1 \u2212 1 \u2212 e\u22122 4 \u2212 3 21 The fraction of incident radiation absorbed by the substrate is As = q = a 4 \u2212 3 D 1 \u2212 2 e\u2212 1 \u2212 a e\u22122a + 1 + a e2a \u2212 2 1 \u2212 3 + e\u22122 \u2212 3 1 \u2212 2e\u2212 4 \u2212 3 22 The volumetric heat source due to radiation absorption is U = \u2212 dQ dz = \u2212 Q0 dq d 23 3 Conductive Heat Transfer In the moving coordinate system shown in Fig. 1 c , the heat conduction equation is JULY 2009, Vol. 131 / 072101-3 ms of Use: http://asme.org/terms w t w r m c a w l d t t T t m t f o t w t a c l w i t w a w c 0 Downloaded Fr H t \u2212 v H x = x k T x + y k T y + z k T z + U 24 here volumetric enthalpy H is related with temperature T by the hermal equation of state T = H/Cs, H CsTm Tm, CsTm H CsTm + Hm Tm + H \u2212 CsTm \u2212 Hm /Cl, H CsTm + Hm 25 here Cs and Cl are the specific heats in solid and liquid phases, espectively, Tm is the melting point, Hm is the latent heat of elting, t is the time, v is the scanning velocity, k is the thermal onductivity, and U is the volumetric heat source given by Eq"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001252_j.ymssp.2007.12.001-Figure11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001252_j.ymssp.2007.12.001-Figure11-1.png",
        "caption": "Fig. 11. Outer race fault insertion and dimensions [14].",
        "texts": [
            " The defect frequencies (frequencies at which the ball passes the defect on the outer race (BPFO) or the inner race (BPFI)) can be estimated for the outer race as 48.9Hz and for the inner race as 71.1Hz. The fundamental train frequency (FTF) is estimated at 4.1Hz, while the speed at which the balls spin (BSF) is 26.5Hz. ARTICLE IN PRESS N. Sawalhi, R.B. Randall / Mechanical Systems and Signal Processing 22 (2008) 1924\u201319511938 A notch fault was introduced into the outer race of the double-row ball bearing. This was performed using electric spark erosion and generated a gap in the outer race with a rectangular cross-section (Fig. 11). The bearings were positioned such that the outer race faults were always in the loaded zone as this is where a fault would most likely occur [25]. The notch fault dimensions were inserted into the bearing\u2019s S-function to simulate the vibration generated from the test rig as a result of this fault. In order to take into account the differences originating from aligning the Line of Action (LoA) of the gear with the vertical direction in the simulation model (pressure angle in the actual case is 201 from the vertical) the position of the spall was shifted by 201 from the centre of the load zone"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001766_j.cirp.2019.05.004-Figure32-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001766_j.cirp.2019.05.004-Figure32-1.png",
        "caption": "Fig. 32. (a) Artefact developed by Fahad et. al. [76]. (b) Micro- and meso-scale artefact developed by Hao et al. [102].",
        "texts": [
            " 22) to evaluate the system\u2019s capabilities in producing lattice structures and to show common metrology problems when inspecting AM parts with lattice structures [22]. Moreover, Bauza et. al. showed that a typical problem with test artefacts is that their form and dimensions can change after removal from the machine build plate or after post treatments, and thus the actual system errors can be masked and changed by these post-built variations (see Fig. 12) [22]. Fahad and Hopkinson focused on the simplicity of the part and the possibility to easily measure it using tactile CMM (see Fig. 32a) [75]. They also showed that the amount of surrounding powder has a significant influence on the final properties of parts and has to be accounted for when testing AM machines [76]. A more complex 0. Examples of geometries of early test artefacts. Left: \u201cuser part\u201d focusing on ating the machine accuracy in the x-y plane [86]. Right: artefact including fic features: overhangs, ramps, cones, hemi-spheres, etc. [175]. art, ogy ain ach tion AM are and to TM e to nal her, and her low rds and hat fted rds rch of ons AM ith AM nal 261: s, (e) le is and much smaller artefact (see Fig. 32b) was proposed by Hao et al. which was especially designed for the evaluation of micro- and meso-scale production capabilities, as well as process stability and repeatability [102]. M\u00f6hring et al. developed a micro-scale test artefact including simple geometrical features as well as freeform structures (Fig. 33) to compare the capabilities of milling and different AM processes in a CIRP round robin experiment involving different micromanufacturing processes. The comparison revealed large form errors especially for thin walled features, caused by insufficient resolution of AM processes, but also showed the advantages of using XCT systems for measuring some of the key features [187]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure12.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure12.6-1.png",
        "caption": "FIGURE 12.6 Example z-loop for inductive coupled voltages.",
        "texts": [
            " They measure the field at the observation point O. We again assume that the sensors are small compared to the smallest wavelength in the spectrum. FAR-FIELD RADIATION MODELS BY USING SENSORS 317 12.2.6 Time Domain Solution for H-Field Sensor We assume that all sides of the sensor arrangements in Fig. 12.5 are the same, \u0394x = \u0394y = \u0394z = w. Since all three loops are fundamentally equivalent, it is sufficient to consider a single loop to illustrate the approach. Our example is a loop in the x\u2013y plane to compute Hz shown in Fig. 12.6. The voltage induced in the loop from the PEEC circuit is vz = d dt\u222b B \u22c5 z\u0302 d \u2248 w2 d dt Bz(t), (12.17) where the normal component of B is integrated over the area w2 of the sensor loop. With B = \ud835\udf07H given by (3.1d), we can solve (12.17) for Hz Hz(t) = 1 \ud835\udf07w2 \u222b t 0 vz(t) dt, (12.18) 318 INCIDENT AND RADIATED FIELD MODELS where the induced unknown voltage time integral of vz in Fig. 12.5 needs to be computed to find Hz(t). Thin filament conductors are used to form the loop in Fig. 12.6 for the computation of the coupled induced voltage. To compute the total induced voltage for the z-loop in the example Fig. 12.6, we need to compute all partial mutual inductances that couple from the inductive PEEC cells in the problem to the loop filament conductors. Of course, the couplings are direction dependent and can be sparse, especially for rectangular Manhattan problems. Then, the voltage induced in test loop in Fig. 12.6 is Vz = 4\u2211 m=1 Nk\u2211 k=1 Lpmk dik(t \u2212 \ud835\udf0fmk) dt , (12.19) where the sum is over all conductors Nk that are not orthogonal to the sensor filament considered. All three components of the field in (12.16) can be computed. We observe that all sensors do not involve partial self-inductances. Hence, sensors do not have an impact on the solution and the above techniques can be applied as a postprocessing step once the solution vector x(t) is known. Rectangular loops can always be used even if the rest of the problem consists of nonorthogonal cells"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure9.31-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure9.31-1.png",
        "caption": "Fig. 9.31. Cost effective AFPM brushless vibration motor. Photo courtesy of Moving Magnet Technologies, SA, Besancon, France.",
        "texts": [
            " There are two types of brushless vibration motors for mobile phones: cylindrical or RFPM motor and coin type or AFPM motor (Fig. 9.30). The unbalanced exciting force generated by an eccentric rotor is F = m\u03b5\u2126 = 2\u03c0m\u03b5n2 (9.7) where m, \u03b5 and n denote the rotor mass, eccentricity and rotational speed respectively. The rotational speed is the most effective design parameter to increase the unbalanced exciting force [62]. 9.9 Vibration Motors 311 312 9 Applications Moving Magnet Technologies single-phase AFPM brushless vibration motors provide a strong vibration feeling and silent alert with a very simple design (Fig. 9.31). Fabrication of MMT vibration motors is cost effective due to contactless design, scalable size and slim shape. The data storage capacity of a computer hard disc drive (HDD) is determined by the aerial recording density and number of discs. It is expected that the aerial density will soon increase from 6 Gb/cm2 = 38.7 Gb/in2 to 15.5 Gb/cm2 = 100 Gb/in2. The mass of the rotor, moment of inertia and vibration increase with the number of discs. Special design features of computer HDD motors are their high starting torque, limited current supply, low vibration and noise, physical constraints on volume and shape, protection against contamination and scaling problems"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure11.16-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure11.16-1.png",
        "caption": "FIGURE 11.16. A passing maneuver.",
        "texts": [
            " Passing and lane-change maneuvers are two other standard tests to examine a vehicle\u2019s dynamic responses. Passing can be expressed by a halfsine or a sine-squared function for steering input. Two examples of such functions are \u03b4 (t) = \u23a7\u23a8\u23a9 \u03b40 sin\u03c9t t1 < t < \u03c0 \u03c9 0 \u03c0 \u03c9 < t < t1 rad (11.276) \u03b4 (t) = \u23a7\u23a8\u23a9 \u03b40 sin 2 \u03c9t t1 < t < \u03c0 \u03c9 0 \u03c0 \u03c9 < t < t1 rad (11.277) \u03c9 = \u03c0L vx . (11.278) where L is the moving length during the passing and vx is the forward speed of the vehicle. The path of a passing car would be similar to Figure 11.16. Let\u2019s examine a vehicle with the characteristics given in (11.226)-(11.229) and a change in half-sine steering input \u03b4 (t). \u03b4 (t) = \u23a7\u23a8\u23a9 0.2 sin \u03c0L vx t 0 < t < vx L 0 vx L < t < 0 rad (11.279) L = 100m (11.280) vx = 30m/ s. (11.281) The equations of motion for zero initial conditions are as given in (11.271)- (11.274). Figures 11.17 to 11.20 show the time responses of the vehicle for the steering function (11.279). Example 420 F Passing with a sine-square steer function. A good driver should change the steer angle as smoothly as possible to minimize undesired roll angle and roll fluctuation"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003657_02783649922066475-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003657_02783649922066475-Figure6-1.png",
        "caption": "Fig. 6. The leg\u2019s structural details. All major connections are yoke and tube style, to minimize reliance on welds.",
        "texts": [
            " The leg must withstand all cases of foot loading (axial to the foot, transverse to the pantograph plane, and transverse normal to the pantograph plane) that might occur throughout the spectrum from flat terrain walking to vertical slope rappelling. There was specific concern for minor falls during rappelling which might result in the entire robot\u2019s dynamic load acting on only a few feet, primarily in a shearing (transverse in-plane) direction. The pantograph leg structure is designed for maximum strength-to-weight performance; all links are thin-wall tubes with tapered wall thickness (to follow moment loading) where possible. The four principal hinge joints (see Fig. 6) experience a variety of forces and moments resulting from axial and transverse foot loads. These joints employ a at East Tennessee State University on June 6, 2015ijr.sagepub.comDownloaded from Table 2. Leg Actuator and Linkage Specifications Actuator (integrated motor, brake, gearing, potentiometer, limit switch) Sussex GB-106L Continuous actuator-force output 11,564 N @ 1.4 cm/sec Resulting force and speed at foot 2,891 N @ 5.6 cm/sec Maximum speed at foot (no load) 7.9 cm/sec Maximum body-lifting speed (assuming 8 legs supporting and lifting) 7"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001017_robot.1986.1087552-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001017_robot.1986.1087552-Figure4-1.png",
        "caption": "Figure 4. Pr inc ipe of the new nota t ion",
        "texts": [],
        "surrounding_texts": [
            "I\nx . -x+ 1 2Fig. 1-b\n-i X\nFigure 1. Denavit and Hartenberg Notation\nand i s f ixed a l so wi th r e spec t t o l i n k (1). This frame i s def ined by some (8,r ,a ,d) parameters but what subsc r ip t s do w e have t o a s s i g n f o r t h e s e parameters ?\nA s o l u t i o n may proposed by the use of double subscripts s u c h t h a t when t r a v e r s i n g f r o m j o i n t 1 t o j o i n t 2 the parameters w i l l be denoted by (81ar12, a.lz,d12) and by (813,~13,~113,d13) when t r a v e r s i n g from j o i n t 1 t o j o i n t 3 , t h e j o i n t v a r i a b l e ql w i l l be 0 1 2 or 813, an addi t iona l cons tan t parameter which s p e c i f i e s the re la t ion be tween 012 and 8 1 3 is to be defined, i.e.\ne12 = e 1 3 + Y13 ( 3 )\nAnother confusion is s t i l l taking place because we have always two f r a m e s f i x e d w i t h r e s p e c t t o l i n k 1. How t o c a l l them R 1 2 and R13 ? A t any case t h e f r a - m e (i) i s no more f i x e d w i t h r e s p e c t t o l i n k i. We see from tha t s imple example that D-H n o t a t i o n w i l l loose one of i ts best advantage which is i t s s implic i t y . And the mathematical formulas used i n t h e r o - bot modeling (geometric-kinematics and especially dynamics) w i l l not be handy.\nFigure 3. Sheth and Uicker parameters\n3. SHETH AND UICKER NOTATION [ 4 ]\nOwing t o the i n e f f i c i e n c y of t h e D-H nota- t ion in r ep resen t ing the c losed loop s t ruc tu re , She th and Uicker have developed another notation system, where each transformation matrix is composed of two p a r t s :\ni) a cons tan t par t spec i fy ing the shape of the l i n k .\nii) a d i s t i n c t v a r i a b l e p a r t r e p r e s e n t i n g the j o i n t motion.\nConsider (Fig.3) which shows two success ive l i nks , each 3oint contains two coordinate sysrems. The f i r s t denoted sj xj gj is an a r b i t r a r i l y c h o s e n\nsystem with 2 , is t h e j o i n t axis, f ixed wi th r e spec t\nt o l i n k ( j ) and may bethought of as Locating the p o s i t i o n o f t h e j o i n t e l e m e n t R.-. The other coor-\ndinate system D . V . W , is a l s o d e f i n e d f i x e d i n t h e\nmat ing jo in t e lement R . + . I t is chosen such t h a t W . lies along the j o i n t a x i s 2 . b u t U . and V . are -7\na r b i t r a r i l y o r i e n t e d .\n-1\n1\n-7 -7 7\nI\n-1' -7 -1",
            "The mot ion of the jo in t ( j ) i s des igna ted by 9 . .\nNow, it is necessary t o d e f i n e the parameters which describe the shape of the l ink and a l so the parameters which describe the joint motion.\n3\n3 . 1 . Shape Matrix\nThe shape of each l ink (such as l i n k ( j ) i n F i g . 3 for example) , is s p e c i f i e d by t h e r e l a t i v e o r i e n t a - t ion between the coordinate system R . + a t t h e\n\"begining\" of the l ink and %- a t the \"following\" end. To determine the constant shape parameters for a l i n k , t h e common perpendicular is found between the two axes E. and 5. This common perpendicular i s ass igned an 'a rb i t ra ry pos i t ive d i rec t ion and is denoted as t ,k. Six parameters are r e q u i r e d t o d e f i n e the &ape of each link. They a r e d e f i n e d f o r l i n k ( j ) as shown i n (Fig. 3 ) according t o the f o l - lowing conventions :\na . = d i s t a n c e from W. t o Z+ measured about t .\n3\ni k -7 Jk a. = angle from p o s i t i v e W . t o p o s i t i v e 7.+ measured\n-7 Ik about t . l k\nb . = distance from t . t o X measured about Z 8 . = angle from t . to p o s i t i v e X+ measured about lk ~k -+ -k.\nlk % 3k\nrjk= d i s t a n c e from U t o t measured along W y , = angle f rom posi t ive U. t o p o s i t i v e t measu-j jk i\nIk red about W -7 j k -1\nF . is the shape matr ix of l ink (j) f o r t h e p a t h \u20ac&I j o i n t ( j ) to j o i n t (k), i ts genera l form is : <+ 'T- = F . k 3k\n3.2. Jo in t Ma t r ix\nThe j o i n t m a t r i x will be denoted by V . (q , ) where q 7 7 j is the jth j o i n t v a r i a b l e . F o r a k inemat ic jo in t (j) having the coordinate systems R . - and R.+ atta-\nched t o its preceeding and following elements respec t ive ly , the t ransformat ion be tween these coordinate systems is given by : V . ( q j ) . Two cases a r e t o be considered :\nR o t a t i o n a l j o i n t : the j o i n t v a r i a b l e q . is given by the angle between X . and U . and is cansidered\nposi t ive conterclockwise about Dosi t ive W . .\nHence,\n3 3\n3\n-7 -7\n-3\ne r 1 er\nP r i s m a t i c j o i n t : t h e j o i n t v a r i a b l e q . is t h e d i s t a n c e between X . and U . measured along W ., and 7\n-3 -7 -1\nis p o s i t i v e i f i n t h e d i r e c t i o n o f p o s i t . i v e Ej. Hence, IT.+ = V . ( q . ) = 1 @ 6 b - 7 3 3\n@ I @ @ (6) o e r 1 q . - - - - - L 1 @ d B 1\n3 . 3 . Transformation Matrix Tk\nThe t ransformation matr ix j T k between the coord ina te\nsystems R . - and %- shown i n (Fig. 3) is given by\nITk = V . ( q . ) F\nThe S-U notat ion has been used t o s tudy the closedloop robots [SI. But as w e see it i s not convenient t o use for a system where D-H can be used i .e. i n the case of a ser ia l robot. The complexity of t h i s no ta t ion has been poin ted out by Roth [81.\nj\n3\nJ J j k\n4. The Modified Notation\n4.1. In t roduct ion\nThe aim of t h e new n o t a t i o n is t o d e f i n e a method which can be used easi ly and without ambigui ty in the closed-loop robots. We t h i n k t h a t t h i s aim has been fu l f i led and w e w i l l show tha t t he g iven no ta - t ion can be used a l so for the open-loop robots as easy and general as that of D-H no ta t ion .\nThe proposed method defines the transformation mat r i x i n the case of two- jo in t l ink by the use Of 4 parameters as i n t h e D-H no ta t ion . In the case Of l i nks wi th more than t w o j o i n t s two addi t iona l parameters may be needed.\nThe proposed notation is def ined such tha t : . The a x i s o f j o i n t ( i) w i l l be gi . The coordinate frame R i (oi,Xi,yi,zi) is f ixed\nwi th respec t t o l i n k (i)\n. The parameters which lead t o d e f i n e frame (i) w i l l have (i) as subscr ip t .\nI176",
            "4.2. Open-loop Robots\nThe system is composed of n+l l inks , l ink (0) is the f ixed base , wh i l e l i nk (n ) is the te rmina l l ink . J o i n t (i) connects l ink ( i - I ) and l ink (i) . L e t :\nR. the f i x e d f rame wi th respec t to l ink (i)\nZ . t h e a x i s o f j o i n t (i)\nx. w i l l be def ined on the common perpendicular of\n1\n-1\n-1 Z . and 2. (Fig. 5 ) . -1 -1+ 1 The fol lowing parameters are required t o def ine the frame (i) w i t h r e s p e c t t o frame (i-1) : (Fig. 5)\nai angle between Z and 2. about zi-l d.. d is tance between 0 and Z i- 1 -i r . d i s t a n c e between 0 and &ii-l i 8. angle between a d X . about gi The v a r i a b l e o f j o i n t (i) denoted by q. is 8 i f (i) is ro ta t iona l and r i f (i) is prismatic. Hence -i- 1 -1 -1\ni i\nqi = 8,(1 - a . ) + ri ai where a , = B i f j o i n t (i) is ro t a t iona l and U.= 1 i f jo i .n t (i) is pr i smat ic .\nThe t ransformation matr ix i - l ~ . is equal t o :\ni-lT. = Rot(X,ai)Trans(X,d.)Rot(Z,8.)Trans(Z,ri)\ncosBi; -sin\u20ac$' I @'I di\nc o s a . s i n 8 . ~ c o s a . c o s e . l -sins. ,-r s i n a 1 1 1 1 1 1 i i - ( 7 )\ns ina . s in8 ' s i na . cose .1 cosa. ' r cosu 1 i l 1 1, 1 1 i i - - - - - + - - - - ----I----\nI -\n0 ' @ I @ ; 1 i\nI t i s to be no ted tha t :\n- R can be defined always such that R I R1 when\n51 = a which means t h a t : a1 = d l = q1 =@,where q. = 8 . 0. + r . ( 1 - a . ) - X can be taken along X i f a= 0, which means\nthat S = s. The proposed notation i s similar t o t h a t o f D-H. I n f a c t it is t o b e n o t e d t h a t ( 8 . , r .\n-\n1 1 1 1\n-n -n- 1\nn\n1 r ' a i + l r d i + l ) of the\nmodi f ied nota t ion a re respec t ive ly (Oi , r i , a . ,d i ) o f D-K . It has been no ted tha t t h i s de f in i t i on o f t he frame R . f i x e d w i t h l i n k i s u c h t h a t Z . is a long the ax i s 0% j o i n t (i) , l e a d s a l s o t o sim&fy t h e Zynamic model o f t he robo t [9 ] .\nThe geometric model of the robo t w i l l be obtained by the success ive mul t ip l i ca t ions o f t he t r ans fo rma t ion mat r ices : O T ~ = O T ~ ' T ~ ... n- 1 Tn ( 8 ) 4.3. Tree-Structure Robots\nThe system t o b e c o n s i d e r e d i n t h i s c a s e is composed of n+l l inks , n j o in t s and m end ef fec tors .\nThe l i n k s and the joints w i l l be numbered as fo l lows (Fig. 6 )\nthe base w i l l be considered as l i n k @ t h e numbers of l i n k s a n d j o i n t s a r e i n c r e a s i n g a t each branch when t ravers ing f rom the base t o an end effector l i n k (i) is a r t i c u l a t e d on j o i n t ( i ) , i .e. j o i n t (i) connects the l ink ( a ( i ) ) and l i nk ( i) , where a ( i ) is the number of the l i n k a n t e c e d e n t t o l i n k (i) when coming from the base frame (i) is de f ined f ixed wi th r e spec t t o l i n k (i) , and Z is the a x i s o f j o i n t (i) .\n-i I n t h e case of a l i nk wi th two j o i n t s t h e frame a t the e n d o f t h i s l i n k , s a y j , w i l l b e d e f i n e d w i t h r e s p e c t t o the frame a t the begin ing of the l i n k , say i = a ( j ) , e x a c t l y a s i n t h e case of open-loop robot descr ibed previous ly , i.e by the a id o f 4 parameters ( a . d O.,r ). X . is t h e common perpendicu-\nl a r on Z and, Z . I n the case o f l i nks w i t h more than two j o i n t s (Fig. 7) , w e def ine the d i f fe ren t f rames as fol lows\n- f i n d t h e common p e r p e n d i c u l a r s t o Z_. and each of the succeeding axis ,on the same l i n k ti), Z j where i = a ( j ) ( j = k. ,R,m ,... ). - l e t cne of these common perpendiculars be the Zi a x i s , it is p r e f e r e d t o t a k e X . as that corresponding t o t h e common perpendicular of the joint on which is a r t i cu la t ed the longes t b ranch , say k ,\nJ ' j ' 1. j T.\ni-j\n-1\n1177"
        ]
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure11.14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure11.14-1.png",
        "caption": "Figure 11.14 The support reactions are found from the V diagram as the shear forces at the member ends.",
        "texts": [
            " VOLUME 1: EQUILIBRIUM In other words: the bending moment is extreme where the shear force is zero. This location can be found from (a): V = 1 6\u03c1gb(h2 \u2212 3x2) = 0 \u21d2 x = 1 3h \u221a 3. Substituting this value of x in the expression for M gives the maximum bending moment: Mmax = M (x= 1 3 h \u221a 3) = 1 6\u03c1gb { h2 ( 1 3h \u221a 3 ) \u2212 ( 1 3h \u221a 3 )3 } = \u221a 3 27 \u03c1gbh3 = 0.064\u03c1gbh3. The support reactions can be found from the V diagram as the shear forces on the beam ends: x = 0 : V = + 1 6\u03c1gbh2, x = h : V = \u2212 1 3\u03c1gbh2. These shear forces on the boundaries of the beam are shown in Figure 11.14. As a check, one can be examine whether the beam as a whole is in equilibrium. The resultant R = 1 2\u03c1gbh2 of the distributed load acts at x = 2 3h. This gives \u2211 Fz = R \u2212 1 6\u03c1gbh2 \u2212 1 3\u03c1gbh2 = 0,\u2211 Ty |B = R \u00b7 1 3h \u2212 1 6\u03c1gbh2 \u00b7 h = 0. Force and moment equilibrium therefore are satisfied. 11 Mathematical Description of the Relationship between Section Forces and Loading 451 Example 4 Cantilever beam ABC in Figure 11.15a is simply supported at A and B, and has an overhang BC at B. The beam is carrying a uniformly distributed load of 40 kN/m over its entire length"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000431_adma.201100131-Figure16-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000431_adma.201100131-Figure16-1.png",
        "caption": "Figure 16 . Photoinduced sophisticated 3D motions of LCE-laminated fi lms based on fl exible polyethylene sheet. A) Schematic illustration of a light-driven plastic motor and photographs of time profi les of the rotation (adapted from Ref. [168a]). B) Photographs of photoinduced inchworm walk and the plausible mechanism. Reproduced with permission from Ref. [159d]. Copyright 2008, RSC.",
        "texts": [
            " Upon simultaneous irradiation with UV light (from the downside right) and visible light (from the upside 2169bH & Co. KGaA, Weinheim wileyonlinelibrary.com 2170 R EV IE W right), the ring rolled intermittently towards the actinic light source, resulting almost in a 360 \u00b0 roll at room temperature. To improve the mechanical properties of AZ-containing LCEs, they prepared a plastic belt of the LCE-laminated fi lms by attaching LCE fi lms on fl exible polyethylene sheet, and then placed the belt on a homemade pulley system as illustrated in Figure 16 A. Upon irradiation of the belt with UV light from top right and visible light from top left simultaneously, a rotation of the belt was induced to drive the two pulleys in a counterclockwise direction at room temperature. This is the fi rst realization of light-driven plastic motors, which directly converts light to mechanically rotational energy with soft materials. It is believed that the sections that were exposed to light expand while those regions away from the light contract, generating an overall rotating moment of the plastic fi lms. In addition to the photodriven rotation, the LCE-laminated fi lms exhibited other photomobility of sophisticated 3D motions like inchworm walk and a fl exible robotic arm (Figure 16 B). [168b] The LCE-laminated fi lm moves forward upon alternate irradiation with UV and visible light at room temperature. As light can be handled remotely, instantly and precisely, these plastic materials can work as main driving parts of light-driven actuators without the aid of batteries, electric wires and gears. LCBCs with at least one block of a photoresponsive LCP in their constitution are among the novel types of macromolecules of industrial and academic interest because they provide novel opportunities to study the nanostructure formation and \u00a9 2011 WILEY-VCH Verlag Gwileyonlinelibrary"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure6.7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure6.7-1.png",
        "caption": "Fig. 6.7. Exploded view of the SEMA AFPM brushless machine. Courtesy of Lynx Motion Technology, Greenville, IN, U.S.A.",
        "texts": [
            " SEMA technology may also find applications in distributed generation systems and energy storage systems, such as flywheel motor\u2013generator systems. Motors with high torque and efficiency are demanded for gearless electromechanical drives. The use of direct drive motors without speed reducers eliminates gear noise, oil leaks, positioning errors from backlash and lack of torsional stiffness. 6.5 Commercial Coreless AFPM Motors 199 200 6 AFPM Machines Without Stator and Rotor Cores The stator coreless and toothless design (Fig. 6.7) not only eliminates the cogging torque but also increases the area available for conductors. It also boosts peak torque capability and allows PMs to be more effectively utilized [151]. The stator coils are fully potted in high-strength, thermally conductive epoxy resin. Such construction gives the machine structural integrity and effectively damps the high frequency vibration when the motor is PWM inverted-fed. Machines with coreless stator winding perform excellently at constant speed, variable speed and at reversals"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure12.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure12.1-1.png",
        "caption": "FIGURE 12.1 Example cell with component of incident field.",
        "texts": [
            " Recently, more progress has been made in the computation of the radiated power [9, 10] and [13\u201315] using PEEC. We start this section with an independent external field applied in the presence of a PEEC model structure. EXTERNAL INCIDENT FIELD APPLIED TO PEEC MODEL 311 The new part of the incident field formulation is the collection of the electric field by the left-hand side of (12.1). Essentially, we need to add the source voltage along the inductive branch of the cell due to the incident electric field Einc(r, t). We integrate the external fields over the cell shown in Fig. 12.1, or Ve(t) = 1 \u222b\u222b\ud835\udcc1 Einc(r, t) \u22c5 d\ud835\udcc1 d , (12.2) where Einc(r, t) = Einc x (r, t) x\u0302 + Einc y (r, t) y\u0302 + Einc z (r, t) z\u0302. (12.3) These sources are simply added to the basic PEEC loop, which is an addition to the circuit based on part of the Kirchhoff\u2019s voltage law (KVL). Hence, it is obvious that the extension of a PEEC model with the electric field is straightforward (Fig. 12.2). The right hand side basic loop in Fig. 12.2 represents the PEEC model for a dielectric with the incident field applied"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002414_s11837-018-3207-3-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002414_s11837-018-3207-3-Figure5-1.png",
        "caption": "Fig. 5. Shrinkage analysis of artifact part, (a) CAD file, (b) green part, and (c) metal part.",
        "texts": [
            " The SLM SS 316L showed slightly higher hardness than AISI type SS 316L, in agreement with the microstructural analysis and discussion of the tensile properties above. Various studies have been carried out on accuracy analysis and error compensation for SLM metal parts. A thermal model and experiments are usually coupled to compensate the geometrical variation of SLM parts.12,13 Therefore, this section only concentrates on the shrinkage of the FDM SS 316L metal part. Shrinkage analysis was performed on an artifact part printed using Ultrafuse filament, as shown in Fig. 5. A set of features of interest (FOIs) were marked on the computer-aided design (CAD) model with a sequence number. The dimension of each FOI on the green part and metal part was measured to compare the size variation and determine the shrinkage rate. The green part was measured using the Keyence VR-3000 at the FOIs before being sent for debinding and sintering. The metal (sintered) part (Fig. 5c) was also measured using the same equipment to compare with the FOIs of the green part and thus estimate the shrinkage. As shown in Fig. 5c, the metal part apparently shrank after the sintering process. The metal part had similar shape characteristics but decreased size due to the shrinkage. However, the major FOIs retained the features of the geometrical configuration, ensuring that the FDM-based metal AM can form metal parts using Ultrafuse filament, as long as the shrinkage is compensated when printing the green part. The size of the FOIs in the x\u2013y plane (diameter and edge length) and along the z-axis (height and depth) are of great interest in this study",
            " A feature at a lower level may thus shrink less than the same feature at a higher level. Also, greater height or depth variation tends to occur at lower levels along the z-axis. Overall, the calculated percentage shrinkage rate along the zaxis was higher than for the geometrical features in the x\u2013y plane, which can be attributed to the effect of gravity on the metal component during the sintering process. In addition, a couple of side holes were designed with multiple geometries to verify that unsupported features could be printed. Figure 5b shows the green part of the artifact which was directly printed using Ultrafuse filament. Note that all the unsupported features were printed without apparent distortion. This demonstrates that the extruded FDM SS 316L SLM SS 316L AISI type SS 316La Yield strength (MPa) 167 541 205 UTS (MPa) 465 648 515 Elongation at break (%) 31 30 60 Young\u2019s modulus (GPa) 152 320 193 ahttp://asm.matweb.com/search/SpecificMaterial.asp?bassnum=mq316q. filament exhibits great shape retention due to its fast phase transformation from viscous to solid state"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000262_s002211207100048x-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000262_s002211207100048x-Figure3-1.png",
        "caption": "FIGURE 3. Hemispherical surface of organism at N = 17, with coefficients given in table 1.7. (- - -) is . t = $T later than (-).",
        "texts": [],
        "surrounding_texts": [
            "In the calculations we have taken two examples a t N = 17 and N = 2 2 , to show the shape of the organism. For this case we have taken u = 25sec-1 and a = 100pm as this is indicative of the angular frequency of movement of the cilium and length of the organism concerned respectively. (Opalina is 200- 300pm long, wave velocity 100-400pm/sec, cilium length 10-15pm, beat frequency 1-4 per see, velocity of propulsion 100-200pm/sec.) For conversion from these dimensional units we, however, note from (20) that is proportional to a u to this order of calculation for the velocity, Velocity (pmlsec) and efficiency for N = 17. E = 0.05 A spherical envelope approach to ciliary propulsion 207 The velocities and efficiencies are shown in tables 1 and 2, having chosen e = 0.05 in both cases. The a, and b, are taken such that the series are of order 1. We may take the a, to be O(n4) and the b, to be O(n%), provided the restriction d6jd6, > 0 is not violated. The velocity of the organism can have both positive and negative sign, this being due to either of two things: (i) the direction in which the wave is progressing or (ii) the effect of longitudinal and transverse oscillations tending to propel the sphere in opposite directions. This second case (ii) corresponds with the observations of Tuck (1968) that longitudinal and transverse oscillations tend to propel an infinite sheet in opposite directions, so that on combination we may have either a positive or negative velocity. Large velocities occur, relatively speaking, when each b, has opposite sign to the corresponding a,. These larger velocities of the order of 100pm/sec are comparable with those observed for Opalina. The efficiency allows us to compare the work done in the ciliary movements to the workdone by an external force pushing an inert organism at the same velocity, and as well enables us to compare the efficiency of various modes. In conclusion, this model for ciliary propulsion of a sphere compares favourably to the velocities experienced in nature and is therefore a quite amenable approach to the problem. This work wa,s carried out while the author was in receipt of a George Murray Scholarship from the University of Adelaide, and a studentship from C.S.I.R.O. 208 J . R. Blake of Australia. The research was carried out at the Department of Applied Mathematics and Theoretical Physics, University of Cambridge. Comments and suggestions from Professor M. J. Lighthill and filmplates and data from Dr M. A. Sleigh are gratefully acknowledged. R E F E R E N C E S GAUNT, J. A. 1929 The triplets of helium. Phil. Trans. Roy. SOC. 228 A, 151-196. GRAY, J. 1928 Ciliary Movement. Macmillan. HAPPEL, J. & BRENNER, H. 1965 Low Reynolds Number Hydrodyfzamics. Prentice-Hall. HOLWILL, M. E. J. & SLEIGH, M. A. 1969 Energetics of ciliary movement in Sahellaria and Mytilus. J. Exp. Biol. 50, 733-743. LIGHTRILL, M. J. 1952 On the squirming motion of nearly spherical deformable bodies through liquids a t very small Reynolds numbers. Comm. Pure Appl. Math. 5,109-118. LIGHTHILL, M. J. 1969 Hydromechanics of aquatic animal propulsion. Ann. Rev. of Fluid Mech. 1, 413-446. SLEIGH, M. A. 1962 The Biology of Cilia and Flagella. Pergamon. SLEIGH, M. A. 1968 Patterns of ciliary beating. In Aspects of Cell Motility, pp. 131-150. TAYLOR, G. I. 1951 Analysis of the swimming of microscopic organisms. Proc. Roy. SOC. TWCE, E. 0. 1968 A note on a swimming problem. J . Fluid Mech. 31, 305-308. Cambridge University Press. A 209, 447-461."
        ]
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure8.22-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure8.22-1.png",
        "caption": "Figure 8.22 (a) The grain stresses on a triangular soil element that is upset (moving upwards). (b) The equations for the force equilibrium in the plane of the drawing can easily be derived from the fact that the force polygon is closed.",
        "texts": [
            " The method with slide planes has the advantage that it is relatively easy to see that a terrain load on ground level increases the horizontal earth pressure on the wall only if it acts on the sliding soil wedge. Comment: For the influence of wall friction, oblique walls, or oblique ground levels, as well as the necessary nuances in the examples shown here, please refer to a text book on the field of soil mechanics. 302 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM The coefficient Kp for passive earth pressure can be derived using the triangular wedge of soil in Figure 8.22a, in the same way as the coefficient for active earth pressure in Section 8.3.1. The equations for the force equilibrium in the plane of the drawing can be found from the closed force polygon in Figure 8.22b. Force equilibrium perpendicular and parallel to PQ: \u03c3g A \u2212 \u03c3g;h( A cos \u03b1) cos \u03b1 \u2212 \u03c3g;v( A sin \u03b1) sin \u03b1 = 0, \u03c4g;max A \u2212 \u03c3g;h( A cos\u03b1) sin \u03b1 + \u03c3g;v( A sin \u03b1) cos \u03b1 = 0 so that \u03c3g = +\u03c3g;h cos2 \u03b1 + \u03c3g;v sin2 \u03b1, \u03c4g;max = +\u03c3g;h sin \u03b1 cos \u03b1 \u2212 \u03c3g;h sin \u03b1 cos \u03b1. The shear stress is a maximum, therefore \u03c4g;max = \u03c3g tan \u03d5. In this expression substitute the expressions found for \u03c3g and \u03c4g;max: +\u03c3g;h sin \u03b1 cos \u03b1 \u2212 \u03c3g;v sin \u03b1 cos \u03b1 = (+\u03c3g;h cos2 \u03b1 + \u03c3g;v sin2 \u03b1) tan \u03d5. 8 Earth Pressures 303 This gives the coefficient for passive earth pressure1: Kp = \u03c3g;h \u03c3g;v = 1 + 2 sin \u03d5 sin(2\u03b1 \u2212 \u03d5) \u2212 sin \u03d5 "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003443_0278364903022001005-Figure12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003443_0278364903022001005-Figure12-1.png",
        "caption": "Fig. 12. 2Txy1Rz 3-DoF 3 \u2212 zRzRzR(iRjR)N PM.",
        "texts": [
            "comDownloaded from The standard base of the mechanism twist system is given by $m1 = (0 0 1 ; 0 0 0) $m2 = (0 0 0 ; 1 0 0) $m3 = (0 0 0 ; 0 1 0). (45) The standard base of the mechanism constraint system is given by $r m1 = (0 0 1 ; 0 0 0) $r m2 = (0 0 0 ; 1 0 0) $r m3 = (0 0 0 ; 0 1 0). (46) When the limb kinematic chain consists of five kinematic pairs, it only exerts one constraint. If the constraint is a couple, no translations can be constrained. So the constraint must be a force and the combined effect of all the p \u00b7 q forces must equal two couples and one force. Figure 12 shows a 3-DoF 3 \u2212 zRzRzR(iRjR)N PM. The three limb central points also form a triangle, which is fixed relative to the moving platform and is parallel to the moving platform plane. Note that each limb constraint force is parallel to the first revolute axis and passes through the corresponding limb central point. It is obvious that the three limb constraint forces are parallel in space. Thus, the standard base of the mechanism constraint system is the same as eq (46) and the moving platform loses one translational DoF in the Z-axis and two rotational DoFs in the XY plane"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001252_j.ymssp.2007.12.001-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001252_j.ymssp.2007.12.001-Figure6-1.png",
        "caption": "Fig. 6. Rolling element spall depth as sensed by the inner race (not to scale).",
        "texts": [
            " Note that the switch values and the periods of switching on (Dfd) will not be the same for both races, which results from the difference in curvature between the two races and this is also a function of the rolling element diameter. The inner race will contact deeper and longer compared to the outer race. Instead of defining two spall geometries, the fault bj switch can be modified to show this variation as sensed by the different races as in Eq. (19). bj \u00bc 0 jak; 1 if 0ofsoDfdo Cdr\u00feCdi Cdr Cdo if piofso\u00f0pi \u00fe Dfdi\u00de ) j \u00bc k; 0 otherwise: 8>>< >>>: (19) The derivation of the switch values is illustrated in Fig. 6 (not to scale), when the spalled rolling element is in touch with the inner race. For this case, the total loss of contact, as a result of creating a slot of width 2x, is the summation of two values. The first is the maximum depth the inner race will enter in the slot region (Cdi as calculated in Eq. (20), which represents the movement of the inner race from its original contact position\u2014 moving downwards). The second represents the movement of the ball from its original contact position as a result of creating the slot (Cdr as presented in Eq"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.31-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.31-1.png",
        "caption": "FIGURE 8.31. A car with a McPherson suspension system.",
        "texts": [
            " The Kennedy theorem states that the instant center of every three relatively moving objects are colinear. Example 316 McPherson suspension roll center. A McPherson suspension is an inverted slider crank mechanism. The instant centers of an example of inverted slider crank mechanism are shown in Figure 8.30. In this figure, the point I12 is the suspension roll center, which is the instant center of rotation for the wheel link number 2 with respect to the chassis link number 1. A car with a McPherson suspension system is shown in Figure 8.31. The kinematic equivalent mechanism is depicted in Figure 8.32. Suspension roll centers along with the body roll center are shown in Figure 8.33. To 8. Suspension Mechanisms 473 find the roll center of the front or rear half of a car, we determine each suspension roll center and then find the intersection of the lines connecting the suspension roll centers to the center of the associated tireprint. Example 317 Roll center of double A-arm suspension. The roll center of an independent suspension such as a double A-arm can be internal or external"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure4.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure4.6-1.png",
        "caption": "Fig. 4.6: A gripper with interchangeable fingers with different fingertips: a) a scheme for a mechanical design; b) an industrial gripper for robots.",
        "texts": [
            " Such design simplicity can help the maintenance of a robotic system, and also its operation both in terms of programming and integrated performance with robotic arm. Indeed, such integration can be considered essential to achieve a suitable versatility of the robotic system when it is completed with a grasping device. Usually, grasping devices and particularly grippers are designed with a certain modularity so they can be easily adjusted to different grasping situations, mostly by interchanging fingers or fingertips as shown in the illustrative cases of Fig. 4.6. The above-mentioned requirements are usually obtained in such a way that industrial grasping devices are not equipped with sensors for controlling the grasp that can be ensured by a previous analysis of the grasp mechanics and grasping operation/strategy. Increasingly, an industrial finger can be sensored through a force sensor or rarely by means of tactile sensors in order to control grasping. However, most of the industrial grippers are made to operate with two fingers, since most of the grasps can be performed with two fingers only and two-finger grippers are the smallest suitable mechanical architectures for grasping hand devices",
            " Therefore, another design problem is related to the question: is it possible to design and properly maintain the grasping mechanical characteristics within finger interchangeable solutions? At the most a multi-task capability for a two-finger gripper consists of gripping several objects with different shape and size, but within limited ranges. This can be achieved by using properly shaped fingers and from a general viewpoint by adopting interchangeable fingers each of which is designed for a specific grasping task, as the case shown in Fig. 4.6. Indeed, fingertips are the only finger part that can be interchangeable as function of the size and shape of the objects. In addition, when a multi-task purpose is concerned with different mechanical operations a solution may be adopted by using a two-finger gripper as an intermediate device to temporally install a required specific end-effector. Another fundamental aspect for gripper design can be identified in the type of finger motion. Two types of finger motion may occur in gripping mechanisms, as shown in Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003977_j.bpj.2010.05.015-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003977_j.bpj.2010.05.015-Figure1-1.png",
        "caption": "FIGURE 1 (a) The sperm structure consists of a spherical head (red), a curved midpiece (yellow to light blue), and a beating tail (dark blue to light blue). A traveling sinusoidal deformation of the tail generates a forward thrust. The curvature plane of the midpiece is tilted by an angle p/3 with respect to the beating plane. (b) Snapshot of a sperm with preferred curvature c0 (m)Lm\u00bc 1 of the midpiece (MCEmodel), moving along a helical trajectory. The trajectory is indicated by the small gray spheres. See also Movie S1, which is published as Supporting Material on the Biophysical Journal web site. Visualization using VMD (24).",
        "texts": [
            " Bending of the tail is generated by motor proteins that move along microtubules and, thereby, generate a local bending moment (11,18). We mimic this sperm structure by a spherical head of radius rh and a flexible flagellum of length Lf. The flagellum is modeled by three semiflexible rods, each consisting of Nf \u00bc 100 monomers; the monomers are connected by bonds of length \u2018b and are arranged in a filamentous structure with a triangular cross section (for details, see Appendix). The flagellum is divided into a passive midpiece of length Lm and an actively beating tail (Fig. 1 a). We model the force generation on the mesoscopic scale by imposing locally a preferred curvature c0(t;s) of the tail at time t, where 0 < s < Lf is a measure of the contour length along the flagellum. Propulsion is induced by a sinusoidal propagating wave c0\u00f0t; s\u00de \u00bc A=Lf sin\u00f0qs ut\u00de (1) along the tail, with Lm < s < Lf. This wave generates an approximately sinusoidal planar beat. Analysis of the flagellar beat of bull sperm showed that 95% of the power consumption is contributed by the first Fourier mode (19), suggesting that a sine wave describes the beat pattern fairly well",
            " To elucidate the impact of various shapes and beating modes on the swimming pattern of sperm from different species, we study two classes of chiral sperm models. The average flagellar shape is determined by the preferred curvature c0\u00bc 1/R0, where R0 is the local radius of curvature. The models differ by the distribution of the preferred curvature c0 along the flagellum. In the first class of models, the midpiece is curved, but the tail is straight. The curvature plane of the midpiece is tilted with respect to the beating plane by an angle p/3 (Fig. 1 a). Moreover, the flagellum is either elastic or stiff. While the bending rigidity is similar for elastic and stiff flagellum, Biophysical Journal 99(4) 1018\u20131026 the torsional rigidity is much larger for the stiff flagellum. This difference has consequences for the dynamic distortion of the flagellum during beating and, thereby, swimming behavior. We denote these two models the \u2018\u2018midpiece curved elastic\u2019\u2019 (MCE) and the \u2018\u2018midpiece curved stiff\u2019\u2019 (MCS). A second class of model is characterized by a preferred curvature c0 along the entire flagellum, not just the midpiece",
            " The friction of the head monomers was estimated by dividing the friction coefficient 6phrh of a sphere of radius rh by the number of monomers. This implies F \u00bc gv with g \u00bc 3 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi kBT m=a2 p for the head monomers. Sperm motion in the bulk with mesoscale hydrodynamics The simulation of MCE sperm with significant asymmetry (midpiece curvature c0 (m)Lm \u00bc 1) produces a pronounced helical trajectory, with the tail pointing away from the helical axis (Fig. 1 b). We analyzed three characteristic parameters of the helical motion of sperm: the curvature C of the trajectory, the tangential velocity v, and the rotation frequency Ub of the beating plane around the flagellar axis. The tangential velocity is fairly constant (v x 0.0034uLf) across all sperm models and preferred curvatures c0; it decreases by ~30% for strongly asymmetric sperm. The curvature C of the trajectory, however, strongly depends on the sperm model and the preferred curvature c0 of the flagellum (Fig",
            " For the TCS model, the curvature C of the trajectory and the preferred curvature c0 (t) are nearly identical (Fig. 2), because the shape of the tail almost perfectly tracks the trajectory of the sperm head. In contrast to the linear relationship between C and c0 (m) of sperm with stiff flagellum (MCS, TCS), sperm with an elastic flagellum (MCE) show a highly nonlinear dependence (Fig. 2). This nonlinearity originates from the deformation of the tail for c0 (m)Lm T 0.5: the flagellum twists and bends due to the forward thrust of the tail and the hydrodynamic friction of the head (Fig. 1 b), which leads to a larger curvature of the trajectory than expected from c0 (m) alone (Fig. 2). Our simulations show that the rotation frequencyUb equals the pitch frequency\u2014as expected for any swimmer with helical trajectory (1). For stiff sperm (MCS and TCS), Ub increases monotonically with c0 (m) (Fig. 3). In contrast, elastic MCE sperm display a more complex behavior: in the low- curvature regime, the beating plane rotates faster than in stiff MCS sperm; initially, Ub rises steeply, reaches a maximum at c \u00f0m\u00de 0 Lmx0:35, then decreases again, and for c0 (m)Lm R 0",
            " The spherical sperm head, which is attached at its center to the flagellum, is constructed from Nh \u00bc 163 monomers. The head has a radius rh \u00bc 4 \u2018b and is held together by springs. The flagellum has a length Lf \u00bc (Nf 1)\u2018b, with Nf \u00bc 100, where 4 \u2018b connect the center of the head to its surface (for stable anchoring), followed by a passive midpiece of length Lm \u00bc 15\u2018b and an active principal piece (tail) of length Lt \u00bc 80\u2018b. We mimic force generation in the flagellum by changing the length of bonds in one of the semiflexible filaments (Fig. 1 a and Fig. 10). A length change (\u20180(t, s) \u2013 \u2018b) of a bond generates a local spontaneous curvature c0\u00f0t; s\u00de \u00bc \u20180\u00f0t; s\u00de \u2018b \u20182bsin\u00f0p=3\u00de : (3) A sinusoidal variation of the bond length as a function of contour length and time then generates the sinusoidal propagating wave of c0(t, s) in Eq. 1. An additional permanent shortening of bonds in one of the rods generates a preferred curvature (Fig. 1 a). This description is similar to but not identical with the sinusoidal beat pattern used in previous models, where the time-dependent flagellar shape is prescribed (3). Five figures, 11 equations, and three movies are available at www.biophys. org/biophysj/supplemental/S0006-3495(10)00615-6. 1. Crenshaw, H. C. 1996. A new look at locomotion in microorganisms: rotating and translating. Am. Zool. 36:608\u2013618. 2. Woolley, D. M. 2003. Motility of spermatozoa at surfaces. Reproduction. 126:259\u2013270. 3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure6.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure6.3-1.png",
        "caption": "FIGURE 6.3 Example cell division with both inductive filaments and capacitive cells.",
        "texts": [
            " For example, a remotely located antenna does not couple at dc. Then, such a solution is appropriate. From a circuit\u2019s point of view, this is similar to a high pass filter where the low frequencies are irrelevant. Hence, this approach is mostly suitable for problems that do not require a dc solution. In the PEEC approach, since the capacitance and inductances are separate, a dc to daylight full-spectrum solution is possible by using the continuity equation (6.16). The continuity equation needs to be satisfied at each PEEC node as shown in Fig. 6.3. We can see four inductive cells joining a capacitance surface area. 140 BUILDING PEEC MODELS First, we apply a Gaussian surface over the area that corresponds to the node at the center of the volume covered with the capacitive cell  on both top and bottom. It is clear that, if we apply Gauss surfaces to each cell in Fig. 6.3, we will wrap up the entire volume of the center part of the conductors. Multiple applications of the continuity equation will cover the entire area. Surface charge is present on both surfaces of the conductive cells of the finite thickness conducting cells shown in Fig. 6.3. The top and bottom surfaces  have capacitive charge. In the following section, it is shown that the charge is on the surface only. The Gaussian volume  over which we integrate the continuity equation also includes the side surface where the current flows. This results in \u222b \u2207 \u22c5 J d = \u222e J \u22c5 n\u0302 d = \u2212 \ud835\udf15 \ud835\udf15t\u222b q(r) d , (6.17) where one of the Gaussian volumes  and the surface  includes all sides as indicated in Fig. 6.3, where the charge is on top and bottom and the current flows through the internal surfaces . By comparing this result with Fig. 6.3, we see that the normal direction current flow occurs through the four partial inductance cross sections  of the conductors. Since they are interior to the structure, the charge density q(r) is zero. We notice that the integrals over the side cross sections are \u222e J \u22c5 n\u0302 d = I, (6.18) where I is the total current flowing into the four partial inductance cells. On the other hand, the conductor top and bottom surfaces are charged with surface charge densities q(r). Integrating over both top and bottom surface will yield \u222b q(r) d = Qtop + Qbottom",
            " The details of the derivation are given in Section 10.4.6. The continuity equation for the rectangular PEEC models is considered in Section 6.3.1. The continuity equation must also be satisfied for the nonorthogonal case at the cell level for the currents and charges. Its differential form is given by (3.3) or \u2207 \u22c5 J + \ud835\udf15q \ud835\udf15t = 0 where again J is the current density and q is the surface charge density. The continuity equation needs to be applied at the location of each node corresponding to Figs. 7.3\u20137.5. Unlike in Fig. 6.3, for the model we only show a quarter of the area in Fig. 7.9 for which the continuity equation is applied. Here, we show only one quarter of the centre cell unlike in Fig. 6.3, which shows the cell for rectangular coordinates. Since only one quarter of the elements surrounding the node is shown, we assume that the surface element in Fig. 7.9 may be connected to other similar surfaces along the a\u2013c and the b\u2013c surfaces. Hence, the volume for which the continuity equation is applied consists of the corners that are involved in the geometry surrounding the node(s). It is sufficient to consider only the corner elements by ignoring the internal surfaces shown in Fig. 7.9 for simplicity"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001514_j.engfailanal.2015.06.004-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001514_j.engfailanal.2015.06.004-Figure4-1.png",
        "caption": "Fig. 4. Model of the cracked gear tooth [30]: (a) schematic of spur gear tooth, (b) schematic of a crack.",
        "texts": [
            " Based on the TVMS calculation model [26], Zhang et al. [29] analyzed vibration signals with or without tooth crack. The signals were acquired at different sampling frequencies, which were determined based on the sampling theorem. Wu\u2019s model [28] assumed the gear tooth to be a variable section cantilever beam, which is clamped on the base circle where the potential energy existing in the part between the base circle and the root circle is ignored. Considering the gear tooth clamped on the dedendum circle (see Fig. 4a), Chaari et al. [30] derived an analytical model to calculate the TVMS using bending, fillet-foundation and contact deflections. Based on the model, the effects of a tooth root crack on the TVMS (see Fig. 4b) were analyzed, and the model was verified by an FE method. On the basis of Charri\u2019s method [30], Ma and Chen [31] computed the TVMS of a cracked gear pair, and analyzed the dynamic responses of the cracked gear system using a four-degree-of-freedom (4-DOF) model. Wan et al. [32] presented different TVMS calculation models for different root circle sizes, and how to select these models need judge whether the number of the teeth is less than (see Fig. 5a) or great than 42 (see Fig. 5b). Based on these models, they also analyzed the influences of the geometric transmission error, TVMS and bearing stiffness on the dynamic responses of a cracked gear rotor system"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure9.28-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure9.28-1.png",
        "caption": "Fig. 9.28. Miniature two-phase, four-coil, 4-pole, single-sided AFPM rotary actuator. Photo courtesy of Moving Magnet Technologies, SA, Besancon, France.",
        "texts": [
            " The stator is mounted on a single-sided printed circuit board. Each part can be manufactured using standard moulding, stamping techniques and automatic coil winding with a specific design for simple and efficient assembly. Three position 9.8 Miniature AFPM Brushless Motors 309 sensing methods for the closed-loop control are used: (a) digital Hall probes located in the stator, (b) EMF signal (sensorless mode) and (c) absolute analogue position sensor (position sensor mode). MMT AFPM miniature rotary actuators (Fig. 9.28) have been designed for automotive applications to provide an efficient, contactless, direct drive rotary motion on a limited stroke. Ring, disc and tile shaped PMs have been used. The main features of this family of actuators are: \u2022 contactless actuation principle; \u2022 constant torque independent of the angular position for a given current; \u2022 linear torque\u2013current characteristic; \u2022 two directions of rotation; \u2022 high torque density. 310 9 Applications Additional functions, such as a magnetic return spring or analogue contactless position sensing can also be implemented"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure1.15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure1.15-1.png",
        "caption": "FIGURE 1.15. Illustration of a wheel and its dimensions.",
        "texts": [
            " Tire and Rim Fundamentals 21 Front tires, especially on front-wheel drive vehicles, wear out more quickly than rear tires. 1.7 Wheel and Rim When a tire is installed on a rim and is inflated, it is called a wheel. A wheel is a combined tire and rim. The rim is the metallic cylindrical part where the tire is installed. Most passenger cars are equipped with steel rims. The steel rim is made by welding a disk to a shell. However, light alloy rims made with light metals such as aluminium and magnesium are also popular. Figure 1.15 illustrates a wheel and the most important dimensional names. A rim has two main parts: flange and spider. The flange or hub is the ring or shell on which the tire is mounted. The spider or center section is the disc section that is attached to the hub. The rim width is also called pan width and measured from inside to inside of the bead seats of the flange. Flange provides lateral support to the tire. A flange has two bead seats providing radial support to the tire. The well is the middle part between the bead seats with sufficient depth and width to enable the tire beads to be mounted and demounted on the rim",
            " The well is built to make mounting and demounting the tire easy. The bead seats are around 5 deg tapered. Wide drop center rims (WDC) are wider than DC rims and are built for low aspect ratio tires. The well of WDC rims are shallower and wider. Today, most passenger cars are equipped with WDC rims. The WDC rims may be manufactured with a hump behind the bead seat area to prevent the bead from slipping down. A sample of rim numbering and its meaning is shown in Figure 1.17. Rim width, rim diameter, and offset are shown in Figure 1.15. Offset is 1. Tire and Rim Fundamentals 23 the distance between the inner plane and the center plane of the rim. A rim may be designed with a negative, zero, or positive offset. A rim has a positive offset if the spider is outward from the center plane. The flange shape code signifies the tire-side profile of the rim and can be B, C, D, E, F , G, J , JJ , JK, and K. Usually the profile code follows the nominal rim width but different arrangements are also used. Figure 1.18 illustrates how a wheel is attached to the spindle axle"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002054_tie.2016.2593683-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002054_tie.2016.2593683-Figure6-1.png",
        "caption": "Fig. 6 Constrained modal shapes and frequencies of stator assembly. (a) 1267 Hz. (b) 2460 Hz. (c) 3259 Hz.",
        "texts": [
            " In addition, it is found by simulation that the relative errors of the stator core frequencies can reach 5% when the material is set to be isotropic. Thus, it is recommended that the material anisotropy should be considered to model the stator more accurately. Due to the difficulty of stator modal test in the test bed, the modal parameters of the stator under actual installation are obtained by FE modal analysis with the corresponding simulated constraints. The stator is clamped by two clamping blocks on two edges. So the contact edges are fixed in the FE model. Fig. 6 shows the constrained modal shapes and frequencies of circumferential modes below 3500 Hz. The coils are hided in order to observe the modal shape more clearly. The motor with 6 poles/9 slots can be regarded as three motors with 2 poles/3 slots according to the spatial periodicity. Thus, 1/3 2-D FE model is built to perform electromagnetic analysis. BEMF is tested to validate the FE model and the comparison is shown in Fig. 7. It can be seen that the BEMF from FEM agrees well with that from test",
            " These vibration harmonics are induced by current harmonics due to PWM and the corresponding current harmonic of each vibration peak around fc can be found in Table II. The other is the vibration harmonics with frequencies 360k, which come from the force harmonics produced by the interaction of permanent magnet field, stator slotting and armature reaction field considering current harmonics with frequencies (6k 1)f1. Dominant vibration peaks appear at 1440 Hz and 2520 Hz in low and medium frequency band, which are induced by structural resonance when the corresponding force harmonics are close to stator modes shown in Fig.6. 0278-0046 (c) 2016 IEEE. Personal use is permitted, but republication/redistribution requires IEEE permission. See http://www.ieee.org/publications_standards/publications/rights/index.html for more information. Boundary element method (BEM) is used to predict the noise radiation after the stator surface vibration is transferred to acoustic mesh. BEM is an efficient noise prediction method and has been used to calculate the noise radiation of motors in [17] and [18]. The noise calculation is performed in LMS Virtual",
            " The nonuniform distribution characteristics of electromagnetic force are weakened when the concentrated force is adopted. Since the vibration and noise are sensitive to the spatial distribution of force, the concentrated force method will induce extra calculation error. It is worse that when the spatial characteristics of force in frequency band close to the modal frequency is affected, the calculation error could be magnified. Fig. 13 shows that considerable calculation errors mainly appear in the frequency band close to the modal frequency shown in Fig. 6. Therefore, the nonuniform distribution of force should be considered to improve the accuracy of vibration and noise prediction. IV. PREDICTION OF ELECTROMAGNETIC VIBRATION AND NOISE UNDER DIFFERENT SUPPLY CURRENT To analyze the changes of vibration and noise peak values due to current harmonics with frequencies (6k 1)f1, vibration and noise under 4 kinds of supply current at 3600 rpm are calculated based on the proposed multiphysics model. The time history of A-phase current under these 4 kinds of supply current are shown in Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure10.12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure10.12-1.png",
        "caption": "FIGURE 10.12 Reflection and image for the four Green\u2019s functions.",
        "texts": [
            "50) The last case is given when both the observation and the source point are region 2 with a dielectric constant \ud835\udf002 g22(r, r\u2032) = 1 4\ud835\udf0b\ud835\udf002 [ 1 [\ud835\udf0c2 + (z \u2212 z\u2032)2]1\u22152 \u2212 \ud835\udefc [\ud835\udf0c2 + (z \u2212 z\u2032 + 2d)2]1\u22152 ] . (10.51) Working out the proper images for the three dielectric layers considered in the next section is quite challenging. However, we can use the two layer results to significantly simplify the process since meeting local boundary conditions is the same for all cases of interest. We summarize the four Green\u2019s function in Fig. 10.12 where we show the location of the image as well as the reflection needed for the next section. Another class of problems which can be solved with the image method consists of three dielectric layers as shown in Fig. 10.13. Even for the more complicated problems, we can take advantage of what we did learn about the two layer dielectric problems. For the three dielectric layer problems, the observation point r as well as the source point r\u2032 can be located in each of the three layer. This results in nine Green\u2019s functions for all the situations",
            " However, fortunately only five Green\u2019s functions are needed to cover the fundamentally different functions. The rest of the equations can be obtained from the formulas in this section by a simple exchange of variables. For the three layer case considered, we have two reflection coefficients, \ud835\udefc1 for the upper dielectric interfaces and \ud835\udefc2 for the lower case. By applying (10.47) to the three layer situation in Fig. 10.13 we obtain \ud835\udefc1 = \ud835\udf001 \u2212 \ud835\udf002 \ud835\udf001 + \ud835\udf002 and \ud835\udefc2 = \ud835\udf002 \u2212 \ud835\udf003 \ud835\udf002 + \ud835\udf003 (10.52) It is noticed from the above example in Fig. 10.12, that in general the regions where the observation point is located does not contain any image charges. INCLUDING DIELECTRIC MODELS IN PEEC SOLUTIONS 267 The locations of images are always equidistant from the interface. The strength of the images requires some thought in that the reflections must be considered carefully. Because there are multi-reflections due to the existence of two parallel dielectric boundaries, the coefficients \ud835\udefci in (10.52) have to be determined by multiple applications of the \ud835\udefc\u2019s in (10.53). A summary of diagrams in Fig. 10.12 is very helpful for the determination of the strength as well as the location of the images. We start with g11 where both the source and the observation points are located in region 1 shown in Fig. 10.14. g11(r, r\u2032) = 1 4\ud835\udf0b\ud835\udf001 [ 1 [\ud835\udf0c2 + (z \u2212 z\u2032)2]1\u22152 + \ud835\udefc1 [\ud835\udf0c2 + (z + z\u2032 \u2212 2d)2]1\u22152 + (1 \u2212 \ud835\udefc2 1) \u221e\u2211 k=0 (\u2212\ud835\udefc1)k(\ud835\udefc2)k+1 [\ud835\udf0c2 + (z + z\u2032 + 2 k d)2]1\u22152 ] (10.53) Each of the situations lead to as new formulation. Also, g33 can be derived from g11 by a change of variables. 268 PEEC MODELS FOR DIELECTRICS For the second Green\u2019s function, the source is still located in layer 1 while the observation point is in the middle layer 2, as shown in Fig",
            " An important issue is how we take care of the singular behavior when x = \ud835\udf14. Describe how you numerically treat the singularity in the Hilbert integrals. 10.2 Debye model evaluation Plot the magnitude of the permittivity and the loss tangent versus frequency for the fourth-order Debye model introduced in Section 10.1.5, (10.15). Do not use the Lorentzian terms in the model. 10.3 Lorentz medium Derive an equivalent circuit for Lorentz medium using a complex pole model. (Hint: Complex pole models are discussed in Chapter 2.) 10.4 Static Green\u2019s function Using Fig. 10.12 explain the contributions to the static Green\u2019s function in (10.53). 10.5 Asymptotic form of the static Green\u2019s function Assume that a problem consists of air above a planar dielectric half space. Assume that the permittivity \ud835\udf00r is close to infinite. Derive the static Green\u2019s function asymptotically based on the results from (10.48)\u2013(10.51). 1. P. G. Huray. The Foundations of Signal Integrity. John Wiley and Sons, Inc., New York, 2010. 2. C. A. Balanis. Advanced Engineering Electromagnetics. John Wiley and Sons, Inc"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure5.12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure5.12-1.png",
        "caption": "FIGURE 5.12. A rigid body with an attached coordinate frame B (oxyz) moving freely in a global coordinate frame G(OXY Z).",
        "texts": [
            "318) Now Br = RT Z,\u03d5 h tI\u0302 i = t cos\u03d5\u0131\u0302\u2212 t sin\u03d5j\u0302 (5.319) that provides Bdr dt = (\u2212t\u03d5\u0307 sin\u03d5+ cos\u03d5) \u0131\u0302\u2212 (sin\u03d5+ t\u03d5\u0307 cos\u03d5) j\u0302 (5.320) and G \u00b5 Bdr dt \u00b6 = G B r\u0307 = RZ,\u03d5 ((\u2212t\u03d5\u0307 sin\u03d5+ cos\u03d5) \u0131\u0302\u2212 (sin\u03d5+ t\u03d5\u0307 cos\u03d5) j\u0302) = \u23a1\u23a3 cos\u03d5 \u2212 sin\u03d5 0 sin\u03d5 cos\u03d5 0 0 0 1 \u23a4\u23a6\u23a1\u23a3 \u2212t\u03d5\u0307 sin\u03d5+ cos\u03d5\u2212 sin\u03d5\u2212 t\u03d5\u0307 cos\u03d5 0 \u23a4\u23a6 = I\u0302 \u2212 t\u03d5\u0307J\u0302 (5.321) which shows Gd dt Bdr dt = \u2212 (\u03d5\u0307+ t\u03d5\u0308) J\u0302 (5.322) 6= Bd dt Gdr dt . 5.9 Rigid Body Velocity Consider a rigid body with an attached local coordinate frame B (oxyz) moving freely in a fixed global coordinate frame G(OXY Z), as shown in Figure 5.12. The rigid body can rotate in the global frame, while the origin of the body frame B can translate relative to the origin of G. The coordinates of a body point P in local and global frames are related by the following equation: GrP = GRB BrP + GdB (5.323) where GdB indicates the position of the moving origin o relative to the fixed origin O. 268 5. Applied Kinematics The velocity of the point P in G is GvP = Gr\u0307P = GR\u0307B BrP + Gd\u0307B = G\u03c9\u0303B G BrP + Gd\u0307B = G\u03c9\u0303B \u00a1 GrP \u2212 GdB \u00a2 + Gd\u0307B = G\u03c9B \u00d7 \u00a1 GrP \u2212 GdB \u00a2 + Gd\u0307B ",
            " The global velocity of point P GvP = G\u03c9B \u00d7 G BrP + Gd\u0307B is a vector addition of rotational and translational velocities, both expressed in the global frame. At the moment, the body frame is assumed to be coincident with the global frame, and the body frame has a velocity Gd\u0307B with respect to the global frame. The translational velocity Gd\u0307B is a common property for every point of the body, but the rotational velocity G\u03c9B \u00d7 G BrP differs for different points of the body. Example 190 Velocity of a moving point in a moving body frame. Assume that point P in Figure 5.12 is moving in frame B, indicated by time varying position vector BrP (t). The global velocity of P is a composition of the velocity of P in B, rotation of B relative to G, and velocity of B relative to G. Gd dt GrP = Gd dt \u00a1 GdB + GRB BrP \u00a2 = Gd dt GdB + Gd dt \u00a1 GRB BrP \u00a2 = Gd\u0307B + G B r\u0307P + G\u03c9B \u00d7 G BrP (5.328) Example 191 Velocity of a body point in multiple coordinate frames. Consider three frames, B0, B1 and B2, as shown in Figure 5.14. The velocity of point P should be measured and expressed in a coordinate frame",
            " The force F and the perpendicular moment M\u22a5 can be replaced by a single force F0 parallel to F. Therefore, the force system is reduced to a force F0 and a moment Mk parallel to each other. A force and a moment about the force axis is called a wrench. The Poinsot theorem is similar to the Chasles theorem that states: Every rigid body motion is equivalent to a screw, which is a translation plus a rotation about the axis of translation. Example 344 F Motion of a moving point in a moving body frame. The velocity and acceleration of a moving point P as shown in Figure 5.12 are found in Example 200. GvP = Gd\u0307B + GRB \u00a1 BvP + B G\u03c9B \u00d7 BrP \u00a2 (9.23) GaP = Gd\u0308B + GRB \u00a1 BaP + 2 B G\u03c9B \u00d7 BvP + B G\u03c9\u0307B \u00d7 BrP \u00a2 +GRB \u00a1 B G\u03c9B \u00d7 \u00a1 B G\u03c9B \u00d7 BrP \u00a2\u00a2 (9.24) Therefore, the equation of motion for the point mass P is GF = mGaP = m \u00b3 Gd\u0308B + GRB \u00a1 BaP + 2 B G\u03c9B \u00d7 BvP + B G\u03c9\u0307B \u00d7 BrP \u00a2\u00b4 +m GRB \u00a1 B G\u03c9B \u00d7 \u00a1 B G\u03c9B \u00d7 BrP \u00a2\u00a2 . (9.25) 526 9. Applied Dynamics Example 345 Newton\u2019s equation in a rotating frame. Consider a spherical rigid body, such as Earth, with a fixed point that is rotating with a constant angular velocity"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure6.15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure6.15-1.png",
        "caption": "FIGURE 6.15. Double A arm suspension in a four-bar linkage mechanism.",
        "texts": [
            " The shorter link of the running dyad, p, must be attached to the motor at P , and the larger link, q, connects joint D to the shorter link at C. The motor will turn the shorter link, PC continuously at an angular speed \u03c9, while the longer link, CD, will run the mechanism and protect the wiper links to go beyond the initial and final angles. Example 230 Application of four-bar linkage in a vehicle. The double A arm suspension is a very popular mechanism for independent suspension of street cars. Figure 6.15 illustrates a double A arm suspension and its equivalent kinematic model. We attach the wheel to a coupler point at C. The double A arm is also called double wishbone suspension. 6. Applied Mechanisms 331 332 6. Applied Mechanisms 6.2 Slider-Crank Mechanism A slider-crank mechanism is shown in Figure 6.16. A slider-crank mechanism is a four-bar linkage. Link number 1 is the ground, which is the base and reference link. Link number 2 \u2261 MA is usually the input link, which is controlled by the input angle \u03b82"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001748_j.jmatprotec.2014.06.002-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001748_j.jmatprotec.2014.06.002-Figure1-1.png",
        "caption": "Fig. 1. Processing strategies of the tensile specimens: (",
        "texts": [
            " The SLM process and measurements A set of SLM processing parameters (laser power P = 180 W, scanning speed V = 900 mm/s, scanning spacing D = 0.06 mm and powder layer thickness H = 0.02 mm) was optimized for the 316L stainless steel powder by our previous research work and used to fabricate all samples. To investigate the effect of build orientations on MPBs and tensile properties of SLM parts, the samples with different build orientations in the X\u2013Y plane and the height direction (Z-axis) were designed, and the laser scans back and forth along the Y-axis. As shown in Fig. 1(a), the specimens were made in the X\u2013Y plane and referred to as horizontal specimens in this study, and the angle between the tensile loading direction and laser scanning direction (Y-axis) was increased from 30\u25e6 to 90\u25e6 by a step size of 15\u25e6. Fig. 1(b) depicts the specimens made along the height direction (Z-axis), referred to as vertical specimens, and the angle between the tensile loading direction and X\u2013Y plane was increased from 45\u25e6 to 90\u25e6 by a step size of 15\u25e6. A non-standard tensile sample was adopted as shown in Fig. 2. Three samples for each build orientation were produced and subjected to the tensile testing in the as-built condition without any heat treatment, and the average tensile strength and elongation are taken. The Zwick/Roell uniaxial tensile apparatus was used to measure the tensile properties"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003916_tro.2009.2020345-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003916_tro.2009.2020345-Figure1-1.png",
        "caption": "Fig. 1. Reaching task while maintaining a maximal angle on the elbow.",
        "texts": [
            " Tasks are generally defined by a set of equalities of reference such as e = 0, where e = s \u2212 s\u2217(t) is an error to be regulated to 0. Each component of a task thus represents a bilateral constraint. On the other hand, there are tasks that would require description through a set of unilateral constraints that are typically represented by inequalities ei \u2264 0. Examples of such constraints are joints limits [15], collision avoidance [16], [17], visibility loss [18], [19], or avoidance of singularities [20], [21]. A typical example is given by Fig. 1: The robot has to reach a position with its right end-effector, while ensuring the joint limit constraints. Such tasks present a strong irregularity at the activation point ei = 0. Due to this irregularity, it is impossible to consider unilateral constraints as classical (bilateral) ones. Specific unilateral constraints have been considered in a large number of works in the past, in particular, via the gradient projection method (GPM) [6], which was revisited recently in humanoid robotics [22]: Unilateral constraint, such as collision avoidance, is embedded in a cost function [23] whose gradient is projected in the space of free motion as a lowest priority task",
            " The first part presents the proposed method in a formal way: in Section II, we first recall the classical inverse-kinematics solutions and emphasize the causes of discontinuity while considering unilateral constraints. We then extend in Section III the solution of [27] for a hierarchy of tasks and unilateral constraints. The proposed solution is proven to appropriately decouple the tasks of the hierarchy in Section IV. We finally extend the proposed solution for inverse-dynamics control in Section V. Based on this theoretical approach, the second part focuses on the intuitive aspect of the method. Section VI presents with all the details a simple case study inspired from Fig. 1. A set of experiments is discussed in Section VII. II. INVERSE-KINEMATICS CONTROL Let q be the vector of robot joint positions and e be the task function. We consider a controller based on joint velocities q\u0307 (we will generalize to torque control in Section V). The Jacobian J of the task e is defined by e\u0307 = \u2202e \u2202q q\u0307 = Jq\u0307. (1) Let n be the number of DOFs (n = dimq) and m be the size of the task (m = dim e). The controller regulates e to 0 according to a reference behavior e\u0307\u2217. The joint motion q\u0307 realizing e\u0307\u2217 is given by the least square inverse of (1) q\u0307 = J+ e\u0307\u2217 (2) where A+ denotes the Moore\u2013Penrose inverse of A [33]",
            " On the other hand, it has been proposed to use the DLS to also smooth the discontinuities arising when a feature gets active [24]. Yet, smoothing such a discontinuity calls for high damping terms, which, in turn, compromise the quality of the execution (in particular, the tasks are not well accomplished and the priority order is not ensured). DLSs are not sufficient in practice to solve this problem [27]. A good explanation of such a discontinuity is given in [31]. This discontinuity is also exemplified in Section VI through the simple case study introduced in Fig. 1. In this section, we build a new control law whose form is similar to (8), yet ensures the continuity even when considering unilateral constraints. This original solution is based on the inversion operator proposed in [27] to solve the discontinuity of control laws similar to (4). Let us first explain the intuition behind the solution. A. Intuition Basically, the control law (4) is based on the inverse matrix( HJ )+H, which is very similar in shape to the inverse of J weighted by H, and is denoted as J#H ",
            " It is a close-form algorithm, where only the main inverse of each stage of the hierarchical set of tasks has to be replaced by the new continuous inverse. Now that all the computations have been realized for any set of tasks, the theory is exemplified with a specific set of tasks through several experiments. We here consider a simple toy example: A 2-DOF planar robot has to reach a desired position with its end-effector while preserving the joint limit of the second joint. This simple example takes its roots from the case presented in Fig. 1. In the following, we consider one unilateral constraint (the joint limits) and one task (the end-effector position). The endeffector positioning task is simply e\u0307r \u2217 = \u2212\u03bbr ( x \u2212 x\u2217) (48) where x and x\u2217 are the current and desired end-effector positions. Theoretically, the gain \u03bbr has to be positive only. However, if considering a discreet controller, care has to be taken to ensure that the gain is restricted with respect to the sample time interval. The unilateral constraint is classically written as q > \u22121 and q < 1 (49) with q being the joint position normalized in [\u22121, 1]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000554_978-3-540-85629-0-Figure3.39-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000554_978-3-540-85629-0-Figure3.39-1.png",
        "caption": "Fig. 3.39. Geometrical construction of the cycloidal (a) and modified cycloidal trajectory with the Alt method (b).",
        "texts": [
            "76 h T , q\u0308max = 5.528 h T 2 , q(3) max = 69.47 h T 3 . Example 3.22 Fig. 3.38 shows the position, velocity, acceleration and jerk for this trajectory with h = 20 and T = 10. 3.9 Modified Cycloidal Trajectory A cycloidal trajectory, defined by eq. (2.22), can be interpreted as the sum of a sinusoidal trajectory and of a constant velocity trajectory, with slope opposite 128 3 Composition of Elementary Trajectories (and equal in amplitude) to the final slope of the sinusoidal profile, as shown in Fig. 3.39(a). In the figure, A and B are the initial and final points, P is the intermediate transition point, APB is the line of the constant velocity motion, M the intermediate point between A and P. The amplitude of the sinusoidal profile, in the cycloidal motion, must be added to the constant velocity trajectory in the direction perpendicular to the axis t. A possible modification of the cycloidal trajectory is obtained by adding the sinusoidal profile in a direction perpendicular to the constant velocity profile, as shown in Fig. 3.39(b). This is the so-called \u201cAlt modification\u201d, after Herman Alt, a German kinematician who first proposed it, [7]. This modification allows to obtain a smoother profile, but implies an higher value of the maximum acceleration. Another modification, which aims at reducing the maximum acceleration value, is shown in Fig. 3.40(a), known as the \u201cWildt modification\u201d (after Paul Wildt, [7]). Point D is located at a distance equal to 0.57T 2 from T 2 , and then connected to M. The segment DM defines the direction along which the sinusoidal profile is added to the constant velocity trajectory",
            " The general expression of the modified cycloidal trajectory, where the base sinusoidal motion is projected along a direction specified by a generic angle \u03b3 (see Fig. 3.40(b)), is q(tm) = h [ tm T \u2212 1 2\u03c0 sin 2\u03c0tm T ] (3.52) where tm is defined by t = tm \u2212 \u03ba T 2\u03c0 sin 2\u03c0tm T (3.53) 3.9 Modified Cycloidal Trajectory 129 with \u03ba = tan \u03b4 tan \u03b3 and tan \u03b4 = h T . The angle \u03b3 determines the direction of the basis of the sinusoidal trajectory (\u03b3 is sometimes called the \u201cdistortion\u201d angle) . For example, for a pure cycloidal trajectory \u03b3 = \u03c0/2 (as shown in Fig. 3.39(a)). The velocity and the acceleration of the trajectory can be computed by differentiating q(t) with respect to the time6 t: q\u0307(t) = h T 1 \u2212 cos 2\u03c0t T 1 \u2212 \u03ba cos 2\u03c0t T q\u0308(t) = h T 2 2\u03c0(1 \u2212 \u03ba) sin 2\u03c0t T[ 1 \u2212 \u03ba cos 2\u03c0t T ]3 . The minimum value of the maximum acceleration is obtained for \u03ba = 1\u2212 \u221a 3 2 = 0.134, case in which the acceleration is q\u0308max = 5.88 h T 2 . The Alt modification is obtained for \u03b3 = \u03c0/2 + atan(h/T ). The velocity value at t = T/2 (note that, in this case, tm = T/2) can be simply computed as q\u0307 ( T 2 ) = 2h T (1 \u2212 \u03ba)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003578_mech-34238-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003578_mech-34238-Figure4-1.png",
        "caption": "Figure 4. Singularities of a 3\u2013RPS Parallel Manipulator.",
        "texts": [
            " (22) with (19) yields q\u0307o J $p (23) where J b1 s2 1 T sT 2 1 b2 s2 2 T sT 2 2 b3 s2 3 T sT 2 3 b1 s1 1 T sT 1 1 b2 s1 2 T sT 1 2 b3 s1 3 T sT 1 3 q\u0307o d\u03072 1 d\u03072 2 d\u03072 3 0 0 0 T 7 oaded From: http://proceedings.asmedigitalcollection.asme.org/ on 02/03/2016 Term Similar to the 3 UPU manipulator, the first three rows of J represent three forces of actuation acting along the three limbs. Obviously, the manipulator will be under singularity if these three forces lie on a common plane and intersect at one point, such as the case when the moving platform of the 3 RPS manipulator shown in Fig. 4a falls on top of the fixed base. On the other hand, the last three rows of J represent three forces of constraint, each being parallel to the revolute joint of a limb. Hence, the manipulator will be under singularity if these three forces lie on a common plane and intersect at one point as shown in Fig. 4b. This singular configuration was discussed by Huang et al. [22]. Singularity also occurs when the six reciprocal screws become linearly dependent as shown in Fig. 4a where the first five reciprocal screws intersect and the sixth reciprocal screw is parallel to the line defined by B1B2. This singular configuration was discussed by Zlatanov et al. [17]. In this case, the six reciprocal screws form a five system. An infinitesimal rotation of the moving platform about the line B1B2 is possible. We have shown that the Jacobian of a limited\u2013DOF parallel manipulator can be derived by making use of the theory of reciprocal screws. The 6 6 Jacobian matrix consists of two submatrices: one associated with the constraints imposed by the joints and the other associated with the actuation effects"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure7.15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure7.15-1.png",
        "caption": "FIGURE 7.15 Orthogonal quadrilateral for partial self-potential term.",
        "texts": [
            "3 for two overlapping inductance half cells which are connected between nodes 0 and 2 and the other one is connected between nodes 0 and 1. Of course the quadrilateral is not rectangular. 182 NONORTHOGONAL PEEC MODELS 7.4 PEEC inductance circuit model for coupled current strips This problem is very similar to the first problem in Chapter 6. However, the conductor shape is trapezoidal rather than rectangular. The thickness T = 1 mm and the corner point is Xc = yc = 30 mm and the width w = 10 mm for the trapezoidal strip in Fig. 7.15. The same way as in the first problem in Chapter 6, split the strip in the middle into two pieces that are connected at the ends. These two half-cell pieces are coupled to each other. Build an inductance\u2013resistance (Lp,R) PEEC model for the strip two partial. Make an R,L equivalent circuit for the problem. Is the model an open-loop model? 1. P. K. Wolff and A. E. Ruehli. Inductance computations for complex three dimensional geometries. In Proceedings of the IEEE International Symposium on Circuits and Systems, pp"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001256_j.ijmachtools.2021.103729-Figure48-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001256_j.ijmachtools.2021.103729-Figure48-1.png",
        "caption": "Fig. 48. Demonstration of the commonly applied layout of powder bed fused specimens with respect to the machine axis [95]. These layouts, or build orientations, affect the subsequent mechanical properties.",
        "texts": [
            " Build direction is yet another important factor that highly affects Almost half of the studies investigated tensile properties, while a third studied hardness properties. Other properties, especially shear and toughness, have been the subject of limited studies. S. Sanchez et al. International Journal of Machine Tools and Manufacture 165 (2021) 103729 specimens mechanical properties and results in microstructural and mechanical anisotropy [417]. Chlebus et al. investigated the tensile properties of specimens built in four directions (Fig. 48) [95]. The results indicated that the specimens built in the 45 \u25e6 \u00d7 45 \u25e6 direction possessed the best tensile strength in both AB and HT conditions. Moreover, specimens built in transverse directions were always stronger than the longitudinally built equivalents. This was explained by the angle between the loading direction and grain growth direction, which can greatly affect specimen tensile behaviour. Ni et al. also compared the tensile strength of longitudinally and transversely built IN718 specimens considering the Schmid factor, which is used to describe the relationship between slip planes and slip direction [259]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003774_j.1469-185x.1974.tb01299.x-Figure19-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003774_j.1469-185x.1974.tb01299.x-Figure19-1.png",
        "caption": "Fig. 19. Diagram illustrating variations of sign of forces (Fl and Fa) during the beating cycle of a cilium.",
        "texts": [
            " In Opalina the dominating feature appears to be the propagation of the bending wave up the cilium during the recovery stroke, during which time the force is large in comparison with that in the rest of the cycle. However, this is not the case in the diaplectic and antiplectic metachronism models, because the dominating feature here is the larger forces in the effective stroke. Over a complete cycle, though, the net force exerted is very small because the larger force in the effective stroke is only for a short time, whereas in the recovery stroke the force component is of opposite sign and longer duration. Fig. 19 gives some indication of the oscillation of sign experienced by Fa. During the effective stroke, Fa changes sign as the cilium moves past the x,-axis. It again changes sign during the propagation of the bending wave along the cilium, and may alter again before the beginning of the effective stroke. The bending moment follows a similar pattern to F , ; that is, it is large and positive during the effective stroke and small and negative during the recovery stroke of cilia, showing antiplectic metachronism"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure4.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure4.3-1.png",
        "caption": "Fig. 4.3. GenSmartTM AFPM synchronous generator with amorphous alloy stator core. Courtesy of LE Incorporated , Indianapolis, IN, U.S.A.",
        "texts": [
            " Again, in the case of a slotless stator with a large air gap between the rotor and stator core the magnets are almost always surface mounted. The stator windings of double-sided AFPM machines can be flat wound (slotted or slotless) as shown in Fig. 2.8 or toroidally wound (normally slotless) as shown in Fig. 2.9. An example of a commercial double-sided AFPM servo motor with ferromagnetic core is shown in Figs. 4.1 and 4.2. External stators have slotted ring-shaped cores made of nonoriented electrotechnical steel ribbon. The inner rotor does not have any ferromagnetic material. PMs are mounted on a nonmagnetic rotating disc. Fig. 4.3 shows a double-sided AFPM synchronous generator with the stator core wound of amorphous alloy ribbon manufactured by LE Incorporated , Indianapolis, IN, U.S.A. The volume of LE AFPM generators is approximately 60% lower than that of classical synchronous generators of the same rating. 4.3 Some Features of Iron-Cored AFPM Machines 125 Iron-cored AFPM machines are distinguished in two ways from coreless (aircored) AFPM machines, namely: (i) iron-cored machines have core losses while coreless machines do not and (ii) the per unit values of the synchronous reactances of iron-cored machines are much higher than those of coreless machines"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002366_j.jmatprotec.2016.08.006-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002366_j.jmatprotec.2016.08.006-Figure6-1.png",
        "caption": "Fig. 6. (a) \u2018IY\u2019 and (b) \u2018Y\u2019 illustrates the set up for the 2D FEA. (c) and (d) shows the displacement for \u2018IY\u2019 support structure and \u2018Y\u2019 support structure respectively. Arbitrarily a",
        "texts": [
            "03 14.80 \u00d7 14.86 9 \u00d7 14.90 15.07 \u00d7 15.08 14.75 \u00d7 14.83 6 \u00d7 14.87 14.94 \u00d7 14.95 14.68 \u00d7 14.70 M.X. Gan, C.H. Wong / Journal of Materials Processing Technology 238 (2016) 474\u2013484 479 w t d \u2018 ( b w 0 T s p w s i h c a t a g d d s o r 6-unit, (e) 25-unit, and (f) 36-unit. Pin support structure and thin plate: (g) d0.6, (h ere oriented vertical, but at an angle for the \u2018Y\u2019 design. To study he effect of the strut\u2019s inclination on the tip displacement, a twoimensional (2D) structural FEA was carried out. Fig. 6 illustrates the 2D beam FE set ups and solution for the IY\u2019 and \u2018Y\u2019 support structure subjected to an arbitrary force F = 10 N Fig. 6 (a) and (b)) at the tip that originated from the shrinkage. Zero displacement boundary conditions were applied at the ase of the support structures and the material properties used ere isotropic elastic modulus of 200 GPa and Poisson\u2019s ratio of .3. Fig. 6(c) illustrates the solution for the \u2018IY\u2019 support structure. he solution showed that the vertical strut, although displaced ubstantially along the x-axis (horizontally), made no distinct dislacement along the y-axis (vertically). Indeed, this agreed well ith the levelled thin plates manufactured. On the contrary, as hown in Fig. 6(d), the \u2018Y\u2019 support structure\u2019s outer strut increased n height, whereas the inner strut decreased. Due to the opposite vertical tip displacements, the difference in eight could have caused the edges or corners of the thin plate to url upwards. Furthermore, according to this FEA, the displacement long the x-axis for the \u2018Y\u2019 support structure was slightly smaller han that of the \u2018IY\u2019 support structure. The smaller displacement long the x-axis indicated lesser shrinkage, which explains the biger thin plates supported by the \u2018Y\u2019 support structure"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003750_(sici)1099-1239(199809)8:11<995::aid-rnc373>3.0.co;2-w-Figure12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003750_(sici)1099-1239(199809)8:11<995::aid-rnc373>3.0.co;2-w-Figure12-1.png",
        "caption": "Figure 12. Ducted fan with stand",
        "texts": [
            " The major problem is that the optimization is a factor 10 too slow for realistic operator sampling rates. Improvement of this optimization is a subject of current research. 4.2. Hover-to-hover experiments To demonstrate the viability of the two-degree-of-freedom approach, we first show some hover-to-hover transitions, both with one and two-degree-of-freedom controllers. These experiments are performed on the Caltech ducted fan experiment. The ducted fan is mounted on ( 1998 John Wiley & Sons, Ltd. Int. J. Robust Nonlinear Control 8, 995\u20141020 (1998) a stand, as shown in Figure 12, and is controlled by an Intel 486,66 MHz PC. It uses a linear current amplifier to regulate the current to the propeller, and PWM servos to steer the paddles. It has a NACA 0015 airofoil to generate lift, although for the experiments presented in this paper, the lift forces are negligible. Horizontal, vertical and pitch position are measured with encoders. Velocities are obtained by numerical differentiation and smoothing. Closed-loop control of the ducted fan is accomplished using the Sparrow real-time kernel"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001767_s00170-011-3395-2-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001767_s00170-011-3395-2-Figure8-1.png",
        "caption": "Fig. 8 a Turbine blade manufactured with and without control. b CMM measurement of the turbine blades. The middle section is along the length of the part. A-A is transverse section close to trailing edge and B-B is transverse section close to leading edge of the blade",
        "texts": [
            " However, the contour temperature had much higher standard deviation than that with a closed loop control. The standard deviation of the melt pool temperature for layer scanning was similar to that with a closed loop control. If taken the temperature drop introduced by height controller for consideration, the melt pool temperature with controller has much smaller variation and is more stable than that without control, especially for the first three layers. The manufactured turbine blades with and without closed loop control are shown in Fig. 8a and the corresponding CMM data are shown in Fig. 8b. The width of the turbine blade decreases gradually from the middle part to the trailing end. With constant laser power, the narrower section toward the trailing edge accumulated more heat than the wider part toward leading edge. This accumulation became worse as the number of deposition layer increased since heat dissipation rate for smaller volume was lower than larger volume. The melt pool temperature near trailing edge thus had higher melt pool temperature (Fig. 9), which reduced the cladding height and generated larger melt pool. Therefore, without closed loop control, the height of the manufactured turbine blade gradually decreased from the middle of the turbine blade to the trailing edge (Fig. 8b) where severe melting down was observed. Without height control, the top surface curved up due to higher powder catchment efficiency at a flat surface than at an edge. Severe distortion due to excessive heat accumulation and geometrically dependent powder catchment efficiency could prevent further deposition on the part. Deposition with both heat input and height control can prevent the process from uneven building up caused by non-uniform heat accumulation and non-uniform powder catchment efficiency"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003811_s0022-5193(87)80201-1-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003811_s0022-5193(87)80201-1-Figure4-1.png",
        "caption": "FIG. 4. Leakiness is defined here as the ratio of the volume of fluid that actually moves between a pair of cylinders of unit length in a unit of time (shown in (a)), to the volume between the cylinders across which they sweep in that unit of time (shown in (b)).",
        "texts": [
            " This suggests that the effect on local shear gradients of the width of an entire array of bristles is an impor t an t area to invest igate (in add i t ion to the effects of bristle d iameter and spacing examined here) if we are to u n d e r s t a n d the func t iona l morpho logy of setulose structures. While the velocity ratio is a measure of the s teepness of the shear grad ien t next to a seta, a bet ter measure of the degree to which the a p p e n d a g e operates as a rake vs. as a paddle is its \" ' leakiness\". We define leakiness here as the rat io of the vo lume of fluid that actual ly moves be tween a pair of setae to the vo lume across which that setal pair sweeps in a uni t of t ime (Fig. 4). We have de t e rmined the vo lume of fluid F L U I D F L O W T H R O U G H B R I S T L E D A P P E N D A G E S 25 moving between a pair of hairs of unit length by taking the area of the velocity profile between them. The leakiness of pairs of hairs of various spacings and Re's are plotted in Fig. 5, which illustrates the following points: (1) Appendages bearing hairs that operate at very low Re's are paddle-like with little fluid leaking between the hairs, whereas those whose bristles operate at Re's approaching one are much leakier and more rake-like"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003922_s11740-009-0197-6-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003922_s11740-009-0197-6-Figure4-1.png",
        "caption": "Fig. 4 Idealisation of the process domain",
        "texts": [
            "2 Idealisation Due to its enormous complexity, hardly any real problem is able to be solved entirely using FEM. Reasonable and suitable results can only be achieved when physical as well as structural conditions are being translated into an appropriate simulation model. Therefore, an idealisation step is required in order to identify relevant part geometries, symmetrical characteristics and decisive properties. Furthermore, internal situations as well as boundary and starting conditions must be defined. Hence, Fig. 4 illustrates a two step approach which allows for the idealisation of the process domain. In the top section of Fig. 4, an exemplary part as well as its subdivision into single layers is displayed. As it is state of the art, arbitrary part cross sections are being filled with so called basic shapes (cf. centre of Fig. 4) so flexible melting sequences can be realized which enable a uniform energy input into the powder layer [11]. The electron beam will consequently be deflected along the scan vectors and thus, the complete surface of the basic shape is being irradiated. The FEM geometry is a representative section of the top layer if one considers that it is very likely that the scan vectors within a basic shape are located in parallel. As a matter of fact, the powder is coated upon a previously solidified layer",
            " As a thermocouple is used for the measurement of the temperature, the accessibility of the domain in terms of powder layer deposition must be provided in order not to interrupt the process sequence. Therefore, the tactile measuring device is attached to the process domain form the reverse side of the platform in such a manner, that the temperature slightly below the surface is captured. Therefore, both diagrams in Fig. 5 reflect the temperature in the centre of the melted area at about 0.3 mm below the powder layer in the real platform as well as in the simulation model (cf. Fig. 4) respectively. Due to a comparatively short simulation period, the abscissae of the given diagrams show different time spans. For a better comparison, two separate measurement runs with equal test conditions are carried out and depicted on the left hand side of Fig. 5. According to Sect. 5.3, the powder layer is preheated before scanning takes place. Thus, 10 s before as well as 5 s after the scanning step recording is conducted at an exemplary position as above mentioned. As the scanning step is considerably transient, very small time steps are chosen within the simulation",
            " In order to achieve a dense coverage of possible parameter combinations at a reasonable effort, the interval between two adjacent scan speeds is 25 mm/s. In terms of the beam power PB, an offset of 30 W was chosen. As a result, a process parameter window could be defined (cf. Fig. 7). In order to ensure comparability between simulation results and experimental series, identical constants as described in Sect. 5.5 are ascertained in terms of the manufacture of sample parts. Moreover, the basic shape for scanning is a square with an edge length of 2.5 mm and the perpendicular distance between two parallel scan vectors is 0.16 mm (cf. Fig. 4). The grey domain in Fig. 7 depicts parameter settings with positive results in terms of layer smoothness and the absence of delamination. Within the adjacent regions on the left as well as on the right hand side, melt ball formation is observed. Furthermore, delamination occurs when the scan speed of the beam is too high. As a possible explanation, it can be considered that the energy input and thus, heat conduction into the preceding layer is insufficient. Consequently, melting temperature is not reached at the interface of two successive layers and therefore, only inadequate binding is achieved"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000846_j.ymssp.2017.05.024-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000846_j.ymssp.2017.05.024-Figure2-1.png",
        "caption": "Fig. 2. Four types of gear trains: (a) simple gear train, (b) compound gear train, (c) reverted gear train and (d) planetary gear train.",
        "texts": [
            " In industrial applications, gearboxes may work under constant operation condition or varying operation condition. This paper focuses on reported studies on the constant operating condition. Limited work in the area of dynamic modeling has been reported on the varying operating condition. According to the arrangement of gear wheels, gear trains can be classified into four categories [2]: simple gear train, compound gear train, reverted gear train and planetary gear train (epicyclic gear train). One example is given in Fig. 2 for each type of gear train. The simple gear train has one gear mounted on each shaft. If there is more than one gear mounted on a shaft, the gear train is called the compound gear train. If the axes of the driving gear shaft and the driven gear shaft are coaxial, the gear train is known as reverted gear train. If one gear rotates on its own axis and also revolves around the axis of another gear, this gear train is termed as planetary gear train. A total of 34 different types of planetary gear sets are described in Ref"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000717_0278364912452673-Figure11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000717_0278364912452673-Figure11-1.png",
        "caption": "Fig. 11. Top view of the 3D-LIPM with finite-sized foot and reaction mass, with a schematic representation of the N-step capture regions. The figure is an extension of Figure 8: state parameters rankle, r and r\u0307 are identical. We have omitted labels that were already shown in Figure 8 to avoid cluttering. The geometric construction is as follows. (1) Find region A as described in Figure 8 and find region B by offsetting region A by \u03b4; (2) use the lines of sight from the base of support to find region C; (3) find the capture regions by offsetting region C by the values of dN from Equation (26a). For this figure, \u03c4hip,max is set to 0.5 and \u03b8max = 1/8, which results in a total lunge time (2 tRM,max) of 1.",
        "texts": [
            " (44) For both 0-step and 1-step capturability, we see that the margin that is gained by the addition of the reaction mass is \u2225\u2225 r\u2032CMP\u2217 \u2225\u2225 max , compared with Equation (26a). Recursively applying the above derivations shows that this trend continues for all N , so that d\u2032N = ( l\u2032max \u2212 r\u2032max + d\u2032N\u22121) e\u2212 t\u2032s + r\u2032max + \u2225\u2225 r\u2032CMP\u2217 \u2225\u2225 max , N \u2265 1 (45a) d\u2032\u221e = l\u2032max e\u2212 t\u2032s 1\u2212 e\u2212 t\u2032s + r\u2032max + \u2225\u2225 r\u2032CMP\u2217 \u2225\u2225 max . (45b) The N-step capture regions for the 3D-LIPM with finitesized foot and reaction mass are shown in Figure 11 and are derived as follows. earliest possible step time Similar to the previous models, the first step to finding the capture regions is to find the set of possible future instantaneous capture point locations at time t\u2032s. The difference that the reaction mass makes is found by rewriting Equation (38) as r\u2032ic( t\u2032s)= r\u2032ic( t\u2032s) \u2223\u2223 \u03c4 \u2032hip=0 \u2212 r\u2032CMP\u2217e t\u2032s (46) where r\u2032ic( t\u2032s) \u2223\u2223 \u03c4 \u2032hip=0 is the instantaneous capture point location at t\u2032s when no hip torque is applied, that is, when r\u2032CMP\u2217 = 0, for which the model reduces to the model without reaction mass",
            " Therefore, the set of possible instantaneous capture point locations at time t\u2032s consists of all points that lie at most \u03b4 away from possible instantaneous capture point locations at t\u2032s for the model without reaction mass (see Section 6.6.1). the earliest possible step time After the application of the hip torque profile, the CMP will coincide with the CoP, and the instantaneous capture point will move on a line through itself and the CoP. Bounds on reachable instantaneous capture point locations are therefore found exactly as in Section 6.6.2, by constructing lines of sight (shown as the dashed lines in Figure 11) from the base of support to the set of possible instantaneous capture point locations at t\u2032s. 7.6.3. Nested regions Finally, we can construct capture regions exactly as in Section 6.6.3. After the first step is taken, no further hip torque is applied and the model essentially reduces to the 3D-LIPM with finite-sized foot. We should hence construct nested regions around the set of possible future instantaneous capture point locations using the values of d\u2032N for the LIPM without reaction mass, i"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure5.19-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure5.19-1.png",
        "caption": "Fig. 5.19 Relation Between Static Stability and Hight of Center of Mass and Supporting Polygon Size",
        "texts": [
            " The robot starts walking again and goes to the rubbish bin and throws the can away. This part is done completely by remote operation. At one point the robot grasped the can too tightly and the can got stuck to it\u2019s hand. However the operator was able to shake the robot\u2019s hand remotely and therefore was able to drop the can. One of the benefits of implementing remote operation is the fact that you get this kind of high level error recovery automatically. Anything that is not anchored to the ground can fall over. As shown in fig. 5.19 an object which, like a humanoid robot, has a high center of mass and a small supporting polygon becomes statically unstable even when tipped over a small angle. Until this point in the book, everything was about how to keep a robot stable to prevent it from falling down. As you can see from the fact that even human beings fall over, In reality it is not possible to prevent robots from falling over altogether. Therefore it is important that humanoid robots are able to fall over gracefully to prevent themselves from being damaged too seriously"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002097_978-1-4939-3017-3-Figure3.15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002097_978-1-4939-3017-3-Figure3.15-1.png",
        "caption": "Figure 3.15.2. Quantities needed for setting up the dynamics of two coupled rigid bodies in the plane.",
        "texts": [
            " In addition it provides a nice application of the energy-momentum method for stability. We begin with two bodies and discuss the situation for three bodies in the Internet Supplement. See Sreenath, Oh, Krishnaprasad, and Marsden [1988], Oh [1987], Grossman, Krishnaprasad, and Marsden [1988], and Patrick [1989] for further details and for the three-dimensional case. Related references are Krishnaprasad [1985], Krishnaprasad and Marsden [1987], and Bloch, Krishnaprasad, Marsden, and Alvarez [1992]. 174 3. Basic Concepts in Geometric Mechanics Summary of Results. Refer to Figure 3.15.1 and define the following quantities, for i = 1, 2: di distance from the hinge to the center of mass of body i; \u03c9i angular velocity of body i; \u03b8 joint angle from body 1 to body 2; \u03bb(\u03b8) = d1d2 cos \u03b8; mi mass of body i; \u03b5 = m1m2/(m1 +m2) = reduced mass; Ii moment of inertia of body i about its center of mass; I\u03031 = I1 + \u03b5d21; I\u03032 = I2 + \u03b5d22 = augmented moments of inertia; \u03b3 = \u03b5\u03bb\u2032/(I\u03031I\u03032 \u2212 \u03b52\u03bb2), \u2032 = d/d\u03b8. As we shall see, the dynamics of the system is given by the Euler\u2013 Lagrange equations for \u03b8, \u03c91, and \u03c92: \u03b8\u0307 = \u03c92 \u2212 \u03c91, \u03c9\u03071 = \u2212\u03b3(I\u03032\u03c9 2 2 + \u03b5\u03bb\u03c92 1), \u03c9\u03072 = \u03b3(I\u03031\u03c9 2 1 + \u03b5\u03bb\u03c92 2)",
            " Casimirs for this bracket are readily checked to be C = \u03a6(\u03bc1 + \u03bc2) (3.15.7) for \u03a6 any smooth function of one variable; that is, {F,C} = 0 for any F . One can also verify directly that, correspondingly, (d\u03bc/dt) = 0, where \u03bc = \u03bc1 + \u03bc2 is the total system angular momentum. The symplectic leaves of the bracket are described by the variables \u03bd = (\u03bc2\u2212\u03bc1)/2, \u03b8, which parametrize a cylinder. The bracket in terms of (\u03b8, \u03bd) is the canonical one on T \u2217S1: {F,H} = \u2202F \u2202\u03b8 \u2202H \u2202\u03bd \u2212 \u2202H \u2202\u03b8 \u2202F \u2202\u03bd . (3.15.8) We now set up the phase space for the dynamics of our problem. Refer to Figure 3.15.2 and define the following quantities: d12 the vector from the center of mass of body 1 to the hinge point in a fixed reference configuration; d21 the vector from the center of mass of body 2 to the hinge point in a fixed reference configuration; R(\u03b8i) = ( cos \u03b8i \u2212 sin \u03b8i sin \u03b8i cos \u03b8i ) , the rotation through angle \u03b8i giving the current orientation of body i (written as a matrix relative to the fixed standard inertial frame); ri current position of the center of mass of body i; r current position of the system center of mass; r0i the vector from the system center of mass to the center of mass of body i; \u03b8 = \u03b82 \u2212 \u03b81 joint angle; R(\u03b8) = R(\u03b82) \u00b7R(\u2212\u03b81) joint rotation; The configuration space we start with is Q, the subset of SE(2)\u00d7 SE(2) (two copies of the special Euclidean group of the plane) consisting of pairs (R(\u03b81), r1), (R(\u03b82), r2)) satisfying the hinge constraint r2 = r1 +R(\u03b81)d12 \u2212R(\u03b82)d21"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003605_027836499601500604-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003605_027836499601500604-Figure2-1.png",
        "caption": "Fig. 2. A robot touching fixed fiducial points, whose relative locations are not necessarily known, in a number of orientations. (With permission from Craig ~19g3J).",
        "texts": [
            " Bennett and Hollerbach (1989; 1991) considered point contact or manipulation of an unsensed ball joint (Sj = 0 and DJ = 3). Fliminating the unsensed DOFs is easy in this case, because the kinematic equations (8) decompose naturally into position and orientation equations. Thus, simply throw away three of the six equations in (8) having to do with orientation. Craig (1990; 1993) developed a variety of closed-loop methods, including the point contact case. A fiducial point on the end effector is touched to a fiducial point in the environment with several different orientations (Fig. 2). In an early instance of a closed-loop method, Tang (1986) proposed a two-stage method in which the angle parameters are calibrated by placing an end effector-mounted block face to face on a planar surface at several positions; this planar contact provides three constraints. Length parameters were calibrated via an open-loop approach by measuring the position of the block on the plate. If only the angle parameters are to be calibrated, then other approaches that constrain or measure orientation are conceivable-for example, using an inclinometer (or triaxial accelerometer or wrist force/torque sensor)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001770_0954405415619883-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001770_0954405415619883-Figure1-1.png",
        "caption": "Figure 1. Schematic of the laser AM technologies: (a) typical SLM machine layout31 and (b) laser net shape manufacturing system.7",
        "texts": [
            "22\u201325 In SLM, a computer-controlled scanning laser beam is applied as the energy source to selectively melt the pre-spread powder particles layer by layer.26\u201328 Furthermore, geometrically complex components with high dimensional precision and good surface integrity can be obtained precisely by the SLM process without subsequent process requirements, with which the conventional methods cannot easily keep pace with.29,30 The SLM process utilizes an inert argon or nitrogen gas as the atmospheric environment. The schematic view of a SLM system is shown in Figure 1(a). The desired microstructures and mechanical behavior of SLM-processed parts are inevitably affected by process parameters.4 Significant research efforts are still required to focus on microstructures and properties of the fabricated parts under various processing conditions. The SLM process will be compared to the other laser technologies, such as laser rapid forming (LRF),32 direct laser deposition (DLD),10 and laser net shape manufacturing (LNSM).7 LRF, which combines laser cladding with rapid prototyping into a solid freeform fabrication process, has an important potential application in the repair of worn or damaged components with lower heat input and local heating"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000015_0020717031000099029-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000015_0020717031000099029-Figure5-1.png",
        "caption": "Figure 5. Kinematic car model.",
        "texts": [
            " The differentiator performance does not significantly depend on the noise frequency. The author found that the second differentiation scheme (15) provides for slightly better accuracies. 7.2. Output-feedback control simulation Consider a simple kinematic model of car control (Murray and Sastry 1993) _x \u00bc v cos\u2019; _y \u00bc v sin\u2019 _\u2019 \u00bc \u00f0v=l\u00de tan _ \u00bc u where x and y are Cartesian coordinates of the rear-axle middle point, \u2019 is the orientation angle, v is the longitudinal velocity, l is the length between the two axles and is the steering angle (figure 5). The task is to steer the car from a given initial position to the trajectory y \u00bc g\u00f0x\u00de, while x; g\u00f0x\u00de and y are assumed to be measured in real time. Note that the actual control here is and _ \u00bc u is used as a new control in order to avoid discontinuities of . Any practical implementation of the developed here controller would require some realtime coordinate transformation with approaching =2. Define \u00bc y g\u00f0x\u00de \u00f024\u00de Let v \u00bc const \u00bc 10m/s, l \u00bc 5m, g\u00f0x\u00de \u00bc 10 sin \u00f00:05x\u00de\u00fe 5, x \u00bc y \u00bc \u2019 \u00bc \u00bc 0 at t \u00bc 0"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure3.41-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure3.41-1.png",
        "caption": "Fig. 3.41: A scheme of a planar RP manipulator and its stiffness-lumped parameters.",
        "texts": [
            " Thus, its entries are the lumped spring parameters and the matrix gives the possibility to compute the vector \u0394v of the corresponding deformations in the manipulator components from the expression vR \u0394= pK (3.7.11) The compliant displacement matrix CF describes the relationship between the deformation vector \u0394v and the vector \u0394S of the corresponding compliant displacement of the manipulator extremity in the form Sv \u0394=\u0394 KC (3.7.12) Therefore, combining Eqs (3.7.10) to (3.7.12) one can obtain the expression of Eq. (3.7.9) to compute Eq. (3.7.1) in order to identify the stiffness matrix K of a manipulator in a prescribed configuration for a given wrench. In Fig. 3.41 a telescopic RP manipulator is modeled for stiffness analysis by identifying the lumped spring parameters K1, K2, KT in the kinematic diagram. The parameter K1 refers to the stiffness of link a1 including the radial compliance of the revolute joint and corresponding actuator. The parameter K2 describes the stiffness of the linear actuator and the link of length s. Chapter 3: Fundamentals of the Mechanics of Robots174 The parameter KT gives the angular compliance characteristic of the revolute joint and corresponding actuator",
            " It can also model the bending deformation of the links when the compliant displacement of link extremity is considered as due to a link rotation because of an angular deformation of the joint. The stiffness matrix of the telescopic RP manipulator is a 3 \u00d7 3 matrix when a planar compliant motion is considered as function of elastic deformations \u0394a1, \u0394s, and \u0394\u03b81 due to the spring parameters K1, K2, KT, respectively. The stiffness matrix can be computed by using the matrix component formulation through a numerical procedure that has been outlined through Eqs (3.7.10)\u2013(3.7.12). In particular, for the case of Fig. 3.41, by means of a static analysis one can obtain ( ) ( ) 1csassa 0ss 0cc C 1111 11 11 F ++ = (3.7.13) The stiffness parameter matrix is computed referring to the model of Fig. 3.41 as a diagonal matrix whose no zero entries are the lumped spring parameters of the manipulator in the form T 2 1 p K00 0K0 00K K = (3.7.14) The kinematic analysis gives the equations of the direct kinematics that can be used to compute the variations of the extremity coordinates that are the components of \u0394v Fundamentals of Mechanics of Robotic Manipulation 175 vector so that it yields ( ) ( ) 100 csass ssacc C 1111 1111 K + +\u2212 = (3.7.15) Performances of robots are evaluated through synthetic measures of main characteristics of their design and operation"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000354_9.119645-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000354_9.119645-Figure3-1.png",
        "caption": "Fig. 3. Simulation results for yd( t ) = A cos x t /5, A = 1 , 2, 3, using the first approximation.",
        "texts": [
            ") at each step, we have E', = x2 - \u20ac 2 = dZ(X) = -BGs inx , + Bx,xi - \u20ac 3 = @,(I) 3 d X ) E'3 = -BGx4 COS x3 54 = d 4 ( d i4 = BGxj sin x3 + ( -BG cos x 3 ) U - ' b ( x ) a( x ) \u20ac 1 = E2 \u20ac 2 = E 3 + J / z ( x ) or E'3 = '$4 E'4 = b ( x ) + ~ ( x ) u =: u ( x , U). As expected, by neglecting the centrifugal term (which is higher order), we obtain an approximate system with a well-defined relative degree. Note that the choice of what to neglect (i.e., $ z ( x ) ) leads to a specification of the coordinate transformation @ ( x ) . In this case, the approximate system has been obtained by a simple modification of the f vector field (i.e., by neglecting & ( a ) ) . The simulation results in Fig. 3 show that the closed-loop system provides good tracking. The tracking error increases in a nonlinear fashion as the amplitude ( A ) of the desired trajectory is increased. This is expected since the neglected term $ , ( x ) is a nonlinear function of the state. A good a priori estimate of the mismatch of the approximate system for a desired trajectov can be calculated using $ ( W 1 ( y d , j d , J d , ~ 2 ) ) ) where @ - I : ( +, x is the inverse of the coordinate transformation. This in turn may be a useful way to define a class of trajectories that the system can track with small error"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002434_tie.2019.2914631-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002434_tie.2019.2914631-Figure1-1.png",
        "caption": "Fig. 1. Navigation and body frames of an USV.",
        "texts": [
            " Section II describes an underactuated USV dynamic and introduces some useful preliminaries. Section III proposes an adaptive NN controller for the underactuated USV with prescribed transient performance. Section IV and V show the simulational and experimental results, respectively. Motivated by [29], the nonlinear dynamics of the USV with unknown disturbances are provided as \u03b7\u0307 = J(\u03c8)\u03bd M\u03bd\u0307 = \u2212C(\u03bd)\u03bd \u2212D(\u03bd)\u03bd + d+ \u03b4(\u03c4) (1) where \u03b7 = [x, y, \u03c8]> denotes the position and yaw angle in the earth-fixed frame, respectively, which are shown in Fig. 1; \u03bd = [u, v, r]> represents velocity states in surge u, sway v, and yaw r in the body-fixed frame; M = M> is a non-diagonal inertia matrix, C and D are total Coriolis and Centripetal acceleration matrix, and damping matrix, respectively; d = [du, dv, dr] > denotes the unknown disturbance; The saturated control vector \u03b4(\u03c4) = [\u03b41(\u03c41), 0, \u03b43(\u03c43)]> = [\u03b41(\u03c4u), 0, \u03b43(\u03c4r)] > is defined as \u03b4i(\u03c4i) =  \u03c4i,max, if \u03c4i > \u03c4i,max \u03c4i, if \u03c4i,min \u2264 \u03c4i \u2264 \u03c4i,max \u03c4i,min, if \u03c4i < \u03c4i,min (2) 0278-0046 (c) 2019 IEEE. Personal use is permitted, but republication/redistribution requires IEEE permission"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.2-1.png",
        "caption": "FIGURE 7.2. A front-wheel-steering vehicle and steer angles of the inner and outer wheels.",
        "texts": [
            " Track w and wheelbase l are considered as kinematic width and length of the vehicle. The mass center of a steered vehicle will turn on a circle with radius R, R = q a22 + l2 cot2 \u03b4 (7.2) where \u03b4 is the cot-average of the inner and outer steer angles. cot \u03b4 = cot \u03b4o + cot \u03b4i 2 . (7.3) The angle \u03b4 is the equivalent steer angle of a bicycle having the same wheelbase l and radius of rotation R. Proof. To have all wheels turning freely on a curved road, the normal line to the center of each tire-plane must intersect at a common point. This is the Ackerman condition. Figure 7.2 illustrates a vehicle turning left. So, the turning center O is on the left, and the inner wheels are the left wheels that are closer to the center of rotation. The inner and outer steer angles \u03b4i and \u03b4o may be calculated 7. Steering Dynamics 381 from the triangles 4OAD and 4OBC as follows: tan \u03b4i = l R1 \u2212 w 2 (7.4) tan \u03b4o = l R1 + w 2 (7.5) Eliminating R1 R1 = 1 2 w + l tan \u03b4i = \u22121 2 w + l tan \u03b4o (7.6) provides the Ackerman condition (7.1), which is a direct relationship between \u03b4i and \u03b4o. cot \u03b4o \u2212 cot \u03b4i = w l (7",
            " The error e is the difference between the outer steer angles calculated by the trapezoidal mechanism and the Ackerman condition at the same inner steer angle \u03b4i. e = \u2206\u03b4o = \u03b4Do \u2212 \u03b4Ao (7.25) Figure 7.10 depicts the error e for a sample steering mechanism using the angle \u03b2 as a parameter. Example 261 F Locked rear axle. Sometimes in a simple design of vehicles, we eliminate the differential and use a locked rear axle in which no relative rotation between the left and right wheels is possible. Such a simple design is usually used in toy cars, or small off-road vehicles such as a mini Baja. Consider the vehicle shown in Figure 7.2. In a slow left turn, the speed of the inner rear wheel should be vri = \u00b3 R1 \u2212 w 2 \u00b4 r = Rw\u03c9ri (7.26) and the speed of the outer rear wheel should be vro = \u00b3 R1 + w 2 \u00b4 r = Rw\u03c9ro (7.27) where, r is the yaw velocity of the vehicle, Rw is rear wheels radius, and \u03c9ri, \u03c9ro should be the angular velocity of the rear inner and outer wheels 388 7. Steering Dynamics about their common axle. If the rear axle is locked, we have \u03c9ri = \u03c9ro = \u03c9 (7.28) however, \u00b3 R1 \u2212 w 2 \u00b4 6= \u00b3 R1 + w 2 \u00b4 (7.29) which shows it is impossible to have a locked axle for a nonzero w"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003908_978-1-84882-730-1-Figure12.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003908_978-1-84882-730-1-Figure12.1-1.png",
        "caption": "Figure 12.1 Motion variables of an underwater vehicle",
        "texts": [
            "1 Control Objective In this chapter, we consider the following mathematical model of an underactuated underwater vehicle when nonlinear hydrodynamic damping terms and roll motion are ignored, see Section 3.4.2.2: 1. Kinematics Px D cos. /cos. /u sin. /vC sin. /cos. /w; Py D sin. /cos. /uC cos. /vC sin. /sin. /w; Pz D sin. /uC cos. /w; P D q; P D r cos. / : (12.1) 2. Kinetics 295 296 12 Trajectory-tracking Control of Underactuated Underwater Vehicles PuD m22 m11 vr m33 m11 wq d11 m11 uC 1 m11 u; Pv D m11 m22 ur d22 m22 v; Pw D m11 m33 uq d33 m33 w; Pq D m33 m11 m55 uw d55 m55 q grGML sin. / m55 C 1 m55 q; Pr D m11 m22 m66 uv d66 m66 rC 1 m66 r : (12.2) In (12.1) and (12.2) the symbols (see Figure 12.1) , , q, and r denote the roll, pitch, and yaw angles and velocities while x, y, z, u, v, and w are the surge, sway, and heave displacements and velocities, respectively. The available control inputs are u, q , and r . Since the sway and heave control forces are not available in the sway and heave dynamics, the underwater vehicle in question is underactuated. Notice that (12.1) is not defined when the pitch angle is equal to \u02d9900. However, during practical operations with underwater vehicles, this problem is unlikely to happen due to the metacentric restoring forces"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure11.11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure11.11-1.png",
        "caption": "Figure 11.11 (a) Small end segment at hinged support A (see Figure 11.10a). (b) Small end segment at free end B (see Figure 11.10b).",
        "texts": [
            " The tangents to the M diagram are also shown at A and B. These intersect in x = 1 2 , at mid-span. In the figure, an important variable p is shown: p = 1 8q 2. We will make use of p in Chapter 12. from which it follows that 11 Mathematical Description of the Relationship between Section Forces and Loading 447 Below we again show how, for two cases, the boundary conditions (end conditions) can be derived from the equilibrium of a small element with length x ( x \u2192 0) at the beam ends. Boundary condition at the hinged support A in case (a) In Figure 11.11a, a member segment with length x ( x \u2192 0) has been isolated at the hinged support at A. The figure shows all the forces acting on it, including the unknown vertical support reaction at A. Moment equilibrium about A requires \u2211 Ty |A = M \u2212 V x \u2212 q x \u00b7 1 2 x = 0. For x \u2192 0 the terms with x disappear and we find the boundary condition at A: M = 0. Boundary conditions at the free member end B in case (b) In Figure 11.11b, the small \u201clast\u201d member segment at the free end B is shown, with all the forces acting on it. The element has a length x. The equations for the equilibrium are \u2211 Fz = \u2212V + q x = 0,\u2211 Ty |B = \u2212M \u2212 V x + q x \u00b7 1 2 x = 0. For x \u2192 0 the terms with x disappear in both equations and we find the boundary conditions at the free member end B: V = 0, M = 0. 448 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Example 3 Figure 11.12a shows a water-retaining slide of width b and height h, which is supported by a hinge at the top and supported against a sill below"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure4.13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure4.13-1.png",
        "caption": "FIGURE 4.13 Three-conductor problem with overlapping cells.",
        "texts": [
            " In this section, we consider a very important meshing issue pertaining to the location of the cells on conductors. Essentially, the mesh subdivision or cells on an object can greatly impact the capacitance results. We consider the impact of badly placed cells and on the partial capacitances and partial inductance values. The impact on the accuracy of capacitance calculations was first presented in Ref. [10] and the potential impact on the quality of the PEEC model solution is considered in Chapter 13. The basic problem is that close conductors that are not well meshed can lead to poor results. In Fig. 4.13, we show a three- conductor problem where the lower plate, after meshing, consists of cells numbered 3\u20135. We chose the geometry of the thin conductors such that two top plates are exactly on top of the lower plate in the x direction with the same length lx = 100 cm. All five cells are subdivided into two cells along y. Both top plates are ly = 50 cm wide with a small spacing S = 10 cm in the middle. The total width of all bottom conductors is y = 110 cm. The lower plate is subdivided into three width cells, 3\u20135, as shown in Fig. 4.13. To emphasize the problem, we chose the width for cells 3 and 5 to be 27.5 cm and cell 4 to be 55 cm. Hence, the top plates clearly do not project onto the lower plate. Finally, the plate-to-plate spacing in the z-direction is chosen to be 2.5 cm to emphasize the projection issue. MESHING RELATED ACCURACY PROBLEMS FOR PEEC MODEL 81 If we compute the capacitances using the approach in Section 4.3, then the matrix solution will be in the form (4.23) Q = Pp\u22121 \ud835\udebd (4.35) for the subdivided system at the cell subdivision level. To reduce the system to the conductor plate level, we simply have to set all the partial potentials \u03a6 on a plate, such as conductor 1 in Fig. 4.13 to be the same, \u03a61. Then by adding all the partial charges Qk from all the cells on each conductor plate, we get the reduced system of short circuit capacitances: \u23a1\u23a2\u23a2\u23a2\u23a2\u23a3 Q1 \u22ee Q4 Q5 \u23a4\u23a5\u23a5\u23a5\u23a5\u23a6 = \u23a1\u23a2\u23a2\u23a2\u23a2\u23a3 Cs1 \u00b7 \u00b7 \u00b7 Cs14 Cs15 \u22ee \u22ee \u22ee \u22ee Cs41 \u00b7 \u00b7 \u00b7 Cs44 Cs45 Cs51 \u00b7 \u00b7 \u00b7 Cs54 Cs55 \u23a4\u23a5\u23a5\u23a5\u23a5\u23a6 \u23a1\u23a2\u23a2\u23a2\u23a2\u23a3 \u03a61 \u22ee \u03a64 \u03a65 \u23a4\u23a5\u23a5\u23a5\u23a5\u23a6 . (4.36) Finally, the capacitance of interest C12 = \u2212Cs12 is found from Cs12 = Q1\u2215\u03a62 for the case when all the other plates are at potentials 0 and \u03a62 = 1. It is clear that for computing C12, in Fig. 4.14 with the non overlapping cells the potentials of all conductors except for conductor 2 are set to zero. The potential on cell 2 is \u03a6 = 1, resulting in positive charges. Note that we created a bad overlap between cells 2 and 4 as shown in Fig. 4.13. This results in an excessive negative charge that is induced to cell 4. This in turn couples from cell 4 to cell 1 with the wrong amount of charge. Hence, C12 = \u2212Cs12 is computed incorrectly. The capacitances computed for this case is a 3 \u00d7 3 symmetric matrix since all the lower plates are at the same potential. Then, the computed capacitances for Fig. 4.13 are in picofarad C11 = C22 = 9.53,C33 = C55 = 21.09,C13 = C25 = 186.79,C12 = \u221237.90. (4.37) where we notice the wrong negative coupling capacitance between the two top conductors. We compare this result with the case where we use the so- called projection meshing. This meshing is presented in Chapter 8. Using projection meshing, we have to subdivide the lower plate such that the cells of the top conductor exactly project onto the lower conductor as shown in Fig. 4.14 where the width of conductor 4 is also ly = 10 cm, which corresponds to S in the top plate in both Figs"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000637_978-3-642-83006-8-Figure1.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000637_978-3-642-83006-8-Figure1.2-1.png",
        "caption": "Fig. 1.2. Longitudinal distribution of pressure on the foot",
        "texts": [
            " However, the same person, with his vestibular system functioning properly but with the locomotor - muscle system of his lower extremities artificially paralysed, moves in the same way as a paralysed paraplegic person. It is quite clear that a healthy man \"feels\" forces, i.e., dynamic reactions, and distributes them in some manner in the form of driving torques along his skeleton activated by numerous muscle groups. Force sensing at the feet of a healthy person is a global feedback which takes care of the overall system behaviour and can be applied to artificial mechanisms, too. Let us suppose the system is in the single-support phase and the contact with the ground is realized by the full foot (Fig. 1.2). Then, it is possible to replace all vertical elementary reaction forces by the resultant Bv. If we reduce it to the centre of supporting area, the reaction force N and moment M will be obtained. As the pressure dia gram is of the same sign, the reaction force RV can be always computed. Obviously, the ZMP in single-sUPPort phase cannot be out of the suppor ting area (area covered by one foot), while in double-support phase it can be anywhere inside the dashed area (Fig. 1.3). Within these areas the ZMP can move in accordance with different laws, continuously or not, depending on which gait type is performed"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure6.27-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure6.27-1.png",
        "caption": "Figure 6.27 (a) The column dimensions and (b) all the forces acting on the isolated column. The dead weight of the column is a uniformly distributed load parallel to the column axis.",
        "texts": [
            " Over the full length, the beam is therefore loaded by a uniformly distributed load of 5.6 kN/m. We also have to include the dead weight of the beam. If we assume a dead weight of 6 kN/m, the total 232 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM load on the beam is (see Figure 6.26) q = (5.6 kN/m) + (6 kN/m) = 11.6 kN/m. The beam is simply supported. The support reactions, which have to be provided by the columns, amount to 1 2 \u00d7 (11.6 kN/m)(8 m) = 46.4 kN. Equal and opposite forces are acting on the columns. Figure 6.27a shows the cross-sectional dimensions of the columns. With mass density \u03c1 = 2400 kg/m, the specific weight of concrete is \u03c1g = (2400 kg/m3) \u00d7 (10 N/kg) = 24000 N/m3) = 24 kN/m3. For the cross-sectional dimensions of the column in Figure 6.27a, the dead weight per length is (0.2 m)(0.4 m)(24 kN/m3) = 1.92 kN/m. This is a uniformly distributed load acting in the direction of the column axis (see Figure 6.27b). The total dead weight of the column is (5 m)(1.92 kN/m) = 9.6 kN. At its base, the column has to be kept in equilibrium by a force of (46.4 kN) + (5 m)(1.92 kN/m) = 56 kN. 6 Loads 233 An equally large, opposite force is acting on the footing of the column. If the footing is square, and has the dimensions given in Figure 6.28a, the dead weight of the footing is (0.8 m)(0.8 m)(0.3 m)(24 kN/m3) = 4.6 kN. The earth pressure on the bottom of the footing has to be in equilibrium with the force of 56 kN from the column, and the footing\u2019s dead weight of 4"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002277_j.prostr.2016.02.039-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002277_j.prostr.2016.02.039-Figure2-1.png",
        "caption": "Fig. 2 \u2013 Original component.",
        "texts": [
            " Conventional manufacturing processes often struggle or even fail to accomplish the designs that result from the use of TO, due to its complex geometries and shapes [Zhou et al. (2002)]. On the other hand, SLM, for its freedom of geometries and lack of manufacture constraints, is a particularly suited manufacturing process for the TO design. There have been several authors combining the use of TO with SLM with the objective of making the most of both technologies [Muir (2013), Emmelmann et al. (2011), Tomlin et al. 2011)]. The methodology followed in this work is illustrated on Figure 1. An initial aircraft component with three static load cases was given. Figure 2 illustrates the original component. Miguel Seabra et al. / Procedia Structural Integrity 1 (2016) 289\u2013296 291 Additve Manufacturing (AM) [Gibson et al. (2010)] is a process of manufacture through which an object is built by material addition in layers. Inside AM\u2019s metal methods, SLM is one of the most promising ones. SLM is the process through which a laser selectively melts particles of metal powder together, in layers, until a 3D object is created, being possible to produce complex geometries and relative densities close to 100% [Aliakbari (2012)]",
            " Conventional manufacturing processes often struggle or even fail to accomplish the designs that result from the use of TO, due to its complex geometries and shapes [Zhou et al. (2002)]. On the other hand, SLM, for its freedom of geometries and lack of manufacture constraints, is a particularly suited manufacturing process for the TO design. There have been several authors combining the use of TO with SLM with the objective of making the most of both technologies [Muir (2013), Emmelmann et al. (2011), Tomlin et al. 2011)]. The methodology followed in this work is illustrated on Figure 1. An initial aircraft component with three static load cases was given. Figure 2 illustrates the original component. Fig. 1 \u2013 Methodology scheme. Fig. 2 \u2013 Original component. Author name / Structural Integrity Procedia 00 (2016) 000\u2013000 3 The component is assembled on its surrounding structures by 12 rivet holes with a 4.1 mm diameter and the load is applied in the larger hole, called the load bearing lug, with a 16 mm diameter. Table 1 shows the loads of each load case. The component was originally made of the aluminum alloy 7050-T7451. The goal is to reduce weight while maintaining the stress levels observed in the original component using the titanium alloy Ti6Al4V"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure6.15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure6.15-1.png",
        "caption": "Fig. 6.15 Propagation of spatial velocity: link2 obtains relative speed with respect to link1 by rotating the joint axis a2 at the speed of q\u03072",
        "texts": [
            " The calculated top motion is simultaneously displayed. 196 6 Dynamic Simulation In this section, we examine the dynamics of humanoid robot regarded as a link system made of multiple rigid bodies. Since each rigid body behaves as we mentioned above, we can obtain the dynamics of the entire system by their consistent integration. First, we examine a pair of rigid bodies connected by a joint, which represents a minimum element of a humanoid robot. Suppose we have the link1 in the space connected with link2 via rotational joint as illustrated in Fig. 6.15. When the link1 is stationary, by rotating the joint a2 at the speed of q\u03072, link2 gets angular velocity of \u03c92 = a2q\u03072. (6.27) In the spatial velocity representation of Section 6.2, the linear velocity of link2 is vo2 = p\u03072 \u2212 \u03c92 \u00d7 p2 where p2 is the origin of the link2. Since we are assuming that link1 is stationary, p\u03072 = 0. By substituting (6.27) into the above equation, we get vo2 = p2 \u00d7 a2q\u03072. (6.28) 6.4 Dynamics of Link System 197 Equations (6.28) and (6.27) give the relative speed (vo2,\u03c92) of link2 with respect to link1"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure3.25-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure3.25-1.png",
        "caption": "Fig. 3.25: A planar 3R manipulator and its parameters.",
        "texts": [
            " Indeed, the algebraic approach can be seen as part both of the matrix formulation and geometric approach. By considering it separately, the algebraic approach is a mathematization of the inverse kinematics problems, in a sense that the attention is focused only on the equations without any reference to their meaning for manipulator behavior. But the kinematic interpretation is again considered when a solution is found. The above-mentioned procedures for inverse kinematics can be better illustrated by referring to an example for a planar 3R manipulator, as shown in Fig. 3.25. From the geometry of the manipulator one can deduce the kinematic equations straightforwardly from equating matrix 0T3 to matrix 0TT, which describes tool frame 3 Chapter 3: Fundamentals of the Mechanics of Robots124 at the end-effector with respect to the base frame 0. Matrix 0T3 is expressed as 1000 0100 sasa0cs caca0sc T 12211123123 12211123123 30 + +\u2212 = (3.2.3) and matrix 0TT can be given as 1000 0100 h0cossin h0sincos T y x T0 \u03d5\u03d5 \u03d5\u2212\u03d5 = (3.2.4) Thus, by equating the two matrices the following set of equations can be determined 123ccos =\u03d5 123ssin =\u03d5 12211x cacah += (3",
            "2.12) Then, a unique value for \u03b81 can be obtained by using solutions for c1 and s1 into the two-argument arctangent function Atan2 as ( )111 c,s2tanA=\u03b8 (3.2.13) The outlined procedure for obtaining \u03b81 is one way to manipulate algebraically the kinematic equations and several other ways can be carried out to gain algebraic solution of the involved equations. Finally, \u03b83 can be solved by using the equality \u03d5=\u03b8+\u03b8+\u03b8 321 (3.2.14) from the first two expressions in Eqs (3.2.5) but also the geometry of Fig. 3.25. Alternatively, looking at the geometry of the manipulator configuration can give the equations and the way to solve them. Thus, once a triangle is determined with a1, a2, and the segment joining O0 with O3, the law of cosines can be applied to give ( )2cos2a1a22 2a2 1a2 yh2 xh \u03b8\u2212\u03c0\u2212+=+ (3.2.15) from which \u03b82 can be solved. Again both elbow-up and elbow-down configurations can be a solution. The angle \u03b81 can be obtained considering the slope angle \u03b3 of h direction and the angle between the line of h and line of a1 link to obtain \u03c8\u2212\u03b3=\u03b81 (3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure5.13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure5.13-1.png",
        "caption": "Fig. 5.13. Eddy currents in the coreless stator winding of an AFPM machine: (a) magnetic field distribution in a coreless stator; (b) eddy currents in a conductor; (c) circulating eddy currents among parallel connected conductors.",
        "texts": [
            "32) where Ie = \u221a I2 ed + I2 eq (5.33) On the basis of equivalent circuits (Fig. 5.5), the following equations can be written in the phasor form: \u2022 for generators Ef = Ei + jIadXad + jIaqXaq (5.34) I\u2032a = Ia + Ie (5.35) \u2022 for motors Ef = Ei \u2212 jIadXad \u2212 jIaqXaq (5.36) I\u2032a = Ia \u2212 Ie (5.37) where I\u2032a is the stator current with eddy currents being accounted for and Ia is the stator current with eddy currents being ignored. In an AFPM machine with a coreless stator the winding is directly exposed to the air gap magnetic field (see Fig. 5.13). Motion of PMs over the coreless winding produces an alternating field through each conductor inducing eddy currents. The loss due to eddy currents in the conductors depends on both the geometry of the wire cross section and the amplitude and waveform of the flux density. In order to minimize the eddy current loss in the conductors, the stator winding should be designed in one of the following ways using: \u2022 parallel wires with smaller cross sections instead of a one thick conductor; \u2022 stranded conductors (Litz wires); \u2022 coils made of copper or aluminum ribbon (foil winding). In an AFPM machine with an ironless winding arrangement (as shown in Fig. 5.13a), in addition to its normal component, the air gap magnetic field has a tangential component, which can lead to serious additional eddy current loss (Fig. 5.13b). The existence of a tangential field component in the air gap discourages the use of ribbon conductors as a low cost arrangement. Litz wires 5.8 Eddy Current Losses in the Stator Windings 173 allow significant reduction of eddy current loss, but they are more expensive and have fairly poor filling factors. As a cost effective solution, a bunch of parallel thin wires can be used. However, this may create a new problem, i.e. unless a complete balance of induced EMFs among the individual conducting paths is achieved, a circulating current between any of these parallel paths [161, 231, 263] may occur as shown in Fig. 5.13c, causing circulating eddy current losses. When operating at relatively high frequency magnetic fields, these eddy current effects may cause a significant increase of winding losses, which are intensified if there are circulating currents among the parallel circuits. These losses will deteriorate the performance of the AFPM brushless machine. Predicting the winding eddy current losses with a good accuracy is therefore very important at the early stage of design of such machines. Eddy current losses may be resistance limited when the flux produced by the eddy currents has a negligible influence on the total field [231]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001234_j.isatra.2014.01.004-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001234_j.isatra.2014.01.004-Figure1-1.png",
        "caption": "Fig. 1. Quadrotor UAV.",
        "texts": [
            " In Section 4, simulation results are performed to highlight the overall validity and the effectiveness of the designed controllers. In Section 5, a discussion, which is based on different synthesis control schemes, is presented to emphasize the performance of the proposed synthesis control method in this work, followed by the concluding remarks in Section 6. 2. Quadrotor model In order to describe the motion situations of the quadrotor model clearly, the position coordinate is to choose. The model of the quadrotor is set up in this work by the body-frame B and the earthframe E as presented in Fig. 1. Let the vector [x,y,z]' denotes the position of the center of the gravity of the quadrotor and the vector [u, v,w]' denotes its linear velocity in the earth-frame. The vector [p,q,r]' represents the quadrotor's angular velocity in the body-frame. m denotes the total mass. g represents the acceleration of gravity. l denotes the distance from the center of each rotor to the center of gravity. The orientation of the quadrotor is given by the rotation matrix R:E-B, where R depends on the three Euler angles [\u03d5,\u03b8,\u03c8]0, which represent the roll, the pitch and the yaw, respectively"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure9.13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure9.13-1.png",
        "caption": "FIGURE 9.13 Volume filament subdivisions for a single conductor bar with 1D current flow.",
        "texts": [
            " This has led to more sophisticated implementation of the VFI, which we consider after introducing more details. 228 SKIN EFFECT MODELING In this section, we treat wide-frequency-band skin-effect models since they are necessary for many applications. In general, several approximations can be made to speed up the solution. Here, we consider approximation based approaches while we will not consider matrix based speed-up techniques, for example, Ref. [41]. The following section is based on these approaches. We start this section with the model in Fig. 9.13. Such a model may be applied for an approximate PEEC model where it is clear that the current flow is in the x-direction. We should note that the models considered in this section are based on the 1D assumptions about the current flow. The cross section is subdivided with a conventional PEEC mesh. In the model in Fig. 9.13, each subbar, which we call a filament, can carry a different current. All surface cells, which excludes node 5, connect to a capacitive part for a complete PEEC model. However, since there are no cross connections, errors may be introduced in the current flow representation especially for nonstraight wires. We should note that the connective surfaces at the ends will not carry capacitive cells since the next section will join at the interface. Of course, for a complete model, capacitive surface cells are connected only to the external surface nodes",
            " The result will be a frequency-dependent 2\u00d72 impedance matrix. PROBLEMS 245 9.2 Construct a Single-conductor VFI skin-effect model Construct a single-conductor skin-effect model similar to the previous problem. Compute the impedance from end to end of the rectangular conductor (copper \ud835\udf0e = 5.8 \u00d7 107 S\u2215m), illustrated in Fig. 9.35. Consider xs = 0, xe = 5 mm, ys = 0, ye = 20 \u03bcm, zs = 0 \u03bcm, ze = 10 \u03bcm. Compute the length impedance in the x-direction of the conductor at 1 MHz, 1 GHz, and 10 GHz. Use the model in Fig. 9.13 with subdivisions along the length x. 9.3 Skin effect using 1D model in Section 9.2.2 Compare the impedance of the previous problem with a skin effect with a model that is constructed with the high-frequency 1D skin-effect model in Section 9.2.2. Hint: Assume that all four side surfaces are represented with such a 1D impedance model. Increase the cross section to ye = 3 mm and ze = 1 mm. Compare the result between the VFI model and the 1D surface impedance model for this larger cross section. 9"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001148_b708013c-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001148_b708013c-Figure2-1.png",
        "caption": "Fig. 2 Electron transfer mechanisms utilized in biofuel cell technology: (a) direct electron transfer and (b) electron transfer carried out by a redox mediator species.",
        "texts": [
            " This method has been attempted at both cathodic and anodic interfaces and has been achieved through solution phase mediators and mediators immobilized in various ways with or near the enzymes themselves. A schematic for this process is presented 1190 | Chem. Soc. Rev., 2008, 37, 1188\u20131196 This journal is c The Royal Society of Chemistry 2008 D ow nl oa de d by U ni ve rs ity o f L ee ds o n 19 /0 5/ 20 13 1 2: 23 :3 0. Pu bl is he d on 0 3 A pr il 20 08 o n ht tp :// pu bs .r sc .o rg | do i:1 0. 10 39 /B 70 80 13 C in Fig. 2. Mediators have been employed in biofuel systems through polymerization on the electrode surface prior to enzyme immobilization, coimmobilization of enzyme and mediator simultaneously or simply allowing the electron transfer mediator to be free in solution. These mediated systems do have drawbacks in that the species utilized to assist electron transfer are often not biocompatible or have short lifetimes themselves. A major advance arose from a research focus of the 1980s which documented several enzymes which are themselves capable of direct electron transfer (DET) via the active site of the enzyme. The first studies involve examining enzymes such as laccase that are capable of catalyzing the four-electron reduction of O2 to water through electron transfer from the electrode surface directly to the active site and through to the substrate. This electron transfer mechanism is also depicted in Fig. 2.19 This system has been utilized in cathodic compartments of biofuel cells; however, enzymes capable of oxidation at the anode surface have also been shown to demonstrate DET. Although these enzymes have been explored in a variety of electrochemical applications, they were not applied to biofuel cell technology until 2005 when laccase was utilized to reduce oxygen at a biocathode through DET and DET of pyrroloquinoline quinone (PQQ) dependent enzymes was demonstrated for anodic compartments.20 DET occurs through the enzyme\u2019s ability to act as a \u2018molecular transducer\u2019 that converts the chemical signal directly to an electrical one through the transfer of charge to a stable redox species which is in turn capable of transferring this charge to another molecule or electrode surface"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003702_s0043-1648(96)07467-4-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003702_s0043-1648(96)07467-4-Figure5-1.png",
        "caption": "Fig. 5. Transmitted load between two tooth pairs.",
        "texts": [
            " A point on one of the interacting surfaces will unlikely have the same x-value as a point on the opposite surface. Therefore, the overlap for a point Pi is calculated relative to an imaginary point Pim, on the opposite interacting flank and with the same x-value as the point Pi and with a position in the y-direction determined by linear interpolation.The sliding velocity for each point is equal to: v sabs[(v qv )y ] (23)i p g i For spur gears, the load is transmitted either by one or two tooth pairs in contact. In the latter case the load is divided between F t1 at y and F t2 at yqp bt , Fig. 5. F sF qF (24) t t1 t2 The division of the transmitted load can be determined by the following equations: F C t 8TF s y [(h (y)qh (y))y(h (yqp ) t1 p g p bt2 2 qh (yqp ))] (25) g bt F C t pTF s y [(h (y)qh (y))y(h (yqp ) t2 p g p bt2 2 qh (yqp ))] (26) g bt where C8T is the tooth deflection constant and h p (y) and h g (y) are the wear depths at the contact points on the pinion and wheel respectively. The wear depths in Eqs. (25) and (26) are calculated as themean value across the contactwidth for each contact"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003754_jsvi.1999.3109-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003754_jsvi.1999.3109-Figure1-1.png",
        "caption": "Figure 1. Ball bearing.",
        "texts": [
            " The excitation is because of the varying compliance vibrations of the bearing which arise because of the geometric and elastic characteristics of the bearing assembly varying according to the cage position. The ball bearing model considered here has equispaced balls rolling on the surfaces of the inner and outer races. For developing the theoretical model it is assumed that the outer race is \"xed rigidly to the support and the inner race is \"xed rigidly to the shaft and there is no bending of races. There is perfect rolling of balls on the races so that the two points of the ball (A and B) touching the outer and inner races have di!erent linear velocities (Figure 1). The center of the ball has a resultant translational velocity. Therefore, u cage \"u ROTOR A R i R i #R O B . (1) The varying compliance frequency is given by ulc\"u cage ]N b , (2) where N b is the number of balls. From equation (1), we can write ulc\"u ROTOR ]BN, (3) where BN\" R i R i #R O ]N b . (4) The number BN depends on the dimensions of the bearing, for SKF6002, BN\"3)6. The damping of a ball bearing is very small; this is because of friction and small amount of lubrication. The estimation of damping of a ball bearing is very di$cult because of the dominant extraneous damping which swamps the damping of the bearing",
            " The &&#'' sign as a subscript in equation (11) signi\"es the above. The total restoring force is the sum of the restoring force from each of the rolling elements. Thus, the total restoring force TABLE 1 Bearing damping SKF6002 (=\"6 N) Fukata (=\"58)8 N) K (N/m) 13)5]1)06 42]106 d (N s/m) 33)75}337)5 105}1050 components in the X and > directions are F x \"C b Nb + i/1 (x cos h i #y sin h i !c 0 )1>5 ` cos h i , (8) F y \"C b Nb + i/1 (x cos h i #y sin h i !c 0 )1>5 ` sinh i , As the shaft rotates, the angle h i changes with time (Figure 1) and is given by h i \" 2n N b (i!1)#u cage ]t, i\"1,2, N b . (9) Here the reference is the vertical axis which is the direction of the constant vertical force. We see that h i is a function of time and this imparts the parametric e!ect to the system. The damping in this system is represented by an equivalent viscous damping C. The value of the damping depends on the linearized bearing sti!ness (5). The value of the damping has been estimated (Table 1). The system governing equations accounting for inertia, restoring and damping force and constant vertical force acting on the inner race are, mxK#CxR #C b Nb + i/1 (x cos h i #y sin h i "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003549_robot.1996.509185-FigureI-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003549_robot.1996.509185-FigureI-1.png",
        "caption": "Fig. It: Example mobile manipulator.",
        "texts": [],
        "surrounding_texts": [
            "ments will be required to perform tasks on uneven terrain which m a y cause the system to approach, or achieve, a dangerous tapover instability. To avoid tipover in an automatic system, or to provide a human operator with an indication of proximity to tipover, it is necessary to define a measure of stability margin. This work presents a new tipover stability measure (the Force-Angle stability measure) which is easily computed and sensitive to topheaviness. The proposed metric is applicable to systems subject to inertial and external forces, operating over euen or uneven terrains. Performance of the measure is demonstrated using a forestry vehicle simulation.\n1 Introduction Mobile machines equipped with manipuhtor arms and controlled by on-board human operators are commonplace systems in the construction, mining, and forestry industries, see for example Fig. 1. When these systems exert large forces, move heavy payloads, or operate over very uneven or sloped terraiin, tipover instabilities may occur which endanger the operator, reduce productivity, and risk damaging the machine. With the introduction of computer control ( j .e. a supervisory control system) the safety and productivity of these mobile manipulators could be improved by automatic detection and prevention of tipover instabilities. In order to accomplish this, an appropriate measure of the tipover stability margin must be defined. Teleoperated or fully autonomous mobile manipulators operating in field environments (as proposed by the nucllear, military and aerospace industries) would also require a similar monitoring of the tipover stability maxgin.\nWork by the vehicular research community has focused on characterizing the lateral rollover propensity of a vehicle [I, 2 , 31. While the proposed static characterizations of machine lateral stability are not appropriate as instantaneous measures of a vehicle's stability, they do highlight the importance of considering vehicle center-of-gravity (c.g.) height and system mass (i.e. heaviness).\nAttempti3 by the robotics research community to solve the motion planning problem for mobile manipulators travelling over sloped terrain, or exerting large forces or moments on the environment, gave rise to various stability constraint definitions [4, 5, 61. By considering the degree to which a stability constraint is satisfied, one obtains a measure of the stability margin. However, the proposed measures are not topheavy sensitive and generally only consider lateral tipover.\nSeveral researchers examined more directly the question of how one should define the instantaneous stability margin for a mobile manipulator. McGhee proposed the use of the shortest, horizontal distance between the c.g. and the support pat,tern boundary projected onto a horizontal plane [7, 81. This measure was refined lby Song and later b,y Sugano, yet it remains insensitive t,o topheaviness and is only an approximation for systems on uneven terrain [9, IO]. Sreenivasan and Wilcox improve cmn the minimum distance measure by considering the minimum of each contact point distance to the net force vector, eliminating the need for a projection plane and thereby making the measure exact [ll]. However, this measure fails in the presence of angular loads and also does not take into consideration topheaviness. Davidson and Ikhweitzer also extend the work of McGhee, this time using screw mechanics to provide a measure wlhich eliminates the need for a projection plane while allowing for angular loads 1121. They recognize however that their measure is not sensitive to topheaviness. Messuri and Klein proposed the use of the rriiriirrium work required to tipover the vehicle, a measur'? which is sensitive tjo c.g. height [13]. Their energy-based approach was extended by\n0-7803-2988-4/96 $4.00 0 1996 I EEE 3111",
            "Ghasempoor and Sepheri to include inertial and external loads, subject however to the same assumption of constant load magnitude and direction throughout the tipover motion [14].\nThis work presents a new tipover stability measure (the Force-Angle stability measure) which has a simple graphical interpretation and is more easily computed than the measure of Ghasempoor and Sepheri yet remains sensitive to topheaviness and applicable to t he general case of systems operating over uneven terrain and subject to inertial and external forces. The simple nature of the proposed measure and the fact t,hat it does not require any integration make it advantageous to previously proposed measures. Performance of the Force-Angle stability measure is demonstrated using a forestry vehicle simulation.\n2 Background In determining the tipover stability margin of a ground vehicle system, one is necessarily concerned with the stability of the central body which generally provides mobility, i.e. the vehicle body or base. It is assumed in this work that the vehicle body is nominally in contact with the ground, as would be the case if mobility is provided via wheels, tracks, alternating (statically stable) legged support, or a combination thereof. A tipover or rollover instability occurs when a nominally upright vehicle body undergoes a rotation which results in a reduction of the number of ground contact points such that all remaining points lie on a single line (the tipover axis). Mobility control is then lost, and finally, if the situation is not reversed, the vehicle is overturned.\nA low c.g. height is always desirable from a stability point-of-view, heaviness on the other hand is stabilizing at low velocities and destabilizing at high velocities. In this work we are concerned with low velocity systems possibly exerting large forces on the environment, hence, heaviness will be considered a stabilizing influence.\n3 Force-Angle Stability Measure To help frame the discussion of the general form of the Force-Angle stability measure we first present a planar cxample which highlights its simple graphical nature. 3.1 Planar Example\nShown in Figure 2 is a two contact point. planar system whose center-of-mass (c.m.) is subject to a net force f, which is the sum of all forces acting on the vehicle body except the supporting reaction forces (which do not contribute to a tipover motion instability). This force vector subtends two angles, B1 and\n0 2 , with the two tip-over axis normals 11 and 12. The Force-Angle stability measure, cy, is given by the minimum of the two angles, weighted by the magnitude of the force vector (as denoted by lifril) for heaviness sensitivity:\n(1) Q = 81 ' llfrll\nCritical tipover stability occurs when 6' goes to zero (and therefore f, coincides with 11 or 12) or, when the magnitude off, goes to zero and even the smallest disturbance may topple the vehicle. Iff , lies outside the cone described by 11 and 12 the angle becomes negat,ive and tipover is in progress. For a vehicle which is capable of adjusting its center-of-mass height, or for a vehicle which carries a variable load, the tipover stability margin should be topheavy sensitive. This is illustrated in Fig. 3 for the Force-Angle stability measure where an increase in c.m. height clearly results in a smaller minimum angle and a reduced measure of tipover stability margin.\n3.2 General Form Geometry\nOf all the vehicle contact points with the ground, it is only necessary to consider those outermost points which form a convex support polygon when projected onto the horizontal plane. These points will simply be referred to as the ground contact points. Let pi represent the location of the ith ground contact point, e.g.\nT Pi = [ P z P y P Z I Z i = (1, \" ' , .) (2)\n3112"
        ]
    },
    {
        "image_filename": "designv10_0_0000078_j.robot.2014.08.014-Figure9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000078_j.robot.2014.08.014-Figure9-1.png",
        "caption": "Fig. 9. Exploded view of the soft robotic glove assembly depicting features of the first and second structural layers that support glove function.",
        "texts": [
            " In addition, forces generated from the soft actuators are distributed along the entire length of the actuator, thus providing good grasping performance during ADL. It is also noted that at higher pressures the multi-segment actuator for the thumb demonstrated some deviation in measurements, largely believed to be due to some twisting of the actuator. The soft wearable robotic glove features an open palm design and consists of a flexible support structure that couples the soft actuators with the user\u2019s hand. The glove has two structural layers (the \u2018\u2018first structural layer\u2019\u2019 and the \u2018\u2018second structural layer\u2019\u2019, as shown in Fig. 9) that together anchor the device to the hand via the wrist and finger tips. The \u2018\u2018first structural layer\u2019\u2019 is a neoprene material covering the dorsal surface of the hand and featuring a strap that wraps around the wrist to form an anchor. The \u2018\u2018second structural layer\u2019\u2019 attaches to the first layer via hook and loops (Velcro) and has \u2018\u2018fingertip pockets\u2019\u2019 and \u2018\u2018actuator pockets\u2019\u2019 at the distal end (see Fig. 9). The hook and loop connection between the two structural layers enables adjustability to hand size, bulk placement and pretension on the finger actuators. The adjustable pretension utilizes the neoprenematerial of the \u2018\u2018second structural layer\u2019\u2019 and the soft actuators as a rubber return spring which upon deactivation, return the fingers to the open hand state. The second structural layer also includes a hook and loop surface for mounting the proximal base of the actuators and strapping (\u2018\u2018actuator alignment loop\u2019\u2019, Fig. 9) maintains actuator alignment with the finger. The glove assembly with pressurized fluid weighs 285 g. To enable portability and minimize additional weight on the hand and arm, the device\u2019s hydraulic power supply and supporting electro-mechanical componentswere integrated into four pouches mounted on a waist belt pack. The pouches house, as shown in Fig. 10: (i) the 5 Ah (14.8 V) lithium polymer battery and power regulators which can support continuous glove operation for approximately 3.8 h, (ii) the microcontroller (Arduino MEGA 2560 R3, Arduino), the hydraulic pressure sensors, and the controller boards managing PWM signals for all valves, (iii) the 9"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002614_j.addma.2019.03.013-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002614_j.addma.2019.03.013-Figure4-1.png",
        "caption": "Fig. 4. Process maps constructed by using support vector machine as functions of scan speeds and (a) current, (b) line energy, (c) area energy, and (d) energy density. A cost parameter of 1.0 was used to construct these process maps.",
        "texts": [
            "\u201d Those that did not satisfy the aforementioned condition were labeled as \u201cbad.\u201d The two\u2013category classification of the cylinders is summarized in Table 3. The line energy Eline, the area energy Earea, and the energy density Evolume were calculated by using the following Eqs. (2), (3), and (4), respectively. =E P V/line scan (2) =E P V L/( )area scan offset (3) =E P V L Z/( )volume scan offset layer (4) where P, Vscan, Loffset, and Zlayer are the power (acceleration voltage\u00d7 current), scan speed, line offset, and layer thickness, respectively. Fig. 4 shows the process maps constructed by classifying the data in Table 3 using SVM. The output of the contour plot in Fig. 4 shows the value of decision function. Fig. 4(a)\u2013(d) shows the process map as a function of: current and scan speed, line energy and scan speed, area energy and scan speed, and energy density and scan speed, respectively. All of these process maps were constructed by setting the cost parameter C=1.0. In these process maps, the blue area showing a positive value of the decision function is the process window in which a fabricated part is predicted to show both a flat surface and a small deviation from the designed shape. Fig. 4(a) predicts that the process condition having a beam current between 6.15 and 15mA and a scan speed between 250 and 1000mm/s provides a near-net shaped part with a flat surface when the line offset, focus offset, layer thickness, and temperature at the bottom of the base plate are set to 250 \u03bcm, 22mA, 70 \u03bcm, and 850 \u00b0C, respectively. It should be noted that the lower limit in both current (6.15 mA) and scan speed (250mm/s) are not a true lower boundary because no data exists below the lower limit. Determining a separating hyperplane by using SVM requires training data of both sides across the separating hyperplane. In the other process maps (Fig. 4(b)\u2013(d)), the process window ranges in a scan speed of between 250 and 1000mm/s. The upper boundaries of the parameters of the vertical axis were not determined in Fig. 4 because no training data exists across the separating hyperplane for the vertical direction (i.e., the data points having a large current for a scan speed less than 1000mm/s). As is known from (2), (3), and (4) with the fact that Loffset and Zlayer are fixed in this study, Fig. 4(b)\u2013(d) shows the same trend, excepting the value of the vertical axis. When the fact that we used the process maps on operating AM machines is considered, the process maps should be a function of the parameters that operators can use to enter into AM machines directly. Among the process maps shown in Fig. 4, accordingly, Fig. 4(a) as a function of current and scan speed is more effective for operators to refer to for industrial use of AM machines. This is because the others require operators to convert the vertical axis into a current value using (2), (3), and (4) whenever they want to change parameters during the process. Although the hyperparameter of the kernel function was optimized by a cross validation technique when constructing the process maps shown in Fig. 4, the cost parameter was fixed to 1.0. To investigate the effect of the cost parameter on the separating boundaries in the process maps, we constructed the process maps as a function of current and scan speed with a varying cost parameter (C=0.1, 0.5, 5.0, 10, 20, and 100 as shown in Fig. 5(a)\u2013(f), respectively). The error rate for each cost parameter, which is the number of misclassified data points divided by the total number of data points, is shown in Table 4. Although using a cost parameter that provides an error rate of 0 would seem to be more effective, in general, experimental results reveal the effect of experimental errors (e"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure14.61-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure14.61-1.png",
        "caption": "Figure 14.61 These trusses are an alternative for the beam subject to bending in Figure 14.56.",
        "texts": [
            " The cable in Figure 14.57 and the bar structure in Figure 14.58a are also kinematically indeterminate. The equilibrium is stable (reliable) in this case as the load makes the structure go back into the original equilibrium position after a disruption. In Figure 14.60a, the bar structure in Figure 14.58a has been changed into a kinematically determinate truss. All the interior members are zero. As in contrast to the cable, the truss has the benefit that the shape does not change when the load changes. In Figure 14.61, the forces on the truss are shifted to the horizontal plane through the supports. The verticals are no longer zero members. These 682 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM trusses, in which all the members are subject to extension (or are zero members), can be an alternative for the beam subject to bending in Figure 14.56. Figure 14.62a shows the bar structure from Figure 14.58b, but now without tension member. This kinematically indeterminate structure can be made kinematically determinate not only by changing it into a truss, but also by replacing the hinged joints between the bars by rigid joints"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure14.33-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure14.33-1.png",
        "caption": "Figure 14.33 The isolated cable CD.",
        "texts": [
            " The vertical component of the tensile force in cable BC is a maximum at B and C, as the cable is steepest here: V BC max = 1 2q BC = 1 2 (250 kN/m)(1624 m) = 203 MN. The cable force is also a maximum here: NBC max = \u221a (H BC)2 + (V BC max) 2 = \u221a (448 MN)2 + (203 MN)2 = 492 MN. c. In Figure 14.32, cables BC and CD have been isolated at C from the tower. If there is no bending in the tower, the resulting horizontal force on the tower must be zero. This means that the horizontal component of the cable force in an end span is equal to that in the middle span: H CD = H BC = 448 MN. In Figure 14.33, cable CD has been isolated. The resultant R of the uniformly distributed load is R = (250 kN/m)(536 m) = 134 MN. Figure 14.34 The forces on tower C and foundation block D. The moment equilibrium of cable CD gives \u2211 T |D = (448 MN)(200 m) \u2212 V CD C (536 m) + (134 MN)(268 m) = 0 so that V CD C = (448 MN)(200 m) + (134 MN)(268 m) 536 m = 234 MN. Next we find from the vertical force equilibrium V CD C = V CD C \u2212 R = (234 MN) \u2212 (134 MN) = 100 MN. In Figure 14.34, cables BC and CD have been isolated from the tower at C and the foundation block at D. The tower is loaded at C by a vertical compressive force: V BC C + V CD C = (203 MN) + (234 MN) = 437 MN. d. The foundation block in D is subject to a horizontal force of 448 MN and an upward vertical force of 100 MN (see Figure 14.34). e. The maximum force in cable CD occurs at C, where the cable is steepest: NCD max = \u221a (H CD)2 + (V CD max) 2 = \u221a (448 MN)2 + (234 MN)2 = 505 MN. Figure 14.33 The isolated cable CD. 14 Cables, Lines of Force and Structural Shapes 665 f. For middle span BC, it applies that pBC k BC = 184 m 1624 m = 0.113. The maximum (vertically measured) distance from cable CD to the chord is pCD k = 1 8q( CD)2 H CD = 1 8 (250 kN/m)(536 m)2 448 \u00d7 103 kN = 20 m so that for the end spans the following applies: pCD k CD = 20 m 536 m = 0.037. Example 3 Cable AB in Figure 14.35 has a span of 60 m and is carrying a number of pipelines with a total weight of 20 kN/m. The difference in elevation of the end supports at A and B is 12 m"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure9.14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure9.14-1.png",
        "caption": "FIGURE 9.14. A controlled compound pendulum.",
        "texts": [
            "306) and therefore, the kinetic and potential energies of the pendulum are K = 1 2 m \u00b3 l2\u03b8\u0307 2 + l2\u03d5\u03072 sin2 \u03b8 \u00b4 (9.307) V = \u2212mgl cos \u03b8. (9.308) The kinetic potential function of this system is then equal to L = 1 2 m \u00b3 l2\u03b8\u0307 2 + l2\u03d5\u03072 sin2 \u03b8 \u00b4 +mgl cos \u03b8 (9.309) which leads to the following equations of motion: \u03b8\u0308 \u2212 \u03d5\u03072 sin \u03b8 cos \u03b8 + g l sin \u03b8 = 0 (9.310) \u03d5\u0308 sin2 \u03b8 + 2\u03d5\u0307\u03b8\u0307 sin \u03b8 cos \u03b8 = 0. (9.311) Example 369 Controlled compound pendulum. A massive arm is attached to a ceiling at a pin joint O as illustrated in Figure 9.14. Assume that there is viscous friction in the joint where an ideal motor can apply a torque Q to move the arm. The rotor of an ideal motor has no moment of inertia by assumption. 9. Applied Dynamics 563 The kinetic and potential energies of the manipulator are K = 1 2 I\u03b8\u0307 2 = 1 2 \u00a1 IC +ml2 \u00a2 \u03b8\u0307 2 (9.312) V = \u2212mg cos \u03b8 (9.313) where m is the mass and I is the moment of inertia of the pendulum about O. The Lagrangean of the manipulator is L = K \u2212 V = 1 2 I\u03b8\u0307 2 +mg cos \u03b8 (9.314) and therefore, the equation of motion of the pendulum is M = d dt \u00b5 \u2202L \u2202\u03b8\u0307 \u00b6 \u2212 \u2202L \u2202\u03b8 = I \u03b8\u0308 +mgl sin \u03b8"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000811_j.jsv.2012.05.039-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000811_j.jsv.2012.05.039-Figure8-1.png",
        "caption": "Fig. 8. Planet gears with manually created pitting damage: (a) baseline, (b) slight, (c) moderate and (d) severe.",
        "texts": [
            " The other one is the signals from the same gearbox with naturally developed gear wear. Fig. 7 shows the planetary gearbox test rig. We did experiments on the 2nd stage planetary gearbox, for both the manually created damage and the naturally developed fault. Table 4 lists the gear parameters of the 2nd stage planetary gearbox. For the manual pitting experiments, we introduced pitting of different levels (i.e. baseline, slight, moderate, and severe pitting) to the planet gears inside of the 2nd planetary gearbox. Fig. 8 shows the picture of planet gears with different pitting levels. In each experiment, a planet gear of different pitting level was used, whereas all the other gears were normal. During experiments, the rotating frequency of the input shaft connecting sun gear of the 2nd planetary gearbox Table 4 Gear parameters. Gear Sun Ring Planet Number of planets Number of teeth 19 81 31 4 was set to 0.7778 Hz, and a load of 10,000 lb-in. was applied to the output shaft connecting planet carrier of the 2nd planetary gearbox"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001049_j.pmatsci.2020.100703-Figure4-42-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001049_j.pmatsci.2020.100703-Figure4-42-1.png",
        "caption": "Fig. 4-42 Longitudinal residual stress distributions in a SS316 deposit printed using (a) DEDGMA. (b) DED-L and (c) PBF-L. The scanning direction is along the positive x-axis. Half of the solution domain is shown because of the symmetry with respect to XZ plane [283].",
        "texts": [
            " For example, Table 4-2 summarizes the reported [309, 328, 332, 394, 483-494] defect formation for aluminum alloys fabricated by different AM processes. Therefore, an alloy is not equally printable by all printing processes. Mechanistic models of different printing processes useful to evaluate the formation of various defects for a particular alloy. For example, Szost et al. [495] found that titanium alloy parts accumulate higher residual stresses when printed by the DED-GMA process compared with the DED-L process. Fig. 4-42 shows that the computed longitudinal stress in the DED-GMA component is the highest among three processes because the component is printed using the thickest layers. In contrast, because of the small molten pool and thin layers in PBF-L, the component printed using this process accumulates the least residual stresses. The effects of composition change are more pronounced during PBF processes compared to DED [20] since the small beam sizes of the heat source and rapid scanning speeds in PBF result in larger surface area to volume ratio, as shown in Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001031_j.jsv.2008.03.038-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001031_j.jsv.2008.03.038-Figure8-1.png",
        "caption": "Fig. 8. A one stage gearbox system [5].",
        "texts": [
            " The total mesh stiffness under each of the four crack sizes has been calculated as a function of the shaft rotation angle and plotted in Fig. 7. From Fig. 7, it can be observed that as the size of the crack grows, the total mesh stiffness when the cracked tooth is in meshing becomes much lower. This is important information for fault detection and assessment. We will adopt the mathematical model with torsional and lateral vibration reported by Bartelmus [5] in 2001. The model of a one stage gearbox system is given in Fig. 8. It is a two-parameter (stiffness and damping) model with torsional and lateral vibration, which means that it includes both the linear and rotational equations of the system\u2019s motion. This model represents a system with six degrees of freedom, which is driven by electric motor moment, M1, and loaded with external moment, M2. This model is simple enough to enable to focus on the effects of crack growth on the vibration response of the system. Thus, in this paper, we have ARTICLE IN PRESS S. Wu et al"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000028_0005-1098(94)90209-7-Figure32-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000028_0005-1098(94)90209-7-Figure32-1.png",
        "caption": "FIG. 32. Nyquist diagrams showing complex valued describing function of an integrated plant/friction model. The family of curves corresponds to different ratios of static: Coulomb friction [from Tou and Schultheiss (1953), courtesy of the publisher].",
        "texts": [
            " The early investigators (Tou and Schultheiss, 1953; Satyendra, 1956; Silverberg, 1957; Shen, 1962; Woodward, 1963) did not employ the memoryless element construction, but developed describing functions for the composite of the elements from the force input to the velocity output, as shown in Fig. 30. With this approach a describing function can be worked out by piecewise integration of the response. This procedure, illustrated in Fig. 31, gives rise to a describing function that is a curve on the complex plane, or a family of curves for the case of Coulomb + static friction, as illustrated in Fig. 32. Here the family of complex valued describing function curves is parameterized by the ratio of the static + Coulomb friction and plotted as - 1 / N ( A , to), as described in equation (16). This analysis predicts that the series compensator will exhibit stick-slip while the parallel c,A TEL#/ \\ x \",t I / \\ - ~ / D \\/\u00b0 / i / ,, \\ b c d f i j FIo. 31. Piecewise evaluation of the output of a system plus friction element with sinusoidal applied force and periods of standstill [from Tou and Schultheiss (1953), courtesy of the publisher]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000028_0005-1098(94)90209-7-Figure45-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000028_0005-1098(94)90209-7-Figure45-1.png",
        "caption": "FIG. 45. Outer race of ball bearing is attached to static outer ring through torsional flexing elements [from Clingman (1991), courtesy of publisher].",
        "texts": [
            " Solids, such as Rulon \u00ae, are often used in conjunction with liquid lubricants as slideway liner materials. 4.2.2. Bearings Bearing friction can be a problem in high-precision positioning, pointing and tracking systems. Several schemes have been proposed to overcome motion errors from ball bearings. One involves active control of the bearing outer race (Bifano and Dow, 1985). In another method, the outer bearing race is not rigidly mounted to the machine frame. Instead, it is connected to the frame through torsional springs as shown in Fig. 45 (Clingman, 1991). At motion initiation and direction reversal, the friction torque and spring torque act in series reducing the effective slope of the friction-displacement curve. A controller can be added to the bearing which drives the torsional springs to null displacement (Clingman, 1991). In order to avoid the nonlinearity of low-velocity friction, oil or air hydrostatic bearings can be used. While producing extremely low friction, air bearings exhibit low stiffness and damping in the normal direction which can make them sensitive to profile errors of guideways and to external disturbance forces"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure10.10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure10.10-1.png",
        "caption": "Figure 10.10 (a) The small force F is the resultant of all the small forces acting on a small but finite area A. (b) The stress vector p is defined as the limit value of F/ A for A \u2192 0.",
        "texts": [
            "9 shows the positive directions of the section forces in various coordinate systems.1 The sign convention given here for the section forces N , V and M is associated with the sign convention for stresses in the cross-section. We look at this in more detail in Section 10.1.2. On a positive sectional plane, consider a small area A. Let F be the resultant of all the small forces that are transferred by the matter via that small area. F is built up by the contributions of a large number of interactions between the particles of matter. Figure 10.10a shows the components Fx ; Fy ; Fz of the small force F . If A is smaller, so is F . It is assumed that the relationship between F and A has a limit when A approaches zero. This limit was defined in Section 6.5 as the stress vector p: p = lim A\u21920 F A . The definition of the stress vector is based on the idealised model of continuous matter. Figure 10.10b shows the components px ; py ; pz of the stress vector p. 1 Note: it is wrong to say that a bending moment is positive when the couple acts on the positive sectional plane in accordance with the positive sense of rotation and on the negative sectional plane in accordance with the negative sense of rotation. This is shown in Figures 10.9a and 10.9d. 10 Section Forces 393 If we look at the same small area A on the negative sectional plane, there is an equal but opposite force, in accordance with the principle of action and reaction"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001498_tie.2010.2045322-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001498_tie.2010.2045322-Figure3-1.png",
        "caption": "Fig. 3. Single-phase open-circuit fault.",
        "texts": [
            " (7) Furthermore, in a practical multiphase drive system, the neutral connection is normally absent. Hence, the summation of the phase currents should be constrained to zero n\u2211 g=1 ig = 0. (8) The aforementioned conditions for fault-tolerant operation (6)\u2013(8) and the condition of mirror symmetry of the faulttolerant currents can be used for any multiphase PM machine. Based on this, the fault-tolerant operations of a five-phase PM machine under various fault conditions are presented as follows. A two-pole five-phase PM machine is shown in Fig. 3. Considering only fundamental and third-harmonic components, the trapezoidal back EMFs (Fig. 1) in the stator phases can be defined as ea =E1 \u00b7 cos(\u03c9t) + E3 \u00b7 cos(3\u03c9t) eb =E1 \u00b7 cos(\u03c9t \u2212 2\u03c0/5) + E3 \u00b7 cos(3\u03c9t + 4\u03c0/5) ec =E1 \u00b7 cos(\u03c9t \u2212 4\u03c0/5) + E3 \u00b7 cos(3\u03c9t \u2212 2\u03c0/5) ed =E1 \u00b7 cos(\u03c9t + 4\u03c0/5) + E3 \u00b7 cos(3\u03c9t + 2\u03c0/5) ee =E1 \u00b7 cos(\u03c9t + 2\u03c0/5) + E3 \u00b7 cos(3\u03c9t \u2212 4\u03c0/5) (9) where E1 and E3 are the amplitudes of the fundamental and third-harmonic back-EMF components, respectively. Assume that an open-circuit fault has occurred in phase a"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000157_j.actamat.2011.12.032-Figure13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000157_j.actamat.2011.12.032-Figure13-1.png",
        "caption": "Fig. 13. Schematic views illustrating the directional solidification-induced columnar grains and c00 precipitate arrays as well as the melt pool/layer structure development in SLM. The coincident plane variants for (100) [200] are shown at [002], [200] and [020].",
        "texts": [
            " 12 the propensity of c00 precipitates are coincident (coherent) with the (100) planes while some are observed to be coincident with the other (01 0) variant. Dense and often irregular arrangements of various geometries and very small (regular) sizes of precipitates (c00) also occur throughout the NiCr (fcc) matrix. In retrospect, comparing and examining Figs. 3, 4\u20136, 8, 9 and 11, as well as Figs. 10 and 12, provides a conceptual mechanistic view of the SLM build process producing columnar (textured) grains and c00 precipitate arrays. Fig. 13 provides a schematic composite showing these features as they relate to the scan geometry and melt pool related layer development, as well as the columnar arrays formed as directional, epitaxial-like solidification phenomena related to the high temperature gradients occurring during the SLM process. The stacks of disc-like or oblate spheroidal c00 precipitates implicit in Fig. 12 appear to be created in successively formed melt pools as a consequence of their short interaction times and high conductive heat transfer, possibly facilitated by the argon or nitrogen gas environments, or their corresponding thermal conductivities"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001885_tac.2011.2157399-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001885_tac.2011.2157399-Figure2-1.png",
        "caption": "Fig. 2. Two inverted pendulums connected by a spring and a damper.",
        "texts": [
            " Consequently, it is not difficult to verify that the recursive design methodology presented in Section IV-B, will result in a controller that guarantees prescribed performance for the outputs, as well as bounded closed loop signals. In this section, we shall present a simulation study on a MIMO system in block triangular form, highlighting the robustness of the proposed RPPC methodology against dynamic uncertainties, measurement errors, unknown nonlinearities and time-varying parameters. Specifically, we consider the control problem of two inverted pendulums connected by a spring and a damper as shown in Fig. 2. Each pendulum is controlled by a torque input applied by a servomotor at its base. The equation of motion in terms of the angular positions of the pendulums , and applied torques , is given by: ! \" # # $ (18) ! \" # # $ % (19) where represents the force applied by the spring and the damper at the connection points , , with denoting the distance between the points and , which is given by In (18) and (19), is defined as follows: while , correspond to friction terms, applied at the servomotor axes, which are assumed to follow the LuGre model [19]: Finally, the time-varying parameter is taken as: if if thus modeling the case where the second servomotor loses 50% of its effectiveness while "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.50-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.50-1.png",
        "caption": "FIGURE 8.50. The coordinate frames of the first and fourth tires of a 4-wheel vehicle with respect to the body frame.",
        "texts": [
            "57) The position of the wheel center W r = 0, for a cambered and steered wheel is at Cr = CTW W r = \u23a1\u23a2\u23a2\u23a3 Rw sin \u03b3 sin \u03b4 \u2212Rw cos \u03b4 sin \u03b3 \u2212Rw(1\u2212 cos \u03b3) 1 \u23a4\u23a5\u23a5\u23a6 (8.58) The zc = Rw (cos \u03b3 \u2212 1) indicates how much the center of the wheel comes down when the wheel cambers. If the wheel is not steerable, then \u03b4 = 0 and the transformation matrix CTW reduces to CTW = \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 cos \u03b3 \u2212 sin \u03b3 \u2212Rw sin \u03b3 0 sin \u03b3 cos \u03b3 Rw (cos \u03b3 \u2212 1) 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 (8.59) that shows CrP = CTW W rP = \u23a1\u23a2\u23a2\u23a3 xP yP cos \u03b3 \u2212Rw sin \u03b3 \u2212 zP sin \u03b3 zP cos \u03b3 + yP sin \u03b3 +Rw (cos \u03b3 \u2212 1) 1 \u23a4\u23a5\u23a5\u23a6 (8.60) Example 333 F Tire to vehicle coordinate frame transformation. Figure 8.50 illustrates the first and fourth tires of a 4-wheel vehicle. There is a body coordinate frame B (x, y, z) attached to the mass center C of the vehicle. There are also two tire coordinate frames T1 (xt1 , yt1 , zt1) and T4 (xt4 , yt4 , zt4) attached to the tires 1 and 4 at the center of their tireprints. The origin of the tire coordinate frame T1 is at Bd1 BdT1 = \u23a1\u23a3 a1 \u2212b1 \u2212h \u23a4\u23a6 (8.61) where, a1 is the longitudinal distance between C and the front axle, b1 is the lateral distance between C and the tireprint of the tire 1, and h is the height of C from the ground level"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002557_s00170-020-05569-3-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002557_s00170-020-05569-3-Figure8-1.png",
        "caption": "Fig. 8 Two-dimensional geometric characteristics of a molten pool obtained by image signals with an infrared spectrum [121]",
        "texts": [
            " [53\u201355] obtained the edge points of the deposited layer width through the Laplacian operator, and then a Hough transform was used to determine the width. Image signals with an infrared spectrumWhen image signals with an infrared spectrum are used to eliminate noises and disturbances, the data process methods are similar to image signals with a visible spectrum [36, 51, 110, 111]. In addition, the geometric characteristics of a molten pool can also be obtained on the basis of the melting point [70, 112, 115, 116, 121, 122], as seen in Fig. 8. Special spectrum signals Except for image signals with a visible or an infrared spectrum, special spectrum signals can also indicate the geometric characteristics of a molten pool. Miyagi et al. [30] investigated the correction between molten pool widths and spectrum signal acquired by photodiodes. This study indicated that the molten pool widths could be obtained by spectrum signals. Moreover, the temperature distribution of a molten pool was determined by the spectrum signals using a spectrometer or hyperspectral cameras; the molten pool boundary could be ensured by the melting point [112, 138, 150, 152, 153]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.14-1.png",
        "caption": "FIGURE 8.14. A Robert suspension mechanism with a Panhard arm.",
        "texts": [],
        "surrounding_texts": [
            "The suspension is what links the wheels to the vehicle body and allows relative motion. This chapter covers the suspension mechanisms, and discusses the possible relative motions between the wheel and the vehicle body. The wheels, through the suspension linkage, must propel, steer, and stop the vehicle, and support the associated forces. 8.1 Solid Axle Suspension The simplest way to attach a pair of wheels to a vehicle is to mount them at opposite ends of a solid axle, such as the one that is shown in Figure 8.1. The solid axle must be attached to the body such that an up and down motion in the z-direction, as well as a roll rotation about the x-axis, is possible. So, no forward and lateral translation, and also no rotation about the axle and the z-axis, is allowed. There are many combinations of links and springs that can provide the kinematic and dynamic requirements. The simplest design is to clamp the axle to the middle of two leaf springs with their ends tied or shackled to the vehicle frame as shown schematically in Figure 8.1. A side view of a multi-leaf spring and solid axle is shown in Figure 8.2. A suspension with a solid connection between the left and right wheels is called dependent suspension. 456 8. Suspension Mechanisms The performance of a solid axle with leaf springs suspension can be improved by adding a linkage to guide the axle kinematically and provide dynamic support to carry the non z-direction forces. The solid axle with leaf spring combination came to vehicle industry from horse-drawn vehicles. Example 299 Hotchkiss drive. When a live solid axle is connected to the body with nothing but two leaf springs, it is called the Hotchkiss drive, which is the name of the car that used it first. The main problems of a Hotchkiss drive, which is shown in Figure 8.2, are locating the axle under lateral and longitudinal forces, and having a low mass ratio \u03b5 = ms/mu, where ms is the sprung mass and mu is the unsprung mass. Sprung mass refers to all masses that are supported by the spring, such as vehicle body. Unsprung mass refers to all masses that are attached to and not supported by the spring, such as wheel, axle, or brakes. Example 300 Leaf spring suspension and flexibility problem. The solid axle suspension systems with longitudinal leaf springs have many drawbacks. The main problem lies in the fact that springs themselves act as locating members. Springs are supposed to flex under load, but their flexibility is needed in only one direction. However, it is the nature of leaf springs to twist and bend laterally and hence, flex also in planes other than the tireplane. Leaf springs are not suited for taking up the driving and braking traction forces. These forces tend to push the springs into an S-shaped profile, as shown in Figure 8.3. The driving and braking flexibility of leaf springs, generates a negative caster and increases instability. Long springs provide better ride. However, long sprigs exaggerate their bending and twisting under different load conditions. Example 301 Leaf spring suspension and flexibility solution. To reduce the effect of a horizontal force and S-shaped profile appearance in a solid axle with leaf springs, the axle may be attached to the chassis by a longitudinal bar as Figure 8.4(a) shows. Such a bar is called an anti-tramp 8. Suspension Mechanisms 457 bar, and the suspension is the simplest cure for longitudinal problems of a Hotchkiss drive. A solid axle with an anti-tramp bar may be kinematically approximated by a four-bar linkage, as shown in Figure 8.4(b). Although an anti-tramp bar may control the shape of the leaf spring, it introduces a twisting angle problem when the axle is moving up and down, as shown in Figure 8.5. Twisting the axle and the wheel about the axle is called caster. The solid axle is frequently used to help keeping the wheels perpendicular to the road. Example 302 Leaf spring location problem. The front wheels need room to steer left and right. Therefore, leaf springs cannot be attached close to the wheel hubs, and must be placed closer to the middle of the axle. That gives a narrow spring-base, which means that a small side force can sway or tilt the body relative to the axle through a considerable roll angle due to weight transfer. This is uncomfortable for the vehicle passengers, and may also produce unwanted steering. The solid axle positively prevents the camber change by body roll. The wheels remain upright and hence, do not roll on a side. However, a solid axle shifts laterally from its static plane and its center does not remain on the vehicle\u2019s longitudinal axis under a lateral force. 458 8. Suspension Mechanisms A solid axle produces bump-camber when single-wheel bump occurs. If the right wheel goes over a bump, the axle is raised at its right end, and that tilts the left wheel hub, putting the left wheel at a camber angle for the duration of deflection. Example 303 Triangular linkage. A triangulated linkage, as shown in Figure 8.6, may be attached to a solid axle to provide lateral and twist resistance during acceleration and braking. Example 304 Panhard arm. High spring rate is a problem of leaf springs. Reducing their stiffness by narrowing them and using fewer leaves, reduces the lateral stiffness and increases the directional stability of the suspension significantly. A Panhard arm is a bar that attaches a solid axle suspension to the chassis laterally. Figure 8.7 illustrates a solid axle and a Panhard arm to guide the axle. Figure 8.8 shows a triangular linkage and a Panhard arm combination for guiding a solid axle. A double triangle mechanism, as shown in Figure 8.9, is an alternative design to guide the axle and support it laterally. Example 305 Straight line linkages. There are many mechanisms that can provide a straight line motion. The simplest mechanisms are four-bar linkages with a coupler point moving straight. Some of the most applied and famous linkages are shown in Figure 8.10. By having proper lengths, the Watt, Robert, Chebyshev, and Evance linkages can make the coupler point C move on a straight line vertically. Such a mechanism and straight motion may be used to guide a solid axle. Two Watt suspension mechanisms with a Panhard arm are shown in Figures 8.11 and 8.12. 460 8. Suspension Mechanisms 8. Suspension Mechanisms 461 Figures 8.13, 8.14, and 8.15 illustrate three combinations of Robert suspension linkages equipped with a Panhard arm. Example 306 Solid axle suspension and unsprung mass problem. A solid axle is counted as an unsprung member, and hence, the unsprung mass is increased where using solid axle suspension. A heavy unsprung mass ruins both, the ride and handling of a vehicle. Lightening the solid axle makes it weaker and increases the most dangerous problem in vehicles: axle breakage. The solid axle must be strong enough to make sure it will not break under any loading conditions at any age. As a rough estimate, 90% of the leaf spring mass may also be counted as unsprung mass, which makes the problem worse. The unsprung mass problem is worse in front, and it is the main reason that they are no longer used in street cars. However, front solid axles are still common on trucks and buses. These are heavy vehicles and solid axle 8. Suspension Mechanisms 463 464 8. Suspension Mechanisms suspension does not reduce the mass ratio \u03b5 = ms/mu very much. When a vehicle is rear-wheel-drive and a solid axle suspension is used in the back, the suspension is called live axle. A live axle is a casing that contains a differential, and two drive shafts. The drive shafts are connected to the wheel hubs. A live axle can be three to four times heavier than a dead I-beam axle. It is called live axle because of rotating gears and shafts inside the axle. Example 307 Solid axle and coil spring. To decrease the unsprung mass and increase vertical flexibility of solid axle suspensions, it is possible to equip them with coil springs. A sample of a solid axle suspension with coil spring is shown in Figure 8.16. The suspension mechanism is made of four longitudinal bars between the axle and chassis. The springs may have some lateral or longitudinal angle to introduce some lateral or longitudinal compliance. Example 308 De Dion axle. When a solid axle is a dead axle with no driving wheels, the connecting beam between the left and right wheels may have different shapes to do different jobs, usually to give the wheels independent flexibility. We may also modify the shape of a live axle to attach the differential to the chassis and reduce the unsprung mass. De Dion design is a modification of a beam axle that may be used as a dead axle or to attach the differential to the chassis and transfer the driving power to the drive wheels by employing universal joints and split shafts. Figure 8.17 illustrates a De Dion suspension. 8.2 Independent Suspension Independent suspensions is introduced to let a wheel to move up and down without affecting the opposite wheel. There are many forms and designs of independent suspensions. However, double A-arm and McPherson strut suspensions are the simplest and the most common designs. Figure 8.18 illustrates a sample of a double A-arm and Figure 8.19 shows a McPherson suspension. Kinematically, a double A-arm suspension mechanism is a four-bar linkage with the chassis as the ground link, and coupler as the wheel carrying link. A McPherson suspension is an inverted slider mechanism that has the chassis as the ground link and the coupler as the wheel carrying link. A double A-arm and a McPherson suspension mechanism on the left and right wheels are schematically shown in Figures 8.20 and 8.21 respectively. Double A-arm, is also called double wishbone, or short/long arm suspension. McPherson also may be written as MacPherson. Example 309 Double A-arm suspension and spring position. Consider a double A-arm suspension mechanism. The coil spring may be between the lower arm and the chassis, as shown in Figure 8.18. It is also possible to install the spring between the upper arm and the chassis, or between the upper and lower arms. In either case, the lower or the upper arm, which supports the spring, is made stronger and the other arm acts as a connecting arm. Example 310 Multi-link suspension mechanism. When the two side bars of an A-arm are attached to each other with a joint, as shown in Figure 8.22, then the double A-arm is called a multi-link mechanism. A multi-link mechanism is a six-bar mechanism that may have a better coupler motion than a double A-arm mechanism. However, multi- 466 8. Suspension Mechanisms 8. Suspension Mechanisms 467 link suspensions are more expensive, less reliable, and more complicated compare to a double A-arm four-bar linkage. There are vehicles with more than six-link suspension with possibly better kinematic performance. Example 311 Swing arm suspension. An independent suspension may be as simple as a triangle shown in Figure 8.23. The base of the triangle is jointed to the chassis and the wheel to the tip point. The base of the triangle is aligned with the longitudinal axis of the vehicle. Such a suspension mechanism is called a swing axle or swing arm. The variation in camber angle for a swing arm suspension is maximum, compared to the other suspension mechanisms. Example 312 Trailing arm suspension. Figure 8.24 illustrates a trailing arm suspension that is a longitudinal arm with a lateral axis of rotation. The camber angle of the wheel, supported by a trailing arm, will not change during the up and down motion. Trailing arm suspension has been successfully using in a variety of frontwheel-drive vehicles, to suspend their rear wheels. Example 313 Semi-trailing arm Semi-trailing arm suspension, as shown in Figure 8.25, is a compromise between the swing arm and trailing arm suspensions. The joint axis may have any angle, however an angle not too far from 45 deg is more applied. Such suspensions have acceptable camber angle change, while they can handle both, the lateral and longitudinal forces. Semi-trailing design has successfully applied to a series of rear-wheel-drive cars for several decades. Example 314 Antiroll bar and roll stiffness. Coil springs are used in vehicles because they are less stiff with better 470 8. Suspension Mechanisms ride comfort compared to leaf springs. Therefore, the roll stiffness of the vehicle with coil springs is usually less than in vehicles with leaf springs. To increase the roll stiffness of such suspensions, an antiroll bar must be used. Leaf springs with reduced layers, uni-leaf, trapezoidal, or nonuniform thickness may also need an antiroll bar to compensate for their reduced roll stiffness. The antiroll bar is also called a stabilizer. Figure 8.26 illustrates an anti-roll bare attached to a solid axle with coil springs. Example 315 Need for longitudinal compliance. A bump is an obstacle on the road that opposes the forward motion of a wheel. When a vehicle goes over a bump, the first action is a force that tends to push the wheel backward relative to the rest of the vehicle. So, the lifting force has a longitudinal component, which will be felt inside the vehicle unless the suspension system has horizontal compliance. There are situations in which the horizontal component of the force is even higher than the vertical component. Leaf springs can somewhat absorb this horizontal force by flattening out and stretching the distance from the forward spring anchor and the axle. Such a stretch is usually less than 1/2 in \u2248 1 cm. 8.3 Roll Center and Roll Axis The roll axis is the instantaneous line about which the body of a vehicle rolls. Roll axis is found by connecting the roll center of the front and rear suspensions of the vehicle. Assume we cut a vehicle laterally to disconnect the front and rear half of the vehicle. Then, the roll center of the front or rear suspension is the instantaneous center of rotation of the body with respect to the ground. 8. Suspension Mechanisms 471 472 8. Suspension Mechanisms Figure 8.27 illustrates a sample of the front suspensions of a car with a double A-arm mechanism. To find the roll center of the body with respect to the ground, we analyze the two-dimensional kinematically equivalent mechanism shown in Figure 8.28. The center of tireprint is the instant center of rotation of the wheel with respect to the ground, so the wheels are jointed links to the ground at their center of tireprints. The instant center I18 is the roll center of the body with respect to the ground. To find I18, we apply the Kennedy theorem and find the intersection of the line I12I28 and I13I38 as shown in Figure 8.29. The point I28 and I38 are the instant center of rotation for the wheels with respect to the body. The instant center of rotation of a wheel with respect to the body is called suspension roll center. So, to find the roll center of the front or rear half of a car, we should determine the suspension roll centers, and find the intersection of the lines connecting the suspension roll centers to the center of their associated tireprints. The Kennedy theorem states that the instant center of every three relatively moving objects are colinear. Example 316 McPherson suspension roll center. A McPherson suspension is an inverted slider crank mechanism. The instant centers of an example of inverted slider crank mechanism are shown in Figure 8.30. In this figure, the point I12 is the suspension roll center, which is the instant center of rotation for the wheel link number 2 with respect to the chassis link number 1. A car with a McPherson suspension system is shown in Figure 8.31. The kinematic equivalent mechanism is depicted in Figure 8.32. Suspension roll centers along with the body roll center are shown in Figure 8.33. To 8. Suspension Mechanisms 473 find the roll center of the front or rear half of a car, we determine each suspension roll center and then find the intersection of the lines connecting the suspension roll centers to the center of the associated tireprint. Example 317 Roll center of double A-arm suspension. The roll center of an independent suspension such as a double A-arm can be internal or external. The kinematic model of a double A-arm suspension for the front left wheel of a car is illustrated in Figure 8.34. The suspension roll center in Figure 8.34(a) is internal, and in Figure 8.34(b) is external. An internal suspension roll center is toward the vehicle body, while an external suspension roll center goes away from the vehicle body. A suspension roll center may be on, above, or below the road surface, as shown in Figure 8.35(a)-(c) for an external suspension roll center. When the suspension roll center is on the ground, above the ground, or below the ground, the vehicle roll center would be on the ground, below the ground, and above the ground, respectively. Example 318 F Camber variation of double A-arm suspension. When a wheel moves up and down with respect to the vehicle body, de- 474 8. Suspension Mechanisms 8. Suspension Mechanisms 475 pending on the suspension mechanism, the wheel may camber. Figure 8.36 illustrates the kinematic model for a double A-arm suspension mechanism. The mechanism is equivalent to a four-bar linkage with the ground link as the vehicle chasis. The wheel is always attached to a coupler point C of the mechanism. We set a local suspension coordinate frame (x, y) with the x-axis indicating the ground link MN . The x-axis makes a constant angle \u03b80 with the vertical direction. The suspension machanism has a length a for the upper A-arm, b for the coupler link, c for the lower A-arm, and d for the ground link. The configuration of the suspension is determined by the angles \u03b82, \u03b83, and \u03b84, all measured from the positive direction of the x-axis. When the suspension is at its equilibrium position, the links of the double A-arm suspension make initial angles \u03b820 \u03b830, and \u03b840 with the x-axis. The equilibrium position of a suspension is called the rest position. To determine the camber angle during the fluctuation of the wheel, we should determine the variation of the coupler angle \u03b83, as a function of vertical motion z of the coupler point C. Using \u03b82 as a parameter, we can find the coordinates (xC , yC) of the 476 8. Suspension Mechanisms coupler point C in the suspension coordinate frame (x, y) as xC = a cos \u03b82 + e cos (p\u2212 q + \u03b1) (8.1) yC = a sin \u03b82 + e sin (p\u2212 q + \u03b1) (8.2) where, q = tan\u22121 a sin \u03b82 d\u2212 a cos \u03b82 (8.3) p = tan\u22121 q 4b2f2 \u2212 (b2 + f2 \u2212 c2) 2 b2 + f2 \u2212 c2 (8.4) f = p a2 + d2 \u2212 2ad cos \u03b82. (8.5) The position vector of the coupler point is uC uC = xC \u0131\u0302+ yC j\u0302 (8.6) and the unit vector in the z-direction is u\u0302z = \u2212 cos \u03b80\u0131\u0302\u2212 sin \u03b80j\u0302. (8.7) Therefore, the displacement z in terms of xC and yC is: z = uC \u00b7 u\u0302z = \u2212xC cos \u03b80 \u2212 yC sin \u03b80 (8.8) 8. Suspension Mechanisms 477 The initial coordinates of the coupler point C and the initial value of z are: xC0 = a cos \u03b820 + e cos (p0 \u2212 q0 + \u03b1) (8.9) yC0 = a sin \u03b820 + e sin (p0 \u2212 q0 + \u03b1) (8.10) z0 = \u2212xC0 cos \u03b80 \u2212 yC0 sin \u03b80 (8.11) and hence, the vertical displacement of the wheel center can be calculated by h = z \u2212 z0 (8.12) The initial angle of the coupler link with the vertical direction is \u03b80 \u2212 \u03b830. Therefore, the camber angle of the wheel would be \u03b3 = (\u03b80 \u2212 \u03b83)\u2212 (\u03b80 \u2212 \u03b830) = \u03b830 \u2212 \u03b83 (8.13) The angle of the coupler link with the x-direction is equal to \u03b83 = 2 tan \u22121 \u00c3 \u2212E \u00b1 \u221a E2 \u2212 4DF 2D ! (8.14) where, D = J5 \u2212 J1 + (1 + J4) cos \u03b82 (8.15) E = \u22122 sin \u03b82 (8.16) F = J5 + J1 \u2212 (1\u2212 J4) cos \u03b82 (8.17) and J1 = d a (8.18) J2 = d c (8.19) J3 = a2 \u2212 b2 + c2 + d2 2ac (8.20) J4 = d b (8.21) J5 = c2 \u2212 d2 \u2212 a2 \u2212 b2 2ab . (8.22) Substituting (8.14) and (8.13), and then, eliminating \u03b82 between (8.13) and (8.8) provides the relationship between the vertical motion of the wheel, z, and the camber angle \u03b3. 478 8. Suspension Mechanisms Example 319 F Camber angle and wheel fluctuations. Consider the double A-arm suspension that is shown in Figure 8.36. The dimensions of the equivalent kinematic model are: a = 22.4 cm b = 22.1 cm c = 27.3 cm d = 17.4 cm \u03b80 = 24.3 deg (8.23) The coupler point C is at: e = 14.8 cm \u03b1 = 54.8 deg (8.24) If the angle \u03b82 at the rest position is at \u03b820 = 121.5 deg (8.25) then the initial angle of the other links are: \u03b830 = 18.36 deg \u03b840 = 107.32 deg (8.26) At the rest position, the coupler point is at: xC0 = \u221222.73 cm yC0 = 9.23 cm z0 = 16.92 cm (8.27) We may calculate h and \u03b3 by varying the parameter \u03b82. Figure 8.37 illustrates h as a function of the camber angle \u03b3. For this suspension mechanism, the wheel gains a positive camber when the wheel moves up, and gains a negative camber when the it moves down. The mechanism is shown in Figure 8.38, when the wheel is at the rest position and has a positive or a negative displacement. 8.4 F Car Tire Relative Angles There are four major wheel alignment parameters that affect vehicle dynamics: toe, camber, caster, and trust angle. 8. Suspension Mechanisms 479 8.4.1 F Toe When a pair of wheels is set so that their leading edges are pointed toward each other, the wheel pair is said to have toe-in. If the leading edges point away from each other, the pair is said to have toe-out. Toe-in and toe-out front wheel configurations of a car are illustrated in Figure 8.39. The amount of toe can be expressed in degrees of the angle to which the wheels are not parallel. However, it is more common to express the toein and toe-out as the difference between the track widths as measured at the leading and trailing edges of the tires. Toe settings affect three major performances: tire wear, straight-line stability, and corner entry handling. For minimum tire wear and power loss, the wheels on a given axle of a car should point directly ahead when the car is running in a straight line. Excessive toe-in causes accelerated wear at the outboard edges of the tires, while too much toe-out causes wear at the inboard edges. Toe-in increases the directional stability of the vehicle, and toe-out increases the steering response. Hence, a toe-in setting makes the steering function lazy, while a toe-out makes the vehicle unstable. With four wheel independent suspensions, the toe may also be set at the rear of the car. Toe settings at the rear have the same effect on wear, directional stability, and turn-in as they do on the front. However, we usually do not set up a rear-drive race car toed out in the rear, because of excessive instability. When driving torque is applied to the wheels, they pull themselves forward and try to create toe-in. Furthermore, when pushed down the road, a non-driven wheel or a braking wheel will tend to toe-out. Example 320 Toe-in and directional stability. Toe settings have an impact on directional stability. When the steering 480 8. Suspension Mechanisms 8. Suspension Mechanisms 481 wheel is centered, toe-in causes the wheels to tent to move along paths that intersect each other in front of the vehicle. However, the wheels are in balance and no turn results. Toe-in setup can increase the directional stability caused by little steering fluctuations and keep the car moving straight. Steering fluctuations may be a result of road disturbances. If a car is set up with toe-out, the front wheels are aligned so that slight disturbances cause the wheel pair to assume rolling directions that approach a turn. Therefore, toe-out encourages the initiation of a turn, while toe-in discourages it. Toe-out makes the steering quicker. So, it may be used in vehicles for a faster response. The toe setting on a particular car becomes a trade-off between the straight-line stability afforded by toe-in and the quick steering response by toe-out. Toe-out is not desirable for street cars, however, race car drivers are willing to drive a car with a little directional instability, for sharper turn-in to the corners. So street cars are generally set up with toe-in, while race cars are often set up with toe-out. Example 321 Toe-in and toe-out in the front and rear axles. Front toe-in: slower steering response, more straight-line stability, greater wear at the outboard edges of the tires. Front toe-zero: medium steering response, minimum power loss, minimum tire wear. Front toe-out: quicker steering response, less straight-line stability, greater wear at the inboard edges of the tires. Rear toe-in: straight-line stability, traction out of the corner, more steerability, higher top speed. 482 8. Suspension Mechanisms 8.4.2 F Caster Angle Caster is the angle to which the steering pivot axis is tilted forward or rearward from vertical, as viewed from the side. Assume the wheel is straight to have the body frame and the wheel frame coincident. If the steering axis is turned about the wheel yw-axis then the wheel has positive caster. If the steering axis is turned about the wheel \u2212yw-axis, then the wheel has negative caster. Positive and negative caster configurations on the front wheel of a car are shown in Figure 8.40. Negative caster aids in centering the steering wheel after a turn and makes the front tires straighten quicker. Most street cars are made with 4\u22126 deg negative caster. Negative caster tends to straighten the wheel when the vehicle is traveling forward, and thus is used to enhance straight-line stability. Example 322 Negative caster of shopping carts. The steering axis of a shopping cart wheel is set forward of where the wheel contacts the ground. As the cart is pushed forward, the steering axis pulls the wheel along, and because the wheel drags along the ground, it falls directly in line behind the steering axis. The force that causes the wheel to follow the steering axis is proportional to the distance between the steering axis and the wheel-to-ground contact point, if the caster is small. This distance is referred to as trail. The cars\u2019 steering axis intersects the ground at a point in front of the tireprint, and thus the same effect as seen in the shopping cart casters is achieved. While greater caster angles improves straight-line stability, they also cause an increase in steering effort. Example 323 Characteristics of caster in front axle. Zero castor provides: easy steering into the corner, low steering out of the corner, low straight-line stability. 8. Suspension Mechanisms 483 Negative caster provides: low steering into the corner, easy steering out of the corner, more straight-line stability, high tireprint area during turn, good turn-in response, good directional stability, good steering feel. When a castered wheel rotates about the steering axis, the wheel gains camber. This camber is generally favorable for cornering. 8.4.3 F Camber Camber is the angle of the wheel relative to vertical line to the road, as viewed from the front or the rear of the car. Figure 8.41 illustrates the wheel number 1 of a vehicle. If the wheel leans in toward the chassis, it is called negative camber and if it leans away from the car, it is called positive camber. The cornering force that a tire can develop is highly dependent on its angle relative to the road surface, and so wheel camber has a major effect on the road holding of a car. A tire develops its maximum lateral force at a small camber angle. This fact is due to the contribution of camber thrust, which is an additional lateral force generated by elastic deformation as the tread rubber pulls through the tire/road interface. To optimize a tire\u2019s performance in a turn, the suspension should provide a slight camber angle in the direction of rotation. As the body rolls in a turn, the suspension deflects vertically. The wheel is connected to the chassis by suspension mechanism, which must rotate to allow for the wheel deflection. Therefore, the wheel can be subject to large camber changes as the suspension moves up and down. So, the more the wheel must deflect from its static position, the more difficult it is to maintain an ideal camber angle. Thus, the relatively large wheel travel and soft roll stiffness needed to provide a smooth ride in passenger cars presents a difficult design challenge, while the small wheel travel and high roll stiffness inherent in racing cars reduces the problem. Example 324 Castor versus camber. Camber doesn\u2019t improve turn-in as the positive caster does. Camber is not generally good for tire wear. Camber in one wheel does not improve directional stability. Camber adversely affects braking and acceleration efforts. 8.4.4 F Trust Angle The trust angle \u03c5 is the angle between vehicle\u2019s centerline and perpendicular to the rear axle. It compares the direction that the rear axle is aimed with the centerline of the vehicle. A nonzero angle configuration is shown in Figure 8.42. Zero angle confirms that the rear axle is parallel to the front axle, and the wheelbase on both sides of the vehicle are the same. A reason for nonzero 484 8. Suspension Mechanisms 8. Suspension Mechanisms 485 trust angle would have unequal toe-in or toe-out on both sides of the axle. Example 325 Torque reaction. There are two kinds of torque reactions in rear-whel-drive: 1\u2212 the reaction of the axle housing to rotate in the opposite direction of the crown wheel rotation, and 2\u2212 the reaction of axle housing to spin about its own center, opposite to the direction of pinion\u2019s rotation. The first reaction leads to a lifting force in the differential causing a wind-up in springs. The second reaction leads to a lifting force on the right wheels. 8.5 Suspension Requirements and Coordinate Frames The suspension mechanism should allow a relative motion between the wheel and the vehicle body. The relative motions are needed to pass the road irregularities and steering. To function properly, a suspension mechanism should have some kinematic and dynamics requirements. 8.5.1 Kinematic Requirements To express the motions of a wheel, we attach a wheel coordinate system W (oxwywzw) to the center of the wheel. A wheel, as a rigid body, has six degrees-of-freedom with respect to the vehicle body: three translations and three rotations, as shown in Figure 8.43. 486 8. Suspension Mechanisms The axes xw, yw, and zw indicate the direction of forward, lateral, and vertical translations and rotations. In the position shown in the figure, the rotation about the xw-axis is the camber angle, about the yw-axis is the spin, and about the zw-axis is the steer angle. Consider a non-steerable wheel. Translation in zw-direction and spin about the yw-axis are the only two DOF allowed for such a wheel. So, we need to take four DOF. If the wheel is steerable, then translation in the zw-direction, spin about the yw-axis, and steer rotation about the zw-axis are the three DOF allowed. So, we must take three DOF of a steerable wheel. Kinematically, non-steerable and steerable wheels should be supported as shown in Figures 8.44 and 8.45 respectively. Providing the required freedom, as well as eliminating the taken DOF, are the kinematic requirements of a suspension mechanism. 8.5.2 Dynamic Requirements Wheels should be able to propel, steer, and stop the vehicle. So, the suspension system must transmit the driving traction and deceleration braking forces between the vehicle body and the ground. The suspension members must also resist lateral forces acting on the vehicle. Hence, the wheel suspension system must make the wheel rigid for the taken DOF. However, there must also be some compliance members to limit the untaken DOF. The most important compliant members are spring and dampers to provide returning and resistance forces in the z-direction. 8. Suspension Mechanisms 487 8.5.3 Wheel, wheel-body, and tire Coordinate Frames Three coordinate frames are employed to express the orientation of a tire and wheel with respect to the vehicle: the wheel frame W , wheel-body frame C, and tire frame T . A wheel coordinate frame W (xw, yw, zw) is attached to the center of a wheel. It follows every translation and rotation of the wheel except the spin. Hence, the xw and zw axes are always in the tire-plane, while the yw-axis is always along the spin axis. A wheel coordinate frame is shown in Figure 8.43. When the wheel is straight and the W frame is parallel to the vehicle coordinate frame, we attach a wheel-body coordinate frame C (xc, yc, zc) at the center of the wheel parallel to the vehicle coordinate axes. The wheelbody frame C is motionless with respect to the vehicle coordinate and does not follow any motion of the wheel. The tire coordinate frame T (xt, yt, zt) is set at the center of the tireprint. The zt-axis is always perpendicular to the ground. The xt-axis is along the intersection line of the tire-plane and the ground. The tire frame does not follows the spin and camber rotations of the tire however, it follows the steer angle rotation about the zc-axis. Figure 8.46 illustrates a tire and a wheel coordinate frames. Example 326 Visualization of the wheel, tire, and wheel-body frames. Figure 8.47 illustrates the relative configuration of a wheel-body frame C, a tire frame T , and a wheel frame W . If the steering axis is along the zc-axis then, the rotation of the wheel about the zc-axis is the steer angle \u03b4. Rotation about the xt-axis is the camber angle \u03b3. 488 8. Suspension Mechanisms Generally speaking, the steering axis may have any angle and may go through any point of the ground plane. Example 327 Wheel to tire coordinate frame transformation. If TdW indicates the T -expression of the position vector of the wheel frame origin relative to the tire frame origin, then having the coordinates of a point P in the wheel frame, we can find its coordinates in the tire frame using the following equation. T rP = TRW W rP + TdW (8.28) If W rP indicates the position vector of a point P in the wheel frame, W rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 (8.29) then the coordinates of the point P in the tire frame T rP are T rP = TRW W rP + Td = TRW W rP + TRW W T dW = \u23a1\u23a3 xP yP cos \u03b3 \u2212Rw sin \u03b3 \u2212 zP sin \u03b3 Rw cos \u03b3 + zP cos \u03b3 + yP sin \u03b3 \u23a4\u23a6 (8.30) 8. Suspension Mechanisms 489 where, WT dW is the W -expression of the position vector of the wheel frame in the tire frame, Rw is the radius of the tire, and TRW is the rotation matrix to go from the wheel frame W to the tire frame T . TRW = \u23a1\u23a3 1 0 0 0 cos \u03b3 \u2212 sin \u03b3 0 sin \u03b3 cos \u03b3 \u23a4\u23a6 (8.31) W T dW = \u23a1\u23a3 0 0 Rw \u23a4\u23a6 . (8.32) As an example, the center of the wheel W rP = W ro = 0 is the origin of the wheel frame W , that is at T ro = TdW = TRW W T dW = \u23a1\u23a3 0 \u2212Rw sin \u03b3 Rw cos \u03b3 \u23a4\u23a6 (8.33) in the tire coordinate frame T . 490 8. Suspension Mechanisms Example 328 F Tire to wheel coordinate frame transformation. If rP indicates the position vector of a point P in the tire coordinate frame, T rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 (8.34) then the position vector W rP of the point P in the wheel coordinate frame is W rP = WRT T rP \u2212 W T dW (8.35) = \u23a1\u23a3 xP yP cos \u03b3 + zP sin \u03b3 zP cos \u03b3 \u2212Rw \u2212 yP sin \u03b3 \u23a4\u23a6 because WRT = \u23a1\u23a3 1 0 0 0 cos \u03b3 sin \u03b3 0 \u2212 sin \u03b3 cos \u03b3 \u23a4\u23a6 (8.36) WdT = \u23a1\u23a3 0 0 Rw \u23a4\u23a6 (8.37) and we may multiply both sides of Equation (8.28) by TRT W to get TRT W T rP = W rP + TRT W TdW (8.38) = W rP + W T dW W rP = WRT T rP \u2212 W T dW . (8.39) As an example, the center of tireprint in the wheel frame is at W rP = \u23a1\u23a3 1 0 0 0 cos \u03b3 \u2212 sin \u03b3 0 sin \u03b3 cos \u03b3 \u23a4\u23a6T \u23a1\u23a3 0 0 0 \u23a4\u23a6\u2212 \u23a1\u23a3 0 0 Rw \u23a4\u23a6 = \u23a1\u23a3 0 0 \u2212Rw \u23a4\u23a6 . (8.40) Example 329 F Wheel to tire homogeneous transformation matrices. The transformation from the wheel to tire coordinate frame may also be expressed by a 4\u00d7 4 homogeneous transformation matrix TTW , T rP = TTW W rP = \u2219 TRW TdW 0 1 \u00b8 W rP (8.41) where TTW = \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 cos \u03b3 \u2212 sin \u03b3 \u2212Rw sin \u03b3 0 sin \u03b3 cos \u03b3 Rw cos \u03b3 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 . (8.42) 8. Suspension Mechanisms 491 The corresponding homogeneous transformation matrix WTT from the tire to wheel frame would be WTT = \u2219 WRT WdT 0 1 \u00b8 = \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 cos \u03b3 \u2212 sin \u03b3 0 0 sin \u03b3 cos \u03b3 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 . (8.43) It can be checked that WTT = TT\u22121W , using the inverse of a homogeneous transformation matrix rule. TT\u22121W = \u2219 TRW TdW 0 1 \u00b8\u22121 = \u2219 TRT W \u2212 TRT W TdW 0 1 \u00b8 = \u2219 WRT \u2212WRT TdW 0 1 \u00b8 (8.44) Example 330 F Tire to wheel-body frame transformation. The origin of the tire frame is at CdT in the wheel-body frame. CdT = \u23a1\u23a3 0 0 \u2212Rw \u23a4\u23a6 (8.45) The tire frame can steer about the zc-axis with respect to the wheel-body frame. The associated rotation matrix is CRT = \u23a1\u23a3 cos \u03b4 \u2212 sin \u03b4 0 sin \u03b4 cos \u03b4 0 0 0 1 \u23a4\u23a6 (8.46) Therefore, the transformation between the tire and wheel-body frames can be expressed by Cr = CRT T r+ CdT (8.47) or equivalently, by a homogeneous transformation matrix CTT . CTT = \u2219 CRT CdT 0 1 \u00b8 = \u23a1\u23a2\u23a2\u23a3 cos \u03b4 \u2212 sin \u03b4 0 0 sin \u03b4 cos \u03b4 0 0 0 0 1 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 (8.48) As an example, the wheel-body coordinates of the point P on the tread of a 492 8. Suspension Mechanisms negatively steered tire at the position shown in Figure 8.48, are: Cr = CTT T rP = \u23a1\u23a2\u23a2\u23a3 cos\u2212\u03b4 \u2212 sin\u2212\u03b4 0 0 sin\u2212\u03b4 cos\u2212\u03b4 0 0 0 0 1 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 \u23a1\u23a2\u23a2\u23a3 Rw 0 Rw 1 \u23a4\u23a5\u23a5\u23a6 = \u23a1\u23a2\u23a2\u23a3 Rw cos \u03b4 \u2212Rw sin \u03b4 0 1 \u23a4\u23a5\u23a5\u23a6 (8.49) The homogeneous transformation matrix for tire to wheel-body frame TTC is: TTC = CT\u22121T = \u2219 CRT CdT 0 1 \u00b8\u22121 = \u2219 CRT T \u2212CRT T CdT 0 1 \u00b8 = \u2219 CRT T \u2212 T CdT 0 1 \u00b8 = \u23a1\u23a2\u23a2\u23a3 cos \u03b4 sin \u03b4 0 0 \u2212 sin \u03b4 cos \u03b4 0 0 0 0 1 Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 (8.50) Example 331 F Cycloid. Assume that the wheel in Figure 8.48 is turning with angular velocity \u03c9 and has no slip on the ground. If the point P is at the center of the tireprint 8. Suspension Mechanisms 493 when t = 0, MrP = \u23a1\u23a3 0 0 \u2212Rw \u23a4\u23a6 (8.51) then we can find its position in the wheel frame at a time t by employing another coordinate frame M . The frame M is called the rim frame and is stuck to the wheel at its center. Because of spin, the M frame turns about the yw-axis, and therefore, the rotation matrix to go from the rim frame to the wheel frame is: WRM = \u23a1\u23a3 cos\u03c9t 0 sin\u03c9t 0 1 0 \u2212 sin\u03c9t 0 cos\u03c9t \u23a4\u23a6 (8.52) So the coordinates of P in the wheel frame are: W rP = WRM MrP = \u23a1\u23a3 \u2212Rw sin t\u03c9 0 \u2212Rw cos t\u03c9 \u23a4\u23a6 (8.53) The center of the wheel is moving with speed vx = Rw\u03c9 and it is at Gr =\u00a3 vxt 0 Rw \u00a4 in the global coordinate frame G on the ground. Hence, the coordinates of point P in the global frame G, would be GrP = W rP + \u23a1\u23a3 vxt 0 Rw \u23a4\u23a6 = \u23a1\u23a3 Rw (\u03c9t\u2212 sin t\u03c9) 0 Rw (1\u2212 cos t\u03c9) \u23a4\u23a6 . (8.54) The path of motion of point P in the (X,Z)-plane can be found by eliminating t between X and Z coordinates. However, it is easier to expressed the path by using \u03c9t as a parameter. Such a path is called cycloid. In general case, point P can be at any distance from the center of the rim frame. If the point is at a distance d 6= Rw, then its path of motion is called the trochoid. A trochoid is called a curtate cycloid when d < Rw and a prolate cycloid when d > Rw. Figure 8.49(a)-(c) illustrate a cycloid, curtate cycloid, and prolate cycloid respectively. Example 332 F Wheel to wheel-body frame transformation. The homogeneous transformation matrix CTW to go from the wheel frame 494 8. Suspension Mechanisms to the wheel-body frame can be found by combined transformation. CTW = CTT TTW (8.55) = \u23a1\u23a2\u23a2\u23a3 c\u03b4 \u2212s\u03b4 0 0 s\u03b4 c\u03b4 0 0 0 0 1 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 c\u03b3 \u2212s\u03b3 \u2212Rw sin \u03b3 0 s\u03b3 c\u03b3 Rw cos \u03b3 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 = \u23a1\u23a2\u23a2\u23a3 cos \u03b4 \u2212 cos \u03b3 sin \u03b4 sin \u03b3 sin \u03b4 Rw sin \u03b3 sin \u03b4 sin \u03b4 cos \u03b3 cos \u03b4 \u2212 cos \u03b4 sin \u03b3 \u2212Rw cos \u03b4 sin \u03b3 0 sin \u03b3 cos \u03b3 Rw cos \u03b3 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 If rP indicates the position vector of a point P in the wheel coordinate frame, W rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 (8.56) then the homogeneous position vector CrP of the point P in the wheel-body 8. Suspension Mechanisms 495 coordinate frame is: CrP = CTW W rP = \u23a1\u23a2\u23a2\u23a3 xP cos \u03b4 \u2212 yP cos \u03b3 sin \u03b4 + (Rw + zP ) sin \u03b3 sin \u03b4 xP sin \u03b4 + yP cos \u03b3 cos \u03b4 \u2212 (Rw + zP ) cos \u03b4 sin \u03b3 \u2212Rw + (Rw + zP ) cos \u03b3 + yP sin \u03b3 1 \u23a4\u23a5\u23a5\u23a6 (8.57) The position of the wheel center W r = 0, for a cambered and steered wheel is at Cr = CTW W r = \u23a1\u23a2\u23a2\u23a3 Rw sin \u03b3 sin \u03b4 \u2212Rw cos \u03b4 sin \u03b3 \u2212Rw(1\u2212 cos \u03b3) 1 \u23a4\u23a5\u23a5\u23a6 (8.58) The zc = Rw (cos \u03b3 \u2212 1) indicates how much the center of the wheel comes down when the wheel cambers. If the wheel is not steerable, then \u03b4 = 0 and the transformation matrix CTW reduces to CTW = \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 cos \u03b3 \u2212 sin \u03b3 \u2212Rw sin \u03b3 0 sin \u03b3 cos \u03b3 Rw (cos \u03b3 \u2212 1) 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 (8.59) that shows CrP = CTW W rP = \u23a1\u23a2\u23a2\u23a3 xP yP cos \u03b3 \u2212Rw sin \u03b3 \u2212 zP sin \u03b3 zP cos \u03b3 + yP sin \u03b3 +Rw (cos \u03b3 \u2212 1) 1 \u23a4\u23a5\u23a5\u23a6 (8.60) Example 333 F Tire to vehicle coordinate frame transformation. Figure 8.50 illustrates the first and fourth tires of a 4-wheel vehicle. There is a body coordinate frame B (x, y, z) attached to the mass center C of the vehicle. There are also two tire coordinate frames T1 (xt1 , yt1 , zt1) and T4 (xt4 , yt4 , zt4) attached to the tires 1 and 4 at the center of their tireprints. The origin of the tire coordinate frame T1 is at Bd1 BdT1 = \u23a1\u23a3 a1 \u2212b1 \u2212h \u23a4\u23a6 (8.61)"
        ]
    },
    {
        "image_filename": "designv10_0_0000656_s0021-9290(01)00245-7-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000656_s0021-9290(01)00245-7-Figure1-1.png",
        "caption": "Fig. 1. The spring-mass model for running. The leg spring is characterized by the stiffness kLEG and the nominal length c0 which is the leg length at touch-down and take-off. In our approach, the leg orientation at touch-down is characterized by a given angle of attack a0: During the flight phase the horizontal velocity is constant. One stride can be defined as the movement from one apex (height yi) to the next (height yi\u00fe1). As the vertical velocity vanishes at these conditions the state of the system is uniquely characterized by the apex height and the horizontal velocity, e.g., as an initial condition \u00f0y0; vX;0\u00de:",
        "texts": [
            " In total 67 contact phases were analyzed in terms of ground reaction forces and kinematic landmarks of the stance leg (hip, knee, ankle and ball of the foot) with a sampling rate of 500Hz. During the contact phase, the leg length cLEG was defined as the distance between the hip and the ball of the foot. The leg stiffness kLEG was defined by kLEG \u00bc FMAX=DcMAX; \u00f01\u00de where FMAX denotes the maximum amount of the ground reaction force and DcMAX the amount of maximum leg shortening Dc \u00bc c0 cLEG\u00f0t\u00de (Seyfarth et al., 1999). Running can be described as a subsequent series of stance and flight phases (Fig. 1). The trajectory of the center of mass m is determined by the gravitational force (during flight and stance phase) and by the force generated by the stance leg during the contact phase. During the flight phase, the horizontal velocity remains constant whereas the vertical velocity crosses zero at the apex. This particular condition is used to define one stride within a periodic running pattern as the movement from one apex to the other includes one stance phase. The operation of the stance leg, landing at a certain angle of attack a0; is represented by a linear spring with the stiffness kLEG and the nominal position c0: The length c0 is equal to the initial leg length to fulfill zero leg force at touch-down"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003717_nme.1620310308-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003717_nme.1620310308-Figure6-1.png",
        "caption": "Figure 6. A six-tooth finite element model of the gear and the pinion",
        "texts": [
            " The finite element description of the surface 536 S. VIJAYAKAR was then created by fitting tenth order truncated Chebyshev series approximations to these data. The interior portions of the finite element were created semi-automatically. Only a sector containing three teeth of each gear was modelled, with each tooth being identical. The gear (gear no. 1) and the pinion (gear no. 2, the smaller gear) were then oriented in space as per the assembly drawings, and the analysis was carried out for each individual time step. Figure 6 shows a six-tooth gear and pinion model. Sectoral symmetry is used to generate stiffness matrices from the stiffness matrix of one tooth. For this particular gear set, a three-tooth model suffices because at the most two teeth contact at a time. Figure 7 shows the surfaces of the three-tooth gear. Figures 8 ,9 and 10 show the contact pattern (which is the locus of the contact zone as the gears roll against each other) for a gear torque of 240,480 and 960 in-lb, respectively. Figures 11 and 13 show views of the contact zone with contact pressure contours on the gear for two particular angular positions"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure9.12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure9.12-1.png",
        "caption": "FIGURE 9.12. A vibrating mass with a hanging pendulum.",
        "texts": [
            " Using x and y for the Cartesian position of m, and using \u03b8 = q as the generalized coordinate, we have x = f(\u03b8) = l sin \u03b8 (9.268) y = g(\u03b8) = l cos \u03b8 (9.269) K = 1 2 m \u00a1 x\u03072 + y\u03072 \u00a2 = 1 2 ml2\u03b8\u0307 2 (9.270) and therefore, d dt \u00b5 \u2202K \u2202\u03b8\u0307 \u00b6 \u2212 \u2202K \u2202\u03b8 = d dt (ml2\u03b8\u0307) = ml2\u03b8\u0308. (9.271) The external force components, acting on m, are Fx = 0 (9.272) Fy = mg (9.273) 558 9. Applied Dynamics and therefore, F\u03b8 = Fx \u2202f \u2202\u03b8 + Fy \u2202g \u2202\u03b8 = \u2212mgl sin \u03b8. (9.274) Hence, the equation of motion for the pendulum is ml2\u03b8\u0308 = \u2212mgl sin \u03b8. (9.275) Example 365 A pendulum attached to an oscillating mass. Figure 9.12 illustrates a vibrating mass with a hanging pendulum. The pendulum can act as a vibration absorber if designed properly. Starting with coordinate relationships xM = fM = x (9.276) yM = gM = 0 (9.277) xm = fm = x+ l sin \u03b8 (9.278) ym = gm = l cos \u03b8 (9.279) we may find the kinetic energy in terms of the generalized coordinates x and \u03b8. K = 1 2 M \u00a1 x\u03072M + y\u03072M \u00a2 + 1 2 m \u00a1 x\u03072m + y\u03072m \u00a2 = 1 2 Mx\u03072 + 1 2 m \u00b3 x\u03072 + l2\u03b8\u0307 2 + 2lx\u0307\u03b8\u0307 cos \u03b8 \u00b4 (9.280) Then, the left-hand side of the Lagrange equations are d dt \u00b5 \u2202K \u2202x\u0307 \u00b6 \u2212 \u2202K \u2202x = (M +m)x\u0308+ml\u03b8\u0308 cos \u03b8 \u2212ml\u03b8\u0307 2 sin \u03b8 (9"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000613_978-94-015-9261-1-Figure5.10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000613_978-94-015-9261-1-Figure5.10-1.png",
        "caption": "Figure 5.10. Zero dynamics submanifold Z', neighborhoods and coordinate transfor  mations in state space",
        "texts": [
            "c~-1 hj } , i = 1, 2, . . . , Po , j = 1, 2, . . . , m and t he com posite mpo-dimensional vector of exte rn al dynamics ~ = {~i} as ~ = H( x) , (5 .127 ) where H(x ) = {H;}. In view of P roposition 5. 1 and under Assumption 5.3 , we establish that for all x E X i) e = y(i - l), i = 1,2 , . . . , po ; ii) rank J{ (x) = mp\u00ab . (5.128) 220 CHAPT ER 5 For th e ease when mpo < n , we introduee the nonlinear equation H( x ) = 0 (5.129) and th e smoot h (n - mpo)-dimensional hypersurfaee (zero dynamies sub manifold, Figure 5.10) Z * = {x EX : H (x ) = O} . Under th e regularity condition (5.128), the set Z * ean be endowed with th e st ruet ure of a smoot h manifold (see Section 5.2) . Here we suppose that Z* is a one-sheet ed submanifold equipped with th e local coordinates z, forming th e (n - mpo) -dim ensional veetor Z = ((x), (5.130) where ( is a loeal homeomorphism from Z* onto Z C IRn - m po such that for all x E Z* the mapping (5 .127), (5 .130) satisfies th e extended regularity condition I \u00e4C/ \u00e4x I det \u00e4H/\u00e4x I =f:"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure9.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure9.1-1.png",
        "caption": "Figure 9.1 Space trusses. (a) Side view and top view of a truncated truss dome. (b) A space truss constructed from plane trusses.",
        "texts": [
            " We will also look at the way in which a structure is modelled as a truss, the nomenclature for the members in a truss, the various types of trusses, and the conventions for labelling the joints and members. A truss is defined as a structure constructed with straight bars (members), which are connected by hinges at so-called joints, and loaded by forces which have their point of application at these joints. There are plane trusses and space trusses. In plane trusses, all the members are in the same plane, and forces only act in the plane of the structure. In space trusses, the members are not all in the same plane (see Figure 9.1). Many space trusses in fact consist of plane trusses, such as the structure in Figure 9.1b. The load shown is transferred to the supports via the plane trusses ABCD and ABEG. From here on, we will look only at plane trusses. The open circles, which indicate the hinged joints, will be omitted since in a truss all joints are hinged by definition. Calculating a plane truss, hereafter referred to as truss, is based on the following assumptions: \u2022 all members are straight; \u2022 all members are connected at hinged joints; \u2022 the load consists of forces that act in the plane of the structure and apply at the joints"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003657_02783649922066475-Figure12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003657_02783649922066475-Figure12-1.png",
        "caption": "Fig. 12. The two-axis range scanner\u2019s internal mechanism.",
        "texts": [
            " 11) with a 350\u25e6 \u00d7 45\u25e6 (vertical) field of view. Elevation maps generated from the range data are used for teleoperation interfaces and planning. Each elevation map covers a circular area with diameter of 15 m, with an average grid resolution of 10 cm. A single scan requires about 24 sec to gather 20,000 range readings (constrained by mirror speed, not laser-ranging rate). The scanner consists of a fast scan (or slew) axis driven by a DC motor and a slow nodding axis driven by a stepper motor (see Fig. 12). The nodding stepper increments the mirror-nod angle once for each scan-axis mirror rotation. The scan motor and coupled encoder are hollow core to permit passage of the driveshaft from the nodding-motor speed reducer to the nodding assembly. The slip ring permits unrestrained rotation of the nodding-motor and speed-reducer assembly. The rangefinder (Acuity 300) has a 10-mW, 780-nm laser with a sample acquisition rate of 38 kHz. The transmitter and receiver are coaxial, with the receiver having an aperture diameter of 7"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure13.30-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure13.30-1.png",
        "caption": "Figure 13.30 Lean-to.",
        "texts": [
            "28c. The magnitude of Mmax can be determined from the moment equilibrium of the isolated part EB in Figure 13.29: Mmax = (2.7 kN)(0.6 m) = 1.62 kNm ( ). Mmax can also be determined from the area of the parabolic V diagram for EB. To do so we have to know that the area of the parabola is equal to two-thirds of the area of the rectangle with a width of 0.9 m and a height of 2.7 kN. This then gives the same value: Mmax = 2 3 \u00d7 (0.9 m)(2.7 kN) = 1.62 kNm ( ). 13 Calculating M, V and N Diagrams 573 In Figure 13.30, the lean-to ACD is modelled as a line element. We want to determine the N diagram and the extreme values of the bending moment due to the three uniformly distributed loads: a. A wind pressure of qw = 5 kN/m (force per length measured along ACD). b. A dead weight of qdw = 5 kN/m (force per length measured along ACD). c. A snow load of qsn = 5 kN/m (force per length measured along the pro- jection of ACD on the horizontal ground plane). Solution: Since the dead weight and the snow load have components transverse to the beam axis (qtr) and parallel to the beam axis (qpa), the N , V and M diagrams are first determined due to the separate loads qtr = 1 kN/m (Figure 13.31a) and qpa = 1 kN/m (Figure 13.31b). By means of superposition, we then determine the final N diagram for each of the given loads and the extreme values of the bending moments. In preparation, the dimensions given in Figure 13.30 are first used to determine the angles \u03b1, \u03b2 and \u03b3 and the lengths AC, CD and ACD of AC, CD and ACD respectively. The angles \u03b1, \u03b2 and \u03b3 we find from tan \u03b1 = 2/5 \u21d2 \u03b1 = 21.8\u25e6, tan \u03b2 = 4/5 \u21d2 \u03b2 = 38.7\u25e6, \u03b3 = \u03b1 + \u03b2 = 60.5\u25e6. The lengths of AC, CD and ACD are AC = (5 m)/ cos\u03b1 = 5.385 m, CD = (2 m)/ cos \u03b1 = 2.154 m, ACD = AC + CD = 7.539 m. 574 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM \u2022 N , V and M diagrams due to qtr = 1 kN/m (Figure 13.31a) In Figure 13.32, ACD has been isolated. The distributed load qtr = 1 kN/m has been replaced by its resultant Rtr"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000637_978-3-642-83006-8-Figure2.43-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000637_978-3-642-83006-8-Figure2.43-1.png",
        "caption": "Fig. 2.43. C0mpensating m0vements of arms in the sagittal plane for the single-support gait upon level ground for",
        "texts": [],
        "surrounding_texts": [
            "move freely (due to the coupling from the powered joints). For unpowe red joints, the only moments considered are those arising from the for ces of viscous friction \u00b718 P 18 = kq , (k=-0.2) where k is the viscous friction coefficient. In this case, the compen sating dynamics comprises the trunk movements both in the sagittal and frontal plane and the arms movements in the sagittal plane only (Fig."
        ]
    },
    {
        "image_filename": "designv10_0_0000845_s0261-3069(99)00078-3-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000845_s0261-3069(99)00078-3-Figure4-1.png",
        "caption": "Fig. 4. Inconel 690 hexagon with seven-hole array made by DLF. Hole diameters are within \"0.05 mm and centered on a bolt circle within \"0.13 mm. Radial distance to hexagonal faces was within \"0.076 mm. Surface roughness is 12 mm, arithmetic average.",
        "texts": [
            " Deposition times ranged from approximately 1 h for the shell type parts to 70 h for the die insert in an unattended process. process optimization. Hence a designer can produce a prototype and modify it quickly after the initial optimization is performed on the first part. Each new material and each new part or feature geometry makes the optimization cycle a prerequisite to building the best part. However, establishment of a knowledge base of part processing expertise over time can eliminate much of the development required. A hexagonal cross-section, seven-hole-array structure shown in Fig. 4, was built to 356 mm in height from Inconel 690, a difficult to process high-temperature, nickel-base alloy. The nominal composition of Inconel 690 in wt.% is 58Ni, 29Cr, 9Fe. The part was deposited at a laser power of 160 W, speed of 12.7 mmrs, vertical layer increment of 0.25 mm, bead overlap of 0.27 mm, and the powder feed rate was approximately 9 grmin. Oxygen was controlled to less than 10 ppm in an argon atmosphere. Deposition time was 172 h at a rate of 0.04 lbrh in one continuous, unattended operation"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure3.21-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure3.21-1.png",
        "caption": "Fig. 3.21: Design parameters for a general n-R manipulator.",
        "texts": [
            "2 seems to be unpractical and numerically too laborious. Therefore, a general design problem for n-R manipulators can be formulated as an optimization problem with prescribed workspace characteristics. The design parameters can be considered as the link time-independent sizes of the chain, i.e. for n revolute-connected manipulators ai, d i, i = 1,...,n, and \u03b1i, i = 1,..., n-1, and the vectors b and k, which represent respectively the position and orientation of the robot base with respect to a fixed frame XYZ, as shown in Fig. 3.21. An optimum design of manipulators can be formulated as an optimization problem in the form min f (3.1.84) subject to Fundamentals of Mechanics of Robotic Manipulation 115 xj \u2265 Xj (j = 1,...,J) (3.1.85) V \u2265 V0 (3.1.86) where f is the objective function; Xj (j=1,...,J) represent given precision points, with respect to the fixed frame, to be reached as workspace points xj ; V0 is a minimum value for a desirable workspace volume. The robot base location can be included into the design parameters through the b vector expressed by means of the radial a0 and axial d0 components, Fig. 3.21. Particularly, the angles \u03b10 and \u03b80 can be considered as describing the orientation of the manipulator base as Zb axis with respect to Z-axis and Xb with respect to X, respectively. In addition, the base frame can be conveniently assumed to be parallel with the frame X1Y1Z1, fixed on the first link of the manipulator chain, at a starting motion configuration. Therefore, the dimensional parameters for the optimal design problem will also include the parameters \u03b10, \u03b80, a0, and d0 as representing the vectors b and k"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002500_j.jfranklin.2019.07.021-Figure9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002500_j.jfranklin.2019.07.021-Figure9-1.png",
        "caption": "Fig. 9. One-link manipulator.",
        "texts": [
            " 5 \u20136 denote the responses of the adaptive laws \u03b8i (i = 1 , 2, 3) and \u03c9 j , respectively; The control input v is plotted in Fig. 7 , and Fig. 8 shows the trajectory of saturation signal u . From Figs. 1\u20138,12 , it is revealed that the system output \u03b7 can follow the given reference signal \u03b7d in a finite time, and all the signals of closed-loop system are bounded. Example 2. In order to further prove the effectiveness of the developed control strategy, an one-link manipular with the inclusion of motor dynamics shown in Fig. 9 is used. The dynamic equation of such system is given by D \u0308q + B \u0307 q + N sin ( q ) = (64a) Please cite this article as: L. Ma, G. Zong and X. Zhao et al., Observed-based adaptive finite-time tracking control for a class of nonstrict-feedback nonlinear systems with input saturation, Journal of the Franklin Institute, https:// doi.org/ 10.1016/ j.jfranklin.2019.07.021 L. Ma, G. Zong and X. Zhao et al. / Journal of the Franklin Institute xxx (xxxx) xxx 19 M w t N v f \u03be \u03be \u02d9 + H = u ( v ) \u2212  \u0307 q (64b) here q\u0308 , \u02d9 q and q denote the link acceleration, velocity and location, respectively"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002569_j.matdes.2020.108642-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002569_j.matdes.2020.108642-Figure2-1.png",
        "caption": "Fig. 2. Three types of samples with different build orientations for mechanical testing and m direction): \u22a5 build direction; DD (diagonal direction): 45\u00b0\u2220 build direction)",
        "texts": [
            " The chemical compositions of Inconel 718 alloy powder employed in this study are listed in Table 3. Fig. 1 shows that the powders are nearly spherical with the particle size distribution of d10 = 18 \u03bcm, d50 = 32 \u03bcm, and d90=54 \u03bcm,whichmeans that 10%, 50% and 90% of the particle size distribution are less than the stated diameter, respectively. The rectangular samples with a size of 50 mm \u00d7 35 mm \u00d7 10 mm were fabricatedwith three different orientationswith regard to the substrate: horizontal (H), diagonal (D), and vertical (V) as given in Fig. 2. They were fabricated on the 316 L stainless steel base plate of 250 mm \u00d7 250 mm by SLM using an iSLM150 machine. The deposition conditions were listed in Table 4. A 67\u00b0 rotation of the laser vector between successive layers was employed. A chessboard scanning strategy with 10\u00d710mmsquareswas used. The tensile testing sampleswith the specified size were cut from the rectangular samples. The samples with different scanning speed (500 mm\u00b7s\u22121, 800 mm\u00b7s\u22121 and 1100 mm\u00b7s\u22121) corresponds to the liner energy density of 0",
            " Layer power Spot size Layer thickness Preheating Scan spacing Scan speed 200 W 100 \u03bcm 30 \u03bcm 100 \u00b0C 80 \u03bcm 500, 800, 1100 mm\u00b7s\u22121 EBSD orientation mapping was performed using Hikari Plus (EDAX Instruments, USA) attached to the SEM (Apreo S) to collect and index Kikuchi patterns. EBSD mapping was conducted with a step size of 0.8 \u03bcm. Each mapping covered an area of 800 \u03bcm \u00d7 600 \u03bcm. To ensure the results of statistical significance for texture analysis, 6 additional EBSDmappingswere used for the examined surfaces, and themappings were merged using OIM software provided by EDAX. The inverse pole figure (IPF) was used to analyze texture in the three directions of BD, TD, and DD as shown in Fig. 2, respectively. For microstructure examinations, the electropolished surfaces were further etched in a mixed solution of HClO4 and CH3COOH (20:80 vol %) under a voltage of 8 V for 5 s and then examined using opticalmicroscopy (OM) and scanning electron microscopy (SEM). The surfaces selected for microstructure characterization were indicated by a letter of h, v, and d, which correspond to the cross-sections having the angle from build direction (BD) of 90\u00b0, 0\u00b0 and 45\u00b0, respectively. In the subsequent analysis of microstructure examination, samples have been labeled such as TDh @0.4 which stands for observation surface h on the transverse sample TD fabricated horizontally at an energy density of 0.4 J\u00b7mm\u22121 (scanning speed 500 mm\u00b7s\u22121) as shown in Fig. 2. For the measurement of the aspect ratio of single columnar grain, the length of the major and minor axis of a fitted ellipse was determined based on the area of the columnar grain. At least 20 columnar grainswere randomly selected for measurements to obtain an averaged aspect ratio for each tested sample. SEM (Quanta 250) was used to analyze the fractography of the tested samples. Phase identification was performed using X-ray diffractometry (Bruker D8 Advance). The tensile testing samples with the dimensions in Fig. 2 were cut from the rectangular samples built at three orientations by electrical dischargemachining (EDM). The quasi-static tensile testingwas carried out on an Instron machine with the cross-head velocity of 8.33 \u00d7 10\u22123mm\u00b7s\u22121 at room temperature. For each condition, identical three testing samples were performed, a total of 27 tensile samples are tested. The average yield strength, ultimate strength, and elongation percentage for each condition were calculated. In the subsequent analysis of tensile testing results, samples have been identified such as TD @0.4 which represents the build orientation (horizontal) at an energy density of 0.4 J\u00b7mm\u22121 as shown in Fig. 2. elt track morphology on the cross-sections perpendicular to build direction; (b), (e) and show the melt pool morphology on the sections 45\u00b0 diagonally to build direction. Fig. 3 shows the surface microstructure after etching the samples. It is clear to see the track segments andfish scalemorphology in the transverse and longitudinal cross-sections, respectively. It was found that the average track width 87 \u03bcm of the samples at a scanning speed of 500 mm\u00b7s\u22121 is slightly larger than the average track width of 82 \u03bcm at higher scanning speed",
            " The extensive precipitation is expected after post aging heat treatment as previously reported [6]. In addition to the phase distribution, the occurrence of different crystal plane peaks also indicated the texture characteristics when observed along different directions at a given energy density as shown in Fig. 5. When measured at the section of TDh @0.4, significant {200} and minor {311} peaks were found, indicating \u2329100\u232a crystal axis preferred to be perpendicular to the examined section (TDh), which corresponds to build direction (see Fig. 2). However, when measured at the section of BDh @0.4 and DDh @0.4, {111} and {110} peaks emerge, suggesting that texture components are changed. A similar trend was also found in the samples @0.25 but quite different from the samples @0.18. The quantitative texture analysis based on EBSD analysis will be present in the Section 3.1.3. Fig. 6 shows the EBSD-reconstructed OIM maps based on the Euler angles and the IPF on the examined cross-sections of the samples. For the TDh samples, the nearly equiaxed grains perpendicular to building directions were found on the cross-sections in the first two columns of Fig",
            " The selection of a particular set of five slip systems is based on the minimum total plastic work. The Taylor factor can be used for determining the grain whether to be easily deformed or not during tensile deformation. The distributions of the Taylor factor on the cross-sections perpendicular to the tensile loading direction for the samples fabricated at the density of 0.4 J\u00b7mm\u22121 are shown in Fig. 10. The cross-sections corresponding to the tensile testing samples built at different orientations are schematically shown in Fig. 2. It is obvious that the statistically highest value of the Taylor factor for the grains on the cross-section of the diagonally built sample, suggesting that most of the grain orientations are relatively hard and require to more stress to initiate slip during tensile testing, i.e., the high yield stress ing results of the samples. is generated. By contrast, most grains have relatively low Taylor factors on the cross-section of the vertically built tensile samples, slip can be initiated at the low level of external force and, therefore, low yield strength was produced"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure9.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure9.6-1.png",
        "caption": "FIGURE 9.6. Intersection of the momentum and energy ellipsoids.",
        "texts": [
            "126) In other words, the dynamics of moment-free motion of a rigid body requires that the corresponding angular velocity \u03c9(t) satisfy both Equations (9.125) and (9.126) and therefore lie on the intersection of the momentum and energy ellipsoids. For clarity, we may define the ellipsoids in the (Lx, Ly, Lz) coordinate system as L2x + L2y + L2z = L2 (9.127) L2x 2I1K + L2y 2I2K + L2z 2I3K = 1. (9.128) Equation (9.127) is a sphere and Equation (9.128) defines an ellipsoid with\u221a 2IiK as semi-axes. To have a meaningful motion, these two shapes must intersect. The intersection may form a trajectory, as shown in Figure 9.6. 540 9. Applied Dynamics It can be deduced that for a certain value of angular momentum there are maximum and minimum limit values for acceptable kinetic energy. Assuming I1 > I3 > I3 (9.129) the limits of possible kinetic energy are Kmin = L2 2I1 (9.130) Kmax = L2 2I3 (9.131) and the corresponding motions are turning about the axes I1 and I3 respectively. Example 353 F Alternative derivation of Euler equations of motion. Assume that the moment of the small force df is shown by dm and a mass element is shown by dm, then, dm = Grdm \u00d7 df = Grdm \u00d7 Gv\u0307dm dm"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003464_027836499301200101-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003464_027836499301200101-Figure1-1.png",
        "caption": "Fig. 1. A planar- manipulator with two revolute and one prismatic joint.",
        "texts": [
            " The Noninvariance of Solutions with Previous Techniques The following examples illustrate how noninvariant results may, in general, be obtained if care is not taken in the definition of metric concepts based on the inner product. The first example employs a planar manipulator to demonstrate how the joint rate solution vector qs to V = Jq, obtained by means of the pseudoinverse J+ of the Jacobian of the mechanism, changes with a change in the units of length. The second example shows how different results are obtained in hybrid control if a translation of the reference frame is performed. Exaniple l.l: Solving the Inverse Velocity Pl0blem for a Planar Manipulator Consider the simple planar manipulator of Figure 1. Two rotational joints and one prismatic joint are present in the kinematic structure of the mechanism. The joint velocity vector q is not homogeneous, being it = [ 91 B2 d 1, If only the linear velocities v are considered in the task space, the resulting Jacobian, expressed in the base frame, is , - - - - n -~ _ _ _ J where al = a2 = 1 [meter] = 1 [m], and s12 = sin(Ol + 02), C12 = cos(6\u2019i + ~2). The problem here is to compute the solution qs = J+V given a velocity vector V. Because J has full column rank, the formal Moore-Penrose inverse J+ _ [JT J]-1 JT"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.1-1.png",
        "caption": "FIGURE 8.1. A solid axle with leaf spring suspension.",
        "texts": [
            " 8 The suspension is what links the wheels to the vehicle body and allows relative motion. This chapter covers the suspension mechanisms, and discusses the possible relative motions between the wheel and the vehicle body. The wheels, through the suspension linkage, must propel, steer, and stop the vehicle, and support the associated forces. 8.1 Solid Axle Suspension The simplest way to attach a pair of wheels to a vehicle is to mount them at opposite ends of a solid axle, such as the one that is shown in Figure 8.1. The solid axle must be attached to the body such that an up and down motion in the z-direction, as well as a roll rotation about the x-axis, is possible. So, no forward and lateral translation, and also no rotation about the axle and the z-axis, is allowed. There are many combinations of links and springs that can provide the kinematic and dynamic requirements. The simplest design is to clamp the axle to the middle of two leaf springs with their ends tied or shackled to the vehicle frame as shown schematically in Figure 8.1. A side view of a multi-leaf spring and solid axle is shown in Figure 8.2. A suspension with a solid connection between the left and right wheels is called dependent suspension. 456 8. Suspension Mechanisms The performance of a solid axle with leaf springs suspension can be improved by adding a linkage to guide the axle kinematically and provide dynamic support to carry the non z-direction forces. The solid axle with leaf spring combination came to vehicle industry from horse-drawn vehicles. Example 299 Hotchkiss drive"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure9.52-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure9.52-1.png",
        "caption": "Figure 9.52 In this section, N14 is found from the moment equilibrium about H.",
        "texts": [
            " VOLUME 1: EQUILIBRIUM with the solution N4 = \u221250 kN, N7 = \u221290 kN. In the same way, we can determine N5 and N8 from the equilibrium of joint A: \u2211 Fx = \u2212(240 kN) + (280 kN) \u2212 0.8N5 = 0,\u2211 Fy = 0.6N5 + N8 = 0 such that N5 = +50 kN, N8 = \u221230 kN. The force in member 13 is the most complicated one to determine. This force is found from the equilibrium of joint D or C. However, we first have to determine one of the member forces N14 and N15, or one of N16 and N17 (see Figure 9.51). With the section in Figure 9.52, N14 is found from the moment equilibrium about H: \u2211 Tz|H = (180 kN)(2 m) \u2212 (240 kN)(3 m) + N14 \u00d7 (3 m) = 0 \u21d2 N14 = +120 kN. At joint D, N13 and N14 are now the only unknowns (see Figure 9.51). The 9 Trusses 351 two equations for the force equilibrium of the joint are: \u2211 Fx = (40 kN) + N14 \u2212 0.8 \u00d7 (50 kN) + 0.8 \u00d7 N15 = 0,\u2211 Fy = \u2212(120 kN) \u2212 N13 \u2212 0.6 \u00d7 (50 kN) \u2212 0.6 \u00d7 N15 = 0. Here substitute N14 = +120 kN to find the solution: N15 = \u2212150 kN, N13 = \u221260 kN. Table 9.3 provides a summary of all the member forces"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure12.7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure12.7-1.png",
        "caption": "FIGURE 12.7 Example of approximation for far-field computation.",
        "texts": [
            " The current I is a vector with a given direction. Assuming that the problem has a total of N inductive current cells and we apply the discretization to (12.28), we have B(r) = \u2212 N\u2211 i=1 \u222b\ud835\udcc1i 1 + j\ud835\udefdR R g(r, r\u2032) \u0302R \u00d7 Ii(r\u2032) d \ud835\udcc1\u2032, (12.32) where \ud835\udcc1\u2032 is along the longitudinal direction of the cell filament. Based on this assumption, the current density is replaced by the current for the constant current density in each cell. Accordingly, the electric field can be computed similar to (12.30) and (12.28). As shown in Fig. 12.7, r is a vector to the distant observation point location. A condition for a far-field approximation is that the observation point is far in comparison to the relative size of the object r = |r|\u226b D. Also, the wavelength \ud835\udf06 is assumed to be smaller than |r|. On the other hand, the point vector |r\u2032| for a point on the object close to the origin is much smaller than r. We call the angle at the origin \ud835\udefc. Under these assumptions, it is possible to use the far-field approximation to simplify the field calculation [12]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure2.16-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure2.16-1.png",
        "caption": "FIGURE 2.16. Optimal traction and braking force distribution between the front and rear wheels.",
        "texts": [
            " For a < 0, the optimal front brake force increases rapidly and the rear brake force goes to zero after a minimum. The deceleration (a/g) = \u2212 (a1/h) is the maximum possible deceleration at which the rear tires lose their ground contact. The graphical representation of the optimal driving and braking forces can be shown better by plotting Fx1/ (mg) versus Fx2/ (mg) using (a/g) as a parameter. Fx1 = a2 \u2212 a g h a1 + a g h Fx2 (2.238) Fx1 Fx2 = a2 \u2212 \u03bcxh a1 + \u03bcxh (2.239) 2. Forward Vehicle Dynamics 71 Such a plot is shown in Figure 2.16. This is a design curve describing the relationship between forces under the front and rear wheels to achieve the maximum acceleration or deceleration. Adjusting the optimal force distribution is not an automatic procedure and needs a force distributor control system to measure and adjust the forces. Example 66 F Slope at zero. The initial optimal traction force distribution is the slope of the optimal curve (Fx1/ (mg) , Fx2/ (mg)) at zero. d Fx1 mg d Fx2 mg = lim a\u21920 \u22121 2 h l \u00b5 a g \u00b62 + 1 2 a2 l a g 1 2 h l \u00b5 a g \u00b62 + 1 2 a1 l a g = a2 a1 (2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000538_acs.chemrev.8b00172-Figure32-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000538_acs.chemrev.8b00172-Figure32-1.png",
        "caption": "Figure 32. Structural composition and functionalization of metal\u2212organic framework (MOF) based composites for sensing applications. Adapted with permission from ref 403. Copyright 2014 Elsevier.",
        "texts": [
            " A composite nanomaterial is made from at least two different materials and bears new characteristics that are not found in the original separate materials. Many examples can be found in sensors and analytical applications, e.g. where hydrogels are doped with CNTs, polymeric nanofibers doped with metal nanoparticles, or metal\u2212organic frameworks (MOFs) of different origins are merged to form a new material. MOFs are an emerging candidate that attract considerable attention for application in electrochemical sensing. They are a hybrid material built from metal ions with well-defined coordination geometry and organic bridging ligands400 as shown in Figure 32. MOFs have been employed as a signal tracer, catalysts, a signal carrier, and an immobilization surface in electrochemical sensors. For example, Au-SH-SiO2 nanoparticles were immobilized on Cu-MOF to construct an electrochemical sensor for Lcysteine.401 Cu-MOF alone possesses electrocatalytic ability toward the oxidation of cysteine, generating a peak at 0.6 V, but introduction of Au-SH-SiO2 nanoparticles dramatically enhanced electroanalytical performance as shown by a lower oxidation potential (0"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000613_978-94-015-9261-1-Figure5.12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000613_978-94-015-9261-1-Figure5.12-1.png",
        "caption": "Figure 5.12. Inv ariant submanifold S' in task-oriented st ate space",
        "texts": [
            " Now we introduee th e extended eoordination eonditions (5.148) whieh , by definition, are satisfied when ei(t ) == O. That induees the following result. Proposition 5.1 6 . Suppose Assumptions 5.2 and 5.3 hold. Th e mo tion of the sys tem (5.111), (5.112) is eoordinated and obey~ the eoordinated eondition (5.104) if and only if the solutions of the system (5.135) satisfy the extended eoordination eonditions (5.148) . We introduee in the set Re. a smooth flpo-dim ensional hypersurfaee S* defined as th e interseetion (Figure 5.12) S\" = niS\"t, i = 1, 2, ... , po (5 .149) NONLI NEAR CONT ROL 01\" MIMO SYSTEMS of th e (m( po - 1) + J.L )-dimension al hyp ersurfaees s: = {~ E R ( : <Pi (6 , 6\" \", ~i ) = O} . 225 As S* C S*l t hen, aecording to Propositi on 5.15, item (i), for all ~ E S* we ean wri te th e extended regular ity conditions (5 .150 ) under whieh th e hyp ersurfaee S* can be eonsidered as a one-sheeted smooth manifold (submanifold of E). - In t he sets R f.. i C jRm we define th e J.L-di mension al hyp ersurfaces whieh, in view of Proposition 5"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.61-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.61-1.png",
        "caption": "FIGURE 8.61.",
        "texts": [],
        "surrounding_texts": [
            "where, a1 is the longitudinal distance between C and the front axle, b1 is the lateral distance between C and the tireprint of the tire 1, and h is the height of C from the ground level. If P is a point in the tire frame at T1rP T1rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 . (8.62) then its coordinates in the body frame are BrP = BRT1 T1rP + BdT1 = \u23a1\u23a3 a1 + xP cos \u03b41 \u2212 yP sin \u03b41 yP cos \u03b41 \u2212 b1 + xP sin \u03b41 zP \u2212 h \u23a4\u23a6 . (8.63) The rotation matrix BRT1 is a result of steering about the z1-axis. BRT1 = \u23a1\u23a3 cos \u03b41 \u2212 sin \u03b41 0 sin \u03b41 cos \u03b41 0 0 0 1 \u23a4\u23a6 (8.64) Employing Equation (8.28), we may examine a wheel point P at W rP W rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 (8.65) 8. Suspension Mechanisms 497 and find the body coordinates of the point BrP = BRT1 T1rP + BdT1 = BRT1 \u00a1 T1RW W rP + T1dW \u00a2 + BdT1 = BRT1 T1RW W rP + BRT1 T1dW + BdT1 = BRW W rP + BRT1 T1dW + BdT1 (8.66) BrP = \u23a1\u23a3 a1 + xP cos \u03b41 \u2212 yP cos \u03b3 sin \u03b41 + (Rw + zP ) sin \u03b3 sin \u03b41 xP sin \u03b41 \u2212 b1 + yP cos \u03b3 cos \u03b41 \u2212 (Rw + zP ) cos \u03b41 sin \u03b3 (Rw + zP ) cos \u03b3 + yP sin \u03b3 \u2212 h \u23a4\u23a6 (8.67) where, BRW = BRT1 T1RW = \u23a1\u23a3 cos \u03b41 \u2212 cos \u03b3 sin \u03b41 sin \u03b3 sin \u03b41 sin \u03b41 cos \u03b3 cos \u03b41 \u2212 cos \u03b41 sin \u03b3 0 sin \u03b3 cos \u03b3 \u23a4\u23a6 (8.68) T1dW = \u23a1\u23a3 0 \u2212Rw sin \u03b3 Rw cos \u03b3 \u23a4\u23a6 . (8.69) Example 334 F Wheel-body to vehicle transformation. The wheel-body coordinate frames are always parallel to the vehicle frame. The origin of the wheel-body coordinate frame of the wheel number 1 is at BdW1 = \u23a1\u23a3 a1 \u2212b1 \u2212h+Rw \u23a4\u23a6 . (8.70) Hence the transformation between the two frames is only a displacement. Br = BIW1 W1r+ BdW1 (8.71) 8.6 F Caster Theory The steer axis may have any angle and any location with respect to the wheel-body coordinate frame. The wheel-body frame C (xc, yc, zc) is a frame at the center of the wheel at its rest position, parallel to the vehicle coordinate frame. The frame C does not follow any motion of the wheel. The steer axis is the kingpin axis of rotation. Figure 8.51 illustrates the front and side views of a wheel and its steering axis. The steering axis has angle \u03d5 with (yc, zc) plane, and angle \u03b8 with (xc, zc) plane. The angles \u03d5 and \u03b8 are measured about the yc and xc axes, 498 8. Suspension Mechanisms respectively. The angle \u03d5 is the caster angle of the wheel, while the angle \u03b8 is the lean angle. The steering axis of the wheel, as shown in Figure 8.51, is at a positive caster and lean angles. The steering axis intersect the ground plane at a point that has coordinates (sa, sb,\u2212Rw) in the wheelbody coordinate frame. If we indicate the steering axis by the unit vector u\u0302, then the components of u\u0302 are functions of the caster and lean angles. C u\u0302 = \u23a1\u23a3 u1 u2 u3 \u23a4\u23a6 = 1p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 \u23a1\u23a3 cos \u03b8 sin\u03d5 \u2212 cos\u03d5 sin \u03b8 cos \u03b8 cos\u03d5 \u23a4\u23a6 (8.72) The position vector of the point that u\u0302 intersects the ground plane, is called the location vector s that in the wheel-body frame has the following coordinates: Cs = \u23a1\u23a3 sa sb \u2212Rw \u23a4\u23a6 (8.73) We express the rotation of the wheel about the steering axis u\u0302 by a zero pitch screw motion s\u030c. CTW = C s\u030cW (0, \u03b4, u\u0302, s) = \u2219 CRW Cs\u2212 CRW Cs 0 1 \u00b8 = \u2219 CRW CdW 0 1 \u00b8 (8.74) Proof. The steering axis is at the intersection of the caster plane \u03c0C and the lean plane \u03c0L, both expressed in the wheel-body coordinate frame. The 8. Suspension Mechanisms 499 two planes can be indicated by their normal unit vectors n\u03021 and n\u03022. C n\u03021 = \u23a1\u23a3 0 cos \u03b8 sin \u03b8 \u23a4\u23a6 (8.75) C n\u03022 = \u23a1\u23a3 \u2212 cos\u03d50 sin\u03d5 \u23a4\u23a6 (8.76) The unit vector u\u0302 on the intersection of the caster and lean planes can be found by u\u0302 = n\u03021 \u00d7 n\u03022 |n\u03021 \u00d7 n\u03022| (8.77) where, n\u03021 \u00d7 n\u03022 = \u23a1\u23a3 cos \u03b8 sin\u03d5 \u2212 cos\u03d5 sin \u03b8 cos \u03b8 cos\u03d5 \u23a4\u23a6 (8.78) |n\u03021 \u00d7 n\u03022| = q cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 (8.79) and therefore, C u\u0302 = \u23a1\u23a3 u1 u2 u3 \u23a4\u23a6 = \u23a1\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3 cos \u03b8 sin\u03d5p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 \u2212 cos\u03d5 sin \u03b8p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 cos \u03b8 cos\u03d5p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 \u23a4\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6 . (8.80) Steering axis does not follow any motion of the wheel except the wheel hop in the z-direction. We assume that the steering axis is a fixed line with respect to the vehicle, and the steer angle \u03b4 is the rotation angle about u\u0302. The intersection point of the steering axis and the ground plane defines the location vector s. Cs = \u23a1\u23a3 sa sb \u2212Rw \u23a4\u23a6 (8.81) The components sa and sb are called the forward and lateral locations respectively. Using the axis-angle rotation (u\u0302, \u03b4), and the location vector s, we can define the steering process as a screw motion s\u030c with zero pitch. Employing Equations (5.473)-(5.477), we find the transformation screw for wheel frame W to wheel-body frame C. CTW = C s\u030cW (0, \u03b4, u\u0302, s) (8.82) = \u2219 CRW Cs\u2212 CRW Cs 0 1 \u00b8 = \u2219 CRW Cd 0 1 \u00b8 500 8. Suspension Mechanisms CRW = I cos \u03b4 + u\u0302u\u0302T vers \u03b4 + u\u0303 sin \u03b4 (8.83) CdW = \u00a1\u00a1 I\u2212 u\u0302u\u0302T \u00a2 vers \u03b4 \u2212 u\u0303 sin \u03b4 \u00a2 Cs. (8.84) u\u0303 = \u23a1\u23a3 0 \u2212u3 u2 u3 0 \u2212u1 \u2212u2 u1 0 \u23a4\u23a6 (8.85) vers \u03b4 = 1\u2212 cos \u03b4 (8.86) Direct substitution shows that CRW and CdW are: CRW = \u23a1\u23a3 u21 vers \u03b4 + c\u03b4 u1u2 vers \u03b4 \u2212 u3s\u03b4 u1u3 vers \u03b4 + u2s\u03b4 u1u2 vers \u03b4 + u3s\u03b4 u22 vers \u03b4 + c\u03b4 u2u3 vers \u03b4 \u2212 u1s\u03b4 u1u3 vers \u03b4 \u2212 u2s\u03b4 u2u3 vers \u03b4 + u1s\u03b4 u23 vers \u03b4 + c\u03b4 \u23a4\u23a6 (8.87) CdW = \u23a1\u23a3 (s1 \u2212 u1 (s3u3 + s2u2 + s1u1)) vers \u03b4 + (s2u3 \u2212 s3u2) sin \u03b4 (s2 \u2212 u2 (s3u3 + s2u2 + s1u1)) vers \u03b4 + (s3u1 \u2212 s1u3) sin \u03b4 (s3 \u2212 u3 (s3u3 + s2u2 + s1u1)) vers \u03b4 + (s1u2 \u2212 s2u1) sin \u03b4 \u23a4\u23a6 (8.88) The vector CdW indicates the position of the wheel center with respect to the wheel-body frame. The matrix CTW is the homogeneous transformation from wheel frame W to wheel-body frame C, when the wheel is steered by the angle \u03b4 about the steering axis u\u0302. Example 335 F Zero steer angle. To examine the screw transformation, we check the zero steering. Substituting \u03b4 = 0 simplifies the rotation matrix CRW and the position vector CdW to I and 0 CRW = \u23a1\u23a3 1 0 0 0 1 0 0 0 1 \u23a4\u23a6 (8.89) CdW= \u23a1\u23a3 0 0 0 \u23a4\u23a6 (8.90) indicating that at zero steering, the wheel frameW and wheel-body frame C are coincident. Example 336 F Steer angle transformation for zero lean and caster. Consider a wheel with a steer axis coincident with zw. Such a wheel has no lean or caster angle. When the wheel is steered by the angle \u03b4, we can find the coordinates of a wheel point P in the wheel-body coordinate frame using transformation method. Figure 8.52 illustrates a 3D view, and Figure 8.53 a top view of such a wheel. 8. Suspension Mechanisms 501 502 8. Suspension Mechanisms Assume W rP = [xw, yw, zw] T is the position vector of a wheel point, then its position vector in the wheel-body coordinate frame C is CrP = CRW W rP = Rz,\u03b4 W rP = \u23a1\u23a3 cos \u03b4 \u2212 sin \u03b4 0 sin \u03b4 cos \u03b4 0 0 0 1 \u23a4\u23a6\u23a1\u23a3 xw yw zw \u23a4\u23a6 = \u23a1\u23a3 xw cos \u03b4 \u2212 yw sin \u03b4 yw cos \u03b4 + xw sin \u03b4 zw \u23a4\u23a6 . (8.91) We assumed that the wheel-body coordinate is installed at the center of the wheel and is parallel to the vehicle coordinate frame. Therefore, the transformation from the frame W to the frame C is a rotation \u03b4 about the wheel-body z-axis. There would be no camber angle when the lean and caster angles are zero and steer axis is on the zw-axis. Example 337 F Zero lean, zero lateral location. The case of zero lean, \u03b8 = 0, and zero lateral location, sb = 0, is important in caster dynamics of bicycle model. The screw transformation for this case will be simplified to C u\u0302 = \u23a1\u23a3 u1 u2 u3 \u23a4\u23a6 = \u23a1\u23a3 sin\u03d5 0 cos\u03d5 \u23a4\u23a6 (8.92) Cs = \u23a1\u23a3 sa 0 \u2212Rw \u23a4\u23a6 (8.93) CRW = \u23a1\u23a3 sin2 \u03d5 vers \u03b4 + cos \u03b4 \u2212 cos\u03d5 sin \u03b4 sin\u03d5 cos\u03d5 vers \u03b4 cos\u03d5 sin \u03b4 cos \u03b4 \u2212 sin\u03d5 sin \u03b4 sin\u03d5 cos\u03d5 vers \u03b4 sin\u03d5 sin \u03b4 cos2 \u03d5 vers \u03b4 + cos \u03b4 \u23a4\u23a6 (8.94) Cd = \u23a1\u23a2\u23a3 cos\u03d5 (sa cos\u03d5+Rw sin\u03d5) vers \u03b4 \u2212 (sa cos\u03d5+Rw sin\u03d5) sin \u03b4 \u22121 2 (Rw \u2212Rw cos 2\u03d5+ sa sin 2\u03d5) vers \u03b4 \u23a4\u23a5\u23a6 . (8.95) Example 338 F Position of the tireprint. The center of tireprint in the wheel coordinate frame is at rT W rT = \u23a1\u23a3 0 0 \u2212Rw \u23a4\u23a6 . (8.96) 8. Suspension Mechanisms 503 If we assume the width of the tire is zero and the wheel is steered, the center of tireprint would be at CrT = CTW W rT = \u23a1\u23a3 xT yT zT \u23a4\u23a6 (8.97) where xT = \u00a1 1\u2212 u21 \u00a2 (1\u2212 cos \u03b4) sa + (u3 sin \u03b4 \u2212 u1u2 (1\u2212 cos \u03b4)) sb (8.98) yT = \u2212 (u3 sin \u03b4 + u1u2 (1\u2212 cos \u03b4)) sa + \u00a1 1\u2212 u22 \u00a2 (1\u2212 cos \u03b4) sb (8.99) zT = (u2 sin \u03b4 \u2212 u1u3 (1\u2212 cos \u03b4)) sa \u2212 (u1 sin \u03b4 + u2u3 (1\u2212 cos \u03b4)) sb \u2212Rw (8.100) or xT = sb \u00c3 cos \u03b8 cos\u03d5 sin \u03b4p cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 + 1 4 sin 2\u03b8 sin 2\u03d5 (1\u2212 cos \u03b4) cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 ! +sa \u00b5 1\u2212 cos2 \u03b8 sin2 \u03d5 cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 \u00b6 (1\u2212 cos \u03b4) (8.101) yT = \u2212sa \u00c3 cos \u03b8 cos\u03d5 sin \u03b4p cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 \u2212 1 4 sin 2\u03b8 sin 2\u03d5 (1\u2212 cos \u03b4) cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 ! +sb \u00b5 1\u2212 cos2 \u03d5 sin2 \u03b8 cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 \u00b6 (1\u2212 cos \u03b4) (8.102) zT = \u2212Rw \u2212 sb cos \u03b8 sin\u03d5+ sa cos\u03d5 sin \u03b8p cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 sin \u03b4 + 1 2 sb cos 2 \u03d5 sin 2\u03b8 \u2212 sa cos 2 \u03b8 sin 2\u03d5 cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 (1\u2212 cos \u03b4) (8.103) Example 339 F Wheel center drop. The zT coordinate in (8.100) or (8.103) indicates the amount that the center of the tireprint will move in the vertical direction with respect to the wheel-body frame when the wheel is steering. If the steer angle is zero, \u03b4 = 0, then zT is at zT = \u2212Rw. (8.104) Because the center of tireprint must be on the ground, H = \u2212Rw \u2212 zT indicated the height that the center of the wheel will drop during steering. H = \u2212Rw \u2212 zT (8.105) = sb cos \u03b8 sin\u03d5+ sa cos\u03d5 sin \u03b8p cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 sin \u03b4 \u22121 2 sb cos 2 \u03d5 sin 2\u03b8 \u2212 sa cos 2 \u03b8 sin 2\u03d5 cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 (1\u2212 cos \u03b4) 504 8. Suspension Mechanisms The zT coordinate of the tireprint may be simplifie for different designs: 1\u2212 If the lean angle is zero, \u03b8 = 0, then zT is at zT = \u2212Rw \u2212 1 2 sa sin 2\u03d5 (1\u2212 cos \u03b4)\u2212 sb sin\u03d5 sin \u03b4. (8.106) 2\u2212 If the lean angle and lateral location are zero, \u03b8 = 0, sb = 0, then zT is at zT = \u2212Rw \u2212 1 2 sa sin 2\u03d5 (1\u2212 cos \u03b4) . (8.107) In this case, the wheel center drop may be expressed by a dimensionless equation. H sa = 1 2 sin 2\u03d5 (1\u2212 cos \u03b4) (8.108) Figure 8.54 illustrates H/sa for the caster angle \u03d5 = 5deg, 0 deg, \u22125 deg, \u221210 deg, \u221215 deg, \u221220 deg, and the steer angle \u03b4 in the range \u221210 deg < \u03b4 < 10 deg. In street cars, we set the steering axis with a positive longitudinal location sa > 0, and a few degrees negative caster angle \u03d5 < 0. In this case the wheel center drops as is shown in the figure. 3\u2212 If the caster angle is zero, \u03d5 = 0, then zT is at zT = \u2212Rw + 1 2 sb sin 2\u03b8 (1\u2212 cos \u03b4)\u2212 sa sin \u03b8 sin \u03b4. (8.109) 4\u2212 If the caster angle and lateral location are zero, \u03d5 = 0, sb = 0, then zT is at zT = \u2212Rw \u2212 sa sin \u03b8 sin \u03b4. (8.110) 8. Suspension Mechanisms 505 In this case, the wheel center drop may be expressed by a dimensionless equation. H sa = \u2212 sin \u03b8 sin \u03b4 (8.111) Figure 8.55 illustrates H/sa for the lean angle \u03b8 = 5deg, 0, \u22125 deg, \u221210 deg, \u221215 deg, \u221220 deg and the steer angle \u03b4 in the range \u221210 deg < \u03b4 < 10 deg. The steering axis of street cars is usually set with a positive longitudinal location sa > 0, and a few degrees positive lean angle \u03b8 > 0. In this case the wheel center lowers when the wheel number 1 turns to the right, and elevates when the wheel turns to the left. Comparison of Figures 8.54 and 8.55 shows that the lean angle has much more affect on the wheel center drop than the caster angle. 5\u2212 If the lateral location is zero, sb = 0, then zT is at zT = \u2212Rw \u2212 sa cos\u03d5 sin \u03b8p cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 sin \u03b4 \u22121 2 sa cos2 \u03b8 sin 2\u03d5 cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 (1\u2212 cos \u03b4) (8.112) and the wheel center drop,H, may be expressed by a dimensionless equation. H sa = \u22121 2 cos2 \u03b8 sin2 \u03d5 (1\u2212 cos \u03b4) cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 \u2212 cos\u03d5 sin \u03b8 sin \u03b4p cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 (8.113) Example 340 F Position of the wheel center. As given in Equation (8.88), the wheel center is at CdW with respect to 506 8. Suspension Mechanisms the wheel-body frame. CdW = \u23a1\u23a3 xW yW zW \u23a4\u23a6 (8.114) Substituting for u\u0302 and s from (8.72) and (8.73) in (8.88) provides the coordinates of the wheel center in the wheel-body frame as xW = (sa \u2212 u1 (\u2212Rwu3 + sbu2 + sau1)) (1\u2212 cos \u03b4) + (sbu3 +Rwu2) sin \u03b4 (8.115) yW = (sb \u2212 u2 (\u2212Rwu3 + sbu2 + sau1)) (1\u2212 cos \u03b4) \u2212 (Rwu1 + sau3) sin \u03b4 (8.116) zW = (\u2212Rw \u2212 u3 (\u2212Rwu3 + sbu2 + sau1)) (1\u2212 cos \u03b4) + (sau2 \u2212 sbu1) sin \u03b4 (8.117) or xW = sa (1\u2212 cos \u03b4) + \u00b5 1 2 Rw sin 2\u03d5\u2212 sa sin 2 \u03d5 \u00b6 cos2 \u03b8 + 1 4 sb sin 2\u03b8 sin 2\u03d5 cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 (1\u2212 cos \u03b4) + (sb cos \u03b8 \u2212Rw sin \u03b8)p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 cos\u03d5 sin \u03b4 (8.118) yW = sb (1\u2212 cos \u03b4) \u2212 1 2 \u00a1 Rw sin 2\u03b8 + sb sin 2 \u03b8 \u00a2 cos2 \u03d5\u2212 1 4 sa sin 2\u03b8 sin 2\u03d5 cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 (1\u2212 cos \u03b4) \u2212 Rw sin\u03d5+ sa cos\u03d5p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 cos \u03b8 sin \u03b4 (8.119) zW = \u2212Rw (1\u2212 cos \u03b4) + \u00b5 Rw cos 2 \u03b8 + 1 2 sb sin 2\u03b8 \u00b6 cos2 \u03d5\u2212 1 2 sa cos 2 \u03b8 sin 2\u03d5 cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 (1\u2212 cos \u03b4) \u2212sa cos\u03d5 sin \u03b8 + sb cos \u03b8 sin\u03d5p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 sin \u03b4 (8.120) The zW coordinate indicates how the center of the wheel will move in the vertical direction with respect to the wheel-body frame, when the wheel is steering. It shows that zW = 0, as long as \u03b4 = 0. 8. Suspension Mechanisms 507 The zW coordinate of the wheel center may be simplified for different designs: 1\u2212 If the lean angle is zero, \u03b8 = 0, then zW is at zW = \u2212Rw \u00a1 1\u2212 cos2 \u03d5 \u00a2 (1\u2212 cos \u03b4)\u2212 sb sin\u03d5 sin \u03b4 \u22121 2 sa sin 2\u03d5 (1\u2212 cos \u03b4) . (8.121) 2\u2212 If the lean angle and lateral location are zero, \u03b8 = 0, sb = 0, then zW is at zW = \u2212Rw \u00a1 1\u2212 cos2 \u03d5 \u00a2 (1\u2212 cos \u03b4)\u2212 1 2 sa sin 2\u03d5 (1\u2212 cos \u03b4) . (8.122) 3\u2212 If the caster angle is zero, \u03d5 = 0, then zW is at zW = \u2212Rw \u00a1 1\u2212 cos2 \u03b8 \u00a2 (1\u2212 cos \u03b4)\u2212 sa sin \u03b8 sin \u03b4 + 1 2 sb sin 2\u03b8 (1\u2212 cos \u03b4) . (8.123) 4\u2212 If the caster angle and lateral location are zero, \u03d5 = 0, sb = 0, then zW is at zW = \u2212Rw \u00a1 1\u2212 cos2 \u03b8 \u00a2 (1\u2212 cos \u03b4)\u2212 sa sin \u03b8 sin \u03b4. (8.124) 5\u2212 If the lateral location is zero, sb = 0, then zT is at zW = \u2212Rw (1\u2212 cos \u03b4)\u2212 sa cos\u03d5 sin \u03b8p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 sin \u03b4 + Rw cos 2 \u03b8 cos2 \u03d5\u2212 1 2 sa cos 2 \u03b8 sin 2\u03d5 cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 (1\u2212 cos \u03b4) (8.125) In each case of the above designs, the height of the wheel center with respect to the ground level can be found by adding H to zW . The equations for calculating H are found in Example 340. Example 341 F Camber theory. Having a non-zero lean and caster angles causes a camber angle \u03b3 for a steered wheel. To find the camber angle of an steered wheel, we may determine the angle between the camber line and the vertical direction zc. The camber line is the line connecting the wheel center and the center of tireprint. The coordinates of the center of tireprint (xT , yT , zT ) are given in Equations (8.101)-(8.103), and the coordinates of the wheel center (xW , yW , zW ) are given in Equations (8.118)-(8.120). The line connecting (xT , yT , zT ) to (xW , yW , zW ) may be indicated by the unit vector l\u0302c l\u0302c = (xW \u2212 xT ) I\u0302 + (yW \u2212 yT ) J\u0302 + (zW \u2212 zT ) K\u0302q (xW \u2212 xT ) 2 + (yW \u2212 yT ) 2 + (zW \u2212 zT ) 2 (8.126) 508 8. Suspension Mechanisms in which I\u0302 , J\u0302 , K\u0302, are the unit vectors of the wheel-body coordinate frame C. The camber angle is the angle between l\u0302c and K\u0302, which can be found by the inner vector product. \u03b3 = cos\u22121 \u00b3 l\u0302c \u00b7 K\u0302 \u00b4 = cos\u22121 (zW \u2212 zT )q (xW \u2212 xT ) 2 + (yW \u2212 yT ) 2 + (zW \u2212 zT ) 2 (8.127) As an special case, let us determine the camber angle when the lean angle and lateral location are zero, \u03b8 = 0, sb = 0. In this case, we have xT = sa \u00a1 1\u2212 sin2 \u03d5 \u00a2 (cos \u03b4 \u2212 1) (8.128) yT = \u2212sa cos\u03d5 sin \u03b4 (8.129) zT = zT = \u2212Rw \u2212 1 2 sa sin 2\u03d5 (1\u2212 cos \u03b4) (8.130) xW = \u00b5 sa + 1 2 Rw sin 2\u03d5\u2212 sa sin 2 \u03d5 \u00b6 (1\u2212 cos \u03b4) (8.131) yW = sb (1\u2212 cos \u03b4)\u2212Rw sin\u03d5+ sa cos\u03d5 sin \u03b4 (8.132) zW = \u00b5 Rw \u00a1 cos2 \u03d5\u2212 1 \u00a2 \u2212 1 2 sa sin 2\u03d5 \u00b6 (1\u2212 cos \u03b4) . (8.133) 8.7 Summary There are two general types of suspensions: dependent, in which the left and right wheels on an axle are rigidly connected, and independent, in which the left and right wheels are disconnected. Solid axle is the most common dependent suspension, while McPherson and double A-arm are the most common independent suspensions. The roll axis is the instantaneous line about which the body of a vehicle rolls. Roll axis is found by connecting the roll center of the front and rear suspensions of the vehicle. The instant center of rotation of a wheel with respect to the body is called suspension roll center. So, to find the roll center of the front or rear half of a car, we should determine the suspension roll centers, and find the intersection of the lines connecting the suspension roll centers to the center of their associated tireprints. Three coordinate frames are employed to express the orientation of a tire and wheel with respect to the vehicle: the wheel frame W , wheel-body frame C, and tire frame T . A wheel coordinate frame W (xw, yw, zw) is attached to the center of a wheel. It follows every translation and rotation of the wheel except the spin. Hence, the xw and zw axes are always in the tire-plane, while the yw-axis is always along the spin axis. When the wheel 8. Suspension Mechanisms 509 is straight and the W frame is parallel to the vehicle coordinate frame, we attach a wheel-body coordinate frame C (xc, yc, zc) at the center of the wheel parallel to the vehicle coordinate axes. The wheel-body frame C is motionless with respect to the vehicle coordinate and does not follow any motion of the wheel. The tire coordinate frame T (xt, yt, zt) is set at the center of the tireprint. The zt-axis is always perpendicular to the ground. The xt-axis is along the intersection line of the tire-plane and the ground. The tire frame does not follows the spin and camber rotations of the tire however, it follows the steer angle rotation about the zc-axis. We define the orientation and position of a steering axis by the caster angle \u03d5, lean angle \u03b8, and the intersection point of the axis with the ground surface at (sa, sb) with respect to the center of tireprint. Because of these parameters, a steered wheel will camber and generates a lateral force. This is called the caster theory. The camber angle \u03b3 of a steered wheel for \u03b8 = 0, and sb = 0 is: \u03b3 = cos\u22121 \u00b3 l\u0302c \u00b7 K\u0302 \u00b4 = cos\u22121 (zW \u2212 zT )q (xW \u2212 xT ) 2 + (yW \u2212 yT ) 2 + (zW \u2212 zT ) 2 (8.134) where xT = sa \u00a1 1\u2212 sin2 \u03d5 \u00a2 (cos \u03b4 \u2212 1) (8.135) yT = \u2212sa cos\u03d5 sin \u03b4 (8.136) zT = zT = \u2212Rw \u2212 1 2 sa sin 2\u03d5 (1\u2212 cos \u03b4) (8.137) xW = \u00b5 sa + 1 2 Rw sin 2\u03d5\u2212 sa sin 2 \u03d5 \u00b6 (1\u2212 cos \u03b4) (8.138) yW = sb (1\u2212 cos \u03b4)\u2212Rw sin\u03d5+ sa cos\u03d5 sin \u03b4 (8.139) zW = \u00b5 Rw \u00a1 cos2 \u03d5\u2212 1 \u00a2 \u2212 1 2 sa sin 2\u03d5 \u00b6 (1\u2212 cos \u03b4) . (8.140) 510 8. Suspension Mechanisms 8.8 Key Symbols a, b, c, d lengths of the links of a four-bar linkage ai distance of the axle number i from the mass center A,B, \u00b7 \u00b7 \u00b7 coefficients in equation for calculating \u03b83 b1 distance of left wheels from mass center B (x, y, z) vehicle coordinate frame C mass center C coupler point C (xc, yc, zc) wheel-body coordinate frame C T dW C expression of the position of W with respect to T e, \u03b1 polar coordinates of a coupler point g overhang h = z \u2212 z0 vertical displacement of the wheel center H wheel center drop Iij instant center of rotation between link i and link j IijImn a line connecting Iij and Imn I\u0302 , J\u0302 , K\u0302 unit vectors of the wheel-body frame C I identity matrix J1, J2, \u00b7 \u00b7 \u00b7 length function for calculating \u03b83 l\u0302c unit vector on the line (xT , yT , zT ) to (xW , yW , zW ) ms sprung mass mu unsprung mass n\u03021 normal unit vectors to \u03c0L n\u03022 normal unit vectors to \u03c0C P point q, p, f parameters for calculating couple point coordinate r position vector Rw tire radius TRW rotation matrix to go from W frame to T frame s position vector of the steer axis sa forward location of the steer axis sb lateral location of the steer axis s\u030cW (0, \u03b4, u\u0302, s) zero pitch screw about the steer axis T (xt, yt, zt) tire coordinate system TTW homogeneous transformation to go from W to T u\u0302 steer axis unit vector u\u0303 skew symmetrix matrix associated to u\u0302 uC position vector of the coupler point u\u0302z unit vector in the z-direction vx forward speed x, y suspension coordinate frame xC , yC coordinate of a couple point (xT , yT , zT ) wheel-body coordinates of the origin of T frame 8. Suspension Mechanisms 511 (xW , yW , zW ) wheel-body coordinates of the origin of W frame vers \u03b4 1\u2212 cos \u03b4 W (xwywzw) wheel coordinate system z vertical position of the wheel center z0 initial vertical position of the wheel center \u03b1 angle of a coupler point with upper A-arm \u03b3 camber angle \u03b4 steer angle \u03b5 = ms/mu sprung to unsprung mass ratio \u03b8 lean angle \u03b80 angle between the ground link and the z-direction \u03b8i angular position of link number i \u03b82 angular position of the upper A-arm \u03b83 angular position of the coupler link \u03b84 angular position of link lower A-arm \u03b8i0 initial angular position of \u03b8i \u03c0C caster plane \u03c0L lean plane \u03c5 trust angle \u03d5 caster angle \u03c9 angular velocity 512 8. Suspension Mechanisms Exercises 1. Roll center. Determine the roll ceneter of the kinematic models of vehicles shown in Figures 8.56 to 8.59. 2 1 Body 4 6 3 5 8 7 8. Suspension Mechanisms 513 514 8. Suspension Mechanisms 8. Suspension Mechanisms 515 4. F Position of the roll center and mass ceneter. Figure 8.66 illustrates the wheels and mass center C of a vehicle. Design a double A-arm suspension such that the roll center of the C 516 8. Suspension Mechanisms c d a 2\u03b8 3\u03b8 4\u03b8 A B M N x y 0\u03b8 \u03b1 e z Cb 8. Suspension Mechanisms 517 Determine CTW for \u03d5 = 8deg, \u03b8 = 12deg, and the location vector Cs Cs = \u23a1\u23a3 3.8 cm 1.8 cm \u2212Rw \u23a4\u23a6 . (a) The vehicle uses a tire 235/35ZR19. (b) The vehicle uses a tire P215/65R15 96H. 10. F Wheel drop. Find the coordinates of the tireprint for \u03d5 = 10deg \u03b8 = 10deg Cs = \u23a1\u23a3 3.8 cm 1.8 cm 38 cm \u23a4\u23a6 if \u03b4 = 18deg. How much is the wheel drop H. 11. F Wheel drop and steer angle. Draw a plot to show the wheel drop H at different steer angle \u03b4 for the given data in Exercise 10. 12. F Camber and steering. Draw a plot to show the camber angle \u03b3 at different steer angle \u03b4 for the following characteristics: \u03d5 = 10deg \u03b8 = 0deg Cs = \u23a1\u23a3 3.8 cm 0 cm 38 cm \u23a4\u23a6 Part III Vehicle Dynamics 9"
        ]
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure13.35-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure13.35-1.png",
        "caption": "Figure 13.35 (a) Isolated beam ACD with a uniformly distributed load of 1 kN/m normal to the roof plane and associated (b) normal force diagram.",
        "texts": [
            "33 (c) Shear force diagram and (d) bending moment diagram due to a uniformly distributed load of 1 kN/m normal to the roof plane. 13 Calculating M, V and N Diagrams 577 \u2022 N , V and M diagrams due to qpa = 1 kN/m (Figure 13.31b) In Figure 13.34, ACD has been isolated and the joining forces acting on ACD are shown. The distributed load qpa = 1 kN/m has been replaced by its resultant Rpa: Rpa = qpa ACD = (1 kN/m)(7.539 m) = 7.539 kN. The moment equilibrium about A gives Ctr = 0 and so Cpa = 0. The force equilibrium of ACD gives Atr = 0 and Apa = Rpa = 7.539 kN. In Figure 13.35a, ACD is shown again with the forces determined. In Figure 13.35b, the associated N diagram is shown: due to a uniformly distributed load the normal force is linear. With this load there are no bending moments and shear forces. Figure 13.36 (a) Lean-to with wind load of 5 kN/m and (b) associated normal force diagram. Figure 13.37 The components of the dead weigh qdw of a member segment with length a. a. Wind load The wind load qw = 5 kN/m in Figure 13.36a is normal to the roof plane ACD. The associated N , V and M diagrams are equal to those in Figure 13.33b to d, but with values that are five times as large: Mw;max = (2",
            "462 as large, so that the extreme values of the bending moments are Mdw;max = (2.558 kNm) \u00d7 4.642 = 11.87 kNm, Mdw;min = (2.320 kNm) \u00d7 4.642 = 10.77 kNm. 13 Calculating M, V and N Diagrams 579 580 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM The N diagram in Figure 13.39a due to the 4.642 kN/m load normal to the beam axis is equal to the N diagram in Figure 13.33b, but with values that are 4.642 times as large. The N diagram in Figure 13.39b due to the load of 1.857 kN/m parallel to the beam axis is equal to the N diagram in Figure 13.35b, but then with values that are 1.857 times as large. Superposing the N diagrams in Figures 13.39a and 13.39b gives the N diagram in Figure 13.39c. This is the requested N diagram due to the dead weight qdw = 5 kN/m. Figure 13.39 (a) The N diagram due to the component normal to the roof plane, superposed on (b) the N diagram due to the component parallel to the roof plane gives (c) the requested N diagram due to the dead weight. 13 Calculating M, V and N Diagrams 581 c. Snow load Over a length a measured horizontally, the resultant of the snow load is aqsn (see Figure 13",
            "310 times as large, so that the extreme values of the bending moment are: Msn;max = (2.558 kNm) \u00d7 4.310 = 11.02 kNm ( ), Msn;min = (2.320 kNm) \u00d7 4.310 = 10 kNm ( ). 582 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM The N diagram in Figure 13.42a due to the load of 4.310 kN/m normal to the beam axis is equal to the N diagram in Figure 13.33b, but with values that are 4.310 times as large. The N diagram in Figure 13.42b due to the load of 1.724 kN/m parallel to the beam axis is equal to the N diagram in Figure 13.35b, but with values that are 1.724 times as large. By superposing the N diagrams in Figures 13.42a and 13.42b on one another we get the N diagram in Figure 13.42c. This is the requested N diagram due to the snow loading qsn = 5 kN/m. With indirectly loaded beams, the load does not act on the beam directly, but is rather transferred to the beam by means of a system of stringer beams and cross beams. Figure 13.43 shows a schematic representation of a bridge constructed as an indirectly loaded beam. Main beam (mb) AB is carrying cross beams (cb) at regular distances which in turn are carrying stringer beams (sb)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001735_tmech.2014.2329945-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001735_tmech.2014.2329945-Figure4-1.png",
        "caption": "Fig. 4. Bifilar torsional method.",
        "texts": [
            " It can be computed analytically by summing the contributions of all the individual elements of quadrotor; however, this method is not easy to apply into composite assembled system. In this paper, we consider experimental technique for measuring the moment of inertia of quadrotor. The bifilar (two-wire) torsional method is an apparatus that has been used for measuring the moment of inertia of aerial vehicles because of its simplicity, safety, and relatively high accuracy [26]. The bifilar torsional method, as shown in Fig. 4, consists of a test object suspended by two thin wires, which have same length, of height h and separation displacements a and b from the quadrotor center of gravity. The quadrotor is rendered to oscillate around a vertical axis through its center by human operator. The period of oscillation is recorded. The bifilar torsional method is easily defined by its kinetic and potential energy via Lagrangian dynamics formulation. The total kinetic energy can be changed into the rotational kinetic energy of the quadrotor by omitting the small amount of kinetic energy in the suspension wires"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure14.41-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure14.41-1.png",
        "caption": "FIGURE 14.41. A McPherson suspension and its equivalent vibrating system.",
        "texts": [
            "6 Key Symbols a, x\u0308 acceleration a, b arm length of displaced spring c damping cF optimum damping ceq equivalent damping d1 road wave length d2 road wave amplitude D dissipation function f, F force f = 1 T cyclic frequency [ Hz] fc damper force fk spring force fn cyclic natural frequency [ Hz] F amplitude of a harmonic force f g gravitational acceleration G0 = |X/Y | absolute displacement frequency response G2 = \u00af\u0304\u0304 X\u0308/Y \u03c92n \u00af\u0304\u0304 absolute acceleration frequency response k stiffness kF optimum stiffness keq equivalent stiffness K kinetic energy L Lagrangean m mass r = \u03c9 \u03c9n frequency ratio S2 = |Z/Y | relative displacement frequency response SZ RMS of S2 SX\u0308 RMS of G2 t time T period v,v, x\u0307, x\u0307 velocity V potential energy x absolute displacement X steady-state amplitude of x y base excitation displacement Y steady-state amplitude of y z relative displacement Z steady-state amplitude of z Zi short notation parameter \u03b1 tilted spring angle \u03b4 spring deflection \u03b4 displacement 926 14. Suspension Optimization \u03be = c 2 \u221a km damping ratio \u03c9 = 2\u03c0f angular frequency [ rad/ s] \u03c9n natural frequency Subscript eq equivalent f front l low r rear s sprung u unsprung u up 14. Suspension Optimization 927 Exercises 1. Equivalent McPherson suspension parameters. Figure 14.41(a) illustrates a McPherson suspension. Its equivalent vibrating system is shown in Figure 14.41(b). (a) Determine keq and ceq if a = 22 cm b = 45 cm k = 10000N/m c = 1000N s/m \u03b1 = 12deg . (b) Determine the stiffness k such that the natural frequency of the vibrating system is fn = 1Hz, if a = 22 cm b = 45 cm m = 1000/4 kg \u03b1 = 12deg . 928 14. Suspension Optimization (c) Determine the damping c such that the damping ratio of the vibrating system is \u03be = 0.4, if a = 22 cm b = 45 cm m = 1000/4 kg \u03b1 = 12deg fn = 1Hz. 2. Equivalent double A-arm suspension parameters. Figure 14.42(a) illustrates a double A-arm suspension"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000206_978-0-387-74315-8-Figure2-3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000206_978-0-387-74315-8-Figure2-3-1.png",
        "caption": "Figure 2-3: Acceleration of a rigid body",
        "texts": [
            "11) it is necessary to replace the transpose operator with the spatial transpose operator describcd in Chapter 3. 30 =vXa. (2.27) . . If O is a moving coordinate systcm and vis the apparent velocity of a in O, then Equation (2.27) gives the apparcnt derivative of a in O. The absolute spatial acceleration of a rigid body is just the absolute derivative of its spatial velocity. However, the spatial accelcration of a rigid body differs somewhat from its conventional acceleration. Consider the rigid body shown in Figure 2-3. It has angular velocity w, and the linear velocity of a point P in the rigid body is ~ , where r is the position vector of P. The spatial velocity of the body in stationary coordinates at O is (2.28) 31 and its absolute spatial acceleration is the component-wise dcrivativc of this, which is (2.29) The angular component of the spatial acceleration is the angular acceleration of the body, but the linear component is the rate of change over time of the velocity of whichever point in the rigid body happens to be passing through the origin"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure7.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure7.2-1.png",
        "caption": "Figure 7.2 A small, rectangular volume element isolated from a gas or fluid, with compressive forces p1; p2; p3 Since there are no shear stresses, the stresses are normal to the sides in question. Here, the arrows should not be interpreted as forces.",
        "texts": [
            " Because of the weak bonding, gas and fluid particles can easily move with respect to one another. As a result, we could (rather boldly) state that no shear stresses can be transmitted in gases and fluids. This is, however, not the case with flowing gases and fluids; because of the differences in speed between adjacent layers, shear stresses can occur, although they are far weaker than in solids. Below, it is assumed that no shear stresses occur in gases and fluids at rest. This means that the stresses in stationary gases and fluids always act normal to any bounding plane. In Figure 7.2, a rectangular volume element has been isolated from a gas or fluid. Compressive stresses p1; p2; p3 act on the boundary of the element. The volume element is so small that, for all the stresses on the boundary, it can be assumed that they are uniformly distributed. In that case, one does not have to draw the entire stress distribution, but a single arrow1 is sufficient. The condition that no shear stresses can act in the material implies that the stresses on the boundary of the volume element have to be of the same magnitude: p1 = p2 = p3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure10.16-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure10.16-1.png",
        "caption": "Figure 10.16 (a) The resultant of the shear stresses on a small area A around a point P is a small shear force, with components Vy and Vz. The forces Vy and Vz in P are statically equivalent to (b) the small forces Vy and Vz in the normal force centre NC (the intersection of the member axis with the cross-sectional plane), together with (c) a small moment Mt in the cross-sectional plane.",
        "texts": [
            " The bending moments My and Mz are positive when a tensile stress (\u03c3xx >0) on a small elemental area A for y >0 makes a positive contribution to My or for z > 0 makes a positive contribution to Mz. Figure 10.15 shows the positive directions of N , My and Mz. These are the section forces that are transferred via normal stresses in the member cross-section. The resultant of the shear stresses on a small area A around a point P is a small shear force V , with components Vy and Vz: Vy = \u03c3xy A, Vz = \u03c3xz A. When assuming these small forces act in the member axis (by shifting them to the origin of the yz coordinate system), we have to add a small moment Mt in the cross-sectional plane (see Figure 10.16): Mt = y \u00b7 Vz \u2212 z \u00b7 Vy = (y\u03c3xz \u2212 z\u03c3xy) A. Summation of the contributions of all the forces Vy and Vz for the entire cross-section results in 398 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Vy = \u222b A \u03c3xy dA, Vz = \u222b A \u03c3xz dA, Mt = \u222b A (y\u03c3xz \u2212 z\u03c3xy) dA. \u2022 Vy and Vz are the components of the shear force V , that are the resultant forces (or rather: pair of forces) due to the shear stresses in the crosssection. \u2022 Mt is a moment (or rather: pair of moments) that acts in the crosssectional plane (the yz plane)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000554_978-3-540-85629-0-Figure8.39-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000554_978-3-540-85629-0-Figure8.39-1.png",
        "caption": "Fig. 8.39. Interpolation of a set of points by means of a Be\u0301zier curve of degree 5; the interpolating cubic spline is also shown with a thin line, see also Fig. 8.36.",
        "texts": [
            " In particular, each segment can be transformed in a standard polynomial form, i.e. bk(u) = a0,k + a1,ku + a2,ku2 + a3,ku3 + a4,ku4 + a5,ku5, 0 \u2264 u \u2264 1 by assuming 402 8 Multidimensional Trajectories and Geometric Path Planning \u23a7\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23a8 \u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23a9 a0,k = p0,k a1,k = \u22125p0,k + 5p1,k a2,k = 10p0,k \u2212 20p1,k + 10p2,k a3,k = \u221210p0,k + 30p1,k \u2212 30p2,k + 10p3,k a4,k = 5p0,k \u2212 20p1,k + 30p2,k \u2212 20p3,k + 5p4,k a5,k = \u2212p0,k + 5p1,k \u2212 10p2,k + 10p3,k \u2212 5p4,k + p5,k. Example 8.19 The points of Example 8.18, interpolated by means of a quintic Be\u0301zier curve, are shown in Fig. 8.39. The curve is obtained by computing the B-spline which crosses all points and calculating tangent and curvature vectors. The B-spline trajectory s(u) obtained as explained in Sec. 8.4 is defined by the knots (assumed with a cord-length distribution) u = [ 0, 0, 0, 0, 0.22, 0.38, 0.56, 0.79, 1, 1, 1, 1 ] and by the control points P = \u23a1 \u23a30 0.33 0.88 1.79 4.57 4.79 5.66 6 0 0.66 1.89 3.26 3.04 2.34 0.66 0 0 0.33 1.71 \u22120.34 \u22120.44 2.68 2.00 2 \u23a4 \u23a6 . By computing the derivatives of s(u), at[ u\u0304k ] = [ 0, 0"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure9.26-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure9.26-1.png",
        "caption": "Fig. 9.26. Construction of penny-motor: 1 \u2014 shaft, 2 \u2014 soft steel yoke cover, 3 \u2014 PM ring, 4 \u2014 ball bearing, 5 \u2014 stator winding, 6 \u2014 flange, 7 \u2014 bottom steel yoke.",
        "texts": [],
        "surrounding_texts": [
            "Ultra-flat PM micromotor, the so called penny-motor is shown in Figs 9.25 and 9.26 [154]. The thickness is 1.4 to 3.0 mm, outer diameter about 12 mm, torque constant up to 0.4 \u00b5Nm/mA and speed up to 60,000 rpm. A 400-\u00b5m eight pole PM and a three-strand, 110-\u00b5m disc shaped, lithographically produced stator winding have been used [154]. Plastic bound NdFeB magnets are a cost effective solution. However, the maximum torque is achieved with sintered NdFeB magnets. A miniature ball bearing has a diameter of 3 mm. Pennymotors find applications in miniaturized hard disc drives, cellular phones as vibration motors, mobile scanners and consumer electronics. Moving Magnet Technologies (MMT) two or three phase, miniature AFPM brushless motors (Fig. 9.27) are well adapted to high volume low cost production. In order to obtain rigidity the stator coils are overmoulded. The stator is mounted on a single-sided printed circuit board. Each part can be manufactured using standard moulding, stamping techniques and automatic coil winding with a specific design for simple and efficient assembly. Three position 9.8 Miniature AFPM Brushless Motors 309 sensing methods for the closed-loop control are used: (a) digital Hall probes located in the stator, (b) EMF signal (sensorless mode) and (c) absolute analogue position sensor (position sensor mode). MMT AFPM miniature rotary actuators (Fig. 9.28) have been designed for automotive applications to provide an efficient, contactless, direct drive rotary motion on a limited stroke. Ring, disc and tile shaped PMs have been used. The main features of this family of actuators are: \u2022 contactless actuation principle; \u2022 constant torque independent of the angular position for a given current; \u2022 linear torque\u2013current characteristic; \u2022 two directions of rotation; \u2022 high torque density. 310 9 Applications Additional functions, such as a magnetic return spring or analogue contactless position sensing can also be implemented. Owing to the linearity of the torquecurrent characteristic and its independence of position, it is possible to operate actuators in an open-loop against a spring or in a simple closed-loop mode by using a position sensor."
        ]
    },
    {
        "image_filename": "designv10_0_0001156_j.addma.2017.10.011-Figure22-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001156_j.addma.2017.10.011-Figure22-1.png",
        "caption": "Fig. 22. Illustration of Revolution powder analyser [62].",
        "texts": [
            " However, HR and AOR measurements may not adequately portray the flow behaviour of feedstock used in SLM systems and would require additional dynamic flow measurements to fully describe the rheological performance of powders distributed by a coater/feeder device [107]. Recently, a powder revolution analyser technique which measures the dynamic state of powders was studied by [62] and was said to be analogous to the delivery conditions in SLM systems. It consists of a rotating drum that illustrates powder motion under repeated roller depositions while powder images are simultaneously captured over multiple revolutions (Refer to Fig. 22). In the study, Gaussian powders which exhibited good flow tend to possess low avalanche angles between 49 \u223c 54\u25e6 and a surface fractal value of \u223c5.0. Other rheology diagnosis tools include Ring Shear Cell [114] and Freeman FT4 Rheometer [115] which measure powder flow under torque and axial loading conditions. The rheometer is a much preferred method for evaluating powder rheology in SLM applications since flow measurements are conducted under relatively lower stress states similar to powder feeding as compared to ring shear testing which is more applicable to hoppers [116]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure6.7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure6.7-1.png",
        "caption": "FIGURE 6.7. Four popular windshield wiper systems.",
        "texts": [
            "86) Example 226 F Designing a four-bar linkage using Freudenstein\u2019s equation. Designing a mechanism can be thought of as determining the required lengths of the links to accomplish a specific task. Freudenstein\u2019s equation (6.24) J1 cos \u03b84 \u2212 J2 cos \u03b82 + J3 = cos (\u03b84 \u2212 \u03b82) (6.87) J1 = d a (6.88) J2 = d c (6.89) J3 = a2 \u2212 b2 + c2 + d2 2ac (6.90) determines the input-output relationship of a four-bar linkage. This equation can be utilized to design a four-bar linkage for three associated inputoutput angles. 322 6. Applied Mechanisms Figure 6.7 illustrates the four popular windshield wiper systems. Doublearm parallel method is the most popular wiping system that serves more than 90% of passenger cars. The double-arm opposing method has been used been using since last century, however, it was never very popular. The single-arm simple method is not very efficient, so the controlled single-arm is designed to maximize the wiped area. Wipers are used on windshields, and headlights. Figure 6.8 illustrates a sample of double-arm parallel windshield wiper mechanism"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure2.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure2.1-1.png",
        "caption": "Fig. 2.1. Single sided disc type machines: (a) for industrial and traction electromechanical drives, (b) for hoist applications. 1 \u2014 laminated stator, 2 \u2014 PM, 3 \u2014 rotor, 4 \u2014 frame, 5 \u2014 shaft, 6 \u2014 sheave.",
        "texts": [
            "35 + 15.90 + 770.40 + 268.93 + 283.21 = $1838.69 2 Principles of AFPM Machines In this chapter the basic principles of the AFPM machine are explained in details. Considerable attention is given to the magnetic circuits, windings, torque production, losses, equivalent circuits, sizing procedure, armature reaction and performance characteristics of AFPM machines. The single-sided construction of an axial flux machine is simpler than the double-sided one, but the torque production capacity is lower. Fig. 2.1 shows typical constructions of single-sided AFPM brushless machines with surface PM rotors and laminated stators wound from electromechanical steel strips. The single-sided motor according to Fig. 2.1a has a standard frame and shaft. It can be used in industrial, traction and servo electromechanical drives. The motor for hoist applications shown in Fig. 2.1b is integrated with a sheave (drum for ropes) and brake (not shown). It is used in gearless elevators [113]. In the double-sided machine with internal PM disc rotor , the armature winding is located on two stator cores. The disc with PMs rotates between two stators. An eight-pole configuration is shown in Fig. 2.2. PMs are embedded or glued in a nonmagnetic rotor skeleton. The nonmagnetic air gap is large, i.e. the total air gap is equal to two mechanical clearances plus the thickness of a PM with its relative magnetic permeability close to unity",
            " The stator core can be fabricated using either laminated steels or SMC powders. General equations given in Chapter 2 for the performance calculations will be developed further and adjusted to the construction of AFPM machines with ferromagnetic cores. Application of the FEM analysis to performance calculations is also emphasized. Single-sided AFPM machines with stator ferromagnetic cores have a single PM rotor disc opposite to a single stator unit consisting of a polyphase winding and ferromagnetic core (Fig. 2.1). The stator ferromagnetic cores can be slotted or slotless. The stator winding is always made of flat wound coils (Fig. 2.8). The PMs can be mounted on the surface of the rotor or embedded (buried) in the rotor disc. In the case of a slotless stator the magnets are almost always surface mounted, while in the case of a slotted stator with a small air gap between the rotor and stator core, the magnets can be either surface mounted on the disc (Fig. 2.1) or buried in the rotor disc (Fig. 2.6). Large axial magnetic forces on bearings are the main drawback of single-sided AFPM machines with ferromagnetic stator cores. In double-sided AFPM machines with ideal mechanical and magnetic symmetry, the axial magnetic forces are balanced. Double-sided AFPM machines with stator ferromagnetic cores have either a single PM rotor disc with ironcored stators on both sides of the discs (Figs 1.4c and 2.3) or outer PM rotor discs with iron-cored stator fixed in the middle (Figs 2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000233_978-1-4614-0460-6-Figure6.39-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000233_978-1-4614-0460-6-Figure6.39-1.png",
        "caption": "Fig. 6.39 Force balance to determine k11",
        "texts": [
            " As anticipated, it matches the m12 result provided in Eq. 6.25. The mass matrix is provided in Eq. 6.30. We note that it is symmetric and, because the off-diagonals are nonzero, the equations of motion are coupled through the mass matrix. fm21 \u00bc m1 \u00fe 9m2 4 \u00fe m1 2 \u00fe 3m2 2 \u00bc m1 3m2 4 (6.29) m \u00bc m11 m12 m21 m22 \u00bc m1 \u00fe m2 4 m1 3m2 4 m1 3m2 4 m1 \u00fe 9m2 4 2 64 3 75 (6.30) Now let\u2019s populate the stiffness matrix. For the k11 on-diagonal term, we need the force required to give a unit displacement to x1 while holding x2 stationary. The forces are shown in Fig. 6.39. The moment sum about the motionless x2 is given by: X M \u00bc fk11 2l 3 \u00fe k1x1 2l 3 \u00bc fk11 2l 3 \u00fe k1 2l 3 \u00bc 0: (6.31) Dividing by 2l/3 and solving for fk11 give the stiffness k11. 2 3x2m2 2 x2m1 fm22 x1 x2 2 3x2m2 2 x2m1 fm22 x1 x2 fm21 Fig. 6.38 Force balance to determine m21 The other on-diagonal term, k22, is determined from the force required to give a unit displacement to x2 while holding x1 fixed. The forces are displayed in Fig. 6.41. Summing the moments about x1 gives: X M \u00bc fk22 2l 3 k2x2 2l 3 \u00bc fk22 2l 3 k2 2l 3 \u00bc 0: (6"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure9.25-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure9.25-1.png",
        "caption": "FIGURE 9.25 Shorted L-shaped conductor over ground-plane model.",
        "texts": [
            " The geometrical data for this example is shown in Table 9.1. The additional data required is w= 10 \u03bcm and the gap width is wc = 0.2 \u03bcm. We used two different models for the contacts area at the inside of the gap. In one case, we use a 0.4 \u03bcm section at the contact surface with a conductivity that is 104 times larger than copper. This leads to contacts at the entire cross-sectional surface. Alternatively, point contacts are used at the front of the gap as shown in Fig. 9.24. The second problem in Fig. 9.25 consists of an L-shaped conductor with a close ground plane. At high frequencies, the current will take the path that minimizes the inductance right 240 SKIN EFFECT MODELING under the L-shaped wire. However, at low frequencies, the current takes the path of least resistance that is across the plane. The lengths of the L conductors are relatively small for this redistribution to be very large. Hence, the change in inductance is relatively modest. Much larger changes in L and R have been obtained for longer lines [45]. The dimensions for the L-shaped problem Fig. 9.25 are also given in Table 9.1 for both the thin conductor LShape5 and the thick conductor LShape18. We chose a very wide conductor of 80 \u03bcm since it leads to a more interesting current behavior. The short-to-ground at the end of the top L-shaped conductor is accomplished with a strip that extends over the entire width from the L-shaped conductor to the ground plane as is shown in Fig. 9.25. The strip thickness is the same as that of the LShape conductor. In this section, we utilize the thin-GSI model in Section 9.2.3. The first problem is the 1 \u03bcm thick and 10 \u03bcm wide HShoe1. We compare thin-GSI with the 3D-VFI solver results. The results are given in Figs. 9.26 and 9.27 for the inductance and resistance. Further, for the thin L-shaped conductor problem, both the ground plane and the L-shaped conductor are chosen to be 5 \u03bcm thick. We again compare the 3D-VFI solution with the thin-GSI solution"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001092_s00604-005-0449-x-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001092_s00604-005-0449-x-Figure7-1.png",
        "caption": "Fig. 7. A schematic diagram showing the edge-plane and basalplane sites on a graphite surface and the corresponding regions along a MWCNT",
        "texts": [
            " Usually researchers obtain good voltammetry and detection limits for CNT modified electrodes, usually with glassy carbon as a substrate, and compare that with the response of the bare electrode and state that the carbon nanotubes are \u2018\u2018electrocatalytic\u2019\u2019 for the molecule studied. This is questionable practice and the actual electrocatalytic sites are usually unexplored. We have explicitly addressed the question of the electrocatalytic sites of carbon nanotubes [159, 192]. Using the examples of the electrochemical reduction and oxidation of ferricyanide and epinephrine, we showed that the electrocatalysis of carbon nanotubes is solely due to edge plane sites and defects occurring on the nanotubes (see Fig. 7). Shown in Fig. 8 is the electrochemical response of an edge plane and basal pyrolytic graphite electrode with a carbon nanotube modified electrode. Note that the oxidation and reduction waves observed at the edge plane are at exactly the same potential as that for the CNT modified electrode. This suggests that all the electrochemistry and electrocatalysis occurs at edge plane sites [159, 192]. This insight has helped Gooding and co-workers establish when their SWCNT modified electrodes are either randomly dispersed or successfully aligned vertically"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure6.9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure6.9-1.png",
        "caption": "FIGURE 6.9. The input and output links of the main four-bar linkage of a windshield wiper at three different positions.",
        "texts": [
            " Wipers are used on windshields, and headlights. Figure 6.8 illustrates a sample of double-arm parallel windshield wiper mechanism. A four-bar linkage makes the main mechanism match the angular positions of the left and right wipers. A dyad or a two-link connects the driving motor to the main four-bar linkage and converts the rotational output of the motor into the back-and-forth motion of the wipers. The input and output links of the main four-bar linkage at three different positions are shown in Figure 6.9. We show the beginning and the end angles for the input link by \u03b821 and \u03b823, and for the output link by \u03b841 and \u03b843 respectively. To design the mechanism we must match the angular positions of the left and right blades at the beginning and at the end positions. Let\u2019s add another match point approximately in the middle of the total sweep angles and design a four-bar linkage to match the angles indicated in Table 6.2. 6. Applied Mechanisms 323 324 6. Applied Mechanisms Substituting the input and output angles in Freudenstein\u2019s equation (6"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure5.4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure5.4-1.png",
        "caption": "FIGURE 5.4 Single loop approximated by four bars.",
        "texts": [
            " But this seems to create a redistribution of the currents. Mainly, the local current at the corner of the conductors is approximated. However, as it is apparent from Chapter 6, the general current flow can be modeled, but the model will be more complex. Probably the most important aspect of the partial inductance concept is the fact that it obeys all the laws of circuit analysis [3]. This is shown to be the case for our simple loop FLUX LOOPS FOR PARTIAL INDUCTANCE 95 formed by four cells or bars that we show in Fig. 5.4. It is clear that the equivalent circuit for this loop is as shown in Fig. 5.5. The approach works for arbitrary connected bars that eventually form closed circuits. Inductance estimations can be performed for simple geometries using approximate models. The loop inductance in (5.16) is a good example. This type of model can often be used to understand fundamental issues. Fortunately, all the modified nodal analysis (MNA) circuit solver techniques in Chapter 2 can be applied for a problem with arbitrary conductor arrangements",
            " However, correctly solving the aspect ratio problem is not trivial. The proper solution of this problem involves two aspects. One aspect that relates to the PEEC model is considered in this section, whereas the second issue that relates to the evaluation of the partial inductances is considered in the following section. Not surprisingly, it is difficult to evaluate partial inductance with a high accuracy for extreme geometries. As an example geometry, it is sufficient to consider a typical loop shown in Fig. 5.4. Here, we consider a physically long version shown in Fig. 5.12. The circuit solution for the loop inductance is given in (5.16). We regroup the terms of the equation in pairs in the form L = (Lp11 \u2212 Lp13) + (Lp22 \u2212 Lp24) + (Lp33 \u2212 Lp31) + (Lp44 \u2212 Lp42). (5.23) It is apparent from Fig. 5.12 that we assume that the length \ud835\udcc1 of the loop is large compared to the spacing d. The length consists of the partial self inductances Lp11 and Lp33, while the orthogonal end branches have the partial self-inductances Lp22 and Lp44"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure6.4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure6.4-1.png",
        "caption": "FIGURE 6.4. Two possible configuration of a four-bar linkage having the same input angle \u03b82.",
        "texts": [
            "2 J3 = a2 \u2212 b2 + c2 + d2 2ac = 2.45 J4 = d b = 1.5 J5 = c2 \u2212 d2 \u2212 a2 \u2212 b2 2ab = \u22121.9375. (6.46) The coefficients of the quadratic equations are then calculated. A = \u22120.6914213562 B = \u22121.414213562 C = 3.894365082 D = \u22123.169733048 E = \u22121.414213562 F = 1.416053390 (6.47) 6. Applied Mechanisms 315 Using the minus sign, the output and coupler angles at \u03b82 = \u03c0/4 rad = 45 deg are \u03b84 \u2248 2 rad \u2248 114.73 deg \u03b83 \u2248 0.897 rad \u2248 51.42 deg (6.48) and using the plus sign, they are \u03b84 \u2248 \u22122.6 rad \u2248 \u2212149 deg \u03b83 \u2248 \u22121.495 rad \u2248 \u221285.7 deg . (6.49) Figure 6.4 depicts the two possible configurations of the linkage for \u03b82 = 45deg. The configuration in Figure 6.4(a) is called convex, non-crossed, or elbow-up, and the configuration in Figure 6.4(b) is called concave, crossed, or elbow-down. Example 219 Velocity analysis of a four-bar linkage. The velocity analysis of a four-bar linkage is possible by taking a time derivative of Equations (6.20) and (6.21), d dt (a sin \u03b82 + b sin \u03b83 \u2212 c sin \u03b84) = a\u03c92 cos \u03b82 + b \u03c93 cos \u03b83 \u2212 c \u03c94 cos \u03b84 = 0 (6.50) d dt (\u2212d+ a cos \u03b82 + b cos \u03b83 \u2212 c cos \u03b84) = \u2212a\u03c92 sin \u03b82 \u2212 b \u03c93 sin \u03b83 + c \u03c94 sin \u03b84 = 0 (6.51) where \u03c92 = \u03b8\u03072 \u03c93 = \u03b8\u03073 \u03c94 = \u03b8\u03074. (6.52) 316 6. Applied Mechanisms Assuming \u03b82 and \u03c92 are given values, and \u03b83, \u03b84 are known from Equations (6"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001534_j.msea.2016.05.046-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001534_j.msea.2016.05.046-Figure1-1.png",
        "caption": "Fig. 1. Scanning strategies cross snake (CS) (a) and cross snake ten (CS10) (b). In snake mode the beam moves back and forth in hatching direction (HD\u00bceither X or Y), perpendicular to the building direction (BD\u00bcZ). In CS10 the hatching direction is changed after every tenth layer. For microstructural investigations specimens were cut parallel to the building (BD) and to the transverse hatching direction (TD) of the last layer, as indicated by the plane shown in grey in (a).",
        "texts": [
            " Particle size analysis performed on a Malvern Mastersizer 3000 shows a particle size distribution mostly between 45 and 105 mm. As starter plate, a 10 mm thick polycrystalline Inconel 718 disk was used with a composition within the specification according to AMS 5662. Experiments consisting of building 9 cuboid shaped samples ( \u00d7 \u00d7 )15 mm 15 mm 10 mm were performed with five deflection speeds v from \u22122.2 ms 1 to \u22128.8 ms 1 and a beam power P of 594 W. For each deflection speed the distance between parallel beam lines, i.e. the line offset Loff (see Fig. 1), was adapted in such a way that the total layer hatching time (time of irradiation by the beam) is kept constant for all experiments resulting in a constant area energy EA: = ( ) E P v L 1 A off The procedure of lowering Loff and increasing v by the same factor will be referred to as multi passing in the following. For the highest deflection speeds three additional powers (P\u00bc627 W and 660 W and 693 W) were investigated. Experiments were performed with a focused beam (spot diameter 350\u2013400 mm) and with a cross snake scanning strategy, where the electron beam meanders in a back and forth pattern during hatching. In addition, the hatching direction is rotated by 90\u00b0 after each layer, see Fig. 1a). In order to investigate the influence of multi passing on the resulting microstructures, two additional scanning strategies have been investigated. The scanning strategy cross snake ten (CS10) was used to study the effect of rotating the scan pattern on grain structure evolution with otherwise identical beam parameters, see Fig. 1b). The scanning strategy was used at a deflection speed of 8.8 m/s. CS10 is similar to cross snake (CS), however, the scan pattern is rotated by 90\u00b0 after ten layers, instead of each layer. In order to analyse the transition between microstructures generated by standard scanning strategies and multi passing, a further experiment (CS40) was conducted, where scanning strategies are changed every 40th layer (equivalent to 2 mm) of the specimen. The beam parameters for standard scanning are: = = = \u03bc\u2212 \u2212E v L1",
            " The preheating parameters were held constant for all building processes and lead to a building temperature of 900 \u00b0C. All specimens were fully dense. The grain structure of all samples was analysed with a scanning electron microscope (SEM) in back-scatter electron (BSE) mode. The grain width and length were measured using the intercept method parallel and transverse to the building direction in a total area of 1.02 mm2. In the plane of observation grains are elongated parallel to the building direction (Fig. 1). The measurements were always conducted at the same building height and position, i.e. 1 mm below the top surface and near the center of the longitudinally cut samples. The grain aspect ratio R is calculated from the average grain length L (in m) and the average grain width W (in m) as follows: = ( )R L W 2 A grain aspect ratio of one indicates an equiaxed grain morphology. With increasing R the columnar character of the microstructure increases. In addition, grain orientation maps were obtained from an equiaxed and two columnar grain structures",
            " Effects such as absorption of the electron beam, phase transitions and wetting and dewetting of single powder particles are captured by the model. In this paper we use a the model in order to simulate hatching processes where several layers are built consisting of many melt lines. The numerical setup is depicted in Fig. 2. An area of 2.5 mm 15 mm is scanned by the electron beam. Due to methodic restrictions, the hatching direction of the electron beam is changed by 180\u00b0 in succeeding layers rather than by 90\u00b0 as in the experiments (see Fig. 1). The parameters for the numerical experiments are equivalent to the experiments. Additionally, a second campaign of simulations has been conducted. Here, the beam irradiates a solid plate only once without powder being applied. These simulations show the idealized solidification conditions without the stochastic influence of powder particles. The EBSD orientation map in Fig. 3a shows a columnar microstructure which is obtained with a standard cross snake scanning strategy at high line offsets and medium beam velocities"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure5.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure5.3-1.png",
        "caption": "FIGURE 5.3. Position vectors of point P before and after rotation of the local frame about the z-axis of the local frame.",
        "texts": [
            "37) and the pitch angle is \u03b2 = \u2212 sin\u22121 (r31) (5.38) and the yaw angle is \u03b3 = tan\u22121 \u00b5 r21 r11 \u00b6 (5.39) provided that cos\u03b2 6= 0. 5.3 Rotation About Local Cartesian Axes Consider a rigid body B with a space-fixed point at point O. The local body coordinate frame B(Oxyz) is coincident with a global coordinate frame G(OXY Z), where the origin of both frames are on the fixed point O. If the body undergoes a rotation \u03d5 about the z-axis of its local coordinate frame, as can be seen in the top view shown in Figure 5.3, then coordinates 226 5. Applied Kinematics of any point of the rigid body in the local and global coordinate frames are related by the equation Br = Rz,\u03d5 Gr. (5.40) The vectors Gr and Br are the position vectors of the point in the global and local frames respectively Gr = \u00a3 X Y Z \u00a4T (5.41) Br = \u00a3 x y z \u00a4T (5.42) and Rz,\u03d5 is the z-rotation matrix Rz,\u03d5 = \u23a1\u23a3 cos\u03d5 sin\u03d5 0 \u2212 sin\u03d5 cos\u03d5 0 0 0 1 \u23a4\u23a6 . (5.43) Similarly, rotation \u03b8 about the y-axis and rotation \u03c8 about the x-axis are described by the y-rotation matrix Ry,\u03b8 and the x-rotation matrix Rx,\u03c8 respectively"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure1.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure1.2-1.png",
        "caption": "Fig. 1.2. AFMPM 8-pole d.c. commutator motor with printed rotor winding: (a) stator with PMs, (b) cross section, (c) rotor (armature) windings and brushes, (d) construction of 2p = 8 winding with 145 bars. 1 \u2014 rotor with double-sided printed winding, 2 \u2014 PMs, 3 \u2014 brushes.",
        "texts": [
            " and synchronous machines; \u2022 induction machines Similar to its RFPM counterpart, the AFPM d.c. commutator machine uses PMs to replace the electromagnetic field excitation system. The rotor (armature) can be designed as a wound rotor or printed winding rotor. In the wound rotor, the armature winding is made of copper wires and moulded with resin. The commutator is similar to that of the conventional type, i.e. it can be either a cylindrical or radial commutator. The disc-type printed armature winding motor is shown in Fig. 1.2. The rotor (armature) does not have a ferromagnetic core and its winding is similar to the wave winding of conventional d.c. commutator machines. The coils are stamped from pieces of sheet copper and then welded, forming a wave winding. 1.4 Types of Axial Flux PM Machines 5 When this motor was invented by J. Henry Baudot [18], the armature was made using a similar method to that by which printed circuit boards are fabricated. Hence, this is called the printed winding motor. The magnetic flux of a d"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure9.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure9.2-1.png",
        "caption": "Fig. 9.2. High speed AFPM synchronous generator integrated with the rotor of a microturbine. 1 \u2014 SmCo PM, 2 \u2014 backing steel ring, 3 \u2014 rotor of microturbine, 4 \u2014 nonmagnetic retaining ring, 5 \u2014 stator winding, 6 \u2014 stator core.",
        "texts": [
            " The low energy density of portable bagpack batteries imposes a major constraint and challenge to future infantry operations. With recent advancement in PM brushless machine technologies, a lightweight miniature generator set can minimize the 9.1 Power Generation 283 mass of heavy batteries and charge them during field operations. A miniature generator can deliver electric power for a long period of time limited only by the fuel availability. Easy and rapid refueling can be done in the field by using, e.g. kerosene. Fig. 9.2 shows a tiny microturbine integrated with a miniature AFPM synchronous generator. At speed 150 000 to 250 000 rpm and outer diameter of PM ring Dout \u2248 50 mm, the generator can deliver about 1 kW electric power. 284 9 Applications A low speed AFPM generator is usually driven by a wind turbine. With wind power rapidly becoming one of the most desirable alternative energy sources world-wide, AFPM generators offer the ultimate low cost solution as compared with e.g. solar panels. Table 9.1 shows specifications of five-phase AFPM synchronous generators manufactured by Kestrel Wind Turbines (Pty) Ltd , Gauteng, South Africa",
            "2) is \u03c3 = 7800\u00d7 3141.62 3 (0.0822 + 0.082\u00d7 0.142 + 0.1422) = 988.6 MPa The energy density according to eqn (9.6) is ek = 480 740 7.25 = 66.3 kJ/kg The shape factor of the ring [3] is ksh = ek \u03c1 \u03c3 = 66 300 7800 988.6\u00d7 106 = 0.523 9.12 Axial Flux Machines with Superconducting Field Excitation System 323 Numerical Example 9.2 Find the main dimensions of a three-phase, 1200 W, 4-pole, 200 000 rpm AFPM synchronous generator for a mobile battery charger. The total nonmagnetic air gap including magnet retaining ring (Fig. 9.2) should be g = 2 mm, number of slots s1 = 12, air gap magnetic flux density Bg = 0.4 T, line current density Am = 16 000 A/m, PM inner\u2013to\u2013outer diameter ratio kd = 0.5, PM temperature 350oC. Vacomax 225 SmCo PMs (Vaccumschmelze, Hanau, Germany) with Br20 = 1.04 T and Hc = 760 kA/m at 20oC are recommended. The temperature coefficient for Br is \u03b1B = \u22120.035 %/oC, leakage flux coefficient \u03c3lM = 1.3, EMF\u2013to\u2013voltage ratio \u03b5 = 1.3 (Ef > V1), pole shoe width\u2013to\u2013pole pitch ratio \u03b1i = 0.72, efficiency \u03b7 = 0"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000233_978-1-4614-0460-6-Figure6.31-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000233_978-1-4614-0460-6-Figure6.31-1.png",
        "caption": "Fig. 6.31 Linearized pendulum and its free body diagram",
        "texts": [
            "7 that the dampers are located at the same physical locations as the springs. Therefore, the damping matrix will have the same form as the stiffness matrix. fk21 fk12 k2x1 fk22 k2x2 6.6 Shortcut Method for Determining Mass, Stiffness, and Damping Matrices 227 Let\u2019s next use the shortcut method for the linearized pendulum. We will begin with the traditional free body diagram analysis and then follow it with the new shortcut method. The pendulum, relevant geometry, and free body diagram are shown in Fig. 6.31. As seen in Sect. 2.5.2, we can sum the moments, M, about O using the force components that are perpendicular to the massless rod. Because we are summing about O, the reaction components at the pivot may be neglected. Also, the free body diagram includes d\u2019Alembert\u2019s inertial moment, J\u20acy \u00bc ml2\u20acy, with the mass moment of inertia, J, so that P MO \u00bc 0. The moment sum about O from Fig. 6.31 is: X MO \u00bc J\u20acy\u00fe kd cos y\u00f0 \u00de a\u00fe mg sin y\u00f0 \u00de l \u00bc 0: (6.14) Substituting J \u00bc ml2 and d \u00bc a sin y\u00f0 \u00de gives: ml2\u20acy\u00fe ka2 cos y\u00f0 \u00de sin y\u00f0 \u00de \u00fe mgl sin y\u00f0 \u00de \u00bc 0: (6.15) For small angles, we can approximate using sin y\u00f0 \u00de y and cos y\u00f0 \u00de 1. Substi- tution yields: ml2\u20acy\u00fe ka2y\u00fe mgly \u00bc ml2\u20acy\u00fe ka2 \u00fe mgl y \u00bc 0: (6.16) 228 6 Model Development by Modal Analysis We can convert to small horizontal displacements, x, of the pendulum mass by again applying the small angle approximation. See Fig. 6.32, where x \u00bc l sin y\u00f0 \u00de ly and, subsequently, \u20acx l\u20acy for a constant rod length"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure3.56-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure3.56-1.png",
        "caption": "Figure 3.56 A structure consisting of a system of mutually perpendicular beams in the horizontal xy plane that is loaded perpendicularly to its plane by a number of forces and couples. The unknown forces Av, Bv and Cv have to be derived from the equilibrium.",
        "texts": [
            " This means that the following requirements have to be met by the forces and moments exerted on a rigid body at rest: \u2211 Fx = 0,\u2211 Fy = 0,\u2211 Fz = 0,\u2211 Tx |A = 0,\u2211 Ty |A = 0,\u2211 Tz|A = 0. The first three equations state that there is force equilibrium in the x, y and z directions respectively, and that the body is therefore not subject to translation acceleration. The latter three equations define that there is moment equilibrium at A about lines parallel to respectively the x, y and z axis, and that the body is not subject to rotational acceleration. The following examples address the equilibrium of a body in space. 94 Example 1 The structure in Figure 3.56 consists of a number of mutually perpendicular beams in the horizontal xy plane that are loaded at the locations shown by three vertical forces of respectively 40, 60 and 100 kN and by two couples of 30 and 50 kNm. The structure is kept in equilibrium by the three vertical forces Av, Bv and Cv. Question: Determine these three unknown forces. Solution: Since all the forces are parallel to the z axis, \u2211 Fx = 0 and \u2211 Fy = 0. The moment vectors of both couples are in the xy plane, so that in addition \u2211 Tz = 0"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003908_978-1-84882-730-1-Figure13.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003908_978-1-84882-730-1-Figure13.2-1.png",
        "caption": "Figure 13.2 Feasible initial conditions",
        "texts": [
            " Roughly speaking, with the above initial conditions, due to the underactuated configuration in the sway and heave, the sway and heave velocities are not able to push the vehicle to the point ae D 0 and de D 0. 13.5 Discussion of the Initial Condition We now discuss how to obtain the initial conditions such that (13.77), (13.93), and (13.95) hold. A close look at these conditions shows that they are always satisfied by selecting the initial value, s.t0/, if the vessel heads toward the conical space containing the initial path to be followed, see Figure 13.2. If the vessel does not, the surge control should be turned off and the yaw and pitch controls should make the vessel turn until (13.77), (13.93), and (13.95) hold before applying the proposed path-following controller. The angle \u01310 (see Figure 13.2) should be increased if the initial velocities v1.t0/ and v2.t0/ are large. Otherwise the vessel might cross the edge-line of the subspace in question, which might result in i D \u02d90:5 and/or D \u02d90:5 . 13.6 Parking and Point-to-point Navigation 13.6.1 Parking Parking Objective. Design the controls 1 and 2 to park the underactuated underwater vehicle (13.1) from the initial position and orientation .x.t0/, y.t0/, z.t0/, .t0/, .t0/, .t0// to the desired parking position and orientation of .xp , yp , zp , p , p , p/ under the following conditions: 1"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000844_bfb0015081-Figure2.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000844_bfb0015081-Figure2.2-1.png",
        "caption": "Fig. 2.2. Cart-pole system",
        "texts": [],
        "surrounding_texts": [
            "Consider the Lagrangian formulation of the dynamics of an n-degree-offreedom mechanical system\nD(q)~ + C(q, (t)O + g(q) = B(q)7- (2.1)\nwhere q E R n is the vector of generalized coordinates, 7- E R m is the input generalized force (m < n), and B(q) E R nx'~ has full rank for all q.\nFor a suitable part i t ion of the vector q of generalized coordinates as qT = (qT, qT), where ql E R ~-m and q2 E R m we may write the system (2.1) as\ndl lql + d12q2 q- hl(ql,Ol,q2,(t2) \"b \u00a2l(ql ,q2) = 0 (2.2)\ndl2ql +d22q2 -'bh2(ql,ql,q2,(~2)+\u00a22(qx,q2) -~ b(qx,q2)7- (2.3)\nwhere hi include Coriolis and centrifugal terms, and \u00a2i contains the terms derived from the potential energy, such as gravitational and elastic generalized forces. The m \u00d7 m matr ix b(ql, q2) is assumed to be invertible.\nExample 2.1. T w o - l i n k r o b o t . Consider the two-link robot shown in Fig. (2.1):\nwhere\ndllql q- di2iiz + hi + \u00a21 = 7-1 (2.4)\nd12ql -~- d22~2 + h2 + \u00a22 = 7-2 (2.5)\ndn = mlg~l + m2(g~ + g2~2 + 2e1~c2 COS(q2)) -k- I1 + I2\nd22 = m292~2 + I 2\nd12 = m2(g~2 + gle~2cos(q2)) +/2 hi = -m2glgc2sin(q2)q~ - 2m~e162sin(q2)q2q1\nh2 = m2glg~2sin(q2)~ 2\n\u00a21 m (mlgcl -k-m2gl)gcos(ql) -}-m2~.c29Cos(ql +q2)\n\u00a22 = m2g~2gcos(qi +q2)\nIf ~-i ---- 0 this system represents the Acrobot [13, 5], while if 72 = 0 the system represents the Pendubot [39]. In addition, with \u00a21 = 0 --- \u00a22 and 7-2 = 0 one has the underactuated manipulator system considered in by several authors, such as [28, 9, 2].\nExample 2.2. C a r t - p o l e s y s t e m . The cart-pole system is one of the classic examples and yet it still holds some interesting challenges from the standpoint of global nonlinear control. Referring to Fig. (2.2) the dynamics are given by:\n(my + me)2 + mpg cosO0 - moo 2 sin0 = F\nrnpgcos02 + mpO- mpggsinO = 0",
            "For simplicity we normalize all constants to unity. To put the system in standard form, we set ql = 0, q2 = x, T = F, and write the equations as\nq l + c o s q l ~ 2 - s i n q l = 0 (2.6)\ncosq l~ l+2~2-01~s inq l = ~- (2.7)\nThe nature of the fixed points of (2.2), (2.3) is closely tied to the controllability of the system. Let ~- = e = constant. Then, since the terms hi are quadratic in the velocities ~, the equilibrium solutions satisfy\n\u00a2l (q l ,q2) = 0 (2.8)\n\u00a22(ql,q2) = b(ql,q2)~ (2.9)\nand may either be isolated fixed points for each fixed e, as in the case of the Acrobot, and Pendubot or they may be higher dimension as happens (for\n= 0) in systems without potential terms. For example, in the absence of gravity, the Pendubot dynamics satisfies\n\u00a2~(ql, q2) = 0 for i = 1, 2 (2.10)\nfor all (ql, q2) E Q, where Q denotes the two dimensional configuration space. In the first case, systems with potential terms are linearly controllable around (almost all) fixed points, i.e. the Taylor series linearization is a controllable linear system. Systems without potential terms are generally not linearly controllable. Their local controllability properties are therefore more subtle to determine.\nWe may interpret Eq. (2.2) as a (dynamic) constraint on the accelerations of the generalized coordinates. It is then interesting to ask whether these constraints are holonomic, i.e. integrable. For many of the most interesting cases, including underactuated robot manipulators [28], the Acrobot [37] and Pendubot [39], the PVTOL system [43], the TORA system, and underwater robots, these constraints turn out to be completely nonintegrable as shown in [30]. An important consequence is that the system (2.2), (2.3) is (strongly) accessible, since nonintegrabiIity of the second order constraint equations means that the dimension of the reachable set is not reduced.\nAccessibility does not imply stabilizability of an equilibrium configuration using time-invariant continuous state feedback (either static or dynamic). In fact, for systems without potential terms it is known [30] that such stabilizability is not possible. The proof of this follows from an application of Broekett 's Theorem [6]. The situation here is, therefore, quite similar to the case of control of nonholonomic mobile robots. Of course, systems with potential terms are exponentially stabilizable by linear time-invariant feedback."
        ]
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure9.60-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure9.60-1.png",
        "caption": "Figure 9.60 (a) The forces in members 5 and 6 follow from the equilibrium of joint C. (b) The closed force polygon for the equilibrium of joint C. FC;2 and FC;3 are known forces. (c) Joint C with all the forces acting on it. From this figure we can see that N5 is compressive and N6 is tensile.",
        "texts": [
            " The two unknowns FB;3 and FB;4 can be determined from the closed force polygon for the equilibrium of joint B (see Figure 9.59b): FB;3 = 4F, FB;4 = 4F \u221a 5. In Figure 9.59c, the forces from the force polygon are shown as they act on joint B in reality. Here we see that FB;3 and FB;4 are both compressive forces. Converted into the normal forces in the members 3 and 4, with the correct sign for tension and compression, we therefore get N3 = \u2212FB;3 = \u22124F, N4 = \u2212FB;4 = \u22124F \u221a 5. The following joint with only two unknowns is C. The forces that the members 2 and 3 exert on the joint are known (see Figure 9.60a): FC;2 = 4F \u221a 5, FC;3 = 4F. The unknown forces FC;5 and FC;6 follow from the force polygon in Figure 9.60b: 9 Trusses 357 FC;5 = F \u221a 5, FC;6 = 3F \u221a 5. In Figure 9.60c, all the forces are shown as they act on joint C in reality. Member 5 presses against the joint and is a compression member, member 6 pulls on the joint and is a tension member: N5 = \u2212F \u221a 5, N6 = +3F \u221a 5. In Figures 9.61 to 9.64, the other member forces are calculated using the same method. Table 9.4 provides a summary of all the member forces. 358 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM 9 Trusses 359 360 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM In Figures 9.63 and 9.64 the support reactions in G and H have also been calculated: Gv = 10F, Hh = 0, Hv = 6F"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure3.28-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure3.28-1.png",
        "caption": "Fig. 3.28. Manipulation assembly task",
        "texts": [
            " The chain becomes closed when contact appears between the surface and the manipulator gripper. Fig. 3.26. Writing task The second example of constrained gripper motion is a grinding task (Fig. 3.27). The necessary theoretical considerations for this type of constraint are demonstrated in Sect. 3.4.3 where the numerical example of the grinding process with a six degrees-of-freedom manipulator is also solved. 92 3 Dynamics and Dynamic Analysis of Manipulation Robots The next example is very typical in industrial practice. This is an assembly task. Figure 3.28 shows the manipulation peg-in-hole assembly task. The as sembly problem is often the subject of the work of many researchers. For instance, in [3] the dynamic analysis and control synthesis using force feed back in the peg-in-hole assembly task are presented. In [4] the complete dy namic analysis for non-stationary constraint is derived and one example of the assembly simulation is given. 3.4.2 Mathematical Model of Manipulator with Constraints of Gripper Motion In order to present the mathematical model of the robot closed configuration we consider first a manipulator as an open kinematic chain"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure4.21-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure4.21-1.png",
        "caption": "Fig. 4.21: An example of a commercial industrial two-finger gripper: a) the mechanical design; b) its kinematic chain.",
        "texts": [
            " In addition, the analysis of the mechanism links can also be not quite evident in terms of numerical determination. In the case of design modeling, the identification procedure can be even more doubtful, when one will think to build a mechanical design corresponding to the kinematic sketch in Fig. 4.20 b), since it can give more than one mechanical solution, even very different form that in Fig. 4.20 a). Indeed, this is an illustrative example that shows that a gripper design can strongly depend on the specific expertise of a designer in mechanisms and grippers. In Fig. 4.21 the mechanical design in a) shows the same geometry as the kinematic diagram in b) so that the modeling is straightforward both for analysis and design purposes. The use of cam connections can be very useful to have small-sized designs with an easy understanding of the gripper operation. Fundamentals of the Mechanics of Robotic Manipulation 265 In Fig. 4.22 the mechanical design of the gripper shows how to obtain a parallel motion of the fingers from a gripping mechanism with finger swinging motion by adding a subchain to the original kinematic chain, when one refers to the case of Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure10.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure10.3-1.png",
        "caption": "Figure 10.3 An arbitrarily shaped cross-section at C. The force transfer is not concentrated in the member axis but is distributed over the section as the sum of many very small interactions between adjacent particles of matter. The stress resultant R is the resultant force due to the stresses in the section.",
        "texts": [
            " The lines of action of the resultant forces at A and B intersect in the line of action of the force of 120 kN (graphical check of the moment equilibrium for a body subjected to three forces, see Section 3.3.2). Figure 10.2 shows the interaction forces that the member has to transfer at C. After introducing a section at C across the member, the interaction forces are found from the equilibrium of one of the isolated parts, to the right or left of C. 10 Section Forces 389 In reality, the member is not one-dimensional, but has cross-sectional dimensions. Figure 10.3 shows an arbitrary section across the member at C. The force transfer is not concentrated in the member axis, but varies over the section as the sum of a large number of very small interactions between adjacent particles of matter. Mathematically, we describe this phenomenon in the section by means of the concept stress (see Section 6.5).1 The distributions in magnitude and direction of the stresses in the section are as yet unknown. The equilibrium, however, shows that the stress resultant R, regardless of the shape of the section, must be 50 kN, and that the line of action of R must coincide with the line of action of the support reaction at A (see Figure 10.3). We usually do not give the section across a member an arbitrary shape, but rather choose one that is straight and normal to the member axis, as shown in Figure 10.4. This type of section is called a normal section or simply cross-section. Hereafter, when we refer to a section, we always mean a normal section. The intersection of the line of action of the stress resultant R and the crosssectional plane is known as the centre of force. The intersection of the member axis with the cross-sectional plane is the normal force centre, or normal centre, of the section"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure10.13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure10.13-1.png",
        "caption": "Figure 10.13 Positive stresses on the sides of a rectangular block. The kernel symbol \u03c3 for stress has two indices. The first index relates to the normal of the plane on which the stress is acting; the second index relates to the direction of the stress. The stress is a normal stress when both indices are equal, and a shear stress when both indices are different.",
        "texts": [
            " The sign convention can be summarised as follows: \u2022 A stress is positive when it acts on a positive plane in the positive direction or on a negative plane in the negative direction. \u2022 A stress is negative when it acts on a positive plane in the negative direction or on a negative plane in the positive direction. 10 Section Forces 395 For more general cases, the stress definition can be summarised in short as follows: \u03c3ij = lim A\u21920 Fj A \u00b7 ni (i, j = x, y, z) in which both i and j can be replaced by x, y or z.1 Figure 10.13 shows the positive stresses acting on the sides of an (infinitesimally) small rectangular block. The block is bounded by six planes, of which three are positive and three are negative. \u03c3ij \u2022 on a small area with the unit normal vector parallel to the i axis (1st index), \u2022 due to a force component parallel to the j axis (2nd index). The stress \u03c3ij is a normal stress when the indices are the same (i = j ) and a shear stress when the indices are different (i = j ). In a member cross-section, there are only normal stresses \u03c3xx and shear stresses \u03c3xy and \u03c3xz (see Figure 10"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003459_0278364905058363-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003459_0278364905058363-Figure3-1.png",
        "caption": "Fig. 3. Foot rotation indicator. The FRI is the point where the ground reaction force would have to act to keep the foot from accelerating. When the foot is stationary, shown in the left figure, the FRI coincides with the ZMP. As the foot starts turning (figure on right), the FRI leaves the support base. Here the distance from the FRI to the ZMP is proportional to the magnitude of the foot moment about the FRI point. Although only two dimensions are depicted in the figure, the FRI definition is applicable to the problem of three-dimensional biped control problems, including foot rotational information for sagittal and coronal planes.",
        "texts": [
            " In legged systems, a loss of rotational equilibrium of the stance foot during single support implies the existence of an unbalanced moment acting on the foot segment, causing foot rotations and movement of the ZMP towards the edge of the footprint boundary. Once the stance foot has rolled to an extreme posture, pushing the ZMP to the very edge of the foot envelope, additional rotational dynamics of the foot, such as different rates of rotational acceleration, are no longer discernible using the ZMP. The FRI point, shown in Figure 3, was introduced by Goswami (1999) in order to specifically address this limitation. Dominant foot rotation has been noted to reflect a loss of balance and an eventual fall in monopods (Lee and Raibert 1991) and bipeds (Arakawa and Fukuda 1997), two classes of legged robots most prone to instabilities. The FRI point extends the concept of the ZMP and quantifies the severity of foot rotational acceleration. A motivation behind its formulation was to achieve a measure of foot rotational acceleration during single support that could be employed by legged control systems as one possible indicator of overall postural instability"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002470_rsos.171628-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002470_rsos.171628-Figure8-1.png",
        "caption": "Figure 8. Surface friction. (a) The forces produced by friction effects between an element of the rod and the substrate are naturally decomposed into a lateral component in the direction uw\u2016 = t \u00d7 uw\u22a5 and a longitudinal one in the direction uw\u00d7 = uw\u22a5 \u00d7 uw\u2016 . We note that in generaluw\u00d7 = t. The notation x\u22a5, x\u2016, x\u00d7 denotes the projection of the vector x in the directionsuw\u22a5,u w \u2016 ,u w \u00d7. (b,c) Kinematic and dynamic quantities at play at any cross section in the case of (b) rolling and slipping and (c) pure rolling motion. Red arrows correspond to forces and torques, green arrows correspond to velocities, and black arrows correspond to geometric quantities.",
        "texts": [
            " This model relates the normal force pushing a body onto a substrate to the friction force through the kinetic \u03bck and static \u03bcs friction coefficients, depending on whether the contact surfaces are in relative motion or not. Despite the simplicity of the model, its formulation and implementation may not necessarily be straightforward, especially in the case of rolling motions. Given the cylindrical geometry of our filaments, F^ Fw ^ v r d uw ^ ' the effect of surface friction can be decomposed into a longitudinal component associated with purely translational displacements, and a lateral component associated with both translational and rotational motions (figure 8). We use the notation x\u22a5, x\u2016, x\u00d7 to denote the projection of the vector x in the directions uw \u22a5 , uw \u2016 , uw\u00d7 , as illustrated in figure 8. The longitudinal friction force Flong is opposite to either the resultant of all forces F\u00d7 acting on an element (static case) or to the translational velocity v\u00d7 (kinetic case) along the direction uw\u00d7 (figure 8). The Amonton\u2013Coulomb model then reads Flong = \u23a7\u23aa\u23aa\u23a8 \u23aa\u23aa\u23a9 \u2212 max(|F\u00d7|,\u03bcs|F\u22a5|) \u00b7 F\u00d7 |F\u00d7| if |v\u00d7| \u2264 v\u03b5, \u2212\u03bck|F\u22a5| \u00b7 v\u00d7 |v\u00d7| if |v\u00d7|> v\u03b5, 17 rsos.royalsocietypublishing.org R.Soc.opensci.5:171628 ................................................ where v\u03b5 \u2192 0 is the absolute velocity threshold value employed to distinguish between static (|v\u00d7| \u2264 v\u03b5) and kinetic (|v\u00d7|> v\u03b5) cases. We define v\u03b5 in a limit form to accommodate the fact that inequalities are numerically evaluated up to a small threshold value. The static friction force is always equal and opposite to F\u00d7 up to a maximum value proportional to the normal force |F\u22a5| through the coefficient \u03bcs. The kinetic friction force is instead opposite to the translational velocity v\u00d7, but does not depend on its actual magnitude and is proportional to |F\u22a5| via \u03bck. In general \u03bcs >\u03bck, so that it is harder to set a body into motion from rest than to drag it. The lateral displacement of a filament in the direction uw \u2016 = uw\u00d7 \u00d7 uw \u22a5 is associated with both translational (v\u2016) and rotational (\u03c9\u00d7 =\u03c9\u00d7uw\u00d7) motions, as illustrated in figure 8b,c. In this case, the distinction between static and kinetic friction does not depend on v\u2016, but on the relative velocity (also referred to as slip velocity) between the rod and the substrate vslip = v\u2016 + vcont, vcont = ruw \u22a5 \u00d7 \u03c9\u00d7, (4.9) where vcont is the local velocity of the filament at the contact point with the substrate, due to the axial component of the angular velocity \u03c9\u00d7. In the static or no-slip scenario (vslip = 0), the linear momentum balance in the direction uw \u2016 , and the angular momentum balance about the axis uw\u00d7 express a kinematic constraint between the linear acceleration auw \u2016 and angular acceleration \u03c9\u00d7 = (uw \u22a5 \u00d7 auw \u2016 )/r, so that (F\u2016 + Froll)u w \u2016 = dm \u00b7 auw \u2016 (4"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure13.81-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure13.81-1.png",
        "caption": "Figure 13.81 (a) Bending moment diagram, (b) shear force diagram and (c) normal force diagram.",
        "texts": [],
        "surrounding_texts": [
            "The statically indeterminate structure in Figure 13.77 has a hinged joint at S. All other joints are rigid. Dimensions and loads are given in the figure. The shear forces directly next to joint C are given: V BC C = 2.5 \u221a 2 kN, V CS C = 10 kN. 13 Calculating M, V and N Diagrams 605 Figure 13.77 Statically indeterminate frame with two given shear forces directly next to joint C. Figure 13.76 (b) Shear force diagram and (c) bending moment diagram. 606 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM The directions follow from the deformation symbols given in the figure. Questions: a. Determine the degree of static indeterminacy of the structure. b. Draw the force polygon for the force equilibrium of joint C. c. Determine the support reactions at A and D. d. Determine the M , V and N diagrams for the entire structure, with the deformation symbols. Solution (units kN and m): a. The two parts ABCS and DS provide e = 2 \u00d7 3 = 6 equilibrium equations (see Figure 13.78). The number of unknown support reactions at A and D is r = 3 + 3 = 6. The number of unknown joining forces at S is v = 2. The degree of static indeterminacy is: n = r + v \u2212 e = 6 + 2 \u2212 6 = 2. The structure is therefore statically indeterminate to the second degree. b. In Figure 13.79a, joint C has been isolated and all forces acting on it are shown. The bending moments acting on the joint are not shown! Figure 13.79b shows the closed force polygon for the force equilibrium of the joint (scale: 1 square = 2.5 kN). The force polygon gives NBC = +7.5 \u221a 2 kN, NCS = +5 kN. c. With NCS = +5 kN the vertical equilibrium of CSD gives Dv = 5 kN (\u2191). 13 Calculating M, V and N Diagrams 607 With V CS = 10 kN the horizontal equilibrium of ABC gives Ah = 10 kN (\u2190). From the force equilibrium of the structure as a whole follows Dh = 10 kN (\u2190), Av = 15 kN (\u2191). Finally, the fixed-end moment reactions at A and D follow from the moment equilibrium about S of DS and ABCS respectively. In Figure 13.80, the support reactions are shown as they act in reality. d. In Figures 13.81a to 13.81c the M , V and N diagrams are shown. At B and C, the bending moment \u201cgoes round the corner\u201d. The slopes of the M diagram are in line with the magnitudes and the deformations symbols of the shear forces."
        ]
    },
    {
        "image_filename": "designv10_0_0002032_205651315x688686-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002032_205651315x688686-Figure4-1.png",
        "caption": "Fig. 4. Schematic of the gas atomisation process for production of AM powders (Courtesy of LPW Technology, UK)",
        "texts": [
            " The process of melting the metal ingots can be the same as described for water atomisation, however for powders produced 247 \u00a9 2015 Johnson Matthey for high end applications such as aerospace, the need to control interstitial elements has led to increased use of vacuum induction melting (VIM) furnaces. A VIM furnace is typically installed directly above the atomisation chamber such that the molten stream of liquid metal enters the atomisation chamber directly from the furnace rather than through a tundish, similar to the set-up shown in Figure 4. The stream of liquid metal is atomised by high pressure jets of gas. Due to the lower heat capacity of the gas (compared to water) the metal droplets have an increased solidification time which results in comparatively more spherical powder particles (i.e. droplet spheroidisation time is shorter than the solidification time). Whilst it is not possible to have complete control over the particle size of asatomised powder, the distribution can be influenced by varying the ratio of the gas to melt flow rate"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001031_j.jsv.2008.03.038-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001031_j.jsv.2008.03.038-Figure6-1.png",
        "caption": "Fig. 6. The cracked tooth model for case 4.",
        "texts": [],
        "surrounding_texts": [
            "The gear mesh stiffness model described in this study was based on the work by Yang and Lin [3] in 1987. They used the potential energy method to analytically model the effective mesh stiffness. The total potential energy stored in the meshing gear system was assumed to include three components: Hertzian energy, bending energy, and axial compressive energy. This model was refined by Tian [4] in 2004 in which shear energy was taken into account as well. Thus, for the single-tooth-pair meshing duration, the total effective mesh stiffness can be expressed as [4] kt \u00bc 1 1=kh \u00fe 1=kb1 \u00fe 1=ks1 \u00fe 1=ka1 \u00fe 1=kb2 \u00fe 1=ks2 \u00fe 1=ka2 , (1) where kh, kb, ks, and ka represent the Hertzian, bending, shear, and axial compressive mesh stiffness, respectively. For the double-tooth-pair meshing duration, the total effective mesh stiffness is the sum of the two pairs\u2019 stiffnesses, which is shown as [4] kt \u00bc X2 i\u00bc1 1 1=kh;i \u00fe 1=kb1;i \u00fe 1=ks1;i \u00fe 1=ka1;i \u00fe 1=kb2;i \u00fe 1=ks2;i \u00fe 1=ka2;i , (2) where i \u00bc 1 represents the first pair of meshing teeth and i \u00bc 2 represents the second. The derivations of these two equations are given in Tian [4]. Calculation of each of the components in these two equations when there are no cracks in any gear is given in Refs. [3,4,15], where Ref. [15] is a shorter version of Ref. [4]. The expressions of these components when cracks are introduced will be provided later in this paper. For typical gear parameters given in Table 1, we wrote simple Matlab programs and obtained numerical values of the total effective mesh stiffness as a function of the gear rotation angle. This total effective mesh stiffness within one shaft period of the gear is plotted in Fig. 1. Fig. 1 represents the total meshing stiffness of the pair of gears when the gear teeth are perfect (that is, have no cracks). On crack development in a gear, Refs. [16\u201319] consider that a crack is developing at the root of a single tooth of the pinion. A tooth root crack typically starts at the point of the largest stress in the material. In Ref. [20], a computational model which applies the principles of linear elastic fracture mechanics is used to simulate gear tooth root crack propagation. Based on the computational results, the crack propagation path shows a slight curve extending from the tooth root as shown in the left side of Fig. 2 [20]. Lewicki [21] also indicates that crack propagation paths are smooth, continuous, and in most cases, rather straight with only a ARTICLE IN PRESS S. Wu et al. / Journal of Sound and Vibration 317 (2008) 608\u2013624610 slight curvature. He has also studied the effects of rim and web thickness on crack propagation path and showed that different paths exist. In this paper, based on the results shown in Ref. [20], we further simplify the crack model. We will consider the crack path to be a straight line as shown in the right side of Fig. 2. The crack starts at the root of the pinion and then proceeds as shown in Fig. 2. Further referring to Figs. 3\u20136, the intersection angle, u, between the crack and the central line of the tooth is set at a constant 45 . The crack length, q1, grows from zero with an increment size of Dq1 \u00bc 0:1 mm until the crack reaches the tooth\u2019s central line. At that point, q1, reaches its maximum value of 3.9mm. After that, the crack then changes direction to q2 (see Fig. 5), which is assumed to be exactly symmetric around the tooth\u2019s central line. Theoretically, the maximum length of q2 should be the ARTICLE IN PRESS S. Wu et al. / Journal of Sound and Vibration 317 (2008) 608\u2013624 611 same as q1, however, the tooth is expected to suffer sudden breakage before the crack runs through the whole tooth. Thus the maximum length of q2 is assumed to be 60% of q1max and the increment size, Dq2, is also 0.1mm. In later reference of the crack growth, we will use a relative length. The highest crack level will be 3:9\u00fe 60% 3:9 \u00bc 6:24mm, or 80% (\u00bc 6:24=7:8) of the theoretical total length of the full through crack. ARTICLE IN PRESS S. Wu et al. / Journal of Sound and Vibration 317 (2008) 608\u2013624612 With the crack introduced as described above, we need to calculate all components of the total mesh stiffness, that is, Hertzian stiffness, axial compressive stiffness, bending stiffness, and shear stiffness. Based on the work documented in Ref. [15], the Hertzian and axial compressive stiffnesses remain the same when a crack is introduced. However, the bending and shear stiffnesses will change due to the appearance of the crack, and their derivation are provided under each of the following four cases. Case 1 Tian [4]: When hc1Xhr & a14ag, where a1 \u00bc 90 \u00f0pressure angle\u00de. ARTICLE IN PRESS S. Wu et al. / Journal of Sound and Vibration 317 (2008) 608\u2013624 613 The potential energy stored in a meshing gear tooth can be calculated by Ub \u00bc Z d 0 M2 2EIx dx \u00bc Z d 0 \u00bdF b\u00f0d x\u00de Fah 2 2EIx dx, (3) Us \u00bc Z d 0 1:2F 2 b 2GAx dx \u00bc Z d 0 \u00bd1:2F cos a1 2 2GAx dx, (4) where Ix and Ax represent the area moment of inertia and area of the section where the distance from the tooth\u2019s root is x, and G represents the shear modulus. They can be obtained by Ix \u00bc 1 12 \u00f0hc1 \u00fe hx\u00de 3L if xpgc; 1 12 \u00f02hx\u00de 3L if x4gc; 8>< >: (5) Ax \u00bc \u00f0hc1 \u00fe hx\u00deL if xpgc; 2hxL if x4gc; ( (6) G \u00bc E 2\u00f01\u00fe n\u00de , (7) where hx represents the distance between the point on the tooth\u2019s curve and the tooth\u2019s central line where the horizontal distance from the tooth\u2019s root is x. The bending mesh stiffness of the cracked tooth is 1 kbcrack \u00bc Z a2 ag 12f1\u00fe cos a1\u00bd\u00f0a2 a\u00de sin a cos a g2\u00f0a2 a\u00de cos a EL\u00bdsin a2 \u00f0q1=Rb1\u00de sin u\u00fe sin a\u00fe \u00f0a2 a\u00de cos a 3 da \u00fe Z ag a1 3f1\u00fe cos a1\u00bd\u00f0a2 a\u00de sin a cos a g2\u00f0a2 a\u00de cos a 2EL\u00bdsin a\u00fe \u00f0a2 a\u00de cos a 3 da (8) and the shear mesh stiffness of the cracked tooth is 1 kscrack \u00bc Z a2 ag 2:4\u00f01\u00fe n\u00de\u00f0a2 a\u00de cos a\u00f0cos a1\u00de 2 EL\u00bdsin a2 \u00f0q1=Rb1\u00de sin u\u00fe sin a\u00fe \u00f0a2 a\u00de cos a da \u00fe Z ag a1 1:2\u00f01\u00fe n\u00de\u00f0a2 a\u00de cos a\u00f0cos a\u00de2 EL\u00bdsin a2 \u00f0q1=Rb1\u00de sin u\u00fe sin a\u00fe \u00f0a2 a\u00de cos a da: (9) Case 2 Tian [4]: When hc1ohr or when hc1Xhr & a1pag. The bending mesh stiffness of the cracked tooth is 1 kbcrack \u00bc Z a2 a1 12f1\u00fe cos a1\u00bd\u00f0a2 a\u00de sin a cos a g2\u00f0a2 a\u00de cos a EL\u00bdsin a2 \u00f0q1=Rb1\u00de sin u\u00fe sin a\u00fe \u00f0a2 a\u00de cos a 3 da (10) and the shear mesh stiffness of the cracked tooth is 1 kscrack \u00bc Z a2 a1 2:4\u00f01\u00fe n\u00de\u00f0a2 a\u00de cos a\u00f0cos a1\u00de 2 EL\u00bdsin a2 \u00f0q1=Rb1\u00de sin u\u00fe sin a\u00fe \u00f0a2 a\u00de cos a da. (11) Case 3: When hc1ohr or when hc1Xhr & a1pag. This case was not covered in Ref. [4]. The bending mesh stiffness of the cracked tooth is 1 kbcrack \u00bc Z a2 a1 12f1\u00fe cos a1\u00bd\u00f0a2 a\u00de sin a cos a g2\u00f0a2 a\u00de cos a EL\u00bdsin a \u00f0q2=Rb1\u00de sin u\u00fe \u00f0a2 a\u00de cos a 3 da (12) ARTICLE IN PRESS S. Wu et al. / Journal of Sound and Vibration 317 (2008) 608\u2013624614 and the shear mesh stiffness of the cracked tooth is 1 kscrack \u00bc Z a2 a1 2:4\u00f01\u00fe n\u00de\u00f0a2 a\u00de cos a\u00f0cos a1\u00de 2 EL\u00bdsin a \u00f0q2=Rb1\u00de sin u\u00fe \u00f0a2 a\u00de cos a da. (13) Case 4: When hc2Xhr & a14ag. This case was not covered in Ref. [4] either. We have found the bending mesh stiffness of the cracked tooth to be 1 kbcrack \u00bc Z a2 ag 12f1\u00fe cos a1\u00bd\u00f0a2 a\u00de sin a cos a g2\u00f0a2 a\u00de cos a EL\u00bdsin a \u00f0q2=Rb1\u00de sin u\u00fe \u00f0a2 a\u00de cos a 3 da (14) ARTICLE IN PRESS S. Wu et al. / Journal of Sound and Vibration 317 (2008) 608\u2013624 615 and the shear mesh stiffness of the cracked tooth is 1 kscrack \u00bc Z a2 ag 2:4\u00f01\u00fe n\u00de\u00f0a2 a\u00de cos a\u00f0cos a1\u00de 2 EL\u00bdsin a \u00f0q2=Rb1\u00de sin u\u00fe \u00f0a2 a\u00de cos a da. (15) With the expressions of the components of the total mesh stiffness provided above, we are able to find the total mesh stiffness value given each shaft rotation angle and each crack size. For a pair of standard steel involute spur teeth whose main parameters are given in Table 1, take four specific crack sizes given in Table 2 as an example. These selected crack sizes cover all the four cases classified in the above derivations. The total mesh stiffness under each of the four crack sizes has been calculated as a function of the shaft rotation angle and plotted in Fig. 7. From Fig. 7, it can be observed that as the size of the crack grows, the total mesh stiffness when the cracked tooth is in meshing becomes much lower. This is important information for fault detection and assessment."
        ]
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure7.24-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure7.24-1.png",
        "caption": "Figure 7.24 The resultant of the water pressure on the circular cylindrical slide passes through C, the centre of arc AB.",
        "texts": [
            " VOLUME 1: EQUILIBRIUM Rx = q\u0302r d \u222b \u03c0/3 0 (a + r sin \u03d5) cos \u03d5 d\u03d5 = q\u0302r d [ a sin \u03d5 + 1 2r sin2 \u03d5 ]\u03d5=\u03c0/3 \u03d5=0 = q\u0302r d (a \u00d7 0.86 + r \u00d7 0.375), Ry = q\u0302r d \u222b \u03c0/3 0 (a + r sin \u03d5) sin \u03d5 d\u03d5 = q\u0302r d [ \u2212a cos \u03d5 + r ( \u2212 1 4 sin 2\u03d5 + 1 2\u03d5 )]\u03d5=\u03c0/3 \u03d5=0 = q\u0302r d (a \u00d7 0.5 + r \u00d7 0.307). By substituting q\u0302 = 40 kN/m, r = 3.0 m, d = 4.0 m and a = 1.4 m, we find Rx = 70.1 kN, Ry = 48.6 kN. The vertical component of the water pressure generates an upward force on the slide. The resulting water pressure R on the 1-metre strip from the slide is shown in Figure 7.24: R = \u221a (70.1 kN)2 + (48.6 kN)2 = 85.3 kN. The line of action, as shown earlier, passes through C and is at an angle of \u03b1 to the horizontal: \u03b1 = arctan ( 48.6 kN 70.1 kN ) = 34.7\u25e6. 7 Gas Pressure and Hydrostatic Pressure 261 Figure 7.25 The resultant of the water pressure on the slide can also be found by looking at the forces acting on the isolated slide together with water mass ADB. Alternative solution: Since the shape of the slide is actually rather simple, the question can also be answered without integrals"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000025_00207179308923053-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000025_00207179308923053-Figure7-1.png",
        "caption": "Figure 7.",
        "texts": [
            " Clearly \u00a3 a = ~ Lua = LuLua + ~Lua, it dt Z dt au Let C = LuLua, K = a/au Li,\u00ab, so that we have a = C + Kit According to Assumptions (3) and (4) ICI :;;; Co, 0 < \u00ab; :;;; K :;;; K M Hence, the following differential inclusion is valid a E [-Co, Co] + [Km, KM]it (22) (23) The real trajectory starting at the axis 0 on the plane a, 0 lies between the extreme solutions of (23). . Figure 6 shows the set of the solutions of (23) and the solution of (21) which are associated with the twisting algorithm (9). An infinite number encircling the origin occurs here within a finite time. The total time duration of the origin encircling is proportional to the total variation of a, which is estimated by a geometric series. The trajectories of the drift algorithm (12) are shown in Fig. 7. A real sliding along the axis 0 = 0 may be seen here. Due to the switching of 0 the motion with ao<O prevails. The duration of any cycle M oMtMzM 3 is proportional to 1:, where 1: is the time interval between the measurements of a, and the drift Order and accuracy in sliding mode control 1259 MoM3 is proportional to r. So, the time that the drift takes to reach a = 0 is proportional to r- 1 with r = constant. With the variable r (14) the convergence time turns out to be bounded (Emelyanov et al"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure7.5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure7.5-1.png",
        "caption": "Figure 7.5 (a) A pneumatic structure consists of a membrane that maintains its shape through internal overpressure. (b) A strip with width b from the circular cylindrical midsection in more detail.",
        "texts": [
            " In the same way, using the equilibrium of a triangular section parallel to the xz plane, we derive p1 = p3 = p. This means that the stress at a particular point is independent of the orientation of the plane that the stress acts on. This characteristic is known as 248 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Pascal\u2019s Law.1 This state of stress is known as isotropic or spherical. With gases and fluids, in which only compressive stresses occur, we also speak of all-round pressure. The type of structure in Figure 7.5a which is sometimes used as a tennis hall, is an air-supported hall or pneu. Pneu is an abbreviation of pneumatic structure. This type of structure consists of a thin, flaccid skin (membrane), which can transfer tensile forces only in its curved plane (see Figure 7.6). The structure maintains its shape through internal overpressure.2 The same holds, for example, for an inflated balloon, or the inner tube of a bike. We will look at three examples for this type of structure. In the first two examples the load is a gas pressure (the overpressure in the pneu). The third example concerns a body subjected to an all-round pressure. Example 1 The pneu in Figure 7.5a consists of a circular cylindrical midsection that is closed by means of spherical ends. The diameter of the circular cylinder is r , the aperture angle is \u03b1, and the internal overpressure is p. Question: Determine the distributed support reactions (forces per length) for the circular cylindrical midsection of the pneu. Solution: In Figure 7.5b, a strip of width b has been isolated. This strip is modelled in Figure 7.7a as a curved line element with a distributed load pb. In addition 1 Blaise Pascal (1623\u20131662), French mathematician, physicist and writer. With Fermat, he was one of the founders of the theory of probability. As a writer he is known for his Pense\u0301es, a collection of loose notes published posthumously. 2 This is the difference between the pressure inside and outside the structure. 7 Gas Pressure and Hydrostatic Pressure 249 to the support reactions at A and B, no forces other than those shown in the figure act in the plane of the drawing"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001098_tsmc.2015.2466194-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001098_tsmc.2015.2466194-Figure4-1.png",
        "caption": "Fig. 4. Diagram of a two-DOF robotic manipulator.",
        "texts": [
            " The inertia matrix M(x1), centripetal and coriolis torques C(x1, x2) and the gravitational force G(x1) are expressed as M(x1) = [ M11 M12 M21 M22 ] (82) C(x1, x2) = [ C11 C12 C21 0 ] (83) G(x1) = [ G11 G21 ] (84) where M11 = m1l21 + m2(l21 + l22 + 2l1l2 cos q2), M12 = M21 = m2(l22 + l1l2 cos q2), M22 = m2l22, C11 = \u2212m2l1l2q\u03072 sin q2, C12 = \u2212m2l1l2(q\u03071 + q\u03072) sin q2, C21 = \u2212m2l1l2q\u03071 sin q2, G11 = (m1l2 + m2l1)g cos q1 + m2l2g cos(q1 + q2), G21 = m2l2g cos(q1 + q2). Mass of link 1 m1 = 1 kg, mass of link 2 m2 = 1 kg, length of link 1 l1 = 0.8 m, length of link 2 l2 = 0.7 m. The initial positions of the robot is given as q0 = [0, 0] and q\u03070 = [0, 0]. The desired trajectory is chosen as qd = [0.1 sin(0.5t), 0.1 sin(0.5t)] where t \u2208 [0, 20]. Digram of a two-DOF robotic manipulator is shown in Fig. 4. The robotic manipulator is under external disturbance composed of Gaussian white noise. fdis is chosen as 1.5[1 \u2212 exp(\u22120.28t)(sin(0.5\u03c0 t))]. The output constraint is set as b = 0.005. The unknown deadzone is defined as br = 2.5 and bl = \u22124.5, with hr(\u03c4 ) = 2(\u03c4 \u2212 br)(sin(\u03c4 )+ 1) (85) hl(\u03c4 ) = (\u03c4 \u2212 bl) 3. (86) Different cases are considered for the simulation. We examined the model-based control with input deadzone and output constraint proposed in (25), the state feedback control law in (28) and the output feedback control law in (54)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure12.23-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure12.23-1.png",
        "caption": "FIGURE 12.23. A harmonically base excitated single DOF system.",
        "texts": [
            "121) \u03a61 = \u03c61 + \u03c62 2 (12.122) \u03a62 = \u03c61 \u2212 \u03c62 2 . (12.123) Figure 12.22 illustrates a sample plot of x(t) for \u03c91 = 10 \u03c92 = 12 \u03c61 = \u03c0 4 \u03c62 = \u03c0 6 (12.124) 756 12. Applied Vibrations The displacement x(t) indicates an oscillation between X1+X2 and X1\u2212 X2, with the higher frequency \u21261 inside an envelope that oscillates, at the lower frequency \u21262. This behavior is called beating. When X1 = X2 = X then x(t) = 2X cos (\u21261t\u2212 \u03a61) cos (\u21262t\u2212 \u03a62) (12.125) which becomes zero at every half period T = 2\u03c0/\u21262. 12.3.2 Base Excitation Figure 12.23 illustrates a one-DOF base excited vibrating system with a mass m supported by a spring k and a damper c. Base excited system is a good model for vehicle suspension system or any equipment the is mounted on a vibrating base. The absolute motion of m with respect to its equilibrium position is measured by the coordinate x. A sinusoidal excitation motion y = Y sin\u03c9t (12.126) is applied to the base of the suspension and makes the system vibrate. The equation of motion for the system can be expressed by either one of the following equations for the absolute displacement x m x\u0308+ c x\u0307+ kx = cY \u03c9 cos\u03c9t+ kY sin\u03c9t (12",
            " Applied Vibrations z\u0308 = \u2212A3\u03c92 sin\u03c9t\u2212B3\u03c9 2 cos\u03c9t = \u2212Z\u03c92 sin (\u03c9t\u2212 \u03d5z) = Z\u0308 sin (\u03c9t\u2212 \u03d5z) (12.188) The amplitude of velocity and acceleration frequency responses, Z\u0307, Z\u0308 Z\u0307 = m\u03c93q (k \u2212m\u03c92) 2 + c2\u03c92 Y (12.189) Z\u0308 = m\u03c94q (k \u2212m\u03c92) 2 + c2\u03c92 Y (12.190) can be written as Z\u0307 \u03c9nY = r3q (1\u2212 r2) 2 + (2\u03ber) 2 (12.191) Z\u0308 \u03c92nY = r4q (1\u2212 r2) 2 + (2\u03ber) 2 . (12.192) Example 443 Transmitted force to the base of a base excited system. The transmitted force fT to the ground by a base excited system, such as is shown in Figure 12.23, is equal to the sum of forces in the spring and damper. fT = fk + fc = k (x\u2212 y) + c (x\u0307\u2212 y\u0307) (12.193) which based on the equation of motion (12.140) is also equal to fT = \u2212mx\u0308. (12.194) Substituting x\u0308 from (12.175) and (12.179) shows that the frequency response of the transmitted force can be written as FT kY = \u03c92 \u221a k2 + c2\u03c92q (k \u2212m\u03c92)2 + c2\u03c92 (12.195) = r2 q 1 + (2\u03ber)2q (1\u2212 r2)2 + (2\u03ber)2 . (12.196) The frequency response of FT kY is the same as is shown in Figure 12.29. 12. Applied Vibrations 767 Example 444 F Line of maxima in X/Y ",
            "334) Z1 = 2\u03ber 1 \u03c92n \u00a1 B2\u03c9 2 +B1\u03c9 +B0 \u00a2 + 1 \u03c92n \u00a1 1\u2212 r2 \u00a2 \u00a1 A2\u03c9 2 +A1\u03c9 +A0 \u00a2 (12.335) Z2 = 1 \u03c92n \u00a1 1\u2212 r2 \u00a2 \u00a1 B2\u03c9 2 +B1\u03c9 +B0 \u00a2 \u22122\u03ber 1 \u03c92n \u00a1 A2\u03c9 2 +A1\u03c9 +A0 \u00a2 (12.336) provides the steady-state solution amplitude Q and phase \u03d5 Q = p A2 +B2 (12.337) tan\u03d5 = \u2212B A . (12.338) We are able to reproduce any of the steady-state responses Si and Gi by setting the coefficients A0, A1, A2, B0, B1, and B2 properly. 786 12. Applied Vibrations Example 452 F Base excited frequency responses. A one DOF base excited vibrating system is shown in Figure 12.23. The equation of relative motion z = x \u2212 y with a harmonic excitation y = Y sin\u03c9t is z\u0308 + 2\u03be\u03c9n z\u0307 + \u03c92n z = \u03c92 Y sin\u03c9t. (12.339) This equation can be found from Equation (12.323) if A0 = 0 A1 = 0 A2 = Y B0 = 0 B1 = 0 B2 = 0. (12.340) So, the frequency response of the system would be Z = Q = p A2 +B2 = r2q (1\u2212 r2) 2 + (2\u03ber) 2 Y (12.341) because, \u2219 A B \u00b8 = \u23a1\u23a2\u23a2\u23a2\u23a3 Z1 (1\u2212 r2) 2 + (2\u03ber) 2 Z2 (1\u2212 r2) 2 + (2\u03ber) 2 \u23a4\u23a5\u23a5\u23a5\u23a6 (12.342) Z1 = r2 \u00a1 1\u2212 r2 \u00a2 Y (12.343) Z2 = 2\u03ber3Y. (12.344) 12.4 Time Response of Vibrating Systems Linear vibrating systems have a general equation of motion as the set of differential equations, [m] x\u0308+ [c] x\u0307+ [k]x = F (12"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002097_978-1-4939-3017-3-Figure1.11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002097_978-1-4939-3017-3-Figure1.11-1.png",
        "caption": "Figure 1.11.1. The geometry for the roller racer.",
        "texts": [
            " Analysis of this system may be a useful guide for modeling and studying the stability of other systems, such as aircraft landing gears and train wheels. The roller racer is modeled as a system of two planar coupled rigid bodies (the main body and the second body) with a pair of wheels attached on each of the bodies at their centers of mass: a nonholonomic generalization of the coupled planar bodies discussed earlier. We assume that the mass and the linear momentum of the second body are negligible, but that the moment of inertia about the vertical axis is not. See Figure 1.11.1. Let (x, y) be the location of the center of mass of the first body and denote the angle between the inertial reference frame and the line passing through the center of mass of the first body by \u03b8, the angle between the bodies by \u03c6, and the distances from the centers of mass to the joint by d1 and d2. The mass of body 1 is denoted by m, and the inertias of the two bodies are written as I1 and I2. The Lagrangian and the constraints are L = 1 2 m(x\u03072 + y\u03072) + 1 2 I1\u03b8\u0307 2 + 1 2 I2(\u03b8\u0307 + \u03c6\u0307)2 and x\u0307 = cos \u03b8 ( d1 cos\u03c6+ d2 sin\u03c6 \u03b8\u0307 + d2 sin\u03c6 \u03c6\u0307 ) ; y\u0307 = sin \u03b8 ( d1 cos\u03c6+ d2 sin\u03c6 \u03b8\u0307 + d2 sin\u03c6 \u03c6\u0307 ) "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001056_s11465-015-0341-2-Figure13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001056_s11465-015-0341-2-Figure13-1.png",
        "caption": "Fig. 13 Complex mould block with conformal cooling channels fabricated by SLM",
        "texts": [
            "2 Manufacture of part with complex cooling channels Similar to other additive manufacturing processes, SLM has the obvious advantage for fabricating complex structures compared with traditional processes. Aircraft blades and molds that often work at high temperatures need complex cooling channels inside to increase their service lives. Traditional processing methods usually obtain a straight hole by drilling. However, the consistent spatial distribution of the hole is difficult to achieve because of the complex surfaces of the parts, thereby reducing the cooling efficiency and uniformity. SLM based on the layer by layer method can realize the rapid manufacture of complex cooling channels. Figure 13 shows the injection mold with conformal cooling channels using SLM fabricated by our HUST-RMC. 4.3 Manufacture of scaffolds and prosthesis with personalized structures Scaffolds and prosthesis require biocompatibility and matched mechanical properties to human tissue [83]. Internal pores with certain sizes are needed for cell parasitism, growth, and tissue regeneration and construction [84]. The traditional methods to fabricate porous structures include casting [85], vapor deposition [86], spraying [87], and sintered powders [88] (shown in Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000350_0005-1098(92)90029-f-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000350_0005-1098(92)90029-f-Figure1-1.png",
        "caption": "FIG. 1. Aircraft coordinate systems (R-- runway, A - - aircraft).",
        "texts": [
            "___~[~6, Q~:Rn\"\">R 6xrn, and u : ~ n x ~ m \" ' ~ m are (con- tinuous) functions. In particular, for each x in the function u(x, \"):Rm--* ~m is one-to-one and hence (algebraically) invertible. The value of the function u(., .) can often be taken to be the components of the force and moment that the actuators were designed to produce. As a example, consider the YAV-8B Harrier produced by McDonnell Aircraft Company (McDonnell Douglas Corporation, 1982; McDonnell Aircraft Company, 1983) depicted in Fig. 1 (aircraft frame--A, runway frame--R). The Harrier is a single-seat transonic light attack V/STOL (vertical/short takeoff and landing) aircraft powered by a single turbo-fan engine. Four exhaust nozzles on the turbo-fan engine provide the gross thrust for the aircraft. These nozzles (two on each side of the fuselage) can be simultaneously rotated from the aft position (used for conventional wing-borne flight) forward approximately 100 degrees allowing jetborne flight and nozzle braking. The throttle and nozzle controls thus provide two degrees of freedom of thrust vectoring within the x - z plane of the aircraft"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002670_j.heliyon.2021.e06892-Figure14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002670_j.heliyon.2021.e06892-Figure14-1.png",
        "caption": "Figure 14. Ti6Al4V tibial stem (knee) prototype with an outer mesh structure manufactured by electron beam melting. a) Computer-aided design (CAD) model, b) prototype fabricated via electron beam melting using the CAD model, and c) CAD model with a 45 rotation.",
        "texts": [
            " Figures 12b) and 12c) display the EBM-manufactured prototypes. Figure 13a) illustrates, by means of an X-ray image, the knee replacement implant consisting of a Ti6Al4V tibial stem and meniscal platform. Figure 13b) shows the incision for knee replacement surgery. In this case, a porous surface is employed in the femoral system to favor femoral bone cell ingrowth. EBM is a promising technique to obtain highly rigid porous structures that can be used as monolithic implants with a lower risk of failure in terms of complexity and mechanical stress. Figure 14 displays the EBM-manufactured Ti6Al4V tibial stem (knee) prototype with an outer mesh structure. In this case, the longer stem provides greater rigidity and stability to the implant. According to Suska et al. [356], EBM has a high potential in the field of maxillofacial reconstruction because it offers innovative designs of mandible prostheses based on the specific (aesthetic or functional) needs of patients. EBM-manufactured jaw prostheses include porous parts that favor bone ingrowth and secondary biological fixation with solid regions, characteristics responsible for an adequate biomechanical functioning"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.13-1.png",
        "caption": "FIGURE 7.13. Kinematic steering condition for a vehicle with different tracks in the front and in the back.",
        "texts": [
            "33) is the yaw rate of the vehicle, which is the vehicle\u2019s angular velocity about the center of rotation. r = Rw \u03c9o R1 + w 2 = Rw \u03c9i R1 \u2212 w 2 (7.35) 7. Steering Dynamics 391 Example 264 F Unequal front and rear tracks. It is possible to design a vehicle with different tracks in the front and rear. It is a common design for race cars, which are usually equipped with wider and larger rear tires to increase traction and stability. For street cars we use the same tires in the front and rear, however, it is common to have a few centimeters of larger track in the back. Such a vehicle is illustrated in Figure 7.13. The angular velocity of the vehicle is r = Rw \u03c9o R1 + wr 2 = Rw \u03c9i R1 \u2212 wr 2 (7.36) and the kinematic steer angles of the front wheels are \u03b4i = tan\u22121 2l (\u03c9o + \u03c9i) wf (\u03c9o \u2212 \u03c9i) + wr (\u03c9o + \u03c9i) (7.37) \u03b4o = tan\u22121 2l (\u03c9o \u2212 \u03c9i) wf (\u03c9o \u2212 \u03c9i) + wr (\u03c9o + \u03c9i) . (7.38) To show these equations, we should find R1 from Equation (7.36) R1 = wr 2 \u03c9o + \u03c9i \u03c9o \u2212 \u03c9i (7.39) 392 7. Steering Dynamics and substitute it in the following equations. tan \u03b4i = l R1 \u2212 wf 2 (7.40) tan \u03b4o = l R1 + wf 2 (7.41) In the above equations, wf is the front track, wr is the rear track, and Rw is the wheel radius"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure5.19-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure5.19-1.png",
        "caption": "FIGURE 5.19. A 2R planar manipulator.",
        "texts": [
            " Applied Kinematics velocity equation (5.326) or (5.327). GaP = Gd dt GvP = G\u03b1B \u00d7 G BrP + G\u03c9B \u00d7 G B r\u0307P + Gd\u0308B = G\u03b1B \u00d7 G BrP + G\u03c9B \u00d7 \u00a1 G\u03c9B \u00d7 G BrP \u00a2 + Gd\u0308B = G\u03b1B \u00d7 \u00a1 GrP \u2212 GdB \u00a2 +G\u03c9B \u00d7 \u00a1 G\u03c9B \u00d7 \u00a1 GrP \u2212 GdB \u00a2\u00a2 + Gd\u0308B. (5.393) The term G\u03c9B \u00d7 \u00a1 G\u03c9B \u00d7 G BrP \u00a2 is called centripetal acceleration and is independent of the angular acceleration. The term G\u03b1B \u00d7 G BrP is called tangential acceleration and is perpendicular to G BrP . Example 199 Acceleration of joint 2 of a 2R planar manipulator. A 2R planar manipulator is illustrated in Figure 5.19. The elbow joint has a circular motion about the base joint. Knowing that 0\u03c91 = \u03b8\u03071 0k\u03020 (5.394) we can write 0\u03b11 = 0\u03c9\u03071 = \u03b8\u03081 0k\u03020 (5.395) 0\u03c9\u03071 \u00d7 0r1 = \u03b8\u03081 0k\u03020 \u00d7 0r1 = \u03b8\u03081RZ,\u03b8+90 0r1 (5.396) 0\u03c91 \u00d7 \u00a1 0\u03c91 \u00d7 0r1 \u00a2 = \u2212\u03b8\u030721 0r1 (5.397) and calculate the acceleration of the elbow joint 0r\u03081 = \u03b8\u03081RZ,\u03b8+90 0r1 \u2212 \u03b8\u0307 2 1 0r1. (5.398) Example 200 Acceleration of a moving point in a moving body frame. Assume the point P in Figure 5.18 is indicated by a time varying local position vector BrP (t). Then, the velocity and acceleration of P can be found by applying the derivative transformation formula (5"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003990_tfuzz.2006.879982-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003990_tfuzz.2006.879982-Figure3-1.png",
        "caption": "Fig. 3. The PUMA-500 series robot manipulator.",
        "texts": [],
        "surrounding_texts": [
            "Theorem 1: Apply the control law (21) to the robot system (11) with the robustifying term (30). Let the weight updating of fuzzy wavelet network be (31) (32) (33) (34) where , and , , , and are positive and diagonal constant matrices. Suppose that the desired trajectory is bounded and double differentiable, and Assumptions 1\u20134 hold. Then, estimation errors , , and are bounded. Proof: Let the Lyapunov-like function candidate to be (35) where . By differentiating with respect to time and using (22), we have (36) Then, substituting (27) and (31)\u2013(33) into (36) yields (37) Consequently, (37) can be reduced to (38) Therefore, (38) satisfies (39) Since (34), , , , and , we can get (40) Clearly, we can obtain the following: (41) Hence, for all , and if , , , , , and are bounded at initial time , they will remain bounded for all . According to bounded and , and are bounded for all , and since is bounded as specified, is bounded as well. All , , , and are bounded indicates all weights , , , and are also bounded. By Lyapunov theory, it is also easy to prove the stability of robotic system (11) with control law (21) and weight update laws (31)\u2013(34). However, in the following lemma, we will emphasize that it is not necessary for to approach zero at infinite time by the NTSM control. According to the theorem, we can make the following assumption. Assumptions 5: The norm and the design constant satisfy and , respectively. Lemma 1: Suppose Assumption 5 holds, then will approach zero within finite time. Proof: We define another Lyapunov function candidate as (42) Similarly, we evaluate the time derivative of along the system (22) to get (43) where . From the above proof, we can assume that for . When , we can show that converges to zero within finite time since (43) satisfies (44) where . If and , then we can rewrite (18) as (45) according to and for . It implies that if and , then for and for , respectively, [21]. In other words, is not an attractor and the surface will be reached in finite time. parameters in Example 1. (a) . (b) . (c) . (d) . (e) . (f) . Remark 1: For each , we can design different odd integers and , which still satisfy for . Thus, the nonsingular sliding surface should be modified as (46) For keeping most equations the same, we redefine , as and (47) Remark 2: Besides the external disturbances and unknown friction forces , perturbations in nominal parameters of practical robotic systems are inevitable, and, therefore, many researches consider both estimated terms ( , and ) and their corresponding uncertain terms ( , and ) [21]. Taking into account these unknown effects, it is convenient to use fuzzy wavelet networks to estimate unknown functions including the external disturbances and unknown friction forces, all estimated terms and their corresponding uncertain terms. Moreover, the reason of emphasis on on-line tuning of parameters is that there is no need to deal with the off-line identification of unknown parameters , and . Although the overall performance of the proposed controller is heavily dependent on the tuning parameters such as , , , , and in (31)\u2013(34), it is still easier than the off-line identification of unknown parameters, which also dominate the performance of model-based controller."
        ]
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure4.23-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure4.23-1.png",
        "caption": "Figure 4.23 (a) The fixed support has no degrees of freedom and generates (b) three support reactions (the two components of a force and a fixed-end moment).",
        "texts": [
            " The steel hinged support in Figure 4.22 can transfer large forces and is an example of what is used in larger bridges. Like roller supports, hinged supports can be made from materials other than steel, or from a combination of materials Although the supports in Figures 4.21 and 4.22 can transfer only compressive forces, it is assumed below that hinged supports can also transfer tensile forces. A fixed support is an infinitely stiff or rigid joint between a body and its environment, see also Section 4.2.2. Figure 4.23a is a model of a fixed support (the dotted line is generally omitted). At A, the fixed support prevents both the displacement and rotation of the body. In fixed supports, all motion is prescribed: fixed supports therefore have no degrees of freedom. A fixed support has three support reactions, see Figure 4.23: two forces and a so-called fixed-end moment. The balcony (cantilever beam) in Figure 4.24a is an example of a fixed supported structure. Another example is the support in Figure 4.24b of a concrete column on a concrete foundation, constructed as a single, monolithic whole. In many cases, a fixed support will not fully prevent rotation. Such a support is incomplete and is referred to as a spring support if the magnitude of the rotation is related to the magnitude of the fixed-end moment. We will always refer to a fully fixed support below"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001190_1.4828755-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001190_1.4828755-Figure8-1.png",
        "caption": "FIG. 8. Spreading and defective positions at twin cantilevers with different bar thicknesses (h) after separating the supports. TV\u00bcRT.",
        "texts": [
            " In the investigations conducted without preheating, a total of 12 defective positions, recognizable macroscopically, appear on the five twin cantilevers, which can lead to cracking. Two of them lead to detachments of the twin cantilever arms directly after the construction process. The appearance of the cracks is disadvantageous for the evaluation of the distortions. The stresses that should manifest themselves as distortions are reduced by an amount that cannot be assessed. In case these macroscopic cracks entail detachment, the spreading can no longer be evaluated for the cracked area. This emerges for the twin cantilever at h\u00bc 1 mm and at h\u00bc 3 mm (Fig. 8). For this reason, several measurement points are lacking for these twin cantilevers. The maximum measured spreading of 8.2 mm appears in the thinnest twin cantilever (h\u00bc 0.5 mm) at the measurement points 1 and 12 (outermost end of the twin cantilever, Fig. 9). On the twin cantilever with h\u00bc 5 mm, a spreading of 1.2 mm is measured at measuring points 1 and 12. As long as there are not any defective positions or cracks within the twin cantilever, the curves of the measuring points run parabollically to the middle axis of the twin cantilever"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.46-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.46-1.png",
        "caption": "FIGURE 8.46. Illustration of tire and wheel coordinate frames.",
        "texts": [
            " When the wheel is straight and the W frame is parallel to the vehicle coordinate frame, we attach a wheel-body coordinate frame C (xc, yc, zc) at the center of the wheel parallel to the vehicle coordinate axes. The wheelbody frame C is motionless with respect to the vehicle coordinate and does not follow any motion of the wheel. The tire coordinate frame T (xt, yt, zt) is set at the center of the tireprint. The zt-axis is always perpendicular to the ground. The xt-axis is along the intersection line of the tire-plane and the ground. The tire frame does not follows the spin and camber rotations of the tire however, it follows the steer angle rotation about the zc-axis. Figure 8.46 illustrates a tire and a wheel coordinate frames. Example 326 Visualization of the wheel, tire, and wheel-body frames. Figure 8.47 illustrates the relative configuration of a wheel-body frame C, a tire frame T , and a wheel frame W . If the steering axis is along the zc-axis then, the rotation of the wheel about the zc-axis is the steer angle \u03b4. Rotation about the xt-axis is the camber angle \u03b3. 488 8. Suspension Mechanisms Generally speaking, the steering axis may have any angle and may go through any point of the ground plane"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000554_978-3-540-85629-0-FigureB.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000554_978-3-540-85629-0-FigureB.2-1.png",
        "caption": "Fig. B.2. Cubic basis functions defined on u = [0, 0, 0, 0, 1, 2, 4, 7, 7, 7, 7].",
        "texts": [
            " , unknot ] be a vector of real numbers (called knots), with uj \u2264 uj+1. The j-th B-spline basis function of degree p (or equivalently of order p + 1) is defined, in a recursive manner, as B0 j (u) = { 1, if uj \u2264 u < uj+1 0, otherwise Bp j (u) = u \u2212 uj uj+p \u2212 uj Bp\u22121 j (u) + uj+p+1 \u2212 u uj+p+1 \u2212 uj+1 Bp\u22121 j+1 (u), p > 0. Note that 468 B B-spline, Nurbs and Be\u0301zier curves 1. Bp j (u) is a piecewise polynomial, defined \u2200u \u2208 [umin, umax]. 2. Bp j (u) is equal to zero everywhere except in the interval u \u2208 [uj , uj+p+1), see Fig. B.1 and Fig. B.2. 3. The interval [ui, ui+1) is called i-th knot span; it can be of zero length, in case knots are coincident1. 4. The B-spline basis functions are normalized so that m\u2211 j=0 Bp j (u) = 1, \u2200u \u2208 [u0, unknot ] (partition of the unity). 5. In every knot span [ui, ui+1) at most p + 1 basis functions Bp j are not null, namely Bp i\u2212p, . . . , B p i (see Fig. B.2); this is illustrated in the graph of Fig. B.3 (truncated triangular table), referred to cubic functions, which shows the dependencies of the 3-rd degree basis functions on B0 3 (which is the only 0-degree term different from zero in the interval [u3, u4)). Given a value of u \u2208 [ui, ui+1), it is possible to evaluate the basis functions with a simple and efficient algorithm. By tacking into account the observation 1 The difference between knots and breakpoints is that breakpoints are the set of distinct knot values",
            " It is therefore necessary to consider separately this case (for u = umax the last basis function has a unit value, while all the other functions are null) or to avoid that this condition occurs by assuming u = umax \u2212 \u03b5, where \u03b5 is a small positive number. B.1 B-spline Functions 471 } } } return mid; } Example B.1 Given p = 3, u = [0, 0, 0, 0, 1, 2, 4, 7, 7, 7, 7] and u = 1.5, then u \u2208 [u4, u5), i = 4, and the nonvanishing basis functions are B3 1 = 0.0313, B3 2 = 0.5885, B3 3 = 0.3733, B3 4 = 0.0069. The shape of all the basis functions defined on u is shown in Fig. B.2. B.1.2 Definition and properties of B-splines Given the B-spline basis functions defined on the nonuniform knot vector (of size nknot) u = [umin, . . . , umin\ufe38 \ufe37\ufe37 \ufe38 p+1 , up+1, . . . , unknot\u2212p\u22121, umax, . . . , umax\ufe38 \ufe37\ufe37 \ufe38 p+1 ] (B.1) a p degree B-spline curve is defined as s(u) = m\u2211 j=0 pjB p j (u), umin \u2264 u \u2264 umax (B.2) where pj , j = 0, . . . ,m are the control points, and form the so-called control polygon. Therefore, to represent a spline curve in the B-form, it is necessary to provide: 1. The integer p, defining the degree of the spline"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure14.59-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure14.59-1.png",
        "caption": "Figure 14.59 By applying an additional bar, the kinematically indeterminate bar structure changes into a kinematically and statically determinate truss.",
        "texts": [],
        "surrounding_texts": [
            "The bending moment and the normal force are the resultants of the normal stresses in a cross-section. Figure 14.54 shows the stress distribution due to a bending moment M and a normal force N .1 A characteristic of stress distribution in bending is that the outermost fibres of the cross-section are most heavily loaded, while the fibres in the environment of the member axis are virtually unloaded. In contrast, the stresses due to a normal force are constant over the cross-section. In extension, all fibres are therefore loaded equally. If we compare the stress distributions due to bending and extension, the material in the cross-section is used far more efficiently in extension than in bending. With bending, the strength capacity of the fibres around the member axis is not used, and the small stresses only marginally contribute to the bending moment. For beams loaded by bending, one often sees an adaptation of the cross-section by omitting the less active material in the cross-section. In this way, a rectangular cross-section may become a tubular section or an I -section (see Figure 14.55). In addition, when designing structures, designers look for shapes in which the bending moments remain as small as possible, and in which the force flow preferably occurs by extension. This is achieved by ensuring the member axis and line of force coincide as much as possible. Since a cable cannot transfer bending moments, it assumes a shape in which the line of force coincides with its axis everywhere. Taking the cable shape and line of force as basis, in the following four examples, we look for 1 In Volume 2, Stresses, Deformations, Displacements, we take a closer look at the exact development of the normal stresses in a cross-section and at the conditions under which the stress distribution in Figure 14.54 applies. 680 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM structural shapes in which the bending moments are as small as possible. Example 1 The beam in Figure 14.56a is subject to bending by the two forces F . The M diagram is shown in Figure 14.56b. In Figure 14.57, the same load is carried by a cable with compression bar. The cable and compression bar transfer normal forces only. The cable has the same shape as the M diagram in Figure 14.56b. With a cable sag the scale factor is H = F = F. H is the (compressive) force in the bar that is equal to the horizontal component of the (tensile) force in the cable. In Figure 14.58a the straight cable parts have been replaced by bars. Plus and minus signs indicate whether the bar forces are tensile or compressive. The structure can be considered a kind of arch under tension that is kept together by a compression bar. If the bar structure in Figure 14.58a is \u201cturned over\u201d with equal loads as shown in Figure 14.58b, all the signs in the bars change. The structure has now changed into an arch under compression with tension bar (tie rod). In the position shown in Figure 14.58b, the bar structure is in equilibrium. However, the equilibrium is unstable (unreliable): a small change in position will cause the equilibrium to fail and the bar structure will collapse.1 The bar structure is kinematically indeterminate. This collapse can be pre- 1 To prove this we have to investigate the equilibrium of the structure in its deformed state. However, this topic is beyond the scope of this book. Here we assume that the reader is acquainted with this phenomenon of instbility on the basis of some practical experience. 14 Cables, Lines of Force and Structural Shapes 681 vented by making the structure kinematically determinate, for example by introducing bracing members, and changing the bar structure into a truss (see Figures 14.59 and 14.60b). If we calculate the member forces for the given load, we find that all the interior members are zero members. The cable in Figure 14.57 and the bar structure in Figure 14.58a are also kinematically indeterminate. The equilibrium is stable (reliable) in this case as the load makes the structure go back into the original equilibrium position after a disruption. In Figure 14.60a, the bar structure in Figure 14.58a has been changed into a kinematically determinate truss. All the interior members are zero. As in contrast to the cable, the truss has the benefit that the shape does not change when the load changes. In Figure 14.61, the forces on the truss are shifted to the horizontal plane through the supports. The verticals are no longer zero members. These 682 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM trusses, in which all the members are subject to extension (or are zero members), can be an alternative for the beam subject to bending in Figure 14.56. Figure 14.62a shows the bar structure from Figure 14.58b, but now without tension member. This kinematically indeterminate structure can be made kinematically determinate not only by changing it into a truss, but also by replacing the hinged joints between the bars by rigid joints. The structure then becomes a bent member, recognisable in Figure 14.62b as a two-hinged frame. One problem is that the frame is statically indeterminate to the first degree. As such, it is not possible to determine the horizontal support reaction H directly from the equilibrium. The deformation of the frame also has to be taken into account. If the deformation by normal forces is ignored (as it was in the cable), it is possible to show that no bending occurs under the given load, and that the normal forces in the frame are equal to the forces in the two-force members in Figure 14.62a. In Figure 14.62c, the frame has been isolated and all the forces acting on it are shown. The line of force coincides everywhere with the bent member axis and there is no bending anywhere. In reality, there is always some axial deformation due to normal forces. As such, the horizontal support reactions are somewhat smaller and, because the vertical support reactions remain equal, the line of force no longer coincides with the bent member axis (see Figure 14.63). Axial deformation therefore induces bending in the two-hinged frame. Since statically indeterminate structures are more sensitive to settling and temperature, statically determinate structures are generally preferable, because the force distribution is more manageable. In Figure 14.64, the statically indeterminate two-hinged frame has been changed into a statically determined three-hinged frame. With the given load, the line of force coincides everywhere with the bent member axis and there is no bending anywhere. 14 Cables, Lines of Force and Structural Shapes 683 Figure 14.64 Three-hinged frames are statically determinate and therefore the force flow is less sensitive to axial deformations, settling and the influence of temperature. Figure 14.65 (a) Three-hinged frame with uniformly distributed load on girder CSD. (b) Support reactions. (c) Bending moment diagram. (d) The line of pressure for girder CSD is a parabola through A, S and B. Example 2 On girder CSD, the three-hinged portal frame in Figure 14.65a is carrying a uniformly distributed load. In Figures 14.65b and 14.65c, the support reactions and the bending moment diagram are shown. The calculation is left to the reader. Question: How can one reduce the bending moment in the frame, without changing the given load? Solution: Figure 14.65d shows the line of force for girder CSD. Cross-section C has to transfer the support reaction at A; the centre of force for cross-section C is therefore at A. In the same way, the centre of force for cross-section D is at B. The line of force passes through hinge S. The line of force for CSD has the same shape as a cable under a uniformly distributed load, which is a parabola. Since the normal force in girder CSD is a compressive force, \u201cthe parabolic cable is upside down\u201d and the line of force is a line of pressure. Check: Regard the line of force as an \u201cupside-down cable\u201d (see Figure 14.65d): H = 1 8q 2 p = 1 8 \u00d7 (4 kN/m)(12 m)2 4 m = 18 kN. This force is indeed equal to the horizontal support reactions at A and B (see Figure 14.65b)."
        ]
    },
    {
        "image_filename": "designv10_0_0003443_0278364903022001005-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003443_0278364903022001005-Figure1-1.png",
        "caption": "Fig. 1. 3-DoF 3-RPS PM.",
        "texts": [
            " For example, two parallel constraint forces not only constrain a translational DoF along the forces, they also constrain a rotational freedom about the normal of the plane determined by the two forces; three coplanar and non-concurrent forces can also constrain a rotational freedom about the normal of the plane formed by these forces; three parallel and non-coplanar forces can constrain two rotational DoFs about the axes perpendicular to the forces (Huang, Tao and Fang 1996). An example is given in what follows. The 3-DoF 3-RPS PM (Hunt 1983) has two rotational and one translational DoFs, as shown in Figure 1, where B denotes the base and M the moving platform. The end of a single RPS limb has five DoFs, including two translational DoFs and three rotational DoFs. The translational DoF along the first revolute axis is constrained. Obviously, the end of every RPS limb has three independent rotational DoFs. From eq (1), we have \u22c23 i=1 x\u0302 l i = x\u0302,\u22c23 i=1 y\u0302 l i = y\u0302, and \u22c23 i=1 z\u0302 l i = z\u0302. Finally, the intersection of U1, U2 and U3 yields UP = [ 0 0 z x\u0302 y\u0302 z\u0302 ] . (2) at OhioLink on October 23, 2014ijr"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000533_j.humov.2007.04.003-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000533_j.humov.2007.04.003-Figure3-1.png",
        "caption": "Fig. 3. Simple work tasks illustrate principles of work production. (a) During cycling, the human performs positive work on the crank to offset energy dissipated by the environment, mainly through aerodynamic drag and friction. At steady speed, the rates of positive work performed by muscle and negative work performed by the environment are equal. If drag and friction are negligible, then no work is necessary for locomotion. (b) As a human bobs up and down vertically, the muscles perform positive work against gravity to lift the COM, and then negative work to actively lower it. Gravity performs zero net work on the body over a bobbing cycle, and it is the human who is responsible for both dissipating and restoring energy in equal amounts. (c) A co-contraction task requires simultaneous positive and negative work to transport a spherical mass horizontally (from left to right) at constant speed. Two telescoping actuators push inward with equal and opposite forces to support the mass. The left-most actuator performs positive work against the mass, and the right-most actuator performs negative work. (d) Co-contraction also occurs in a different configuration, where the mass is supported above the actuators, each of which pivots freely about a hinge joint. Both actuators push inward and upward to support the mass, and again one performs positive and the other negative work. Traditional views of walking are often implicitly based on analogies to cycling or bobbing, concerned with work performed to propel or lift the COM. From the dynamic walking perspective, a better analogy might be the co-contraction tasks, because the two legs perform positive and negative work against each other simultaneously during double support.",
        "texts": [
            " Dynamic walking offers a different perspective than that afforded through more traditional approaches, by considering conceptually why positive work must be performed at all, and how negative work dictates the positive work. Two principles of work provide the basis for examining the two major theories of walking from the dynamic walking perspective. In most forms of propulsive locomotion, the human contribution is primarily to positive work, performed against an external resistive load. Examples include pedaling (Fig. 3a) and rowing, in which positive work is performed against drag from water or air. In climbing stairs or a reasonably steep hill, positive work is performed against the resistance of gravity, resulting in an increase in gravitational potential energy. In all of these cases, the environment performs negative work on the COM. The body then performs an equal amount of positive work, so that zero net work is performed on the COM over a cycle. Quantifying the energy expended during such tasks is relatively simple, because the cost is both dominated by and proportional to the positive work performed by muscle",
            " The likely explanation is that biomechanical or task constraints unavoidably require that negative work be performed. It is therefore especially important to understand why this negative work is performed and in what amount, and to understand the underlying constraints responsible for it. Self-resistive tasks occur commonly in human movement, and their consideration may contribute to the study of walking. One example of a non-locomotory, self-resistive task is bobbing the body up and down (see Fig. 3b), where equal amounts of positive and negative work are performed in succession. Starting from the lower position with the legs bent, the muscles perform positive work to raise the body center of mass against the gravitational load. This work is converted into gravitational potential energy. The next task of lowering the COM therefore requires the steady dissipation of this energy. The muscles perform negative work on the COM to bring the body back to its initial position. Metabolic energy is expended for the positive work, and to a lesser extent, the negative work as well. An obvious means of reducing energetic cost would simply be to remain at one height. If no net change in height is desired over an integral number of cycles, periodic bobbing accomplishes nothing other than to waste energy. Self-resistance can also occur with positive and negative work performed simultaneously, for example in the horizontal transport of a mass by antagonistic actuators (see Fig. 3c). Here we consider two linear (sliding) actuators, both pushing inward to support a spherical mass. No net work is needed for the mass to move horizontally at constant speed. However, because the actuators are pushing against each other with equal and opposite forces, one actuator performs positive work, and the other negative work. The negative work dissipates the mechanical energy of the positive work. Metabolic energy will be expended to perform positive work and (assuming no means of re-generation) negative work. This may be classified as a form of inter-limb co-activation or co-contraction, extending a term usually applied to antagonists located on opposite sides of a joint to refer to opposite limbs. Co-contraction can also apply to other configurations (see Fig. 3d), for example with the linear actuators both pushing inward and upward simultaneously. Again, the load can be supported as it moves horizontally, with simultaneous positive and negative work performed by the actuators. The actuators function partially as agonists, because they cooperate to resist the force of gravity. However, they also function as antagonists, with the horizontal component of forces in opposition. Most researchers view co-contraction as energetically unfavorable, employed to satisfy task constraints at the expense of energy expenditure",
            " But if task constraints do necessitate raising and lowering of the COM, this leads naturally to the question of what the constraints are and why some displacement must occur. The propulsion and bobbing analogies, despite their apparent applicability to walking, raise more questions than they answer. A possible explanation for these issues is simply that a different analogy may apply better to human walking. We propose that inter-limb co-contraction in fact serves as a superior analogy compared to bobbing and propulsion, with the human legs during double support taking the place of the sliding actuators used in the example of Fig. 3d. The analogy applies well to human walking because the legs perform simultaneous positive and negative work during double support (Donelan, Kram, & Kuo, 2002b), with substantial metabolic cost (Donelan, Kram, & Kuo, 2002a). It is also apt because co-contraction is generally understood to be energetically costly and to be performed only when task constraints make it otherwise favorable. There nevertheless remain the questions of why the negative work cannot be avoided, and what task constraints require that it be performed",
            "5 acting as the respective normalization factors). As with any zero-work task, there is an equal amount of negative work performed. During single support, this negative work is performed by the stance leg prior to mid-stance, followed by positive work after mid-stance. During double support, negative work is performed by the leading leg, simultaneous with the positive work of the trailing leg. The level COM path therefore requires positive and negative work that cancel both in succession (similar to bobbing in Fig. 3b) and simultaneously through inter-limb co-contraction (similar to Fig. 3d). This mechanical work rate can also predict a minimum metabolic rate. At a nominal walking speed of 1.25, humans prefer a normalized step length of about 0.7. Assuming that positive work is performed at an efficiency of 25% and that negative work is performed at 120%, the resulting energetic cost of transport is MR = 0.18. This is more than double that observed in humans. Of course, not only do humans walk with lower energetic cost, they also clearly do not transport the COM on a level path. The simple model merely illustrates the obvious fact that a level path is energetically uneconomical",
            " This activity, however, would still be expected to cost energy. Net economy may be improved by combining push-off with active emulation of a hip spring in an optimal manner. This depends on the energetic cost of emulating a spring. One obvious candidate for the cost of emulating a spring is the within-cycle work performed on the leg. Even though zero net work is required to move the legs through a complete cycle, positive and negative work must be performed on the leg within a cycle, somewhat analogous to the bobbing motion (Fig. 3b). There may be a metabolic cost associated with this within-cycle work. Keeping the amplitude of leg motion fixed, the rate of within-cycle, positive work performed on the leg is predicted to increase with f 3s2 (Kuo, 2001), where f is the normalized frequency of swinging. This cost, however, would not appear to explain the actual, energetically optimum combination of step length and frequency preferred by humans (Elftman, 1966; Grieve & Gear, 1966). Summed with stepto-step transition costs, it would tend to favor a smaller change in step length as a function of speed than is observed (Kuo, 2001)",
            " The single support phase would occur normally, but the step-to-step redirection could be performed by the spring rather than the legs, with the spring resisting and then reversing the downward motion of the body. Here, negative and then positive work is performed, but unlike the two separate legs, the spring by itself can re-use negative work. Walking has often been interpreted in terms of COM propulsion in the forward direction, or lifting in the upward direction. The propulsion interpretation appears to be rooted in an assumption that walking resembles tasks with an external load, such as pedaling or rowing (Fig. 3a). The lifting interpretation appears to be based on a different assumption, that walking resembles active bobbing of the body up and down (Fig. 3b). In contrast, the dynamic walking perspective is that walking bears greater resemblance to an inter-limb cocontraction task (Fig. 3c and d), as exemplified by the flying ball analogy (Fig. 7). The propulsion interpretation is superficially appealing, because humans are certainly conscious of the exertion of pushing off during walking. Experience with skating or skiing also seems to apply, so that push-off is interpreted in terms of propelling or horizontally accelerating the COM. The propulsion interpretation is not technically wrong, because push-off does include a component of force in the forward direction. Propulsion is, however, somewhat misleading, because in human walking, the propulsion is not performed against external drag",
            " However, the average vertical force produced by the trailing leg during push-off is actually below body weight, and it is only in combination with the leading leg that the total vertical force exceeds body weight. It is therefore only with both legs that the COM can be lifted, if that interpretation were based on the amount of force. However, even this interpretation is suspect, because over the duration of double support, the net change in COM height is nearly zero. As with the flying ball analogy, the COM ends single support with an appropriate height, momentum, and energy. It requires only redirection and not lifting. The lifting interpretation is apparently based on the vertical bobbing analogy (Fig. 3b). A natural conclusion would be that energy expenditure could be reduced simply by reducing the amount of vertical displacement. The lifting interpretation therefore appears to favor a flattened COM trajectory, just as recommended by the six determinants of gait theory but contradicted by experimental evidence. It is better to interpret push-off in terms of the net action of the two legs together. The step-to-step transition can produce redirection of COM velocity with no net height change, acceleration, or work",
            ", 1963; Donelan et al., 2002b) explain that the inverted pendulum conserves energy by exchanging kinetic and gravitational potential energy. Closer inspection reveals this to be a correct description, but nevertheless a poor explanation. This is because all rigid bodies can exchange kinetic and gravitational potential energy when moving. For example, a car rolling up a hill, or any linkage of bodies flying through the air, will also exchange the two types of energy. A person bobbing up and down (Fig. 3b) will not only exchange the two types of mechanical energy, but also require expenditure of considerable metabolic energy. Most relevant is therefore whether work must be performed on a system in order to make it move. Fortunately, the inverted pendulum is indeed one means of conserving mechanical energy without need for work, but it is also not the only means. Mechanical energy may also be conserved without exchanging different types of energy, as in a frictionless car rolling on the level, where kinetic energy is constant"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure3.29-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure3.29-1.png",
        "caption": "Fig. 3.29. Open kinematic chain under action of external force FA and moment M A",
        "texts": [
            " For instance, in [3] the dynamic analysis and control synthesis using force feed back in the peg-in-hole assembly task are presented. In [4] the complete dy namic analysis for non-stationary constraint is derived and one example of the assembly simulation is given. 3.4.2 Mathematical Model of Manipulator with Constraints of Gripper Motion In order to present the mathematical model of the robot closed configuration we consider first a manipulator as an open kinematic chain. Let us apply an external force F A acting on the gripper at point A and external moment if A (Fig. 3.29). We assume that the mechanism is a set of n segments with n d~grees of freedom. Also, we assume that the mechanism position is determined by a vector of generalized coordinates, q. In a completely analogous way as 3.4 Dynamics of Robots Under Action of External Reaction Forces 93 presented in Sect. (3.2) the dynamic model of such a mechanism can be presented in the following matrix form (3.64) where H =H(q):(n x n) is the inertial matrix, h = h(q, 4):(n x 1) the vector con sisting of gravity, centrifugal and Coriolis forces, P=P(t):(n x 1) the vector of driving torques (forces) in joints"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003786_1045389x10369718-Figure19-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003786_1045389x10369718-Figure19-1.png",
        "caption": "Figure 19. Schematics of the migration of hydrated counter-ions within ionic polymer metal composites due to an imposed electric field. Reproduced with permission from Institute of Physics Publishing Ltd, http://dx.doi.org/10.1088/0964-1726/10/4/327.",
        "texts": [
            " Perfluorinated alkenes and styrene-divinylbenzene copolymers are the typically used ion exchange polymers which can exchange ions of an environmental charge with their own ions. Styrene-divinylbenzene copolymers are highly cross-linked and rigid, which severely limit their applications as electroactive polymers. Perfluorinated alkenes have large polymer backbones and side chains. The side chains are terminated by sulfonate or carboxylate (for cation exchange) or ammonium cations (for anion exchange) to interact with a solvent to produce the electro-active effect. As shown in Figure 19, in the case of cation exchange polymer, the ionic polymer-metal composite bends toward the anode if the ionomeric polymer-metal strip is subject to DC voltage bias across the strip followed by a slow relaxation in the opposite direction (towards the cathode) (Bar-Cohen and Leary, 2000; Bao et al., 2002; Lavu et al., 2005; Lee et al., 2005; Oh and Jeon, 2006). The bending force of ionic polymer metal composites is from redistribution of hydrated ions and water. Nafion , (DuPont, perfluorosulfonic acid polymer) and Flemion (Asahi Glass Co"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure5.15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure5.15-1.png",
        "caption": "FIGURE 5.15. A rotating rigid body B(Oxyz) with a fixed point O in a reference frame G(OXY Z).",
        "texts": [
            "344) and solving for the position of the zero velocity point in the body coordinate frame Br0. Br 0 = \u2212GRT B G\u03c9\u0303 \u22121 B Gd\u0307B = \u2212GRT B h GR\u0307B GRT B i\u22121 Gd\u0307B = \u2212GRT B h GRB GR\u0307\u22121B i Gd\u0307B = \u2212GR\u0307\u22121B Gd\u0307B (5.345) Therefore, Gr0 indicates the path of motion of the pole in the global frame, while Br0 indicates the same path in the body frame. The Gr0 refers to the Lagrangian centroid and Br0 refers to the Eulerian centroid. 5.10 Angular Acceleration Consider a rotating rigid body B(Oxyz) with a fixed point O in a reference frame G(OXY Z) as shown in Figure 5.15. Equation (5.172), for the velocity vector of a point in a fixed origin body frame, Gr\u0307(t) = Gv(t) = G\u03c9\u0303B Gr(t) = G\u03c9B \u00d7 Gr(t) (5.346) 5. Applied Kinematics 273 can be utilized to find the acceleration vector of the body point Gr\u0308 = Gd dt Gr\u0307(t) = G\u03b1B \u00d7 Gr+ G\u03c9B \u00d7 \u00a1 G\u03c9B \u00d7 Gr \u00a2 (5.347) = \u00b3 \u03c6\u0308u\u0302+ \u03c6\u0307u\u0307 \u00b4 \u00d7 Gr+ \u03c6\u0307 2 u\u0302\u00d7 \u00a1 u\u0302\u00d7 Gr \u00a2 . (5.348) G\u03b1B is the angular acceleration vector of the body with respect to the G frame. G\u03b1B = Gd dt G\u03c9B (5.349) Proof. Differentiating Equation (5.346) gives Gr\u0308 = G\u03c9\u0307B \u00d7 Gr+ G\u03c9B \u00d7 Gr\u0307 = G\u03b1B \u00d7 Gr+ G\u03c9B \u00d7 \u00a1 G\u03c9B \u00d7 Gr \u00a2 (5"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.15-1.png",
        "caption": "FIGURE 8.15. A Robert suspension mechanism with a Panhard arm.",
        "texts": [],
        "surrounding_texts": [
            "The suspension is what links the wheels to the vehicle body and allows relative motion. This chapter covers the suspension mechanisms, and discusses the possible relative motions between the wheel and the vehicle body. The wheels, through the suspension linkage, must propel, steer, and stop the vehicle, and support the associated forces. 8.1 Solid Axle Suspension The simplest way to attach a pair of wheels to a vehicle is to mount them at opposite ends of a solid axle, such as the one that is shown in Figure 8.1. The solid axle must be attached to the body such that an up and down motion in the z-direction, as well as a roll rotation about the x-axis, is possible. So, no forward and lateral translation, and also no rotation about the axle and the z-axis, is allowed. There are many combinations of links and springs that can provide the kinematic and dynamic requirements. The simplest design is to clamp the axle to the middle of two leaf springs with their ends tied or shackled to the vehicle frame as shown schematically in Figure 8.1. A side view of a multi-leaf spring and solid axle is shown in Figure 8.2. A suspension with a solid connection between the left and right wheels is called dependent suspension. 456 8. Suspension Mechanisms The performance of a solid axle with leaf springs suspension can be improved by adding a linkage to guide the axle kinematically and provide dynamic support to carry the non z-direction forces. The solid axle with leaf spring combination came to vehicle industry from horse-drawn vehicles. Example 299 Hotchkiss drive. When a live solid axle is connected to the body with nothing but two leaf springs, it is called the Hotchkiss drive, which is the name of the car that used it first. The main problems of a Hotchkiss drive, which is shown in Figure 8.2, are locating the axle under lateral and longitudinal forces, and having a low mass ratio \u03b5 = ms/mu, where ms is the sprung mass and mu is the unsprung mass. Sprung mass refers to all masses that are supported by the spring, such as vehicle body. Unsprung mass refers to all masses that are attached to and not supported by the spring, such as wheel, axle, or brakes. Example 300 Leaf spring suspension and flexibility problem. The solid axle suspension systems with longitudinal leaf springs have many drawbacks. The main problem lies in the fact that springs themselves act as locating members. Springs are supposed to flex under load, but their flexibility is needed in only one direction. However, it is the nature of leaf springs to twist and bend laterally and hence, flex also in planes other than the tireplane. Leaf springs are not suited for taking up the driving and braking traction forces. These forces tend to push the springs into an S-shaped profile, as shown in Figure 8.3. The driving and braking flexibility of leaf springs, generates a negative caster and increases instability. Long springs provide better ride. However, long sprigs exaggerate their bending and twisting under different load conditions. Example 301 Leaf spring suspension and flexibility solution. To reduce the effect of a horizontal force and S-shaped profile appearance in a solid axle with leaf springs, the axle may be attached to the chassis by a longitudinal bar as Figure 8.4(a) shows. Such a bar is called an anti-tramp 8. Suspension Mechanisms 457 bar, and the suspension is the simplest cure for longitudinal problems of a Hotchkiss drive. A solid axle with an anti-tramp bar may be kinematically approximated by a four-bar linkage, as shown in Figure 8.4(b). Although an anti-tramp bar may control the shape of the leaf spring, it introduces a twisting angle problem when the axle is moving up and down, as shown in Figure 8.5. Twisting the axle and the wheel about the axle is called caster. The solid axle is frequently used to help keeping the wheels perpendicular to the road. Example 302 Leaf spring location problem. The front wheels need room to steer left and right. Therefore, leaf springs cannot be attached close to the wheel hubs, and must be placed closer to the middle of the axle. That gives a narrow spring-base, which means that a small side force can sway or tilt the body relative to the axle through a considerable roll angle due to weight transfer. This is uncomfortable for the vehicle passengers, and may also produce unwanted steering. The solid axle positively prevents the camber change by body roll. The wheels remain upright and hence, do not roll on a side. However, a solid axle shifts laterally from its static plane and its center does not remain on the vehicle\u2019s longitudinal axis under a lateral force. 458 8. Suspension Mechanisms A solid axle produces bump-camber when single-wheel bump occurs. If the right wheel goes over a bump, the axle is raised at its right end, and that tilts the left wheel hub, putting the left wheel at a camber angle for the duration of deflection. Example 303 Triangular linkage. A triangulated linkage, as shown in Figure 8.6, may be attached to a solid axle to provide lateral and twist resistance during acceleration and braking. Example 304 Panhard arm. High spring rate is a problem of leaf springs. Reducing their stiffness by narrowing them and using fewer leaves, reduces the lateral stiffness and increases the directional stability of the suspension significantly. A Panhard arm is a bar that attaches a solid axle suspension to the chassis laterally. Figure 8.7 illustrates a solid axle and a Panhard arm to guide the axle. Figure 8.8 shows a triangular linkage and a Panhard arm combination for guiding a solid axle. A double triangle mechanism, as shown in Figure 8.9, is an alternative design to guide the axle and support it laterally. Example 305 Straight line linkages. There are many mechanisms that can provide a straight line motion. The simplest mechanisms are four-bar linkages with a coupler point moving straight. Some of the most applied and famous linkages are shown in Figure 8.10. By having proper lengths, the Watt, Robert, Chebyshev, and Evance linkages can make the coupler point C move on a straight line vertically. Such a mechanism and straight motion may be used to guide a solid axle. Two Watt suspension mechanisms with a Panhard arm are shown in Figures 8.11 and 8.12. 460 8. Suspension Mechanisms 8. Suspension Mechanisms 461 Figures 8.13, 8.14, and 8.15 illustrate three combinations of Robert suspension linkages equipped with a Panhard arm. Example 306 Solid axle suspension and unsprung mass problem. A solid axle is counted as an unsprung member, and hence, the unsprung mass is increased where using solid axle suspension. A heavy unsprung mass ruins both, the ride and handling of a vehicle. Lightening the solid axle makes it weaker and increases the most dangerous problem in vehicles: axle breakage. The solid axle must be strong enough to make sure it will not break under any loading conditions at any age. As a rough estimate, 90% of the leaf spring mass may also be counted as unsprung mass, which makes the problem worse. The unsprung mass problem is worse in front, and it is the main reason that they are no longer used in street cars. However, front solid axles are still common on trucks and buses. These are heavy vehicles and solid axle 8. Suspension Mechanisms 463 464 8. Suspension Mechanisms suspension does not reduce the mass ratio \u03b5 = ms/mu very much. When a vehicle is rear-wheel-drive and a solid axle suspension is used in the back, the suspension is called live axle. A live axle is a casing that contains a differential, and two drive shafts. The drive shafts are connected to the wheel hubs. A live axle can be three to four times heavier than a dead I-beam axle. It is called live axle because of rotating gears and shafts inside the axle. Example 307 Solid axle and coil spring. To decrease the unsprung mass and increase vertical flexibility of solid axle suspensions, it is possible to equip them with coil springs. A sample of a solid axle suspension with coil spring is shown in Figure 8.16. The suspension mechanism is made of four longitudinal bars between the axle and chassis. The springs may have some lateral or longitudinal angle to introduce some lateral or longitudinal compliance. Example 308 De Dion axle. When a solid axle is a dead axle with no driving wheels, the connecting beam between the left and right wheels may have different shapes to do different jobs, usually to give the wheels independent flexibility. We may also modify the shape of a live axle to attach the differential to the chassis and reduce the unsprung mass. De Dion design is a modification of a beam axle that may be used as a dead axle or to attach the differential to the chassis and transfer the driving power to the drive wheels by employing universal joints and split shafts. Figure 8.17 illustrates a De Dion suspension. 8.2 Independent Suspension Independent suspensions is introduced to let a wheel to move up and down without affecting the opposite wheel. There are many forms and designs of independent suspensions. However, double A-arm and McPherson strut suspensions are the simplest and the most common designs. Figure 8.18 illustrates a sample of a double A-arm and Figure 8.19 shows a McPherson suspension. Kinematically, a double A-arm suspension mechanism is a four-bar linkage with the chassis as the ground link, and coupler as the wheel carrying link. A McPherson suspension is an inverted slider mechanism that has the chassis as the ground link and the coupler as the wheel carrying link. A double A-arm and a McPherson suspension mechanism on the left and right wheels are schematically shown in Figures 8.20 and 8.21 respectively. Double A-arm, is also called double wishbone, or short/long arm suspension. McPherson also may be written as MacPherson. Example 309 Double A-arm suspension and spring position. Consider a double A-arm suspension mechanism. The coil spring may be between the lower arm and the chassis, as shown in Figure 8.18. It is also possible to install the spring between the upper arm and the chassis, or between the upper and lower arms. In either case, the lower or the upper arm, which supports the spring, is made stronger and the other arm acts as a connecting arm. Example 310 Multi-link suspension mechanism. When the two side bars of an A-arm are attached to each other with a joint, as shown in Figure 8.22, then the double A-arm is called a multi-link mechanism. A multi-link mechanism is a six-bar mechanism that may have a better coupler motion than a double A-arm mechanism. However, multi- 466 8. Suspension Mechanisms 8. Suspension Mechanisms 467 link suspensions are more expensive, less reliable, and more complicated compare to a double A-arm four-bar linkage. There are vehicles with more than six-link suspension with possibly better kinematic performance. Example 311 Swing arm suspension. An independent suspension may be as simple as a triangle shown in Figure 8.23. The base of the triangle is jointed to the chassis and the wheel to the tip point. The base of the triangle is aligned with the longitudinal axis of the vehicle. Such a suspension mechanism is called a swing axle or swing arm. The variation in camber angle for a swing arm suspension is maximum, compared to the other suspension mechanisms. Example 312 Trailing arm suspension. Figure 8.24 illustrates a trailing arm suspension that is a longitudinal arm with a lateral axis of rotation. The camber angle of the wheel, supported by a trailing arm, will not change during the up and down motion. Trailing arm suspension has been successfully using in a variety of frontwheel-drive vehicles, to suspend their rear wheels. Example 313 Semi-trailing arm Semi-trailing arm suspension, as shown in Figure 8.25, is a compromise between the swing arm and trailing arm suspensions. The joint axis may have any angle, however an angle not too far from 45 deg is more applied. Such suspensions have acceptable camber angle change, while they can handle both, the lateral and longitudinal forces. Semi-trailing design has successfully applied to a series of rear-wheel-drive cars for several decades. Example 314 Antiroll bar and roll stiffness. Coil springs are used in vehicles because they are less stiff with better 470 8. Suspension Mechanisms ride comfort compared to leaf springs. Therefore, the roll stiffness of the vehicle with coil springs is usually less than in vehicles with leaf springs. To increase the roll stiffness of such suspensions, an antiroll bar must be used. Leaf springs with reduced layers, uni-leaf, trapezoidal, or nonuniform thickness may also need an antiroll bar to compensate for their reduced roll stiffness. The antiroll bar is also called a stabilizer. Figure 8.26 illustrates an anti-roll bare attached to a solid axle with coil springs. Example 315 Need for longitudinal compliance. A bump is an obstacle on the road that opposes the forward motion of a wheel. When a vehicle goes over a bump, the first action is a force that tends to push the wheel backward relative to the rest of the vehicle. So, the lifting force has a longitudinal component, which will be felt inside the vehicle unless the suspension system has horizontal compliance. There are situations in which the horizontal component of the force is even higher than the vertical component. Leaf springs can somewhat absorb this horizontal force by flattening out and stretching the distance from the forward spring anchor and the axle. Such a stretch is usually less than 1/2 in \u2248 1 cm. 8.3 Roll Center and Roll Axis The roll axis is the instantaneous line about which the body of a vehicle rolls. Roll axis is found by connecting the roll center of the front and rear suspensions of the vehicle. Assume we cut a vehicle laterally to disconnect the front and rear half of the vehicle. Then, the roll center of the front or rear suspension is the instantaneous center of rotation of the body with respect to the ground. 8. Suspension Mechanisms 471 472 8. Suspension Mechanisms Figure 8.27 illustrates a sample of the front suspensions of a car with a double A-arm mechanism. To find the roll center of the body with respect to the ground, we analyze the two-dimensional kinematically equivalent mechanism shown in Figure 8.28. The center of tireprint is the instant center of rotation of the wheel with respect to the ground, so the wheels are jointed links to the ground at their center of tireprints. The instant center I18 is the roll center of the body with respect to the ground. To find I18, we apply the Kennedy theorem and find the intersection of the line I12I28 and I13I38 as shown in Figure 8.29. The point I28 and I38 are the instant center of rotation for the wheels with respect to the body. The instant center of rotation of a wheel with respect to the body is called suspension roll center. So, to find the roll center of the front or rear half of a car, we should determine the suspension roll centers, and find the intersection of the lines connecting the suspension roll centers to the center of their associated tireprints. The Kennedy theorem states that the instant center of every three relatively moving objects are colinear. Example 316 McPherson suspension roll center. A McPherson suspension is an inverted slider crank mechanism. The instant centers of an example of inverted slider crank mechanism are shown in Figure 8.30. In this figure, the point I12 is the suspension roll center, which is the instant center of rotation for the wheel link number 2 with respect to the chassis link number 1. A car with a McPherson suspension system is shown in Figure 8.31. The kinematic equivalent mechanism is depicted in Figure 8.32. Suspension roll centers along with the body roll center are shown in Figure 8.33. To 8. Suspension Mechanisms 473 find the roll center of the front or rear half of a car, we determine each suspension roll center and then find the intersection of the lines connecting the suspension roll centers to the center of the associated tireprint. Example 317 Roll center of double A-arm suspension. The roll center of an independent suspension such as a double A-arm can be internal or external. The kinematic model of a double A-arm suspension for the front left wheel of a car is illustrated in Figure 8.34. The suspension roll center in Figure 8.34(a) is internal, and in Figure 8.34(b) is external. An internal suspension roll center is toward the vehicle body, while an external suspension roll center goes away from the vehicle body. A suspension roll center may be on, above, or below the road surface, as shown in Figure 8.35(a)-(c) for an external suspension roll center. When the suspension roll center is on the ground, above the ground, or below the ground, the vehicle roll center would be on the ground, below the ground, and above the ground, respectively. Example 318 F Camber variation of double A-arm suspension. When a wheel moves up and down with respect to the vehicle body, de- 474 8. Suspension Mechanisms 8. Suspension Mechanisms 475 pending on the suspension mechanism, the wheel may camber. Figure 8.36 illustrates the kinematic model for a double A-arm suspension mechanism. The mechanism is equivalent to a four-bar linkage with the ground link as the vehicle chasis. The wheel is always attached to a coupler point C of the mechanism. We set a local suspension coordinate frame (x, y) with the x-axis indicating the ground link MN . The x-axis makes a constant angle \u03b80 with the vertical direction. The suspension machanism has a length a for the upper A-arm, b for the coupler link, c for the lower A-arm, and d for the ground link. The configuration of the suspension is determined by the angles \u03b82, \u03b83, and \u03b84, all measured from the positive direction of the x-axis. When the suspension is at its equilibrium position, the links of the double A-arm suspension make initial angles \u03b820 \u03b830, and \u03b840 with the x-axis. The equilibrium position of a suspension is called the rest position. To determine the camber angle during the fluctuation of the wheel, we should determine the variation of the coupler angle \u03b83, as a function of vertical motion z of the coupler point C. Using \u03b82 as a parameter, we can find the coordinates (xC , yC) of the 476 8. Suspension Mechanisms coupler point C in the suspension coordinate frame (x, y) as xC = a cos \u03b82 + e cos (p\u2212 q + \u03b1) (8.1) yC = a sin \u03b82 + e sin (p\u2212 q + \u03b1) (8.2) where, q = tan\u22121 a sin \u03b82 d\u2212 a cos \u03b82 (8.3) p = tan\u22121 q 4b2f2 \u2212 (b2 + f2 \u2212 c2) 2 b2 + f2 \u2212 c2 (8.4) f = p a2 + d2 \u2212 2ad cos \u03b82. (8.5) The position vector of the coupler point is uC uC = xC \u0131\u0302+ yC j\u0302 (8.6) and the unit vector in the z-direction is u\u0302z = \u2212 cos \u03b80\u0131\u0302\u2212 sin \u03b80j\u0302. (8.7) Therefore, the displacement z in terms of xC and yC is: z = uC \u00b7 u\u0302z = \u2212xC cos \u03b80 \u2212 yC sin \u03b80 (8.8) 8. Suspension Mechanisms 477 The initial coordinates of the coupler point C and the initial value of z are: xC0 = a cos \u03b820 + e cos (p0 \u2212 q0 + \u03b1) (8.9) yC0 = a sin \u03b820 + e sin (p0 \u2212 q0 + \u03b1) (8.10) z0 = \u2212xC0 cos \u03b80 \u2212 yC0 sin \u03b80 (8.11) and hence, the vertical displacement of the wheel center can be calculated by h = z \u2212 z0 (8.12) The initial angle of the coupler link with the vertical direction is \u03b80 \u2212 \u03b830. Therefore, the camber angle of the wheel would be \u03b3 = (\u03b80 \u2212 \u03b83)\u2212 (\u03b80 \u2212 \u03b830) = \u03b830 \u2212 \u03b83 (8.13) The angle of the coupler link with the x-direction is equal to \u03b83 = 2 tan \u22121 \u00c3 \u2212E \u00b1 \u221a E2 \u2212 4DF 2D ! (8.14) where, D = J5 \u2212 J1 + (1 + J4) cos \u03b82 (8.15) E = \u22122 sin \u03b82 (8.16) F = J5 + J1 \u2212 (1\u2212 J4) cos \u03b82 (8.17) and J1 = d a (8.18) J2 = d c (8.19) J3 = a2 \u2212 b2 + c2 + d2 2ac (8.20) J4 = d b (8.21) J5 = c2 \u2212 d2 \u2212 a2 \u2212 b2 2ab . (8.22) Substituting (8.14) and (8.13), and then, eliminating \u03b82 between (8.13) and (8.8) provides the relationship between the vertical motion of the wheel, z, and the camber angle \u03b3. 478 8. Suspension Mechanisms Example 319 F Camber angle and wheel fluctuations. Consider the double A-arm suspension that is shown in Figure 8.36. The dimensions of the equivalent kinematic model are: a = 22.4 cm b = 22.1 cm c = 27.3 cm d = 17.4 cm \u03b80 = 24.3 deg (8.23) The coupler point C is at: e = 14.8 cm \u03b1 = 54.8 deg (8.24) If the angle \u03b82 at the rest position is at \u03b820 = 121.5 deg (8.25) then the initial angle of the other links are: \u03b830 = 18.36 deg \u03b840 = 107.32 deg (8.26) At the rest position, the coupler point is at: xC0 = \u221222.73 cm yC0 = 9.23 cm z0 = 16.92 cm (8.27) We may calculate h and \u03b3 by varying the parameter \u03b82. Figure 8.37 illustrates h as a function of the camber angle \u03b3. For this suspension mechanism, the wheel gains a positive camber when the wheel moves up, and gains a negative camber when the it moves down. The mechanism is shown in Figure 8.38, when the wheel is at the rest position and has a positive or a negative displacement. 8.4 F Car Tire Relative Angles There are four major wheel alignment parameters that affect vehicle dynamics: toe, camber, caster, and trust angle. 8. Suspension Mechanisms 479 8.4.1 F Toe When a pair of wheels is set so that their leading edges are pointed toward each other, the wheel pair is said to have toe-in. If the leading edges point away from each other, the pair is said to have toe-out. Toe-in and toe-out front wheel configurations of a car are illustrated in Figure 8.39. The amount of toe can be expressed in degrees of the angle to which the wheels are not parallel. However, it is more common to express the toein and toe-out as the difference between the track widths as measured at the leading and trailing edges of the tires. Toe settings affect three major performances: tire wear, straight-line stability, and corner entry handling. For minimum tire wear and power loss, the wheels on a given axle of a car should point directly ahead when the car is running in a straight line. Excessive toe-in causes accelerated wear at the outboard edges of the tires, while too much toe-out causes wear at the inboard edges. Toe-in increases the directional stability of the vehicle, and toe-out increases the steering response. Hence, a toe-in setting makes the steering function lazy, while a toe-out makes the vehicle unstable. With four wheel independent suspensions, the toe may also be set at the rear of the car. Toe settings at the rear have the same effect on wear, directional stability, and turn-in as they do on the front. However, we usually do not set up a rear-drive race car toed out in the rear, because of excessive instability. When driving torque is applied to the wheels, they pull themselves forward and try to create toe-in. Furthermore, when pushed down the road, a non-driven wheel or a braking wheel will tend to toe-out. Example 320 Toe-in and directional stability. Toe settings have an impact on directional stability. When the steering 480 8. Suspension Mechanisms 8. Suspension Mechanisms 481 wheel is centered, toe-in causes the wheels to tent to move along paths that intersect each other in front of the vehicle. However, the wheels are in balance and no turn results. Toe-in setup can increase the directional stability caused by little steering fluctuations and keep the car moving straight. Steering fluctuations may be a result of road disturbances. If a car is set up with toe-out, the front wheels are aligned so that slight disturbances cause the wheel pair to assume rolling directions that approach a turn. Therefore, toe-out encourages the initiation of a turn, while toe-in discourages it. Toe-out makes the steering quicker. So, it may be used in vehicles for a faster response. The toe setting on a particular car becomes a trade-off between the straight-line stability afforded by toe-in and the quick steering response by toe-out. Toe-out is not desirable for street cars, however, race car drivers are willing to drive a car with a little directional instability, for sharper turn-in to the corners. So street cars are generally set up with toe-in, while race cars are often set up with toe-out. Example 321 Toe-in and toe-out in the front and rear axles. Front toe-in: slower steering response, more straight-line stability, greater wear at the outboard edges of the tires. Front toe-zero: medium steering response, minimum power loss, minimum tire wear. Front toe-out: quicker steering response, less straight-line stability, greater wear at the inboard edges of the tires. Rear toe-in: straight-line stability, traction out of the corner, more steerability, higher top speed. 482 8. Suspension Mechanisms 8.4.2 F Caster Angle Caster is the angle to which the steering pivot axis is tilted forward or rearward from vertical, as viewed from the side. Assume the wheel is straight to have the body frame and the wheel frame coincident. If the steering axis is turned about the wheel yw-axis then the wheel has positive caster. If the steering axis is turned about the wheel \u2212yw-axis, then the wheel has negative caster. Positive and negative caster configurations on the front wheel of a car are shown in Figure 8.40. Negative caster aids in centering the steering wheel after a turn and makes the front tires straighten quicker. Most street cars are made with 4\u22126 deg negative caster. Negative caster tends to straighten the wheel when the vehicle is traveling forward, and thus is used to enhance straight-line stability. Example 322 Negative caster of shopping carts. The steering axis of a shopping cart wheel is set forward of where the wheel contacts the ground. As the cart is pushed forward, the steering axis pulls the wheel along, and because the wheel drags along the ground, it falls directly in line behind the steering axis. The force that causes the wheel to follow the steering axis is proportional to the distance between the steering axis and the wheel-to-ground contact point, if the caster is small. This distance is referred to as trail. The cars\u2019 steering axis intersects the ground at a point in front of the tireprint, and thus the same effect as seen in the shopping cart casters is achieved. While greater caster angles improves straight-line stability, they also cause an increase in steering effort. Example 323 Characteristics of caster in front axle. Zero castor provides: easy steering into the corner, low steering out of the corner, low straight-line stability. 8. Suspension Mechanisms 483 Negative caster provides: low steering into the corner, easy steering out of the corner, more straight-line stability, high tireprint area during turn, good turn-in response, good directional stability, good steering feel. When a castered wheel rotates about the steering axis, the wheel gains camber. This camber is generally favorable for cornering. 8.4.3 F Camber Camber is the angle of the wheel relative to vertical line to the road, as viewed from the front or the rear of the car. Figure 8.41 illustrates the wheel number 1 of a vehicle. If the wheel leans in toward the chassis, it is called negative camber and if it leans away from the car, it is called positive camber. The cornering force that a tire can develop is highly dependent on its angle relative to the road surface, and so wheel camber has a major effect on the road holding of a car. A tire develops its maximum lateral force at a small camber angle. This fact is due to the contribution of camber thrust, which is an additional lateral force generated by elastic deformation as the tread rubber pulls through the tire/road interface. To optimize a tire\u2019s performance in a turn, the suspension should provide a slight camber angle in the direction of rotation. As the body rolls in a turn, the suspension deflects vertically. The wheel is connected to the chassis by suspension mechanism, which must rotate to allow for the wheel deflection. Therefore, the wheel can be subject to large camber changes as the suspension moves up and down. So, the more the wheel must deflect from its static position, the more difficult it is to maintain an ideal camber angle. Thus, the relatively large wheel travel and soft roll stiffness needed to provide a smooth ride in passenger cars presents a difficult design challenge, while the small wheel travel and high roll stiffness inherent in racing cars reduces the problem. Example 324 Castor versus camber. Camber doesn\u2019t improve turn-in as the positive caster does. Camber is not generally good for tire wear. Camber in one wheel does not improve directional stability. Camber adversely affects braking and acceleration efforts. 8.4.4 F Trust Angle The trust angle \u03c5 is the angle between vehicle\u2019s centerline and perpendicular to the rear axle. It compares the direction that the rear axle is aimed with the centerline of the vehicle. A nonzero angle configuration is shown in Figure 8.42. Zero angle confirms that the rear axle is parallel to the front axle, and the wheelbase on both sides of the vehicle are the same. A reason for nonzero 484 8. Suspension Mechanisms 8. Suspension Mechanisms 485 trust angle would have unequal toe-in or toe-out on both sides of the axle. Example 325 Torque reaction. There are two kinds of torque reactions in rear-whel-drive: 1\u2212 the reaction of the axle housing to rotate in the opposite direction of the crown wheel rotation, and 2\u2212 the reaction of axle housing to spin about its own center, opposite to the direction of pinion\u2019s rotation. The first reaction leads to a lifting force in the differential causing a wind-up in springs. The second reaction leads to a lifting force on the right wheels. 8.5 Suspension Requirements and Coordinate Frames The suspension mechanism should allow a relative motion between the wheel and the vehicle body. The relative motions are needed to pass the road irregularities and steering. To function properly, a suspension mechanism should have some kinematic and dynamics requirements. 8.5.1 Kinematic Requirements To express the motions of a wheel, we attach a wheel coordinate system W (oxwywzw) to the center of the wheel. A wheel, as a rigid body, has six degrees-of-freedom with respect to the vehicle body: three translations and three rotations, as shown in Figure 8.43. 486 8. Suspension Mechanisms The axes xw, yw, and zw indicate the direction of forward, lateral, and vertical translations and rotations. In the position shown in the figure, the rotation about the xw-axis is the camber angle, about the yw-axis is the spin, and about the zw-axis is the steer angle. Consider a non-steerable wheel. Translation in zw-direction and spin about the yw-axis are the only two DOF allowed for such a wheel. So, we need to take four DOF. If the wheel is steerable, then translation in the zw-direction, spin about the yw-axis, and steer rotation about the zw-axis are the three DOF allowed. So, we must take three DOF of a steerable wheel. Kinematically, non-steerable and steerable wheels should be supported as shown in Figures 8.44 and 8.45 respectively. Providing the required freedom, as well as eliminating the taken DOF, are the kinematic requirements of a suspension mechanism. 8.5.2 Dynamic Requirements Wheels should be able to propel, steer, and stop the vehicle. So, the suspension system must transmit the driving traction and deceleration braking forces between the vehicle body and the ground. The suspension members must also resist lateral forces acting on the vehicle. Hence, the wheel suspension system must make the wheel rigid for the taken DOF. However, there must also be some compliance members to limit the untaken DOF. The most important compliant members are spring and dampers to provide returning and resistance forces in the z-direction. 8. Suspension Mechanisms 487 8.5.3 Wheel, wheel-body, and tire Coordinate Frames Three coordinate frames are employed to express the orientation of a tire and wheel with respect to the vehicle: the wheel frame W , wheel-body frame C, and tire frame T . A wheel coordinate frame W (xw, yw, zw) is attached to the center of a wheel. It follows every translation and rotation of the wheel except the spin. Hence, the xw and zw axes are always in the tire-plane, while the yw-axis is always along the spin axis. A wheel coordinate frame is shown in Figure 8.43. When the wheel is straight and the W frame is parallel to the vehicle coordinate frame, we attach a wheel-body coordinate frame C (xc, yc, zc) at the center of the wheel parallel to the vehicle coordinate axes. The wheelbody frame C is motionless with respect to the vehicle coordinate and does not follow any motion of the wheel. The tire coordinate frame T (xt, yt, zt) is set at the center of the tireprint. The zt-axis is always perpendicular to the ground. The xt-axis is along the intersection line of the tire-plane and the ground. The tire frame does not follows the spin and camber rotations of the tire however, it follows the steer angle rotation about the zc-axis. Figure 8.46 illustrates a tire and a wheel coordinate frames. Example 326 Visualization of the wheel, tire, and wheel-body frames. Figure 8.47 illustrates the relative configuration of a wheel-body frame C, a tire frame T , and a wheel frame W . If the steering axis is along the zc-axis then, the rotation of the wheel about the zc-axis is the steer angle \u03b4. Rotation about the xt-axis is the camber angle \u03b3. 488 8. Suspension Mechanisms Generally speaking, the steering axis may have any angle and may go through any point of the ground plane. Example 327 Wheel to tire coordinate frame transformation. If TdW indicates the T -expression of the position vector of the wheel frame origin relative to the tire frame origin, then having the coordinates of a point P in the wheel frame, we can find its coordinates in the tire frame using the following equation. T rP = TRW W rP + TdW (8.28) If W rP indicates the position vector of a point P in the wheel frame, W rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 (8.29) then the coordinates of the point P in the tire frame T rP are T rP = TRW W rP + Td = TRW W rP + TRW W T dW = \u23a1\u23a3 xP yP cos \u03b3 \u2212Rw sin \u03b3 \u2212 zP sin \u03b3 Rw cos \u03b3 + zP cos \u03b3 + yP sin \u03b3 \u23a4\u23a6 (8.30) 8. Suspension Mechanisms 489 where, WT dW is the W -expression of the position vector of the wheel frame in the tire frame, Rw is the radius of the tire, and TRW is the rotation matrix to go from the wheel frame W to the tire frame T . TRW = \u23a1\u23a3 1 0 0 0 cos \u03b3 \u2212 sin \u03b3 0 sin \u03b3 cos \u03b3 \u23a4\u23a6 (8.31) W T dW = \u23a1\u23a3 0 0 Rw \u23a4\u23a6 . (8.32) As an example, the center of the wheel W rP = W ro = 0 is the origin of the wheel frame W , that is at T ro = TdW = TRW W T dW = \u23a1\u23a3 0 \u2212Rw sin \u03b3 Rw cos \u03b3 \u23a4\u23a6 (8.33) in the tire coordinate frame T . 490 8. Suspension Mechanisms Example 328 F Tire to wheel coordinate frame transformation. If rP indicates the position vector of a point P in the tire coordinate frame, T rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 (8.34) then the position vector W rP of the point P in the wheel coordinate frame is W rP = WRT T rP \u2212 W T dW (8.35) = \u23a1\u23a3 xP yP cos \u03b3 + zP sin \u03b3 zP cos \u03b3 \u2212Rw \u2212 yP sin \u03b3 \u23a4\u23a6 because WRT = \u23a1\u23a3 1 0 0 0 cos \u03b3 sin \u03b3 0 \u2212 sin \u03b3 cos \u03b3 \u23a4\u23a6 (8.36) WdT = \u23a1\u23a3 0 0 Rw \u23a4\u23a6 (8.37) and we may multiply both sides of Equation (8.28) by TRT W to get TRT W T rP = W rP + TRT W TdW (8.38) = W rP + W T dW W rP = WRT T rP \u2212 W T dW . (8.39) As an example, the center of tireprint in the wheel frame is at W rP = \u23a1\u23a3 1 0 0 0 cos \u03b3 \u2212 sin \u03b3 0 sin \u03b3 cos \u03b3 \u23a4\u23a6T \u23a1\u23a3 0 0 0 \u23a4\u23a6\u2212 \u23a1\u23a3 0 0 Rw \u23a4\u23a6 = \u23a1\u23a3 0 0 \u2212Rw \u23a4\u23a6 . (8.40) Example 329 F Wheel to tire homogeneous transformation matrices. The transformation from the wheel to tire coordinate frame may also be expressed by a 4\u00d7 4 homogeneous transformation matrix TTW , T rP = TTW W rP = \u2219 TRW TdW 0 1 \u00b8 W rP (8.41) where TTW = \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 cos \u03b3 \u2212 sin \u03b3 \u2212Rw sin \u03b3 0 sin \u03b3 cos \u03b3 Rw cos \u03b3 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 . (8.42) 8. Suspension Mechanisms 491 The corresponding homogeneous transformation matrix WTT from the tire to wheel frame would be WTT = \u2219 WRT WdT 0 1 \u00b8 = \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 cos \u03b3 \u2212 sin \u03b3 0 0 sin \u03b3 cos \u03b3 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 . (8.43) It can be checked that WTT = TT\u22121W , using the inverse of a homogeneous transformation matrix rule. TT\u22121W = \u2219 TRW TdW 0 1 \u00b8\u22121 = \u2219 TRT W \u2212 TRT W TdW 0 1 \u00b8 = \u2219 WRT \u2212WRT TdW 0 1 \u00b8 (8.44) Example 330 F Tire to wheel-body frame transformation. The origin of the tire frame is at CdT in the wheel-body frame. CdT = \u23a1\u23a3 0 0 \u2212Rw \u23a4\u23a6 (8.45) The tire frame can steer about the zc-axis with respect to the wheel-body frame. The associated rotation matrix is CRT = \u23a1\u23a3 cos \u03b4 \u2212 sin \u03b4 0 sin \u03b4 cos \u03b4 0 0 0 1 \u23a4\u23a6 (8.46) Therefore, the transformation between the tire and wheel-body frames can be expressed by Cr = CRT T r+ CdT (8.47) or equivalently, by a homogeneous transformation matrix CTT . CTT = \u2219 CRT CdT 0 1 \u00b8 = \u23a1\u23a2\u23a2\u23a3 cos \u03b4 \u2212 sin \u03b4 0 0 sin \u03b4 cos \u03b4 0 0 0 0 1 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 (8.48) As an example, the wheel-body coordinates of the point P on the tread of a 492 8. Suspension Mechanisms negatively steered tire at the position shown in Figure 8.48, are: Cr = CTT T rP = \u23a1\u23a2\u23a2\u23a3 cos\u2212\u03b4 \u2212 sin\u2212\u03b4 0 0 sin\u2212\u03b4 cos\u2212\u03b4 0 0 0 0 1 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 \u23a1\u23a2\u23a2\u23a3 Rw 0 Rw 1 \u23a4\u23a5\u23a5\u23a6 = \u23a1\u23a2\u23a2\u23a3 Rw cos \u03b4 \u2212Rw sin \u03b4 0 1 \u23a4\u23a5\u23a5\u23a6 (8.49) The homogeneous transformation matrix for tire to wheel-body frame TTC is: TTC = CT\u22121T = \u2219 CRT CdT 0 1 \u00b8\u22121 = \u2219 CRT T \u2212CRT T CdT 0 1 \u00b8 = \u2219 CRT T \u2212 T CdT 0 1 \u00b8 = \u23a1\u23a2\u23a2\u23a3 cos \u03b4 sin \u03b4 0 0 \u2212 sin \u03b4 cos \u03b4 0 0 0 0 1 Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 (8.50) Example 331 F Cycloid. Assume that the wheel in Figure 8.48 is turning with angular velocity \u03c9 and has no slip on the ground. If the point P is at the center of the tireprint 8. Suspension Mechanisms 493 when t = 0, MrP = \u23a1\u23a3 0 0 \u2212Rw \u23a4\u23a6 (8.51) then we can find its position in the wheel frame at a time t by employing another coordinate frame M . The frame M is called the rim frame and is stuck to the wheel at its center. Because of spin, the M frame turns about the yw-axis, and therefore, the rotation matrix to go from the rim frame to the wheel frame is: WRM = \u23a1\u23a3 cos\u03c9t 0 sin\u03c9t 0 1 0 \u2212 sin\u03c9t 0 cos\u03c9t \u23a4\u23a6 (8.52) So the coordinates of P in the wheel frame are: W rP = WRM MrP = \u23a1\u23a3 \u2212Rw sin t\u03c9 0 \u2212Rw cos t\u03c9 \u23a4\u23a6 (8.53) The center of the wheel is moving with speed vx = Rw\u03c9 and it is at Gr =\u00a3 vxt 0 Rw \u00a4 in the global coordinate frame G on the ground. Hence, the coordinates of point P in the global frame G, would be GrP = W rP + \u23a1\u23a3 vxt 0 Rw \u23a4\u23a6 = \u23a1\u23a3 Rw (\u03c9t\u2212 sin t\u03c9) 0 Rw (1\u2212 cos t\u03c9) \u23a4\u23a6 . (8.54) The path of motion of point P in the (X,Z)-plane can be found by eliminating t between X and Z coordinates. However, it is easier to expressed the path by using \u03c9t as a parameter. Such a path is called cycloid. In general case, point P can be at any distance from the center of the rim frame. If the point is at a distance d 6= Rw, then its path of motion is called the trochoid. A trochoid is called a curtate cycloid when d < Rw and a prolate cycloid when d > Rw. Figure 8.49(a)-(c) illustrate a cycloid, curtate cycloid, and prolate cycloid respectively. Example 332 F Wheel to wheel-body frame transformation. The homogeneous transformation matrix CTW to go from the wheel frame 494 8. Suspension Mechanisms to the wheel-body frame can be found by combined transformation. CTW = CTT TTW (8.55) = \u23a1\u23a2\u23a2\u23a3 c\u03b4 \u2212s\u03b4 0 0 s\u03b4 c\u03b4 0 0 0 0 1 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 c\u03b3 \u2212s\u03b3 \u2212Rw sin \u03b3 0 s\u03b3 c\u03b3 Rw cos \u03b3 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 = \u23a1\u23a2\u23a2\u23a3 cos \u03b4 \u2212 cos \u03b3 sin \u03b4 sin \u03b3 sin \u03b4 Rw sin \u03b3 sin \u03b4 sin \u03b4 cos \u03b3 cos \u03b4 \u2212 cos \u03b4 sin \u03b3 \u2212Rw cos \u03b4 sin \u03b3 0 sin \u03b3 cos \u03b3 Rw cos \u03b3 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 If rP indicates the position vector of a point P in the wheel coordinate frame, W rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 (8.56) then the homogeneous position vector CrP of the point P in the wheel-body 8. Suspension Mechanisms 495 coordinate frame is: CrP = CTW W rP = \u23a1\u23a2\u23a2\u23a3 xP cos \u03b4 \u2212 yP cos \u03b3 sin \u03b4 + (Rw + zP ) sin \u03b3 sin \u03b4 xP sin \u03b4 + yP cos \u03b3 cos \u03b4 \u2212 (Rw + zP ) cos \u03b4 sin \u03b3 \u2212Rw + (Rw + zP ) cos \u03b3 + yP sin \u03b3 1 \u23a4\u23a5\u23a5\u23a6 (8.57) The position of the wheel center W r = 0, for a cambered and steered wheel is at Cr = CTW W r = \u23a1\u23a2\u23a2\u23a3 Rw sin \u03b3 sin \u03b4 \u2212Rw cos \u03b4 sin \u03b3 \u2212Rw(1\u2212 cos \u03b3) 1 \u23a4\u23a5\u23a5\u23a6 (8.58) The zc = Rw (cos \u03b3 \u2212 1) indicates how much the center of the wheel comes down when the wheel cambers. If the wheel is not steerable, then \u03b4 = 0 and the transformation matrix CTW reduces to CTW = \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 cos \u03b3 \u2212 sin \u03b3 \u2212Rw sin \u03b3 0 sin \u03b3 cos \u03b3 Rw (cos \u03b3 \u2212 1) 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 (8.59) that shows CrP = CTW W rP = \u23a1\u23a2\u23a2\u23a3 xP yP cos \u03b3 \u2212Rw sin \u03b3 \u2212 zP sin \u03b3 zP cos \u03b3 + yP sin \u03b3 +Rw (cos \u03b3 \u2212 1) 1 \u23a4\u23a5\u23a5\u23a6 (8.60) Example 333 F Tire to vehicle coordinate frame transformation. Figure 8.50 illustrates the first and fourth tires of a 4-wheel vehicle. There is a body coordinate frame B (x, y, z) attached to the mass center C of the vehicle. There are also two tire coordinate frames T1 (xt1 , yt1 , zt1) and T4 (xt4 , yt4 , zt4) attached to the tires 1 and 4 at the center of their tireprints. The origin of the tire coordinate frame T1 is at Bd1 BdT1 = \u23a1\u23a3 a1 \u2212b1 \u2212h \u23a4\u23a6 (8.61)"
        ]
    },
    {
        "image_filename": "designv10_0_0003445_027836498900800301-Figure9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003445_027836498900800301-Figure9-1.png",
        "caption": "Fig. 9. Optical tactile element : surface re.flectance.",
        "texts": [
            " The finger can sense contacts perpendicular to the finger axis, radially, and also axial contact at the fingertip. Radial contact is detected using a mobile ring around the finger. Movement of the ring caused by contact is sensed using fiber-optic transmitter/receiver pairs as described above. Fingertip contact is also measured this way. Six fiber-optic cable pairs are evenly spaced around the mobile ring. A contact force of 10 g (0.098 N) is the threshold for detection. Three of these fingers were built and incorporated into a gripping system based upon contact data from the sensors. Figure 9 illustrates this transduction method. A planar tactile array sensor using a similar technique for contact sensing is described in Schneiter and Sheridan ( 1984). Fiber-optic transmitter/receiver pairs are arranged to form a matrix of sensing sites. On top of the fiber ends is a layer of clear Sylgard elastomer topped with a white silicon rubber layer. The amount of light reflected from the silicon layer is determined by its height above the fiber-optic cable ends. Any contacting objects will press into this elastomer and alter its height"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-FigureD.5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-FigureD.5-1.png",
        "caption": "FIGURE D.5 Single zero thickness conductor sheets.",
        "texts": [
            "20d) This formula is useful directly if one conductor can be approximated by a thin wire and also for the case where the second conductor is of a different shape. Please note that the \ud835\udf16 inside the brackets in equation (D.19) represents a very small number to prevent the singular behavior of the numerical evaluation of the equation. It is also useful for the case of a wide and and a narrow capacitive cell. D.1.6 Pp11 for Rectangular Sheet Cell The analytical Pp for a zero thickness rectangular cell shown in Fig. D.5 leads to a relatively simple formula. PARTIAL POTENTIAL COEFFICIENTS FOR ORTHOGONAL GEOMETRIES 415 The result is Pp11 = 1 6 \ud835\udf0b\ud835\udcc1\ud835\udf16 [ 3 log (u + \u221a u2 + 1) + u2 + 1 u + 3 u log ( 1 u + \u221a 1 u2 + 1 ) \u2212 ( u4\u22153 + 1 u2\u22153 )3\u22152 ] , (D.21) where u = \ud835\udcc1\u2215(ye1 \u2212 ys1) and \ud835\udcc1 = xe1 \u2212 xs1. Importantly, this result is computationally useful since it is simple. It also eliminates the singularity problem for the partial self-potential term. D.1.7 Pp12 for Two Parallel Rectangular Sheet Cells The geometry for two parallel conducting sheets is shown in Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure2.4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure2.4-1.png",
        "caption": "Fig. 2.4 a-d. Schematic representation of typical mechanical structures of robots",
        "texts": [
            " In prismatic joints the motion is translational (Figure 2.3). The mechanical structure of the mechanism depends on the type of the joints that are included in the given robot and the disposition of the joints. For example, the robot shown in Fig. 2.1 has six revolute degrees of freedom, with the second and the third always being in parallel (anthropomor phic manipulator). 2.2 Definitions 21 Different schematic representations have been introduced in order to de scribe manipulator configurations simply. Figure 2.4{aHd) shows several types of robots. Here, revolute joints are denoted by cylinders, while prismatic joints are represented by parallelepipeds. Manipulator link is a rigid body described by its kinematic and dynamic parameters. Figure 2.5 shows one typical manipulator link. Kinematic par ameters describing the link may be defined in different ways. Generally speaking, these parameters are the length of the manipulator link, the angle between the axes lying at two ends of the link, etc. Precise definitions of link parameters for the Denavit-Hartenberg kinematic notation will be presented in Sect"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure2.12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure2.12-1.png",
        "caption": "Fig. 2.12. Three-phase, nine-coil stator of a single-sided AFPM brushless machine with salient-pole stator. Photo-courtesy of Mii Technologies, Lebanon, NH, U.S.A.",
        "texts": [
            "1 \u03b8m, the number of winding sections is F = 6, the number of coils distributed in a coil group is z = 1, the angular coil span is 1.333\u03c0, the distribution factor is kd1 = 1 and the pitch factor is kp1 = sin[(1.333 \u2212 0.1333)\u03c0/2]sin(0.1333\u03c0/2)/(0.1333\u03c0/2) = 0.944. Fig. 1.5b shows a double layer, three-phase, 12-coil non-overlap stator winding of a double-sided AFPM machine with 2p = 8 rotor poles. Figures 1.6 and 2.11 show a single layer, 9-coil (Qc = 9, s1 = 2Qc = 18) non-overlap winding, while Fig. 2.12 shows a double layer 9-coil (s1 = Qc = 9) non-overlap stator winding. The difference in the number of the stator coils and rotor poles provides the starting torque for motors and the reduction of torque pulsations. Since the dimensions of AFPM machines are functions of the radius, the electromagnetic torque is produced over a continuum of radii, not just at a constant radius as in cylindrical machines. The pole pitch \u03c4(r) and pole width bp(r) of an axial flux machine are functions of the radius r, i"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure10.10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure10.10-1.png",
        "caption": "FIGURE 10.10. A front-wheel-steering four-wheel vehicle and the forces in the xy-plane acting at the trireprints.",
        "texts": [
            " However, because of tire flexibility, the velocity vector of the tire is lazier than the heading and turns by a \u03b2 angle, where \u03b2 < \u03b4. So, a positive steer angle generates a negative sideslip angle. Analysis of Figure 10.9(b) and using the definition for positive direction of the angles, shows that under a practical situation we have the same relation (10.104). According to (3.131), the existence of a sideslip angle is sufficient to generate a lateral force Fy, which is proportional to \u03b1 when the angle is small. Fy = \u2212C\u03b1 \u03b1 (10.107) 10.3.3 Two-wheel Model and Body Force Components Figure 10.10 illustrates the forces in the xy-plane acting at the tireprints of a front-wheel-steering four-wheel vehicle. When we ignore the roll motion of the vehicle, the xy-plane remains parallel to the road\u2019s XY -plane, and we may use a two-wheel model for the vehicle. Figure 10.11 illustrates a two-wheel model for a vehicle with no roll motion. The two-wheel model is also called a bicycle model, although a two-wheel model does not act similar to a bicycle. The force system applied on a two-wheel vehicle, in which only the front wheel is steerable, is Fx = Fxf cos \u03b4 + Fxr \u2212 Fyf sin \u03b4 (10"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.30-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.30-1.png",
        "caption": "FIGURE 8.30. Instant center or rotation for an example of inverted slider crank mechanism.",
        "texts": [
            " So, to find the roll center of the front or rear half of a car, we should determine the suspension roll centers, and find the intersection of the lines connecting the suspension roll centers to the center of their associated tireprints. The Kennedy theorem states that the instant center of every three relatively moving objects are colinear. Example 316 McPherson suspension roll center. A McPherson suspension is an inverted slider crank mechanism. The instant centers of an example of inverted slider crank mechanism are shown in Figure 8.30. In this figure, the point I12 is the suspension roll center, which is the instant center of rotation for the wheel link number 2 with respect to the chassis link number 1. A car with a McPherson suspension system is shown in Figure 8.31. The kinematic equivalent mechanism is depicted in Figure 8.32. Suspension roll centers along with the body roll center are shown in Figure 8.33. To 8. Suspension Mechanisms 473 find the roll center of the front or rear half of a car, we determine each suspension roll center and then find the intersection of the lines connecting the suspension roll centers to the center of the associated tireprint"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001609_j.ijmecsci.2010.05.005-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001609_j.ijmecsci.2010.05.005-Figure2-1.png",
        "caption": "Fig. 2. (a) Contact in the plane normal to plane of rotation and (b) contact in the plane of rotation.",
        "texts": [
            " The contact force is calculated using the Hertzian contact deformation theory. According to the Hertzian contact deformation theory, the non-linear relation load\u2013deformation is given by [10] F \u00bc Kdn r \u00f01\u00de where K is the load\u2013deflection factor or constant for Hertzian contact elastic deformation, dr the radial deflection or contact deformation and n the load\u2013deflection exponent; n\u00bc3/2 for ball bearing and 10/9 for roller bearing. The load\u2013deflection factor K depends on the contact geometry. The ball and the raceway contact are as shown in Fig. 2(a) and (b). Total deflection between two raceways is the sum of the approaches between the rolling elements and each raceway. Using this we get K \u00bc 1 1=Ki 1=n \u00fe 1=Ko 1=n \" #n \u00f02\u00de Ki and Ko inner and outer raceways to ball contact stiffness, respectively, which are obtained using Kp \u00bc 2:15 105 X r 1=2\u00f0d \u00de 3=2 \u00f03\u00de where P r is the curvature sum which is calculated using the radii of curvature in a pair of principal planes passing through the point contact. dn is the dimensionless contact deformation obtained using curvature difference (Refer Table 6"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure5.26-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure5.26-1.png",
        "caption": "FIGURE 5.26 Two-bar example for a retarded partial inductance computation.",
        "texts": [
            " In this section, we consider the retardation issue from a partial inductance perspective. The retarded potential coefficient issues are quite similar, but they are not considered. 120 INDUCTANCE COMPUTATIONS As an intermediate solution between the quasistatic and the high- frequency solution, it may be sufficient to include the retardation from cell center to cell center. The maximum cell size of \ud835\udf06min\u221520 may limit the error to be sufficiently small if the fmax is small or if the conductors are small in size, unlike the example shown in Fig. 5.26. In the time domain, we mostly utilize the center-to-center delay. For the two conductors example in Fig. 5.26, we may have to further subdivide the conductors such that the cell size is small enough. The retardation time td in the time domain is simply a delay between two points in space r\ud835\udfcf and r\ud835\udfd0 as given in (2.11.1) as \ud835\udf0fd = R12 \u2215v, (5.55) where v is the velocity in the material and where in Fig. 5.26 R12 = \u221a (x1 \u2212 x2)2 + (y1 \u2212 y2)2 + (z1 \u2212 z2)2. (5.56) In air v = c, the speed of light, and in a lossless dielectric it is v = c\u2215 \u221a \ud835\udf16r, where \ud835\udf16r is the relative dielectric constant. The time domain solver will have to take the delay of the couplings into account, a feature that is missing in a conventional SPICE circuit solver in both the time and the frequency domains. For FW frequency domain solutions, the accuracy is improved if we also take the retardation inside the integral into account. The retardation is distributed over the areas of the cells",
            "28 that the starting points of the two wires are in general much farther apart, leading to a large delay factor e\u2212j\ud835\udefdb2 , which is for most cases much larger than the small ones for the evaluation due to the wires segments with a length of \ud835\udf06\u221520 or smaller. Also, we can view the second term as a correction factor to the conventional solution where the quasistatic solution is multiplied by the delay factor e\u2212j\ud835\udefdb2 Lp12. An example where this approach is used is given in Section 13.5.6. We again refer to the example in Fig. 5.26, where we want to compute the partial inductance with retardation. If the two conductors are sufficiently separated, another integration method based on local expansion and far-field translation can be employed. In general, rewrite the partial inductance in the following simplified format: LpR 1,2(s) = \ud835\udf070 4\ud835\udf0b12 \u222b ze1 zs1 \u222b ye1 ys1 \u222b xe1 xs1 \u222b ze2 zs2 \u222b ye2 ys2 \u222b xe2 xs2 e\u2212j\ud835\udefdR12 R12 dx2 dy2 dz2 dx1 dy1 dz1. (5.69) 124 INDUCTANCE COMPUTATIONS If the center of conductor 1 is ro1, the center of conductor 2 is ro2, the vector D = ro2 \u2212 ro1"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001205_s10846-016-0442-0-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001205_s10846-016-0442-0-Figure5-1.png",
        "caption": "Fig. 5 Reduction of full vehicle model into bicycle model and its associated kinematic properties",
        "texts": [
            " [ vX vY ] = [ cos \u03c8 \u2212 sin \u03c8 sin \u03c8 cos \u03c8 ] [ vx vy ] where vx = v cos \u03b8 vy = v sin \u03b8 (2) A simpler kinematic model can be found in which the vehicle is modelled a bicycle. With this, responses from each wheel is simplified to be only one wheel per axle [10, 20\u201323]. By assuming only one wheel per axle, this model neglect the effect of slip from each wheel to the direction of the vehicle. Therefore, for this model, direction of vehicle\u2019s velocity is the same as vehicle heading direction. Referring to notation in Fig. 5, \u03b8 is the vehicle heading with respect to the local coordinated and \u03c8 is the vehicle\u2019s orientation with respect to the global coordinates and r is the vehicle\u2019s yaw rate or \u03c8\u0307 as depicted kinematic model for the bicycle model is shown in Eq. 3.\u23a1 \u23a3 vX vY r \u23a4 \u23a6 = \u23a1 \u23a3 cos \u03c8 0 sin \u03c8 0 0 1 \u23a4 \u23a6 [ v \u03c8\u0307 ] (3) Another studies [24, 25] extended the kinematic bicycle model by including side slip angle of both front and rear wheel to account for slippery terrain the study needs to address. The equation is as given in Eq"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000741_iros.2003.1248880-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000741_iros.2003.1248880-Figure2-1.png",
        "caption": "Figure 2: Link configulation",
        "texts": [
            " We set the reference momentum as Pmf = 0, Le' = 0 taking adequate reference frame. By using a selection matrix S = E , (10) becomes where F;: is the desired foot speed in space. For a space robot with one leg of 6 D.O.F, matrix A of (11) becomes square. Therefore, (12) is CB = A-'., where E B is the base link speed obtained by the reaction of the leg motion. Finally, the joint speed is given by the following equation (same as (3)). 4 Calculation of the inertia matrices In this section we describe a method to calculate the inertia matrices Mg and Hg, which appears in Figure 2 shows a part of the robot links containing the joint j - 1 and joint j. We assume the robot can rotate its joints with specified speed under the proper control law like PD feedback. The joint j's rotation with speed b j yields additional linear and angular mw menta of (1). % P j = w j x (63 - r j ) k j , (13) Lj = sj X P , + l j W j , (14) wj U j 6 , , (15) - where ~j and uj are the position vector and the rotation axis vector of the joint j . h j , E, a n d i j mean mass, center of mass and inertia tensor of all link structure driven by the joint j"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001514_j.engfailanal.2015.06.004-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001514_j.engfailanal.2015.06.004-Figure5-1.png",
        "caption": "Fig. 5. Schematic of spur gear [32]: (a) z < 42, (b) z > 42.",
        "texts": [
            " Based on the model, the effects of a tooth root crack on the TVMS (see Fig. 4b) were analyzed, and the model was verified by an FE method. On the basis of Charri\u2019s method [30], Ma and Chen [31] computed the TVMS of a cracked gear pair, and analyzed the dynamic responses of the cracked gear system using a four-degree-of-freedom (4-DOF) model. Wan et al. [32] presented different TVMS calculation models for different root circle sizes, and how to select these models need judge whether the number of the teeth is less than (see Fig. 5a) or great than 42 (see Fig. 5b). Based on these models, they also analyzed the influences of the geometric transmission error, TVMS and bearing stiffness on the dynamic responses of a cracked gear rotor system. Liang et al. [33,34] proposed different mesh models for different root circle sizes, and these models are selected by judging the root circle diameter smaller (see Fig. 6a) or greater than the base circle (see Fig. 6b). Moreover they also analyzed the TVMS of a cracked planetary gear pair based on the potential energy method"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure10.4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure10.4-1.png",
        "caption": "FIGURE 10.4. A six-wheel passenger car and its wheel numbering.",
        "texts": [
            " We start numbering from the front left wheel as number 1, and then the front right wheel would be number 2. Numbering increases sequentially on the right wheels going to the back of the vehicle up to the rear right wheel. Then, we go to the left of the vehicle and continue numbering the wheels from the rear left toward the front. Each wheel is indicated by a position vector ri Bri = xii+ yij + zik (10.6) expressed in the body coordinate frame B. Numbering of a four wheel vehicle is shown in Figure 10.3. Example 375 Wheel numbers and their position vectors. Figure 10.4 depicts a six-wheel passenger car. The wheel numbers are indicated besides each wheel. The front left wheel is wheel number 1, and the front right wheel is number 2. Moving to the back on the right side, we count the wheels numbered 3 and 4. The back left wheel gets number 5, 586 10. Vehicle Planar Dynamics and then moving forward on the left side, the only unnumbered wheel is the wheel number 6. If the global position vector of the car\u2019s mass center is given by Gd = \u2219 XC YC \u00b8 (10.7) and the body position vectors of the wheels are Br1 = \u2219 a1 w/2 \u00b8 (10"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001236_j.euromechsol.2007.11.005-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001236_j.euromechsol.2007.11.005-Figure6-1.png",
        "caption": "Fig. 6. Schematic graph of breakage.",
        "texts": [
            " (1) describing the bending deflection expression, the computation of the summation have to be done taking into account the changed shape of the elementary cross sectional areas Ai and area moment of inertia Ii as shown in Fig. 7(A). Tooth deflections and gearmesh stiffness are then computed. A broken tooth on the addendum circle is also considered. To simplify the modelling of this type of damage, at each section Sb of the tooth, the shape of the breakage is approximated with straight line as shown in Fig. 6 defined by the height hb , the thickness tb , the width wb and the thickness {t1, t2, t3, . . . , etc.} of the tooth. The width of contact changes as the line of contact moves and is evaluated instantaneously by: W1 = W \u2212 wb (14) The tooth and gearmesh stiffness of the gear pair is also calculated according to Eqs. (1), (6), (9) and (11) by taking into account the geometric changes due to the tooth breakage and updating the values of width, cross sectional area and area moment of inertia (Fig. 7(B))"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003887_bf02721775-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003887_bf02721775-Figure1-1.png",
        "caption": "Fig . 1.",
        "texts": [
            "L]~, where 1 L = - r x V i is the angular-momentum operator. Since (7) is an equation of motion of a threedimensional harmonic oscillator with a strong spin-orbit coupling, it is easily solved by the usual polynomial method, and we get the following eigenvalues: (s) (1} E~u~ = M 2 - - + 2 n o ) 4- ~oJ , 2 for J = t d - 89 5w = M 24- ~ - 4- 2n~o + 3~oJ, for J = 1 89 . Thus the squared mass E 2 of our system is a linear function of the total angular momentum J and the quantum number n of the radial vibration. The results are illustrated in Fig. 1. As is seen from Fig. 1, the harmonic-oscillator levels for J = ld- 89 and J = l - - 89 are widely separated by the strong spin-orbit coupling, and along each ((trajectory >) the parities of the successive states are alternating as is indicated by \u2022 in Fig. 1. These level characters, the presence of strong spin-orbit coupling as is suggcsted by the Kycia-Riley relations (4), and the Gaussian nature of the wavc functions seem to make our model suitable as a starting point for developing models of particles and resonances. On such attcmpts we would like to discuss in anotller occasion. (~) T. I~YOIA and K. RILEy: Phys. ReV. Left . , 10, 266 (1963)."
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure9.15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure9.15-1.png",
        "caption": "Fig. 9.15. GM\u2019s Sequel fuel cell concept electric vehicle: (a) Front view of the AFPM wheel motor, (b) Back view of the AFPM wheel motor, (c) Sequel FCEV, (d) Rear AFPM wheel motor with electromechanical brake, rear steering actuator and integrated controlled damper. Courtesy of General Motors Corporation, U.S.A.",
        "texts": [
            " The high cost of producing hydrogen, as compared to hydrocarbon fuels, has been a major barrier to its commercialisation, which would also require major infrastructural changes such as build-up of a hydrogen production and distribution infrastructure. However, with the continuing technology development in this fast-moving field and increasing possibility for introduction of renewable energy sources into fuel supply path, costs are expected to decline to an extent within the next decade. 9.3 Ship Propulsion 295 Recently developed GM Sequel FCEV is powered by an induction motor located in the front compartment and two AFPM wheel motor on the rear. Figure 9.15 shows front and back view of the AFPM wheel motor and the mechanical integration detail. This powertrain arrangement demonstrated excellent vehicle performance, driveability and safety as well as potential of axial flux motor technology. GM Sequel FCEV achieved a record-breaking range of 300 mile on a single tank of hydrogen. Stators of large AFPM brushless motors with disc type rotors usually have three basic parts [48]: \u2022 aluminum cold plate; \u2022 bolted ferromagnetic core; \u2022 polyphase winding. 296 9 Applications The cold plate is a part of the frame and transfers heat from the stator to the heat exchange surface"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002769_j.cirpj.2021.04.010-Figure26-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002769_j.cirpj.2021.04.010-Figure26-1.png",
        "caption": "Fig. 26. Schematic representation of rolling process: (a) Vertical rolling, (b) Pinch rolling, (c) Profiled roller-type A, (d) curved-slotted roller-type B, (e) flat-slotted roller-type C. All dimensions are in mm [122].",
        "texts": [
            " 25) changed as per the layer top and lateral surface in this approach. It was reported that the maximum absolute error in width was reduced from 0.45 mm (without rolling) to 0.12 mm (rolling applied). On the other hand, 413 S. Pattanayak and S.K. Sahoo CIRP Journal of Manufacturing Science and Technology 33 (2021) 398\u2013442 the YS increased to 547.9 MPa from 526.7 MPa, UTS increased to 816.7 MPa from 765.9 MPa, and elongation increased to 40.7% from 37.9%. Both pinch and vertical rolling approach (Fig. 26) was reported [122] for studying the effect of rolling parameters (roller depth, curvature depth, roller shape, rolling direction, and roller thickness) on the residual stress distribution. Higher will be the magnitude of compressive residual stress when roller depth and curvature depth are high. That is why a curve-slotted roller A schematic representation of forging-coupled GMAW-AM is depicted in Fig. 27. In this approach, after deposition of each bead, a reciprocating hammer applies a definite magnitude of force on these layers"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.5-1.png",
        "caption": "FIGURE 8.5. An anti-tramp bar introduces a twistng angle problem. (a) The wheel moves up and (b) The wheel moves down.",
        "texts": [
            " To reduce the effect of a horizontal force and S-shaped profile appearance in a solid axle with leaf springs, the axle may be attached to the chassis by a longitudinal bar as Figure 8.4(a) shows. Such a bar is called an anti-tramp 8. Suspension Mechanisms 457 bar, and the suspension is the simplest cure for longitudinal problems of a Hotchkiss drive. A solid axle with an anti-tramp bar may be kinematically approximated by a four-bar linkage, as shown in Figure 8.4(b). Although an anti-tramp bar may control the shape of the leaf spring, it introduces a twisting angle problem when the axle is moving up and down, as shown in Figure 8.5. Twisting the axle and the wheel about the axle is called caster. The solid axle is frequently used to help keeping the wheels perpendicular to the road. Example 302 Leaf spring location problem. The front wheels need room to steer left and right. Therefore, leaf springs cannot be attached close to the wheel hubs, and must be placed closer to the middle of the axle. That gives a narrow spring-base, which means that a small side force can sway or tilt the body relative to the axle through a considerable roll angle due to weight transfer"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure11.11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure11.11-1.png",
        "caption": "FIGURE 11.11 Basic PEEC loop that includes the coupling source for the magnetic coupling.",
        "texts": [
            " At this point, we can set up the augmented MNA system of equations by starting with the conventional form that also includes conductors as is shown in Fig. 11.8. Hence, we include the formulation in (6.55) with the additional magnetic inductance Lm and the source equation (11.65). The unknowns are \ud835\udebd, I, and M \u23a1\u23a2\u23a2\u23a2\u23a3 s Pp\u22121 \u2212A\ud835\udcc1 \ud835\udfce AT \ud835\udcc1 \u2212R \u2212 s Lp \u2212s Lm 0 Bc Bm \u2212 \ud835\udefc\u0302I \u23a4\u23a5\u23a5\u23a5\u23a6 \u23a1\u23a2\u23a2\u23a2\u23a3 \ud835\udebd I M \u23a4\u23a5\u23a5\u23a5\u23a6 = \u23a1\u23a2\u23a2\u23a2\u23a3 I0 \u2212V0 \u2212Binc \u23a4\u23a5\u23a5\u23a5\u23a6 . (11.66) where \u0302I is again the identity matrix to distinguish it from a current. We finally consider the equivalent circuit that corresponds to (11.66). The standard PEEC basic loop equivalent circuit in Fig. 11.11 includes an additional dependent voltage source. Further, the electrical and magnetic current densities equations are included, which control the dependent sources. Also, independent magnetic field sources are included. It is to be pointed out that we present a quasistatic version. However, it takes into account the electrical field coupling. This makes it suitable for a large variety of applications ranging from power systems to radio frequency integrated circuits (RFIC). Further, the model is well suited to be extended to incorporate nonlinear and materials with hysteresis"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure3.33-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure3.33-1.png",
        "caption": "Figure 3.33 The left section of the three-hinged frame ABC, as an example of a body subject to two forces at two points.",
        "texts": [
            " For a good insight into the behaviour of a structure, it is important to be able to quickly recognise three more or less particular cases of equilibrium. They are covered below. 1. A body subject to two forces at two points (see Figure 3.32). A body subject to two forces can be in equilibrium only if both forces: \u2022 have the same line of action, \u2022 have the same magnitude, and \u2022 have opposite directions. If these three conditions are not all met, the two forces together form either a resultant force or a resultant couple, and the system will not be in equilibrium. Figure 3.33 shows the left part AB of a so-called three-hinged frame. The foundation exerts a force FA at A on AB, while the right part BC of the frame exerts a force FB at B on AB. If we neglect the weight of the frame, the part AB of the frame is in equilibrium only if both forces FA and FB are equal and opposite, with AB as the common line of action. Certain construction elements are intentionally designed to this type of force transfer. These are straight bars, only subject to a force at both ends (see Figure 3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure3.60-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure3.60-1.png",
        "caption": "Figure 3.60 The forces (in kN) as they are acting on the cube.",
        "texts": [
            " Using this, one finds from equation (e) F2 = 5 6G = 20 kN and then from equation (a) F6 = \u2212 5 6G = \u221220 kN. ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM 3 Statics of a Rigid Body 97 Determining the forces F1, F2 and F3 from the three remaining equations (b), (d) and (f) demands some arithmetic. Sometimes one can reduce the amount of calculation by looking at the moment equilibrium about another point. Also here: \u2211 Tz|P = \u2212 ( F1 + 4 5F2 + 4 5F6 ) \u00b7 a = 0 (g) so that F1 = 0. Equation (d) now gives F3 = \u2212 1 6G = \u22124 kN. Finally, equation (f) gives F5 = \u2212 7 6G = \u221228 kN. Figure 3.60 depicts the forces (in kN) as they are acting on the cube in reality. The forces F3, F5 and F6 act in directions opposite to those shown in Figure 3.58. By using alternative equilibrium equation (g), equation (b) for the force equilibrium in y direction was not used, and can be used as a check. With the forces expressed in kN this gives \u2211 Fy = F1 + 4 5F2 + F3 + 4 5F4 + F5 + 4 5F6 = 0 + 4 5 \u00d7 20 \u2212 4 + 4 5 \u00d7 40 \u2212 28 \u2212 4 5 \u00d7 20 = 0. The conditions for force equilibrium in y direction are met. 98 Example 3 The cube that has been halved diagonally in Figure 3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002638_j.conengprac.2020.104560-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002638_j.conengprac.2020.104560-Figure2-1.png",
        "caption": "Fig. 2. The quadrotor and the cable suspended payload in wind environment.",
        "texts": [
            " For a matrix \ud835\udc68, notation \ud835\udc68\u22a4 denotes its transpose. For a vector \ud835\udc99, notation \u2016\ud835\udc99\u2016 denotes its Euclidean norm which is defined as \u2016\ud835\udc99\u2016= \u221a ( \ud835\udc99\ud835\udc47 \ud835\udc99 ) . \ud835\udc60\ud835\udc65 and \ud835\udc50\ud835\udc65 denote sin \ud835\udc65 and cos \ud835\udc65, respectively. The quadrotor dynamic system is a nonlinear, underactuated, and strong coupled system. The quadrotor UAV has six degrees of freedom and is actuated by four motors. Fig. 1 shows the quadrotor UAV flying under both cable suspended payload disturbance and wind disturbance. The configuration of a quadrotor UAV is illustrated in Fig. 2. Two coordinate systems are defined in this paper, namely, world-fixed frame  = [\ud835\udc861, \ud835\udc862, \ud835\udc863] and body-fixed frame  = [\ud835\udc831, \ud835\udc832, \ud835\udc833]. In addition, it is supposed that the origin of the body-fixed frame is located at the center of the quadrotor\u2019s mass. \ud835\udf38 = [\ud835\udc65, \ud835\udc66, \ud835\udc67]\u22a4 denotes the position of the quadrotor in the world-fixed frame and its linear velocity is represented by \ud835\udf42 = [?\u0307?, ?\u0307?, ?\u0307?]\u22a4 = [\ud835\udc62, \ud835\udc63,\ud835\udc64]\u22a4. \ud835\udf3c = [\ud835\udf19, \ud835\udf03, \ud835\udf13]\u22a4 represents the Euler angular between the world frame and the body frame, while its corresponding Euler angles rate is denoted by \u2126 = [",
            " (1) Thrust \ud835\udc53 \u2208 R and moment \ud835\udf49 \u2208 R3, as the control inputs of the quadrotor UAV, can be presented as \u23a1 \u23a2 \u23a2 \u23a2 \u23a2 \u23a3 \ud835\udc53 \ud835\udf0f\ud835\udf19 \ud835\udf0f\ud835\udf03 \ud835\udf0f\ud835\udf13 \u23a4 \u23a5 \u23a5 \u23a5 \u23a5 \u23a6 = \u23a1 \u23a2 \u23a2 \u23a2 \u23a2 \u23a3 1 1 1 1 \u2212\ud835\udc51\ud835\udf19 \u2212\ud835\udc51\ud835\udf19 \ud835\udc51\ud835\udf19 \ud835\udc51\ud835\udf19 \ud835\udc51\ud835\udf03 \u2212\ud835\udc51\ud835\udf03 \ud835\udc51\ud835\udf03 \u2212\ud835\udc51\ud835\udf03 \ud835\udc50\ud835\udf0f\ud835\udc53 \u2212\ud835\udc50\ud835\udf0f\ud835\udc53 \u2212\ud835\udc50\ud835\udf0f\ud835\udc53 \ud835\udc50\ud835\udf0f\ud835\udc53 \u23a4 \u23a5 \u23a5 \u23a5 \u23a5 \u23a6 \u23a1 \u23a2 \u23a2 \u23a2 \u23a2 \u23a3 \ud835\udc531 \ud835\udc532 \ud835\udc533 \ud835\udc534 \u23a4 \u23a5 \u23a5 \u23a5 \u23a5 \u23a6 , (2) where \ud835\udc53\ud835\udc56, \ud835\udc56 = 1, 2, 3, 4 is the thrust force generated by the \ud835\udc56th propeller, \ud835\udf49 = [\ud835\udf0f\ud835\udf19, \ud835\udf0f\ud835\udf03 , \ud835\udf0f\ud835\udf13 ]\u22a4 represents the equivalent control moment of the four thrust forces on the quadrotor, \ud835\udc51\ud835\udf19 is the half of roll motor-to-motor distance, \ud835\udc51\ud835\udf03 is the half of pitch motor-to-motor distance, and \ud835\udc50\ud835\udf0f\ud835\udc53 is a fixed constant reflecting the relationship between the thrust force \ud835\udc53\ud835\udc56 and its corresponding torque. The thrust force \ud835\udc53\ud835\udc56 produced by the rotating propeller has a squared relationship with the applied corresponding motor command \ud835\udc62\ud835\udc56, \ud835\udc62\ud835\udc56 = \ud835\udc50\ud835\udc62\ud835\udc53 \u221a \ud835\udc53\ud835\udc56 \u2212 \ud835\udc53\ud835\udc4f \u2212 \ud835\udc62\ud835\udc4f, (3) where \ud835\udc50\ud835\udc62\ud835\udc53 , \ud835\udc53\ud835\udc4f, and \ud835\udc62\ud835\udc4f are relevant proportion and offset coefficients. These coefficients can be obtained by fitting a quadratic polynomial to thrust and applied motor command data collected from hover flights with a varying payload and a dynamometer. As can be seen from Fig. 2, the quadrotor UAV\u2019s translational and rotational motions are driven by adjusting the speed of four motors, while the four lifts (\ud835\udc531, \ud835\udc532, \ud835\udc533, \ud835\udc534) are proportional to the square of the corresponding rotor speeds. The pitching motion is achieved by decreasing the forces \ud835\udc532 and \ud835\udc534 and increasing the forces \ud835\udc531 and \ud835\udc533, leading to a forward movement. In a similar manner, the rolling motion is achieved by increasing the forces \ud835\udc533 and \ud835\udc534 and decreasing the forces \ud835\udc531 and \ud835\udc532, resulting in a lateral movement"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001478_j.conengprac.2011.04.005-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001478_j.conengprac.2011.04.005-Figure1-1.png",
        "caption": "Fig. 1. STARMAC II quadrotor helicopter.",
        "texts": [
            " Controllers are derived using these models and implemented on the Stanford Testbed of Autonomous Rotorcraft for Multi-Agent Control (STARMAC), demonstrating significant improvements over existing methods. The design of the STARMAC platform is described, and flight results are presented demonstrating improved accuracy over commercially available quadrotors. & 2011 Elsevier Ltd. All rights reserved. Quadrotor helicopters are an increasingly popular rotorcraft concept for unmanned aerial vehicle (UAV) platforms. These vehicles use two pairs of counter-rotating, fixed-pitch rotors located at the four corners of the aircraft, as shown in Fig. 1. Their use as autonomous platforms has been envisaged in a variety of applications, both as individual vehicles and in multiple vehicle teams, including surveillance, search and rescue, and mobile sensor networks (Hoffmann, Waslander, & Tomlin, 2006b). Recent interest in the quadrotor design from numerous communities, including research, surveillance, construction and police use (Hull, 2010), can be linked to two main advantages over comparable vertical take off and landing (VTOL) UAVs, such as helicopters"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003881_mssp.2001.1414-Figure10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003881_mssp.2001.1414-Figure10-1.png",
        "caption": "Figure 10. The contact geometry of two spur gears.",
        "texts": [
            " Sirichai [6] developed a \"nite element model of spur gears in mesh from which the proportion of loading at each position throughout the mesh cycle was obtained, as shown in Fig. 9. The frictional force acts in a direction perpendicular to the normal contact force and will apply both translational forces and moments to the gears, and to include the e!ects of these additional loadings, the geometry of the gear teeth must be considered. The location of the contact point c, as a function of shaft rotation angle, is required so that the moment arms about both the gear and pinion centres, qh and of , are known. Figure 10 shows how the contact point, c, traverses from point a to e through the mesh cycle, passing through point p at the pitch point. To relate the angle a to the shaft rotation angle, the involute pro\"le of the gear teeth must be taken into account. Figure 11 shows an involute gear tooth, with the angle b representing the angle along the involute between the initial point of contact, a, and the current point of contact, c. The shaft rotation angle h is then equal to the sum of angles a plus b giving h\"a#b. (25) The angle between the involute origin at radius i and a point on the involute at radius r is given by the following equation: /\"CA r iB 2 !1D 1@2 !tan~1CA r iB 2 !1D 1@2 . (26) Therefore the angle b is expressed as b\"SCA rb r p B 2 !1D!tan~1SCA rb r p B 2 !1D!Laoi. (27) At the point of contact ra\"rb and referring to Fig. 10, ra is expressed as rb\"ra\" oa sinLoap sin(Lbao!a) . (28) Using equations (25), (27) and (28), the values of a, b, and ra can be found iteratively for each required shaft rotation angle h. The iterative procedure involves \"rst assuming an initial value of a and then calculating ra and b. The value of a is then adjusted using equation (25), and the process is repeated until a suitable degree of accuracy is obtained. Once the values of a and ra are found, the value of of can be calculated from of \"ra sin(Lboa#a)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003619_bf02665211-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003619_bf02665211-Figure7-1.png",
        "caption": "Fig. 7 - - A n g l e between the liquid-gas interface and the laser beam: (a) in the xz plane and (b) in the yz plane.",
        "texts": [
            " Thus, as the powder-feed rate is increased, with all the other parameters remaining constant, the clad height will increase, leading to a greater angle between the normal to the liquid-air interface and the laser beam (Figure 5), thus increasing the absorption of the workpiece. A model has therefore been developed for computing the workpiece absorption and is presented subsequently. As described in Figure 6, the laser beam illuminates g,liq either the liquid surface o~ or the solid surface S~112 (either the substrate or the previous track). Assuming, as described in Figure 7, that the liquid-gas interface is planar and that the cross section of the previous track is triangular, the liquid surface illuminated by the beam makes an angle 0~q with the horizontal, given by 01iq = arctan [5] Xrnax - - Xmi n whereas the previous track illuminated by the beam makes an angle with the horizontal. Let fl~(0) be the workpiece absorption of a planar surface making an angle 0 with the horizontal (0 < 0 < 60 deg). As indicated previously (assumption A7 in Section II) flw(O) has the following form: /3w(0) = r (1 + a~0) [61 where a~ is a proportionality coefficient which has to be measured for each material 9 As the laser beam illuminates the substrate with absorption flw(0), the previously deposited track with absorption fl~(OsoO, and the liquid pool with absorption flw(O~iq) (Figure 6), the workpiece absorption [3w is given by l [ Slas~w(O) + Slas[~w( Osol) ] -}- Slasflw( Oliq) _ sol sol liq 2 fl~ = [7] a las c~iq = 1rr2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003908_978-1-84882-730-1-Figure14.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003908_978-1-84882-730-1-Figure14.1-1.png",
        "caption": "Figure 14.1 Basic motion tasks for a mobile robot",
        "texts": [
            " Output feedback simultaneous stabilization and trajectory-tracking, and path-following controllers are then developed using the control design techniques proposed for underactuated ships in Chapters 6 and 11. For VTOL aircraft, the observer and control design strategies used for underactuated ships in Chapters 5 and 6 are utilized to design a global output feedback trajectorytracking controller. 14.1 Mobile Robots 14.1.1 Basic Motion Tasks In order to derive the most suitable feedback controllers for each case, it is convenient to classify the possible motion tasks as follows: Point-to-point motion: The robot must reach a desired goal configuration starting from a given initial configuration, see Figure 14.1a. Path-following: The robot must reach and follow a geometric path in the Cartesian space starting from a given initial configuration (on or off the path), see Figure 14.1b. Trajectory-tracking: The robot must reach and follow a trajectory in the Cartesian space (i.e., a geometric path with an associated timing law) starting from a given initial configuration (on or off the trajectory), see Figure 14.1c. The three tasks are sketched in Figure 14.1with reference to a three wheel car-like robot. Execution of these tasks can be achieved using either feedforward commands, or feedback control, or a combination of the two. Indeed, feedback solutions exhibit 341 342 14 Control of Other Underactuated Mechanical Systems an intrinsic degree of robustness. Explanations of the above motion tasks are similar to those of ocean vehicles, see Section 3.2. 14.1.2 Modeling and Control Properties 14.1.2.1 Modeling We consider a unicycle-type mobile robot, which under the assumption of no wheel slips has the following dynamics [142]: 14"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure15.14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure15.14-1.png",
        "caption": "Figure 15.14 The displacement of point i due to a translation ux0; uy0 and a small rotation \u03d5z0.",
        "texts": [
            " The displacement of point i is (see Figure 15.13) uxi = ux0 \u2212 a, uyi = uy0 + b. a and b are the result of the rotation \u03d5z0. Due to the rotation, point i moves through an arc length \u03d5z0ri along the circle with radius ri and centre O\u2032. For a and b it applies that a = ri { cos \u03b1i \u2212 cos(\u03b1i + \u03d5z0) } , b = ri { sin(\u03b1i + \u03d5z0) \u2212 sin \u03b1i } . The expressions are much simplified when the rotation is small. If \u03d5z0 1 722 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM the circle can be replaced by its tangent (see Figure 15.14). The displacement of point i due to the rotation is then \u03d5z0ri with the following components: a = \u03d5z0ri sin \u03b1i = \u03d5z0yi, b = \u03d5z0ri cos \u03b1i = \u03d5z0xi. For small rotations the following applies (ignoring the signs): \u2022 the horizontal displacement is equal to \u201crotation \u00d7 vertical distance to the centre of rotation\u201d; \u2022 the vertical displacement is equal to \u201crotation \u00d7 horizontal distance to the centre of rotation\u201d. Conclusion: Due to a translation ux0; uy0 and a small rotation \u03d5z0 of the uxi = ux0 \u2212 a = ux0 \u2212 \u03d5z0yi, uyi = ux0 + b = uy0 + \u03d5z0xi"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure1.28-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure1.28-1.png",
        "caption": "Fig. 1.28. Mechanism of a four-legged loco motion machine",
        "texts": [
            "20 as an example, it is possible to isolate the chains of \"legs\", \"body\" and \"arms\"; here, the chain consisting of \"legs\" is independent since the chains of \"body\" and \"arms\" impose no kinematic constraints on it. For the mechanism presented in Fig. 1.26, if the motion of member 1 is known, the chain I is independent, with all the remaining ones being satellites. The sequence to be followed in performing kinematic and dynamic analyses has thus been determined; namely, the basic independent chains should be analysed first, and the satellite (guided) chains second. Figure 1.27 illustrates the kinematic chains of a six-legged walking machine, while in Fig. 1.28 the scheme of a four-legged locomotion machine is presented. [IJ Vukobratovic, M., \"Applied Dynamics of Manipulation Robots\", Springer-Verlag, 1988. [2J Vukobratovic, M., Potkonjak V., \"Applied Dynamic and CAD of Manipulation Robots\", Monograph, Springer-Verlag, 1985. [3J Robototechnika (in Russian), edited by E. P. Popov and E. I. Yureevitch, Mashinostroenie, Moscow, 1984. [4J Stepanenko, Y., \"Dynamics of Spatial Mechanisms\", (in Russian), Mathematical Institute, Beograd, 1974. Chapter 2 Manipulator Kinematic Model Kinematic modelling of manipulators plays an important role in contemporary robot control"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000013_j.actamat.2016.07.019-Figure20-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000013_j.actamat.2016.07.019-Figure20-1.png",
        "caption": "Fig. 20. LBM-produced Ti-6Al-4V bracket for Airbus A350 (from LZN Laser Zentrum Nord GmbH and Airbus), with topology optimized bionic design resulting in ~30% weight saving cp. to conventional milled bracket. Three brackets with support structures on a build plate after LBM process (left) and finished part (right).",
        "texts": [
            " These three steps are iterated until the part is completed. After completion unmelted metal powder can be sieved and reintroduced into a subsequent LBM process. The part itself is fixed to a build plate, quite often connected by so-called support structures. Support structures are lattice-like structures which are necessary for heat dissipation and fixation of the part in the powder bed, especially for supporting its horizontally oriented and overhanging surfaces, cf. Fig. 1, Detail A, and also cf. Fig. 20 (left). That way deformation of the part is prevented. The support structures need to be removed later on to finish the part. In addition to support structures, pre-heating of the build plate can be used to avoid distortion of parts by lowering thermal gradients, resulting in a reduction of residual stresses which evolve during the LBM process [27]. For LBM fabrication of Ti-6Al-4V parts, typical pre-heating temperatures are 200 C [25] up to 500 C [26]. The LBM process is carried out in a closed process chamber in which an inert gas atmosphere is continuously maintained so that the residual oxygen content is less than 0",
            " as cooling channels close to the surface that are not producible by conventional technologies can shorten heat and cooling cycles. Ti-6Al-4V has been the material of choice for a variety of biomedical applications, e.g. hip endoprosthesis [159]. The process offers the possibility to manufacture osseointegrative structures such as lattices and thus improves functionality of implants. Ti-6Al4V is also in the focus of the aerospace industry, under discussion for serial production of brackets (cf. Fig. 20), parts of the fuel system and many other parts [1]. The aerospace sector is probably the one most radically affected by AM in the near future due to the enormous lightweight potential; besides structural elements, this also includes engine parts. Fuel nozzles for the GE LEAP aero engine (cf. Fig. 21) is a prominent examplewith a planned production volume of 25,000 parts [9]. The manufactured part is said to be 25% lighter and stronger than the previous conventional design. An overview on the current state of AM of metals was presented with a focus on the interrelationship between process, microstructure and properties"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure7.4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure7.4-1.png",
        "caption": "Figure 7.4 Forces on the sides of the triangular volume element.",
        "texts": [
            " 7 Gas Pressure and Hydrostatic Pressure 247 To demonstrate this, a small triangular part has been isolated from the material parallel to the xy plane in Figure 7.3. The oblique side has an area A. The area of the vertical side is therefore A cos \u03b1, while that of the horizontal side is A sin \u03b1. The triangular part is so small that, for all the stresses on the boundary, it can be assumed that they are uniformly distributed. Assume that a compressive stress p is acting on the oblique side. This stress acts normal to the side as there is no shear stress. In Figure 7.4, the forces (force = stress \u00d7 area) on the edges of the triangular part are shown. The lines of action of the forces pass through a single point. This means that there is moment equilibrium in the xy plane. Here it is assumed that the element is so small that its dead weight can be neglected. The equations for the force equilibrium in respectively the x and y direction are \u2211 Fx = p1 A cos \u03b1 \u2212 p A cos \u03b1 = 0,\u2211 Fy = p2 A sin \u03b1 \u2212 p A sin \u03b1 = 0, so that p1 = p2 = p3. The result is independent of angle \u03b1"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure2.12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure2.12-1.png",
        "caption": "Fig. 2.12. Euler angles between the hand coordinate frame and the fixed reference frame",
        "texts": [
            "21) Once the numerical values of the \u00b0A3 matrix are evaluated, the other three external coordinates describing the orientation (Euler angles) may also be determined. Then, the direct kinematic problem is completely solved. Manipulator hand orientation with respect to the reference frame is usually specified by three Euler angles (Fig. 2.7). One method for evaluation of Euler angles is the following. Let us consider two coordinate frames and describe rotation between them. The first system Qn is assigned to the last manipulator link, and its unit axes are denoted by qnl' qnz, qn3 (Fig. 2.12). The transformation matrix \u00b0An which maps vectors from system Qn to system \u00a2xyz is (2.22) 2.3 Direct Kinematic Problem 33 The rotation of system On Xn Yn Zn with respect to Oxyz can be described by the following sequence of rotations: a rotation by t/! about z axis (yaw), followed by 'a rotation by () about the new Y axis (pitch), and rotation by q> about the new x axis (roll) (Fig. 2.12). This sequence of rotations corresponds to the following product of rotation matrices -sint/! cost/! o 0] [COS() ~ ~Sin() ~ ~in()] [~ ~osq> o cos () 0 sin q> ~ sin q>] (2.23) cosq> or [ COS t/! cos() \u00b0An = sint/! cos() -sin () cos t/! sin () sin q> - sin t/! cos q> sin t/! sin () sin q> + cos t/! cos q> cos ()sin q> cos t/! sin () cos q> + sin t/! sin q> ] sin t/! sin () cos q> - cos t/! sin q> cos ()cos q> (2.24) By equating the numerical values of ajk elements (Eq. (2.22)), obtained by a kinematic modelling procedure, with the elements from Eq"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000554_978-3-540-85629-0-Figure1.8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000554_978-3-540-85629-0-Figure1.8-1.png",
        "caption": "Fig. 1.8. A multi-dimensional trajectory defined in the working space of an industrial robot (courtesy COMAU).",
        "texts": [
            " The curve followed by the end effector must be designed on the basis of the constraints imposed by the task (e.g. the interpolation of a given set of viapoints), while the determination of the motion law descends from other constraints, such as the imposition of the conditions on the maximum velocities, accelerations, and torques that the actuation system is able to provide. From the composition of the geometric path and of the motion law the complete trajectory is obtained p\u0303(t) = p(u(t)) as shown in Fig. 1.8. Once the desired movement is specified, the inverse kinematics3 of the mechanism is employed to obtain the corresponding trajectory in the actuation (joint) space, where the motion is generated and controlled. 3 The direct kinematics of a mechanical device is a (nonlinear) function q \u2192 p = f (q) mapping the joint positions q = [q1, q2, . . . , qn]T (i.e. the actuators\u2019 positions) to the corresponding position/orientation p of a specific point of the 8 1 Trajectory Planning 1.5 Contents and Structure of this Book A relevant, detailed bibliography is available for the problem of moving parts of automatic machines by means of mechanical cams, and in particular for the problem of the determination of the best cam profile in order to obtain the desired motion at the load"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001351_j.ijfatigue.2019.03.025-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001351_j.ijfatigue.2019.03.025-Figure2-1.png",
        "caption": "Fig. 2. Fatigue test bar dimensions in mm.",
        "texts": [
            " The asbuilt and cylindrical bars were placed closely together at the grip sections with rows of bars perpendicular to the recoating direction. The rows were staggered such that the recoater only contacted the edge of one component at any given time. Triangular supports were added to support the end of each row. In total, 15 as-built bars were included in the assembly. One of the bars with parameter set A was deleted midbuild, because of a recoater jam, so 14 as-built fatigue bars were fully fabricated. The build layout can be seen in Fig. 1 and the fatigue bar geometry is shown in Fig. 2. The fatigue bars were heat treated post-process in order to relieve the residual stress and homogenize the microstructure. This will allow consistent assessment of the as-built surface roughness. The bars were stress relieved at 1065 \u00b0C for 90min and furnace cooled. The bars were removed from the build plate using wire EDM and post-processed through the standard solution and aging heat treatment for AM Alloy 718 [24,25]. No additional heat treatment was used to reduce the porosity because it has been shown that the surface roughness will dominate the fatigue failure [12]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002362_j.ymssp.2020.106640-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002362_j.ymssp.2020.106640-Figure6-1.png",
        "caption": "Fig. 6. Spalled gears obtained from the fatigue experiment.",
        "texts": [],
        "surrounding_texts": [
            "The contact stress along the potential contact line can be expressed as:\nrH \u00bc\nffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi \u00f0nL 1\u00deFi\npL\n1 qp \u00fe 1 qg\n1 m2p Ep \u00fe 1 m2g Eg\nvuuut \u00f05\u00de\nwhere q is the radius of curvature. m is the Poisson\u2019s ratio. Subscript p and g represent the driving gear and driven gear, respectively. nL is the number of contact points along the potential contact line.\nThe spalling pattern is usually assumed as regular shape in traditional study. Moreover, in many experimental studies of the gear spalling, the tooth surface is damaged deliberately. While in this paper, the pitting and spalling faults are obtained through fatigue experiment rather than man-made destruction.\nBoth the fatigue experiment and vibration test are based on a power circulation test rig (see Fig. 5). The \u2018back to back\u2019 configuration is adopted and the test rig consists of two gear boxes: test gear box and slave gear box. By adjusting the mass of the weights, various levels of the torque can be applied on the load clutch. The rotational speed of the driving motor is 1492 r/min (constant speed).\nThe fatigue experiment is performed to obtain the spalling morphology. High torque (500 N m) is applied to accelerate the failure process. After running for 120 h under high torque, five tooth pairs suffered from various degrees of surface damage (see Figs. 6 and 7).\nFive spalled gear pairs obtained from the fatigue experiment are used to carry out the vibration test. The torque and the rotational speed are 200 N m and 1492 r/min, respectively. Two accelerometers with 2 mV/g sensitivity are mounted on the bearing of the driving shaft (near the damaged gear). The data acquisition system is YE6232B and the sampling frequency is 8000 Hz. The vibration signals of the spalled gear pairs are collected in both horizontal and vertical direction (sampling time is 60 s for each spalled gear pair).",
            "Generally, the spalling morphology obtained from the fatigue experiment is irregular and intricate. All the geometric information of the surface spalling is stored in the profile error matrix. The profile error vector (e in Eq. (1)) is one row of the profile error matrix. The procedure of the acquisition of the profile error matrix is as follows (see Fig. 8): (1) The close-up image of the spalled tooth surface is shot after the fatigue experiment; (2) The boundary of the tooth surface is determined according to the geometrical parameters of the gear pair and the edge of the spalling area is recognized through image processing technique; (3) The spalling morphology is converted into binary images (grey level of 255 means healthy area, and 0 means spalling area); (4) The binary images are transformed into profile error matrix. The profile error matrix is then imported into the three-dimensional LTCA method for the calculation of the mesh stiffness and contact stress. It should be noted that in the first step of the above-mentioned steps, the image distortion is inevitable during the process of projection because the involute surface is a curved surface, while the plane of the projection is a flat surface. If the spalling area is not far away from the pitch line, the error caused by the distortion is not obvious. Generally, spalling occurs in a narrow band (single tooth contact zone) around the pitch line. Therefore, the distortion in the transfer process is neglected in this paper. The accurate spalling morphology should be measured using professional equipment (such as the three-dimensional profile measuring apparatus).",
            "The main parameters of the gear pair are listed in Table 2. Finite element method is applied to test the effectiveness of the proposed method. The three dimensional model of the spalled gear is established in SolidWorks and then meshed in HyperMesh. Finally, the finite element model is established in ANSYS (see Fig. 9). It should be noted that the finite element model in Fig. 9 is only used to verify the proposed model. But for the proposed method, the calculation of the mesh stiffness and the contact stress are all performed in MATLAB environment.\nConsidering the spalling pattern in Fig. 8, a comparison among the proposed method, the finite element method and Luo\u2019s method [22] has been carried out (see Fig. 10). The relative error at moments A and B (see Fig. 10) has been listed in Table 3. The maximum errors of the proposed method and the Luo\u2019s method are 1.08% and 3.40% (see Table 3), respectively. Meanwhile, the extended tooth contact is ignored in Luo\u2019s method. Therefore, sudden change can be found between the double tooth contact area and single tooth contact area. It can be concluded that the computation accuracy of the proposed method is higher than the Luo\u2019s method.\nThe proposed method and the finite element method take 2 min and 5 h for the mesh stiffness calculation over a mesh cycle, respectively. The computational efficiency of the proposed method is much higher than the finite element method owing to the simplification of the contact iteration. The non-linear iteration is quite time-consuming in commercial finite element software (such as ANSYS). However, in the proposed method, the contact compliance matrix is calculated using analytical formula and there is no nonlinear iteration in the calculation process. The global compliance matrix is obtained"
        ]
    },
    {
        "image_filename": "designv10_0_0003717_nme.1620310308-Figure16-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003717_nme.1620310308-Figure16-1.png",
        "caption": "Figure 16. The effect of an X translation on the contact pattern",
        "texts": [],
        "surrounding_texts": [
            "The case chosen as a numerical example is that of contacting hypoid gears. The surfaces of the teeth of these gears are created by a complicated cutting process. The cutting machines used have many kinematic settings. The settings are chosen such that the contact zone remains in the centre of the tooth surfaces as the gears roll against each other. A heuristic procedure is available to select the settings, but, in practice, these settings have to be selected after a tedious iterative process involving cutting and experimentally testing actual gears. Even so, it is very difficult to predict the actual contact stresses, fatigue life, kinematic errors and other design criteria, especially when not installed in ideal conditions. The contact stresses are so sensitive to the actual surface profile that conventional 3-D contact analysis is not feasible. A sample 90\" hypoid gear set from the rear axle of a commercial vehicle was selected. The gear ratio of this set was 41: 11 and the axial offset was 1-5 in. The gear surfaces had been experimentally shown to be ideal for this particular gear ratio and axial offset. In other words, the contact zone was found to remain in the central portion of the gear teeth in the operational torque range. The object of this numerical study is to verify this by looking at the manner in which the contact pattern shifts when the gears are moved around from their ideal locations. Kinematic errors were also calculated. The model was constructed by first generating values of co-ordinate normal vectors for points on the surface by simulating the cutting machines. The finite element description of the surface 536 S. VIJAYAKAR was then created by fitting tenth order truncated Chebyshev series approximations to these data. The interior portions of the finite element were created semi-automatically. Only a sector containing three teeth of each gear was modelled, with each tooth being identical. The gear (gear no. 1) and the pinion (gear no. 2, the smaller gear) were then oriented in space as per the assembly drawings, and the analysis was carried out for each individual time step. Figure 6 shows a six-tooth gear and pinion model. Sectoral symmetry is used to generate stiffness matrices from the stiffness matrix of one tooth. For this particular gear set, a three-tooth model suffices because at the most two teeth contact at a time. Figure 7 shows the surfaces of the three-tooth gear. Figures 8 ,9 and 10 show the contact pattern (which is the locus of the contact zone as the gears roll against each other) for a gear torque of 240,480 and 960 in-lb, respectively. Figures 11 and 13 show views of the contact zone with contact pressure contours on the gear for two particular angular positions. Figures 12 and 14 show magnified views of the contact zone for these two positions. They show contours of normal contact pressures on the surfaces. Computational grids of 11 x 25 cells were used on these surfaces to obtain the pressure distributions. Another useful piece of information that can be obtained is the kinematic transmission error, which is the deviation of the motion of the gears from their ideal angular motion. Figure 15 shows the transmission error in radians for three different torque levels (0, 480 and 960in-lb). These A THREE-DIMENSIONAL CONTACT PROBLEM 531 538 S . VIJAYAKAR A THREE-DIMENSIONAL CONTACT PROBLEM 539 540 S. VIJAYAKAR 542 S . VIJAYAKAR transmission error curves are remarkably similar to those measured experimentally by other researcher^.^ Finally, the position of the pinion was perturbed slightly from the design location, and Figures 16 to 19 show the contact patterns that were obtained. When compared to the contact pattern for the unperturbed position in Figure 7, it shows that the best contact pattern does indeed occur at the designed position, lending credence to the notion that an analysis of the kind described in this paper has the potential to be used in the design process itself."
        ]
    },
    {
        "image_filename": "designv10_0_0003787_ac981376m-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003787_ac981376m-Figure6-1.png",
        "caption": "Figure 6. Schematic diagram of the diamond/ADH biosensor assembly, together with an expanded view, showing the processes involved in the determination of ethanol. In the present study, alcohol dehydrogenase (ADH) was immobilized on nylon mesh disks (50 \u00d7 50 \u00b5m openings), with a loading of \u223c33 activity units.",
        "texts": [
            " One hundred fifty microliters of this solution was mixed with 200 \u00b5L of BSA (10 wt %) and 200 \u00b5L of glutaraldehyde (2.5 vol %). Five microliters of this mixture was spread uniformly on the precleaned nylon disk and allowed to dry at room temperature for 30 min. The nylon net modified with alcohol dehydrogenase (ADH) was placed over the diamond electrode (0.14 cm2), which had been mounted in a Teflon holder, leaving the front (diamond) side exposed. The nylon net was held in place by a Teflon cap with a circular hole. A schematic diagram of the biosensor is shown in Figure 6. The assembly of this sensor is similar to that previously reported by Abrun\u0303a and co-workers for modified GC electrodes.25 (24) Pariente, F.; Lorenzo, E.; Tobalina, F.; Abrun\u0303a, H. D. Anal. Chem. 1995, 67, 3936. (25) Wu, Q.; Maskus, M.; Pariente, F.; Tobalina, F.; Ferna\u0301ndez, V. M.; Lorenzo, E.; Abrun\u0303a, H. D. Anal. Chem. 1996, 68, 3688. (26) Pariente, F.; Tobalina, F.; Moreno, G.; Herna\u0301ndez, L.; Lorenzo, E.; Abrun\u0303a, H. D. Anal. Chem. 1997, 69, 4065. (27) Huang, T.; Warsinke, A.; Kuwana, T.; Scheller, F",
            "4 V vs SCE) where the current due to NADH oxidation was small. The plots for NADH were obtained at 0.58 V, where the contribution of NADH oxidation was the largest. These results indicated that the NADH oxidation current can be corrected for AA interference when both compounds are present at comparable concentrations. Such corrections for AA interference have previously been reported for modified GC electrodes.25 Ethanol Biosensor Response. To demonstrate the operation of the diamond electrode in a dehydrogenase-based sensor, the assembly depicted in Figure 6 was constructed. The assembly of the sensor is similar to that previously reported by Pariente et al. for modified GC electrodes.24 The enzyme catalyst, alcohol dehydrogenese (ADH), was immobilized on a nylon mesh disk, as already mentioned. In an ethanol biosensor assembly, the immobilized ADH oxidizes ethanol to acetaldehyde in the presence of NAD+. The cofactor, NAD+, is converted to its reduced form (NADH) by accepting the electrons generated in the enzymatic reaction. Oxidation of the NADH at the electrode generates an electrical signal the magnitude of which is a measure of the ethanol present in the electrolyte solution"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure7.19-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure7.19-1.png",
        "caption": "Figure 7.19 The 4-metre wide flap modelled as a line element with (a) the water pressure normal to the flap and (b) the resulting load diagram.",
        "texts": [
            " Question: Determine the support reactions at A and B due to the total water pressure. The dead weight of the flap should be ignored. Solution: The linear distribution of the water pressure on both sides of the flap is shown in Figure 7.18. To the left of the flap, the water pressure at A is (1000 kg/m3)(10 N/kg)(1 m) = 10 kN/m2 and at B it is (1000 kg/m3)(10 N/kg)(4 m) = 40 kN/m2. To the right of the flap, the water pressure at B is (1000 kg/m3)(10 N/kg)(2 m) = 20 kN/m2. 7 Gas Pressure and Hydrostatic Pressure 257 In Figure 7.19a, the 4-metre wide flap is modelled as a line element, with line loads due to the water pressures normal to it. To the left of the flap, the distributed load varies linearly from (4 m)(10 kN/m2) = 40 kN/m at A, to (4 m)(40 kN/m2) = 160 kN/m at B. To the right of the flap, the load increases linearly from 0 at C to (4 m)(20 kN/m2) = 80 kN/m at B. Figure 7.19b represents the load diagram for the resulting water pressure. The length of flap AB is (see Figure 7.18) \u221a (3 m)2 + (0.96 m)2 = 3.15 m. The distances BC and CA are respectively 2.10 m and 1.05 m. To work quickly, the load diagram in Figure 7.20 has been placed horizontally and is split up into a number of areas for which the resultants can be easily calculated: R1 = (2.10 m(80 kN/m) = 168 kN, R2 = 1 2 \u00d7 (1.05 m)(80 kN/m) = 42 kN, R3 = 1 2 \u00d7 (1.05 m)(40 kN/m) = 21 kN. 258 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM The support reaction Ar at A is found from the moment equilibrium of the flap about B: Ar = (1"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000790_j.engfailanal.2011.07.006-Figure10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000790_j.engfailanal.2011.07.006-Figure10-1.png",
        "caption": "Fig. 10. Scheme of spur gear system with six degrees of freedom.",
        "texts": [
            " 8 and 9, respectively. The same phenomenon is also found that the maximum stiffness reduction during the mesh process for a fixed crack appears at the beginning of the engagement when the cracked tooth of the pinion is getting into engagement. Based on the mesh stiffness model of gear pair with or/and without a tooth root crack, a dynamic lumped parameter model of a spur gearbox system comprising of six degrees of freedom (DOF) is established. A schematic of the gear dynamic model is shown in Fig. 10 where the y axis is parallel to the line of action (LOA) of the gear pair. Tp/Tg is external/braking torque; Xp/Xg is nominal operational speeds of pinion/gear; Jp/Jg is inertial moment of pinion/gear; mp/mg is mass of pinion/gear; KiBj/CiBj is stiffness/damping of the supporting bearings and i = p, g for pinion and gear, respectively, j = x, y for x and y direction. k(t) is time varying mesh stiffness by which the influence of gear tooth crack is incorporated and Cm is the damping between gear teeth in mesh",
            " yi= _yi=\u20acyi is lateral displacement/velocity/ acceleration along y direction. MiN and Mif are the force moments induced, respectively by the normal mesh force along LOA and the corresponding surface friction force based on the method used by He et al. [29]. The supporting bearing stiffness KiBj and damping CiBj are assumed to be constant although some time-varying models have been developed like [30\u201332] because its beyond the scope of this paper. The subscript i = p, g are for pinion and gear, respectively and j = x, y for x, y directions shown in Fig. 10. Statistical features which are commonly used to provide a measurement of the vibration level are widely used in mechanical fault detection [33\u201336]. And Wu et al. [1] studied the performances of some of these statistical indicators when gear tooth cracks with different size were in presence. A conclusion was drawn that RMS shows the best performance when Method 2 for generating residual signals is used and kurtosis is the most robust indicator no matter what signals are used. The Method 2 is also used in this paper to produce the residual signal and the RMS and kurtosis indicators are also used to predict the severity of the crack propagation"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000206_978-0-387-74315-8-Figure5-1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000206_978-0-387-74315-8-Figure5-1-1.png",
        "caption": "Figure 5-1: Acceleration about joint i only",
        "texts": [
            "e inertia of the composite body to calculate the forces involved. From Equation (5.5) and the discussion in the previous section, it is evident that II q is the vector of joint forces required to impart a joint space acceleration of q to a stationary robot which is free from 85 gravitational or othcr extern al influences. In particular, column i of II (which is H oi) is the vector of joint forces required to produce a unit acceleration about joint i and zero acceleration about aH other joints (see Figure 5-1). Since no motion occurs about any of the joints i+l .. n, they can aH be replaced by rigid connections without having any effect on the dynamics lower down the chainj so the moving part of the robot simplifies to a single composite rigid body. Using the system model described in Chapter 4, let \u00ee~ be the inertia of , the composite rigid body comprising links i. . n, then n -c L-1. = 1., I J j=i which expressed as a recurrence relation is -c -c -1. =1. 1 +1 .. , ,+ , (5.7) The force required to give \u00eef an acceleration of ;i (i"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000228_ac202878q-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000228_ac202878q-Figure1-1.png",
        "caption": "Figure 1. Schematics of graphene sheet orientation in MWCNTs (a) and stacked graphene nanofibers (SGNF), b). The highly electroactive edge portion of the sheets are highlighted in yellow. Reproduced by permission of the PCCP Owner Societies.62 http://dx.doi.org/10.1039/C0CP00213E.",
        "texts": [
            " The anodized epitaxial graphene was then shown to be a superior electroanalytical platform for a wide range of biomolecules compared to other carbon based electrodes, with the ability to resolve the anodic peaks of all four nucleic acids in both single and double stranded nucleic acids. The importance of accessible graphene sheet edges is further demonstrated by Ambrosi and Pumera62 who demonstrated that stacked graphene nanofibers have superior electrochemical performance for the oxidation of DNA bases than CNTs, edge plane pyrolytic graphite, graphite microparticles, and GC. The difference in the number of edge sites can be seen in Figure 1. Chang et al.63 were able to control the density of oxygen-containing functional groups on nanoplatelets of graphitic oxide by varying the temperature of microwave-assisted hydrothermal elimination. They determined that the edgeplane-like sites of the electrode were the electroactive sites, which were demonstrated to be useful for increasing the current response and peak shift between uric acid and ascorbic acid. This effect was attributed to ability for these molecules to form hydrogen bonds with the graphitic oxide"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003988_j.snb.2009.12.046-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003988_j.snb.2009.12.046-Figure1-1.png",
        "caption": "Fig. 1. Schematic diagram of the cylinder array.",
        "texts": [
            " / Sensors and Actuators B 145 (2010) 417\u2013427 Throughout Tables 1\u20133: DA = dopamine, AA = ascorbic acid, UA = uric acid, Ep = epinephrine, Trp = L-tryptophan, NE = norepinephrine, S = serotonin, Arg = L-arginine, Ur = urea, Cr = creatinine, Ox = oxalate, Asp = aspirin, Cys = L-cysteine, Glu = glutathione, RTIL = room temperature ionic liquid, GC = glassy carbon, SWCNT = single-walled CNT, DWCNT = double-walled CNT, MWCNT = multiwalled CNT. M.C. Henstridge et al. / Sensors and Actuators B 145 (2010) 417\u2013427 419 g t c [ m a t s m f t a t t a t k s p s t s a t a p o m i o c r p k d b fi p e l redox process: A \u2212 e\u2212 B (1) occurs at a \u2018film\u2019 approximated by a regular array of vertically aligned cylinders (Fig. 1) protruding from a conducting electrode et species undergo oxidation or reduction becomes shifted from hat required to electrolyse the interfering species. The range of hemically modified electrodes is truly huge and highly diverse 150\u2013159]. In the case of dopamine and its selective measure- ent we have attempted to categorise the various electroanalytical pproaches to the problem in Tables 1\u20133. Table 1 reports elecrodes chemically modified with polymer films or coatings. Table 2 ummarises reports of dopamine determination using nanotube odified electrodes, while Table 3 lists other strategies including a ew uses of bare, that is unmodified, electrodes"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001391_bf02133524-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001391_bf02133524-Figure1-1.png",
        "caption": "Fig. 1.",
        "texts": [
            "22) with L = [&, -- &], & > O, & > O, we obtain the analytical description of the behaviour illustrated by fig. 2-a (where tg -1 is omitted for brevity and dashed lines define the secondary hardening; /Co. < 0): a) b) L) K1 K2 - ~ H n - h t v p ! \" ~ H.-h p ~H2~'h The familiar picture of fig. 2-b corresponds to the assumption (2.23), i. e. to hi = & = h > 0. Although in direct opposition to the concept of Bauschinger effect, the concept of a uniformly expanding yield locus without shape changes (isotropie hardening, fig. 1-c) has been frequently accepted and even found to be in general agreement with certain experiments (see e. g. [24]). Inequality (2.19) defines an elastic range homothetic (with respect to the origin of the axes) to that defined by (2.20), when it can be expressed in the form: I~IQ-- aK ~< 0 (2.25) where a is a scalar depending on the yielding history. This condition is fulfilled if H complies with the requirement: HX ------- ( a - - 1)K for any }. (2.26) which supplies the following 3,(y - - 1 ) equations between the "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001073_b978-0-12-384050-9.50007-6-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001073_b978-0-12-384050-9.50007-6-Figure6-1.png",
        "caption": "Fig. 6. M a m m a l on vertical support . C , Center of gravity; W, weight of animal . \u039b a n d \u0392 are lateral projections of axes defined in text. Horizontal (Ah, Bh) and vertical {Av, Bv) components of forces through A and \u0392 are indicated. At equi l ib rium, Wd + Avt = Ahk, a n d W(d + t) = Bvt + Bhk.",
        "texts": [
            " The adducted hind limbs provide much or most of the propulsive thrust in climbing. If the hind feet subtend a small central angle, adduction can generate little force normal to the contact surface. Frictional forces between the hind feet and the support must therefore be slight, and the feet will tend to slip downward. This presents particular difficulties for a large arboreal mammal which cannot rely on claw grip, e.g., for a large anthropoid. The normal force at the lower contact points (Bh: Fig. 6) can be augmented by bringing A and \u0392 closer together and leaning away from the tree; since W(d + t) = Bvt + Bhk (Fig . 6 ) , an increase in d and decrease in k implies an increase in Bh. The result is an equal increase in Ah, however, which cannot be tolerated unless the upper grip is excep tionally secure\u2014implying that the upper limbs ( if clawless ) must subtend a large central angle. Therefore, habitually tree-climbing mammals of very large size might be expected to have either claws (e.g., Fielarctos, Pfeffer, 1969) or relatively long and powerful forelimbs (e"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001766_j.cirp.2019.05.004-Figure16-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001766_j.cirp.2019.05.004-Figure16-1.png",
        "caption": "Fig. 16. 3D reconstructed volume of an AM part measured by: (a) reference lab metrology system and (b) fast method applied to an airport XCT scanner. scanning time was 1570s for the first case and only 1 s for the second case influence on reconstruction accuracy of the fast scan is clearly visible [297].",
        "texts": [
            " Therefore, forcoordinate metrology of AM parts, XCT is sometimes preferred to tactile and optical probing methods, even when measuring external surfaces only. For example, Fig.15 illustrates the external thread measurement of a dental implant produced by laser sintering of titanium alloy (Ti6Al4V) [316], where XCT was preferred to conventional thread measuring techniques [50], to overcomethelimitationsarisingwhenmeasuringAMpartswithsm features, large form errors and complex surface texture. Research efforts are going in the direction of reducing scanning time. For example, Warnett et al. recently studied XCT reconstruction of an AM part (see Fig. 16) using a techni called real-time tomography (RTT) originally developed for airp baggage inspection [297]. This setup allowed a reduction scanning time for a single part by a factor of 1500 compared conventional XCT system. However, the smallest obtainable vo size was 1 mm, which is still too large for conventional us manufacturing metrology. Uncertainties in XCT coordinate measurements are affected large number of influence factors [51]. In particular, when scann metal parts, dimensional and form measurements are stron influenced by the material and its penetrated thickness, wh directly affect the X-ray attenuation [51]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure8.7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure8.7-1.png",
        "caption": "FIGURE 8.7 Contact and macromodel for contact.",
        "texts": [
            " Rather, they represent integer coordinates that enumerate the nodes along each side of a body, which may consist of multiple cells. As an example, A = 0 means the first node along the A side. The (A,B,C) notation is used instead of (X,Y ,Z) since the overall body may not be Manhattan brick shaped. An example of the joining of nodes is shown in Fig. 8.10. It is still not trivial to correctly connect the electrical circuit to the electromagnetic model. Inaccuracies may occur near the connections. For example, a wire contact attached to a conducting body as shown in Fig. 8.7a may create a large number of cells and unknowns. However, this can be avoided with clever approximations. If we attempt to replace the contact with a single wire macromodel, as shown in the example in Fig 8.7b, then the contact area must be chosen to be large enough as is shown by the dashed area in Fig. 8.7b. Essentially, the contact modeling is not trivial. Often an insufficient number or too many cells are used in the models of the details of the connections. The size of the contact is not defined by the conductor and the size and the resistance will increase with cell refinements. To avoid this, we have to use a cell with a fixed size with the wire cross section shown by dashed lines in Fig. 8.7b. Hence, the connection size could be locally defined independent of the number of surrounding cells used for the mesh. This is an example that shows that care must be taken when more efficient, simplified models are used. A similar simplifying technique can be used for a wire which is a via connection through a plane which does not make a connection to this plane as is considered in Fig. 8.17. In fact, for a very simple model, a wire can connect through a ground plane where we completely ignore the presence of a hole in the plane through which the wire is connected",
            " 8.2 is assumed which is 4 cm long and 2 cm wide. Two 5 mm by 5 mm contacts are placed at 1 cm and 3 cm along the length in the middle, 1 cm from each side. Compute the open-loop inductance between the two contacts for different mesh densities for the plate. How dense does the meshing have to be to get a good answer? Asking in a different way, we would like to find the smallest number of node for which we get a good value of the inductance. This problem relates to the contact issue presented in Fig. 8.7. 8.2 Point located on a quadrilateral element Construct a quadrilateral element as shown in Fig. 8.8, which is not parallel to the global (x, y, z) coordinates. Note that the element must have a planar surface. How do you choose the four corner points such that the surface is planar? Use the equations from Section 7.1 to compute the matrix of global coordinates (x, y, z) that correspond to the uniform steps in \u22121 \u2265 a \u2265 1 and \u22121 \u2265 b \u2265 1, where each of the two local variables (a, b) are incremented by steps size of 0"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure3.69-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure3.69-1.png",
        "caption": "Fig. 3.69: A model of CaPaMan architecture for static and dynamic analysis.",
        "texts": [
            "68 shows the singular configurations of CaPaMan, as they have been determined with the above-mentioned Jacobian analysis. The static behavior of CaPaMan can be investigated in term of force transmission capability through a static analysis. Force transmission is characterized by the parallelogram linkages in the sense that, because of the prismatic joint and assuming friction as negligible, the only force applied to a leg by the movable plate is the one of Rk which is contained in the plane of the mechanism, as shown in Fig. 3.69. Force Rk is given by the components Rky and Rkz with respect to the k-frame fixed with the linkage plane since Rkx \u2260 0 will determine the sliding of the prismatic joint to a static equilibrium. Thus, the components Rky and Rkz will balance the force F and the torque N acting on the movable plate. The force FP and the torque NP with respect to the frame fixed to the movable plate can be obtained by using the rotation matrix R as R= PFF (3.9.64) R= PNN A static equilibrium of the movable plate can be expressed for the triangular assembly with the spherical joints in the form ++= 321P RRRF (3",
            "69) when the trk overall gear ratio is taken into account. Equations (3.9.67) to (3.9.69) can be useful to evaluate the static performance of CaPaMan as well as to properly select the actuators for given applications. Figures 3.70 and 3.71 show results for illustrative numerical examples. An analysis of dynamic behavior of CaPaMan requires the evaluation of inertia actions and computation of the force F and torque N acting on the moving plate, according to Chapter 3: Fundamentals of the Mechanics of Robots222 the model of Fig. 3.69 and Eq. (3.9.65), but considering += exinP FFF (3.9.70) += exinP NNN in which Fin and Nin are the inertia force and torque; Fex and Nex are the external actions due to the payload, the application, or the interaction with the environment. The inertia actions Fin and Nin can be formulated through the mass M and the inertia matrix IH of the movable plate and payload as Fundamentals of Mechanics of Robotic Manipulation 223 M= Hin aF (3.9.71) I+I= HH inN \u00d7 in which aH is the acceleration of H center point of MP; \u03c9 is the angular velocity and \u03c9 dot is the angular acceleration of the MP",
            " In fact generally the coupler link can be considered not deformable since it is massive in order to ensure no flexion deformation for payload capability and high precision positioning of CaPaMan. However kc can be used to conveniently consider the stiffness of the coupler link and the transversal compliance of the sliding joint. Because of the ball joint SJ in the mechanical design of the CaPaMan mobile plate, only force acts on each CaPaMan leg and particularly this force lies in the plane of the parallelogram with components Ry and Rz, according to the scheme of Fig. 3.69. Therefore kh describes primarily the axial stiffness of the connecting rod and radial stiffness of the joints, although it may take into account the compliance of the prismatic guide when the sliding joint works at its extremity. A stiffness evaluation of CaPaMan can be obtained by computing in a general way displacements \u0394x, \u0394y, \u0394z, \u0394\u03d5, \u0394\u03c8, \u0394\u03b8, occurring at the movable plate at a static configuration when a force F = (Fx , Fy , Fz ) and/or a torque T = (Tx , Ty , Tz ) act upon it, according to the procedure outlined in 3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003684_ias.1997.643010-Figure12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003684_ias.1997.643010-Figure12-1.png",
        "caption": "Figure 12 The locus of the stator flux linkage cps qd, Wb",
        "texts": [
            " Figure 1 l(a-c) shows the dynamic responses of torque, IqsI and speed of the drive system with respect to a step change in speed reference from 700 to 2200 rpm. The motor is driven at light load (load torque<0.4Nm). It is seen from these figures that transitions between the constant torque and constant power operations are very smooth. It must also be mentioned here that these responses are just as fast, if not faster, than the response under an optiniised conventional pwm current regulated torque controller. Other than the initial position determination, the encoder was not used for torque control. It is used only for the outer speed control Figure 12 shows the locus of the stator flux linkage which is ncarly a circle in both constant torque and constant powcr operations. Figure 13 shows the experimcntal data on torque and stator flux in the T-lq,] plane (solid trace) superimposed on the calculatcd MPTA, CL and VL trajectorics (faint traces) of figure 6. Transitions between constant torque and constant power operations clearly occur at the intersections of maximum torque-pcr-ampere and current limit trajectories and maximum torque-per-ampere and voltage limit trajectories as indicated in figure 6"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001446_j.oceaneng.2016.06.041-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001446_j.oceaneng.2016.06.041-Figure2-1.png",
        "caption": "Fig. 2. Non-symmetric dead-zone of the rudder.",
        "texts": [
            " It is necessary to design a controller that takes these nonlinearities into account to achieve the desired performance. In this section, we redesign the adaptive controller based on the results in the previous section. We divide the controller design into two steps. We only consider the dead-zone in the first step, and then we consider both the input saturation and dead-zone nonlinearities. 4.1. Dead-zone model First, we consider the case with a dead-zone, ignoring the input saturation based on the given dynamics. The control input with a dead-zone, as shown in Fig. 2, is described as \u03b4 \u03c3= + ( )( ) ( )m u 20r u u where u is the control signal to be designed: = > \u2264 ( ) ( ) \u23a7\u23a8\u23a9m m u m u , 0 , 0 21 u r l \u03c3 = \u2212 \u2264 \u2212 \u2212 \u2212 < \u2264 \u2265 ( ) ( ) \u23a7 \u23a8\u23aa \u23a9\u23aa m b u b m u b u b m b u b , , , 22 u l l l u l r r r r where mr and ml denote the right and the left slopes of the deadzone characteristic, respectively. br and bl represent the widths of the dead-zone. For convenience, in the following paragraph, we simplify ( )m u and ( )v u as m, and s, respectively. Based on the dead-zone model (20), the nonlinear dynamic system (5) can be rewritten as \u03c8 \u03c9 \u03c9 \u03a6 \u03c3 \u0307 = \u0307 = + + + ( ) \u22a4h c mu c D 23 y y 3 3 The following assumption is made for this problem: Assumption 3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003403_978-1-4615-5633-6-Figure3.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003403_978-1-4615-5633-6-Figure3.2-1.png",
        "caption": "Figure 3.2. Phasor diagram showing relationship between d-q and D-Q variables",
        "texts": [
            "10)' we have cosd/d + sind/q } -sind/d + COSd/q From (3.20), we can obtain From (3.21), we can express (3.19) as [L di' .dd L ~I (R . L )~/l (\"' \"') 1 dt + J dt 1 Z + I + JW o 1 Z = VI - V2 where (3.18) (3.19) (3.20) (3.21) (3.22) Simplifying (3.22) we get where d~1 L Z (R . L )~I 'I 'I I dt + I + JW I Z = VI - V2 do W = Wo + dt (3.23) Eq. (3.23) can also be derived directly from applying Park's transformation. Eq. (3.21) is a very useful relation and can be represented by a phasor diagram shown in Fig. 3.2. Stationary three phase symmetric networks can be decoupled through transfor mations involving constant real matrices. The most well known among these is Clarke's transformation using Q' - f3 variables. Using a power invariant trans formation given by where o [Gc] = 1 J3 1 J3 1 73 (3.24) MODELLING OF THE ELECTRIC NETWORK 69 Eq. (3.1) is transformed to three decoupled equations given below (3.25) (3.26) (3.27) The advantage of using Clarke's a, (3, 0 components is that a three phase network is transformed to three decoupled networks, 'a',' (3' and zero sequence"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002286_j.actamat.2015.07.014-Figure14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002286_j.actamat.2015.07.014-Figure14-1.png",
        "caption": "Fig. 14. (a) Reconstructed 3D morphology of Defect No. 1. (b) SEM image of the same kind of defect at the top surface. (c) Cross sectional images viewed from front, (d) side and (e) top surfaces. Some powder particles trapped in the cavity can be observed and verified, red arrow. (For interpretation of the references to color in this figure legend, the reader is referred to the web version of this article.)",
        "texts": [
            " At the same time partial melted particles might weld onto the surface, see SEM image in Fig. 13(c). The depth variation of the melt along the track can be up to \u00b127% [18] and a number of voids are observed in this case. Finally, defects between adjacent tracks can be generated due to poor overlapping, flow instability or surface oscillation. Particles trapped within a void could clearly be visualized in 3D reconstructed images. One example is Defect No. 1 and its SEM image together with 2D cross sectional images from different view directions are presented in Fig. 14(a\u2013e). A rough surface and stochastic particles may deteriorate the wetting behavior of the next layer and act as the origin of continued layer instability. Defect No. 4 contains a main larger defect and a second smaller with arrested parallel contact lines. (b) This track will be stable when contact angle ior and local contact angle and break the stable condition. (c) Formed surface waves e is necking and leave long and shallow single layer defects. defect and can be attributed to the latter, see Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure9.34-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure9.34-1.png",
        "caption": "FIGURE 9.34 Two rectangular conductors.",
        "texts": [
            " COMPARISONS OF DIFFERENT SKIN-EFFECT MODELS 243 Finally, the surface equivalence GIBC solution that uses appropriate boundary conditions compares well with the VFI model in Section 9.3.7. However, the GIBC model requires a sufficiently large number of surface cells to form a good closed surface for a reasonably accurate result. This aspect is different for the VFI model that works well with a reduced accuracy even for a moderate number of cells or filaments. 244 SKIN EFFECT MODELING PROBLEMS 9.1 Volume filament approach Using the VFI approach, compute the impedance matrix of two parallel conductors (copper \ud835\udf0e = 5.8 \u00d7 107 S\u2215m), illustrated in Fig. 9.34. Use the subdivisions as shown in Fig. 9.15 and the shorted ends as in Fig. 9.16. Consider the dimensions xs1 = xs2 = 0, xe1 = xe2 = 2 cm, ys1 = 0, ye1 = 0.1 cm, ys2 = ye1 + 0.2 cm, ye2 = ys2 + 0.1 cm, zs1 = zs2 = 0, ze1 = ze2 = 50 \u03bcm. The result will be a frequency-dependent 2\u00d72 impedance matrix. PROBLEMS 245 9.2 Construct a Single-conductor VFI skin-effect model Construct a single-conductor skin-effect model similar to the previous problem. Compute the impedance from end to end of the rectangular conductor (copper \ud835\udf0e = 5"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000554_978-3-540-85629-0-Figure5.14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000554_978-3-540-85629-0-Figure5.14-1.png",
        "caption": "Fig. 5.14. (a) Mechanical cam (C) with a follower (F); (b) function defining an electronic cam.",
        "texts": [
            " As a matter of fact, it is possible to define the position profile of an actuator as a function of a generic variable \u03b8 instead of the time t. For example, in master-slave applications the motion profiles of the slave axis are defined with respect to the configuration of the master, which can be either real, that is an axis of motion of the machine, or virtual, that is a simple signal in the controller, implementing the so-called electronic cams. This idea derives from the mechanical cams, see Fig. 5.14, used in automatic machines with the goal of transferring, coordinating and changing the type of motion from a master device to one or more slave systems. With reference to Fig. 5.14 the body C, the cam, is supposed to rotate at a constant angular velocity, and therefore its angular position \u03b8 is a linear function of time, while body the F, the follower, has an alternative motion q(\u03b8) defined by the profile of the cam3. In the same manner, an electronic cam is defined by providing the function q(\u03b8) which describes the position of the slave with respect to \u03b8. The trajectory of the axis of motion is therefore q\u0303(t) = (q \u25e6 \u03b8)(t), with the velocity and the acceleration given by 3 The design of mechanical cams have been extensively and carefully investigated, and on this argument a wide literature is available in the mechanical field, [4, 5, 6, 7, 8, 9]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure3.62-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure3.62-1.png",
        "caption": "Figure 3.62 (a) If there is moment equilibrium, the moment vectors form a closed polygon. (b) Top view of the diagonally-halved cube with the moment vectors acting on it.",
        "texts": [
            " 98 Example 3 The cube that has been halved diagonally in Figure 3.61 is subject to a couple of 80 kNm in plane ABEF and a couple of 60 kNm in plane BCDE. The directions are shown in the figure. The body is kept in equilibrium by a couple on the diagonal plane ACDF. Question: Determine the magnitude of that couple and resolve it into a component in plane ACDF and a component perpendicular to plane ACDF. Solution: There is moment equilibrium if the moment vectors form a closed polygon. The polygon in Figure 3.62a shows that a couple of 100 kNm is acting on plane ACDF. Of this couple, the moment vector has a component perpendicular to plane ACDF of 10 \u221a 2 kNm and a component along plane ACDF of 70 \u221a 2 kNm. Figure 3.62b shows the top view for the halved cube, with all the moment vectors that act on it. When interpreting these results, one should remember that the moment vector is perpendicular to the plane on which the couple is exerted. The component of the couple that is acting in the diagonal plane has a moment vector perpendicular to that plane and is 10 \u221a 2 kNm. The component of the couple that is acting perpendicular to the diagonal plane has its moment vector in that plane and is 70 \u221a 2 kNm. ENGINEERING MECHANICS"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001409_j.engfailanal.2014.05.018-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001409_j.engfailanal.2014.05.018-Figure8-1.png",
        "caption": "Fig. 8. Schematic of crack propagation for method 2: (a) q 6 q1max, (b) q > q1max.",
        "texts": [
            " 7 increases slightly with the increase of crack propagation direction angle t, this is because a crack with a pretty large t has little effect on the rigidity of the transition part. So when the transition part of the gear tooth is considered, not only the TVMS of the healthy gear pair will change (see Fig. 4), but the TVMS of the cracked gear pair will vary with the change of crack propagation direction angles. Instead of resorting to a straight line (see straight line PG in Fig. 5) as a limiting line for reducing the tooth thickness in method 1, a parabolic curve between the crack tip and the tooth profile tip is chosen as the limiting line in method 2 (see Fig. 8), which is proposed by Mohammed et al. [1] and the crack propagation path is assumed to be the same as that in method 1. Similarly, once kb_crack and ks_crack are determined, the single-tooth-pair mesh stiffness for cracked gears can be obtained by Eq. (17). Effective area moment of inertia and area of the cross section at the position of y in method 2 can be calculated as follows: When q \u00bc q1 6 q1 max (see Fig. 8a), I \u00bc 1 12 \u00f02x\u00de3L \u00bc 2 3 x3L; y 6 yP 1 12 xB xP \u00f0yB yP \u00de2 \u00f0y yP\u00de 2 \u00fe xP \u00fe x h i3 L; y > yP 8< : ; \u00f023\u00de A \u00bc 2xL; y 6 yP xB xP \u00f0yB yP \u00de2 \u00f0y yP\u00de 2 \u00fe xP \u00fe x h i L; y > yP ( : \u00f024\u00de When q = q1max + q2 > q1max (see Fig. 8b), I \u00bc 2 3 x3L; y < ymax 1 12 x \u00f0y ymax\u00dexP yP ymax h i3 L; ymax 6 y 6 yP 1 12 xB\u00fexP \u00f0yB yP \u00de2 \u00f0y yP\u00de 2 xP \u00fe x h i3 L; y > yP 8>>< >>>: ; \u00f025\u00de A \u00bc 2xL; y < ymax x \u00f0y ymax\u00dexP yP ymax h i L; ymax 6 y 6 yP xB\u00fexP \u00f0yB yP\u00de2 \u00f0y yP\u00de 2 xP \u00fe x h i L; y > yP 8>>< >>: ; \u00f026\u00de The schematic of the calculation process for method 2 is presented in Fig. 9. Likewise, TVMS obtained by method 2 under 10 crack cases shown in Table 3 are displayed in Fig. 10 and the variation of TVMS is the same as that by method 1, except that the reduction of TVMS using method 2 is smaller than that from method 1 when the crack depth is larger and a detailed comparison will be done in Section 3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure5.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure5.2-1.png",
        "caption": "FIGURE 5.2. Position vectors of point P before and after the rotation of the local frame about the Z-axis of the global frame.",
        "texts": [
            "4) Similarly, rotation \u03b2 degrees about the Y -axis, and \u03b3 degrees about the X-axis of the global frame relate the local and global coordinates of point P by the following equations: Gr = RY,\u03b2 Br (5.5) Gr = RX,\u03b3 Br (5.6) where RY,\u03b2 = \u23a1\u23a3 cos\u03b2 0 sin\u03b2 0 1 0 \u2212 sin\u03b2 0 cos\u03b2 \u23a4\u23a6 (5.7) RX,\u03b3 = \u23a1\u23a3 1 0 0 0 cos \u03b3 \u2212 sin \u03b3 0 sin \u03b3 cos \u03b3 \u23a4\u23a6 . (5.8) Proof. Let (\u0302\u0131, j\u0302, k\u0302) and (I\u0302 , J\u0302 , K\u0302) be the unit vectors along the coordinate axes of Oxyz and OXY Z respectively. The rigid body has a space fixed point at O, which is the common origin of Oxyz and OXY Z. The dashed lines in Figure 5.2 illustrate the top view of the coordinate frames at initial position. The initial position of a body point P is indicated by P1. The position vector r1 of P1 can be expressed in body and global coordinate frames by Br1 = x1\u0131\u0302+ y1j\u0302+ z1k\u0302 (5.9) Gr1 = X1I\u0302 + Y1J\u0302 + Z1K\u0302 (5.10) where Br1 refers to the position vector r1 expressed in the body coordinate frame B, and Gr1 refers to the position vector r1 expressed in the global coordinate frame G. If the rigid body undergoes a rotation \u03b1 about the Z-axis, then the local frame Oxyz, and point P will be seen in a second position, as shown by 5. Applied Kinematics 221 the solid lines in Figure 5.2. Now the position vector r2 of P2 is expressed in both coordinate frames by Br2 = x2\u0131\u0302+ y2j\u0302+ z2k\u0302 (5.11) Gr2 = X2I\u0302 + Y2J\u0302 + Z2K\u0302. (5.12) Using Equation (5.11) and the definition of the inner product, we may write X2 = I\u0302 \u00b7 r2 = I\u0302 \u00b7 x2\u0131\u0302+ I\u0302 \u00b7 y2j\u0302+ I\u0302 \u00b7 z2k\u0302 (5.13) Y2 = J\u0302 \u00b7 r2 = J\u0302 \u00b7 x2\u0131\u0302+ J\u0302 \u00b7 y2j\u0302+ J\u0302 \u00b7 z2k\u0302 (5.14) Z2 = K\u0302 \u00b7 r2 = K\u0302 \u00b7 x2\u0131\u0302+ K\u0302 \u00b7 y2j\u0302+ K\u0302 \u00b7 z2k\u0302 (5.15) or equivalently\u23a1\u23a3 X2 Y2 Z2 \u23a4\u23a6 = \u23a1\u23a3 I\u0302 \u00b7 \u0131\u0302 I\u0302 \u00b7 j\u0302 I\u0302 \u00b7 k\u0302 J\u0302 \u00b7 \u0131\u0302 J\u0302 \u00b7 j\u0302 J\u0302 \u00b7 k\u0302 K\u0302 \u00b7 \u0131\u0302 K\u0302 \u00b7 j\u0302 K\u0302 \u00b7 k\u0302 \u23a4\u23a6\u23a1\u23a3 x2 y2 z2 \u23a4\u23a6 . (5.16) The elements of the Z-rotation matrix, RZ,\u03b1, are called the direction cosines of Br2 with respect to OXY Z. Figure 5.2 shows the top view of the initial and final configurations of r in both coordinate systems Oxyz 222 5. Applied Kinematics and OXY Z. Analyzing Figure 5.2 indicates that I\u0302 \u00b7 \u0131\u0302 = cos\u03b1, I\u0302 \u00b7 j\u0302 = \u2212 sin\u03b1, I\u0302 \u00b7 k\u0302 = 0 J\u0302 \u00b7 \u0131\u0302 = sin\u03b1, J\u0302 \u00b7 j\u0302 = cos\u03b1, J\u0302 \u00b7 k\u0302 = 0 K\u0302 \u00b7 \u0131\u0302 = 0, K\u0302 \u00b7 j\u0302 = 0, K\u0302 \u00b7 k\u0302 = 1. (5.17) Combining Equations (5.16) and (5.17) shows that\u23a1\u23a3 X2 Y2 Z2 \u23a4\u23a6 = \u23a1\u23a3 cos\u03b1 \u2212 sin\u03b1 0 sin\u03b1 cos\u03b1 0 0 0 1 \u23a4\u23a6\u23a1\u23a3 x2 y2 z2 \u23a4\u23a6 (5.18) which can also be shown in the following short notation: Gr2 = RZ,\u03b1 Br2 (5.19) RZ,\u03b1 = \u23a1\u23a3 cos\u03b1 \u2212 sin\u03b1 0 sin\u03b1 cos\u03b1 0 0 0 1 \u23a4\u23a6 . (5.20) Equation (5.19) states that the vector r at the second position in the global coordinate frame is equal to RZ times the position vector in the local coordinate frame"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure10.18-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure10.18-1.png",
        "caption": "FIGURE 10.18 Images for g22 where the observation and source points are in region 2.",
        "texts": [
            " The fundamental formulation is same for both situations. Similar to the orthogonal case, the dielectrics are represented with the same circuit elements. Using the nonorthogonal excess capacitance of a dielectric cell that is equivalent to (10.68) as Ce = \ud835\udf160(\ud835\udf16r \u2212 1) / \u222ba\u222bb\u222bc ha 2|||| \ud835\udf15r \ud835\udf15a \u22c5 ( \ud835\udf15r \ud835\udf15b \u00d7 \ud835\udf15r \ud835\udf15c )|||| da db dc, (7.24) where \ud835\udf16r is the dielectric constant of the dielectric cell. The equivalent circuit shown in Fig. 7.8 for the dielectric excess capacitance model is copied for convenience from 168 NONORTHOGONAL PEEC MODELS Fig. 10.18. Importantly, we simply can take the nonorthogonal partial inductance in (7.17) and the series resistance in (7.14). The details of the derivation are given in Section 10.4.6. The continuity equation for the rectangular PEEC models is considered in Section 6.3.1. The continuity equation must also be satisfied for the nonorthogonal case at the cell level for the currents and charges. Its differential form is given by (3.3) or \u2207 \u22c5 J + \ud835\udf15q \ud835\udf15t = 0 where again J is the current density and q is the surface charge density",
            " We should observe that this solution can also used to obtain g32 since the observation point is just in the other outer layer while the source point is in the same layer. g12(r, r\u2032) = (1 + \ud835\udefc1) 4\ud835\udf0b\ud835\udf001 [ \u221e\u2211 k=0 (\u2212\ud835\udefc1\ud835\udefc2)k [\ud835\udf0c2 + (z \u2212 z\u2032 + 2 k d)2]1\u22152 + \u221e\u2211 k=0 (\u2212\ud835\udefc1)k(\ud835\udefc2)k+1 [\ud835\udf0c2 + (z + z\u2032 + 2 k d)2]1\u22152 ] (10.56) We give the reflection diagrams for g12 in Fig. 10.17. The resultant Green\u2019s function is given in (10.57). Finally, the Green\u2019s function for the case where both the source and observation point are in layer 2 is g22. In this case, images occur on both sides. The refections are shown in Fig. 10.18. g22(r, r\u2032) = 1 4\ud835\udf0b\ud835\udf002 [ 1 [\ud835\udf0c2 + (z \u2212 z\u2032)2]1\u22152 + \u221e\u2211 k=1 (\u2212\ud835\udefc1)k\ud835\udefck\u22121 2 [\ud835\udf0c2 + (z + z\u2032 \u2212 2kd)2]1\u22152 + \u221e\u2211 k=1 (\u2212\ud835\udefc1\ud835\udefc2)k [\ud835\udf0c2 + (z \u2212 z\u2032 + 2kd)2]1\u22152 270 PEEC MODELS FOR DIELECTRICS + \u221e\u2211 k=1 (\u2212\ud835\udefc1)k\ud835\udefck+1 2 [\ud835\udf0c2 + (z + z\u2032 + 2kd)2]1\u22152 + \u221e\u2211 k=1 (\u2212\ud835\udefc1\ud835\udefc2)k [\ud835\udf0c2 + (z \u2212 z\u2032 \u2212 2kd)2]1\u22152 ] (10.57) This includes the images which are needed to construct all cases required. Note that we also included the permitivtiy needed for each Green\u2019s function for clarity. The technique presented in this section is based on the volume equivalence theorem in Section 3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure10.12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure10.12-1.png",
        "caption": "FIGURE 10.12. A two-wheel model for a vehicle moving with no roll.",
        "texts": [
            " Load transfer also occurs because of acceleration, however, assuming a linear relationship between the wheel load Fz and the cornering stiffness C\u03b1 makes the two-wheel model valid, because any increase in cornering stiffness for the more loaded wheel compensates for the decrease in the cornering stiffness of the unloaded wheel. However, when the acceleration is high, 608 10. Vehicle Planar Dynamics the load transfer is higher than the linear limit and C\u03b1 is a descending nonlinear function of Fz. Hence, at high acceleration, the load transfer causes a decrease in cornering stiffness. Example 383 Kinematic steering of a two-wheel vehicle. For the two-wheel vehicle shown in Figure 10.12, we use the cot-average (7.3) of the outer and inner steer angles as the input steer angle, cot \u03b4 = cot \u03b4o + cot \u03b4i 2 . (10.155) where, tan \u03b4i = l R1 \u2212 w 2 (10.156) tan \u03b4o = l R1 + w 2 . (10.157) The radius of rotation R for the two-wheel vehicle is given by (7.2) R = q a22 + l2 cot2 \u03b4. (10.158) and is measured at the mass center of the steered vehicle. 10. Vehicle Planar Dynamics 609 10.4 Two-wheel Rigid Vehicle Dynamics We can approximate the planar equations of motion (10.26)-(10.28) along with (10"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001777_j.optlastec.2016.04.009-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001777_j.optlastec.2016.04.009-Figure1-1.png",
        "caption": "Fig. 1. Schematic overview of the interaction zone between laser radiation and powder bed.",
        "texts": [
            " The effects of laser power and scan speed on the thermal behavior and the forming mechanism of cross-sectional configuration of the molten pool were analyzed. The formation mechanisms of processing defects, such as \u201cballing effect\u201d , warpage, pores, microcracks, delamination, etc. were discussed in order to optimize the processing parameters and obtain desired SLM-produced parts. Meanwhile, the corresponding experiments were also implemented to investigate the microstructure of the SLM-produced components with different laser processing conditions for verifying the reliability and accuracy of the model proposed in this paper. Fig. 1 depicts the schematic overview of the interaction zone between laser radiation and powder. As the top surface of the powder bed is irradiated by the incident laser beam, a small fraction of laser energy is dissipated by radiation and convection. The remainder, the vast majority of laser energy, is absorbed by powder particles, leading to the rapid heating and resultant localized melting. After the moving Gaussian laser heat source leaves the melt region, which is called molten pool, rapid consolidation of the melt occurs",
            " On the other hand, the processing conditions play a crucial role in influencing thermal behavior and attendant mechanical properties of as-fabricated parts. When the laser beam scans through the powder bed with a high speed (300 mm/s) or a relatively high laser power (150 W), the particles within the laser radiation region undergo an extremely rapid heating process with the maximum heating rate of 1.30 107 \u00b0C/s, which potentially results in thermo-capillary instability in the molten pool and droplets splashing from the melt pool. It is responsible for the balling effect on the top surface of every fabricating layer (Fig. 1). Meanwhile, local residual stress accumulation maybe occurs for the enormous cooling rate of 1.26 107 \u00b0C/s under scan speed of 300 mm/s or laser power of 150 W. It possibly gives rise to the increment of cracking susceptibility, distortion of SLM-processed parts and delamination between neighbor layers. It is worth noting that the combination of the lower heating rate of 4.52 106 \u00b0C/s and cooling rate of 4.38 106 \u00b0C/s has a tendency to reduce the possibility of these above metallurgical defects under a relatively low scan speed of 100 mm/s and a laser power of 125 W",
            " However, An low scan speed (50 mm/s) or an intense laser power (150 W) applied in the SLM process yields an extremely long liquid lifetime g SLM process using different processing parameters. (a) different laser powers (5.15 ms), maybe, resulting in the formation of gas since the gas atoms release from the lattice in heat affected zone, as proposed by Weingarten [34]. This phenomenon causes the appearance of spherical pores and the attendant high porosity in SLM-produced parts. Moreover, these pores appear generally in the back-end of the molten pool, as revealed by Fig. 1. Meanwhile, when an overwhelmingly high scan speed (300 mm/s) or an exceedingly low laser power (75 W) is used, an extremely short liquid lifetime (0.30 ms) and low temperature (1524 \u00b0C) occur, causing the formation of a small amount of liquid phase with a relatively high viscosity. It is detrimental for the wetting of the liquid phase among the gaps of powders, generating the appearance of irregular pores and the attendant high porosity in SLM-produced parts. In conclusion, an appropriate scan speed (100 mm/s) combined with a suitable laser power (125 W) plays a paramount role in SLM process of TiC/Inconel 718 with an appropriate temperature of 2291 \u00b0C and a proper liquid lifetime of 1"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure5.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure5.6-1.png",
        "caption": "Figure 5.6 Graphical method for finding the pile forces: (a) line of action figure and (b) force polygon.",
        "texts": [
            "8): The pile forces can also be found graphically. Imagine F (a), F (b) and F (c) are the forces that the piles exert on the structure. These forces have to be in equilibrium with the load F , so that: F (a) + F (b) + F (c) = F or F (a) + F (b) = F \u2212 F (c). In addition to the force equilibrium there also has to be moment equilibrium. Therefore ( F (a)+ F (b)) and ( F \u2212 F (c)) have a common line of action. The line of action of ( F (a)+ F (b)) passes through Sab and that of ( F \u2212 F (c)) passes through P, see the line of action figure in Figure 5.6a. The common line of action is therefore PSab. Since ( F (a)+ F (b)) and F (c) are in equilibrium with F in P, ( F (a)+ F (b)) and F (c) can be obtained from a force polygon (see Figure 5.6b). ( F (a) + F (b)) can then be resolved in Sab into F (a) and F (b). The force polygon in Figure 5.6b now shows: F (a) = 150 kN \u2193; F (b) = 60 \u221a 10 kN \u2191 and F (c) = 30 kN \u2191. These are the forces that the piles exert on the retaining wall. Translated into the pile forces with the correct sign for tension and compression one now finds: N(a) = +150 kN; N(b) = \u221260 \u221a 10 kN and N(c) = \u221230 kN. 160 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Example 4 The simply supported beam ABC in Figure 5.7a consists of the members AB and BC that are connected rigidly in joint B. The beam is loaded at joint B by a couple of 30 kNm"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001963_j.bios.2013.11.007-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001963_j.bios.2013.11.007-Figure6-1.png",
        "caption": "Fig. 6. (a) A photograph of the fully-assembled paper-based MFC and (b) schematic diagram of the device (Fraiwan et al., 2013a) .",
        "texts": [
            " Microorganisms oxidize organic matter in the anodic chamber, completing respiration by transferring electrons to the anode. During this process, chemical energy is captured throughout the electron transport chain. Nicotinamide adenine dinucleotide (NAD\u00fe) and nicotinamide adenine dinucleotide dehydrogenase (NADH) function as coenzymes for the reactions, repeatedly oxidizing and reducing to synthesize adenosine triphosphate (ATP), the biological energy unit (Logan and Regan, 2006). The first paper-based MFC was developed by our group (Fraiwan et al., 2013a, 2013b, 2013c) (Fig. 6). The MFC featured (i) a paper-based proton exchange membrane by infiltrating sodium polystyrene sulfonate, (ii) micro-fabricated paper chambers by patterning hydrophobic barriers of photoresist and (iii) paper reservoirs for holding the anolyte and catholyte for an extended period of time. The rapid electricity generation should be noted since conventional MFCs require long start-up times (typically several days to a week), which is attributed to the accumulation and acclimation of bacteria on the anode of MFCs (Qian et al"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001272_j.apsusc.2007.08.046-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001272_j.apsusc.2007.08.046-Figure1-1.png",
        "caption": "Fig. 1. Laser sintered vector from Inconel 625 powder on steel substrate (thickness of powder layer is 50 mm) for V = 0.13 m/s.",
        "texts": [
            " At laser power 50 W, laser spot diameter 70 mm, scanning speed 0.13 m/s and powder layer thickness 50 mm, the width of an individual vector is about 120 mm, i.e. a track of consolidated powder is larger than the laser spot diameter. Besides, denuded from powder zones of about 30 mm appear on the substrate (or on the previous remelted layer) on both sides of the vector. At a hatch distance less than 120 mm a partial second remelting of the previous vector occurs. Because of the fact that track of consolidated powder is larger than the laser spot size (Fig. 1), after a certain number of vectors breaks appear in the remelted layer. This results in a repetitive fault in the subsequent SLM cycle and, therefore, in a defective porous structure. At a hatch distance of 60 mm (Fig. 2a), the crosssection of the sample represents a regular structure with pores. The angle of slope a of the pore chains on the laser scanning planes is about 508 (Fig. 2c), porosity of the samples is about d). Vectors are oriented perpendicularly to the cross-section plane. View of the f the manufactured samples; angle a is measured between the scanning direction 6"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000459_1521-4109(200108)13:12<983::aid-elan983>3.0.co;2-#-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000459_1521-4109(200108)13:12<983::aid-elan983>3.0.co;2-#-Figure1-1.png",
        "caption": "Fig. 1. Schematic of a \u2018\u2018first-generation\u2019\u2019 glucose biosensor (based on a probe manufactured by YSI Inc.).",
        "texts": [
            " First-generation devices have relied on the use of the natural oxygen cosubstrate, and the production and detection of hydrogen peroxide (Equations 1 and 2). Such measurements of Electroanalysis 2001, 13, No. 12 # WILEY-VCH Verlag GmbH, D-69469 Weinheim, 2001 1040-0397/01/1208\u20130983 $17.50\u00fe.50=0 peroxide formation has the advantage of being simpler, especially when miniaturized sensors are concerned. A very common configuration is the YSI probe, involving the entrapment of GOx between an outer diffusion-limiting=biocompatible polycarbonate membrane and an inner anti-interference cellulose acetate one (Fig. 1). The amperometric measurement of hydrogen peroxide requires application of a potential at which coexisting species, such as ascorbic and uric acids or acetaminophen, are also electroactive. The anodic contributions of these and other oxidizable constituents of biological fluids can compromise the selectivity and hence the overall accuracy. Extensive efforts during the 1980s were devoted for minimizing the error of electroactive interferences in glucose electrodes. One useful strategy is to employ a permselective coating that minimizes access of such constituent to the transducer surface"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure7.8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure7.8-1.png",
        "caption": "FIGURE 7.8 Model for volume capacitance from Section 10.4.6.",
        "texts": [
            " Using the excess capacitance model in Section 10.4.6, we can also treat dielectric hexagonal blocks. The fundamental formulation is same for both situations. Similar to the orthogonal case, the dielectrics are represented with the same circuit elements. Using the nonorthogonal excess capacitance of a dielectric cell that is equivalent to (10.68) as Ce = \ud835\udf160(\ud835\udf16r \u2212 1) / \u222ba\u222bb\u222bc ha 2|||| \ud835\udf15r \ud835\udf15a \u22c5 ( \ud835\udf15r \ud835\udf15b \u00d7 \ud835\udf15r \ud835\udf15c )|||| da db dc, (7.24) where \ud835\udf16r is the dielectric constant of the dielectric cell. The equivalent circuit shown in Fig. 7.8 for the dielectric excess capacitance model is copied for convenience from 168 NONORTHOGONAL PEEC MODELS Fig. 10.18. Importantly, we simply can take the nonorthogonal partial inductance in (7.17) and the series resistance in (7.14). The details of the derivation are given in Section 10.4.6. The continuity equation for the rectangular PEEC models is considered in Section 6.3.1. The continuity equation must also be satisfied for the nonorthogonal case at the cell level for the currents and charges"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003443_0278364903022001005-Figure9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003443_0278364903022001005-Figure9-1.png",
        "caption": "Fig. 9. 3-DoF 3 \u2212 u1 Ru2 Pu1 R(iRjR)N rotational PM.",
        "texts": [
            " The standard base of the mechanism twist system is at OhioLink on October 23, 2014ijr.sagepub.comDownloaded from $m1 = (1 0 0 ; 0 0 0) $m2 = (0 1 0 ; 0 0 0) $m3 = (0 0 1 ; 0 0 0). (32) The standard base of the mechanism constraint system is $r m1 = (1 0 0 ; 0 0 0) $r m2 = (0 1 0 ; 0 0 0) $r m3 = (0 0 1 ; 0 0 0), (33) which includes three non-coplanar constraint forces passing through a common point. To form such a mechanism constraint system, the limb constraint system may only contain one constraint force or two constraint forces or three constraint forces. Figure 9 shows a 3-DoF 3 \u2212 u1 Ru2 Pu1 R(iRjR)N spherical PM, which was presented by Karouia and Herv\u00e9 (2002). Counting from the base, the first revolute axis $i1 and the third revolute axis $i3 are parallel to each other and bevel with the base plane. The second prismatic pair $i2 is perpendicular to the two adjacent revolute axes. The fourth revolute axis $i4 and the fifth revolute axis $i5 intersect at a common point and form a 2R spherical subchain. The three limb central points coincide with each other"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure5.10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure5.10-1.png",
        "caption": "Fig. 5.10. Coreless AFPM machine stators with (a) normal overlap winding, (b) non-overlap winding, (c) phase-group non-overlap winding, and (d) testing set-up of AFPM machine as a generator [146].",
        "texts": [
            " In coreless non-overlap winding AFPM machines this is almost completely absent due to the low armature reaction effect. There are, thus, no disadvantages in using non-overlap coils with concentrated parameters in coreless stator AFPM machines, except for the lower torque performance in the case of some winding and PM configurations. In Fig. 5.9 the layouts of overlap and non-overlap windings are shown. Congruent with these layouts, three air-cored AFPM stators (1 kW) with normal overlap, non-overlap and phase-group windings are shown in Fig. 5.10(a)-(c). These stators have been designed and built for the PM rotor discs of the AFPM machine under test shown in Fig. 5.10(d). For the tests an a.c. electromechanical drive has been used as prime mover and a torque transducer to measure the shaft torque as shown in Fig. 5.10(d). For the numeric modeling of the AFPM machine the 2D FEM analysis has been used. The 2D FEM models of the normal three-phase overlap winding and two types of non-overlap coil AFPM machines are shown in Fig. 5.11. Owing to the axial symmetry, it is only necessary to model half of the machine in the FEM analysis, i.e., one rotor disc and half a stator. For the normal overlap winding AFPM machine, it is possible to model just one polepitch of the machine by applying negative periodical condition on the left and right boundaries, as shown in Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003786_1045389x10369718-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003786_1045389x10369718-Figure5-1.png",
        "caption": "Figure 5. New light on nanocomposites. Bimodal and reversible, actuation of a CNT/elastomer nanocomposite induced by infrared irradiation. Reversible expansion occurs at small pre-strains (top) and reversible contraction at large pre-strains (bottom). Reproduced with permission from Nature Publishing Group, http://dx.doi.org/10.1038/nmat1400.",
        "texts": [
            " The coolinginduced crystallization of cross-linked PCO films under tension (not deformation to obtain a temporary shape) led to significant elongation, and subsequent heating to melt the network resulted in contraction of the polymer film. They found that the two-way SME was affected by the cross-linking density of the PCO network and the stress applied to stretch the PCO chains at high temperature. Finite external stress was required for the two-way SME. The two-way SME was also observed on a multiwalled carbon nanotube (CNT)/elastomer composite by Ahir and Vaia as shown in Figure 5 (Ahir and Terentjev, 2005; Vaia, 2005). Infrared light was used as the heat source to heat the CNT/elastomer composite. They found that the expansion or contraction of the composite depends on the extent to which the composite is strained. If the material is slightly pulled, it expands if it is exposed to infrared light. Conversely, if the material is subject to a strain greater than 10%, it contracts under identical exposure to infrared light. This process is completely reversible and persists after numerous cycles"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002595_978-3-030-19416-1-Figure3.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002595_978-3-030-19416-1-Figure3.2-1.png",
        "caption": "Fig. 3.2 (a) Scheme of spray drier Buchi B-290 with 3-fluid nozzle (1) compressor; (2) inlet filter; (3) heating coil; (4) nozzle; (5) stock solutions; (6) drying chamber; (7) cyclone; (8) dry product; (9) outlet filter; (10) aspirator. (b) Scheme of droplets",
        "texts": [
            " Some of the advantages of using the spray drying technique are: (i) versatility of the applications for several raw materials and sensitive or temperature resistant materials; (ii) its reproducible production of particles; and (iii) process flexibility (Oliveira 2005). The particle properties of the final product will depend on the process parameters which involves four main steps: the feed through an atomizer nozzle, followed by the atomization of the solution and contact of the droplets produced with the pressurized air, the drying process, and the collection of the obtained particles (Fig.\u00a03.2). He et\u00a0al. (1999) evaluated the influence of atomization process parameters, such as the type of atomizer nozzle, the inlet and outlet temperature, the flow rate of the nozzle peristaltic pump and drying air flow, on the properties of the particles (size, morphology, zeta potential, among others). Particle size is mainly affected but not the zeta potential. They considered the type of spray nozzle and the inlet temperature used as the major influencers. A variety of atomizing nozzles, e.g. the centrifugal, pressure or 2-fluid nozzle (2FN) have been applied. Recently, the 3-fluid nozzle (3FN) (Fig.\u00a03.2) has been used especially for the encapsulation of immiscible liquids such as oils or aromas (Legako and Dunford Spray drying technique applied to agricultural purposes has few articles in the literature. Fran\u00e7a et\u00a0 al. (2018) obtained microspheres and microcapsules of controlled release fertilizer by a swelling mechanism. They used the 2-fluid nozzle to obtain microsphere where the nutrient was dispersed on the polymeric matrix of Cs. To obtain microcapsules, they used the 3-fluid nozzle and the nutrient was concentrated in the core and covered by a polymeric shell of Cs (Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000845_s0261-3069(99)00078-3-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000845_s0261-3069(99)00078-3-Figure2-1.png",
        "caption": "Fig. 2. Schematic representation of a plate being deposited by overlapping traverses of the molten pool across the width dimension of the plate. As-deposited properties of the solidified material depend on the molten pool characteristics produced by the process parameters chosen and the heat flow characteristics away from the pool.",
        "texts": [
            " Successive layers are stacked and the entire part or specific feature of a part assembly is built additively. Features or assembly components that require deposition in planar layers at angled orientations to the parent or base feature can be deposited by tilting the laser beam-powder delivery head or tilting the work piece so that the beam axis is normal to the deposition plane. This capability provides a means of producing overhanging features without building underlying support structure typically required in plastic rapidprototyping processes. Fig. 2 schematically shows a plate type of structure being built. A small molten pool is created at the focal zone of the laser beam that can vary in size from one-half to five times the focal-spot diameter of the laser beam, depending on the power and velocity of the moving spot. For example, the Nd-YAG laser used in DLF processing is focused to 0.5 mm and has been used to produce solidified features 0.3]2.5 mm thick with a single pass by varying the laser power and translation speed. Thicker components are made with multiple over-lapping beads"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure1.11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure1.11-1.png",
        "caption": "FIGURE 1.11. Ground-sticking behavior of radial and non-radial tires in the presence of a lateral force.",
        "texts": [
            " One ply is set on a bias in one direction as succeeding plies are set alternately in opposing directions as they cross each other. The ends of the plies are wrapped around the bead wires, anchoring them to the rim of the wheel. Figure 1.10 shows the interior structure and the carcass arrangement of a nonradial tire. The most important difference in the dynamics of radial and non-radial tires is their different ground sticking behavior when a lateral force is applied on the wheel. This behavior is shown in Figure 1.11. The radial tire, shown in Figure 1.11(a), flexes mostly in the sidewall and keeps the tread flat on the road. The bias-ply tire, shown in Figure 1.11(b) has less contact with the road as both tread and sidewalls distort under a lateral load. The radial arrangement of carcass in a radial tire allows the tread and sidewall act independently. The sidewall flexes more easily under the weight 16 1. Tire and Rim Fundamentals of the vehicle. So, more vertical deflection is achieved with radial tires. As the sidewall flexes under the load, the belts hold the tread firmly and evenly on the ground and reduces tread scrub. In a cornering maneuver, the independent action of the tread and sidewalls keeps the tread flat on the road"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure3.30-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure3.30-1.png",
        "caption": "Figure 3.30 Corner joint in a frame; the three unknown section forces M , V and N can be deduced from the equilibrium.",
        "texts": [
            " The resultant force is zero if the condition for force equilibrium is met in the direction AB, or in another direction that is not perpendicular to AB. The equilibrium conditions can therefore be formulated in various ways. For a manual calculation, one always has to look for equilibrium equations that are as simple as possible in order to limit the amount of calculation. When using a computer for the calculation, the systematics and the general applicability of the set up of the calculation (the program) are more important than the number of calculations involved and the laborious character of the calculations. Example Figure 3.30 shows the corner joint of a frame. The joint is loaded at C by a vertical force of 5 kN. So-called section forces act on the crosssectional planes a and b. They act in the centre lines shown. The system is in equilibrium. Question: Determine the three unknown section forces M , V and N (with the correct sign for the directions shown).1 1 M (bending moment), V (shear force) and N (normal force) are section forces. Their nomenclature and sign conventions will be revealed in Chapter 10. Figure 3.29 The relationships \u2211 Tz|A = 0 and \u2211 Tz|B = 0 imply that there is no resultant couple and that, if there is a resultant force R, its line of action is along AB",
            " 3 Statics of a Rigid Body 75 Solution: The two unknown section forces V and N are determined using the two equations for the force equilibrium. For the coordinate system shown, \u2211 Fx = +(4 kN) \u2212 2 5 \u221a 5 \u00d7 V + 1 5 \u221a 5 \u00d7 N = 0,\u2211 Fy = +(2 kN) \u2212 (5 kN) \u2212 1 5 \u221a 5 \u00d7 V \u2212 2 5 \u221a 5 \u00d7 N = 0. These are two equations with two unknowns. The solution is V = +\u221a 5 kN and N = \u22122 \u221a 5 kN. It would also be possible to construct a closed force polygon and to derive the forces from there. This is shown in Figure 3.31. The force of 2 \u221a 5 kN on the line of action of N is active in an opposite direction to that shown in Figure 3.30. That is why there is a minus sign in the expression for N . M is found using the equation for the moment equilibrium about an arbitrary point. If A is selected, the contribution of V and N to the moment is zero, and M can be found even if V and N are still unknown: \u2211 Tz|A = \u2212M \u2212 (4 kN)(1 m) \u2212 (2 kN)(1.5 m) + +(5 kN)(0.5 m) + (16 kNm) = 0 \u21d2 M = 11.5 kNm. If M had been calculated first, one would be able to derive V directly afterwards from, for example, the moment equilibrium about C: \u2211 Tz|C= \u2212(11"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure3.8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure3.8-1.png",
        "caption": "Fig. 3.8 Ground Reaction Forces and Equivalent Force and Moment for 3D Models",
        "texts": [
            " Let \u03c3x(\u03be, \u03b7) and \u03c3y(\u03be, \u03b7) be the x and y components, respectively, of the horizontal ground reaction forces. The sum of them can be expressed as fx = \u222b S \u03c3x(\u03be, \u03b7)dS (3.14) fy = \u222b S \u03c3y(\u03be, \u03b7)dS. (3.15) The moment \u03c4 t(p) of the horizontal ground reaction force about a point p on the ground surface is expressed as \u03c4 t(p) \u2261 [\u03c4tx \u03c4ty \u03c4tz] T (3.16) \u03c4tx = 0 \u03c4ty = 0 \u03c4tz = \u222b S {(\u03be \u2212 px)\u03c3y(\u03be, \u03b7)\u2212 (\u03b7 \u2212 py)\u03c3x(\u03be, \u03b7)}dS. These equations mean that the horizontal ground reaction forces generate the vertical component of the moment. From the above discussions, we can see that, as shown in Fig. 3.8, the ground reaction forces distributed over the surface of the sole can be replaced by the force f = [fx fy fz] T , and the moment \u03c4 p = \u03c4n(p) + \u03c4 t(p) = [0 0 \u03c4tz] T , about the ZMP p. When a robot moves, \u03c4tz = 0 is not generally satisfied. Thus, the ZMP is not a point where all components of the moment becomes zero for 3D cases. The ZMP is defined as the point where the horizontal component of the moment of the ground reaction forces becomes zero for 3D cases. 76 3 ZMP and Dynamics 2 Region of ZMP in 3D Let us define the region of the ZMP for 3D cases"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000402_s00170-011-3566-1-Figure13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000402_s00170-011-3566-1-Figure13-1.png",
        "caption": "Fig. 13 Schematic diagram showing the effect of layer thickness on the wetting condition",
        "texts": [
            " Therefore, the thicker layer thickness is unfavorable for the wetting ability of molten pool regardless of the single track or the multi-layers process. The above results can be explained as follows. Firstly, at a thicker thickness, the layer energy absorbed in unit volume of powder is insufficient; hence, the temperature of molten pool is low, resulting in a weak flow ability and balling phenomenon. Secondly, although a thicker layer thickness could enable a big molten pool, the molten pool is far away from the substrate, leading to a relatively small contact area betweenmolten pool and substrate, as schematically illustrated in Fig 13. In this condition, the small wetting area could not support a big molten pool, thereby the molten track tends to break up into balls. The above balling discipline was established based on 316L stainless steel. However, whether the above balling discipline was appropriate for the other metal powder was still not clear. In order to disclose this, the pure nickel powder was selected as representative material (Fig. 1b) for laser scanning experiment to investigate its balling behavior. The experiment of single track scanning of pure nickel material was also performed on the sloped steel substrate (Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure3.29-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure3.29-1.png",
        "caption": "Fig. 3.29. A numerical example: free-collision path of the end-effector of a PUMA-like industrial manipulator among spherical modeled obstacles.",
        "texts": [
            " Chapter 3: Fundamentals of the Mechanics of Robots136 However, the procedure does not seem to be greatly affected by small variations in the discretization size \u0394 of the work area and the number of d.o.f.s of the manipulator. Finally, the efficiency of the numerical procedure is greatly affected by the sizes b, h, and w of the work area. By enlarging the work area the computational time has grown considerably. This is because the procedure has been considered with more adjacent configurations in the Map generation. Thus, it is desirable to restrict the work area as much as possible, to minimize the computational time. The example of Fig. 3.29 illustrates the free-collision path obtained with 64 configurations for an industrial PUMA 560-like robot (m = 5) with l1 = 255 u, l2 = 450 u, l3 = 100 u, l4 = 433 u l5 = 420 u, from P5 START (800, -200, 600) to P5 END (800, 350, - 400) assuming a Ci = (104, 232, 0, 97, 236, 450, 32, 164, 450, 451, -32, 437, 800, -200, 600). The work area with Hw (-200, -200, -400) and b = h = w = 1,000 u has been discretized by \u0394 = 10 u. The obstacles have been considered as a union of 25 equal circles with rj = 100 u; to avoid interference with the robot trunk, it has been modeled as an obstacle through a further 8 circles with rj = 150 u: in total there are v = 33 obstacles in the work area environment",
            "33) have been considered to take into account the specific geometry of the robot and the real bounds on the revolute joints. However, the procedure has been worked essentially with the same features of planar cases. A further practical feasibility of the proposed path planning procedure can be deduced by observing that the obtained trajectory can be performed with an easy programming development in industrial robots, because of the imposition in Eqs (3.2.33) of straightline displacements. Fundamentals of Mechanics of Robotic Manipulation 137 Thus, for the example of Fig. 3.29 with a PUMA 560-like robot, it may be very easy to obtain a program by using few instructions for linear movements of the reference point on the end-effector. In fact, it is possible to program the computed free-collision path, for example, through the MOVES motion instruction of VAL-II, which can be used to connect the extreme points of straight-line segments. The adjacent condition of the configuration sequence will ensure that during the MOVES action the robot will move along the computed path configurations"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000857_j.matdes.2014.07.043-Figure11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000857_j.matdes.2014.07.043-Figure11-1.png",
        "caption": "Fig. 11. (a) Convex radii and (b) overhang of 15 mm. Overhang of 9 mm: (c) titanium and (d) aluminum.",
        "texts": [
            " Regarding the concave radii, in the top of the radius, the tangent decreases to 0 as the part grows in the z-axis, resulting in excessive layer individual overhangs. It was identified that as the tangent angle a (Fig. 10a) increases, the warping distortion decreases and the accuracy of the curve was improved (Fig. 10e). In Fig. 10b, it is possible to see that titanium has a higher definition than aluminum (Fig. 10c). The pieces with a greater overhang of 9 mm have presented problems in construction 1 For interpretation of color in Fig. 11, the reader is referred to the web version of this article. compared to the aluminum ones. In fact, the aluminum parts, although with the deformations of the radius of curvature, are not collapsed unlike of the pieces in titanium (Fig. 10d). The lowest point of a traditional convex radius is vulnerable to warp. The titanium samples with an overhang of 15 mm are collapsed unlike of the aluminum samples. However, the accuracy of the radii in the aluminum sample is poor (Fig. 11b) and the radius did not resemble the original CAD data for tangent angle b (Fig. 11a) lower than 40 . In Fig. 11, it is possible to see that titanium sample (Fig. 11c) has a higher definition than aluminum sample (Fig. 10d) for the convex radii sample: maximum distance between CAD model and the built part is of 1.23 mm for aluminum (red1 color in Fig. 11d), instead the distance maximum is 0.90 mm for titanium The standard deviation of the part respect to the CAD model is 0.1654 for titanium and 0.223 for aluminum. The complete set of experiments is shown in Table 3. All titanium samples showed a greater ease of removal from the platform with respect to the aluminum pieces. For this reason, only for aluminum samples, a column was inserted with the result on the difficulty (\u2013) or ease (x) to detach the samples of the building platform. The difficulty or the ease to detach a sample are measured by the time necessary to remove it from the platform: easy corresponds to less than 3 min, difficult to more than 5 min"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure12.16-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure12.16-1.png",
        "caption": "Figure 12.16 From the moment equilibrium of joint B it follows that the bending moment in B \u201cgoes round the corner\u201d.",
        "texts": [
            " We now have to determine only the bending moment at B. Without resolving F , we can calculate MB from the moment equilibrium of the isolated segment BC: MB = (6 kN)(1.5 m) = 9 kNm. Of course it is also possible to determine MB from the equilibrium of the isolated segment AB. The M diagram is shown in Figure 12.15c. All values are plotted normal to the member axis. The bending moment at joint B is the same magnitude on both sides of the joint and is also plotted at the same side. This follows directly from the moment equilibrium of joint B (see Figure 12.16, which shows only the bending moments). It is said that the bending moment at B \u201cgoes round the corner\u201d, which is further emphasised in Figure 12.15c by the dotted arc (normally not drawn). Figure 12.15 (a) The support reactions at A of a lighting mast loaded by a vertical force at C. (b) The force at C resolved into components normal to and parallel to member axis BC. (c) Bending moment diagram. The bending moment \u201cgoes round the corner\u201d at B. (d) Shear force diagram. (e) Normal force diagram. 12 Bending Moment, Shear Force and Normal Force Diagrams 497 The shear forces can be calculated directly from the slopes of the M diagram (rule 5): V (AB) = M(AB) (AB) = 0 kNm 3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.65-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.65-1.png",
        "caption": "FIGURE 8.65.",
        "texts": [],
        "surrounding_texts": [
            "where, a1 is the longitudinal distance between C and the front axle, b1 is the lateral distance between C and the tireprint of the tire 1, and h is the height of C from the ground level. If P is a point in the tire frame at T1rP T1rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 . (8.62) then its coordinates in the body frame are BrP = BRT1 T1rP + BdT1 = \u23a1\u23a3 a1 + xP cos \u03b41 \u2212 yP sin \u03b41 yP cos \u03b41 \u2212 b1 + xP sin \u03b41 zP \u2212 h \u23a4\u23a6 . (8.63) The rotation matrix BRT1 is a result of steering about the z1-axis. BRT1 = \u23a1\u23a3 cos \u03b41 \u2212 sin \u03b41 0 sin \u03b41 cos \u03b41 0 0 0 1 \u23a4\u23a6 (8.64) Employing Equation (8.28), we may examine a wheel point P at W rP W rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 (8.65) 8. Suspension Mechanisms 497 and find the body coordinates of the point BrP = BRT1 T1rP + BdT1 = BRT1 \u00a1 T1RW W rP + T1dW \u00a2 + BdT1 = BRT1 T1RW W rP + BRT1 T1dW + BdT1 = BRW W rP + BRT1 T1dW + BdT1 (8.66) BrP = \u23a1\u23a3 a1 + xP cos \u03b41 \u2212 yP cos \u03b3 sin \u03b41 + (Rw + zP ) sin \u03b3 sin \u03b41 xP sin \u03b41 \u2212 b1 + yP cos \u03b3 cos \u03b41 \u2212 (Rw + zP ) cos \u03b41 sin \u03b3 (Rw + zP ) cos \u03b3 + yP sin \u03b3 \u2212 h \u23a4\u23a6 (8.67) where, BRW = BRT1 T1RW = \u23a1\u23a3 cos \u03b41 \u2212 cos \u03b3 sin \u03b41 sin \u03b3 sin \u03b41 sin \u03b41 cos \u03b3 cos \u03b41 \u2212 cos \u03b41 sin \u03b3 0 sin \u03b3 cos \u03b3 \u23a4\u23a6 (8.68) T1dW = \u23a1\u23a3 0 \u2212Rw sin \u03b3 Rw cos \u03b3 \u23a4\u23a6 . (8.69) Example 334 F Wheel-body to vehicle transformation. The wheel-body coordinate frames are always parallel to the vehicle frame. The origin of the wheel-body coordinate frame of the wheel number 1 is at BdW1 = \u23a1\u23a3 a1 \u2212b1 \u2212h+Rw \u23a4\u23a6 . (8.70) Hence the transformation between the two frames is only a displacement. Br = BIW1 W1r+ BdW1 (8.71) 8.6 F Caster Theory The steer axis may have any angle and any location with respect to the wheel-body coordinate frame. The wheel-body frame C (xc, yc, zc) is a frame at the center of the wheel at its rest position, parallel to the vehicle coordinate frame. The frame C does not follow any motion of the wheel. The steer axis is the kingpin axis of rotation. Figure 8.51 illustrates the front and side views of a wheel and its steering axis. The steering axis has angle \u03d5 with (yc, zc) plane, and angle \u03b8 with (xc, zc) plane. The angles \u03d5 and \u03b8 are measured about the yc and xc axes, 498 8. Suspension Mechanisms respectively. The angle \u03d5 is the caster angle of the wheel, while the angle \u03b8 is the lean angle. The steering axis of the wheel, as shown in Figure 8.51, is at a positive caster and lean angles. The steering axis intersect the ground plane at a point that has coordinates (sa, sb,\u2212Rw) in the wheelbody coordinate frame. If we indicate the steering axis by the unit vector u\u0302, then the components of u\u0302 are functions of the caster and lean angles. C u\u0302 = \u23a1\u23a3 u1 u2 u3 \u23a4\u23a6 = 1p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 \u23a1\u23a3 cos \u03b8 sin\u03d5 \u2212 cos\u03d5 sin \u03b8 cos \u03b8 cos\u03d5 \u23a4\u23a6 (8.72) The position vector of the point that u\u0302 intersects the ground plane, is called the location vector s that in the wheel-body frame has the following coordinates: Cs = \u23a1\u23a3 sa sb \u2212Rw \u23a4\u23a6 (8.73) We express the rotation of the wheel about the steering axis u\u0302 by a zero pitch screw motion s\u030c. CTW = C s\u030cW (0, \u03b4, u\u0302, s) = \u2219 CRW Cs\u2212 CRW Cs 0 1 \u00b8 = \u2219 CRW CdW 0 1 \u00b8 (8.74) Proof. The steering axis is at the intersection of the caster plane \u03c0C and the lean plane \u03c0L, both expressed in the wheel-body coordinate frame. The 8. Suspension Mechanisms 499 two planes can be indicated by their normal unit vectors n\u03021 and n\u03022. C n\u03021 = \u23a1\u23a3 0 cos \u03b8 sin \u03b8 \u23a4\u23a6 (8.75) C n\u03022 = \u23a1\u23a3 \u2212 cos\u03d50 sin\u03d5 \u23a4\u23a6 (8.76) The unit vector u\u0302 on the intersection of the caster and lean planes can be found by u\u0302 = n\u03021 \u00d7 n\u03022 |n\u03021 \u00d7 n\u03022| (8.77) where, n\u03021 \u00d7 n\u03022 = \u23a1\u23a3 cos \u03b8 sin\u03d5 \u2212 cos\u03d5 sin \u03b8 cos \u03b8 cos\u03d5 \u23a4\u23a6 (8.78) |n\u03021 \u00d7 n\u03022| = q cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 (8.79) and therefore, C u\u0302 = \u23a1\u23a3 u1 u2 u3 \u23a4\u23a6 = \u23a1\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3 cos \u03b8 sin\u03d5p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 \u2212 cos\u03d5 sin \u03b8p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 cos \u03b8 cos\u03d5p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 \u23a4\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6 . (8.80) Steering axis does not follow any motion of the wheel except the wheel hop in the z-direction. We assume that the steering axis is a fixed line with respect to the vehicle, and the steer angle \u03b4 is the rotation angle about u\u0302. The intersection point of the steering axis and the ground plane defines the location vector s. Cs = \u23a1\u23a3 sa sb \u2212Rw \u23a4\u23a6 (8.81) The components sa and sb are called the forward and lateral locations respectively. Using the axis-angle rotation (u\u0302, \u03b4), and the location vector s, we can define the steering process as a screw motion s\u030c with zero pitch. Employing Equations (5.473)-(5.477), we find the transformation screw for wheel frame W to wheel-body frame C. CTW = C s\u030cW (0, \u03b4, u\u0302, s) (8.82) = \u2219 CRW Cs\u2212 CRW Cs 0 1 \u00b8 = \u2219 CRW Cd 0 1 \u00b8 500 8. Suspension Mechanisms CRW = I cos \u03b4 + u\u0302u\u0302T vers \u03b4 + u\u0303 sin \u03b4 (8.83) CdW = \u00a1\u00a1 I\u2212 u\u0302u\u0302T \u00a2 vers \u03b4 \u2212 u\u0303 sin \u03b4 \u00a2 Cs. (8.84) u\u0303 = \u23a1\u23a3 0 \u2212u3 u2 u3 0 \u2212u1 \u2212u2 u1 0 \u23a4\u23a6 (8.85) vers \u03b4 = 1\u2212 cos \u03b4 (8.86) Direct substitution shows that CRW and CdW are: CRW = \u23a1\u23a3 u21 vers \u03b4 + c\u03b4 u1u2 vers \u03b4 \u2212 u3s\u03b4 u1u3 vers \u03b4 + u2s\u03b4 u1u2 vers \u03b4 + u3s\u03b4 u22 vers \u03b4 + c\u03b4 u2u3 vers \u03b4 \u2212 u1s\u03b4 u1u3 vers \u03b4 \u2212 u2s\u03b4 u2u3 vers \u03b4 + u1s\u03b4 u23 vers \u03b4 + c\u03b4 \u23a4\u23a6 (8.87) CdW = \u23a1\u23a3 (s1 \u2212 u1 (s3u3 + s2u2 + s1u1)) vers \u03b4 + (s2u3 \u2212 s3u2) sin \u03b4 (s2 \u2212 u2 (s3u3 + s2u2 + s1u1)) vers \u03b4 + (s3u1 \u2212 s1u3) sin \u03b4 (s3 \u2212 u3 (s3u3 + s2u2 + s1u1)) vers \u03b4 + (s1u2 \u2212 s2u1) sin \u03b4 \u23a4\u23a6 (8.88) The vector CdW indicates the position of the wheel center with respect to the wheel-body frame. The matrix CTW is the homogeneous transformation from wheel frame W to wheel-body frame C, when the wheel is steered by the angle \u03b4 about the steering axis u\u0302. Example 335 F Zero steer angle. To examine the screw transformation, we check the zero steering. Substituting \u03b4 = 0 simplifies the rotation matrix CRW and the position vector CdW to I and 0 CRW = \u23a1\u23a3 1 0 0 0 1 0 0 0 1 \u23a4\u23a6 (8.89) CdW= \u23a1\u23a3 0 0 0 \u23a4\u23a6 (8.90) indicating that at zero steering, the wheel frameW and wheel-body frame C are coincident. Example 336 F Steer angle transformation for zero lean and caster. Consider a wheel with a steer axis coincident with zw. Such a wheel has no lean or caster angle. When the wheel is steered by the angle \u03b4, we can find the coordinates of a wheel point P in the wheel-body coordinate frame using transformation method. Figure 8.52 illustrates a 3D view, and Figure 8.53 a top view of such a wheel. 8. Suspension Mechanisms 501 502 8. Suspension Mechanisms Assume W rP = [xw, yw, zw] T is the position vector of a wheel point, then its position vector in the wheel-body coordinate frame C is CrP = CRW W rP = Rz,\u03b4 W rP = \u23a1\u23a3 cos \u03b4 \u2212 sin \u03b4 0 sin \u03b4 cos \u03b4 0 0 0 1 \u23a4\u23a6\u23a1\u23a3 xw yw zw \u23a4\u23a6 = \u23a1\u23a3 xw cos \u03b4 \u2212 yw sin \u03b4 yw cos \u03b4 + xw sin \u03b4 zw \u23a4\u23a6 . (8.91) We assumed that the wheel-body coordinate is installed at the center of the wheel and is parallel to the vehicle coordinate frame. Therefore, the transformation from the frame W to the frame C is a rotation \u03b4 about the wheel-body z-axis. There would be no camber angle when the lean and caster angles are zero and steer axis is on the zw-axis. Example 337 F Zero lean, zero lateral location. The case of zero lean, \u03b8 = 0, and zero lateral location, sb = 0, is important in caster dynamics of bicycle model. The screw transformation for this case will be simplified to C u\u0302 = \u23a1\u23a3 u1 u2 u3 \u23a4\u23a6 = \u23a1\u23a3 sin\u03d5 0 cos\u03d5 \u23a4\u23a6 (8.92) Cs = \u23a1\u23a3 sa 0 \u2212Rw \u23a4\u23a6 (8.93) CRW = \u23a1\u23a3 sin2 \u03d5 vers \u03b4 + cos \u03b4 \u2212 cos\u03d5 sin \u03b4 sin\u03d5 cos\u03d5 vers \u03b4 cos\u03d5 sin \u03b4 cos \u03b4 \u2212 sin\u03d5 sin \u03b4 sin\u03d5 cos\u03d5 vers \u03b4 sin\u03d5 sin \u03b4 cos2 \u03d5 vers \u03b4 + cos \u03b4 \u23a4\u23a6 (8.94) Cd = \u23a1\u23a2\u23a3 cos\u03d5 (sa cos\u03d5+Rw sin\u03d5) vers \u03b4 \u2212 (sa cos\u03d5+Rw sin\u03d5) sin \u03b4 \u22121 2 (Rw \u2212Rw cos 2\u03d5+ sa sin 2\u03d5) vers \u03b4 \u23a4\u23a5\u23a6 . (8.95) Example 338 F Position of the tireprint. The center of tireprint in the wheel coordinate frame is at rT W rT = \u23a1\u23a3 0 0 \u2212Rw \u23a4\u23a6 . (8.96) 8. Suspension Mechanisms 503 If we assume the width of the tire is zero and the wheel is steered, the center of tireprint would be at CrT = CTW W rT = \u23a1\u23a3 xT yT zT \u23a4\u23a6 (8.97) where xT = \u00a1 1\u2212 u21 \u00a2 (1\u2212 cos \u03b4) sa + (u3 sin \u03b4 \u2212 u1u2 (1\u2212 cos \u03b4)) sb (8.98) yT = \u2212 (u3 sin \u03b4 + u1u2 (1\u2212 cos \u03b4)) sa + \u00a1 1\u2212 u22 \u00a2 (1\u2212 cos \u03b4) sb (8.99) zT = (u2 sin \u03b4 \u2212 u1u3 (1\u2212 cos \u03b4)) sa \u2212 (u1 sin \u03b4 + u2u3 (1\u2212 cos \u03b4)) sb \u2212Rw (8.100) or xT = sb \u00c3 cos \u03b8 cos\u03d5 sin \u03b4p cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 + 1 4 sin 2\u03b8 sin 2\u03d5 (1\u2212 cos \u03b4) cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 ! +sa \u00b5 1\u2212 cos2 \u03b8 sin2 \u03d5 cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 \u00b6 (1\u2212 cos \u03b4) (8.101) yT = \u2212sa \u00c3 cos \u03b8 cos\u03d5 sin \u03b4p cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 \u2212 1 4 sin 2\u03b8 sin 2\u03d5 (1\u2212 cos \u03b4) cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 ! +sb \u00b5 1\u2212 cos2 \u03d5 sin2 \u03b8 cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 \u00b6 (1\u2212 cos \u03b4) (8.102) zT = \u2212Rw \u2212 sb cos \u03b8 sin\u03d5+ sa cos\u03d5 sin \u03b8p cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 sin \u03b4 + 1 2 sb cos 2 \u03d5 sin 2\u03b8 \u2212 sa cos 2 \u03b8 sin 2\u03d5 cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 (1\u2212 cos \u03b4) (8.103) Example 339 F Wheel center drop. The zT coordinate in (8.100) or (8.103) indicates the amount that the center of the tireprint will move in the vertical direction with respect to the wheel-body frame when the wheel is steering. If the steer angle is zero, \u03b4 = 0, then zT is at zT = \u2212Rw. (8.104) Because the center of tireprint must be on the ground, H = \u2212Rw \u2212 zT indicated the height that the center of the wheel will drop during steering. H = \u2212Rw \u2212 zT (8.105) = sb cos \u03b8 sin\u03d5+ sa cos\u03d5 sin \u03b8p cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 sin \u03b4 \u22121 2 sb cos 2 \u03d5 sin 2\u03b8 \u2212 sa cos 2 \u03b8 sin 2\u03d5 cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 (1\u2212 cos \u03b4) 504 8. Suspension Mechanisms The zT coordinate of the tireprint may be simplifie for different designs: 1\u2212 If the lean angle is zero, \u03b8 = 0, then zT is at zT = \u2212Rw \u2212 1 2 sa sin 2\u03d5 (1\u2212 cos \u03b4)\u2212 sb sin\u03d5 sin \u03b4. (8.106) 2\u2212 If the lean angle and lateral location are zero, \u03b8 = 0, sb = 0, then zT is at zT = \u2212Rw \u2212 1 2 sa sin 2\u03d5 (1\u2212 cos \u03b4) . (8.107) In this case, the wheel center drop may be expressed by a dimensionless equation. H sa = 1 2 sin 2\u03d5 (1\u2212 cos \u03b4) (8.108) Figure 8.54 illustrates H/sa for the caster angle \u03d5 = 5deg, 0 deg, \u22125 deg, \u221210 deg, \u221215 deg, \u221220 deg, and the steer angle \u03b4 in the range \u221210 deg < \u03b4 < 10 deg. In street cars, we set the steering axis with a positive longitudinal location sa > 0, and a few degrees negative caster angle \u03d5 < 0. In this case the wheel center drops as is shown in the figure. 3\u2212 If the caster angle is zero, \u03d5 = 0, then zT is at zT = \u2212Rw + 1 2 sb sin 2\u03b8 (1\u2212 cos \u03b4)\u2212 sa sin \u03b8 sin \u03b4. (8.109) 4\u2212 If the caster angle and lateral location are zero, \u03d5 = 0, sb = 0, then zT is at zT = \u2212Rw \u2212 sa sin \u03b8 sin \u03b4. (8.110) 8. Suspension Mechanisms 505 In this case, the wheel center drop may be expressed by a dimensionless equation. H sa = \u2212 sin \u03b8 sin \u03b4 (8.111) Figure 8.55 illustrates H/sa for the lean angle \u03b8 = 5deg, 0, \u22125 deg, \u221210 deg, \u221215 deg, \u221220 deg and the steer angle \u03b4 in the range \u221210 deg < \u03b4 < 10 deg. The steering axis of street cars is usually set with a positive longitudinal location sa > 0, and a few degrees positive lean angle \u03b8 > 0. In this case the wheel center lowers when the wheel number 1 turns to the right, and elevates when the wheel turns to the left. Comparison of Figures 8.54 and 8.55 shows that the lean angle has much more affect on the wheel center drop than the caster angle. 5\u2212 If the lateral location is zero, sb = 0, then zT is at zT = \u2212Rw \u2212 sa cos\u03d5 sin \u03b8p cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 sin \u03b4 \u22121 2 sa cos2 \u03b8 sin 2\u03d5 cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 (1\u2212 cos \u03b4) (8.112) and the wheel center drop,H, may be expressed by a dimensionless equation. H sa = \u22121 2 cos2 \u03b8 sin2 \u03d5 (1\u2212 cos \u03b4) cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 \u2212 cos\u03d5 sin \u03b8 sin \u03b4p cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 (8.113) Example 340 F Position of the wheel center. As given in Equation (8.88), the wheel center is at CdW with respect to 506 8. Suspension Mechanisms the wheel-body frame. CdW = \u23a1\u23a3 xW yW zW \u23a4\u23a6 (8.114) Substituting for u\u0302 and s from (8.72) and (8.73) in (8.88) provides the coordinates of the wheel center in the wheel-body frame as xW = (sa \u2212 u1 (\u2212Rwu3 + sbu2 + sau1)) (1\u2212 cos \u03b4) + (sbu3 +Rwu2) sin \u03b4 (8.115) yW = (sb \u2212 u2 (\u2212Rwu3 + sbu2 + sau1)) (1\u2212 cos \u03b4) \u2212 (Rwu1 + sau3) sin \u03b4 (8.116) zW = (\u2212Rw \u2212 u3 (\u2212Rwu3 + sbu2 + sau1)) (1\u2212 cos \u03b4) + (sau2 \u2212 sbu1) sin \u03b4 (8.117) or xW = sa (1\u2212 cos \u03b4) + \u00b5 1 2 Rw sin 2\u03d5\u2212 sa sin 2 \u03d5 \u00b6 cos2 \u03b8 + 1 4 sb sin 2\u03b8 sin 2\u03d5 cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 (1\u2212 cos \u03b4) + (sb cos \u03b8 \u2212Rw sin \u03b8)p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 cos\u03d5 sin \u03b4 (8.118) yW = sb (1\u2212 cos \u03b4) \u2212 1 2 \u00a1 Rw sin 2\u03b8 + sb sin 2 \u03b8 \u00a2 cos2 \u03d5\u2212 1 4 sa sin 2\u03b8 sin 2\u03d5 cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 (1\u2212 cos \u03b4) \u2212 Rw sin\u03d5+ sa cos\u03d5p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 cos \u03b8 sin \u03b4 (8.119) zW = \u2212Rw (1\u2212 cos \u03b4) + \u00b5 Rw cos 2 \u03b8 + 1 2 sb sin 2\u03b8 \u00b6 cos2 \u03d5\u2212 1 2 sa cos 2 \u03b8 sin 2\u03d5 cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 (1\u2212 cos \u03b4) \u2212sa cos\u03d5 sin \u03b8 + sb cos \u03b8 sin\u03d5p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 sin \u03b4 (8.120) The zW coordinate indicates how the center of the wheel will move in the vertical direction with respect to the wheel-body frame, when the wheel is steering. It shows that zW = 0, as long as \u03b4 = 0. 8. Suspension Mechanisms 507 The zW coordinate of the wheel center may be simplified for different designs: 1\u2212 If the lean angle is zero, \u03b8 = 0, then zW is at zW = \u2212Rw \u00a1 1\u2212 cos2 \u03d5 \u00a2 (1\u2212 cos \u03b4)\u2212 sb sin\u03d5 sin \u03b4 \u22121 2 sa sin 2\u03d5 (1\u2212 cos \u03b4) . (8.121) 2\u2212 If the lean angle and lateral location are zero, \u03b8 = 0, sb = 0, then zW is at zW = \u2212Rw \u00a1 1\u2212 cos2 \u03d5 \u00a2 (1\u2212 cos \u03b4)\u2212 1 2 sa sin 2\u03d5 (1\u2212 cos \u03b4) . (8.122) 3\u2212 If the caster angle is zero, \u03d5 = 0, then zW is at zW = \u2212Rw \u00a1 1\u2212 cos2 \u03b8 \u00a2 (1\u2212 cos \u03b4)\u2212 sa sin \u03b8 sin \u03b4 + 1 2 sb sin 2\u03b8 (1\u2212 cos \u03b4) . (8.123) 4\u2212 If the caster angle and lateral location are zero, \u03d5 = 0, sb = 0, then zW is at zW = \u2212Rw \u00a1 1\u2212 cos2 \u03b8 \u00a2 (1\u2212 cos \u03b4)\u2212 sa sin \u03b8 sin \u03b4. (8.124) 5\u2212 If the lateral location is zero, sb = 0, then zT is at zW = \u2212Rw (1\u2212 cos \u03b4)\u2212 sa cos\u03d5 sin \u03b8p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 sin \u03b4 + Rw cos 2 \u03b8 cos2 \u03d5\u2212 1 2 sa cos 2 \u03b8 sin 2\u03d5 cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 (1\u2212 cos \u03b4) (8.125) In each case of the above designs, the height of the wheel center with respect to the ground level can be found by adding H to zW . The equations for calculating H are found in Example 340. Example 341 F Camber theory. Having a non-zero lean and caster angles causes a camber angle \u03b3 for a steered wheel. To find the camber angle of an steered wheel, we may determine the angle between the camber line and the vertical direction zc. The camber line is the line connecting the wheel center and the center of tireprint. The coordinates of the center of tireprint (xT , yT , zT ) are given in Equations (8.101)-(8.103), and the coordinates of the wheel center (xW , yW , zW ) are given in Equations (8.118)-(8.120). The line connecting (xT , yT , zT ) to (xW , yW , zW ) may be indicated by the unit vector l\u0302c l\u0302c = (xW \u2212 xT ) I\u0302 + (yW \u2212 yT ) J\u0302 + (zW \u2212 zT ) K\u0302q (xW \u2212 xT ) 2 + (yW \u2212 yT ) 2 + (zW \u2212 zT ) 2 (8.126) 508 8. Suspension Mechanisms in which I\u0302 , J\u0302 , K\u0302, are the unit vectors of the wheel-body coordinate frame C. The camber angle is the angle between l\u0302c and K\u0302, which can be found by the inner vector product. \u03b3 = cos\u22121 \u00b3 l\u0302c \u00b7 K\u0302 \u00b4 = cos\u22121 (zW \u2212 zT )q (xW \u2212 xT ) 2 + (yW \u2212 yT ) 2 + (zW \u2212 zT ) 2 (8.127) As an special case, let us determine the camber angle when the lean angle and lateral location are zero, \u03b8 = 0, sb = 0. In this case, we have xT = sa \u00a1 1\u2212 sin2 \u03d5 \u00a2 (cos \u03b4 \u2212 1) (8.128) yT = \u2212sa cos\u03d5 sin \u03b4 (8.129) zT = zT = \u2212Rw \u2212 1 2 sa sin 2\u03d5 (1\u2212 cos \u03b4) (8.130) xW = \u00b5 sa + 1 2 Rw sin 2\u03d5\u2212 sa sin 2 \u03d5 \u00b6 (1\u2212 cos \u03b4) (8.131) yW = sb (1\u2212 cos \u03b4)\u2212Rw sin\u03d5+ sa cos\u03d5 sin \u03b4 (8.132) zW = \u00b5 Rw \u00a1 cos2 \u03d5\u2212 1 \u00a2 \u2212 1 2 sa sin 2\u03d5 \u00b6 (1\u2212 cos \u03b4) . (8.133) 8.7 Summary There are two general types of suspensions: dependent, in which the left and right wheels on an axle are rigidly connected, and independent, in which the left and right wheels are disconnected. Solid axle is the most common dependent suspension, while McPherson and double A-arm are the most common independent suspensions. The roll axis is the instantaneous line about which the body of a vehicle rolls. Roll axis is found by connecting the roll center of the front and rear suspensions of the vehicle. The instant center of rotation of a wheel with respect to the body is called suspension roll center. So, to find the roll center of the front or rear half of a car, we should determine the suspension roll centers, and find the intersection of the lines connecting the suspension roll centers to the center of their associated tireprints. Three coordinate frames are employed to express the orientation of a tire and wheel with respect to the vehicle: the wheel frame W , wheel-body frame C, and tire frame T . A wheel coordinate frame W (xw, yw, zw) is attached to the center of a wheel. It follows every translation and rotation of the wheel except the spin. Hence, the xw and zw axes are always in the tire-plane, while the yw-axis is always along the spin axis. When the wheel 8. Suspension Mechanisms 509 is straight and the W frame is parallel to the vehicle coordinate frame, we attach a wheel-body coordinate frame C (xc, yc, zc) at the center of the wheel parallel to the vehicle coordinate axes. The wheel-body frame C is motionless with respect to the vehicle coordinate and does not follow any motion of the wheel. The tire coordinate frame T (xt, yt, zt) is set at the center of the tireprint. The zt-axis is always perpendicular to the ground. The xt-axis is along the intersection line of the tire-plane and the ground. The tire frame does not follows the spin and camber rotations of the tire however, it follows the steer angle rotation about the zc-axis. We define the orientation and position of a steering axis by the caster angle \u03d5, lean angle \u03b8, and the intersection point of the axis with the ground surface at (sa, sb) with respect to the center of tireprint. Because of these parameters, a steered wheel will camber and generates a lateral force. This is called the caster theory. The camber angle \u03b3 of a steered wheel for \u03b8 = 0, and sb = 0 is: \u03b3 = cos\u22121 \u00b3 l\u0302c \u00b7 K\u0302 \u00b4 = cos\u22121 (zW \u2212 zT )q (xW \u2212 xT ) 2 + (yW \u2212 yT ) 2 + (zW \u2212 zT ) 2 (8.134) where xT = sa \u00a1 1\u2212 sin2 \u03d5 \u00a2 (cos \u03b4 \u2212 1) (8.135) yT = \u2212sa cos\u03d5 sin \u03b4 (8.136) zT = zT = \u2212Rw \u2212 1 2 sa sin 2\u03d5 (1\u2212 cos \u03b4) (8.137) xW = \u00b5 sa + 1 2 Rw sin 2\u03d5\u2212 sa sin 2 \u03d5 \u00b6 (1\u2212 cos \u03b4) (8.138) yW = sb (1\u2212 cos \u03b4)\u2212Rw sin\u03d5+ sa cos\u03d5 sin \u03b4 (8.139) zW = \u00b5 Rw \u00a1 cos2 \u03d5\u2212 1 \u00a2 \u2212 1 2 sa sin 2\u03d5 \u00b6 (1\u2212 cos \u03b4) . (8.140) 510 8. Suspension Mechanisms 8.8 Key Symbols a, b, c, d lengths of the links of a four-bar linkage ai distance of the axle number i from the mass center A,B, \u00b7 \u00b7 \u00b7 coefficients in equation for calculating \u03b83 b1 distance of left wheels from mass center B (x, y, z) vehicle coordinate frame C mass center C coupler point C (xc, yc, zc) wheel-body coordinate frame C T dW C expression of the position of W with respect to T e, \u03b1 polar coordinates of a coupler point g overhang h = z \u2212 z0 vertical displacement of the wheel center H wheel center drop Iij instant center of rotation between link i and link j IijImn a line connecting Iij and Imn I\u0302 , J\u0302 , K\u0302 unit vectors of the wheel-body frame C I identity matrix J1, J2, \u00b7 \u00b7 \u00b7 length function for calculating \u03b83 l\u0302c unit vector on the line (xT , yT , zT ) to (xW , yW , zW ) ms sprung mass mu unsprung mass n\u03021 normal unit vectors to \u03c0L n\u03022 normal unit vectors to \u03c0C P point q, p, f parameters for calculating couple point coordinate r position vector Rw tire radius TRW rotation matrix to go from W frame to T frame s position vector of the steer axis sa forward location of the steer axis sb lateral location of the steer axis s\u030cW (0, \u03b4, u\u0302, s) zero pitch screw about the steer axis T (xt, yt, zt) tire coordinate system TTW homogeneous transformation to go from W to T u\u0302 steer axis unit vector u\u0303 skew symmetrix matrix associated to u\u0302 uC position vector of the coupler point u\u0302z unit vector in the z-direction vx forward speed x, y suspension coordinate frame xC , yC coordinate of a couple point (xT , yT , zT ) wheel-body coordinates of the origin of T frame 8. Suspension Mechanisms 511 (xW , yW , zW ) wheel-body coordinates of the origin of W frame vers \u03b4 1\u2212 cos \u03b4 W (xwywzw) wheel coordinate system z vertical position of the wheel center z0 initial vertical position of the wheel center \u03b1 angle of a coupler point with upper A-arm \u03b3 camber angle \u03b4 steer angle \u03b5 = ms/mu sprung to unsprung mass ratio \u03b8 lean angle \u03b80 angle between the ground link and the z-direction \u03b8i angular position of link number i \u03b82 angular position of the upper A-arm \u03b83 angular position of the coupler link \u03b84 angular position of link lower A-arm \u03b8i0 initial angular position of \u03b8i \u03c0C caster plane \u03c0L lean plane \u03c5 trust angle \u03d5 caster angle \u03c9 angular velocity 512 8. Suspension Mechanisms Exercises 1. Roll center. Determine the roll ceneter of the kinematic models of vehicles shown in Figures 8.56 to 8.59. 2 1 Body 4 6 3 5 8 7 8. Suspension Mechanisms 513 514 8. Suspension Mechanisms 8. Suspension Mechanisms 515 4. F Position of the roll center and mass ceneter. Figure 8.66 illustrates the wheels and mass center C of a vehicle. Design a double A-arm suspension such that the roll center of the C 516 8. Suspension Mechanisms c d a 2\u03b8 3\u03b8 4\u03b8 A B M N x y 0\u03b8 \u03b1 e z Cb 8. Suspension Mechanisms 517 Determine CTW for \u03d5 = 8deg, \u03b8 = 12deg, and the location vector Cs Cs = \u23a1\u23a3 3.8 cm 1.8 cm \u2212Rw \u23a4\u23a6 . (a) The vehicle uses a tire 235/35ZR19. (b) The vehicle uses a tire P215/65R15 96H. 10. F Wheel drop. Find the coordinates of the tireprint for \u03d5 = 10deg \u03b8 = 10deg Cs = \u23a1\u23a3 3.8 cm 1.8 cm 38 cm \u23a4\u23a6 if \u03b4 = 18deg. How much is the wheel drop H. 11. F Wheel drop and steer angle. Draw a plot to show the wheel drop H at different steer angle \u03b4 for the given data in Exercise 10. 12. F Camber and steering. Draw a plot to show the camber angle \u03b3 at different steer angle \u03b4 for the following characteristics: \u03d5 = 10deg \u03b8 = 0deg Cs = \u23a1\u23a3 3.8 cm 0 cm 38 cm \u23a4\u23a6 Part III Vehicle Dynamics 9"
        ]
    },
    {
        "image_filename": "designv10_0_0001199_b978-0-08-100433-3.00013-0-Figure13.13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001199_b978-0-08-100433-3.00013-0-Figure13.13-1.png",
        "caption": "Figure 13.13 T700 blisk repair using the laser engineering net-shape process: (a) in-process repair of the leading edge of a Ti64 airfoil, (b) the blisk after deposition, and (c) the blisk after finishing. Courtesy Optomec.",
        "texts": [
            " LENS-based repair solutions can precisely add material to worn components to restore their geometry. Integrally bladed rotors, also known as blisks, are high-value components that may cost tens or hundreds of thousands of dollars to fabricate. Without repair, any single airfoil that is worn out causes entire blisks to be scrapped. For example, a T700 blisk that was made from AM355 steel and suffered from erosion on the airfoil\u2019s leading edges during service was repaired with the LENS process. Fig. 13.13 illustrates the repair of the leading edge of a Ti64 airfoil, the blisk after deposition, and the blisk after finishing. Optomec used a cobalt-based wear-resistant material to repair the leading edge. The unique capabilities of the LENS repair system include [45]: \u2022 Low heat input: The airfoil does not distort beyond dimensional limits. \u2022 Small heat-affected zone: This minimizes degradation of the mechanical properties of the airfoil. \u2022 Superior mechanical properties: LENS Ti-6Al-4V fatigue properties are equivalent to those of the wrought material"
        ],
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    },
    {
        "image_filename": "designv10_0_0000801_j.finel.2014.04.003-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000801_j.finel.2014.04.003-Figure3-1.png",
        "caption": "Fig. 3. Illustration of interface between active and inactive (or quiet) elements.",
        "texts": [
            " If the temperature at node 2 is not equal to initial temperature 0T , when element 2 is activated, the initial temperature across the element as interpolated by Eq. (12) is not equal to the initial temperature 0T which results into artificial numerical thermal energy introduced during element activation. A remedy to this problem is to set n 1T \u00bc 0T in Eqs. (10) and (11) for the inactive elements before their activation. n 1T is not reset for the active elements, otherwise, energy would be lost. Fig. 3 illustrates a finite element mesh for the simulation of building a thin wall by powder fed metal deposition additive manufacturing. The active element region is shown as contour shaded with temperature results and the inactive (or quiet) region as translucent. The interface between active and inactive (or quiet) elements is also illustrated in the figure. Surface convection and radiation conditions (Eq. (5)) need to be applied to this interface since it is an external free surface. The interface continuously changes as the metal is deposited during processing. Therefore, algorithms need to be developed to calculate the location of the evolving interface and apply surface convection and radiation as needed. As seen in Fig. 3, the interface is an internal surface in the mesh. When using general purpose FEA codes, it is difficult to compute the interface between active and inactive (or quiet) elements using user subroutines. Thus, it is common practice to neglect the surface convection and radiation on this interface. This simplification may be appropriate in weld modeling since the filler metal and therefore the size of the interface is negligible compared to the base metal. However, in additive manufacturing, the size of the deposited material could be significant compared to the substrate and thus neglecting surface convection and radiation on the interface between active and inactive elements can be a source of error as demonstrated in the examples section"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002011_1.4008346-Figure9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002011_1.4008346-Figure9-1.png",
        "caption": "Fig. 9",
        "texts": [
            " If /3 is zero and there is no spin at a race contact, these points and all points on lines through these points parallel to the direction of rolling, roll without slip. Let these points lie at x = \u00b1ca. The velocities of slip at other points within the pressure area are given by Equations (46) or (49) with (3 = 0 and a defined by Equations (159) or (160) and distribute themselves as shown in Fig. 8. Between the lines \u00b1cct the velocity of slip of the race on the ball Vv is in one direction and without the lines it is in the opposite direction. Consequently the frictional forces act in opposite directions within and without the lines as shown in Fig. 9. The net force in the F-direction is: / .l J l - q4 \u2014 V V l + m2( = s s * f J 9 A * 1 7T J - 1 m2 (154) The q integrations in Equations (151), (153), and (154) are easily evaluated by use of a planimeter or Simpson's Rule. In evaluating Equations (151), (153), and (154) it will be found that e2 \u2014\u00bb\u2022 1 for some values of the variable q. Values of K are not readily obtained from existing tables when t2 -+ 1 and the following relation, which is obtained through Landen's Transformation, may be used to compute K and E"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001771_s11661-018-4607-2-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001771_s11661-018-4607-2-Figure1-1.png",
        "caption": "Fig. 1\u2014(a) The as-built components. The vertical and horizontal built samples are shown in (b) and (c), respectively.",
        "texts": [
            "[24] Bulk material was fabricated using an EOS EOSINT M280 AM system at the Naval Surface Warfare Center, Dahlgren Division, Virginia, equipped with a 400-W continuous wave Nd:YAG fiber laser using the EOS MS-1 speed parameter set, which includes specifications of a bidirectional scanning strategy wherein the scanning vector is rotated by 70 deg between successive layers. To minimize oxidation, laser-melting operations were conducted in a nitrogen environment. A stainless steel base plate and a layer thickness of 20 lm were maintained throughout the process. Figure 1 shows the component built on a base plate for this study: (a) illustrates how the tensile samples, in accordance with ASTM E8,[25] were directly fabricated; (b) and (c) show the X axis of the specimen is defined as the long transverse direction (TD) and the Y axis as parallel to the building direction (BD). Selected AM samples were annealed in a vacuum furnace at 1100 C for 1 hour (HT1) and 13 hours (HT13), and cooled at a rate of 100 C/min. Samples were observed using a Leica light optical microscope and an FEI Quanta 200 SEM equipped with an AZtec energy dispersive spectrometer (EDS)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure11.5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure11.5-1.png",
        "caption": "FIGURE 11.5. A two-wheel model for a vehicle with roll and yaw rotations.",
        "texts": [
            "103) Assuming small angles for slip angles \u03b2f , \u03b2, and \u03b2r, the tire sideslip angles for the front and rear wheels, \u03b1f and \u03b1r, may be approximated as \u03b1f = 1 vx (vy + a1r \u2212 zfp)\u2212 \u03b4 \u2212 \u03b4\u03d5f = \u03b2 + a1 r vx \u2212 C\u03b2f p vx \u2212 \u03b4 \u2212 C\u03b4\u03d5f\u03d5 (11.104) \u03b1r = 1 vx (vy \u2212 a2r \u2212 zrp)\u2212 \u03b4\u03d5r = \u03b2 \u2212 a2 r vx \u2212 C\u03b2r p vx \u2212 C\u03b4\u03d5r\u03d5. (11.105) 11.3.3 F Body Force Components on a Two-wheel Model Figure 11.4 illustrates a top view of a car and the force systems acting at the tireprints of a front-wheel-steering four-wheel vehicle. When we consider the roll motion of the vehicle, the xy-plane does not remain parallel to the road\u2019s XY -plane, however, we may still use a two-wheel model for the vehicle. Figure 11.5 illustrates the force system and Figure 11.6 illustrates the kinematics of a two-wheel model for a vehicle with roll and yaw rotations. The rolling two-wheel model is also called the bicycle model. The force system applied on the bicycle vehicle, having only the front wheel steerable, is Fx = 2X i=1 (Fxi cos \u03b4 \u2212 Fyi sin \u03b4) (11.106) Fy = 2X i=1 Fyi (11.107) Mx = Mxf +Mxr \u2212 wcf \u03d5\u0307\u2212 wkf\u03d5 (11.108) Mz = a1Fyf \u2212 a2Fyr (11.109) where \u00a1 Fxf , Fxr \u00a2 and \u00a1 Fyf , Fyr \u00a2 are the planar forces on the tireprint of the front and rear wheels"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure2.21-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure2.21-1.png",
        "caption": "Fig. 2.21 The Position, Attitude and Rotation of a single link aj , bj each specify the joint axis vector and location of the origin viewed from the parent coordinate system",
        "texts": [
            " First we will start off by calculating the homogeneous 14 There is a well known method you can use to describe the link structure of a robot called the Denavit-Hartenberg (DH) method [18]. We also used this method at first. However this method has a restriction which requires you to change the orientation of the link coordinates with each link. We found the implementation with this restriction to be rather error-prone, so we instead adopted the method outlined above. 48 2 Kinematics transform of a single link as shown in Fig. 2.21. We need to set a local coordinate system \u03a3j which has it\u2019s origin on the joint axis. The joint axis vector seen from the parent coordinates is aj and the origin of \u03a3j is bj . The joint angle is qj and the attitude of the link when the joint angle is 0 is, E. The homogeneous transform relative to the parent link is: iT j = [ ea\u0302jqj bj 0 0 0 1 ] . (2.56) Next let us assume there are two links as shown in Fig. 2.22. We will assume that the absolute position and attitude of the parent link pi,Ri is known"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure3.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure3.6-1.png",
        "caption": "Fig. 3.6: A kinematic scheme for manipulator link parameters according to the H-\u2013D notation.",
        "texts": [
            " These parameters determine the relative position and orientation of a link with respect to a neighborhood one as a function of the Fundamentals of Mechanics of Robotic Manipulation 77 variable coordinates of the joints that are connected through the link. Usually the variable coordinate of a joint is named as \u2018joint variable\u2019. Chapter 3: Fundamentals of the Mechanics of Robots78 In order to determine those geometrical sizes and kinematic parameters, one can usually refer to a scheme like that in Fig. 3.6 by using a H\u2013D notation, in agreement with a procedure proposed by Hartenberg and Denavit in 1955. This scheme gives the minimum number of parameters that are needed to describe the geometry of a link between two joints, but also indicates the joint variables. The joints in Fig. 3.6 are indicated as big black points in order to stress attention to the link geometry and H\u2013D parameters. In particular, referring to Fig. 3.6 for the j-link, the j-frame XjYjZj is assumed as fixed to the j-link, with the Zj axis coinciding with the joint axis, with the Xj axis lying on the common normal between Zj and Zj+1 pointing to Zj+1. The origin Oj and the Yj axis are determined to obtain a Cartesian frame as shown in Fig. 3.6. The kinematic parameters of a manipulator can be defined according to the H\u2013D notation in Fig. 3.6 as: - aj, which is named as the link length that is measured as the distance between the Zj and Zj+1 axes along Xj; - \u03b1j, which is named as the twist angle that is measured as the angle between the Zj and Zj+1 axes about Xj; - d j+1, which is named as the link offset that is measured as the distance between the Xj and X j+1 axes along Z j+1; - \u03b8j+1, which is named as the joint angle that is measured as the angle between the Xj Fundamentals of Mechanics of Robotic Manipulation 79 and X j+1 axes about Zj j+1 When a joint can be modeled as a rotation pair, the angle \u03b8j+1 is the corresponding kinematic variable",
            " Finally, the location of the extremity link N+1 can be computed with respect to the base frame as Fundamentals of Mechanics of Robotic Manipulation 83 NN00 T XX = (3.1.13) and therefore the resultant transformation matrix can be obtained as TT...TTT 1N 0k 1kkN1-N2110N0 \u220f== \u2212 = + (3.1.14) The last expression can be considered the fundamental typical formulation for the direct kinematic of manipulators. In particular, by using the above-mentioned properties and formulation for the transformation matrix, a general expression for kTk+1 can be given by referring to the general manipulator scheme of Fig. 3.6, in the form 1000 dcoscossincossinsin dsinsincoscoscossin a0sincos T 1kkkk1kk1k 1kkkk1kk1k k1k1k 1kk +++ +++ ++ + \u03b1\u03b1\u03b1\u03b8\u03b1\u03b8 \u03b1\u2212\u03b1\u2212\u03b1\u03b8\u03b1\u03b8 \u03b8\u2212\u03b8 = (3.1.15) Chapter 3: Fundamentals of the Mechanics of Robots84 as a composition of elementary matrices that describe elementary differences between reference frames that are described by the H\u2013D parameters. An elementary matrix corresponds to a rotation about a reference axis or a translation along such an axis. In addition, the combination of a rotation with a translation with respect to the same axis can be described as an elementary screw motion that can be characterized by the axis itself (which is usually referred to as \u2018Mozzi\u2019s\u2019 axis or \u2018Screw axis\u2019) and its parameters"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002513_j.oceaneng.2018.04.026-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002513_j.oceaneng.2018.04.026-Figure3-1.png",
        "caption": "Fig. 3. The geometrical relationship between course \u03c7, heading angle \u03c8 and sideslip angle \u03b2.",
        "texts": [
            " In the case of undereducated vessels, the controller can not reject the disturbance force component in the sway direction as there is no sway control force generated by the actuators of the vessel; even if the disturbance value is known. Although the terms course and heading are used interchangeably in much of the literature on guidance, navigation and control of marine craft (Fossen, 2011), they are not equivalent and this leads to confusion. Here they are going to be used differently, hence the following definitions, adopted from (Fossen, 2011; Breivik and Fossen, 2004), are required to avoid this confusion (see Fig. 3): Definition 1. (Velocity Vector). It is the specification of the vessels's Table 1 Surface vessel parameters. Parameter Value Unit m1 120:0 103 kg m2 172:9 103 kg m3 636:0 105 kg d1 215:0 102 kg\u22c5s 1 d2 97:0 103 kg\u22c5s 1 d3 802:0 104 kg\u22c5m2\u22c5s 1 Table 2 States and Control variables weights. Variable Weight X 5 Y 5 \u0425 5 U 1 \u03a5 1 R 1 \u03c4u 0.001 \u03c4r 0.001 speed U \u00bc ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u2 \u00fe \u03c52 p , and orientation \u03b2 \u00bc tan 1 \u03c5 u with respect to the vessel's body reference frame. Definition 2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002400_tcyb.2017.2692767-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002400_tcyb.2017.2692767-Figure2-1.png",
        "caption": "Fig. 2. Single-link manipulator actuated by a dc motor.",
        "texts": [
            " By following (14), (26), and (35)\u2013(37), it is not difficult to develop an analogous control scheme for the MIMO nonlinear systems (60). V. SIMULATION STUDY In this section, the effectiveness and merits of the proposed controller is demonstrated via two practical examples. Results within both scenarios are compared with the results obtained by applying the existing methods in [34] and [37]. A popular benchmark of practical example, i.e., the tracking control of a single-link manipulator, is performed (see Fig. 2). In modern industrial process, one-link manipulators have been widely applied in multifarious mechanical platforms or mobile robots, due to the advantages of high load-to-weight-ratio, low power consumption, lightweight, etc. As a result, they play an important role in the areas of resource exploration, military reconnaissance, and home service. In engineering applications, owing to work environments becoming complex gradually and the presence of geometrical errors, ambient noise, and vibration interferences, some technical challenges need to be conquered, some of which are stated as follows"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure10.20-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure10.20-1.png",
        "caption": "FIGURE 10.20 One section example of equivalent circuit for dielectric.",
        "texts": [],
        "surrounding_texts": [
            "Some observations help to understand this important model. \u2022 The excess capacitance Ce takes the dielectric constant above air \ud835\udf00r = 1 into account. \u2022 A parallel resistance could be used to include a model for losses due to finite conductivity of the dielectric material. This would include dielectrics with a dc resistance. 274 PEEC MODELS FOR DIELECTRICS \u2022 Surprisingly, a partial inductance Lp11 has to be computed for the dielectric cell. Also, all the partial inductances in a PEEC model are coupled including the dielectric ones. \u2022 Capacitive cells are only placed on surfaces with conductor\u2013dielectric interfaces. Interfaces with dielectric cells only do not connect to the free-charge model. We take a specific example to illustrate the last point. For example, node 5 in Fig. 10.19 includes several faces. The free side faces do not have free charge associated with them, whereas the top face does have a free charge associated with the conductor surface. However, bound charge or the excess capacitance Ce is associated with all faces. This happens automatically when we compute Ce for the model."
        ]
    },
    {
        "image_filename": "designv10_0_0000091_s00170-015-7077-3-Figure15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000091_s00170-015-7077-3-Figure15-1.png",
        "caption": "Fig. 15 Approximate surface\u2014\u201cstair stepping\u201d effect [89] Fig. 16 Sketch of the traditional flat-top overlapping model (FOM) [66]",
        "texts": [
            " The adaptive slicing strategy which involves non-constant layer thickness is an effective way to address this issue. In the adaptive slicing strategy [62, 63, 88], the layer thickness can be adjusted automatically within a certain range to fit the shape of the model, as shown in Fig. 14c. The \u201cstair stepping\u201d effect comes from the approximate construction of surfaces using deposition layers with a certain layer thickness. It exhibits dimensional errors normal to the build (deposition) direction as shown in Fig. 15. For a given part surface, the thicker the layer thickness, the larger the error of the produced part would be [89]. Therefore, the accuracy of the part manufactured by wire-feed AM technology is typically 10 times lower than that made by powder bead/feed technology, due to the thicker layer thickness of the wire-feed AM technology (about 1.5 mm), as summarised in Table 2. With high deposition rate, the WAAM process is effective for simpler geometries. However, when high-accuracy parts are desired, a milling process is necessary to be integrated in the WAAM process"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure9.8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure9.8-1.png",
        "caption": "FIGURE 9.8. A central coordinate frame B1 and a translated frame B2.",
        "texts": [
            " The angular velocity and angular momentum of a rigid body transform from the frame B1 to the frame B2 by vector transformation rule B2\u03c9 = B2RB1 B1\u03c9 (9.161) B2L = B2RB1 B1L. (9.162) However, L and \u03c9 are related according to Equation (9.52) B1L = B1I B1\u03c9 (9.163) and therefore, B2L = B2RB1 B1I B2RT B1 B2\u03c9 = B2I B2\u03c9 (9.164) which shows how to transfer the moment of inertia from the coordinate frame B1 to a rotated frame B2 B2I = B2RB1 B1I B2RT B1 . (9.165) Now consider a central frame B1, shown in Figure 9.8, at B2rC , which rotates about the origin of a fixed frame B2 such that their axes remain parallel. The angular velocity and angular momentum of the rigid body transform from frame B1 to frame B2 by B2\u03c9 = B1\u03c9 (9.166) B2L = B1L+ (rC \u00d7mvC) . (9.167) 9. Applied Dynamics 545 Therefore, B2L = B1L+mB2rC \u00d7 \u00a1 B2\u03c9\u00d7B2rC \u00a2 = B1L+ \u00a1 m B2 r\u0303C B2 r\u0303TC \u00a2 B2\u03c9 = \u00a1 B1I +m B2 r\u0303C B2 r\u0303TC \u00a2 B2\u03c9 (9.168) which shows how to transfer the moment of inertia from frame B1 to a parallel frame B2 B2I = B1I +mr\u0303C r\u0303TC . (9.169) The parallel-axes theorem is also called the Huygens-Steiner theorem"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure13.75-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure13.75-1.png",
        "caption": "Figure 13.75 To determine the normal force and shear force in CD, the support reactions at C can be resolved into components normal to and parallel to member CD.",
        "texts": [
            "5 = 0 \u21d2 Bv = 21.5 kN (\u2191). The support reactions at A and C are found from the equilibrium of the structure as a whole: \u2211 Fvert = 0 \u21d2 Cv = 24.5 kN (\u2191),\u2211 T|A = 0 \u21d2 Ch = 32.5 kN (\u2192),\u2211 Fhor = 0 \u21d2 Ah = 37.5 kN (\u2190). In Figure 13.74, the support reactions are shown as they act in reality. c. By resolving the horizontal and vertical support reaction at C into components parallel to and normal to CD we find the normal force NCD and the 604 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM shear force V CD (see Figure 13.75): NCD = 1 2 \u221a 2 \u00d7 (32.5 + 24.5) = 28.5 \u221a 2 kN, V CD = 1 2 \u221a 2 \u00d7 (32.5 \u2212 24.5) = 4 \u221a 2 kN. The normal force is a tensile force; the deformation symbol for the shear force is given in Figure 13.75. d. In Figures 13.76a to 13.76c, the N , V and M diagrams are shown. We provide a number of comments about the M and V diagrams below. At E and D, the bending moment \u201cgoes round the corner\u201d. At G, the tangent to the M diagram is parallel to the chord k of the parabola. The tangents at E and D are formed by the dashed M diagram due to the resultant of the distributed load on DE. The slope of this dashed M diagram gives the magnitude and the deformation symbol for the shear forces at E and D. The shear force in DE varies linearly between the values at D and E"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003737_tnn.2005.863458-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003737_tnn.2005.863458-Figure2-1.png",
        "caption": "Fig. 2. Schematic of flexible-joint manipulator modeled by torsional spring.",
        "texts": [
            " On the other hand, a harmonic drive transmission is much more flexible than a conventional gear transmission. The flexibility of the joint causes difficulty in modeling manipulator dynamics and becomes a potential source of uncertainty that can degrade the performance of a manipulator and in some cases can even destabilize the system [4]. Consequently, addressing this issue is important for calibration as well asmodelingandcontrolofrobotmanipulators.Jointelasticitycan be modeled as a torsional spring between the input shaft (motor) and the output shaft (link) of the manipulator, as shown in Fig. 2. Due to the presence of joint flexibility, there are twice as many degrees of freedom compared to the rigid joint case. To compensate for joint flexibility, many sophisticated control algorithms have been proposed both in constrained [23], [24] and unconstrained motions [25]\u2013[29]. Most of these schemes, however, assume the availability of both the link and the motor positions, a condition that may not always be satisfied. Luenberger observers, reduced-order high-gain observers, and Kalman filter based observers have been used to relax the requirement of measurement from both sides of the transmission device [1], [2], [30]\u2013[33]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000354_9.119645-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000354_9.119645-Figure4-1.png",
        "caption": "Fig. 4. Simulation results for yd(t ) = A cos r t / 5 , A = 1 , 2, 3, using the second approximation.",
        "texts": [
            "3) is that it is easy to identify the terms in the g vector field that should be dropped (in this case, $3). However, when this modification is expressed completely in x (or t ) coordinates it is, in fact, very complicated. For example, in x coordinates, dropping G3 corresponds to subtracting from the original g vector field. The system with g modified in this manner is input-output linearizable and is an approximation to the original system since A g u is higher order in x 2 , x 3 , x 4 , and U (for every x l , in fact). The simulation results in Fig. 4 show that the tracking error is somewhat smaller than that obtained by Approximation 1 . Approximation 3 (Jacobian Approximation) To provide a basis for comparison, we calculate a linear control law based on the standard Jacobian approximation of the original system (2.1) at x = 0, 7 = 0. As with the previous approximations, define t , = I#I1(x) = h ( x ) = x 1 and write (2.1) as a( x ) The Jacobian approximation is, of course, obtained by replacing the f vector field by its linear approximation and the g vector field by its constant approximation, that is, by neglecting $2 and G4"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.44-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.44-1.png",
        "caption": "FIGURE 7.44. The multi-link steering is a 6-link mechanism that may be treeted as two combined 4-bar linkages.",
        "texts": [
            "25176 rad \u2248 14.425 deg (7.158) \u03b4o = tan\u22121 l R1 + w 2 = 0.22408 rad \u2248 12.839 deg (7.159) Because the mechanism is symmetric, each wheel of the steering mechanism 7. Steering Dynamics 429 in Figure 7.43 must be able to turn at least 14.425 deg. To be safe, we try to optimize the mechanism for \u03b4 = \u00b115 deg. The multi-link steering mechanism is a six-link Watt linkage. Let us divide the mechanism into two four-bar linkages. The linkage 1 is on the left and the linkage 2 is on the right, as shown in Figure 7.44. We may assume that MA is the input link of the left linkage and PB is its output link. Link PB is rigidly attached to PC, which is the input of the right linkage. The output of the right linkage is ND. To find the inner-outer steer angles relationship, we need to find the angle of ND as a function of the angle of MA. The steer angles can be calculated based on the angle of these two links. \u03b41 = \u03b82 \u2212 (90\u2212 54.6) deg (7.160) \u03b42 = \u03d54 \u2212 (90 + 54.6) deg (7.161) Figure 7.44 illustrates the link numbers, and the input-output angles of the four-bar linkages. The length of the links for the mechanisms are collected in Table 7.1. Equation (6.1) that is repeated below, provides the angle \u03b84 as a function of \u03b82. \u03b84 = 2 tan \u22121 \u00c3 \u2212B \u00b1 \u221a B2 \u2212 4AC 2A ! (7.162) A = J3 \u2212 J1 + (1\u2212 J2) cos \u03b82 (7.163) B = \u22122 sin \u03b82 (7.164) C = J1 + J3 \u2212 (1 + J2) cos \u03b82 (7.165) 7. Steering Dynamics 431 J1 = d1 a1 (7.166) J2 = d1 c1 (7.167) J3 = a21 \u2212 b21 + c21 + d21 2a1c1 (7.168) J4 = d1 b1 (7.169) J5 = c21 \u2212 d21 \u2212 a21 \u2212 b21 2a1b1 (7"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure3.12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure3.12-1.png",
        "caption": "Figure 3.12 Three times the same couple: (a) Tz = \u221212 kNm, (b) Ty = +12 kNm, (c) the couple using visual notation.",
        "texts": [
            " Conclusion: The effect of a couple on the equilibrium of a body does not change if you replace it by another couple with the same moment and the same direction of rotation. The magnitude of the moment of a couple determines the state of rotation of the body. In addition to a magnitude, the moment also has a direction of rotation. The sign for the direction of rotation is linked to the coordinate system, see the sign convention in Section 1.3.2. In the xy coordinate system shown, the moment of the couple in Figure 3.12a is Tz = \u2212Fa = \u221212 kNm. The letter T is given the index z, which indicates the normal of the plane in which the couple acts. In Figure 3.12b the moment of the same couple is in another coordinate system: Ty = +Fa = +12 kNm. In Figure 3.12c, the couple is represented by a curved arrow. In this visual notation the arrow indicates the direction of rotation of the moment and includes a value. The same conventions apply as for the visual notation of 60 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM a force (see Section 1.3.6). Couples can be compounded in many different ways. In Figure 3.13, the couples operating in the xy plane, T1, T2 and T3, have been compounded by replacing them by equal couples of which the forces have common lines of action"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000456_tac.2005.858646-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000456_tac.2005.858646-Figure4-1.png",
        "caption": "Fig. 4. Performance with the input noise magnitude 0.1 m.",
        "texts": [
            " 3(c) that the control u remains continuous until the entrance into the 3-sliding mode. The steering angle remains rather smooth and is quite feasible. In the presence of output noise with the magnitude 0.01 m the tracking accuracies j j 0:02; j _ j 0:14; j j 1:3 were obtained. With the measurement noise of the magnitude 0.1 the accuracies changed to j j 0:20; j _ j 0:62; j j 2:8 which corresponds to the asymptotics stated by Theorem 4. The performance of the controller with the measurement error magnitude 0.1 m is shown in Fig. 4. It is seen from Fig. 4(c) that the control u is a continuous function of t. The steering angle vibrations have the magnitude of about 7 and the frequency 1, which is also quite feasible. The performance does not change, when the frequency of the noise varies in the range 100\u2013100 000. The advantages of the new controller are obvious (compare Figs. 2 and 3). Simulation shows that the previous controller [15], [16] is also much more sensitive to the parameter choice and noises. VII. CONCLUSION A new arbitrary-order sliding mode controller is proposed"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.8-1.png",
        "caption": "FIGURE 8.8. A triangle mechanism and a Panhard arm to guide a solid axle.",
        "texts": [
            "6, may be attached to a solid axle to provide lateral and twist resistance during acceleration and braking. Example 304 Panhard arm. High spring rate is a problem of leaf springs. Reducing their stiffness by narrowing them and using fewer leaves, reduces the lateral stiffness and increases the directional stability of the suspension significantly. A Panhard arm is a bar that attaches a solid axle suspension to the chassis laterally. Figure 8.7 illustrates a solid axle and a Panhard arm to guide the axle. Figure 8.8 shows a triangular linkage and a Panhard arm combination for guiding a solid axle. A double triangle mechanism, as shown in Figure 8.9, is an alternative design to guide the axle and support it laterally. Example 305 Straight line linkages. There are many mechanisms that can provide a straight line motion. The simplest mechanisms are four-bar linkages with a coupler point moving straight. Some of the most applied and famous linkages are shown in Figure 8.10. By having proper lengths, the Watt, Robert, Chebyshev, and Evance linkages can make the coupler point C move on a straight line vertically"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure11.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure11.6-1.png",
        "caption": "FIGURE 11.6. Kinematics of a two-wheel model for a vehicle with roll and yaw rotations.",
        "texts": [
            "103) Assuming small angles for slip angles \u03b2f , \u03b2, and \u03b2r, the tire sideslip angles for the front and rear wheels, \u03b1f and \u03b1r, may be approximated as \u03b1f = 1 vx (vy + a1r \u2212 zfp)\u2212 \u03b4 \u2212 \u03b4\u03d5f = \u03b2 + a1 r vx \u2212 C\u03b2f p vx \u2212 \u03b4 \u2212 C\u03b4\u03d5f\u03d5 (11.104) \u03b1r = 1 vx (vy \u2212 a2r \u2212 zrp)\u2212 \u03b4\u03d5r = \u03b2 \u2212 a2 r vx \u2212 C\u03b2r p vx \u2212 C\u03b4\u03d5r\u03d5. (11.105) 11.3.3 F Body Force Components on a Two-wheel Model Figure 11.4 illustrates a top view of a car and the force systems acting at the tireprints of a front-wheel-steering four-wheel vehicle. When we consider the roll motion of the vehicle, the xy-plane does not remain parallel to the road\u2019s XY -plane, however, we may still use a two-wheel model for the vehicle. Figure 11.5 illustrates the force system and Figure 11.6 illustrates the kinematics of a two-wheel model for a vehicle with roll and yaw rotations. The rolling two-wheel model is also called the bicycle model. The force system applied on the bicycle vehicle, having only the front wheel steerable, is Fx = 2X i=1 (Fxi cos \u03b4 \u2212 Fyi sin \u03b4) (11.106) Fy = 2X i=1 Fyi (11.107) Mx = Mxf +Mxr \u2212 wcf \u03d5\u0307\u2212 wkf\u03d5 (11.108) Mz = a1Fyf \u2212 a2Fyr (11.109) where \u00a1 Fxf , Fxr \u00a2 and \u00a1 Fyf , Fyr \u00a2 are the planar forces on the tireprint of the front and rear wheels. The force system may be approximated by the 678 11"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001975_tro.2013.2281564-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001975_tro.2013.2281564-Figure3-1.png",
        "caption": "Fig. 3. Actuation cabling between disks. Each cable exerts a force at the routing hole based on the relative location of the adjacent disks\u2019 routing holes.",
        "texts": [
            " The resulting force at the disks\u2019 cable routing holes are reformulated as a resultant force and moment acting at the disk center of mass. When considering cable-actuated robots, it is important to ensure that the model does not allow for compressive forces by the actuation cabling. In this model, the tensions of the actuation cables at the base are the input parameters to the model. By ensuring that these inputs are always positive tension inputs, the cabling cannot apply a compressive force. A geometric analysis is used to determine the loading on each disk. Fig. 3 shows the cable routing between three disks. Because the cables follow a linear path between holes, the cable routing may be by determined by calculating the hole coordinates. At the continuum robot base and within each local disk frame, the three cable routing hole position vectors\u2014rlcl,1 , rlcl,2 , and rlcl,3\u2014are defined by (28), shown below, where rh is the radial distance of the holes from the center rlcl,1 = rh [ 1, 0, 0 ] , rlcl,2 = rh [\u22121/2, \u221a 3/2 0 ] rlcl,3 = rh [\u22121/2, \u2212 \u221a 3/2, 0 ] . (28) Using the pi,lcl , the position pi,j,hl of the jth cable in the ith subsegment may be calculated using (29), shown below. As a result, the cable force directions f i,j may be found as the unit vector of pi,j,hl , defined in (30), shown below, and shown in Fig. 3. pi,j,hl = { pi,lcl + Rirlcl,j \u2212 rlcl,j , i = 1 Ri\u22121pi,lcl + Rirlcl,j \u2212 Ri\u22121rlcl,j , i > 1 (29) f i,j = pi,j,hl /\u2225\u2225pi,j,hl \u2225\u2225. (30) The coupling of the cable tension along the continuum robot and the frictional forces at each disk complicates the analysis. In order to compute the friction at each disk, the average tension of the cable before and after the disk is needed. However, in order to compute the cable tensions before and after the disk, the magnitude of the frictional force is needed. Therefore, an iterative approach is required",
            " (38) With these updated values of Ti,j , (33) can be used to calculate more accurate estimates of the frictional force magnitude at each disk, which then may be used to calculate more accurate cable tensions using (37). Once a sufficient number of iterations has occurred (the determination of a sufficient number is addressed in Section V), the contact force F i,j,con which includes frictional effects may be determined for each cable and disk, as in (39), shown below, with the two forces shown in Fig. 3 F i,j,con = { Ti+1,jf i(i+1),j \u2212 Ti,jf (i\u22121)i,j , i < N \u2212Ti,jf (i\u22121)i,j , i = N. (39) Using the calculated contact and friction forces applied at each disk\u2019s cable routing hole, the resulting actuation force F i,act and moment M i,act on the disk\u2019s center of mass may be computed, as follows: F i,act = \u2211 j F i,j,con , M i,act = \u2211 j ri,j \u00d7 F i,j,con . (40) In order to formulate the net external force F i,ex and moment M i,ex applied to each disk i, the calculated forces and moments are added together, as in F i,ex = F i,inr + F i,gr + F i,act , M i,ex = M i,inr + M i,el + M i,act "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002262_asjc.1758-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002262_asjc.1758-Figure4-1.png",
        "caption": "Fig. 4. Boundary layer solution.",
        "texts": [
            " This is why, a lot of theoretical and simulation based works have been investigated by various researchers on this particular approach for quadrotor control [61\u201364]. Fig. 3 illustrates the reaching and sliding phases of system states on sliding surface S=0. Sliding mode control can generally be categorized into various classes, however, classical first order, second or higher order, and terminal sliding mode control (TSMC) have extensively been used for flight control purposes. Since chattering is an alarming issue in SMC approach, therefore there exist multiple solutions to cope with this problem. Boundary layer solution, as depicted in Fig. 4, is the most common among all, however, the thickness of the boundary layer puts serious constraints on controller performance. Another solution is the incorporation of fuzzy-based adaptive gains in sliding surface. Since the chattering intensity depends on sliding gains, therefore, these gains can be made adaptive using fuzzy rules and chattering can be reduced; see [66] for a detailed study on the comparison of both solutions. Unfortunately, both these discussed solutions deprive the SMC of invariance property and therefore, degradation in performance arises"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure5.41-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure5.41-1.png",
        "caption": "Figure 5.41 The forces acting on SD are found from the equilibrium of the isolated parts.",
        "texts": [
            " In the same way, one can find N(2) from the moment equilibrium of DB about D: \u2211 T (BD) z |D = +(10 \u221a 2 kN)(3 \u221a 2 m) \u2212 N(2) \u00d7 ( \u221a 2 m) = 0 so that N(2) = 30 \u221a 2 kN. Shoring bar 2 is a tension bar. To demonstrate clearly how the shoring bars act on frame ASB, the frame and the shoring bars have been isolated from one another in Figure 5.40. 180 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM c. The force acting at S on SD is equal to the support reaction at A. The force that shoring bar (2) exerts on SD is also known. Still unknown are the components of the force exerted on SD at D. These are found via the force equilibrium of column BD (see Figure 5.41). Check: SD must be in equilibrium. Example 2 The structure in Figure 5.42a is loaded on rafter ACE by a vertical force F = 40 kN. Questions: a. Determine the support reactions. b. Determine the force in bar CD (with the correct sign for tension and compression). c. Determine the hinge force at E. Solution: a. The support reactions follow directly from the equilibrium of the structure as a whole. There are only vertical support reactions. They are shown in Figure 5.42b. b. Suppose the tensile force in CD is N(CD)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000736_s00170-010-2631-5-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000736_s00170-010-2631-5-Figure4-1.png",
        "caption": "Fig. 4 Orientation of samples with base geometry in X\u2013Y plane, case A",
        "texts": [
            " In Fig. 3, dimension and geometry of the testing samples are shown. All samples are grown in the z axis of the SLM equipment working area; the difference, in each case, is in which plane and direction is contained the base geometry of the sample as may be seen in Fig. 3. Case A: Samples manufactured with base geometry in X\u2013Y machine plane (with different rotation angles). In this case, it has to be considered that melting angle is different in the calibrated zone, depending on the scanning angle (Fig. 4). Samples have been manufactured at 0\u00ba, 30\u00ba, 45\u00ba, and 90\u00ba with respect to X axis. Manufactured samples by SLM may be observed in Fig. 5. Case B: Samples manufactured with base geometry in Z\u2013X to Z\u2013Y plane (with different rotation To avoid the use of supports from samples to the platform, samples have been manufactured with prismatic geometry and then slightly machined to the final geometry (Fig. 6). Case C: Samples made with base geometry in the Z\u2013X/Y plane (with different rotation angle) and main axis of the sample in Z direction"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001620_j.addma.2017.11.001-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001620_j.addma.2017.11.001-Figure1-1.png",
        "caption": "Fig. 1. Components of an Arcam machine.",
        "texts": [
            " Paricular attention is being paid to the production of fully dense ear-net-shaped metal parts made of g-TiAl alloys, because these aterials have shown interesting properties for aerospace appliations, such as low density, high specific strength, high specific tiffness, fracture toughness [26,56\u201358], ductility, fatigue strength, reep [26,56\u201358], and corrosion and oxidation resistance at high emperatures [26,46,50]. As far as the technologies used to produce an electron beam re concerned, the EBM system is considered similar to a weldng machine, while the operating principle is similar to an electron icroscope. Arcam, which was founded 1997, developed the first dditive technology based on an electron beam (EB) [8]. Fig. 1 shows he main components of an Arcam machine: the electron beam unit nd the build chamber. The electron beam unit consists of an upper olumn that contains the electron generating part, and a lower colmn that contains the magnetic lenses used to form and deflect the eam. A heated filament, the cathode, emits electrons in the upper olumn. The potential between the cathode and anode is usually bout 60 kV. The electrons are accelerated to a velocity of between .1 and 0.4 times the speed of light [59]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001533_j.ymssp.2012.06.021-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001533_j.ymssp.2012.06.021-Figure3-1.png",
        "caption": "Fig. 3. Damaged sun gears: (a) having a chipped tooth and (b) having a missing tooth.",
        "texts": [
            " The planetary carrier, in turn, transmits torque to the output shaft. In a planetary gearbox, sun gear teeth easily suffer damage because their multiplicity of meshes with the planet gears increases the potential for damage on the sun gear. In addition, the second stage of the planetary gearbox in our test rig undertakes a larger load than the first stage. For these reasons, faults on the sun gear in the second stage planetary gearbox are created in our experiments. The fault modes include a chipped tooth and a missing tooth, which are illustrated in Fig. 3. There are 4 planet gears in the second stage of the planetary gearbox test rig, and 28, 36 and 100 teeth on the sun gear, planet gears and ring gear, respectively. Since vibration-based analysis is one of the principal tools for diagnosing mechanical faults [27\u201328], the vibration is measured for a normal gear, the gears with a chipped tooth and a missing tooth using a tri-axial accelerometer. The accelerometer was produced by PCB Electronics with the model number 356A12. It is mounted on the planetary gearbox casing",
            " Discussions 1) A planetary gearbox is more complicated than a fixed-axis gearbox. It is a compound assembly of various parts that constantly interact. Accordingly the vibration transmission paths from gear meshing points to transducers are complicated. The complicated transmission paths may deteriorate or attenuate vibration response of faulty gears through dissipation and interference effects, though the gear fault is quite severe (a missing tooth). This point can be verified by contrasting the damaged gears displayed in Fig. 3 and the vibration signals and their spectra shown in Fig. 8, from which both cyclic impacts and characteristic frequencies cannot be observed at all. It is concluded that the vibration signals depend on not only damage itself but also the machine system. In addition, maybe because the second stage planetary gear-pair works under bigger load and smaller speed than the first stage, the cyclic impacts of the second stage are constrained in time domain and are not excited as large as those of the first stage"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.3-1.png",
        "caption": "FIGURE 8.3. A driving and braking trust, force leaf springs into an S shaped profile.",
        "texts": [
            " The solid axle suspension systems with longitudinal leaf springs have many drawbacks. The main problem lies in the fact that springs themselves act as locating members. Springs are supposed to flex under load, but their flexibility is needed in only one direction. However, it is the nature of leaf springs to twist and bend laterally and hence, flex also in planes other than the tireplane. Leaf springs are not suited for taking up the driving and braking traction forces. These forces tend to push the springs into an S-shaped profile, as shown in Figure 8.3. The driving and braking flexibility of leaf springs, generates a negative caster and increases instability. Long springs provide better ride. However, long sprigs exaggerate their bending and twisting under different load conditions. Example 301 Leaf spring suspension and flexibility solution. To reduce the effect of a horizontal force and S-shaped profile appearance in a solid axle with leaf springs, the axle may be attached to the chassis by a longitudinal bar as Figure 8.4(a) shows. Such a bar is called an anti-tramp 8"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure13.39-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure13.39-1.png",
        "caption": "Figure 13.39 (a) The N diagram due to the component normal to the roof plane, superposed on (b) the N diagram due to the component parallel to the roof plane gives (c) the requested N diagram due to the dead weight.",
        "texts": [
            "642 kN/m normal to the beam axis (Figure 13.38b) and 1.857 kN/m parallel to the beam axis (Figure 13.38c). The bending moment in ACD is caused by the load of 4.642 kN/m normal to the beam axis. The M diagram is equal to that in Figure 13.33d, but 4.462 as large, so that the extreme values of the bending moments are Mdw;max = (2.558 kNm) \u00d7 4.642 = 11.87 kNm, Mdw;min = (2.320 kNm) \u00d7 4.642 = 10.77 kNm. 13 Calculating M, V and N Diagrams 579 580 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM The N diagram in Figure 13.39a due to the 4.642 kN/m load normal to the beam axis is equal to the N diagram in Figure 13.33b, but with values that are 4.642 times as large. The N diagram in Figure 13.39b due to the load of 1.857 kN/m parallel to the beam axis is equal to the N diagram in Figure 13.35b, but then with values that are 1.857 times as large. Superposing the N diagrams in Figures 13.39a and 13.39b gives the N diagram in Figure 13.39c. This is the requested N diagram due to the dead weight qdw = 5 kN/m. Figure 13.39 (a) The N diagram due to the component normal to the roof plane, superposed on (b) the N diagram due to the component parallel to the roof plane gives (c) the requested N diagram due to the dead weight. 13 Calculating M, V and N Diagrams 581 c. Snow load Over a length a measured horizontally, the resultant of the snow load is aqsn (see Figure 13.40). The components of this force normal to and parallel to the axis are respectively aqsn cos \u03b1 and aqsn sin \u03b1. They act on a member segment with length a/cos \u03b1"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002179_j.acme.2014.02.003-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002179_j.acme.2014.02.003-Figure1-1.png",
        "caption": "Fig. 1 \u2013 Schematic of SLM device [12].",
        "texts": [
            " Selective Laser Melting is a technology based on the processing of metal powder. In the process metal powder is locally melted, layer by layer, using a high-power laser. Subsequent layers are built with a newly applied powder, directly on previous layers, which provides a permanent connection in the whole build. The metal powder is administered by a movable tray, which also aligns the applied layer. Particle size of AISI 316L powder used in the process is less than 63 mm. A schematic diagram of a SLM device is shown in Fig. 1. All samples had the same dimensions illustrated in Fig. 2. The thinnest part of the samples had a diameter of 4 mm, which provides the same resolution measurement as XCT. Technological parameters of the SLM process, which were used to manufacture the specimens (A, B and C), are presented in Table 1. All of the process parameters were the same for the three presented samples. The only difference was the distribution (building orientation) of the specimens within the chamber of the SLM machine (see Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure8.10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure8.10-1.png",
        "caption": "Fig. 8.10. AFPM machine with external air cooling.",
        "texts": [
            " For medium to large power AFPM machines, the loss per unit heat dissipation area increases almost linearly with the power ratings. Thus the forced-cooling with the aid of external devices may be necessary. Some common techniques are described as follows. External Fans Large AFPM machines may require a substantial amount of air flow per unit time in order to bring out the heat generated in the stator windings. Depending on the operation conditions obtained on site, either an air-blast or a suction fan may be used as shown in Fig. 8.10. In both cases, intake and/or discharge ducts are needed to direct and condition the air flow. Since the inlet air temperature, for a given volumetric flow rate, has a significant effect on the machine temperature, this cooling arrangement can also help prevent recirculation of hot air should the machine operate in a confined space (e.g. small machine room). For high speed AFPM machines, a shaft-integral fan may be 266 8 Cooling and Heat Transfer a good option. Fig. 8.11 shows the assembly of a large power AFPM machine developed in the Department of Electrical and Electronic Engineering at the University of Stellenbosch, South Africa, in which the rotor hub part serves as both cooling fan and supporting structure for the rotor discs"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001358_cdc.1978.268148-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001358_cdc.1978.268148-Figure4-1.png",
        "caption": "Figure 4. Coordinate frames f o r the Stanford A n .",
        "texts": [
            " The end o f the man ipu la to r u i th respec t to the base i s g i v e n by T6: T6 = A1 + A2 + A + A4 + A s + A6 3 (4 ) I f the manipulator i s r e l a t e d t o a reference coord ina te frame by a t ransformat ion 2, and has a t o o l a t t a c h e d t o i t s end described by E, ue have the d e s c r i p t i o n o f t h e end o f t h e t o o l u i t h r e s p e c t t o the reference coordinate system described by X as f o l lous C53: X = Z + T 6 + E (5 1 uhich may be so lved fo r T6 as: T 6 = - Z + X - E (6 uhere the - s i g n i n d i c a t e s m u l t i p l i c a t i o n by the matr ix inverse. I n Figure 4 the Stanford arm i s shoun u i t h coordinate frames a signed t o h e l i n k s . The parameters are shoun i n Table 1. L ink Var iab le a a d Ca Sa 1 el -90\u00b0 0 0 0 -1 2 -90' 0 d2 0 1 4 e4 -!ao 0 0 0 -1 5 e5 9 o O O O 0 1 3 5 O 0 0 % 1 0 6 e6 O O O O 1 0 s2c2 cl T 3 = 0 0 0 -sl :: 0 1 ClS2 -S1d2 s l s 2 c2 0 ...1 1 uhere: OX = -C(C1S2C4-SlS4)C5-C1s*s51s6-cclc2s4+sls41c6 (12) (19) a = (S1C2C4+ClS4)S5+SlS2C5 aZ = -S2C4S5+C2C5 p x = -Sld2+C1S2d3 (13) Y pY = CldZ+S1S2% The f i r s t column o f T6 can be obtained as the vec- (\") to r c ross product o f t h e second and t h i r d columns"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001766_j.cirp.2019.05.004-Figure33-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001766_j.cirp.2019.05.004-Figure33-1.png",
        "caption": "Fig. 33. Micro test artefact for interdisciplinary analyses in micro manufacturing [187].",
        "texts": [
            " Right: artefact including fic features: overhangs, ramps, cones, hemi-spheres, etc. [175]. art, ogy ain ach tion AM are and to TM e to nal her, and her low rds and hat fted rds rch of ons AM ith AM nal 261: s, (e) le is and much smaller artefact (see Fig. 32b) was proposed by Hao et al. which was especially designed for the evaluation of micro- and meso-scale production capabilities, as well as process stability and repeatability [102]. M\u00f6hring et al. developed a micro-scale test artefact including simple geometrical features as well as freeform structures (Fig. 33) to compare the capabilities of milling and different AM processes in a CIRP round robin experiment involving different micromanufacturing processes. The comparison revealed large form errors especially for thin walled features, caused by insufficient resolution of AM processes, but also showed the advantages of using XCT systems for measuring some of the key features [187]. Following the trend towards XCT inspection, Shah et al. presented their test artefact especially tailored for inspection with XCT (see Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure7.15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure7.15-1.png",
        "caption": "Figure 7.15 (a) As a result of the distributed load q, a small force F = q s is acting on a small boundary element with length s. (b) The horizontal components of the load on the boundary elements of a horizontal strip are equal and opposite. Together they form an equilibrium system with zero resultant.",
        "texts": [
            " Comment: This force is relatively large for a concrete cross-section of 5.6 m2. The compressive force in the ring may therefore call for a larger cross-section. Example 3 A uniformly distributed load q is acting on the plane body in Figure 7.14. The load acts in the plane of the body along the entire outline and normal to the body. Questions: a. Show that the resultant of the distributed load on the body is zero, regardless of the shape of the body. b. Determine the resultant of the load above section AB. Solution: a. In Figure 7.15a, a minor force F is acting perpendicular to the given boundary element with small length s: F = q s. The horizontal and vertical components of F are respectively Fx = q s cos \u03d5, 1 The convention is that N as tensile force is positive. The prime for a switch in sign indicates that compressive forces are now positive (see Section 6.5). 254 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Fy = q s sin \u03d5. Since s cos \u03d5 = y and s sin \u03d5 = x, we can also write Fx = q y, Fy = q x. The components Fx and Fy of force F on boundary element s are equal to the product of the distributed load q and the projection of s on the y axis and the x axis respectively. Figure 7.15b shows a horizontal strip from the body with a small width y. The horizontal components of the load on the boundary elements are equal and opposite. They form an equilibrium system with resultant zero. Since this applies to all the horizontal strips of which the body is composed, the resulting horizontal load on the body is zero. By dividing the body into vertical strips, and looking at the vertical component of the load on the boundary elements, we can similarly deduce that the resulting vertical load on the body is zero",
            " Conclusion: If a uniformly distributed load acts on a plane body in the plane of the body along its entire outline, and everywhere normal to the body, the load forms an equilibrium system with resultant zero. One can show that this is also true in three-dimensional cases: If a uniformly distributed load acts on a body in space on its entire surface, and everywhere normal to the body, the load forms an equilibrium system with resultant zero. b. In Figure 7.16a, the part of the body above section AB has been isolated. Assume the resultant of the distributed load on the outside between A and B is R. Figure 7.15 (a) As a result of the distributed load q, a small force F = q s is acting on a small boundary element with length s. (b) The horizontal components of the load on the boundary elements of a horizontal strip are equal and opposite. Together they form an equilibrium system with zero resultant. 7 Gas Pressure and Hydrostatic Pressure 255 If a uniformly distributed load q is also applied to section AB, as in Figure 7.16b, the total load on the isolated part of the body forms an equilibrium system: the resultant R of the load on the outside of the body is equal and opposite to the resultant Rsection of the load on the section"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003991_s00604-009-0153-3-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003991_s00604-009-0153-3-Figure2-1.png",
        "caption": "Fig. 2 Stress-Wrinkles in a polyethylene sheet (reprinted with permission from ref. [22]) of length L=25 cm, width W=10 cm, and thickness t= 0.01 cm under uniaxial tensile strain \u03b5=10%, schematically shown in (1.). Wrinkles relax as the stress is released (2.)",
        "texts": [
            " In most of the work cited above, the polydimethylsiloxane (PDMS) was used as elastomeric substrate and covered by a hard film. This coupled membrane system is then exposed to strain \u03b5, which causes wrinkling. Using this approach, both transient wrinkles, which only exist when macroscopic strains are applied and \u201cpermanent\u201d wrinkles, which remain in the absence of macroscopic strains can be produced. Both cases are illustrated in Figs. 2 and 3 for the simplest situation of a uniaxial deformation: In Fig. 2, the sheet is macroscopically stretched as indicated by the arrows. As a consequence, transversal contraction takes place perpendicular to the stretching direction. Wrinkles will appear in the macroscopically stretched state and\u2014provided the system is linearly elastic and no plastic deformations occur\u2014disappear upon relaxation. An example of such a situation is the PE sheet from ref. [22]. In Fig. 3, the film was wrinkle-free in the macroscopically stretched state (e.g. it could be prepared on a stretched substrate, as in some of the cases explained below)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003928_tpel.2006.872371-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003928_tpel.2006.872371-Figure1-1.png",
        "caption": "Fig. 1. Cross section of a 12/8 SRM along with the flux lines at aligned position.",
        "texts": [
            " These technologies are explained in detail and are supported with simulation and experimental results for validation purposes. Experimental results are obtained from two different switched reluctance machine (SRM) prototypes that have been primarily designed and developed for electrically assisted power steering (12/8, 2[kW], 42[V]) and electric coolant pump (8/6, 2[kW], 42[V]). Electromagnetic torque in SRMs is produced by polarized rotor poles tending to align with the excited stator poles (see Fig. 1). The resulting mechanical work in motoring is provided by the magnetic field via excitation of the machine coils from an electric source. During generation, on the other hand, mechanical energy supplied by prime mover is mostly responsible for build up of magnetic field and generated electric power. The electromagnetic torque can be expressed in terms of co-energy as follows: (1) The electromechanical exchange of energy results in an induced voltage in machine coils that opposes the increase of current during motoring"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001996_j.rcim.2015.12.004-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001996_j.rcim.2015.12.004-Figure3-1.png",
        "caption": "Fig. 3. Weld bead geometry.",
        "texts": [
            " The model is used to predict the combination of optimal welding process parameters, namely wirefeed rate and travel speed, which will produce the desired bead geometry. A multi-bead overlapping model that has been introduced in a previous publication will also be briefly reviewed and then incorporated into the methodology, to determine the optimum spacing between adjacent weld beads. 2.1. Single bead profile Experiments were carried out for a total of 81 combinations of 9 different wire-feed rates (F) and 9 different travel speeds (S), as summarised in Table.1. The details of welding process, wire diameter, and material used will be introduced in Section 5. Fig. 3 schematically shows a deposited bead and the cross-section of the bead geometry with a width, w and a height, h. All 81 bead profiles are fitted with a parabola function using curve fitting method described in Ref. [21]. The relative error of area prediction, E, is defined as the percentage of the area difference between the predicted and the actual bead area over the actual bead area, = \u2212 \u00d7 ( ) E A A A 100% 1 p a a where Ap is the prediction of the bead area by the parabola model. The actual area, Aa of a weld bead cross-section, namely metal deposition rate per unit length, can be calculated as \u03c0 \u03c0\u03bb = = ( )A FD S D 4 4 2a w w 2 2 where, Dw is the diameter of the wire electrode, \u03bb is the ratio of wire-feed rate to travel speed"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000692_9.1273-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000692_9.1273-Figure1-1.png",
        "caption": "Fig. 1.",
        "texts": [
            "1) applied to the AOCF (7.5) can be shown to be globally stable (i.e., the conditions S.1-S.3 and SI1, SI2 hold) by combining appropriately : This leads to the following AOCF: a trivial extension of Lemma 1 in [20] Theorem 3.2 of chapter 4 in [ 141 Corollary 4.2 and Theorem 6.2 of [15] [ z ; ] = [ \u2018\u2018z] + [ :2] [ Theorem 4.2 of this paper. (7.5) VIII. APPLICATION TO A ROBOT MANIPULATOR We consider an application to a telescopic arm in a vertical plane which performs a \u201cpick and place\u201d operation; see Fig. 1. We call Mthe mass of the load, f ( t ) the variable length of the arm, friction coefficients, CY$ and k, the stiffness coefficients, u1 and u2 the voltages applied to the electrical motors in the joint and the arm, respectively. Assuming that the time constants of these Note that the transformation (7.4) is uniquely invertible as y ( t ) the angle with vertical axis, CY^ and kF the viscous follows: 1 y=x1. P3= -03 (7.6a) I - T r - ~ 1 1 -- -11 1 1 656 IEEE TRANSACTIONS ON AUTOMATIC CONTROL, VOL"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure3.16-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure3.16-1.png",
        "caption": "Fig. 3.16. Disc-type coreless winding assembled of coils of the same shape according to U.S. Patent No. 5 744 896 [P83]: (a) single coil; (b) three adjacent coils. 1 \u2014 coil side, 2 \u2014 inner offsetting bend, 3 \u2014 outer offsetting bend.",
        "texts": [
            " There are two types of windings: (a) winding comprised of multi-turn coils wound with turns of insulated conductor of round or rectangular cross section; (b) printed winding also called film coil winding . Coils are connected in groups to form the phase windings typically connected in star or delta. Coils or groups of coils of the same phase can be connected in parallel to form parallel paths. 112 3 Materials and Fabrication To assemble the winding of the same coils and obtain high density packing, coils should be formed with offsetting bends, as shown in Fig. 3.16. The space between two sides of the same coil is filled with coil sides from each of the adjacent coils. Coils can be placed in a slotted structure of the mould (Fig. 3.17). With all the coils in position, the winding (often with a supporting structure or hub) is moulded into a mixture of epoxy resin and hardener and then cured in a heated oven. Because of the difficulty of releasing the cured stator from the slotted structure of the mould (Fig. 3.17a), each spacing block that forms a guide slot consists of several removable pins of different size (Fig",
            "127 m, the average diameter D = 0.5(Dout + Din) = 0.5(0.22 + 0.127) = 0.1735 m, the average pole pitch \u03c4 = \u03c00.1735/6 = 0.091 m, the length of conductor (equal to the radial length lM of the PM) Li = lM = 0.5(Dout\u2212Din) = 0.5(0.22\u22120.127) = 0.0465 m, the length of shorter end connection without inner bends l1emin = (wc/\u03c4c)\u03c0Din/(2p) = (7/9)\u03c0\u00d7 0.127/6 = 0.052 m and the length of the longer end connection without outer bends l1emax = 0.052\u00d7 0.22/0.127 = 0.0896 m. The average length of the stator turn with 15-mm bends (Fig. 3.16) is l1av \u2248 2Li + l1emin + l1emax + 4\u00d7 0.015 = 2\u00d7 0.0465 + 0.052 + 0.0896 + 0.06 = 0.2943 m The stator winding resistance at 75oC according to eqn (2.40) is R1 = 234\u00d7 0.2943 47\u00d7 106 \u00d7 2\u00d7 \u03c0 \u00d7 (0.912\u00d7 10\u22123)2/4 = 1.122 \u2126 The maximum width of the coil at the diameter Din is ww = \u03c0Din/s1 = \u03c00.127/54 = 0.0074 m = 7.4 mm. The thickness of the coil is tw = 8 mm. The number of conductors per coil is Nc = aw\u00d7Nct = 2\u00d726 = 52. The maximum value of the coil packing factor is at Din, i.e. kfmax = d2 w \u00d7Nc twww = 0",
            " The disc-type PM brushless machines without stator and rotor cores were first manufactured commercially in the late 1990s for servo mechanisms and industrial electromechanical drives [150], solar powered electrical vehicles [225] as well as micromotors for computer peripherals and vibration motors for mobile phones [93]. The AFPM brushless machine without any ferromagnetic core is shown in Fig. 6.1. The machine consists of a twin rotor (3) with rare earth PMs (2) and nonmagnetic supporting structure. The steel-free stator (armature) winding (1) is located between two identical parts of the rotor. The stator polyphase winding fixed to the frame (6) is assembled as \u201cflower petals\u201d (Fig. 3.16) [P83]. Multi-turn coils are arranged in overlapping layers around the shaftaxis of the machine. The whole winding is then embedded in a high mechanical integrity plastic or resin. Since the topology shown in Fig 6.1 does not use any ferromagnetic core with slots, the machine is free of cogging (detent) torque 194 6 AFPM Machines Without Stator and Rotor Cores 6.3 Air Gap Magnetic Flux Density 195 and core losses. The only eddy current losses are losses in the stator winding conductors and metallic parts (if they exist) that reinforce the stator coreless winding",
            "5 Nm/A The EMF at 1000 rpm is Ef = kEn = 0.0873\u00d7 1000 = 87.34 V The electromagnetic torque at Ia = Iaq = 8.2 A is Td = kT Ia = 2.5\u00d7 8.2 = 20.5 Nm The electromagnetic power is Pelm = 2\u03c0nsTd = 2\u03c0 1000 60 \u00d7 20.5 = 2148.6 W The length of the stator shorter end connection without inner bends l1emin = (wc/\u03c4c)\u03c0Din/(2p) = (3/3)\u03c0\u00d70.127/24 = 0.017 m and the length of the stator 6.8 Characteristics of Coreless AFPM Machines 211 longer end connection without outer bends l1emax = 0.017 \u00d7 0.22/0.127 = 0.0288 m (Fig. 3.16). The average length of the stator turn with 15-mm bends is l1av \u2248 2Li + l1emin + l1emax + 4\u00d7 0.015 = 2\u00d7 0.0465 + 0.017 + 0.0288 + 0.06 = 0.1984 m The stator winding resistance at 75oC according to eqn (2.40) is R1 = 240\u00d7 0.1984 47\u00d7 106 \u00d7 8\u00d7 \u03c0 \u00d7 (0.455\u00d7 10\u22123)2/4 = 0.7789 \u2126 The maximum width of the coil at the diameter Din is ww = \u03c0Din/s1 = \u03c00.127/72 = 0.0055 m = 5.5 mm. The thickness of the coil is tw = 8 mm. The number of conductors per coil is Nc = aw \u00d7 Nct = 8 \u00d7 20 = 160. The maximum value of the coil packing factor is at Din, i"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000619_scirobotics.aav1488-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000619_scirobotics.aav1488-Figure1-1.png",
        "caption": "Fig. 1. Soft actuator design. (A) Side view of the computer-aided design (CAD) of the soft actuator with infrared-reflective balls for tracking the motion of the tip. Embedded sensors are used to estimate the coordinates of the tip and the forces applied by the actuator when in contact. The plots we present in this paper describe the position of the marker at the tip relative to the marker at the base. (B) Physical actuator with embedded soft sensors.",
        "texts": [
            " The experimental setup was not varied in this case except for adding a load cell to the external contact environment for obtaining the ground truth. The case of predicting forces applied at the tip is then presented. Last, we present simulation studies that investigate how the redundant architecture could be exploited by the learned network to be more robust to noise even in the case of the complete loss of some sensors. The proposed approach was validated primarily on a pneumatically actuated planar soft finger with three embedded soft resistive strain sensors (Fig.\u00a01). The soft finger was composed of a series of channels and chambers surrounded by a soft elastomer. On pressurization, the finger deformed according to the internal stress distribution along the elastomer (42). A single pneumatic actuator was used for driving the finger. cPDMS, with a resistance that increases with strain, encased in a nonconductive elastomer, served as the soft strain sensor. The sensors were manually manufactured with varying lengths and implanted in the finger by randomly placing them roughly aligned with finger length during the curing process of the finger",
            " We used two OptiTrack cameras to provide ground truth validation for the position of the finger in the real world. We placed infrared tracking balls for the motion capture system at the base and tip of our pneumatic actuator with embedded sensors. All the coordinates of the fingertip were measured with respect to the base coordinates. We mounted the actuator onto the front of a metal stand (80/20, McMaster-Carr) such that two reflective markers affixed to the base and tip of the actuator were visible from the front (Fig.\u00a01). For the force sampling, we used a compression load cell (FX1901, TE Connectivity). Then, we connected the finger to a volumetric control system designed to apply commanded pressures to the internal chamber of the actuator (43). A randomly generated sequence of pressure inputs was sent serially to the control board, which regulated the pressure inside the finger using a low-level PD controller running at 1000\u00a0Hz. The pseudorandom sequence was in the form of square wave with a range varying from 0 to 3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure12.19-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure12.19-1.png",
        "caption": "Figure 12.19 (a) A structure of which parts AC, BC and DC are rigidly joined at C. (b) Bending moment diagram. (c) Shear force diagram. (d) Normal force diagram.",
        "texts": [
            " 12 Bending Moment, Shear Force and Normal Force Diagrams 499 Figure 12.18 The shape of the V diagram can also be found by plotting the successive step changes due to the point loads at the beam: (a) from left to right or (b) from right to left. Note that, between the two successive zero moments, in AS and SC, M is zero, and therefore the corresponding area of the V diagram is also zero (rule 12). It is left to the reader to check this. Example 3 The support reactions and the interaction forces at joint C for the structure in Figure 12.19a were calculated in Section 5.1, Example 5 (see Figure 12.20). Question: Determine the M , V and N diagrams. 500 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Solution: If we determine the interaction forces at joint E, it is possible to plot the M , V and N diagrams for the entire structure. As there are no distributed loads, the bending moment varies linearly along all members, and the shear force and normal force are constant in all members (rule 1). The M diagram is shown in Figure 12.19b. Since the variation of M along all members is linear, it is sufficient to determine the bending moments at the member ends to get the M diagram. The bending moments M (CA) C , M (CB) C and M (CE) C at joint C have already been calculated1 (see Figure 12.20). M (ED) E follows from the moment equilibrium of the isolated part ED (see Figure 12.21): M (CE) E = M (ED) E with tension on the upper side of ED. This value in the M diagram is therefore plotted at the upper side. The moment equilibrium of joint E in Figure 12.21 gives M (CE) E = M (ED) E . The bending moment \u201cgoes round the corner\u201d. This is emphasised in the M diagram in Figure 12.19b by means of a dotted arc at joint E. 1 The upper index refers to the member in which the bending moment acts and the lower index refers to the location. 12 Bending Moment, Shear Force and Normal Force Diagrams 501 The magnitude of the shear force and the associated deformation symbol follow from the slope of the M diagram.1 The V diagram is shown in Figure 12.19c. The N diagram is shown in Figure 12.19d. Example 4 We are given the trussed beam in Figure 12.22a. All the forces acting on the isolated hinged beam ASB shown in Figure 12.22b were calculated in Section 5.6. The N , V and M diagrams are shown in Figures 12.22c to 12.22e. Determining and drawing the V and M diagrams is done in the same way as for the hinged beam in Example 2. To draw the M diagram, we have to calculate only the bending moments at C and D. The bending moment varies linearly between D and E, so that the M diagram must pass through S where the bending moment is zero. This also fixes the value at E. The V diagram can subsequently be calculated from the M diagram. We can change the order: first draw the V diagram and calculate the values at C, E and D from the areas of the V diagram. Note: The shear forces at A and B are not equal to the support reactions at A and B! Why not? 1 In Figure 12.19b, the \u201csteps\u201d in the M diagrams are no longer shown. 502 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Example 5 The support reactions for the three-hinged portal frame in Figure 12.23a were calculated in Section 5.3, Example 1. As there are no distributed loads, the normal forces and shear forces in each field are constant, and the bending moment varies linearly (rule 1). N and V diagrams To draw the N and V diagrams, we have to investigate the force equilibrium of the separate parts. The necessary calculations are left to the reader"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure2.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure2.3-1.png",
        "caption": "FIGURE 2.3. Measuring the force under the wheels to find the height of the mass center.",
        "texts": [
            " Most cars are approximately symmetrical about the longitudinal center plane passing the middle of the wheels, and therefore, the lateral position of the mass center C is close to the center plane. However, the lateral position of C may be calculated by weighing one side of the car. Example 43 Height mass center determination. To determine the height of mass center C, we should measure the force under the front or rear wheels while the car is on an inclined surface. Experimentally, we use a device such as is shown in Figure 2.3. The car is parked on a level surface such that the front wheels are on a scale jack. The front wheels will be locked and anchored to the jack, while the rear wheels will be left free to turn. The jack lifts the front wheels and the required vertical force applied by the jacks is measured by a load cell. Assume that we have the longitudinal position of C and the jack is lifted such that the car makes an angle \u03c6 with the horizontal plane. The slope angle \u03c6 is measurable using level meters. Assuming the force under the front wheels is 2Fz1 , the height of the mass center can be calculated by 2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002448_j.matdes.2017.06.040-Figure13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002448_j.matdes.2017.06.040-Figure13-1.png",
        "caption": "Fig. 13. Schematic representations of the inclusion creation mechanism by spatter particles mixed into the metallic powder during SLM process. (a) Spatter particles A and B are mixed with the metallic powder. (b) The laser melts the spatter A which has a particle size close to the layer thickness, but cannot melt the spatter B because of its size being much larger than the layer thickness. (c) New powder layer is deposited on the building platform. (d) Laser beam creates the next layer and the spatter B remains as an un-melted region in the final product.",
        "texts": [
            " The spherical shape of the spatter is due to the fact that molten metal is solidifying before impinging on the powder bed. Fig. 12d also shows that the surface of the spatter contains dark patches (marked with red circle), suggesting a difference in chemical composition. The EDS analysis reveals that these areas are surface oxides. This is because Mg alloy elements, due to their high affinity to oxygen, react with the remaining oxygen in the building chamber during flight process. These findings are in agreement with other studies on the spatter compositions [15]. Fig. 13a shows the schematic illustration of the spatter particles mixed with the metallic powder which is exposed to the laser beam afterward. Depending on the size of spatter particles (see Fig. 9), themelting point of the material, and the amount of energy input of the beam, the laser beam may or may not melt the spatter particles during SLM process (see Fig. 13b). As shown in Fig. 13a, the size of spatter A is close to the powder layer thickness and it can be completely melted by the laser beam as shown in Fig. 13b. However, since the size of the spatter B is larger than the powder layer thickness, the laser cannot melt it (see Fig. 13.b) and as the next layer is formed (Fig. 13.c and d), the spatter particle B creates an inclusion in the part. A similar trend can be observed when the spatter particles fall into the layers which are just consolidated by the laser beam(s). Fig. 14a shows a representative spatter called \u201cSpatter C\u201d which is redeposited on the surface of the solidified layers. As the recoating devices drag powder across the build surface (Fig. 14b) and laser beam melts the next layer, the spatter C is not melted and stays on the build surface (Fig. 14c). Subsequently, a new powder layer is coated on the build surface andwhile the laser beamcreates the next layer, the spatter remains as a non-melted region into the part (Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003504_9.802914-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003504_9.802914-Figure1-1.png",
        "caption": "Fig. 1. Two inverted pendulums connected by a spring.",
        "texts": [
            " Because of the functional approximation properties of RBNN\u2019s, the functional form of the subsystem dynamics does not need to be known. Within this section, we will present illustrative examples for both the direct and indirect approaches. While the approach could be applied to intervehicle spacing regulation in a platoon of an automated highway system, since that system fits the assumptions of our framework [16], instead we study the control of two inverted pendulums connected by a spring as shown in Fig. 1. Each pendulum may be positioned by a torque input ui applied by a servomotor at its base. It is assumed that both i and _ i (angular position and rate) are available to the ith controller for i = 1; 2. The equations which describe the motion of the pendulums are defined by _x1; 1 =x1; 2 (45) _x1; 2 = m1gr J1 kr2 4J1 sin(x1; 1) + kr 2J1 (l b) + u1 J1 + kr2 4J1 sin(x2; 1) (46) _x2; 1 =x2; 2 (47) _x2; 2 = m2gr J2 kr2 4J2 sin(x2; 1) kr 2J2 (l b) + u2 J2 + kr2 4J2 sin(x1; 2) (48) where 1 = x1; 1 and 2 = x2; 1 are the angular displacements of the pendulums from vertical"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure3.27-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure3.27-1.png",
        "caption": "Fig. 3.27: Natural reference points Pi i = 0,1, ... , m (m > N+1) fixed on manipulator links, and a spherical model for an obstacle.",
        "texts": [
            " The most used aspects can be recognized in: - mechanical energy in joint actuation; - mechanical energy of performing a task; - velocity of performing a path; - acceleration of performing a path; - obstacle avoidance; - time duration of the path. The above-mentioned aspects can be used separately but also in conjunction through suitable weighted formulation. However, a common frame can be considered for the formulation of an optimization problem whose solution gives an optimal path in terms of the used index of merit. In the following an optimal path planning is presented as related to basic kinematic formulations that also include a solution for the inverse kinematic problem. In Fig. 3.27 a robotic manipulator is sketched as an open kinematic chain with N+1 rigid links connecting consecutive revolute or prismatic joints. The manipulator chain can be modeled by using the so-called \u2018natural coordinates\u2019 as a natural way to describe mechanisms through some reference points in Cartesian coordinates. These points can be selected according to a few basic rules, i.e. a reference point in each revolute joint and two reference points for each prismatic joint to give the stroke length and the actual position of the joint",
            " Fundamentals of Mechanics of Robotic Manipulation 131 Thus, a path planning problem is concerned with determining a suitable sequence of possible Cj configurations. However, the Cartesian coordinates of the joint reference points must comply with congruence constraints due to the rigidity of links and kinematic characteristics of joints. Particularly, a rigid body condition for a link i can be expressed as P P - P P = l0 i-1 0 i i (3.2.28) where | - | is the module operator for vectors and l i is the link length. A correct model for a joint on link li needs, Fig. 3.27, that point Pi+1 be on link li and it is expressed as P P x P P = 0i-1 i+1 i-1 i (3.2.29) and the orientation of link li+1 be maintained as formulated in the form P P P P = l l cosi-1 i i+1 i+2 i i+1 i\u22c5 \u03c8 (3.2.30) where \u03c8 i is a constant structural angle between links li and li+1 (generally it is assumed equal to \u03c0/2). Moreover, a prismatic joint may slide along the link segment li between a Chapter 3: Fundamentals of the Mechanics of Robots132 minimum value li m and a maximum value li M so that the rigid body condition (3.2.8) for a link l i with a prismatic joint must be replaced by M ii01-i0 m i lPP-PPl \u2264\u2264 (3.2.31) An obstacle within the manipulator workspace can be modeled as a sphere with radius r j and center Qj, as shown in Fig. 3.27, that can be determined as a container of the obstacle itself. A free-collision configuration may be evaluated by looking at the distances among the obstacles and the manipulator links. These distances may be computed from the geometry of Fig. 3.27 as sij (i = 1,...,m and j = 1,...,v) given as \u03a6<\u03a6 \u03a6\u2265\u03a6\u03d5 minj1-iji minji ij )Q P,Q(Pmin sinQP =s (3.2.32) with \u03d5 = cos P Q - P Q + l 2 P Q l -1 i j 2 i-1 j 2 i 2 i j i \u03a6min -1 i i j i-1 j = sin l min( P Q , P Q ) where \u03a6 is the angle at Qj under which link li is seen. In order to generate a path to a given end configuration Cf it can be convenient to calculate a suitable Map of Feasible Configurations in which the problem of obstacle avoidance path will then be solved. The feasible configurations for the Map can be computed by assuming that the movements of end-effector point H be parallel to the axes of a Cartesian base frame and the configurations during the motion be adjacent to each other",
            "33) by assuming a new Cr from the computed configurations with the aim of filling up the workspace area. Then, a suitable numbering can also be arranged so that a path will be synthetically described by a sequence of numbered configurations joining a start one to a final one. If a configuration will correspond to a point out of the workspace it will not be considered. The work area within the workspace is assumed to be given through a left bottom corner point Hw and a parallelepiped region with size b, h, and w, as shown in Fig. 3.27. A free-collision path configuration may be formulated by means of a refinement of Eqs (3.2.33) and (3.2.34) for adjacent configurations by considering the obstacle avoidance through Eqs (3.2.32) and the link constraints (3.2.28) to (3.2.31). Summarizing the problem of path planning for manipulators can be formulated as a sequence of manipulator free-collision configurations, with the s one obtained from a previous r one by solving the problem min P P - P P 0 i r 0 i s i=1 m .2 (3.2.35) subject to P P - P P = l0 i-1 0 i i (3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure15.17-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure15.17-1.png",
        "caption": "Figure 15.17 (a) A slide or a so-called shear force hinge with (b) the interaction forces after a kinematically admissible virtual displacement.",
        "texts": [
            " The virtual work equation now includes only the work performed by the external forces: \u03b4A = \u03b4A(1) e + \u03b4A(2) e = 0. Conclusion: Due to a kinematically admissible virtual displacement of a mechanism in equilibrium, the virtual work performed by the (external) load equals zero. The fact that the work performed by the load is zero is a necessary and sufficient condition for the equilibrium of a mechanism. The approach can easily be expanded to include mechanisms of more than two bodies, as well as other than hinged joints. Figure 15.17 shows a slide or a so-called shear force hinge with the interaction forces M; N , and a kinematically admissible virtual displacement \u03b4u; \u03b4\u03d5. We can see immediately that the pair of bending moments M does not perform any work. How different is it for the normal force pair N , that as a couple N\u03b4u undergoes a (small) rotation \u03b4\u03d5 and therefore, taking into account the directions shown in the figure, performs the work \u2212N\u03b4u\u03b4\u03d5. The geometric significance of \u03b4u\u03b4\u03d5 can be seen on the figure. When deriving the virtual work equation for a rigid body, the demand arose that the geometric relationships have to be linear in the virtual displacements"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure5.17-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure5.17-1.png",
        "caption": "FIGURE 5.17. The motion of a vehicle at 30 deg latitude and heading north on the Earth.",
        "texts": [
            "365) Its velocity is therefore given by B Gv = B r\u0307+ B G\u03c9B \u00d7 B Gr = 0 + \u03c6\u0307k\u0302 \u00d7 l\u0131\u0302 = l \u03c6\u0307j\u0302 (5.366) Gv = GRB Bv = \u23a1\u23a3 l \u03c6\u0307 cos\u03c6 l \u03c6\u0307 sin\u03c6 0 \u23a4\u23a6 . (5.367) The acceleration of the bob is then equal to B Ga = B Gv\u0307 + B G\u03c9B \u00d7 B Gv = l \u03c6\u0308j\u0302+ \u03c6\u0307k\u0302 \u00d7 l \u03c6\u0307j\u0302 = l \u03c6\u0308j\u0302\u2212 l \u03c6\u0307 2 \u0131\u0302 (5.368) Ga = GRB Ba = \u23a1\u23a2\u23a3 l \u03c6\u0308 cos\u03c6\u2212 l \u03c6\u0307 2 sin\u03c6 l \u03c6\u0308 sin\u03c6+ l \u03c6\u0307 2 cos\u03c6 0 \u23a4\u23a5\u23a6 . (5.369) Example 196 Motion of a vehicle on the Earth. Consider the motion of a vehicle on the Earth at latitude 30 deg and heading north, as shown in Figure 5.17. The vehicle has the velocity v = B E r\u0307 = 80 km/h = 22.22m/ s and acceleration a = B E r\u0308 = 0.1m/ s 2, both with respect to the road. The radius of the Earth is R, and hence, the vehicle\u2019s kinematics are B Er = Rk\u0302 m (5.370) B E r\u0307 = 22.22\u0131\u0302 m/ s (5.371) B E r\u0308 = 0.1\u0131\u0302 m/ s2 (5.372) \u03b8\u0307 = v R rad/ s (5.373) \u03b8\u0308 = a R rad/ s2. (5.374) There are three coordinate frames involved. A body coordinate frame B is attached to the vehicle as shown in the figure. A global coordinate G is set up at the center of the Earth"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure9.4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure9.4-1.png",
        "caption": "Fig. 9.4. Stator and rotor of a medium power AFPM generator with inner coreless stator and twin external rotor: (a) stator coils, (b) rotor disc with NdFeB PMs.",
        "texts": [
            "1 shows specifications of five-phase AFPM synchronous generators manufactured by Kestrel Wind Turbines (Pty) Ltd , Gauteng, South Africa. All three types of Kestrel generators use laminated cores. The KestreL 2000 is a double-sided AFPM generator with twin rotors. The KestreL 600 generator rated at 400 W is in fact capable of delivering power in excess of 600 W. Controlled by a shunt regulator the maximum charging current is 50 A into a 12 V d.c. battery bank. At wind speeds of 52.2 km/h, which translates into 14.5 m/s (28.2 knots) and rotational speed of 1100 rpm, the power output is 600 W (Fig. 9.3). Figure 9.4 shows a medium-power AFPM generator with twin external rotor and inner coreless stator for a 5-m blade diameter wind turbine. The stator system consists of 15 non-overlapping concentrated coils, one coil per phase. The thickness of coil is 12 mm and diameter of wire 0.6 mm. Each coil is individually rectified to d.c., to reduce the commutation torque ripple effect and provide better control over the voltage. The rotor has 2p = 16 poles. Performance characteristics are shown in Fig. 9.5. Figure 9"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003908_978-1-84882-730-1-Figure8.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003908_978-1-84882-730-1-Figure8.6-1.png",
        "caption": "Figure 8.6 Interpretation of path-tracking errors",
        "texts": [
            " This filter and the derivative of the path parameter used as an additional control allow a global controller and desired path with arbitrary curvature. 5. Add projection integrators to the path-tracking error dynamics\u2019 output to compensate for constant bias of the disturbances. 186 8 Path-tracking Control of Underactuated Ships 8.2.2 Coordinate Transformations 8.2.2.1 Choosing the Body-fixed Frame Origin To avoid the yaw moment control r acting directly on the sway dynamics, we choose the body-fixed frame origin such that it is on the center line of the ship (see Figure 8.6), i.e., yg D 0 and that: xg D Y Pr m : (8.70) With this choice, the ship model (8.64) can be written as: P D J . /v; NM Pv D NC .v/v . ND C NDn.v//v C ; (8.71) where NM D 2 4 m11 0 0 0 m22 0 0 0 m33 3 5 ; NDn.v/D 2 4 dn1 juj 0 0 0 dn2 jvj 0 0 0 dn3 jr j 3 5 ; ND D 2 4 d11 0 0 0 d22 d23 0 d32 d33 3 5 ; NC .v/D 2 64 0 0 m22v 0 0 m11u m22v m11u 0 3 75 ; (8.72) with d11 D Xu; d22 D Yv; d23 D Yr ; d32 D Nv; d33 D Nr ; dn1 D Xjuju; dn2 D Yjvjv; dn3 DNjr jr : It is noted that the ship position .x;y/ is the coordinates of the body-fixed frame origin (not the ship\u2019s center of gravity) with respect to the earth-fixed frame",
            "71) and (8.82) as: Px D . OuC Qu/cos. / .vC Qv/sin. /; Py D . OuC Qu/sin. /C .vC Qv/cos. /; P D OrC Qr; POuD m22 m11 Ov Or d11 m11 Ou dn1 m11 j Ouj OuC 1 m11 uC m22 m11 Ov Qr; (8.86) POv D m11 m22 Ou Or d22 m22 Ov d23 m22 Or dn2 m22 j Ovj Ov m11 m22 Ou Qr; POr D m11 m22 m33 Ou Ov d22 m33 Ov d33 m33 Or dn3 m33 j Or j OrC 1 m33 r C m11 m33 Ou Qv m22 m33 Ov Qu: Transformation of Path-tracking Errors We now interpret the path-tracking errors in a frame attached to the reference path \u02dd as follows (see Figure 8.6): 2 64 xe ye e 3 75D J T . / 2 64 x xd y yd d 3 75 ; (8.87) where d is the angle between the path and the X-axis defined by d D arctan y 0 d .s/ x 0 d .s/ ! ; (8.88) with x 0 d .s/ and y 0 d .s/ being defined in (8.69). In Figure 8.6, OEXEYE is the earth-fixed frame; OpXpYp is a frame attached to the path \u02dd such that OpXp and OpYp are parallel to the surge and sway axes of the ship, respectively, ud is tangential to the path, CG is the center of gravity of the ship; and ObXbYb is the body-fixed frame. Therefore xe , ye , and e can be referred to as tangential, cross and heading errors, respectively. Differentiating both sides of (8.87) along the solutions of the first three equations of (8.86) results in the kinematic path-tracking errors: 190 8 Path-tracking Control of Underactuated Ships Pxe D Ou ud cos"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure5.29-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure5.29-1.png",
        "caption": "Figure 5.29 (a) The assumed directions of the support reactions; (b) the support reactions as they really act.",
        "texts": [
            " Vertical loading now generates exclusively vertical support reactions. Whether rod AB is subject to tension or compression depends on the loading. The name tie-rod indicates that such a structure is used only if tension can be expected in the rod. Example In Figure 5.28, a vertical and a horizontal load is acting on a three-hinged frame with tie-rod. Questions: a. Determine the support reactions. b. Determine the force in rod AB. c. Determine the interaction forces at S. d. Determine the forces acting on joint A. Solution (units in kN and m): a. In Figure 5.29a, the structure has been isolated from its supports. The support reactions follow from the equilibrium of the structure as a whole. For the directions assumed for Ah, Av and Bv we find 174 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM \u2211 F (ASB) x = 40 \u2212 Ah = 0 \u21d2 Ah = 40 kN, \u2211 F (ASB) y = \u221260 \u2212 60 + Av + Bv = \u221260 \u2212 60 + Av + 70 = 0 \u21d2 Av = 50 kN, \u2211 T (ASB) z |A = \u221240 \u00d7 2 \u2212 60 \u00d7 2 \u221260 \u00d7 6 + Bv \u00d7 8 = 0 \u21d2 Bv = 70 kN. The support reactions are shown in Figure 5.29b. b. To calculate the force in rod AB, it is isolated from ASB in Figure 5.30. We can immediately recognise a two-force member in rod AB: the rod is loaded only by forces at its ends A and B and can therefore be in equilibrium only if these forces are equal and opposite with AB as common line of action. It is assumed that a tensile force N acts in rod AB. The magnitude of N follows from the moment equilibrium about S of one of the parts AS or BS. In Figure 5.31a both parts have been isolated at S"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000957_robot.2009.5152561-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000957_robot.2009.5152561-Figure2-1.png",
        "caption": "Fig. 2. Effect of blade flapping in forward flight: the deflection of the rotor plane due to flapping causes an effective deflection of the thrust vector, generating moments about the center of gravity.",
        "texts": [
            " Both effects will be discussed here in sufficient detail as to understand their impact on the vehicle\u2019s flight characteristics. A rotor in translational flight undergoes an effect known as blade flapping. The advancing blade of the rotor has a higher velocity relative to the free-stream, while the retreating blade sees a lower effective airspeed. This causes an imbalance in lift, inducing an up and down oscillation of the rotor blades [25]. In steady state, this causes the effective rotor plane to tilt at some angle off of vertical, causing a deflection of the thrust vector (see Figure 2). If the rotor plane is not aligned with the vehicle\u2019s center of gravity, this will create a moment about the center of gravity (c.g.) that can degrade attitude controller performance [9]. For stiff rotors without hinges at the hub, there is also a moment generated directly at the rotor hub from the deflection of the blades. The full analysis of blade flapping is beyond the scope of this paper, but is presented in more detail in the helicopter literature and in previous work [25], [26], [21]. Due to the quadrotor\u2019s bilateral symmetries, moments generated by lateral deflections of the rotor plane cancel"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure13.78-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure13.78-1.png",
        "caption": "Figure 13.78 The frame consists of the two singly-cohesive sub-structures ABCS and DS.",
        "texts": [
            " 606 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM The directions follow from the deformation symbols given in the figure. Questions: a. Determine the degree of static indeterminacy of the structure. b. Draw the force polygon for the force equilibrium of joint C. c. Determine the support reactions at A and D. d. Determine the M , V and N diagrams for the entire structure, with the deformation symbols. Solution (units kN and m): a. The two parts ABCS and DS provide e = 2 \u00d7 3 = 6 equilibrium equations (see Figure 13.78). The number of unknown support reactions at A and D is r = 3 + 3 = 6. The number of unknown joining forces at S is v = 2. The degree of static indeterminacy is: n = r + v \u2212 e = 6 + 2 \u2212 6 = 2. The structure is therefore statically indeterminate to the second degree. b. In Figure 13.79a, joint C has been isolated and all forces acting on it are shown. The bending moments acting on the joint are not shown! Figure 13.79b shows the closed force polygon for the force equilibrium of the joint (scale: 1 square = 2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure8.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure8.3-1.png",
        "caption": "Fig. 8.3. Velocity triangles for a PM channel.",
        "texts": [
            " The Ideal Radial Channel According to the theory of an ideal impeller, a number of assumptions have to be made to establish the one-dimensional model of the ideal radial channel [78, 232]: (a) there are no tangential components in the flow through the channel; (b) the velocity variation across the width or depth of the channel is zero; (c) the inlet flow is radial, which means that air enters the impeller without pre-whirl; 258 8 Cooling and Heat Transfer (d) the pressure across the blades can be replaced by tangential forces acting on the fluid; (e) the flow is treated as incompressible and frictionless. Figure 8.3 shows a radial channel with the velocity triangles drawn at the inlet and the outlet. It can be observed that the pressures at the inlet p1 and the outlet p2, and friction Ffr make no contribution to the sum of the momentum, \u2211 M0. If gravity is ignored, the general representation of conservation of momentum takes the following form [266]:\u2211 M0 = \u2202 \u2202t [ \u222b cv (r\u00d7 v)\u03c1dV ] + \u222b cs (r\u00d7 v)\u03c1(v \u00b7 n)dA (8.16) where r is the position vector from 0 to the elemental control volume dV and v is the velocity of the element",
            "19) Based on the principle of conservation of energy, the input shaft power may be given as: Pin = m\u0307( p2 \u2212 p1 \u03c1 + w2 2 \u2212 w2 1 2 + z2 \u2212 z1 + U2 \u2212 U1) (8.20) If the potential (z2\u2212 z1) and internal energy (U2\u2212U1) (friction) are ignored, eqn (8.20) may be written in the same units as eqn (8.19) as: Pin Q = (p2 \u2212 p1) + \u03c1 w2 2 \u2212 w2 1 2 (8.21) If equations (8.19) and (8.21) are equated and noting that w1 = Q/A1 and w2 = Q/A2, where A1 and A2 are the cross-section areas of the inlet and outlet of the channel respectively, the pressure difference \u2206p between the entrance and exit of the radial channel (shown in Fig. 8.3) may be expressed as: \u2206p = p2 \u2212 p1 = \u03c1\u21262(R2 out \u2212R2 in)\u2212 \u03c1 2 ( 1 A2 2 \u2212 1 A2 1 )Q2 (8.22) Eqn (8.22) may be termed the ideal equation describing the air flow through the radial channel. 260 8 Cooling and Heat Transfer The Actual Radial Channel The actual characteristics of a hydraulic machine differ from the ideal case owing to two reasons: (i) the uneven spatial distribution of velocities in the blade passages, and (ii) the leakage and recirculation of flow and hydraulic losses such as friction and shock losses"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001031_j.jsv.2008.03.038-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001031_j.jsv.2008.03.038-Figure3-1.png",
        "caption": "Fig. 3. The cracked tooth model for case 1.",
        "texts": [],
        "surrounding_texts": [
            "The gear mesh stiffness model described in this study was based on the work by Yang and Lin [3] in 1987. They used the potential energy method to analytically model the effective mesh stiffness. The total potential energy stored in the meshing gear system was assumed to include three components: Hertzian energy, bending energy, and axial compressive energy. This model was refined by Tian [4] in 2004 in which shear energy was taken into account as well. Thus, for the single-tooth-pair meshing duration, the total effective mesh stiffness can be expressed as [4] kt \u00bc 1 1=kh \u00fe 1=kb1 \u00fe 1=ks1 \u00fe 1=ka1 \u00fe 1=kb2 \u00fe 1=ks2 \u00fe 1=ka2 , (1) where kh, kb, ks, and ka represent the Hertzian, bending, shear, and axial compressive mesh stiffness, respectively. For the double-tooth-pair meshing duration, the total effective mesh stiffness is the sum of the two pairs\u2019 stiffnesses, which is shown as [4] kt \u00bc X2 i\u00bc1 1 1=kh;i \u00fe 1=kb1;i \u00fe 1=ks1;i \u00fe 1=ka1;i \u00fe 1=kb2;i \u00fe 1=ks2;i \u00fe 1=ka2;i , (2) where i \u00bc 1 represents the first pair of meshing teeth and i \u00bc 2 represents the second. The derivations of these two equations are given in Tian [4]. Calculation of each of the components in these two equations when there are no cracks in any gear is given in Refs. [3,4,15], where Ref. [15] is a shorter version of Ref. [4]. The expressions of these components when cracks are introduced will be provided later in this paper. For typical gear parameters given in Table 1, we wrote simple Matlab programs and obtained numerical values of the total effective mesh stiffness as a function of the gear rotation angle. This total effective mesh stiffness within one shaft period of the gear is plotted in Fig. 1. Fig. 1 represents the total meshing stiffness of the pair of gears when the gear teeth are perfect (that is, have no cracks). On crack development in a gear, Refs. [16\u201319] consider that a crack is developing at the root of a single tooth of the pinion. A tooth root crack typically starts at the point of the largest stress in the material. In Ref. [20], a computational model which applies the principles of linear elastic fracture mechanics is used to simulate gear tooth root crack propagation. Based on the computational results, the crack propagation path shows a slight curve extending from the tooth root as shown in the left side of Fig. 2 [20]. Lewicki [21] also indicates that crack propagation paths are smooth, continuous, and in most cases, rather straight with only a ARTICLE IN PRESS S. Wu et al. / Journal of Sound and Vibration 317 (2008) 608\u2013624610 slight curvature. He has also studied the effects of rim and web thickness on crack propagation path and showed that different paths exist. In this paper, based on the results shown in Ref. [20], we further simplify the crack model. We will consider the crack path to be a straight line as shown in the right side of Fig. 2. The crack starts at the root of the pinion and then proceeds as shown in Fig. 2. Further referring to Figs. 3\u20136, the intersection angle, u, between the crack and the central line of the tooth is set at a constant 45 . The crack length, q1, grows from zero with an increment size of Dq1 \u00bc 0:1 mm until the crack reaches the tooth\u2019s central line. At that point, q1, reaches its maximum value of 3.9mm. After that, the crack then changes direction to q2 (see Fig. 5), which is assumed to be exactly symmetric around the tooth\u2019s central line. Theoretically, the maximum length of q2 should be the ARTICLE IN PRESS S. Wu et al. / Journal of Sound and Vibration 317 (2008) 608\u2013624 611 same as q1, however, the tooth is expected to suffer sudden breakage before the crack runs through the whole tooth. Thus the maximum length of q2 is assumed to be 60% of q1max and the increment size, Dq2, is also 0.1mm. In later reference of the crack growth, we will use a relative length. The highest crack level will be 3:9\u00fe 60% 3:9 \u00bc 6:24mm, or 80% (\u00bc 6:24=7:8) of the theoretical total length of the full through crack. ARTICLE IN PRESS S. Wu et al. / Journal of Sound and Vibration 317 (2008) 608\u2013624612 With the crack introduced as described above, we need to calculate all components of the total mesh stiffness, that is, Hertzian stiffness, axial compressive stiffness, bending stiffness, and shear stiffness. Based on the work documented in Ref. [15], the Hertzian and axial compressive stiffnesses remain the same when a crack is introduced. However, the bending and shear stiffnesses will change due to the appearance of the crack, and their derivation are provided under each of the following four cases. Case 1 Tian [4]: When hc1Xhr & a14ag, where a1 \u00bc 90 \u00f0pressure angle\u00de. ARTICLE IN PRESS S. Wu et al. / Journal of Sound and Vibration 317 (2008) 608\u2013624 613 The potential energy stored in a meshing gear tooth can be calculated by Ub \u00bc Z d 0 M2 2EIx dx \u00bc Z d 0 \u00bdF b\u00f0d x\u00de Fah 2 2EIx dx, (3) Us \u00bc Z d 0 1:2F 2 b 2GAx dx \u00bc Z d 0 \u00bd1:2F cos a1 2 2GAx dx, (4) where Ix and Ax represent the area moment of inertia and area of the section where the distance from the tooth\u2019s root is x, and G represents the shear modulus. They can be obtained by Ix \u00bc 1 12 \u00f0hc1 \u00fe hx\u00de 3L if xpgc; 1 12 \u00f02hx\u00de 3L if x4gc; 8>< >: (5) Ax \u00bc \u00f0hc1 \u00fe hx\u00deL if xpgc; 2hxL if x4gc; ( (6) G \u00bc E 2\u00f01\u00fe n\u00de , (7) where hx represents the distance between the point on the tooth\u2019s curve and the tooth\u2019s central line where the horizontal distance from the tooth\u2019s root is x. The bending mesh stiffness of the cracked tooth is 1 kbcrack \u00bc Z a2 ag 12f1\u00fe cos a1\u00bd\u00f0a2 a\u00de sin a cos a g2\u00f0a2 a\u00de cos a EL\u00bdsin a2 \u00f0q1=Rb1\u00de sin u\u00fe sin a\u00fe \u00f0a2 a\u00de cos a 3 da \u00fe Z ag a1 3f1\u00fe cos a1\u00bd\u00f0a2 a\u00de sin a cos a g2\u00f0a2 a\u00de cos a 2EL\u00bdsin a\u00fe \u00f0a2 a\u00de cos a 3 da (8) and the shear mesh stiffness of the cracked tooth is 1 kscrack \u00bc Z a2 ag 2:4\u00f01\u00fe n\u00de\u00f0a2 a\u00de cos a\u00f0cos a1\u00de 2 EL\u00bdsin a2 \u00f0q1=Rb1\u00de sin u\u00fe sin a\u00fe \u00f0a2 a\u00de cos a da \u00fe Z ag a1 1:2\u00f01\u00fe n\u00de\u00f0a2 a\u00de cos a\u00f0cos a\u00de2 EL\u00bdsin a2 \u00f0q1=Rb1\u00de sin u\u00fe sin a\u00fe \u00f0a2 a\u00de cos a da: (9) Case 2 Tian [4]: When hc1ohr or when hc1Xhr & a1pag. The bending mesh stiffness of the cracked tooth is 1 kbcrack \u00bc Z a2 a1 12f1\u00fe cos a1\u00bd\u00f0a2 a\u00de sin a cos a g2\u00f0a2 a\u00de cos a EL\u00bdsin a2 \u00f0q1=Rb1\u00de sin u\u00fe sin a\u00fe \u00f0a2 a\u00de cos a 3 da (10) and the shear mesh stiffness of the cracked tooth is 1 kscrack \u00bc Z a2 a1 2:4\u00f01\u00fe n\u00de\u00f0a2 a\u00de cos a\u00f0cos a1\u00de 2 EL\u00bdsin a2 \u00f0q1=Rb1\u00de sin u\u00fe sin a\u00fe \u00f0a2 a\u00de cos a da. (11) Case 3: When hc1ohr or when hc1Xhr & a1pag. This case was not covered in Ref. [4]. The bending mesh stiffness of the cracked tooth is 1 kbcrack \u00bc Z a2 a1 12f1\u00fe cos a1\u00bd\u00f0a2 a\u00de sin a cos a g2\u00f0a2 a\u00de cos a EL\u00bdsin a \u00f0q2=Rb1\u00de sin u\u00fe \u00f0a2 a\u00de cos a 3 da (12) ARTICLE IN PRESS S. Wu et al. / Journal of Sound and Vibration 317 (2008) 608\u2013624614 and the shear mesh stiffness of the cracked tooth is 1 kscrack \u00bc Z a2 a1 2:4\u00f01\u00fe n\u00de\u00f0a2 a\u00de cos a\u00f0cos a1\u00de 2 EL\u00bdsin a \u00f0q2=Rb1\u00de sin u\u00fe \u00f0a2 a\u00de cos a da. (13) Case 4: When hc2Xhr & a14ag. This case was not covered in Ref. [4] either. We have found the bending mesh stiffness of the cracked tooth to be 1 kbcrack \u00bc Z a2 ag 12f1\u00fe cos a1\u00bd\u00f0a2 a\u00de sin a cos a g2\u00f0a2 a\u00de cos a EL\u00bdsin a \u00f0q2=Rb1\u00de sin u\u00fe \u00f0a2 a\u00de cos a 3 da (14) ARTICLE IN PRESS S. Wu et al. / Journal of Sound and Vibration 317 (2008) 608\u2013624 615 and the shear mesh stiffness of the cracked tooth is 1 kscrack \u00bc Z a2 ag 2:4\u00f01\u00fe n\u00de\u00f0a2 a\u00de cos a\u00f0cos a1\u00de 2 EL\u00bdsin a \u00f0q2=Rb1\u00de sin u\u00fe \u00f0a2 a\u00de cos a da. (15) With the expressions of the components of the total mesh stiffness provided above, we are able to find the total mesh stiffness value given each shaft rotation angle and each crack size. For a pair of standard steel involute spur teeth whose main parameters are given in Table 1, take four specific crack sizes given in Table 2 as an example. These selected crack sizes cover all the four cases classified in the above derivations. The total mesh stiffness under each of the four crack sizes has been calculated as a function of the shaft rotation angle and plotted in Fig. 7. From Fig. 7, it can be observed that as the size of the crack grows, the total mesh stiffness when the cracked tooth is in meshing becomes much lower. This is important information for fault detection and assessment."
        ]
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure8.15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure8.15-1.png",
        "caption": "Figure 8.15 (a) The grain stresses on a triangular soil element that is sliding (moving downwards). (b) The equations for the force equilibrium in the plane of the drawing can be derived easily from the fact that the force polygon is closed.",
        "texts": [
            " The coefficients mentioned here for active, passive and neutral earth pressure relate to the normal stresses in dry soil, or in other words, the normal stresses in the grain skeleton (grain pressures). The earth pressure in soil saturated with water is found by superposing the water pressure on the normal stress in the grain skeleton. The following sections address the magnitude of the coefficients for active, passive, and neutral earth pressure. The coefficient Ka for active earth pressure is derived using the triangular slice of soil in Figure 8.15a. If the area of side PQ is equal to A, then the area of side OP is equal to A cos \u03b1 and that of side OQ is equal to A sin \u03b1. The forces on the three sides of the soil element are therefore OP: \u03c3g;h( A cos \u03b1), OQ: \u03c3g;v( A sin \u03b1), PQ: \u03c3g A and \u03c4g;max A. The equations for the force equilibrium in the plane of the drawing are easy to derive from the closed force polygon in Figure 8.15b. Force equilibrium normal to PQ: \u03c3g A \u2212 \u03c3g;h( A cos\u03b1) cos \u03b1 \u2212 \u03c3g;v( A sin \u03b1) sin \u03b1 = 0. 296 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Force equilibrium parallel to PQ: \u03c4g;max A + \u03c3g;h( A cos\u03b1) sin \u03b1 \u2212 \u03c3g;v( A sin \u03b1) cos \u03b1 = 0. From the two equations above, one can find the stresses in the slide plane PQ: \u03c3g = +\u03c3g;h cos2 \u03b1 + \u03c3g;v sin2 \u03b1, \u03c4g;max = \u2212\u03c3g;h sin \u03b1 cos \u03b1 + \u03c3g;v sin \u03b1 cos \u03b1. Since the shear stress has its maximum, also \u03c4g;max = \u03c3g tan \u03d5. In this expression, substitute those found before for \u03c3g and \u03c4g;max: \u2212\u03c3g;h sin \u03b1 cos \u03b1 + \u03c3g;v sin \u03b1 cos \u03b1 = (+\u03c3g;h cos2 \u03b1 + \u03c3g;v sin2 \u03b1) tan \u03d5"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001514_j.engfailanal.2015.06.004-Figure27-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001514_j.engfailanal.2015.06.004-Figure27-1.png",
        "caption": "Fig. 27. (a) 16 DOF model [46,50], (b) 16 DOF model [59].",
        "texts": [
            " [70] presented a two stage gearbox model with 12 DOFs (see Fig. 26a). Mohammed et al. [71] developed a 12 DOF gear dynamic model including the gyroscopic and friction effects (see Fig. 26b). In order to investigate dynamic characteristics of the system when the crack in the pinion grows, Zhou et al. [46] adopted a 16 DOF mathematic model, which is developed by Howard et al. [50] for a one-stage gear system. In the model, the gear engagement is described by a TVMS function k(t) and a viscous damping c(t) (see Fig. 27a). Using a lumped parameter technique, Endo et al. [59] established a 16 DOF LMM (Fig. 27b) of a gear test rig to reproduce fault signals. Jia and Howard [51] presented a 26 DOF mathematic model including two pairs of meshing gears (see Fig. 28a) in which the vertical direction (xi) is parallel to the line of action in order to facilitate modeling. Liang et al. [34], Chen and Shao [39,40], Chen et al. [41] as well as Shao and Chen [45] presented a mathematic model of a one stage planetary-gear set with 21 DOFs (see Fig. 28b) to study the influences of the tooth root crack on the system vibration"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003981_rob.4620080405-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003981_rob.4620080405-Figure2-1.png",
        "caption": "Figure 2. HITACHI-HPR Robot.",
        "texts": [
            " MXR2 = MX2 + D3 (M3 + M4 + M5 + M6) XXR3 = XX3 - YY3 + YY4 + 2 R4 MZ4 + (R42 - D4\u2019) (M4 + M5 + M6) XYR3 = XY3 - D4 MZ4 - D4 R4 (M4 + MS + M6) ZZR3 = ZZ3 + YY4 + 2 R4 MZ4 + (D4\u2019 + R4\u2019) (M4 + M5 + Ms) MXR3 1 MX3 + D4 (M4 + MS + M6) MYR3 = MY3 + MZ4 + R4 (M4 + MS + M6) X X k = xx4 - YY4 + YYS Z Z k = zz4 + YY5 MY& = MY4 - MZS XXR5 XXs - YYs + YY6 ZZRS = ZZS + YY6 MYR5 = MY5 + MZ6 xx&j = xx6 - YY6 504 Journal of Robotic Systems-1 991 This method has been applied to the five degrees of freedom HITACHI-HPR robot (Fig. 2). Details for the description of this robot can be found in ref. 29. We have: q, = [el e3 e4 06 e 7 i ~ 1 e3-e4 3r-e4 -lI[] 1 The results completely agree with those obtained using the symbolic approach given in ref. 29. a s CONCLUSIONS s Two numerical methods are given to determine the set of base parameters of the dynamic model of robots. Starting from the standard inertial parameters, Gautier: Numerical Calculation of Base Inertial Parameters 505 each method gives the number, the choice and the numerical values of the base parameters, and are based on QR or SVD decompositions"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001185_j.ymssp.2017.08.002-Figure16-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001185_j.ymssp.2017.08.002-Figure16-1.png",
        "caption": "Fig. 16",
        "texts": [],
        "surrounding_texts": [
            "Totally different from the traditional methods, the proposed method focuses on the bearing fault diagnosis without any manual feature extraction. For comparison, several other methods are also used for analyzing the same bearing dataset, including the standard DBN [37,49], standard CNN [38,50], standard deep auto-encoder (DAE) [31,51], BP neural network and SVM. Two important points need to be emphasized: (1) The inputs of all the deep learning methods are the compressed data. (2) The inputs of BP and SVM have five types. The first is the compressed data, and the second is 22 statistical parameters manually extracted from the raw signal. More details about the 22 statistical parameters can be seen in paper [29]. Another three kinds of inputs are all feature parameters extracted by three advanced signal processing techniques, which are called EMD features, WPD features and LMD features for short, respectively. Specifically, calculate the 22 feature parameters mentioned above from the most sensitive frequency-band signals (contain highest energy distribution) decomposed by EMD, WPD and LMD, respectively. In this case study, Experiment 1 and Experiment 2 are carried out to prove the effectiveness of the proposed method. Experiment 1 is mainly used to investigate the diagnosis performance of different methods using the compressed data (raw vibration data), and the proposed method is compared with some commonly used fault feature extraction methods (EMD, WPD, LMD) in Experiment 2. The detailed descriptions of Experiment 1 and Experiment 2 are as follows Experiment 1: The proposed method is used for comparison with the standard DBN, standard CNN, standard DAE, BP with compressed data, BP with 22 features, SVM with compressed data and SVM with 22 features, respectively. Experiment 2: The proposed method is compared with BP with EMD features, BP with WPD features, BP with LMD features, SVM with EMD features, SVM with WPD features and SVM with LMD features, respectively. Ten trials are carried out for diagnosing each dataset to show the stability of different methods. Table 4 lists the average testing accuracy and standard deviation of all methods in Experiment 1, and Fig. 7 shows their corresponding diagnosis results in each trial. It can be observed from Table 4 that the average testing accuracy of the proposed method is 94.80% (1422/1500) using the compressed data, which is slightly higher than the standard DBN (1294/1500), CNN (1348/1500) and DAE (1327/1500). Compared with BP (774/1500) and SVM (869/1500) using the compressed data, the proposed method shows obvious advantages. After extracting the 22 features, although the accuracies of BP and SVM reach 83.07% (1246/1500) and 84.53% (1268/1500), respectively, their diagnosis performance still cannot be compared with the proposed method. The standard deviation of the proposed method is 0.53, and it is smaller than other methods, which are 1.18, 0.93, 1.44, 6.40, 2.16, 3.42 and 1.77, respectively. Fig. 8 is the multi-class confusion matrix of the proposed method for the first trial in Experiment 1. The multi-class confusion matrix records the classification results of all the conditions in detail, which contains both classification information and misclassification information. The ordinate axis of the confusion matrix represents actual label of classification, and the horizontal axis represents predicted label. Therefore, the element on the main diagonal of the confusion matrix represents the classification accuracy of each condition. From Fig. 8, we can find that the lowest accuracy happens in condition 2. Table 5 lists the average testing accuracy and standard deviation of all methods in Experiment 2, and Fig. 9 shows their corresponding diagnosis results in each trial. From Tables 4 and 5, it is clearly that the accuracies of BP and SVM are further improved after extracting features based on EMD, WPD and LMD, however, they are still lower than the proposed method. From the comparison mentioned above, it can be concluded that: (1) The diagnosis performance of the traditional methods such as SVM and BP depends heavily on manual feature extraction and advanced signal processing techniques. Maybe their diagnosis accuracies will be further improved after selecting the most sensitive features or designing new features. However, it is a very time-consuming and labor-intensive task [31,52]. (2) Deep learning methods show obvious higher accuracies than the traditional methods when dealing with the raw vibration data. The main reason is that deep learning methods can adaptively learn the valuable information from the input data through multiple feature transformations [36]. (3) The proposed method is more effective than the standard DBN, CNN and DAE. Compared with the standard DAE, the proposed method considers the 2-D structure and periodic characteristics of the input data. Compared with the standard DBN and CNN, the effectiveness of the proposed method arises from two main respects. One is that the proposed deep model combines the advantages of the standard DBN and CNN. The other is that Gaussian visible units and the modified learning algorithm further enhance the feature learning ability of the proposed deep model. It is meaningful to further compare the classification performance of the proposed method and the standard DBN or the standard CNN. In the evaluation system of classifiers, F-measure is a widely used criterion [53], which contains both the precision rate and recall rate. The calculations of F-measure, precision rate and recall rate can be expressed as precision \u00bc TP TP \u00fe FP 100 \u00f015\u00de 5. Two dimensional visualization of different features using PCA. (a) Compressed data features, (b) extracted 22 features, and (c) deep features. PC 3 (a) Fig. 17. Roller bearing experimental setup. recall \u00bc TP TP \u00fe FN 100 \u00f016\u00de F-measure \u00bc 2TP 2TP \u00fe FP \u00fe FN \u00f017\u00de where TP represents the number of true positives, FP the number of false positives, and FN the number of false negatives. Each condition can have its own F-measure value by doing the same calculation, and F-measure reaches its best value at 1 and the worst at 0. Table 6 lists the precision and recall rates using different deep learning methods at the first trial, and the corresponding Fmeasure values are shown in Fig. 10. The results show the proposed method has higher precision, recall and F-measure compared with traditional deep learning methods such as the standard DBN and CNN. In this case study, the main parameters of the proposed method in Experiment 1 are listed in Table 7. The structure design of deep learning models remains a great challenge, and there is not a systematic method at present [31]. In this paper, the architecture of the proposed model is determined by experiments. It can be found from Table 7 that the proposed model contains two Gaussian CRBMs (improved CRBMs), i.e., a Gaussian visible layer (input layer), two hidden layers and two max pooling layers. In this paper, the regularization constant k and sparse coefficient q are determined through crossvalidation, as show in Fig. 11. The candidate sets of k and q are both [0.01, 0.02, 0.03, 0.04, 0.05, 0.06, 0.07, 0.08, 0.09, 0.10]. It can be found that the accuracy is mainly influenced by the sparse coefficient q, and small sparse coefficient seems to be good choice. The main reason is that larger sparse coefficient (higher activity) of the hidden units will probably result in the overcompleteness of the learned features [40]. The parameters of other methods in Experiment 1 are described as follows. (1) Standard DBN: The architecture is 1024- 400-400-200-6, which is determined by experimentation. The pre-training of each RBM is completed using 100 iterations. The learning rate and momentum are set to 0.1 and 0.9, respectively. (2) Standard CNN: The architecture of the CNN contains input layer, convolutional layer C1, pooling layer P2, convolutional layer C3, pooling layer P4 and output layer (from lowest to highest). The size of the input feature map is 32 32, C1 layer contains 6 kernels, C3 contains 12 kernels, and the scales of P2 layer and P4 layer are both set to 2. The learning rate is 0.1 and iteration number is 150. (3) Standard DAE: The architecture is 1024-400-100-100-6. The sparsity parameter and sparse penalty factor are 0.3 and 5, respectively. The learning rate is 0.1 and iteration number is 100. (4) BP with compressed data: The architecture is 1024-1500-6, which is decided by the guiding principles and experiences. The learning rate is 0.1 and iteration number is 500. (5) BP with 22 features: The architecture is 22-45-6, the learning rate is 0.1 and iteration number is 500. (6) SVM with compressed data: RBF kernel is applied. Two main parameters, the penalty factor and the radius of the kernel function, are set to 30 and 0.08, respectively. Each of them is determined through a 10-fold cross validation. (7) SVM with 22 features: RBF kernel is applied. The penalty factor and the radius of the kernel function are set to 50 and 0.01, respectively."
        ]
    },
    {
        "image_filename": "designv10_0_0000384_adem.201500419-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000384_adem.201500419-Figure1-1.png",
        "caption": "Fig. 1. Schematic of selective laser melting (SLM) parameters (modified from[50]). Reproduced with permission from Wiley [ref. no. 3690340540650].",
        "texts": [
            " As the formation of a fully molten is essential for manufacturing fully-dense parts, sufficient laser energy density applied to powder materials is required for obtaining high density ADVANCED ENGINEERING MATERIALS 2016, 18, No. 4 \u00a9 2015 WILEY-VCH Ve parts. In general, a minimum critical laser energy density is required to produce parts with maximum density.[34,35,46\u201349] For example, the critical laser energy density for SLM of fully dense commercially pure titanium (CP\u2013Ti), Ti\u20136Al\u20134V and Ti\u201324Nb\u20134Zr\u20138Sn are around 120,[47,48] 120,[49] and 40 Jmm 3,[46] respectively. The schematic of these SLM processing parameters is shown in Figure 1.[50] During SLM, laser beam moves across the powder bed with a constant speed known as scan speed (v), which controls the time of production in SLM. In other words, higher scan speeds are required if short production time is needed. However, the maximum laser power of a specific SLM device should be considered when increasing scan speed. Layer thickness (t) defines the amount of energy and production time required to melt/consolidate a layer of powder. Layer thickness is very important as good connectivity between layers can only be possible when previously processed layers are re-melted too"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure4.31-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure4.31-1.png",
        "caption": "Figure 4.31 A beam grillage.",
        "texts": [
            "28b here, with the vertical floor loading and the horizontal wind loading. Sometimes the stiffnesses of the joints are accentuated by thickenings in the connections, but generally they are omitted. If there are also hinged joints in a frame, they have to be clearly depicted by means of open circles. This is the case in Figure 4.30, which could represent a building made of concrete, on which a steel floor was placed at a later stage. Beam grillages are planar structures that are loaded normal to their plane, see Figure 4.31. A beam grillage consists of two cooperative beam layers: beams and girders. The beams and girders are generally placed in two mutually perpendicular directions. Beams grillages are often used as floor structures in bridges and buildings. Lock doors are also sometimes built as a system of beams and girders. A 1 In Chapter 9, which addresses calculations related to trusses, another demand is covered, namely that the load has to be exerted only at connections. 4 Structures 129 fa\u00e7ade made of posts and girders (columns and beams), with perpendicular wind loading, can sometimes also be seen as a beam grillage"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure2.5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure2.5-1.png",
        "caption": "Fig. 2.5. Manipulator link",
        "texts": [
            " 2.1 has six revolute degrees of freedom, with the second and the third always being in parallel (anthropomor phic manipulator). 2.2 Definitions 21 Different schematic representations have been introduced in order to de scribe manipulator configurations simply. Figure 2.4{aHd) shows several types of robots. Here, revolute joints are denoted by cylinders, while prismatic joints are represented by parallelepipeds. Manipulator link is a rigid body described by its kinematic and dynamic parameters. Figure 2.5 shows one typical manipulator link. Kinematic par ameters describing the link may be defined in different ways. Generally speaking, these parameters are the length of the manipulator link, the angle between the axes lying at two ends of the link, etc. Precise definitions of link parameters for the Denavit-Hartenberg kinematic notation will be presented in Sect. 2.3. Dynamic parameters describing the link are its mass, the location of its centre of mass and the inertia tensor. These parameters are used in dynamic modelling of the mechanism"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003728_1.3099008-Figure25-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003728_1.3099008-Figure25-1.png",
        "caption": "Fig 25. A belt-driven and\u2022forced oscillator, and its dynamics. The friction is modeled by the Coulomb law. The three-dimensional phase portrait, in cylindrical coordinates, plots displacement radially, velocity axially, and time circumferentially yielding a toms. A cross-section of the phase portrait defines a Poincar6 section, and is shown on the right (from Poppet al, 1995).",
        "texts": [
            "org/about-asme/terms-of-use Appl Mech Rev vol 51, no 5, May 1998 Feeny et al: Historical review of dry friction and stick-slip phenomena 335 tors whose only nonlinearity is due to dry friction. More complete discussions are included in chapters of a book on nonlinear dynamics with dry friction ( P o p p e t al, 1996; Feeny, 1996). In these systems, the motion collapses during stick-slip, leading to a one-dimensional map in the Poincar6 section. In the former case, the system was both belt driven and harmonically excited. The underlying map was a one-dimensional circle map. lntermittency was the predominant route to chaos. The strange attractor was on a folded toroidal structure (Fig 25). The latter study was on a forced oscillator with displacement-dependent dry damping. The underlying map was a one-dimensional single-humped map. The strange attractor was a stretching and folding branched manifold (Fig 26). The mechanism of dimensional collapse is as follows. The discontinuity of the friction law occurs on a surface D in the phase space. In regions R of D in which the vector field points towards D from both sides, trajectories will stick to this surface for an interval of time"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.18-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.18-1.png",
        "caption": "FIGURE 8.18. A double A-arm suspension.",
        "texts": [
            " De Dion design is a modification of a beam axle that may be used as a dead axle or to attach the differential to the chassis and transfer the driving power to the drive wheels by employing universal joints and split shafts. Figure 8.17 illustrates a De Dion suspension. 8.2 Independent Suspension Independent suspensions is introduced to let a wheel to move up and down without affecting the opposite wheel. There are many forms and designs of independent suspensions. However, double A-arm and McPherson strut suspensions are the simplest and the most common designs. Figure 8.18 illustrates a sample of a double A-arm and Figure 8.19 shows a McPherson suspension. Kinematically, a double A-arm suspension mechanism is a four-bar linkage with the chassis as the ground link, and coupler as the wheel carrying link. A McPherson suspension is an inverted slider mechanism that has the chassis as the ground link and the coupler as the wheel carrying link. A double A-arm and a McPherson suspension mechanism on the left and right wheels are schematically shown in Figures 8.20 and 8.21 respectively. Double A-arm, is also called double wishbone, or short/long arm suspension. McPherson also may be written as MacPherson. Example 309 Double A-arm suspension and spring position. Consider a double A-arm suspension mechanism. The coil spring may be between the lower arm and the chassis, as shown in Figure 8.18. It is also possible to install the spring between the upper arm and the chassis, or between the upper and lower arms. In either case, the lower or the upper arm, which supports the spring, is made stronger and the other arm acts as a connecting arm. Example 310 Multi-link suspension mechanism. When the two side bars of an A-arm are attached to each other with a joint, as shown in Figure 8.22, then the double A-arm is called a multi-link mechanism. A multi-link mechanism is a six-bar mechanism that may have a better coupler motion than a double A-arm mechanism"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003898_tbme.1980.326583-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003898_tbme.1980.326583-Figure1-1.png",
        "caption": "Fig. 1. Four-mass sagittal plane model for the body.",
        "texts": [
            " The present work derives largely from that of Gurfinkel and extends his model to encompass the dynamics of a multilink inverted pendulum including both the effects of joint torques produced by muscular contractions and of disturbing torques resulting from external influences acting on the body of a subject. The results presented are intended to clarify the mechanisms involved in the maintenance of stable posture and also to suggest a potential new methodology for clinical posturography [13] The experimental results regarding center of pressure and center of gravity motion presented later in this paper were obtained using the four-mass sagittal plane linkage model illustrated in Fig. 1. This model results from adding the three mass model of Hemami and Jaswa [9] to the triangular shaped foot model suggested by Gurfinkel [12]. However, rather than using generalized coordinates in the differential equations for this model as was done in the referenced works, the method of free body analysis (D'Alembert's principle) with hard kinematic constraints has been employed [4] . This decision was made in the interest of obtaining the governing equations in a form better suited for digital computation [2], [14]",
            " Numerical differentiation of the smoothed body segment angles yields tables of values for the first and second derivatives needed for evaluation of (25). This differentiation is accomplished by means of a five-point method included as part of the SPARTA program provided by the computer manufacturer [13], [181. Examination of the mathematical model which has been proposed shows that numerical values are required for many body segment parameters. The standardized coefficients of Braune and Fischer [191 were used for this purpose. Since all subjects stood with arms folded as illustrated in Fig. 1, some further measurements and approximations were used to permit the head, arms, and trunk to be treated as a single unit [13]. When analysis of a data record has been completed, the results obtained can be presented either numerically, or graphically. For example, Fig. 3(a) and (b) shows strip chart recordings of center of pressure location (stabilogram) obtained both directly from the force plate and, subsequent to data analysis, from evaluation of (25). These recordings were produced by using two channels of digital to analog conversion synchronized to a real-time clock in the PDP-11/10 computer"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure6.24-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure6.24-1.png",
        "caption": "Fig. 6.24 A person is applying a moment \u03c4 gripping the handle of a bar. To prevent the rotation of the bar, it might be enough to apply an appropriate force f at the far side of the bar. Assuming this point to be the origin O, doesn\u2019t this violate the rule 2, which asserts the same moment will appear at the origin? (The answer is on page 210).",
        "texts": [
            " Rule 1: A force f acting on a point p is converted to a force f and a torque p\u00d7 f with respect to the world frame (Fig. 6.23(a)). Rule 2: A moment \u03c4 acting on a point p is the same \u03c4 with respect to the world frame (Fig. 6.23(b)). In general, when a force fp and a moment \u03c4 p are simultaneously acting on a point p, they are converted a force and moment fo, \u03c4 o around the origin given by [ fo \u03c4 o ] = [ fp p\u00d7 fp + \u03c4 p ] . (6.50) One might think them obvious, we tend to make a curious mistake for this sort of calculation. The quiz shown in Fig. 6.24 might help to understand this problem. 6.6 Appendix 207 6.6.2 Subroutines 210 6 Dynamic Simulation 1. Aldebaran robotics, http://www.aldebaran-robotics.com 2. Carnegie mellon university TARTAN RESCUE, http://www.rec.ri.cmu.edu/projects/tartanrescue/ 3. DARPA Robotics Challenge, http://www.theroboticschallenge.org 4. Products of Kondo Kagaku Co. Ltd. (in Japanese), http://kondo-robot.com/product/ 5. Broad Agency Announcement DARPA Robotics Challenge, DARPA-BAA-1239 (April 2012) 6. Takanishi, A., Ishida, M"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure4.28-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure4.28-1.png",
        "caption": "Figure 4.28 (a) An apartment building constructed of only surface elements; (b) the same building with the main load-bearing structure constructed of beams and columns.",
        "texts": [
            " Surface elements (plates) can be used for the road deck and the walls; and together they form a so-called trough bridge. If the transverse measurements of the bridge are small compared to the span, the bridge can be modelled as a line element, or in other words, a bar with a U-section. In order to limit the use of material and thereby reduce the self-weight that has to be carried, the surface elements can be replaced by planar structures made of line elements, as has been done in Figure 4.27b for the vertical walls. The second example is the apartment building in Figure 4.28a. The structure consists of only surface elements. The vertical floor loading is transferred to the vertical walls and from there is transferred to the foundation. The horizontal wind loading is also distributed across the floors via the walls to the foundation. Figure 4.28b represents the same building, but now all the horizontal and vertical surface elements in the main load-bearing structure have been replaced by planar structures made up of beams and columns. Although the structure now consists of only line elements, the transfer of forces is mostly unchanged and occurs through the same planes as in Figure 4.28a. These examples illustrate that spatial structures can be composed of planar structures that consist of line elements. It is therefore certainly worth the effort of further investigating these types of planar structures. 128 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Planar trusses and frames are planar structures that are loaded in their plane (see Figure 4.29). The difference between a truss and a frame is determined by the nature of the joints in the connections. \u2022 in a truss, the bars are joined together by hinges at all the connections;1 \u2022 in frames, all the joints are fixed and entirely stiff. The truss in Figure 4.29a appeared in the bridge in Figure 4.27b. The open circles, which represent the hinged joints, are generally omitted as in a truss all the joints are by definition hinged. The structure in Figure 4.29b is a frame. You will recognise part of the building in Figure 4.28b here, with the vertical floor loading and the horizontal wind loading. Sometimes the stiffnesses of the joints are accentuated by thickenings in the connections, but generally they are omitted. If there are also hinged joints in a frame, they have to be clearly depicted by means of open circles. This is the case in Figure 4.30, which could represent a building made of concrete, on which a steel floor was placed at a later stage. Beam grillages are planar structures that are loaded normal to their plane, see Figure 4"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003483_s0890-6955(02)00163-3-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003483_s0890-6955(02)00163-3-Figure4-1.png",
        "caption": "Fig. 4. Infrared image acquisition system. (a) Schematic. (b) Experimental setup.",
        "texts": [
            " Sending the commands through Ethernet or RS485 serial connections, the desired delivery rate of each kind of powder is controlled by the central PC in real time. Up to four powder feeders can be controlled simultaneously by the central PC. Functionally graded material can be fabricated by synchronizing the motion, laser operating parameters, and powder delivery rates according to the design of the material. Fig. 3 shows the schematic and real setup of multiple powder feeding systems. A coaxial infrared image sensing setup with respect to the metal powder delivery nozzle has been developed. As shown in Fig. 4 (a), the laser head consists of a partial reflective mirror and lens set. The Nd: YAG laser beam inducted by the optical fiber is reflected from the partial reflective mirror, and it is focused on the substrate by the set of lenses arranged in the laser head. The optical part of the laser head forms another optical path for the observation of the laser processing. A high frame-rate (up to 800 frames/s) camera installed at the top of the laser head takes gray images with a 128 \u00d7 128 resolution. The radiation from the molten pool produced by the laser beam passes through the partial reflective mirror and forms an image on the CCD chip of the camera",
            " In order to get the infrared image of the molten pool to reduce the high intensity light from the molten pool and eliminate the image noise from the metal powder, an infrared filter is selected ( 700 nm) and installed between the iris and the camera. During the LBAM process, the camera acquires images of the build up process at a constant frame rate. Images are transferred to a frame grabber installed on a PC that carries out the image processing and control process. The real setup of the infrared image acquisition system is shown in Fig. 4 (b). Fig. 5 (a) provides the original infrared image acquired by the coaxially installed camera. Because the observation is directly from the top of the molten pool, a full field of view of the molten pool can be acquired without any blocking. With a right combination of the Nd: YAG filter and the IR filter as well as a right adjustment of the iris, a clear image of the molten pool and the surrounding thermal area can be obtained without the noise from the metal powder. The radiation wavelength received by the camera is between 0"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure10.17-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure10.17-1.png",
        "caption": "FIGURE 10.17 Images for g12 where the source point is in region 2 and the observation point is in region 1.",
        "texts": [
            " g31(r, r\u2032) = (1 \u2212 \ud835\udefc1)(1 \u2212 \ud835\udefc2) 4\ud835\udf0b\ud835\udf003 [ \u221e\u2211 k=0 (\u2212\ud835\udefc1\ud835\udefc2)k [\ud835\udf0c2 + (z \u2212 z\u2032 \u2212 2 k d)2]1\u22152 ] (10.55) For the case where the source point is in layer 2 and the observation point is in Layer 1 which is g12 is treated next. We should observe that this solution can also used to obtain g32 since the observation point is just in the other outer layer while the source point is in the same layer. g12(r, r\u2032) = (1 + \ud835\udefc1) 4\ud835\udf0b\ud835\udf001 [ \u221e\u2211 k=0 (\u2212\ud835\udefc1\ud835\udefc2)k [\ud835\udf0c2 + (z \u2212 z\u2032 + 2 k d)2]1\u22152 + \u221e\u2211 k=0 (\u2212\ud835\udefc1)k(\ud835\udefc2)k+1 [\ud835\udf0c2 + (z + z\u2032 + 2 k d)2]1\u22152 ] (10.56) We give the reflection diagrams for g12 in Fig. 10.17. The resultant Green\u2019s function is given in (10.57). Finally, the Green\u2019s function for the case where both the source and observation point are in layer 2 is g22. In this case, images occur on both sides. The refections are shown in Fig. 10.18. g22(r, r\u2032) = 1 4\ud835\udf0b\ud835\udf002 [ 1 [\ud835\udf0c2 + (z \u2212 z\u2032)2]1\u22152 + \u221e\u2211 k=1 (\u2212\ud835\udefc1)k\ud835\udefck\u22121 2 [\ud835\udf0c2 + (z + z\u2032 \u2212 2kd)2]1\u22152 + \u221e\u2211 k=1 (\u2212\ud835\udefc1\ud835\udefc2)k [\ud835\udf0c2 + (z \u2212 z\u2032 + 2kd)2]1\u22152 270 PEEC MODELS FOR DIELECTRICS + \u221e\u2211 k=1 (\u2212\ud835\udefc1)k\ud835\udefck+1 2 [\ud835\udf0c2 + (z + z\u2032 + 2kd)2]1\u22152 + \u221e\u2211 k=1 (\u2212\ud835\udefc1\ud835\udefc2)k [\ud835\udf0c2 + (z \u2212 z\u2032 \u2212 2kd)2]1\u22152 ] (10"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003505_j.jtbi.2004.08.015-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003505_j.jtbi.2004.08.015-Figure1-1.png",
        "caption": "Fig. 1. Spring-mass model for running. Parameters: m\u2014point mass, \u20180\u2014 rest length, a0\u2014leg angle of attack during flight, g\u2014gravitational acceleration, k\u2014spring stiffness, r\u2014radial and j\u2014angular position of the point mass, Dj\u2014angle swept during stance.",
        "texts": [
            " The former by considering a special case, the latter by comparing a return-map analysis based on the approximation with numerical results throughout the range of the parameters spring stiffness, angle of attack, and system energy. In both situations, model parameters relevant to human locomotion are addressed. Planar spring-mass running is characterized by alternating flight and contact phases. As described previously (Seyfarth et al., 2002), during flight the center of mass trajectory is influenced by the gravitational force. Here, a virtual leg of length l0 and a constant angle of attack a0 are assumed (Fig. 1). When the leg strikes the ground, the dynamic behavior of spring-mass running is further influenced by the force exerted by the leg spring (stiffness k; rest length l0) attached to the center of mass. The transition from stance-to-flight occurs if the spring reaches its rest length again during lengthening. To investigate periodicity for this running model, it suffices to consider the apex height yi of two subsequent flight phases. This holds since (i) at apex the vertical velocity _yi equals zero, (ii) the forward velocity _xi can be expressed in terms of the apex height due to the constant system energy Es; and (iii) the forward position xi has no influence on the further system dynamics",
            " In terms of the apex return map, a periodic movement trajectory in spring-mass running is represented by a fixed point yi\u00fe1\u00f0yi\u00de \u00bc yi: Moreover, as a sufficient condition, a slope dyi\u00fe1\u00f0yi\u00de=dyi within a range of \u00f0 1; 1\u00de in the neighborhood of the fixed point indicates the stability of the movement pattern (higher than period 1 stability, which corresponds to symmetric contacts with time reflection symmetry about midstance, is not considered). The size of the neighborhood defines the basin of attraction of the stable trajectory. The analytical solution for the center of mass motion during flight is well known (ballistic flight trajectory), but a different situation applies to the stance phase. Using polar coordinates (r;j), the Lagrange function of ARTICLE IN PRESS H. Geyer et al. / Journal of Theoretical Biology 232 (2005) 315\u2013328 317 the contact phase is given by (see Fig. 1 for notation) L \u00bc m 2 \u00f0_r2 \u00fe r2 _j2\u00de k 2 \u00f0\u20180 r\u00de2 mgr sin j: (1) From the Lagrange function, the derived center of mass dynamics are characterized by a set of coupled nonlinear differential equations. As of today, the analytical solution for the contact phase remains open. For such situations, a common approach is to ask for simplifications, which could provide an approximate solution. In the case of Eq. (1), for sufficiently small angles Dj swept during stance, the sine term on the right-hand side can assumed to be sin j 1 (2) and the equations of motion simplify to m\u20acr \u00bc k\u00f0\u20180 r\u00de \u00fe mr _j2 mg (3) and d dt \u00f0mr2 _j\u00de \u00bc 0 (4) transforming the spring-mass model into an integrable central force system, where the mechanical energy E and the angular momentum P \u00bc mr2 _j are conserved",
            "2) can be written as P \u00bc PTD \u00fe mg tc Z tmid 0 Z t 0 jx\u00f0t0\u00dejdt0 dt \u00fe Z tc tmid Z tmid 0 jx\u00f0t0\u00dejdt0 dt Z tc tmid Z t tmid jx\u00f0t0\u00dejdt0 dt : \u00f0A:3\u00de The time reflection symmetry about midstance yieldsR tc tmid \u00bc R tmid 0 ; and (A.3) simplifies to P \u00bc PTD \u00fe mg tc Z tc tmid Z tmid 0 jx\u00f0t0\u00dejdt0 dt \u00bc PTD \u00fe mg 2 Z tmid 0 jx\u00f0t0\u00dejdt0: \u00f0A:4\u00de ARTICLE IN PRESS H. Geyer et al. / Journal of Theoretical Biology 232 (2005) 315\u2013328328 The mean angular momentum is increased compared to the initial value, which, however, means that the amount of mean angular momentum decreases j PjojPTDj since the initial value PTD is negative (according to the definition of the coordinate system in Fig. 1 the angular velocity _j is defined negative for forward motion). The miscalculation of angular momentum in the central force approach (P PTD) has a more profound effect on the angular motion (P _j) than on the radial (P r2). Considering that, due to the alignment of the gravitational force with the radial axis ( mg sin j ! mg in (3)), the spring compression is increased, this leads to a clear overestimation of the angular velocity. Here, an approximation of the central force system dynamics with an error decreasing this inherent overestimation may result in a better performance when compared to the actual spring-mass dynamics"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure9.5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure9.5-1.png",
        "caption": "FIGURE 9.5. A two-link manipulator.",
        "texts": [
            " However, the Euler equation reduces to \u03c9\u03071 = I2 \u2212 I3 I1 \u03c92\u03c93 (9.107) \u03c9\u03072 = I3 \u2212 I1 I22 \u03c93\u03c91 (9.108) \u03c9\u03073 = I1 \u2212 I2 I3 \u03c91\u03c92 (9.109) that show the angular velocity can be constant if I1 = I2 = I3 (9.110) or if two principal moments of inertia, say I1 and I2, are zero and the third angular velocity, in this case \u03c93, is initially zero, or if the angular velocity vector is initially parallel to a principal axis. Example 351 Angular momentum of a two-link manipulator. A two-link manipulator is shown in Figure 9.5. Link A rotates with angular velocity \u03d5\u0307 about the z-axis of its local coordinate frame. Link B is attached to link A and has angular velocity \u03c8\u0307 with respect to A about the xA-axis. We assume that A and G were coincident at \u03d5 = 0, therefore, the rotation matrix between A and G is GRA = \u23a1\u23a3 cos\u03d5(t) \u2212 sin\u03d5(t) 0 sin\u03d5(t) cos\u03d5(t) 0 0 0 1 \u23a4\u23a6 . (9.111) Frame B is related to frame A by Euler angles \u03d5 = 90deg, \u03b8 = 90deg, and \u03c8, hence, ARB = \u23a1\u23a3 c\u03c0c\u03c8 \u2212 c\u03c0s\u03c0s\u03c8 \u2212c\u03c0s\u03c8 \u2212 c\u03c0c\u03c8s\u03c0 s\u03c0s\u03c0 c\u03c8s\u03c0 + c\u03c0c\u03c0s\u03c8 \u2212s\u03c0s\u03c8 + c\u03c0c\u03c0c\u03c8 \u2212c\u03c0s\u03c0 s\u03c0s\u03c8 s\u03c0c\u03c8 c\u03c0 \u23a4\u23a6 \u23a1\u23a3 \u2212 cos\u03c8 sin\u03c8 0 sin\u03c8 cos\u03c8 0 0 0 \u22121 \u23a4\u23a6 (9"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure7.18-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure7.18-1.png",
        "caption": "Figure 7.18 Longitudinal section of a channel with the 4-metre wide flap AB. The distribution of the water pressure is shown on both sides of the flap.",
        "texts": [
            " VOLUME 1: EQUILIBRIUM The isotropic compressive stress p increases linearly with depth z. This is referred to as a hydrostatic pressure distribution. Below you will find a number of examples covering loads due to a hydrostatic pressure. We assume that in all cases the fluid is at rest and that the pressure distribution is hydrostatic. At any point the hydrostatic pressure is equally large in all directions (isotropic state of stress) and always acts normal to the plane in question (as there are no shear stresses). Example 1 Figure 7.18 shows the longitudinal section of a channel with a 4-metre wide flap AB. The flap is supported at A by a hinge and is resting at B on a sill. The support in B can be seen as a roller support. The water level on both sides of the flap is shown in the figure. The density of water is 1000 kg/m3. The gravitational field intensity is 10 N/kg. Question: Determine the support reactions at A and B due to the total water pressure. The dead weight of the flap should be ignored. Solution: The linear distribution of the water pressure on both sides of the flap is shown in Figure 7.18. To the left of the flap, the water pressure at A is (1000 kg/m3)(10 N/kg)(1 m) = 10 kN/m2 and at B it is (1000 kg/m3)(10 N/kg)(4 m) = 40 kN/m2. To the right of the flap, the water pressure at B is (1000 kg/m3)(10 N/kg)(2 m) = 20 kN/m2. 7 Gas Pressure and Hydrostatic Pressure 257 In Figure 7.19a, the 4-metre wide flap is modelled as a line element, with line loads due to the water pressures normal to it. To the left of the flap, the distributed load varies linearly from (4 m)(10 kN/m2) = 40 kN/m at A, to (4 m)(40 kN/m2) = 160 kN/m at B. To the right of the flap, the load increases linearly from 0 at C to (4 m)(20 kN/m2) = 80 kN/m at B. Figure 7.19b represents the load diagram for the resulting water pressure. The length of flap AB is (see Figure 7.18) \u221a (3 m)2 + (0.96 m)2 = 3.15 m. The distances BC and CA are respectively 2.10 m and 1.05 m. To work quickly, the load diagram in Figure 7.20 has been placed horizontally and is split up into a number of areas for which the resultants can be easily calculated: R1 = (2.10 m(80 kN/m) = 168 kN, R2 = 1 2 \u00d7 (1.05 m)(80 kN/m) = 42 kN, R3 = 1 2 \u00d7 (1.05 m)(40 kN/m) = 21 kN. 258 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM The support reaction Ar at A is found from the moment equilibrium of the flap about B: Ar = (1"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure2.7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure2.7-1.png",
        "caption": "Figure 2.7 The forces drawn to scale in (a) a parallelogram and (b) force polygon.",
        "texts": [
            " If you look closely, you will recognise the force polygon in Figure 2.2. 26 Figure 2.5 shows all the forces in changing order in a force polygon. The order clearly does not influence the end result (the vector addition is associative and commutative, see Section 1.3.7). Example 2 Two forces F1 and F2 are acting on the ring in Figure 2.6. Their directions are shown in the figure. The forces are not shown to scale. Question: Find the magnitude and direction of the resultant force on the ring if F1 = 1000 N, F2 = 750 N. Solution: In Figure 2.7, the forces have been drawn to scale, with 1 cm =\u0302 250 N. Using the parallelogram rule in Figure 2.7a or the force polygon in Figure 2.7b, we can construct the resultant R. Through measuring we find that R has a length of approximately 5.95 cm, so that R \u2248 5.95 \u00d7 250 N = 1488 N. With a protractor, we find that the line of action of R makes an angle \u03b3 of approximately 22.5\u25e6 with the vertical. Check: The magnitude and direction of the resultant R can also be calculated from the force triangle ABC. In doing so, we use the cosine rule and the sine rule, as shown in Figure 2.8. In the triangle ABC in Figure 2.7b \u03b1 = 180\u25e6 \u2212 (50\u25e6 + 15\u25e6) = 115\u25e6. A ring subject to two forces. ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM 2 Statics of a Particle 27 Using the cosine rule we find that R = \u221a F 2 1 + F 2 2 \u2212 2F1F2 cos \u03b1 = \u221a (1000 N)2 + (750 N)2 \u2212 2 \u00d7 (1000 N)(750 N) cos 115\u25e6 = \u221a 2.196 \u00d7 106 N2 = 1482 N. The angle \u03b2 in triangle ABC can be calculated using the sine rule: R sin \u03b1 = F2 sin \u03b2 , sin \u03b2 = F2 R sin \u03b1 = 750 kN 1482 kN sin 115\u25e6 = 0.459 so that \u03b2 = 27.3\u25e6. The angle \u03b3 that the resultant R makes with the vertical is therefore \u03b3 = 50\u25e6 \u2212 \u03b2 = 22.7\u25e6. The values measured in Figure 2.7 correspond well with the values we have calculated. In a plane, we can resolve a force F into two components with given lines of action. Example The force F = \u221a 34 kN in Figure 2.9a has to be resolved into two forces Fa and Fb with the given lines of action a and b. 28 Solution: Figure 2.9b shows the forces Fa and Fb on the lines of action a and b. Along a and b the directions of the forces can be chosen freely. In the analytical approach, we calculate Fa and Fb on the basis of the condition that the sum of the components from Fa and Fb is equal to the corresponding component of F in each of the coordinate directions Fx;a + Fx;b = Fx, Fy;a + Fy;b = Fy"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003720_j.electacta.2004.10.080-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003720_j.electacta.2004.10.080-Figure1-1.png",
        "caption": "Fig. 1. Schematic of the biofuel cell (not to scale).",
        "texts": [
            " The salt-extracted films were then resuspended n 1.0 mL of lower aliphatic alcohols. Enzyme/Nafion casting solutions with an enzyme to BAB ratio of 2:1 (usually 1200 L of 1.0 M enyme:600 L) and 0.03 g NAD+ were vortexed in preparation or coating on electrode. The solution was pipeted onto the lectrode, allowed to soak into the carbon felt electrode and ried. .3. Physical cell apparatus The physical test cell consisted of custom fabricated U\u201d shaped cylindrical glass tubing with 2.6 cm diameter, 14.8 cm height and 7.6 cm length, as shown in Fig. 1. Approximately, 50 mL of solution was contained on both sides of a Nafion membrane (Aldrich). The cathode side of the test cell contained pH 7.15 phosphate buffer solution saturated with dissolved oxygen bubbled in via external source (AirGas). The cathode material is an ELAT electrode with 20% Pt on Vulcan XC-72 (E-Tek). The anode side of the test cell is filled with pH 7.15 buffer containing 1.0 mM NAD+ and 1.0 mM fuel (ethanol or methanol). The modified electrode with enzyme acts as the anode"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002504_j.compositesb.2018.12.146-Figure13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002504_j.compositesb.2018.12.146-Figure13-1.png",
        "caption": "Fig. 13. (a) The CAD model, (b) support structures on platform, (c) the processing period, (d) complex shaped Ni-based components fabricated by SLM.",
        "texts": [
            " Thirdly, the wetting characteristics of TiC surrounded by nickel liquid were improved with the elevating E, thus forming a strong interfacial bonding between TiC and Inconel 718 matrix owing to the diffusion of Ti and C into substrate during SLM process [32]. Consequently, the load applied to the composites can be transferred from the Inconel 718 matrix, across the TiC/Inconel 718 interface, to hard TiC particles [33,34]. The combination of these three mechanisms results in high performances on the TiC/Inconel 718 nanocomposites. The complex shaped and large sized TiC/Inconel 718 component was processed under the optimal processing parameters (E=300 J/m) for application of aerospace device under severe aerospace environment, as shown in Fig. 13. The design from the CAD model illustrates the geometry and complicated features with a diameter of 220mm and height of 52mm (Fig. 13a). Fig. 13b shows a support structure used in the building process. Noticeably, there are only a half of supports used in the suspended thin-wall blades. On the one hand, these supports could ensure successful fabrication of components. On the other hand, they could provide avenues for heat dissipation during solidification and reduce the thermal distortions. The SLM-processed component exhibited a relatively high quality with a middle axial diameter of 38mm. For details, the SLM-fabricated part had a spinning blade with thickness of 1mm, height of 35mm and rotation angle of 40\u00b0 (Fig. 13d). The detailed characterization of the component will be given in our future work. It comes to conclusion that SLM technology has a great potential in processing high-temperature nickel-based superalloys with tailored microstructures and performances as well as the net-shaping configurations. The SLM process was applied to fabricate nano-TiC reinforced Inconel 718 nanocomposites, the conclusions are summarized as follows: (1) The densification behavior of the SLM-processed TiC/Inconel 718 nanocomposites was influenced by laser energy linear density (E)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000595_1.3209132-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000595_1.3209132-Figure4-1.png",
        "caption": "Fig. 4 Stressed volume in the Lundberg\u2013Palmgren theory \u202033\u2021",
        "texts": [
            " The Lundberg\u2013Palmgren heory 33 states that for bearing rings subjected to N cycles of epeated loading the probability of survival S is given by ln 1 S = A Ne 0 cV z0 h 1 here 0 is the maximum orthogonal shear stress in the contact, z0 s the corresponding depth at which this stress occurs, and V is the tressed volume of material. The parameters A, c, and h are maerial characteristics that are determined experimentally. The paameter e is the Weibull slope for the experimental life data ploted on a Weibull probability paper. The stressed volume of aterial V was assumed to be V = a z0 2 rr 2 here the dimensions a, z0, and rr correspond to the width, depth, nd length of the volume as shown in Fig. 4. The following loadife equation for the bearing was obtained by substituting for 0, 0, and V in terms of the bearing dimensions and contact load in q. 1 , L10 = C P p 3 ere, L10 is the life for 10% probability of failure, C is the bearing asic dynamic load rating, and P is the equivalent load on the earing. The exponent p is 3 for ball bearings having an elliptical ontact area, 10/3 for roller bearings having modified line contact reas, and 4 for pure line contacts. Equation 3 formed the basis of the first bearing life standards sed in the industry ISO 281 35 , and the Lundberg\u2013Palmgren heory 33 has been extensively used since the 1950s"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001038_j.matdes.2015.10.002-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001038_j.matdes.2015.10.002-Figure1-1.png",
        "caption": "Figure 1: Selective Laser Melting (SLM) process overview [5]",
        "texts": [
            " This concept was used to simplify the process modeling in which reduces the computational costs. Keywords: Selective Laser Melting, Finite Element Simulation, Optical Penetration Depth, Melt pool size, AISI 316L 1. Introduction Additive Manufacturing (AM) is referred to technologies that fabricate directly threedimensional objects in layer-by-layer fashion. Selective Laser Melting (SLM), as one of the powder bed fusion (PBF) AM processes, enables production of complex metallic parts from a CAD model. A schematic of a typical SLM process is depicted in Fig. 1. In this process, in order to deposit powder layers with predefined thickness, an amount of powder comes up to the build table and a roller or blade spreads powder at the build platform. Full dense parts created by scanning a high intensity laser beam in a special pattern and local consolidation of the powder bed in successive layers. A neutral gas flow, usually nitrogen or argon gas, protects molten pool * Corresponding Author; Tel: (+98) 31 3391 5238, Fax: (+98) 31 3391 2628, E-mail address: eforoozmehr@cc"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000019_cr068123a-Figure10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000019_cr068123a-Figure10-1.png",
        "caption": "Figure 10. Integrated needle-type glucose/insulin microsensor based on electrocatalytic (RuOx) insulin detection and biocatalytic (GOx) glucose sensing. (Reprinted with permission from ref 103. Copyright 2001 American Chemical Society.)",
        "texts": [
            " reported on a dual (side-by-side) Pt electrode amperometric biosensor for the simultaneous monitoring of glucose and lactate.102 The surface coating (based on electropolymerized overoxidized polypyrrole film) resulted in excellent selectivity and no cross talk. Wang and Zhang developed a needle-type sensor for the simultaneous continuous monitoring of glucose and insulin.103 The integrated microsensor consisted of dual electrocatalytic (RuOx) and biocatalytic (GOx) modified carbon electrodes inserted into a needle (Figure 10) and responded independently to nanomolar and millimolar concentrations of insulin and glucose, respectively. The enormous activity in the field of glucose biosensors is a reflection of the major clinical importance of the topic. Such huge demands for effective management of diabetes have made the disease a model in developing novel approaches for biosensors. Accordingly, for nearly 50 years we have witnessed tremendous progress in the development of electrochemical glucose biosensors. Elegant research on new sensing concepts, coupled with numerous technological innovations, has thus opened the door to widespread applications of electrochemical glucose biosensors"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000613_978-94-015-9261-1-Figure5.9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000613_978-94-015-9261-1-Figure5.9-1.png",
        "caption": "Figure 5.9 . Subma nifold S ; I neighborh oods and cco rdinate transforrn ation s in output space",
        "texts": [
            " To investigate sys te m beh avior in the dornain Y, we define the error vector E= {Ej}, j = 1,2 , . .. , m - u, (5 .106) 214 CHAPTER 5 characterizing violation of the coordination condit ion (5 .104) . The coordi nated mode of system evolut ion corresponds to the identi ty E( t) == 0 that impli es, at least , t hat for ini ti al states Xo of t he system (5 .102) it holds E( O) = O. Stability of t he mode mentioned depends on t he system proper ty of being asy rnptotica lly stable wit h respect to t he ou tput equilibrium We consider the geometrie object (Figure 5.9) s; = {y E Y: epy(Y) = O}. Under Assum ption 5.2, t he set S; is a smooth j.l-dimensional hyp ersurface t hat ca n be en dowed wit h the structure of a mul t ishee ted smo ot h man i fold (see Sect ion 4.3). The coordination condit ion (5 .104) and t he identi ty E( t ) == 0 correspond to t he syst em traj ectaries y(t) lying ent irely on t he hypersurface S;. Here we restriet ourselves to consideration of a case when in the dornain Y the coordination conditions (5.104) define a one-sheeted manifold S; equipped with local coordinat es Yj defined in a connecte d open set Sy C IR"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000980_pnas.0709765105-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000980_pnas.0709765105-Figure1-1.png",
        "caption": "Fig. 1. Microchannels used to probe the chemotactic response of bacteria to nutrient patches and plumes. (a) Microchannel for diffusing nutrient pulse experiments. The two inlets (\u2018\u2018in\u2019\u2019) are for bacteria (yellow) and chemoattractant (white). (b) Microinjector tip, corresponding to the red dashed box in a. The 300- m-wide injected nutrient band was visualized by using 100 M fluorescein. (c) Cylindrical particle (R 250 m) used in the nutrient plume experiments.",
        "texts": [
            " Here, we have fabricated microchannels that create patches with spatial and temporal scales consistent with those expected in the ocean. By simultaneously measuring the spatiotemporal distribution of nutrients and cells, we quantified the nutrient exposure experienced by chemotactic marine bacteria. Diffusing Nutrient Pulse. We examined the chemotactic response to a diffusing nutrient patch of the marine -proteobacterium Pseudoalteromonas haloplanktis, which has previously served as a model of chemotaxis (11). We simulated a nutrient pulse by using a microinjector (Fig. 1a), which generated a onedimensional (1D), 300- m wide band of nutrients (Fig. 1b), mimicking the sudden release of organic matter associated with viral lysis or sloppy predation of phytoplankton cells (2). The chemoattractant consisted of culture filtrates from the marine phytoplankton Dunaliella tertiolecta. Although patch composition in the ocean is poorly characterized, phytoplankton-derived DOM was chosen because it is likely a significant nutrient component of many patches (4, 5, 16, 17). Algal-derived dissolved organic carbon (DOC) concentration in the culture filtrates (567 M) was on the high end of measured bulk DOC concentrations during phytoplankton blooms (e",
            " Microscale sources of DOM in the ocean are often in motion relative to their ambient environment, and advection can stretch released solutes into elongated plumes. Nutrient-enriched plumes are predicted to arise from rapid dissolution of fresh fecal pellets (20) and in the wake of marine snow particles (6), where DOM is released because of polymer hydrolysis by particle-attached bacteria (21). Nutrient plumes could represent an important resource for chemotactic bacteria, provided they respond rapidly. We investigated the response of P. haloplanktis to a nutrient plume composed of culture filtrates from D. tertiolecta by using the microchannel in Fig. 1a with a microfabricated polydimethylsiloxane (PDMS) cylinder in front of the injector to model a particle (Fig. 1c). Because P. haloplanktis is superior in using patchy resources, we focus here on the effect of particle sinking speed, rather than a comparison with E. coli. To simulate sinking at speed U, the particle was stationary and immersed in a steady flow of mean velocity U, varied from 66 to 660 m s 1 (5.7\u201357 m day 1), consistent with empirically predicted values in the ocean [U 50\u2013500 m s 1 for particles of radius R 250 m (22, 23)]. To mimic DOM leakage from the particle, nutrients were added immediately upstream via the microinjector and advected around the particle by the flow, forming a plume in its wake (Fig. 4a). The shape of the plume is dictated by the Peclet number Pe 2UR/D, a larger Pe corresponding to more slender plumes. Here D is the diffusivity of chemoattractants, assumed to be 0.5 10 9 m2 s 1, a value typical of low molecular weight substances. The radius of the particle (R 250 m) is consistent with the size of a small marine snow aggregate or a large fecal pellet (17, 22). Bacteria introduced upstream (Fig. 1a) were advected past the particle and exposed to the nutrient plume. For moderate sinking speeds, P. haloplanktis cells aggregated rapidly in the plume (e.g., Fig. 4a, U 110 m s 1). The accumulation was quantified (Fig. 4 b2, c2, and d2) for three sinking speeds U (66, 220, and 660 m s 1), corresponding to increasing Pe (66, 220, and 660) and thus narrower plumes (Fig. 4 b1, c1, and d1). We confined our measurements to the first 24 mm of the plume (y 96 R), where the highest chemoattractant concentrations occur and the disruptive effects of turbulence will be smallest in the ocean (24)",
            " The results presented here strongly suggest that marine bacteria are capable of responding to ephemeral patches of organic matter before these patches dissipate, indicating that bacterial chemotactic behavior in the ocean may impact planktonic trophodynamics and biogeochemical transformation rates. Microfluidics. Microfluidic channels were fabricated by using PDMS and soft lithography, as described previously (38, 39). The microchannel used in the \u2018\u2018diffusing nutrient patch\u2019\u2019 experiments was 45 mm long (with a working section of 25 mm), 3 mm wide, and 50 m deep, with two in-line inlets to separately introducebacteriaandnutrients (Fig.1a). Thenutrient inlet led toa100- m-wide PDMS microinjector (Fig. 1b), which generated a coherent nutrient band. The microchannel used in the nutrient plume experiments was identical to the first, with the exception of an additional microfabricated PDMS cylinder to simulate a sinking particle. The cylinder had radius R 250 m and was centered 290 m downstream of the tip of the microinjector (Fig. 1c). Bacteria. A culture of P. haloplanktis ATCC 700530 was obtained from the American Type Culture Collection and grown overnight to mid-exponential phase in 1% tryptic soy broth at room temperature while agitated on a shaker (175 rpm). Cultures were diluted 1:20 in artificial seawater and starved at room temperature for 72 h (12). E. coli HCB1 (provided by H. Berg) was grown in tryptone broth at 34\u00b0C on an orbital shaker (220 rpm) to an optical density of 0.4 and then washed thrice by centrifuging at 2,000 g for 5 min, and the pellet was resuspended in Howard Berg motility medium (10 mM potassium phosphate/100 M EDTA/10 mM NaCl, pH 7"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure9.7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure9.7-1.png",
        "caption": "Figure 9.7 The members along the chord or circumference of the truss are chord members (ch), the others are known as bracing members (br). Chord members can be divided into top chord members (t) and bottom chord members (b), while for bracing members we distinguish between verticals (v) and diagonals (d). A vertical chord member is also referred to as a vertical.",
        "texts": [
            " For (c) and (d), the use of hinged joints generates a mechanism; they cannot be considered trusses. The force flow in (c) and (d) occurs mainly by bending. 324 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM One also often assumes that the dead weight of a truss applies at the joints. The total dead weight Fdw of a truss member is split up into two equal forces in both adjacent joints (see Figure 9.6). This is a rough model of reality, but since the dead weight is generally small with respect to the other loads that the truss has to bear, the deviations that occur are relatively small. Figure 9.7 shows part of a truss. The letters show the names of the members in the truss. The members along the chord or perimeter of the truss are called chord members (ch), the others are referred to as bracing members (br). Chord members can be divided into top chord members (t) and bottom chord members (b). For bracing members, we distinguish between verticals (v) and diagonals (d), depending on whether the members are positioned vertically or obliquely. Vertical chord members are also referred to as verticals"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000981_c2cs35475f-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000981_c2cs35475f-Figure8-1.png",
        "caption": "Fig. 8 A biosensor comprising an immobilized enzyme (formaldehyde dehydrogenase) in mesoporous silica, an electrochemical mediator, and an electrochemical cell.93",
        "texts": [
            "90 Recently, Shimomura and co-workers proposed the use of hybrid mesoporous membranes in which highly regular mesochannels are formed inside the linear pores (with a diameter generally around 200 nm) of porous anodic alumina membranes (PAAMs). For synthesis, surfactant-based precursor solutions are penetrated into the matrices and dried until the solvents are completely evaporated.91,92 Shimomura et al. fabricated biosensors comprised of an enzyme (formaldehyde dehydrogenase) immobilized in a mesoporous silica material, an electrochemical mediator (i.e., quinone), and an electrochemical cell (Fig. 8).93 Rapid response and high sensitivity were confirmed for detection of formaldehyde in aqueous solution. In addition, high selectivity, reusability, and remarkable storage stability were confirmed, suggesting that formaldehyde dehydrogenase retains its highly ordered structure in these mesoporous silica materials. Thus, hybrid mesoporous membranes are very promising for effective immobilization of enzymes in their applications as electrochemical biosensors. There are several reports of synthesis of mesoporous materials within the regular, straight channels of PAAMs"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002362_j.ymssp.2020.106640-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002362_j.ymssp.2020.106640-Figure1-1.png",
        "caption": "Fig. 1. Finite element model in MATLAB environment.",
        "texts": [
            " (3) Vibration experiment is carried out to verify the proposed dynamic model. Dynamic behaviors of the gear pairs under different spalling patterns are discussed. The meshing deformation during the engagement process can be divided into two components: contact deformation and global deformation. The global compliance, including the deformation of the fillet-foundation and gear tooth, is acquired using the finite element method. To this end, a finite element model of the gear pair is established in MATLAB environment (see Fig. 1). The method for obtaining the global compliance is illustrated in Fig. 2. The degrees of freedom of the nodes at inner hub are fully constrained. The unitary force is applied at the contact position along the line of action. Local distorted deformation will inevitably appear at the load point (see Fig. 2a). A partial model is used to compensate the distorted deformation in Refs. [36\u201338]. However, a simpler approach is adopted in this paper. Local distorted deformation only influences the displacement field near the load point"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003690_1.1802311-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003690_1.1802311-Figure7-1.png",
        "caption": "Fig. 7 The forces acting on the bearing ball",
        "texts": [
            "D (29) When the bearing is deformed under the load, the distances between the curvature center of the inner ring groove and the final position of the ball center, the center of the outer ring groove and the final position of the ball center are, respectively; D ik5ri2D/21d ik5~ f i20.5!D1d ik (30) Dok5ro2D/21dok5~ f o20.5!D1dok The changes to the curvature center for the inner ring are: D icu5Ddx2Dgzr ic cos wk1Dgyric sin wk (31) D icv5Ddy cos wk1Ddz sin wk The following relations can be derived from Fig. 7: sin uok5 Uk ~ f o20.5!D1dok cos uok5 Vk ~ f o20.5!D1dok (32) sin u ik5 Uik2Uk ~ f i20.5!D1d ik cos u ik5 Vik2Vk ~ f i20.5!D1d ik where Uik5BD sin u1Ddx2Dgzr ic cos wk1Dgyric sin wk (33) Vik5BD cos u1Ddy cos wk1Ddz sin wk ric5(1/2)Dm1( f i20.5)D cos u. If ric5(1/2)Dm , the tangential stiffness matrix of the bearing is symmetric. From Figs. 6 and 7, the following equations can be obtained. Displacement equations: ~Uik2Uk!21~Vik2Vk!22D ik 2 50 (34) Uk 21Vk 22Dok 2 50 Journal of Mechanical Design rom: http://mechanicaldesign"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure14.16-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure14.16-1.png",
        "caption": "Figure 14.16 The forces acting in B on the trolley.",
        "texts": [
            " This is at A, as shown in Figure 14.15. With (5) N = H \u221a 1 + (tan \u03b1)2 we find Nmax = NA = (16 kN) \u221a 1 + (3/2)2 = 28.84 kN. The cable force is a minimum at D, where the cable is horizontal and V = 0: Nmin = H = 16 kN e. The horizontal support reaction at A is equal to H : Ah = H = 16 kN (\u2190). Since the cable supports are at equal elevations, the vertical support reaction 14 Cables, Lines of Force and Structural Shapes 647 at A is equal to the support reaction of the beam in Figure 14.14a: Av = 24 kN (\u2191). Check: Figure 14.16 shows the forces acting on the pulley at B. The forces NCB and G, both 20 kN, are known. The force equilibrium can be used to find the support reactions at B: Bh = H = 16 kN (\u2192), Bv = (12 kN) + (20 kN) = 32 kN) (\u2191). Check: The horizontal support reaction at B is equal to H . The vertical support reaction is equal to the support reaction at B of the beam in Figure 14.14a, increased with the vertical cable force G. The support reactions at A and B are shown in Figure 14.17. Figure 14.18 shows a cable subject to a distributed load qz = qz(x)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003908_978-1-84882-730-1-Figure10.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003908_978-1-84882-730-1-Figure10.1-1.png",
        "caption": "Figure 10.1 General framework of ship path-following",
        "texts": [
            " The positive constant terms dv , dr , dvi , and dri ; i D 2;3 represent the hydrodynamic damping in surge, sway and yaw. The bounded time-varying terms, wv.t/ and wr .t/, are the environmental disturbances induced by wave, wind, and ocean current with j wv.t/j wvmax <1 and j wr .t/j wrmax <1. The available control is the yaw moment r . Since the sway control force is not available in the sway dynamics, the ship model (10.1) is again underactuated. The control objective of this chapter is to design the yaw moment r to force the underactuated ship (10.1) to follow a specified path \u02dd , see Figure 10.1, where M is the orthogonal projection of the ship point P on \u02dd , xn and xt are the normal and the tangent unit vectors to the path at M , respectively. We assume that this point is uniquely defined. This assumption holds if the interior of any circle tangenting \u02dd at two or more points does not contain any point of the path and the distance between the ship and \u02dd is not too large. Furthermore, it is assumed that the radius of any osculating circle of the path is larger than or equal to Rmin which is feasible for the ship to follow"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000789_1.1372322-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000789_1.1372322-Figure2-1.png",
        "caption": "Fig. 2 Three models of walking: \u201ea\u2026 The Idealized Simplest Model \u201eISM\u2026 consists of point mass pelvis and feet connected by massless legs, after Garcia et al. \u202020\u2021. It employs a linearized analytic solution to the equations of motion. \u201eb\u2026 The Simplest Model \u201eSM\u2026 is identical to the ISM except that the equations of motion are solved numerically. \u201ec\u2026 The Anthropomorphic Model \u201eAM\u2026 is similar to that of McGeer \u202018\u2021, with legs with more realistic inertial parameters as well as curved feet to improve efficiency. All three models are powered on level ground by an impulsive push P along the stance leg applied at toe-off, as well as a springlike hip torque between the legs \u202022\u2021. The SM is used to test the linearizing assumptions of the ISM, while the AM tests the idealized inertial parameters.",
        "texts": [
            " The extreme simplicity of the model should also clearly reveal underlying principles of the energetic cost of transport, a task difficult to achieve when using a large number of empirical parameters. We use three variations of walking models to study hypothetical metabolic costs. All are based on the principles of passive dynamic walking, but with the addition of actuation for walking on level ground. These quasi-passive models include two variations of @22#, termed the Idealized Simple Model ~ISM, see Fig. 2~a!! and the Simple Model ~SM, see Fig. 2~b!!, as well as a model more similar to that of McGeer @18# with more realistic inertial characteristics, termed the Anthropomorphic Model ~AM, see Fig. 2~c!!. The Idealized Simple Model provides analytical predictions but relies on linearization and other approximations. The Simple Model is identical except that it retains all nonlinearities. It requires numerical computations but makes it possible to test the consequences of the ISM approximations. Finally, the Anthropomorphic Model is used to test whether the fundamental principles of the ISM apply to a nonlinear model that is more physically realistic. We will use dimensionless variables throughout, with the following base units; overall mass M, leg length l, and the gravitational constant g. Time is therefore normalized by Al/g . All three models include rigid stance and swing legs connected by a hinge joint at the pelvis and constrained to planar motion ~see Fig. 2!. Actuation is provided in the form of an impulse P directed along the stance leg and a springlike hip torque acting between the legs. The toe-off impulse P is applied instantaneously before heel strike, which is modeled as an instantaneous and perfectly inelastic collision that sets the initial conditions for the following step. The hip torque can either be produced by a torisional spring of stiffness k or by impulsive torques that occur at the beginning and end of the swing phase, as long as both yield the equivalent swing leg natural frequency v[Ak11. The ISM and SM are irreducibly simple, with legs of zero mass connecting a massive pelvis of mass M and point mass feet of mass m, taking the limit as m/M approaches zero @22#. Journal of Biomechanical Engineering Analysis of the ISM yields three simple multiplicative power laws relating actuation variables P and v to the resulting speed v , step period t, and step length described by the initial leg angle a ~see Fig. 2!. From Kuo @22#, v;v1/2P1/2 (1a) t;v21 (1b) a;v21/2P1/2. (1c) These approximations were found to apply to both the SM and AM. \u201ea\u2026 Two Components of Metabolic Cost. We propose a general model of metabolic cost of transport ~cost normalized by body weight and distance traveled!, E, that has a component corresponding to the metabolic cost of pushing off with the stance leg ~i.e., toe-off!, E toe , and a component corresponding to forced motion of the swing leg, Eswing : E;E toe1Eswing . (2) This general model can be used to examine the implications of several physiological hypotheses for the cost of tuning the swing leg"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure7.16-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure7.16-1.png",
        "caption": "Figure 7.16 (a) The resultant R of the uniformly distributed load on the outside AB is equal and opposite to (b) the resultant Rsection of an equally large uniformly distributed load on section AB.",
        "texts": [
            " By dividing the body into vertical strips, and looking at the vertical component of the load on the boundary elements, we can similarly deduce that the resulting vertical load on the body is zero. Conclusion: If a uniformly distributed load acts on a plane body in the plane of the body along its entire outline, and everywhere normal to the body, the load forms an equilibrium system with resultant zero. One can show that this is also true in three-dimensional cases: If a uniformly distributed load acts on a body in space on its entire surface, and everywhere normal to the body, the load forms an equilibrium system with resultant zero. b. In Figure 7.16a, the part of the body above section AB has been isolated. Assume the resultant of the distributed load on the outside between A and B is R. Figure 7.15 (a) As a result of the distributed load q, a small force F = q s is acting on a small boundary element with length s. (b) The horizontal components of the load on the boundary elements of a horizontal strip are equal and opposite. Together they form an equilibrium system with zero resultant. 7 Gas Pressure and Hydrostatic Pressure 255 If a uniformly distributed load q is also applied to section AB, as in Figure 7.16b, the total load on the isolated part of the body forms an equilibrium system: the resultant R of the load on the outside of the body is equal and opposite to the resultant Rsection of the load on the section. Therefore R = Rsection = qa, in which a is the length of the section. The line of action of R coincides with the perpendicular bisector of AB. In a fluid at rest, the (all-round or isotropic) pressure increases linearly with depth. This can be derived from the vertical force equilibrium of the fluid column in Figure 7"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003464_027836499301200101-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003464_027836499301200101-Figure2-1.png",
        "caption": "Fig. 2. Peg-in-the-hole geometry. Frame PI is at the peg center of mass and parallel to the principal axes of the",
        "texts": [
            " The point here is that a different solution for q is obtained without physically changing the problem itself! Example 1.2: Defining the Position/Force Controlled Degrees of Freedom Within the Hybrid Control Scheme In the hybrid control scheme (Mason 1978; Asada and Slotine 1986), a central role is played by the &dquo;selection matrices&dquo; S and (16 - S), which define the task space degrees of freedom that have to be position (velocity) or force controlled. For example, in the peg-in-the-hole task, given the reference frame Fi of Figure 2, the motion along and the rotation about the zl-axis are position controlled, while forces/torques along the other axes have to be controlled (presumably to a zero value). This problem may be formulated as a geometrical projection problem. The twist vector V in space V must be projected onto the subspace of twists-of-freedom V~ with basis B and the frame force (wrench) vector W in space W into a wrench-of-constraint subspace F, with basis A. For the peg\u2019s inertia tensor peg-in-the-hole task these bases for frame F, are In the reference frame Fi, the velocity projection operator matrix P2, and the wrench projection operator matrix P f, perform the &dquo;correct&dquo; selection of the motion and force frame vectors along the directions they can be controlled. The projection operator matrix for frame velocities or twists is the well-known filter matrix If a translation of the reference frame is performed along the xl-axes (Fig. 2), a new reference frame F2 is obtained. In this new frame the base matrices B and A transform as with at MCMASTER UNIV LIBRARY on April 2, 2015ijr.sagepub.comDownloaded from 4 and therefore, The projection operator matrix 2p v for velocities computed by - - - - . - hence, performing a different, physically inconsistent projection on the frame velocities expressed in F2. The &dquo;selected&dquo; twist vectors are not the same as before, and the actual numerical result depends on the value of the translation r, (Lipkin and Duffy 1988; Melchiorri 1990a)",
            " The kinetic energy of the body computed in F\u2019 is The kinetic energy of the body with respect to an inertial frame is gauge invariant, since one can show that the kinetic energy metric satisfies the condition MG, = G~MvG. Thus, for all frames F and F\u2019 related by a translation and a rotation. Examples The following examples illustrate some of the ideas presented in this article. They are deliberately simple to clarify points of the theory without confusing the issues with numerical complexity. Example 5.1: Transformation of the Kinetic Energy Metric Consider the peg in Figure 2. The kinetic energy metric Mz, for the frame Fi , located at the center mass and parallel to the peg\u2019s principal axes, is given by the matrix of (83), where m is the total mass of the peg, 13 is the 3 x 3 identity matrix, and 1m the diagonal inertia matrix of the peg. This same metric, when calculated at the origin of frame F2, which is parallel to Fj and at a distance r along the z i -axes of Fl, transforms to where G is given by (87) with R = I and The kinetic energy metric in the new frame, therefore, is Observe that, in general, the kinetic energy metric of a rigid body contains zero and first- and second-order mass moments",
            " Unfortunately, this set of &dquo;eigenvalues,&dquo; which is the union of the eigenvalues of the matrix mI3 and the inertia matrix I&dquo;,, considered separately, cannot be eigenvalues of the coupled system (91) when L # 4~3, Therefore, such &dquo;eigenvalues&dquo; are not invariants under general rigid-body transformations where translation of the reference frame occurs. at MCMASTER UNIV LIBRARY on April 2, 2015ijr.sagepub.comDownloaded from 16 Example 5.2: Peg-in-the-Hole: Inverse Velocit)\u00b0 at the Center of Mass With reference to Figure 2, F, is assumed to be the center-of-mass frame of the peg along its principal axes as in Example 5.1. The expression for the peg-in-the-hole Jacobian (Doty et al. 1990) in this frame computes to and equils, as it should, the matrix \u2019B in (4). The motion V of the peg is constrained by V = Jq, where q = [ d B ]T are the permitted degrees of freedom in generalized coordinates. Because this Jacobian has fullcolumn rank, solutions will be invariant to the kinetic energy metric Mq used for the joint rates (Fact 2)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002549_j.matdes.2019.108076-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002549_j.matdes.2019.108076-Figure2-1.png",
        "caption": "Fig. 2. Topology optimization of satellite antenna bracket structure.",
        "texts": [
            " In 2015, Airbus Defence and Space's UK division used the laser additive manufacturing technology to manufacture a structural bracket (Fig. 1) for Eurostar E3000, an aluminium made telecommunications satellites. It is a single laser melted part weighing 35% less than the previous bracket, which comprised with four pieces and 44 rivets. The additive layer manufactured (ALM) bracket is also 40% stiffer than the previous heavier version and does it not generate waste produced under conventional machining process. Fig. 2 shows the Sentinel 1 Upper S-Band Antenna bracket which was developed by RUAG Space and the European Space Agency (ESA). By using topological optimization and laser additive manufacturing process, the weight of antenna bracket was reduced from 1.626 kg to 0.936 kg, achieving a weight reduction of 42%. However, compared with the general topology optimized structure, the lattice structure exhibits a combination of high performance features such as high strength, low mass, low thermal conductivity and excellent toughness, sound absorption and vibration damping properties [36], which can meet the design and versatility requirements of ultra-light components [13,14]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000957_robot.2009.5152561-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000957_robot.2009.5152561-Figure4-1.png",
        "caption": "Fig. 4. Free body diagram showing forces and moments on a quadrotor helicopter relative to both the body and inertial frames of reference.",
        "texts": [
            " This section will discuss the inertial dynamics of the quadrotor, the existing attitude and altitude control scheme on STARMAC, and present the adjustments made to compensate for the aerodynamic effects discussed above. A. Inertial Dynamics The thrust produced by the jth rotor acts perpendicularly to the rotor plane along the zRj axis, which defines the axis of the rotor relative to the vehicle (note that this may change in flight). The vehicle body drag force is Db # v2 !, vehicle mass is m, acceleration due to gravity is g, and the inertia matrix is IB $ R3\"3. A free body diagram is depicted in Figure 4. The total force F on the vehicle is F = !Dbev + mgeD + 4# j=1 $ !TjRRj ,IzRj % (7) where RRj ,I is the rotation matrix from the plane of rotor j to inertial coordinates and ev and eD are the body velocity and down directions. The total moment on the vehicle M, is M = 4# j=1 $ Mj + Mbf,j + rj % (!TjRRj ,BzRj ) % (8) where RRj ,B is the rotation matrix from the plane of rotor j to body coordinates and rj is the vector from the c.g. to each rotor. Mj and Mbf,j are the reaction torque and flapping moment from each rotor, respectively"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003578_mech-34238-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003578_mech-34238-Figure2-1.png",
        "caption": "Figure 2. Singularities of a 3\u2013UPU Parallel Manipulator.",
        "texts": [
            "org/ on 02/03/2016 Term where Jx b1 s3 1 T sT 3 1 b2 s3 2 T sT 3 2 b3 s3 3 T sT 3 3 q\u0307 d\u03073 1 d\u03073 2 d\u03073 3 T Equation (14) represents the actuation effect. Complementing Eq. (14) with (11) yields q\u0307o J $p (15) where J b1 s3 1 T sT 3 1 b2 s3 2 T sT 3 2 b3 s3 3 T sT 3 3 nT 1 01 3 nT 2 01 3 nT 3 01 3 q\u0307o d\u03073 1 d\u03073 2 d\u03073 3 0 0 0 T The first three rows of J represent three forces of actuation acting along the three limbs. Obviously, the manipulator will be under singularity if these three forces lie on a common plane, such as the case when the moving platform falls on top of the fixed base, or are parallel to each other as shown in Fig. 2a. This singular configuration was discussed by Tsai and Joshi [12]. On the other hand, the last three rows of J represent three moments of constraint, each being perpendicular to the universal joint of a limb. Hence, the manipulator will be under singularity if these three moments either lie on a common plane or are parallel to one another as shown in Fig. 2a and 2b. These conditions were presented by Di Gregorio and Parenti-Castelli [2] and Zlatanov et al. [17]. THE 3\u2013RPS PARALLEL MANIPULATOR Figure 3 shows the 3\u2013RPS parallel manipulator which consists of a moving platform, a fixed base, and three limbs of identical kinematic structure. Each limb connects the fixed base to the moving platform by a revolute joint followed by a prismatic joint and a spherical joint. A linear actuator drives each of the three prismatic joints [9]. The connectivity of each limb is equal to 5"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000448_s11740-009-0192-y-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000448_s11740-009-0192-y-Figure4-1.png",
        "caption": "Fig. 4 Base-plate, supports and cantilever geometry for the simulation",
        "texts": [
            "1 Simulation model For the geometry generation, a specific interface between the SLM manufacturing system and the numerical simulation in ANSYS 11.0 is used. Thereby, the scanning pattern serves as a base for the selection of corresponding nodes and elements. Accordingly, the underlying method allows the representation of the exact part orientation and the single layers due to the direct access to the manufacturing systems control unit. Hence, supports are recognized as well through the algorithm. Figure 4 displays the geometry consisting of a t-shaped cantilever, accordant supports and the structural panel. The geometry in the simulation model and the produced parts in the validation (cp. chapter 3) is nearly identical with the exception of the supports. Since there is currently no method to represent supports by means of single finite elements, they are modelled as small pillars. Within every pillar, the material properties are matched in correlation with the real support parameters (e.g. thermal conductivity, density and stiffness)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure6.42-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure6.42-1.png",
        "caption": "FIGURE 6.42. A separate illustration of the input, output, and the cross links for a universal joint.",
        "texts": [
            " The yokes are connected by a rigid cross-link. The ends of the cross-link are set in bearings in the yokes. When the driver yoke turns, the cross-link rotates relative to the yoke about its axis AB. Similarly, the cross-link rotates about the axis CD and relative to the driven yoke. Although the driver and driven shafts make a complete revolution at the same time, the velocity ratio is not constant throughout the revolution. A separate illustration of the input, output, and the cross links are shown in Figure 6.42. The angular velocity of the cross-link may be shown by 1\u03c93 = 1\u03c92 + 1 2\u03c93 = 1\u03c94 + 1 4\u03c93 (6.289) 366 6. Applied Mechanisms where, 1\u03c92 is the angular velocity of the driver yoke about the x2-axis and 12\u03c93 is the angular velocity of the cross-link about the axis AB relative to the drive yoke expressed in the ground coordinate frame. Figure 6.43 shows that the unit vectors j\u03022 and j\u03023 are along the arms of the cross link, and the unit vectors \u0131\u03022 and \u0131\u03024 are along the shafts. Having the angular velocity vectors, 1\u03c92 = \u23a1\u23a3 \u03c921 \u0131\u03021 j\u03021 k\u03021 \u23a4\u23a6 = \u23a1\u23a3 \u03c921 0 0 \u23a4\u23a6 (6"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure5.13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure5.13-1.png",
        "caption": "FIGURE 5.13. Geometric interpretation of rigid body velocity.",
        "texts": [
            " Applied Kinematics The velocity of the point P in G is GvP = Gr\u0307P = GR\u0307B BrP + Gd\u0307B = G\u03c9\u0303B G BrP + Gd\u0307B = G\u03c9\u0303B \u00a1 GrP \u2212 GdB \u00a2 + Gd\u0307B = G\u03c9B \u00d7 \u00a1 GrP \u2212 GdB \u00a2 + Gd\u0307B . (5.324) Proof. Direct differentiating shows GvP = Gd dt GrP = Gr\u0307P = Gd dt \u00a1 GRB BrP + GdB \u00a2 = GR\u0307B BrP + Gd\u0307B. (5.325) The local position vector BrP can be substituted from (5.323) to obtain GvP = GR\u0307B GRT B \u00a1 GrP \u2212 GdB \u00a2 + Gd\u0307B = G\u03c9\u0303B \u00a1 GrP \u2212 GdB \u00a2 + Gd\u0307B = G\u03c9B \u00d7 \u00a1 GrP \u2212 GdB \u00a2 + Gd\u0307B. (5.326) It may also be written using relative position vector GvP = G\u03c9B \u00d7 G BrP + Gd\u0307B . (5.327) 5. Applied Kinematics 269 Example 189 Geometric interpretation of rigid body velocity. Figure 5.13 illustrates a body point P of a moving rigid body. The global velocity of point P GvP = G\u03c9B \u00d7 G BrP + Gd\u0307B is a vector addition of rotational and translational velocities, both expressed in the global frame. At the moment, the body frame is assumed to be coincident with the global frame, and the body frame has a velocity Gd\u0307B with respect to the global frame. The translational velocity Gd\u0307B is a common property for every point of the body, but the rotational velocity G\u03c9B \u00d7 G BrP differs for different points of the body"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003657_02783649922066475-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003657_02783649922066475-Figure7-1.png",
        "caption": "Fig. 7. Robot frames, translation, and turning mechanisms (plan view). The robot body consists of three \u201cframes\u201d: an outer frame comprised of two leg towers, a mid-frame that houses the translation drive, and an inner frame that houses the turning drive as well as the winch and electronics enclosure.",
        "texts": [
            " Though the legs are attached to the frames in fixed rectangular constellations, each leg can individually move its foot up and down with its pantograph actuator. All of the legs can thus adjust lengths to best suit the terrain and objectives. For instance, if walking on a side slope, the legs adjust their lengths differentially so the body remains level. In another common situation, where the terrain slope changes, legs individually change length to adjust the robot\u2019s pitch as it crosses the region of slope change. Decoupled translation and rotation of the inner and outer frames is achieved with the mechanism shown in Figure 7. The outer frame is composed of a transverse middle frame section, as well as left and right leg towers, each of which provide attachment for a pair of legs. The inner frame rotates on a bearing interface to the mid-frame. The turning actuator is attached to the inner frame, and drives an output pinon gear along a spur gear sector fixed to the mid-frame for a total turning motion of \u00b111\u25e6. Yaw position of the inner frame is measured by a potentiometer mounted within the central rotary bearing. The leg towers (on the outer frame) translate with respect to the mid-frame on bearing rails attached to the underside of the towers"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000047_5.4400-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000047_5.4400-Figure7-1.png",
        "caption": "Fig. 7. Phase plane plot of system with control when a sliding mode exists on the switching surface.",
        "texts": [
            " The \"unstable\" equilibrium point (0,O) i s now a saddle point with asymptotes x2 = 3x1 and x, = -xl, as shown in Fig. 5(b). Observe from the dotted line trajectory that if the state vec- tor i s perturbed below the surface, ul(xl, x,) = slxl + x, = 0, at time to, it circles to the point tl before intercepting the surface again. On the other hand, if the switching surface is u2(xl, x,) = slxl + x, = 0 with s1 < 1, then a perturbation off the surface i s always immediatelyforced back to the surface since the phase-plane velocity vectors always point towards the surface. Fig. 7 illustrates the phenomena. As suggested by Figs. 3, 6, and 7, different choices of switching surfaces produce radically different system responses.The richnessof variablestructurecontrol comes from this ability to choose various controller structures at different points in time. The above example also illustrates an important notion 214 PROCEEDINGS OF THE IEEE, VOL. 76, NO. 3, MARCH 1988 in VSC. For the switching surface, u2(xl, x,) = slxl + x2 = 0 of Fig. 7, once the state trajectory intercepts the surface it remains on the surface for all subsequent time. This property of remainingon the switching surfaceonce intercepted i s called a sliding mode. A sliding mode will exist for a system i f in the vicinity o f the switching surface, the state velocity vector (the derivative o f the state vector) is directed towards the surface. The lack of a sliding mode for the \"a = u1 scenario\" described in the second example disappears when using a full state feedback control law: u(x) = kl(xl, x2)x1 + k,(xl, x2)x2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002360_j.ijheatmasstransfer.2019.05.003-Figure11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002360_j.ijheatmasstransfer.2019.05.003-Figure11-1.png",
        "caption": "Fig. 11. Temperature contour and melt region during the scanning of the first layer of the powder.",
        "texts": [
            " The latter is used as the initial thermal and fluid condition for the next CFD (fusion) model. This procedure is repeated for three layers, based on Fig. 9. The process parameters and the laser specifications used in this study are given in Table 3. The base metal is nickle alloy IN718. The thermal field at t = 1910 ms along with the applied coordinate system for the computational domain are shown in Fig. 10. The domain size is 600 lm 400 lm 150 lm and the hatch spacing is 100 lm. The temperature field during the scanning of the first layer along with the melt region are both shown in Fig. 11 for different times. According to Fig. 11(a), the maximum temperature zone is formed very close to the centre of the moving laser beam. At 400 ls, most of the powder and base-plate are still at their initial temperature of 300 K. Due to the relatively high scanning velocity of the laser and the presence of air between the powders, the flow (propagation) of the heat in the transverse direction is much slower. On the other hand, the thermal resistance of the material (at t >300 ls) on the right side of the scanning beam, is much lower than that of the unmelted powders due to the dense morphology of the metal. This means that the fused material conducts heat much better than the unsintered powder. The melt region contour during the scanning of the first layer is also shown in Fig. 11(b). At t = 860 ls, it can be clearly observed that the melt region of the two neighbouring tracks do not reach each other and accordingly, lack of fusion sites and unsintered particles are formed between the tracks. Corresponding to Fig. 11(c) at t = 1400 ls, a hole is formed close to the position of the centre of the laser beam. When the temperature in the region goes above the saturation temperature of the metal, the recoil pressure becomes dominant and pushes the deformable liquid down and makes a depression zone (hole) [34]. The size of the keyhole increases with further increase in temperature due to greater recoil pressure. Fig. 12 shows the temperature history in the middle of the four scanning tracks, where according to this figure, the temperature of the powder particles, right before coming in direct laser contact, would not be increased",
            " However, while melted, the region\u2019s temperature will instantly get affected by the hot neighbouring tracks. What is also interesting is that the maximum temperature in the next tracks will increase while the laser starts to irradiate them, which is also shown in Fig. 12. This is largely due to the heat accumulating coming from the previously-scanned tracks. Also, that is the main underlying reason behind the fact that the unsintered zones formed in between the tracks, get smaller and shorter in the next tracks [16], which is observed in Fig. 11. In other words, as the laser moves to the next tracks, the unsintered regions formed between the tracks get shorter and start from a further distance from the starting edge, where bigger melt pools are formed [17]. Formation of a bigger melt pool is also noticed for a double-track L-PBF, during the scanning of the second track [27]. The sharper change of temperature-time curves of C and D seen in Fig. 12 is because the larger and deeper melt pool has reached these investigation points. This, once again highlights the fact that the overall temperature of each track increases while the laser moves to the subsequent tracks. This event can fully justify why the unsintered regions get shorter in the subsequent tracks as also shown in Fig. 11. Themelting of powder particles during L-PBF is caused either via direct contactwith the laser raysor via contactwithahot returnflow formed due to the negative pressure gradient present on the fusion front. Startingwith the former, when the laser irradiates the powder for the first time, the powder begins tomelt, as observed in Fig. 13(a) and (b), and gains a high flowability. According to Fig. 13(a) and (b), the velocitymagnitude can even reach 5 m.s 1 duringmelting of the powder (about the same rangewas also reported in [25]), which is a significant value, especiallywhen compared to the overall size of the considered domain"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000637_978-3-642-83006-8-Figure2.46-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000637_978-3-642-83006-8-Figure2.46-1.png",
        "caption": "Fig. 2.46. Compensating movements for the double-support gait upon level ground for ZMP law I and different dura tion of double-support phase",
        "texts": [
            " T e 1 8 = ( 1 , 0, 0) ct - T e 1 9 = (- 1 ,0 ,0 ) ~ T e 2 0 = (- 1 ,0 ,0 ) ~ '\" '\" 127 sagittal (b) plane 128 second step is to calculate the right-hand sides of the differential equations describing the system's equilibrium in both the sagittal and frontal plane. 129 This procedure enables numerical integration of the system in both phases of the gait. As in the previous examples, an iterative procedu re is to be employed for the synthesis of nominal dynamics. Compensating movements for different duration periods p, and ZI1P law (Fig. 2.23) of the double-support phase are presented in Fig. 2.46-2.47. Example 5. This example is related to the modelling of a two-link foot Fig. 2.19. The mechanism consists of 19 links and 19 revolute jOints (Fig. frontal (2.49a) and sagittal (2.49b) plane, respectively. The right foot is modelled by links 2 and 3, the left one by links 13 and 14. The arms are fi xed, i.e. q16=q18=1.862253 [radj and q17=q19=1.5707 [rad]. The conditi ons q8: = q8_q 4 and q9: = q9+q 4 ensure the pelvis (link 8) is in the horizontal position and the legs are parallel. in the frontal plane"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003488_tia.2004.836222-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003488_tia.2004.836222-Figure6-1.png",
        "caption": "Fig. 6. Physical location of windings and flux paths in two-pole SMPM machine.",
        "texts": [
            " From the figure, it can be known that the inductance component is dominant in high-frequency impedance because the high-frequency impedance is increased according to the frequency of the injected signal with the same magnitude. The figure also shows that the variation of the -axis high-frequency impedances is much larger than the variation of the -axis highfrequency impedances when the magnitude of the injected highfrequency signal is varied. These can be explained using the high-frequency flux path and the physical location of windings as in Fig. 6. Fig. 6 shows the physical locations of windings and flux paths in a two-pole SMPM machine. Actual -axis windings are located in the axis, and -axis windings are located in the axis in the rotor reference frame. Because the high-frequency flux passes through the stator leakage path, resultant flux from high-frequency voltage injected on the axis passes as flux B in the figure, and resultant flux from high-frequency voltage injected on the axis passes as flux A in the figure. Therefore, -axis inductance is larger than -axis inductance because flux A passes through the highly saturated part of the stator",
            " The detailed procedure of high-frequency impedance measurement is described in [13]. Fig. 7 shows the measured high-frequency impedance characteristics. Injection conditions of 200 Hz and 80 V\u2013400 Hz have not been performed due to the mechanical resonance problem. Even though these injection conditions are not investigated, the provided measurement results can show a tendency similar to the FEA results. The high-frequency impedances in Fig. 7 are somewhat different from the high-frequency impedances in Fig. 6 in absolute value. This is from the nonideal effect of the PWM inverter system and inevitable errors in digital signal processing. However, the impedance characteristics are the same as the FEA results. As the frequency of injected high-frequency voltage increases, the impedances are also increased. This supports the view that the inductance component is dominant in the high-frequency impedance. At the same frequency, the -axis high-frequency impedances are larger than the -axis high-frequency impedances"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000779_j.phpro.2014.08.120-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000779_j.phpro.2014.08.120-Figure1-1.png",
        "caption": "Fig. 1. Hour-glass shaped specimen for HCF tests (left) and CT-specimen for crack propagation tests (right).",
        "texts": [
            " For relaxation of residual stresses, a post-process heat treatment for 3h at 650\u00b0C in vacuum with subsequent argon cooling was performed. The specimens were built in upright standing position (90\u00b0) with manufacturing layers being perpendicularly orientated to the latter loading direction. Additionally HCF specimens were built with a tilted orientation of 45\u00b0 to the manufacturing plane. To investigate the HCF properties of laser additive manufactured Ti-6Al-4V, hour-glass shaped specimens derived from ASTM E 466 [AST07] (Fig. 1) were produced as blanks as well as net-shaped specimens (\u201cas-built\u201d condition). The blanks were machined and polished according to specification, the net-shaped specimens were kept in untreated surface condition with an average surface roughness of Ra = 12 \u03bcm. For the determination of the fatigue crack growth rates as well as the threshold value of stress intensity range, compact-tension specimens (CT-specimen) according to ASTM E 647 [AST00] were manufactured. Additive manufactured blanks were machined to final geometry according to the ASTM specifications with an initial crack length of 10 mm (Fig. 1). High cycle fatigue tests were performed on a servo hydraulic test rig from MTS equipped with a 100 kN load cell and hydraulic clamping jaws. The tests were undertaken in alignment with DIN 50100 [DIN78] at a load ratio of = 0.1 and a frequency of 50 Hz. Polished and \u201cas-built\u201d specimens in untreated surface conditions were tested to identify the influence of residual porosity and bonding defects as well as surface roughness on the fatigue life of laser additive manufactured Ti-6Al-4V. To create the Kitagawa-Takahashi diagram (see section 3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure14.5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure14.5-1.png",
        "caption": "FIGURE 14.5. A MacPherson suspenssion and its equivalent vibrating system.",
        "texts": [
            " When the mass oscillates with displacement x << 1, the elongation \u03b4 of the spring is \u03b4 \u2248 a b x. (14.32) We may substitute the system with a translational mass-spring system such as shown in Figure 14.4(b). The new system has the same mass m and an 888 14. Suspension Optimization equivalent spring keq. keq = \u00b3a b \u00b42 k (14.33) The equivalent spring provides the same potential energy as the original spring, when the mass moves. V = 1 2 keqx 2 = 1 2 k\u03b42 = 1 2 k \u00b3a b x \u00b42 = 1 2 k \u00b3a b \u00b42 x2 (14.34) Example 498 Equivalent spring and damper for a McPherson suspension. Figure 14.5 illustrates a McPherson strut mechanism and its equivalent vibrating system. We assume the tire is stiff and therefore, the wheel center gets the same motion y. Furthermore, we assume the wheel and and body of the vehicle move only vertically. To find the equivalent parameters for a 1/8 vibrating model, we use m equal to 1/4 of the body mass. The spring k and damper c make an angle \u03b1 with the direction of wheel motion. They are also displaced b\u2212 a from the wheel center. So, the equivalent spring keq and damper ceq are keq = k \u00b3a b cos\u03b1 \u00b42 (14"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure8.9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure8.9-1.png",
        "caption": "Figure 8.9 An extensive, uniformly distributed terrain load p on the ground level increases both the vertical earth pressure and the vertical grain pressure by an amount p. The water pressure does not change.",
        "texts": [
            "7, the contribution of the vertical grain pressure \u03c3g;v in the vertical earth pressure is shown by means of a hatching. If, with fully saturated soil, the water level is raised to above the ground level, this influences the vertical earth pressures, but not the vertical grain pressures (see Figure 8.8): \u03c3e;v = \u03b3wd + \u03b3ez, \u03c3g;v = \u03c3e;v \u2212 \u03c3w = \u03b3wd + \u03b3ez \u2212 \u03b3w(d + z) = (\u03b3e \u2212 \u03b3w)z. (9) An extensive, uniformly distributed load p on the ground level increases both the vertical earth pressure and the vertical grain pressure by an amount p. The water pressure does not change (see Figure 8.9): \u03c3e;v = p + \u03b3ez, \u03c3g;v = \u03c3e;v \u2212 \u03c3w = p + (\u03b3e \u2212 \u03b3w)z. (10) Example Figure 8.10a shows a package with three soil layers. In the figure, the specific weight of the soil is given for each layer. The water level is 3 metres below ground level. A uniformly distributed load of 15 kN/m2 is acting on the ground level. Question: Determine the distribution of the vertical earth pressure and grain pressure. Solution (units in kN and m): First the vertical earth pressure \u03c3e;v = p + \u03b3ez is determined. Second, we determine the water pressure \u03c3w = \u03b3wz"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003848_tnn.2005.863416-Figure9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003848_tnn.2005.863416-Figure9-1.png",
        "caption": "Fig. 9. Two-link robot manipulator.",
        "texts": [
            " Tracking loop controller parameters are chosen as , . Initial conditions are , and the desired trajectory is given by , . The position tracking errors and are shown in Fig. 4 without the saturation compensator. The control input signal is shown in Fig. 5. The NN saturation compensator weights tuning parameters are chosen as (6.3) When the NN saturation compensation is included, the tracking errors are given in Fig. 6, control signal in Fig. 7 and NN output in Fig. 8. Two-Link Robot Arm A planar two-serial link arm is shown in Fig. 9. The model is given in [20], [31]. In order to focus on the effects of actuator saturation nonlinearity, gravity and friction are not included in the system model. Yet, the model contains all nonlinear terms arising in general n-link manipulators [19], [20], [31] [see equation (6.4) at the bottom of the page] where , are angles of joint 1 and joint 2; , are masses of link 1 and link 2; , are lengths of link 1 and link 2. The system parameters are chosen as (6.5) while the parameters of the saturation nonlinearity are (6"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002021_j.addma.2017.08.014-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002021_j.addma.2017.08.014-Figure4-1.png",
        "caption": "Fig. 4 Representative mesh design and heat input modeling of the microscale laser scan model.",
        "texts": [
            " When the maximum temperature of the material is higher than the melting point, but the current temperature is lower than the melting point, the material state is identified as solid. An ABAQUS subroutine USDFLD has been developed to simulate the material state transition (powder-liquid-solid) of the SLM process. \ud835\udc53\ud835\udc5a = { 1, \ud835\udc47\ud835\udc5a\ud835\udc4e\ud835\udc65 < \ud835\udc47\ud835\udc5a\ud835\udc52\ud835\udc59\ud835\udc61 2, \ud835\udc47\ud835\udc50 > \ud835\udc47\ud835\udc5a 3, \ud835\udc47\ud835\udc5a\ud835\udc4e\ud835\udc65 > \ud835\udc47\ud835\udc5a \ud835\udc4e\ud835\udc5b\ud835\udc51 \ud835\udc47\ud835\udc50 < \ud835\udc47\ud835\udc5a (1) 2.4 Microscale laser scan model Model dimensions and mesh: In the microscale laser scan model, the temperature field of the melt pool was calculated using the commercial FEA package ABAQUS/Standard. Fig. 4 shows the mesh design. A half symmetrical model with respect to the X-Z plane was developed to reduce the computational time. The model consists of two parts: the powder layer and the substrate. The powder layer dimensions are 5 mm (length) \u00d7 1 mm (width) \u00d7 0.03 mm (thickness). The substrate Powder fm = 1 Liquid fm = 2 Solid fm = 3 Melt Solidify Re-melt (a) (b) Powder: T max < T melt Liquid: T C > T melt Solid: T max > T melt and T C < T melt T max : Maximum temperature T melt : Melting temperature T C : Current temperature Page 10 of 27 dimensions are 5 mm (length) \u00d7 1 mm (width) \u00d7 2 mm (height)",
            " = \u2212\ud835\udc58 \u2206\ud835\udc47 (2) The heat loss due to heat convection \ud835\udc5e\ud835\udc50 can be given by Eq. 3, where \u210e\ud835\udc50 is the heat convection coefficient, \ud835\udc47 is the melt pool temperature, and \ud835\udc470 is the ambient temperature. \ud835\udc5e\ud835\udc50 = \u210e\ud835\udc50 (\ud835\udc47 \u2212 \ud835\udc470) (3) Heat loss caused by radiation \ud835\udc5e\ud835\udc5f is also included in this model and is determined by Eq. 4, where \ud835\udf0e is the Stefan-Boltzman constant, \ud835\udf16 is the emissivity of the powder bed, and \ud835\udc470 is the ambient temperature. \ud835\udc5e\ud835\udc5f = \ud835\udf0e\ud835\udf16 (\ud835\udc474 \u2212 \ud835\udc470 4) (4) Heat input modeling: The Gaussian distributed moving heat flux (Fig. 4) has been developed to model the heat input of the scanning laser in the microscale model. The heat flux was applied on the top surface of the powder layer via an ABAQUS subroutine DFLUX. The moving track of the heat flux was on the symmetrical plane in X direction as shown in Fig. 4. The assumptions for the symmetrical model were: (a) laser absorption rate and heat conductivity are same for the materials on two sides of the symmetrical plane, (b) half of the heat flux energy is transferred to the symmetrical model. Power intensity of the heat source \ud835\udc39 is determined by laser absorption coefficient \ud835\udc34 (0.09) of the powder materials, laser power \ud835\udc43, laser spot radius \ud835\udc5f, and the coordinates Page 11 of 27 of the laser spot center (x, y). A direct determination of absorption of metal powders is highly dependent on the local powder density, particle size, and laser source [27]. A laser absorption ratio of 0.09 [28] was used for the materials. The heat flux moving along the X direction (Fig. 4) is given by Eq. 5 [29]. \ud835\udc39 = \ud835\udc34\ud835\udc43 \ud835\udf0b \ud835\udc5f2 \ud835\udc52 [\u22122 ( \ud835\udc652+ \ud835\udc662) \ud835\udc5f2 ] (5) Temperature history output: The thermal history of the red dot located at the center of the scan track is shown in Fig. 5. The transient (in millisecond scale) heating and cooling process of the materials is achieved. The temperature of powder materials rises from room temperature to the highest temperature within 0.4 milliseconds which is defined as the total heating time by the moving heat flux (powder\u2019s exposure time to laser). However, a lower temperature peak is expected if the heat source passing a neighboring point of this red dot"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure3.49-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure3.49-1.png",
        "caption": "Figure 3.49 A curved beam AB is loaded at B by the three components of a force.",
        "texts": [
            " For the components of T |A, the moment of F about point A in a xyz coordinate system, we have earlier derived that Tx |A = ryFz \u2212 rzFy, Ty |A = rzFx \u2212 rxFz, Tz|A = rxFy \u2212 ryFx. Here one recognises the moment about three lines through A, parallel to the ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM 3 Statics of a Rigid Body 87 x, y and z axis respectively. Comment: For a moment about the origin O of the coordinate system or a moment about one of the coordinate axes, the point O is generally omitted in the representation of the moment. Example The curved beam AB in Figure 3.49 is loaded at B by a force of which the components are defined with respect to magnitude and direction in the figure. Question: Find the moment about the x, y and z axis respectively of the force(s) at B. Solution: Tx = +(25 kN)(3 m) \u2212 (50 kN)(1 m) = +25 kNm, Ty = +(40 kN)(1 m) \u2212 (25 kN)(2 m) = \u221210 kNm, Tz = +(50 kN)(2 m) \u2212 (40 kN)(3 m) = \u221220 kNm. Two parallel forces that are equal and opposite form a couple (see Section 3.1.4). Figure 3.50 shows two forces F1 = F and F2 = \u2212 F , forming a couple in space"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure4.17-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure4.17-1.png",
        "caption": "Fig. 4.17 3D linear inverted pendulum. The center of mass moves on the constraint plane by properly controlling the kick force. Inclination of the constraint plane does not affect to the horizontal motion of the CoM.",
        "texts": [
            " For a 3D inverted pendulum, we introduce a constraint plane defined as z = kxx+ kyy + zc, (4.33) where kx, ky determine the slope and zc determines the height of the constraint plane. To let the CoM move along this plane, we need its acceleration be orthogonal to the normal vector of the constraint. Therefore, we need [f( x r ) f( y r ) f( z r )\u2212Mg] \u23a1 \u23a3 \u2212kx \u2212ky 1 \u23a4 \u23a6 = 0. (4.34) By solving this equation for f , and substituting into (4.33), we obtain f = Mgr zc . (4.35) The center of mass moves on the constraint plane by applying the kick force f in proportion to the leg length r. Figure 4.17 shows an image of this motion. The horizontal dynamics of the CoM can be derived from (4.30) and (4.31) by substituting the kick force of (4.35) to obtain x\u0308 = g zc x, (4.36) y\u0308 = g zc y. (4.37) These are linear equations having zc, the intersection of the constraint plane as the only parameter. The inclination parameters kx, ky of the constraint plane do not affect the horizontal motion of the CoM since they are not part of (4.36) and (4.37). We call such pendulum as 3D linear inverted pendulum4"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001085_jor.1100060208-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001085_jor.1100060208-Figure2-1.png",
        "caption": "FIG. 2. Stride length versus cadence at slow, normal, and fast walking speeds. lsovelocity curves of the different possible combinations of cadence and stride length to maintain a constant velocity (- - -). Children select a more rapid cadence than do teen, adult, or senior subjects to compensate for their shorter stride.",
        "texts": [
            " As might be expected because of their smaller size, children had a shorter stride length at the customary slow, normal, and fast speeds than those of teens, young adult, or senior subjects (Table 2). Children's average cadences at the three speeds were higher than in any of the other age groups to J Orthop Res, Vol. 6 , No. 2 , 1988 ENERG Y-SPEED RELATIONSHIP 100- - c - .- E . 80- c - > 60- - > V 9 w 217 40- There were no significant differences in the values for rate of oxygen uptake between male and I I I IZ0T P ADULT compensate for their shorter stride (Table 2). Fig. 2 depicts the relationship of cadence to stride length at the slow, normal, and fast walking speeds. This relationship was similar for the groups of teenage, adult, and senior subjects, but was different for children. For any velocity, children assumed a more rapid cadence than those of the other three groups. Rate of Oxygen Uptake i J Orthop Res, Vol. 6 , No. 2, 1988 218 R. L. WATERS ETAL. males walked faster than the females, contributing to the higher oxygen uptake. With respect to age, there was no significant difference at their customary normal speed in the rate of energy expenditure between the groups of young adult and senior subjects, which averaged 12"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003702_s0043-1648(96)07467-4-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003702_s0043-1648(96)07467-4-Figure4-1.png",
        "caption": "Fig. 4. Gear and pinion tooth indented into each other.",
        "texts": [
            " The spring stiffness then becomes: UE C s0.72 (17) S a H where: U4F R ta s (18) H Uy pE 2T1F s (19) t d b b1 y1 1 1UR s q (20)\u017e /R R1 2 d w1R s sin a qy (21)1 12 d w2R s sin a yy (22)2 12 This means that the spring stiffness varies with respect to curvature and tooth force. When the gear teeth are pressed together they deform so that they conformwithin the contact. When determining the contact pressure the deformation of each spring, uzi, is calculated from the overlap of freely interpenetrating surfaces see Fig. 4. A point on one of the interacting surfaces will unlikely have the same x-value as a point on the opposite surface. Therefore, the overlap for a point Pi is calculated relative to an imaginary point Pim, on the opposite interacting flank and with the same x-value as the point Pi and with a position in the y-direction determined by linear interpolation.The sliding velocity for each point is equal to: v sabs[(v qv )y ] (23)i p g i For spur gears, the load is transmitted either by one or two tooth pairs in contact"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure2.19-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure2.19-1.png",
        "caption": "FIGURE 2.19. 180deg sliding rotation of a rear-wheel-locked car.",
        "texts": [
            " When the rear wheels lock, they slide on the road and they lose their capacity to support lateral force. The resultant shear force at the tireprint of the rear wheels reduces to a dynamic friction force in the opposite direction of the sliding. A slight lateral motion of the rear wheels, by any disturbance, develops a yaw motion because of unbalanced lateral forces on the front and rear wheels. The yaw moment turns the vehicle about the z-axis until the rear end leads the front end and the vehicle turns 180 deg. Figure 2.19 illustrates a 180 deg sliding rotation of a rear-wheel-locked car. 74 2. Forward Vehicle Dynamics The lock-up of the front tires does not cause a directional instability, although the car would not be steerable and the driver would lose control. 2.7 F Vehicles With More Than Two Axles If a vehicle has more than two axles, such as the three-axle car shown in Figure 2.20, then the vehicle will be statically indeterminate and the normal forces under the tires cannot be determined by static equilibrium equations"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000958_j.matdes.2013.03.056-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000958_j.matdes.2013.03.056-Figure2-1.png",
        "caption": "Fig. 2. The schematic of selective laser melting process.",
        "texts": [
            " The processes of SLM include: (i) a thin layer of metal powder is deposited by the recoater; (ii) the metal powder layer is melted by the incident laser beam to bond with the already solidified areas of the part underneath; (iii) the building platform is lowered by one layer thickness for the next powder deposition and laser exposure. With the repeat of above actions, the full part can be built layer by layer until the required shape is obtained. The contents of oxygen and H2O in the building chamber are monitored by an oxygen sensor and a H2O sensor in real time, respectively. During SLM experiments, the contents of oxygen and water are both controlled below 50 ppm in an argon environment. The schematic of selective laser melting process is shown in Fig. 2. The range of SLM parameters used in this study is presented in Table 2. Tensile testpieces were designed according to Chinese GB/T 228-2002 standard [16] (nearly equivalent with ISO 6892-1998 [17]), as shown in Fig. 3a. During SLM experiments, the 3D model of tensile sample was first imported and sliced by processing software to produce laser scanning paths. Then the samples were fabricated layer by layer by laser melting of 304 steel powders selectively based on Q235 steel substrate in the building container"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure12.20-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure12.20-1.png",
        "caption": "Figure 12.20 The interaction forces between joint C and the isolated parts AC, BC and DC.",
        "texts": [
            "18 The shape of the V diagram can also be found by plotting the successive step changes due to the point loads at the beam: (a) from left to right or (b) from right to left. Note that, between the two successive zero moments, in AS and SC, M is zero, and therefore the corresponding area of the V diagram is also zero (rule 12). It is left to the reader to check this. Example 3 The support reactions and the interaction forces at joint C for the structure in Figure 12.19a were calculated in Section 5.1, Example 5 (see Figure 12.20). Question: Determine the M , V and N diagrams. 500 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Solution: If we determine the interaction forces at joint E, it is possible to plot the M , V and N diagrams for the entire structure. As there are no distributed loads, the bending moment varies linearly along all members, and the shear force and normal force are constant in all members (rule 1). The M diagram is shown in Figure 12.19b. Since the variation of M along all members is linear, it is sufficient to determine the bending moments at the member ends to get the M diagram. The bending moments M (CA) C , M (CB) C and M (CE) C at joint C have already been calculated1 (see Figure 12.20). M (ED) E follows from the moment equilibrium of the isolated part ED (see Figure 12.21): M (CE) E = M (ED) E with tension on the upper side of ED. This value in the M diagram is therefore plotted at the upper side. The moment equilibrium of joint E in Figure 12.21 gives M (CE) E = M (ED) E . The bending moment \u201cgoes round the corner\u201d. This is emphasised in the M diagram in Figure 12.19b by means of a dotted arc at joint E. 1 The upper index refers to the member in which the bending moment acts and the lower index refers to the location"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure3.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure3.2-1.png",
        "caption": "Fig. 3.2 Support Polygon",
        "texts": [
            "1007/978-3-642-54536-8_3, c\u00a9 Springer-Verlag Berlin Heidelberg 2014 70 3 ZMP and Dynamics In Fig. 3.1 an example of force distribution across the foot is given. As the load has the same sign all over the surface, it can be reduced to the resultant force R, the point of attack of which will be in the boundaries of the foot. Let the point on the surface of the foot, where the resultant R passed, be denoted as the zero-moment point, or ZMP in short. 2 ZMP and Support Polygon We explain the support polygon which is another important concept related to the ZMP. As shown in Fig. 3.2, let us consider the region formed by enclosing all the contact points between the robot and the ground by using an elastic cord braid. We call this region as the support polygon. Mathematically the support polygon is defined as a convex hull, which is the smallest convex set including all contact points. Definitions of the convex set and the convex hull are explained in the appendix of this chapter. Rather than detailed discussions, we first show a simple and important relationship between the ZMP and the support polygon, i"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002415_tac.2019.2906931-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002415_tac.2019.2906931-Figure4-1.png",
        "caption": "Fig. 4. Pendulum",
        "texts": [
            " The lower bound of k\u03b8 can be calculated with \u03b4 being known and therefore the designed adaptive controller is implementable. For simplicity, we let \u0393 = \u03b3I . The upper bound of the overall stabilization errors in the mean square sense of (54) can be decreased by increasing \u03b3 and cn. Remark 11: The obtained bounds in Lemma 1 and Theorem 1 depend on the quantization bound \u03b4. To reduce the conservatism of the results, we can design a quantizer by choosing a small quantization density. In this section we consider a pendulum system from [19] as shown in Figure 4. The equation of the motion for the pendulum system is represented as ml\u03b8\u0308 +mgsin(\u03b8) + kl\u03b8\u0307 = u(t) (72) where \u03b8 denotes the angle of the pendulum, m, l and g are the mass [kg], the length of the robe [m], and the acceleration due to the gravity, k is an unknown friction coefficient, u represents an input torque provided by a DC motor. The states \u03b8 and \u03b8\u0307 are quantized by a quantizer satisfying the bounding property (6). The objective is to design a control input for u to make the output \u03b8 track a reference signal \u03b8r(t) = sin(t)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.43-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.43-1.png",
        "caption": "FIGURE 7.43. A multi-link steering mechanism that must be optimized by varying x.",
        "texts": [
            " An isosceles trapezoidal linkage is 7. Steering Dynamics 427 not as exact as the Ackerman steering at every arbitrary turning radius, however, it is simple enough to be mass produced, and exact enough work in street cars. Example 295 F Optimization of a multi-link steering mechanism. Assume that we want to design a multi-link steering mechanism for a vehicle with the following dimensions. w = 2.4m l = 4.8m (7.153) a2 = 0.45l Due to space constraints, the position of some joints of the mechanism are determined as shown in Figure 7.43. However, we may vary the length x to design the best mechanism according to the Ackerman condition. cot \u03b42 \u2212 cot \u03b41 = w l = 1 2 (7.154) The steering wheel input \u03b4s turns the triangle PBC which turns both the left and the right wheels. The vehicle must be able to turn in a circle with radius Rm. Rm = 10m (7.155) 428 7. Steering Dynamics The minimum turning radius determines the maximum steer angle \u03b4 Rm = q a22 + l2 cot2 \u03b4M 10 = q (0.45\u00d7 4.8)2 + 4.82 cot2 \u03b4M \u03b4M = 0.23713 rad \u2248 13.587 deg (7.156) where \u03b4 is the cot-average of the inner and outer steer angles. Having R and \u03b4 is enough to determine \u03b4o and \u03b4i. R1 = l cot \u03b4M = 4.8 cot 0.23713 = 19.861m (7.157) \u03b4i = tan\u22121 l R1 \u2212 w 2 = 0.25176 rad \u2248 14.425 deg (7.158) \u03b4o = tan\u22121 l R1 + w 2 = 0.22408 rad \u2248 12.839 deg (7.159) Because the mechanism is symmetric, each wheel of the steering mechanism 7. Steering Dynamics 429 in Figure 7.43 must be able to turn at least 14.425 deg. To be safe, we try to optimize the mechanism for \u03b4 = \u00b115 deg. The multi-link steering mechanism is a six-link Watt linkage. Let us divide the mechanism into two four-bar linkages. The linkage 1 is on the left and the linkage 2 is on the right, as shown in Figure 7.44. We may assume that MA is the input link of the left linkage and PB is its output link. Link PB is rigidly attached to PC, which is the input of the right linkage. The output of the right linkage is ND"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002727_j.matdes.2021.109940-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002727_j.matdes.2021.109940-Figure3-1.png",
        "caption": "Fig. 3. Points measured on the XZ plane of the samples (a) SS-67 and SS-90, (b) SS-0.",
        "texts": [
            " Based on the cosa method, two-dimensional surface-detection technology was used. The diameter of the two-dimensional surface probe was 140 mm. A Cr target ray source, voltage of 30 kV, current of 2 mA, and X-ray incident angle of 35 were used and the diameter of the irradiation spot was 3 mm. As the penetration depth of this method was 15 lm, the obtained residual stress was surface residual stress. For the SS-90 and SS-67 samples, nine equidistant points were selected in the XZ plane for residual stress measurements (Fig. 3a). For the SS-0 samples, six equidistant points were selected in the XZ plane for residual stress measurements owing to dimensional limitations (Fig. 3b). Tensile samples were produced using electrical discharge machining (EDM) from blocks fabricated by SLM according to the standard ASTM E8; the directions of the SLM-built samples for tensile testing are illustrated in Fig. 2e. The samples for tensile testing were machined with a cross-section of 2 6 mm2 and gauge length of 25 mm. The tensile tests were conducted at room temperature using an AG-X100 kN testing machine with loading veloc- ity of 0.2 mm/min. The tensile tests were performed 3 times for each sample under the same conditions, and the average values were calculated and reported"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.43-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.43-1.png",
        "caption": "FIGURE 8.43. Six degrees of freedom of a wheel with respect to a vehicle body.",
        "texts": [
            "5 Suspension Requirements and Coordinate Frames The suspension mechanism should allow a relative motion between the wheel and the vehicle body. The relative motions are needed to pass the road irregularities and steering. To function properly, a suspension mechanism should have some kinematic and dynamics requirements. 8.5.1 Kinematic Requirements To express the motions of a wheel, we attach a wheel coordinate system W (oxwywzw) to the center of the wheel. A wheel, as a rigid body, has six degrees-of-freedom with respect to the vehicle body: three translations and three rotations, as shown in Figure 8.43. 486 8. Suspension Mechanisms The axes xw, yw, and zw indicate the direction of forward, lateral, and vertical translations and rotations. In the position shown in the figure, the rotation about the xw-axis is the camber angle, about the yw-axis is the spin, and about the zw-axis is the steer angle. Consider a non-steerable wheel. Translation in zw-direction and spin about the yw-axis are the only two DOF allowed for such a wheel. So, we need to take four DOF. If the wheel is steerable, then translation in the zw-direction, spin about the yw-axis, and steer rotation about the zw-axis are the three DOF allowed",
            "3 Wheel, wheel-body, and tire Coordinate Frames Three coordinate frames are employed to express the orientation of a tire and wheel with respect to the vehicle: the wheel frame W , wheel-body frame C, and tire frame T . A wheel coordinate frame W (xw, yw, zw) is attached to the center of a wheel. It follows every translation and rotation of the wheel except the spin. Hence, the xw and zw axes are always in the tire-plane, while the yw-axis is always along the spin axis. A wheel coordinate frame is shown in Figure 8.43. When the wheel is straight and the W frame is parallel to the vehicle coordinate frame, we attach a wheel-body coordinate frame C (xc, yc, zc) at the center of the wheel parallel to the vehicle coordinate axes. The wheelbody frame C is motionless with respect to the vehicle coordinate and does not follow any motion of the wheel. The tire coordinate frame T (xt, yt, zt) is set at the center of the tireprint. The zt-axis is always perpendicular to the ground. The xt-axis is along the intersection line of the tire-plane and the ground"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure15.25-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure15.25-1.png",
        "caption": "Figure 15.25 (a) Hinged beam with the mechanisms for determining (b) the support moment at B, (c) the shear force directly to the left of B and (d) the shear force directly to the right of B.",
        "texts": [
            " After replacing the distributed loads on AC and CB by their resultants, we 15 Virtual Work 735 find the following for the virtual work equation: \u03b4A = \u2212VC \u00b7 a\u03b4\u03b8 \u2212 VC \u00b7 2a\u03b4\u03b8 + qa \u00b7 1 2a\u03b4\u03b8 \u2212 2qa \u00b7 a\u03b4\u03b8 = 0 so that VC = \u2212 1 2qa. The minus sign indicates that the direction of VC is opposite to the direction assumed in Figure 15.24b. In the virtual work equation, the contribution of the shear force VC, regardless of the sign, is equal to \u03b4A(due to VC) = \u201cshear force \u00d7 displacement in the shear force hinge\u201d = VC \u00b7 \u03b4u. The sign is determined by the directions in which we assume the shear force and the virtual displacement. Example 3 \u2013 Forces in a hinged beam For the hinged beam in Figure 15.25a we will look for the mechanisms to determine the support moment at B, the shear force directly to the left of B and the shear force directly to the right of B. Solution: In order to find the support at B we convert the beam into a mechanism by introducing a hinge at B (Figure 15.25b). The bending moment is allowed to act on the mechanism as a load (as a pair of moments). The direction of MB and the direction of the virtual displacement \u03b4\u03b8 can be chosen arbitrarily. Figure 15.25c shows the mechanism for the shear force directly to the left of B. At this point, a slide or shear force hinge has been fitted into the beam. The shear force VB has been applied on the mechanism as a load (a pair of 736 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM forces). At the shear force hinge, segments SB and BC can displace only with respect to one another, and cannot turn with respect to one another. Segments SB and BC therefore remain parallel. BC is fixed in a horizontal position due to the supports at B and C. With the vertical displacement \u03b4u, SB therefore remains horizontal. Figure 15.25d gives the mechanism for the shear force directly to the right of B. Here too, SB and BC remain parallel. Due to the displacement \u03b4u at the shear force hinge, BC undergoes a rotation \u03b4\u03b8 = \u03b4u/2a. SB undergoes the same rotation. If the beam carries a uniformly distributed load q over its entire length (Figure 15.25a) the support moment at B is MB = \u2212qa2. The shear force directly to the left of B is VB = + 3 2qa. The shear force directly to the right of B is VB = \u2212 3 2qa. Check the answers using these mechanisms. Example 4 \u2013 Normal force Here we will derive the normal force in the truss in Figure 15.26a for the member DE using the principle of virtual work. Solution: Convert the truss into a mechanism by introducing a connection in member DE that cannot transfer normal forces. Such a telescopic connection is normal force N is applied to the mechanism as a load (a pair of forces)"
        ],
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    },
    {
        "image_filename": "designv10_0_0001190_1.4828755-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001190_1.4828755-Figure2-1.png",
        "caption": "FIG. 2. Systematic approach: (a) Manufactured twin cantilever with support on the substrate. (b) Separating the twin cantilever from the supports with a saw. (c) Spreading after separation of the supports.",
        "texts": [
            " By choosing a smaller surface than 110 10 mm2, the scan strategy (scan vector length, overlap of stripes, etc.) would not represent the typical scanning process and therefore not be representative for conventionally manufactured SLM components. The distortions that occur through the thermally induced residual stresses should be checked and visualized by separating the twin cantilever from the supports after the SLM process. Therefore, twin cantilevers are built with support structures in the overhang area by means of SLM on a substrate plate [Fig. 2(a)]. The substrate plate consists of the material AlSi1MgMn with a linear thermal expansion coefficient of aAlSi1MgMn\u00bc 23.4 10 6/K. For the substrate plate, a material is selected which has a similar thermal expansion coefficient as the material AlSi10Mg does, so that the distortions of the twin cantilever are not influenced by the shrinkage when the system cools. (The material AlSi10Mg is not available as semifinished product.) After the construction process ends, the twin cantilevers are marked at 12 measurement points along the middle axis of the component at a length of 110 mm and a distance of 10 mm to each other and the z positions are measured [Fig. 2(b), measurement points]. The supports are separated from the substrate plate [Fig. 2(b)], and the measurement points are measured again. The difference of the measured values before and after separation of the supports results in the This article is copyrighted as indicated in the article. Reuse of AIP content is subject to the terms at: http://scitation.aip.org/termsconditions. Downloaded to IP: 147.143.2.5 On: Tue, 23 Dec 2014 13:39:43 amount of spreading in the z direction. From the values identified, the curves can be created in a diagram that illustrates the spreading (distortion) along the geometry of the twin cantilever"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure3.23-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure3.23-1.png",
        "caption": "FIGURE 3.23. Damping structure of a tire.",
        "texts": [],
        "surrounding_texts": [
            "The tire is the main component interacting with the road. The performance of a vehicle is mainly influenced by the characteristics of its tires. Tires affect a vehicle\u2019s handling, traction, ride comfort, and fuel consumption. To understand its importance, it is enough to remember that a vehicle can maneuver only by longitudinal, vertical, and lateral force systems generated under the tires. Figure 3.1 illustrates a model of a vertically loaded stationary tire. To model the tire-road interactions, we determine the tireprint and describe the forces distributed on the tireprint. 3.1 Tire Coordinate Frame and Tire Force System To describe the tire-road interaction and force system, we attach a Cartesian coordinate frame at the center of the tireprint, as shown in Figure 3.2, assuming a flat and horizontal ground. The x-axis is along the intersection line of the tire-plane and the ground. Tire plane is the plane made by narrowing the tire to a flat disk. The z-axis is perpendicular to the ground, opposite to the gravitational acceleration g, and the y-axis makes the coordinate system a right-hand triad. To show the tire orientation, we use two angles: camber angle \u03b3 and sideslip angle \u03b1. The camber angle is the angle between the tire-plane and 96 3. Tire Dynamics the vertical plane measured about the x-axis. The camber angle can be recognized better in a front view as shown in Figure 3.3. The sideslip angle \u03b1, or simply sideslip, is the angle between the velocity vector v and the x-axis measured about the z-axis. The sideslip can be recognized better in a top view, as shown in Figure 3.4. The force system that a tire receives from the ground is assumed to be located at the center of the tireprint and can be decomposed along x, y, and z axes. Therefore, the interaction of a tire with the road generates a 3D force system including three forces and three moments, as shown in Figure 3.2. 1. Longitudinal force Fx. It is a force acting along the x-axis. The resultant longitudinal force Fx > 0 if the car is accelerating, and Fx < 0 if the car is braking. Longitudinal force is also called forward force. 2. Normal force Fz. It is a vertical force, normal to the ground plane. The resultant normal force Fz > 0 if it is upward. Normal force is also called vertical force or wheel load. 3. Lateral force Fy. It is a force, tangent to the ground and orthogonal to both Fx and Fz. The resultant lateral force Fy > 0 if it is in the y-direction. 4. Roll moment Mx. It is a longitudinal moment about the x-axis. The resultant roll moment Mx > 0 if it tends to turn the tire about the x-axis. The roll moment is also called the bank moment, tilting torque, or overturning moment. 3. Tire Dynamics 97 98 3. Tire Dynamics 5. Pitch moment My. It is a lateral moment about the y-axis. The resultant pitch momentMy > 0 if it tends to turn the tire about the y-axis and move forward. The pitch moment is also called rolling resistance torque. 6. Yaw moment Mz. It is an upward moment about the z-axis. The resultant yaw moment Mz > 0 if it tends to turn the tire about the z-axis. The yaw moment is also called the aligning moment, self aligning moment, or bore torque. The moment applied to the tire from the vehicle about the tire axis is called wheel torque T . Example 75 Origin of tire coordinate frame. For a cambered tire, it is not always possible to find or define a center point for the tireprint to be used as the origin of the tire coordinate frame. It is more practical to set the origin of the tire coordinate frame at the center of the intersection line between the tire-plane and the ground. So, the origin of the tire coordinate frame is at the center of the tireprint when the tire is standing upright and stationary on a flat road. Example 76 SAE tire coordinate system. The tire coordinate system adopted by the Society of Automotive Engineers (SAE) is shown in Figure 3.5. The origin of the coordinate system is at the center of the tireprint when the tire is standing stationary. The x-axis is at the intersection of the tire-plane and the ground plane. The z-axis is downward and perpendicular to the tireprint. The y-axis is on the ground plane and goes to the right to make the coordinate frame a righthand frame. The sideslip angle \u03b1 is considered positive if the tire is slipping to the right, and the camber angle \u03b3 is positive when the tire leans to the right. The SAE coordinate system is as good as the coordinate system in Figure 3.2 and may be used alternatively. However, having the z-axis directed downward is sometimes inefficient and confusing. Furthermore, in SAE convention, the camber angle for the left and right tires of a vehicle have opposite signs. So, the camber angle of the left tire is positive when the tire leans to the right and the camber angle of the right tire is positive when the tire leans to the left. 3.2 Tire Stiffness As an applied approximation, the vertical tire force Fz can be calculated as a linear function of the normal tire deflection 4z measured at the tire center. Fz = kz4z (3.1) 3. Tire Dynamics 99 The coefficient kz is called tire stiffness in the z-direction. Similarly, the reaction of a tire to a lateral and a longitudinal force can be approximated by Fx = kx4x (3.2) Fy = ky4y (3.3) where the coefficient kx and ky are called tire stiffness in the x and y directions. Proof. The deformation behavior of tires to the applied forces in any three directions x, y, and z are the first important tire characteristics in tire dynamics. Calculating the tire stiffness is generally based on experiment and therefore, they are dependent on the tire\u2019s mechanical properties, as well as environmental characteristics. Consider a vertically loaded tire on a stiff and flat ground as shown in Figure 3.6. The tire will deflect under the load and generate a pressurized contact area to balance the vertical load. Figure 3.7 depicts a sample of experimental stiffness curve in the (Fz,4z) plane. The curve can be expressed by a mathematical function Fz = f (4z) (3.4) however, we may use a linear approximation for the range of the usual application. Fz = \u2202f \u2202 (4z) 4z (3.5) 100 3. Tire Dynamics F1 < F2 < F3 3. Tire Dynamics 101 The coefficient \u2202f \u2202(4z) is the slope of the experimental stiffness curve at zero and is shown by a stiffness coefficient kz kz = tan \u03b8 = lim 4z\u21920 \u2202f \u2202 (4z) . (3.6) Therefore, the normal tire deflection 4z remains proportional to the vertical tire force Fz. Fz = kz4z (3.7) The tire can apply only pressure forces to the road, so normal force is restricted to Fz > 0. The stiffness curve can be influenced by many parameters. The most effective one is the tire inflation pressure. Lateral and longitudinal force/deflection behavior is also determined experimentally by applying a force in the appropriate direction. The lateral and longitudinal forces are limited by the sliding force when the tire is vertically loaded. Figure 3.8 depicts a sample of longitudinal and lateral stiffness curves compared to a vertical stiffness curve. The practical part of a tire\u2019s longitudinal and lateral stiffness curves is the linear part and may be estimated by linear equations. Fx = kx4x (3.8) Fy = ky4y (3.9) The coefficients kx and ky are called the tire stiffness in the x and y directions. They are measured by the slope of the experimental stiffness curves 102 3. Tire Dynamics in the (Fx,4x) and (Fy,4y) planes. kx = lim 4x\u21920 \u2202f \u2202 (4x) (3.10) ky = lim 4y\u21920 \u2202f \u2202 (4y) (3.11) When the longitudinal and lateral forces increase, parts of the tireprint creep and slide on the ground until the whole tireprint starts sliding. At this point, the applied force saturates and reaches its maximum supportable value. Generally, a tire is most stiff in the longitudinal direction and least stiff in the lateral direction. kx > kz > ky (3.12) Figure 3.9 illustrates tire deformation under a lateral and a longitudinal force. Example 77 F Nonlinear tire stiffness. In a better modeling, the vertical tire force Fz is a function of the normal tire deflection 4z, and deflection velocity 4z\u0307. Fz = Fz (4z,4z\u0307) (3.13) = Fzs + Fzd (3.14) In a first approximation we may assume Fz is a combination of a static and a dynamic part. The static part is a nonlinear function of the vertical tire deflection and the dynamic part is proportional to the vertical speed of the tire. Fzs = k14z + k2 (4z)2 (3.15) Fzd = k3z\u0307 (3.16) 3. Tire Dynamics 103 The constants k1 and k2 are calculated from the first and second slopes of the experimental stiffness curve in the (Fz,4z) plane, and k3 is the first slope of the curve in the (Fz, z\u0307) plane, which indicates the tire damping. k1 = \u2202 Fz \u22024z \u00af\u0304\u0304\u0304 4z=0 (3.17) k2 = 1 2 \u22022 Fz \u2202 (4z) 2 \u00af\u0304\u0304\u0304 \u00af 4z=0 (3.18) k3 = \u2202 Fz \u2202 z\u0307 \u00af\u0304\u0304\u0304 z\u0307=0 (3.19) The value of k1 = 200000N/m is a good approximation for a 205/50R15 passenger car tire, and k1 = 1200000N/m is a good approximation for a X31580R22.5 truck tire. Tires with a larger number of plies have higher damping, because the plies\u2019 internal friction generates the damping. Tire damping decreases by increasing speeds. Example 78 F Hysteresis effect. Because tires are made from rubber, which is a viscoelastic material, the loading and unloading stiffness curves are not exactly the same. They are similar to those in Figure 3.10, which make a loop with the unloading curve below the loading. The area within the loop is the amount of dissipated energy during loading and unloading. As a tire rotates under the weight of a vehicle, it experiences repeated cycles of deformation and recovery, and it dissipates energy loss as heat. Such a behavior is a common property of hysteretic material and is called hysteresis. So, hysteresis is a characteristic of a deformable material such as rubber, that the energy of deformation is greater than the energy of recovery. The amount of dissipated energy depends on the mechanical characteristics of the tire. Hysteretic energy loss in rubber decreases as temperature increases. The hysteresis effect causes a loaded rubber not to rebound fully after load removal. Consider a high hysteresis race car tire turning over road irregularities. The deformed tire recovers slowly, and therefore, it cannot push the tireprint tail on the road as hard as the tireprint head. The difference in head and tail pressures causes a resistance force, which is called rolling resistance. Race cars have high hysteresis tires to increase friction and limit traction. Street cars have low hysteresis tires to reduce the rolling resistance and low operating temperature. Hysteresis level of tires inversely affect the stopping distance. A high hysteresis tire makes the stopping shorter, however, it wears rapidly and has a shorter life time. 104 3. Tire Dynamics 3.3 Tireprint Forces The force per unit area applied on a tire in a tireprint can be decomposed into a component normal to the ground and a tangential component on the ground. The normal component is the contact pressure \u03c3z, while the tangential component can be further decomposed in the x and y directions to make the longitudinal and lateral shear stresses \u03c4x and \u03c4y. For a stationary tire under normal load, the tireprint is symmetrical. Due to equilibrium conditions, the overall integral of the normal stress over the tireprint area AP must be equal to the normal load Fz, and the integral of shear stresses must be equal to zero. Z AP \u03c3z(x, y) dA = Fz (3.20)Z AP \u03c4x(x, y) dA = 0 (3.21)Z AP \u03c4y(x, y) dA = 0 (3.22) 3.3.1 Static Tire, Normal Stress Figure 3.11 illustrates a stationary tire under a normal load Fz along with the generated normal stress \u03c3z applied on the ground. The applied loads on the tire are illustrated in the side view shown in Figure 3.12. For a stationary tire, the shape of normal stress \u03c3z(x, y) over the tireprint area depends on tire and load conditions, however its distribution over the tireprint is generally in the shape shown in Figure 3.13. 3. Tire Dynamics 105 106 3. Tire Dynamics The normal stress \u03c3z(x, y) may be approximated by the function \u03c3z(x, y) = \u03c3zM \u00b5 1\u2212 x6 a6 \u2212 y6 b6 \u00b6 (3.23) where a and b indicate the dimensions of the tireprint, as shown in Figure 3.14. The tireprints may approximately be modeled by a mathematical function x2n a2n + y2n b2n = 1 n \u2208 N. (3.24) For radial tires, n = 3 or n = 2 may be used, x6 a6 + y6 b6 = 1 (3.25) while for non-radial tires n = 1 is a better approximation. x2 a2 = y2 b2 = 1. (3.26) Example 79 Normal stress in tireprint. A car weighs 800 kg. If the tireprint of each radial tire is AP = 4\u00d7a\u00d7b = 4 \u00d7 5 cm \u00d7 12 cm, then the normal stress distribution under each tire, \u03c3z, 3. Tire Dynamics 107 must satisfy the equilibrium equation. Fz = 1 4 800\u00d7 9.81 = Z AP \u03c3z(x, y) dA = Z 0.05 \u22120.05 Z 0.12 \u22120.12 \u03c3zM \u00b5 1\u2212 x6 0.056 \u2212 y6 0.126 \u00b6 dy dx = 1.7143\u00d7 10\u22122\u03c3zM (3.27) Therefore, the maximum normal stress is \u03c3zM = Fz 1.7143\u00d7 10\u22122 = 1.1445\u00d7 10 5 Pa (3.28) and the stress distribution over the tireprint is \u03c3z(x, y) = 1.1445\u00d7 105 \u00b5 1\u2212 x6 0.056 \u2212 y6 0.126 \u00b6 Pa. (3.29) Example 80 Normal stress in tireprint for n = 2. The maximum normal stress \u03c3zM for an 800 kg car having an AP = 4\u00d7 a\u00d7 b = 4\u00d7 5 cm\u00d7 12 cm, can be found for n = 2 as Fz = 1 4 800\u00d7 9.81 = Z AP \u03c3z(x, y) dA = Z 0.05 \u22120.05 Z 0.12 \u22120.12 \u03c3zM \u00b5 1\u2212 x4 0.054 \u2212 y4 0.124 \u00b6 dy dx = 1.44\u00d7 10\u22122\u03c3zM (3.30) 108 3. Tire Dynamics \u03c3zM = Fz 1.44\u00d7 10\u22122 = 1.3625\u00d7 10 5 Pa. (3.31) Comparing \u03c3zM = 1.3625 \u00d7 105 Pa for n = 2 to \u03c3zM = 1.1445 \u00d7 105 Pa for n = 3 shows that maximum stress for n = 2 is \u00a1 1\u2212 1.1445 1.3625 \u00a2 \u00d7100 = 16% more than n = 3. 3.3.2 Static Tire, Tangential Stresses Because of geometry changes to a circular tire in contact with the ground, a three-dimensional stress distribution will appear in the tireprint even for a stationary tire. The tangential stress \u03c4 on the tireprint can be decomposed in x and y directions. The tangential stress is also called shear stress or friction stress. The tangential stress on a tire is inward in x direction and outward in y direction. Hence, the tire tries to stretch the ground in the x-axis and compact the ground on the y-axis. Figure 3.15 depicts the shear stresses on a vertically loaded stationary tire. The force distribution on the tireprint is not constant and is influenced by tire structure, load, inflation pressure, and environmental conditions: The tangential stress \u03c4x in the x-direction may be modeled by the following equation. \u03c4x(x, y) = \u2212\u03c4xM \u00b5 x2n+1 a2n+1 \u00b6 sin2 \u00b3x a \u03c0 \u00b4 cos \u00b3 y 2b \u03c0 \u00b4 n \u2208 N (3.32) 3. Tire Dynamics 109 \u03c4x is negative for x > 0 and is positive for x < 0, showing an inward longitudinal stress. Figure 3.16 illustrates the absolute value of a \u03c4x distribution for n = 1. The y-direction tangential stress \u03c4y may be modeled by the equation \u03c4y(x, y) = \u2212\u03c4yM \u00b5 x2n a2n \u2212 1 \u00b6 sin \u00b3y b \u03c0 \u00b4 n \u2208 N (3.33) where \u03c4y is positive for y > 0 and negative for y < 0, showing an outward lateral stress. Figure 3.17 illustrates the absolute value of a \u03c4y distribution for n = 1. 3.4 Effective Radius Consider a vertically loaded wheel that is turning on a flat surface as shown in Figure 3.18. The effective radius of the wheel Rw, which is also called a rolling radius, is defined by Rw = vx \u03c9w (3.34) where, vx is the forward velocity, and \u03c9w is the angular velocity of the wheel. The effective radius Rw is approximately equal to Rw \u2248 Rg \u2212 Rg \u2212Rh 3 (3.35) and is a number between the unloaded or geometric radius Rg and the loaded height Rh. Rh < Rw < Rg (3.36) 110 3. Tire Dynamics Proof. An effective radius Rw = vx/\u03c9w is defined by measuring a wheel\u2019s angular velocity \u03c9w and forward velocity vx. As the tire turns forward, each part of the circumference is flattened as it passes through the contact area. A practical estimate of the effective radius can be made by substituting the arc with the straight length of tireprint. The tire vertical deflection is Rg \u2212Rh = Rg (1\u2212 cos\u03d5) (3.37) and therefore Rh = Rg cos\u03d5 (3.38) a = Rg sin\u03d5. (3.39) If the motion of the tire is compared to the rolling of a rigid disk with radius Rw, then the tire must move a distance a = Rw\u03d5 for an angular rotation \u03d5. a = Rg sin\u03d5 = Rw\u03d5 (3.40) Hence, Rw = Rg sin\u03d5 \u03d5 . (3.41) Expanding sin\u03d5 \u03d5 in a Taylor series show that Rw = Rg \u00b5 1\u2212 1 6 \u03d52 +O \u00a1 \u03d54 \u00a2\u00b6 . (3.42) 3. Tire Dynamics 111 Using Equation (3.37) we may approximate cos\u03d5 \u2248 1\u2212 1 2 \u03d52 (3.43) \u03d52 \u2248 2 (1\u2212 cos\u03d5) \u2248 2 \u00b5 1\u2212 Rh Rg \u00b6 (3.44) and therefore, Rw \u2248 Rg \u00b5 1\u2212 1 3 \u00b5 1\u2212 Rh Rg \u00b6\u00b6 = 2 3 Rg + 1 3 Rh. (3.45) Because Rh is a function of tire load Fz, Rh = Rh (Fz) = Rg \u2212 Fz kz (3.46) the effective radius Rw is also a function of the tire load. The angle \u03d5 is called tireprint angle or tire contact angle. The vertical stiffness of radial tires is less than non-radial tires under the same conditions. So, the loaded height of radial tires, Rh, is less than the non-radials\u2019. However, the effective radius of radial tires Rw, is closer to their unloaded radius Rg. As a good estimate, for a non-radial tire, Rw \u2248 0.96Rg, and Rh \u2248 0.94Rg, while for a radial tire, Rw \u2248 0.98Rg, and Rh \u2248 0.92Rg. 112 3. Tire Dynamics Generally speaking, the effective radius Rw depends on the type of tire, stiffness, load conditions, inflation pressure, and wheel\u2019s forward velocity. Example 81 Compression and expansion of tires in the tireprint zone. Because of longitudinal deformation, the peripheral velocity of any point of the tread varies periodically. When it gets close to the starting point of the tireprint, it slows down and a circumferential compression results. The tire treads are compressed in the first half of the tireprint and gradually expanded in the second half. The treads in the tireprint contact zone almost stick to the ground, and therefore their circumferential velocity is close to the forward velocity of the tire center vx. The treads regain their initial circumferential velocity Rg\u03c9w after expanding and leaving the contact zone. Example 82 Tire rotation. The geometric radius of a tire P235/75R15 is Rg = 366.9mm, because hT = 235\u00d7 75% = 176.25mm \u2248 6.94 in (3.47) and therefore, Rg = 2hT + 15 2 = 2\u00d7 6.94 + 15 2 = 14.44 in \u2248 366.9mm. (3.48) Consider a vehicle with such a tire is traveling at a high speed such as v = 50m/ s = 180 km/h \u2248 111.8mi/h. The tire is radial, and therefore the effective tire radius Rw is approximately equal to Rw \u2248 0.98Rg \u2248 359.6mm. (3.49) After moving a distance d = 100 km, this tire must have been turned n1 = 44259 times because n1 = d \u03c0D = 100\u00d7 103 2\u03c0 \u00d7 359.6\u00d7 10\u22123 = 44259. (3.50) Now assume the vehicle travels the same distance d = 100 km at a low inflation pressure, such that the effective radius of the tire remained close to at the loaded radius Rw \u2248 Rh \u2248 0.92Rg = 330.8mm. (3.51) 3. Tire Dynamics 113 This tire must turn n2 = 48112 times to travel d = 100km, because, n2 = d \u03c0D = 100\u00d7 103 2\u03c0 \u00d7 330.8\u00d7 10\u22123 = 48112. (3.52) Example 83 F Radial motion of tire\u2019s peripheral points in the tireprint. The radial displacement of a tire\u2019s peripheral points during road contact may be modeled by a function d = d (x, \u03b8) . (3.53) We assume that a peripheral point of the tire moves along only the radial direction during contact with the ground, as shown in Figure 3.19. Let\u2019s show a radius at an angle \u03b8, by r = r (x, \u03b8). Knowing that cos \u03b8 = Rh r (3.54) cos\u03c6 = Rh Rg (3.55) we can find r = Rg cos\u03c6 cos \u03b8 . (3.56) 114 3. Tire Dynamics Thus, the displacement function is d = Rg \u2212 r (x, \u03b8) = Rg \u00b5 1\u2212 Rh Rg cos \u03b8 \u00b6 \u2212 \u03c6 < \u03b8 < \u03c6. (3.57) Example 84 Tread travel. Let\u2019s follow a piece of tire tread in its travel around the spin axis when the vehicle moves forward at a constant speed. Although the wheel is turning at constant angular velocity \u03c9w, the tread does not travel at constant speed. At the top of the tire, the radius is equal to the unloaded radius Rg and the speed of the tread is Rg\u03c9w relative to the wheel center. As the tire turns, the tread approaches the leading edge of tireprint, and slows down. The tread is compacted radially, and gets squeezed in the heading part of the tireprint area. Then, it is stretched out and unpacked in the tail part of the tireprint as it moves to the tail edge. In the middle of the tireprint, the tread speed is Rh\u03c9w relative to the wheel center. The variable radius of a tire during the motion through the tireprint is r = Rg cos\u03c6 cos \u03b8 \u2212 \u03c6 < \u03b8 < \u03c6 (3.58) where \u03c6 is the half of the contact angle, and \u03b8 is the angular rotation of the tire, as shown in Figure 3.19. The angular velocity of the tire is \u03c9w = \u03b8\u0307 and is assumed to be constant. Then, the radial velocity r\u0307 and acceleration r\u0308 of the tread with respect to the wheel center are r\u0307 = Rg\u03c9w cos\u03c6 sin \u03b8 cos2 \u03b8 (3.59) r\u0308 = 1 2 Rg\u03c9 2 w cos\u03c6 cos3 \u03b8 (3\u2212 cos 2\u03b8) . (3.60) Figure 3.20 depicts r, r\u0307, and r\u0308 for a sample car with the following data: Rg = 0.5m (3.61) \u03c6 = 15deg (3.62) \u03c9w = 60 rad/ s (3.63) 3.5 Rolling Resistance A turning tire on the ground generates a longitudinal force called rolling resistance. The force is opposite to the direction of motion and is proportional to the normal force on the tireprint. Fr = \u2212Fr \u0131\u0302 (3.64) Fr = \u03bcr Fz (3.65) 3. Tire Dynamics 115 The parameter \u03bcr is called the rolling friction coefficient. \u03bcr is not constant and mainly depends on tire speed, inflation pressure, sideslip and camber angles. It also depends on mechanical properties, speed, wear, temperature, load, size, driving and braking forces, and road condition. Proof.When a tire is turning on the road, that portion of the tire\u2019s circumference that passes over the pavement undergoes a deflection. Part of the energy that is spent in deformation will not be restored in the following relaxation. Hence, a change in the distribution of the contact pressure makes normal stress \u03c3z in the heading part of the tireprint be higher than the tailing part. The dissipated energy and stress distortion cause the rolling resistance. Figures 3.21 and 3.22 illustrate a model of normal stress distribution across the tireprint and their resultant force Fz for a turning tire. Because of higher normal stress in the front part of the tireprint, the resultant normal force moves forward. Forward shift of the normal force makes a resistance moment in the \u2212y direction, opposing the forward rotation. Mr = \u2212Mr j\u0302 (3.66) Mr = Fz\u2206x (3.67) The rolling resistance momentMr can be substituted by a rolling resistance 116 3. Tire Dynamics 3. Tire Dynamics 117 force Fr parallel to the x-axis. Fr = \u2212Fr \u0131\u0302 (3.68) Fr = 1 Rh Mr = \u2206x Rh Fz (3.69) Practically the rolling resistance force can be defined using a rolling friction coefficient \u03bcr. Fr = \u03bcr Fz (3.70) Example 85 A model for normal stress of a turning tire. We may assume that the normal stress of a turning tire is expressed by \u03c3z = \u03c3zm \u00b5 1\u2212 x2n a2n \u2212 y2n b2n + x 4a \u00b6 (3.71) where n = 3 or n = 2 for radial tires and n = 1 for non-radial tires. We may determine the stress mean value \u03c3zm by knowing the total load on the tire. As an example, using n = 3 for an 800 kg car with a tireprint AP = 4\u00d7 a\u00d7 b = 4\u00d7 5 cm\u00d7 12 cm, we have Fz = 1 4 800\u00d7 9.81 = Z AP \u03c3z(x, y) dA = Z 0.05 \u22120.05 Z 0.12 \u22120.12 \u03c3zm \u00b5 1\u2212 x6 0.056 \u2212 y6 0.126 + x 4\u00d7 0.05 \u00b6 dy dx = 1.7143\u00d7 10\u22122\u03c3zm (3.72) and therefore, \u03c3zm = Fz 1.7143\u00d7 10\u22122 = 1.1445\u00d7 10 5 Pa (3.73) Example 86 Deformation and rolling resistance. The distortion of stress distribution is proportional to the tire-road deformation that is the reason for shifting the resultant force forward. Hence, the rolling resistance increases with increasing deformation. A high pressure tire on concrete has lower rolling resistance than a low pressure tire on soil. To model the mechanism of dissipation energy for a turning tire, we assume there are many small dampers and springs in the tire structure. Pairs of parallel dampers and springs are installed radially and circumstantially. Figures 3.23 and 3.24 illustrate the damping and spring structure of a tire. 118 3. Tire Dynamics 3. Tire Dynamics 119 3.5.1 F Effect of Speed on the Rolling Friction Coefficient The rolling friction coefficient \u03bcr increases with a second degree of speed. It is possible to express \u03bcr = \u03bcr(vx) by the function \u03bcr = \u03bc0 + \u03bc1 v 2 x. (3.74) Proof. The rolling friction coefficient increases by increasing speed experimentally. We can use a polynomial function \u03bcr = nX i=0 \u03bci v i x (3.75) to fit the experimental data. Practically, two or three terms of the polynomial would be enough. The function \u03bcr = \u03bc0 + \u03bc1 v 2 x (3.76) is simple and good enough for representing experimental data and analytic calculation. The values of \u03bc0 = 0.015 (3.77) \u03bc1 = 7\u00d7 10\u22126 s2/m2 (3.78) are reasonable values for most passenger car tires. However, \u03bc0 and \u03bc1 should be determined experimentally for any individual tire. Figure 3.25 depicts a comparison between Equation (3.74) and experimental data for a radial tire. Generally speaking, the rolling friction coefficient of radial tires show to be less than non-radials. Figure 3.26 illustrates a sample comparison. Equation (3.74) is applied when the speed is below the tire\u2019s critical speed. Critical speed is the speed at which standing circumferential waves appear and the rolling friction increases rapidly. The wavelength of the standing waves are close to the length of the tireprint. Above the critical speed, overheating happens and tire fails very soon. Figure 3.27 illustrates the circumferential waves in a rolling tire at its critical speed. Example 87 Rolling resistance force and vehicle velocity. For computer simulation purposes, a fourth degree equation is presented to evaluate the rolling resistance force Fr Fr = C0 + C1 vx + C2 v 4 x. (3.79) The coefficients Ci are dependent on the tire characteristics, however, the following values can be used for a typical raided passenger car tire: C0 = 9.91\u00d7 10\u22123 C1 = 1.95\u00d7 10\u22125 (3.80) C2 = 1.76\u00d7 10\u22129 120 3. Tire Dynamics Example 88 Road pavement and rolling resistance. The effect of the pavement and road condition is introduced by assigning a value for \u03bc0 in equation \u03bcr = \u03bc0 + \u03bc1 v 2 x. Table 3.1 is a good reference. 122 3. Tire Dynamics Example 89 Tire information tips. A new front tire with a worn rear tire can cause instability. Tires stored in direct sunlight for long periods of time will harden and age more quickly than those kept in a dark area. Prolonged contact with oil or gasoline causes contamination of the rubber compound, making the tire life short. Example 90 F Wave occurrence justification. The normal stress will move forward when the tire is turning on a road. By increasing the speed, the normal stress will shift more and concentrate in the first half of the tireprint, causing low stress in the second half of the tireprint. High stress in the first half along with no stress in the second half is similar to hammering the tire repeatedly. Example 91 F Race car tires. Racecars have very smooth tires, known as slicks. Smooth tires reduce the rolling friction and maximize straight line speed. The slick racing tires are also pumped up to high pressure. High pressure reduces the tireprint area. Hence, the normal stress shift reduces and the rolling resistance decreases. Example 92 F Effect of tire structure, size, wear, and temperature on the rolling friction coefficient. The tire material and the arrangement of tire plies affect the rolling friction coefficient and the critical speed. Radial tires have around 20% lower \u03bcr, and 20% higher critical speed. Tire radius Rg and aspect ratio hT /wT are the two size parameters that affect the rolling resistance coefficient. A tire with larger Rg and smaller hT /wT has lower rolling resistance and higher critical speed. Generally speaking, the rolling friction coefficient decreases with wear in both radial and non-radial tires, and increases by increasing temperature. 3.5.2 F Effect of Inflation Pressure and Load on the Rolling Friction Coefficient The rolling friction coefficient \u03bcr decreases by increasing the inflation pressure p. The effect of increasing pressure is equivalent to decreasing normal load Fz. The following empirical equation has been suggested to show the effects of both pressure p and load Fz on the rolling friction coefficient. \u03bcr = K 1000 \u00b5 5.1 + 5.5\u00d7 105 + 90Fz p + 1100 + 0.0388Fz p v2x \u00b6 (3.81) The parameter K is equal to 0.8 for radial tires, and is equal to 1.0 for nonradial tires. The value of Fz, p, and vx must be in [ N], [ Pa], and [m/ s] respectively. 3. Tire Dynamics 123 Example 93 Motorcycle rolling friction coefficient. The following equations are suggested for calculating rolling friction coefficient \u03bcr applicable to motorcycles. They can be only used as a rough lower estimate for passenger cars. The equations consider the inflation pressure and forward velocity of the motorcycle. \u03bcr = \u23a7\u23aa\u23aa\u23a8\u23aa\u23aa\u23a9 0.0085 + 1800 p + 2.0606 p v2x vx \u2264 46m/ s (\u2248 165 km/h) 1800 p + 3.7714 p v2x vx > 46m/ s (\u2248 165 km/h) (3.82) The speed vx must be expressed in m/ s and the pressure p must be in Pa. Figure 3.28 illustrates this equation for vx \u2264 46m/ s (\u2248 165 km/h). Increasing the inflation pressure p decreases the rolling friction coefficient \u03bcr. Example 94 Dissipated power because of rolling friction. Rolling friction reduces the vehicle\u2019s power. The dissipated power because of rolling friction is equal to the rolling friction force Fr times the forward velocity vx. Using Equation (3.81), the rolling resistance power is P = Fr vx = \u2212\u03bcr vx Fz = \u2212K vx 1000 \u00b5 5.1 + 5.5\u00d7 105 + 90Fz p + 1100 + 0.0388Fz p v2x \u00b6 Fz. (3.83) 124 3. Tire Dynamics The resultant power P is in [W] when the normal force Fz is expressed in [ N], velocity vx in [m/ s], and pressure p in [ Pa]. The rolling resistance dissipated power for motorcycles can be found based on Equation (3.82). P = \u23a7\u23aa\u23aa\u23a8\u23aa\u23aa\u23aa\u23a9 \u00b5 0.0085 + 1800 p + 2.0606 p v2x \u00b6 vxFz vx \u2264 46m/ s (\u2248 165 km/h)\u00b5 1800 p + 3.7714 p v2x \u00b6 vxFz vx > 46m/ s (\u2248 165 km/h) (3.84) Example 95 Rolling resistance dissipated power. If a vehicle is moving at 100 km/h \u2248 27.78m/ s \u2248 62mi/h and each radial tire of the vehicle is pressurized up to 220 kPa \u2248 32 psi and loaded by 220 kg, then the dissipated power, because of rolling resistance, is P = 4\u00d7 K vx 1000 \u00b5 5.1 + 5.5\u00d7 105 + 90Fz p + 1100 + 0.0388Fz p v2x \u00b6 Fz = 2424.1W \u2248 2.4 kW. (3.85) To compare the given equations, assume the vehicle has motorcycle tires with power loss given by Equation (3.84). P = \u00b5 0.0085 + 1800 p + 2.0606 p v2x \u00b6 vxFz = 5734.1W \u2248 5.7 kW. (3.86) It shows that if the vehicle uses motorcycle tires, it dissipates more power. Example 96 Effects of improper inflation pressure. High inflation pressure increases stiffness, which reduces ride comfort and generates vibration. Tireprint and traction are reduced when tires are over inflated. Over-inflation causes the tire to transmit shock loads to the suspension, and reduces the tire\u2019s ability to support the required load for cornerability, braking, and acceleration. Under-inflation results in cracking and tire component separation. It also increases sidewall flexing and rolling resistance that causes heat and mechanical failure. A tire\u2019s load capacity is largely determined by its inflation pressure. Therefore, under-inflation results in an overloaded tire that operates at high deflection with a low fuel economy, and low handling. Figure 3.29 illustrates the effect of over and under inflation on tire-road contact compared to a proper inflated tire. Proper inflation pressure is necessary for optimum tire performance, safety, and fuel economy. Correct inflation is especially significant to the endurance and performance of radial tires because it may not be possible to find a 5 psi \u2248 35 kPa under-inflation in a radial tire just by looking. 3. Tire Dynamics 125 However, under-inflation of 5 psi \u2248 35 kPa can reduce up to 25% of the tire performance and life. A tire may lose 1 to 2 psi (\u2248 7 to 14 kPa) every month. The inflation pressure can also change by 1 psi\u2248 7 kPa for every 10 \u25e6F \u2248 5 \u25e6C of temperature change. As an example, if a tire is inflated to 35 psi \u2248 240 kPa on an 80 \u25e6F \u2248 26 \u25e6C summer day, it could have an inflation pressure of 23 psi \u2248 160 kPa on a 20 \u25e6F \u2248 \u22126 \u25e6C day in winter. This represents a normal loss of 6 psi \u2248 40 kPa over the six months and an additional loss of 6 psi \u2248 40 kPa due to the 60 \u25e6F \u2248 30 \u25e6C change. At 23 psi \u2248 160 kPa, this tire is functioning under-inflated. Example 97 Small / large and soft / hard tires. If the driving tires are small, the vehicle becomes twitchy with low traction and low top speed. However, when the driving tires are big, then the vehicle has slow steering response and high tire distortion in turns, decreasing the stability. Softer front tires show more steerability, less stability, and more wear while hard front tires show the opposite. Soft rear tires have more rear traction, but they make the vehicle less steerable, more bouncy, and less stable. Hard rear tires have less rear traction, but they make the vehicle more steerable, less bouncy, and more stable. 3.5.3 F Effect of Sideslip Angle on Rolling Resistance When a tire is turning on the road with a sideslip angle \u03b1, a significant increase in rolling resistance occurs. The rolling resistance force Fr would 126 3. Tire Dynamics then be Fr = Fx cos\u03b1+ Fy sin\u03b1 (3.87) \u2248 Fx \u2212 C\u03b1\u03b1 2 (3.88) where, Fx is the longitudinal force opposing the motion, and Fy is the lateral force. Proof. Figure 3.30 illustrates the top view of a turning tire on the ground under a sideslip angle \u03b1. The rolling resistance force is defined as the force opposite to the velocity vector of the tire, which has angle \u03b1 with the xaxis. Assume a longitudinal force Fx in \u2212x-direction is applied on the tire. Sideslip \u03b1 increases Fx and generates a lateral force Fy. The sum of the components of the longitudinal force Fx and the lateral force Fy makes the rolling resistance force Fr. Fr = Fx cos\u03b1+ Fy sin\u03b1 (3.89) For small values of the sideslip \u03b1, the lateral force is proportional to \u2212\u03b1 and therefore, Fr \u2248 Fx \u2212 C\u03b1\u03b1 2. (3.90) 3. Tire Dynamics 127 3.5.4 F Effect of Camber Angle on Rolling Resistance When a tire travels with a camber angle \u03b3, the component of rolling moment Mr on rolling resistance Fr will be reduced, however, a component of aligning moment Mz on rolling resistance will appear. Fr = \u2212Fr \u0131\u0302 (3.91) Fr = 1 Rh Mr cos \u03b3 + 1 Rh Mz sin \u03b3 (3.92) Proof. Rolling moment Mr appears when the normal force Fz shifts forward. However, only the component Mr cos \u03b3 is perpendicular to the tireplane and prevents the tire\u2019s spin. Furthermore, when a moment in zdirection is applied on the tire, only the component Mz sin \u03b3 will prevent the tire\u2019s spin. Therefore, the camber angle \u03b3 will affect the rolling resistance according to Fr = \u2212Fr \u0131\u0302 Fr = 1 h Mr cos \u03b3 1 h Mz sin \u03b3 (3.93) where Mr may be substituted by Equation (3.66) to show the effect of normal force Fz. Fr = \u2206x h Fz cos \u03b3 1 h Mz sin \u03b3 (3.94) 3.6 Longitudinal Force The longitudinal slip ratio of a tire is s = Rg\u03c9w vx \u2212 1 (3.95) where, Rg is the tire\u2019s geometric and unloaded radius, \u03c9w is the tire\u2019s angular velocity, and vx is the tire\u2019s forward velocity. Slip ratio is positive for driving and is negative for braking. To accelerate or brake a vehicle, longitudinal forces must develop between the tire and the ground. When a moment is applied to the spin axis of the tire, slip ratio occurs and a longitudinal force Fx is generated at the tireprint. The force Fx is proportional to the normal force, Fx = Fx \u0131\u0302 (3.96) Fx = \u03bcx (s) Fz (3.97) where the coefficient \u03bcx (s) is called the longitudinal friction coefficient and is a function of slip ratio s as shown in Figure 3.31. The friction coefficient 128 3. Tire Dynamics reaches a driving peak value \u03bcdp at s \u2248 0.1, before dropping to an almost steady-state value \u03bcds. The friction coefficient \u03bcx (s) may be assumed proportional to s when s is very small \u03bcx (s) = Cs s s << 1 (3.98) where Cs is called the longitudinal slip coefficient. The tire will spin when s & 0.1 and the friction coefficient remains almost constant. The same phenomena happens in braking at the values \u03bcbp and \u03bcbs. Proof. Slip ratio, or simply slip, is defined as the difference between the actual speed of the tire vx and the equivalent tire speeds Rw\u03c9w. Figure 3.32 illustrates a turning tire on the ground. The ideal distance that the tire would freely travel with no slip is denoted by dF , while the actual distance the tire travels is denoted by dA. Thus, for a slipping tire, dA > dF , and for a spinning tire, dA < dF . The difference dF \u2212 dA is the tire slip and therefore, the slip ratio of the tire is s = dF \u2212 dA dA . (3.99) To have the instant value of s, we must measure the travel distances in an infinitesimal time length, and therefore, s \u2261 d\u0307F \u2212 d\u0307A d\u0307A . (3.100) 3. Tire Dynamics 129 If the angular velocity of the tire is \u03c9w then, d\u0307F = Rg\u03c9w and d\u0307A = Rw\u03c9w where, Rg is the geometric tire radius and Rw is the effective radius. Therefore, the slip ratio s can be defined based on the actual speed vx = Rw\u03c9w, and the free speed Rg\u03c9w s = Rg\u03c9w \u2212Rw\u03c9w Rw\u03c9w = Rg\u03c9w vx \u2212 1 (3.101) A tire can exert longitudinal force only if a longitudinal slip is present. Longitudinal slip is also called circumferential or tangential slip. During acceleration, the actual velocity vx is less than the free velocity Rg\u03c9w, and therefore, s > 0. However, during braking, the actual velocity vx is higher than the free velocity Rg\u03c9w and therefore, s < 0. The frictional force Fx between a tire and the road surface is a function of normal load Fz, vehicle speed vx, and wheel angular speed \u03c9w. In addition to these variables there are a number of parameters that affect Fx, such as tire pressure, tread design, wear, and road surface. It has been determined empirically that a contact friction of the form Fx = \u03bcx(\u03c9w, vx)Fz can model experimental measurements obtained with constant vx, \u03c9w. Example 98 Slip ratio based on equivalent angular velocity \u03c9eq. It is possible to define an effective angular velocity \u03c9eq as an equivalent angular velocity for a tire with radius Rg to proceed with the actual speed vx = Rg\u03c9eq. Using \u03c9eq we have vx = Rg\u03c9eq = Rw\u03c9w (3.102) 130 3. Tire Dynamics and therefore, s = Rg\u03c9w \u2212Rg\u03c9eq Rg\u03c9eq = \u03c9w \u03c9eq \u2212 1. (3.103) Example 99 Slip ratio is \u22121 < s < 0 in braking. When we brake, a braking moment is applied to the wheel axis. The tread of the tire will be stretched circumstantially in the tireprint zone. Hence, the tire is moving faster than a free tire Rw\u03c9w > Rg\u03c9w (3.104) and therefore, s < 0. The equivalent radius for a braking tire is more than the free radius Rw > Rg. (3.105) Equivalently, we may express the condition using the equivalent angular velocity \u03c9e and deduce that a braking tire turns slower than a free tire Rg\u03c9eq > Rg\u03c9w. (3.106) The brake moment can be high enough to lock the tire. In this case \u03c9w = 0 and therefore, s = \u22121. It shows that the longitudinal slip would be between \u22121 < s < 0 when braking. \u22121 < s < 0 for a < 0 (3.107) Example 100 Slip ratio is 0 < s <\u221e in driving. When we drive, a driving moment is applied to the tire axis. The tread of the tire will be compressed circumstantially in the tireprint zone. Hence, the tire is moving slower than a free tire Rw\u03c9w < Rg\u03c9w (3.108) and therefore s > 0. The equivalent radius for a driving tire is less than the free radius Rw < Rg. (3.109) Equivalently, we may express the condition using the equivalent angular velocity \u03c9e and deduce that a driving tire turns faster than a free tire Rg\u03c9eq < Rg\u03c9w. (3.110) The driving moment can be high enough to overcome the friction and turn the tire on pavement while the car is not moving. In this case vx = 0 3. Tire Dynamics 131 and therefore, s = \u221e. It shows that the longitudinal slip would be between 0 < s <\u221e when accelerating. 0 < s <\u221e for a > 0 (3.111) The tire speed Rw\u03c9w equals vehicle speed vx only if acceleration is zero. In this case, the normal force acting on the tire and the size of the tireprint are constant in time. No element of the tireprint is slipping on the road. Example 101 Power and maximum velocity. Consider a moving car with power P = 100 kW \u2248 134 hp can attain 279 km/h \u2248 77.5m/ s \u2248 173.3mi/h. The total driving force must be Fx = P vx = 100\u00d7 103 77.5 = 1290.3N. (3.112) If we assume that the car is rear-wheel-drive and the rear wheels are driving at the maximum traction under the load 1600N, then the longitudinal friction coefficient \u03bcx is \u03bcx = Fx Fz = 1290.3 1600 \u2248 0.806. (3.113) Example 102 Slip of hard tire on hard road. A tire with no slip cannot create any tangential force. Assume a toy car equipped with steel tires is moving on a glass table. Such a car cannot accelerate or steer easily. If the car can steer at very low speeds, it is because there is sufficient microscopic slip to generate forces to steer or drive. The glass table and the small contact area of the small metallic tires deform and stretch each other, although such a deformation is very small. If there is any friction between the tire and the surface, there must be slip to maneuver. Example 103 Samples for longitudinal friction coefficients \u03bcdp and \u03bcds. Table 3.2 shows the average values of longitudinal friction coefficients \u03bcdp and \u03bcds for a passenger car tire 215/65R15. It is practical to assume \u03bcdp = \u03bcbp, and \u03bcds = \u03bcbs. 132 3. Tire Dynamics Example 104 Friction mechanisms. Rubber tires generate friction in three mechanisms: 1\u2212 adhesion, 2\u2212 deformation, and 3\u2212 wear. Fx = Fad + Fde + Fwe. (3.114) Adhesion friction is equivalent to sticking. The rubber resists sliding on the road because adhesion causes it stick to the road surface. Adhesion occurs as a result of molecular binding between the rubber and surfaces. Because the real contact area is much less than the observed contact area, high local pressure make molecular binding, as shown in Figure 3.33. Bound occurs at the points of contact and welds the surfaces together. The adhesion friction is equal to the required force to break these molecular bounds and separate the surfaces. The adhesion is also called cold welding and is attributed to pressure rather than heat. Higher load increases the contact area, makes more bounds, and increases the friction force. So the adhesion friction confirms the friction equation Fx = \u03bcx (s) Fz. (3.115) The main contribution to tire traction force on a dry road is the adhesion friction. The adhesion friction decreases considerably on a road covered by water, ice, dust, or lubricant. Water on a wet road prevents direct contact between the tire and road and reduces the formation of adhesion friction. The main contribution to tire friction when it slides on a road surface is the viscoelastic energy dissipation in the tireprint area. This dissipative energy is velocity and is time-history dependent. Deformation friction is the result of deforming rubber and filling the microscopic irregularities on the road surface. The surface of the road has many peaks and valleys called asperities. Movement of a tire on a rough 3. Tire Dynamics 133 surface results in the deformation of the rubber by peaks and high points on the surface. A load on the tire causes the peaks of irregularities to penetrate the tire and the tire drapes over the peaks. The deformation friction force, needed to move the irregularities in the rubber, comes from the local high pressure across the irregularities. Higher load increases the penetration of the irregularities in the tire and therefore increases the friction force. So the deformation friction confirms the friction equation (3.115). The main contribution to the tire traction force on a wet road is the deformation friction. The adhesion friction decreases considerably on a road covered by water, ice, dust, or lubricant. Deformation friction exists in relative movement between any contacted surfaces. No matter how much care is taken to form a smooth surface, the surfaces are irregular with microscopic peaks and valleys. Opposite peaks interact with each other and cause damage to both surfaces. Wear friction is the result of excessive local stress over the tensile strength of the rubber. High local stresses deform the structure of the tire surface past the elastic point. The polymer bonds break, and the tire surface tears in microscopic scale. This tearing makes the wear friction mechanism. Wear results in separation of material. Higher load eases the tire wear and therefore increases the wear friction force. So the wear friction confirms the friction equation (3.115). Example 105 Empirical slip models. Based on experimental data and curve fitting methods, some mathematical equations are presented to simulate the longitudinal tire force as a function of longitudinal slip s. Most of these models are too complicated to be useful in vehicle dynamics. However, a few of them are simple and accurate enough to be applied. The Pacejka model, which was presented in 1991, has the form Fx (s) = c1 sin \u00a1 c2 tan \u22121 \u00a1c3s\u2212 c4 \u00a1 c3s\u2212 tan\u22121 (c3s) \u00a2\u00a2\u00a2 (3.116) where c1, c2, and c3 are three constants based on the tire experimental data. The 1987 Burckhardt model is a simpler equation that needs three numbers. Fx (s) = c1 \u00a1 1\u2212 e\u2212c2s \u00a2 \u2212 c3s (3.117) There is another Burckhardt model that includes the velocity dependency. Fx (s) = \u00a1 c1 \u00a1 1\u2212 e\u2212c2s \u00a2 \u2212 c3s \u00a2 e\u2212c4v (3.118) This model needs four numbers to be measured from experiment. By expanding and approximating the 1987 Burckhardt model, the simpler model by Kiencke and Daviss was suggested in 1994. This model is Fx (s) = ks s 1 + c1s+ c2s2 (3.119) 134 3. Tire Dynamics where ks is the slope of Fx (s) versus s at s = 0 ks = lim s\u21920 4Fs 4s (3.120) and c1, c2 are two experimental numbers. Another simple model is the 2002 De-Wit model Fx (s) = c1 \u221a s\u2212 c2s (3.121) that is based on two numbers c1, c2. In either case, we need at least one experimental curve such as shown in Figure 3.31 to find the constant numbers ci. The constants ci are the numbers that best fit the associated equation with the experimental curve. The 1997 Burckhardt model (3.118) needs at least two similar tests at two different speeds. Example 106 F Alternative slip ratio. An alternative method for defining the slip ratio is s = \u23a7\u23aa\u23aa\u23a8\u23aa\u23a9 1\u2212 vx Rg\u03c9w Rg\u03c9w > vx driving Rg\u03c9w vx \u2212 1 Rg\u03c9w < vx braking (3.122) where vx is the speed of the wheel center, \u03c9w is the angular velocity of the wheel, and Rg is the tire radius. In another alternative definition, the following equation is used for longitudinal slip: s = 1\u2212 \u00b5 Rg\u03c9w vx \u00b6n where n = \u00bd +1 Rg\u03c9w \u2264 vx \u22121 Rg\u03c9w > vx (3.123) s \u2208 [0, 1] In this definition s is always between zero and one. When s = 1, then the tire is either locked while the car is sliding, or the tire is spinning while the car is not moving. Example 107 F Tire on soft sand. Figure 3.34 illustrates a tire turning on sand. The sand will be packed when the tire passes. The applied stresses from the sand on the tire are developed during the angle \u03b81 < \u03b8 < \u03b82 measured counterclockwise from vertical direction. It is possible to define a relationship between the normal stress \u03c3 and tangential stress \u03c4 under the tire \u03c4 = (c+ \u03c3 tan \u03b8) \u00b3 1\u2212 e r k [\u03b81\u2212\u03b8+(1\u2212s)(sin \u03b8)\u2212sin \u03b81] \u00b4 (3.124) 3. Tire Dynamics 135 where s is the slip ratio defined in Equation (3.122), and \u03c4M = c+ \u03c3 tan \u03b8 (3.125) is the maximum shear stress in the sand applied on the tire. In this equation, c is the cohesion stress of the sand, and k is a constant. Example 108 F Lateral slip ratio. Analytical expressions can be established for the force contributions in x and y directions using adhesive and sliding concept by defining longitudinal and lateral slip ratios sx and sy sx = Rg\u03c9w vx \u2212 1 (3.126) sy = Rg\u03c9w vy (3.127) where vx is the longitudinal speed of the wheel and vy is the lateral speed of the wheel. The unloaded geometric radius of the tire is denoted by Rg and \u03c9w is the rotation velocity of the wheel. At very low slips, the resulting tire forces are proportional to the slip Fx = Csx sx (3.128) Fy = Csy sy (3.129) where Csx is the longitudinal slip coefficient and Csy is the lateral slip coefficient. 3.7 Lateral Force When a turning tire is under a vertical force Fz and a lateral force Fy, its path of motion makes an angle \u03b1 with respect to the tire-plane. The angle 136 3. Tire Dynamics is called sideslip angle and is proportional to the lateral force Fy = Fy j\u0302 (3.130) Fy = \u2212C\u03b1 \u03b1 (3.131) where C\u03b1 is called the cornering stiffness of the tire. C\u03b1 = lim \u03b1\u21920 \u2202 (\u2212Fy) \u2202\u03b1 = \u00af\u0304\u0304\u0304 lim \u03b1\u21920 \u2202Fy \u2202\u03b1 \u00af\u0304\u0304\u0304 (3.132) The lateral force Fy is at a distance ax\u03b1 behind the centerline of the tireprint and makes a moment Mz called aligning moment. Mz = Mz k\u0302 (3.133) Mz = Fy ax\u03b1 (3.134) For small \u03b1, the aligning moment Mz tends to turn the tire about the z-axis and make the x-axis align with the velocity vector v. The aligning moment always tends to reduce \u03b1. Proof. When a wheel is under a constant load Fz and then a lateral force is applied on the rim, the tire will deflect laterally as shown in Figure 3.35. The tire acts as a linear spring under small lateral forces Fy = ky\u2206y (3.135) with a lateral stiffness ky. The wheel will start sliding laterally when the lateral force reaches a maximum value FyM . At this point, the lateral force approximately remains constant and is proportional to the vertical load FyM = \u03bcy Fz (3.136) 3. Tire Dynamics 137 where, \u03bcy is the tire friction coefficient in the y-direction. A bottom view of the tireprint of a laterally deflected tire is shown in Figure 3.36. If the laterally deflected tire is turning forward on the road, the tireprint will also flex longitudinally. A bottom view of the tireprint for such a laterally deflected and turning tire is shown in Figure 3.37. Although the tire-plane remains perpendicular to the road, the path of the wheel makes an angle \u03b1 with tire-plane. As the wheel turns forward, undeflected treads enter the tireprint region and deflect laterally as well as longitudinally. When a tread moves toward the end of the tireprint, its lateral deflection increases until it approaches the tailing edge of the tireprint. The normal load decreases at the tail of the tireprint, so the friction force is lessened and the tread can slide back to its original position when leaving the tireprint region. The point where the laterally deflected tread slides back is called sliding line. A turning tire under lateral force and the associated sideslip angle \u03b1 are shown in Figure 3.38. Lateral distortion of the tire treads is a result of a tangential stress distribution \u03c4y over the tireprint. Assuming that the tangential stress \u03c4y is proportional to the distortion, the resultant lateral force Fy Fy = Z AP \u03c4y dAp (3.137) is at a distance ax\u03b1 behind the center line. ax\u03b1 = 1 Fy Z AP x \u03c4y dAp (3.138) The distance ax\u03b1 is called the pneumatic trail, and the resultant moment 138 3. Tire Dynamics Mz is called the aligning moment. Mz = Mz k\u0302 (3.139) Mz = Fy ax\u03b1 (3.140) The aligning moment tends to turn the tire about the z-axis and make it align with the direction of tire velocity vector v. A stress distribution \u03c4y, the resultant lateral force Fy, and the pneumatic trail ax\u03b1 are illustrated in Figure 3.38. There is also a lateral shift in the tire vertical force Fz because of slip angle \u03b1, which generates a slip moment Mx about the forward x-axis. Mx = \u2212Mx \u0131\u0302 (3.141) Mx = Fz ay\u03b1 (3.142) The slip angle \u03b1 always increases by increasing the lateral force Fy. However, the sliding line moves toward the tail at first and then moves forward by increasing the lateral force Fy. Slip angle \u03b1 and lateral force Fy work as action and reaction. A lateral force generates a slip angle, and a slip angle generates a lateral force. Hence, we can steer the tires of a car to make a slip angle and produce a lateral force to turn the car. Steering causes a slip angle in the tires and creates a lateral force. The slip angle \u03b1 > 0 if the tire should be turned about the z-axis to be aligned with the velocity vector v. A positive slip angle \u03b1 generates a negative lateral force Fy. Hence, steering to the right about the \u2212z-axis makes a positive slip angle and produces a negative lateral force to move the tire to the right. A sample of measured lateral force Fy as a function of slip angle \u03b1 for a constant vertical load is plotted in Figure 3.39. The lateral force Fy is linear for small slip angles, however the rate of increasing Fy decreases 3. Tire Dynamics 139 for higher \u03b1. The lateral force remains constant or drops slightly when \u03b1 reaches a critical value at which the tire slides on the road. Therefore, we may assume the lateral force Fy is proportional to the slip angle \u03b1 for low values of \u03b1. Fy = \u2212C\u03b1 \u03b1 (3.143) C\u03b1 = lim \u03b1\u21920 \u2202 (\u2212Fy) \u2202\u03b1 (3.144) The cornering stiffness C\u03b1 of radial tires are higher than C\u03b1 for non-radial tires. This is because radial tires need a smaller slip angle \u03b1 to produce the same amount of lateral force Fy. Examples of aligning moments for radial and non-radial tires are illustrated in Figure 3.40. The pneumatic trail ax\u03b1 increases for small slip angles up to a maximum value, and decreases to zero and even negative values for high slip angles. Therefore, the behavior of aligning momentMz is similar to what is shown in Figure 3.40. The lateral force Fy = \u2212C\u03b1 \u03b1 can be decomposed to Fy cos\u03b1, parallel to the path of motion v, and Fy sin\u03b1, perpendicular to v as shown in Figure 3.41. The component Fy cos\u03b1, normal to the path of motion, is called cornering force, and the component Fy sin\u03b1, along the path of motion, is called drag force. Lateral force Fy is also called side force or grip. We may combine the lateral forces of all a vehicle\u2019s tires and have them acting at the car\u2019s mass center C. 140 3. Tire Dynamics 3. Tire Dynamics 141 Example 109 Effect of tire load on lateral force curve. When the wheel load Fz increases, the tire treads can stick to the road better. Hence, the lateral force increases at a constant slip angle \u03b1, and the slippage occurs at the higher slip angles. Figure 3.42 illustrates the lateral force behavior of a sample tire for different normal loads. Increasing the load not only increases the maximum attainable lateral force, it also pushes the maximum of the lateral force to higher slip angles. Sometimes the effect of load on lateral force is presented in a dimensionless variable to make it more practical. Figure 3.43 depicts a sample. Example 110 Gough diagram. The slip angle \u03b1 is the main affective parameter on the lateral force Fy and aligning moment Mz = Fyax\u03b1. However, Fz and Mz depend on many other parameters such as speed v, pressure p, temperature, humidity, and road conditions. A better method to show Fz and Mz is to plot them versus each other for a set of parameters. Such a graph is called a Gough diagram. Figure 3.44 depicts a sample Gough diagram for a radial passenger car tire. Every tire has its own Gough diagram, although we may use an average diagram for radial or non-radial tires. Example 111 Effect of velocity. The curve of lateral force as a function of the slip angle Fy (\u03b1) decreases as velocity increases. Hence, we need to increase the sideslip angle at higher velocities to generate the same lateral force. Sideslip angle increases by 142 3. Tire Dynamics 3. Tire Dynamics 143 increasing the steer angle. Figure 3.45 illustrates the effect of velocity on Fy for a radial passenger tire. Because of this behavior, a fixed steer angle, the curvature of a one-wheel-car trajectory, increases by increasing the driving speed. Example 112 F A model for lateral force. When the sideslip angle is not small, the linear approximation (3.131) cannot model the tire behavior. Based on a parabolic normal stress distribution on the tireprint, the following third-degree function was presented in the 1950s to calculate the lateral force at high sideslips Fy = \u2212C\u03b1 \u03b1 \u00c3 1\u2212 1 3 \u00af\u0304\u0304\u0304 C\u03b1 \u03b1 FyM \u00af\u0304\u0304\u0304 + 1 27 \u00b5 C\u03b1 \u03b1 FyM \u00b62! (3.145) where FyM is the maximum lateral force that the tire can support. FyM is set by the tire load and the lateral friction coefficient \u03bcy. Let\u2019s show the sideslip angle at which the lateral force Fy reaches its maximum value FyM by \u03b1M . Equation (3.145) shows that \u03b1M = 3FyM C\u03b1 (3.146) and therefore, Fy = \u2212C\u03b1 \u03b1 \u00c3 1\u2212 \u03b1 \u03b1M + 1 3 \u00b5 \u03b1 \u03b1M \u00b62! (3.147) Fy FyM = 3\u03b1 \u03b1M \u00c3 1\u2212 \u03b1 \u03b1M + 1 3 \u00b5 \u03b1 \u03b1M \u00b62! . (3.148) 144 3. Tire Dynamics Figure 3.46 shows the cubic curve model for lateral force as a function of the sideslip angle. The Equation is applicable only for 0 \u2264 \u03b1 \u2264 \u03b1M . Example 113 F A model for lateral stress. Consider a tire turning on a dry road at a low sideslip angle \u03b1. Assume the developed lateral stress on tireprint can be expressed by the following equation: \u03c4y(x, y) = c\u03c4yM \u00b3 1\u2212 x a \u00b4\u00b5 1\u2212 x3 a3 \u00b6 cos2 \u00b3 y 2b \u03c0 \u00b4 (3.149) The coefficient c is proportional to the tire load Fz sideslip \u03b1, and longitudinal slip s. If the tireprint AP = 4 \u00d7 a \u00d7 b = 4 \u00d7 5 cm \u00d7 12 cm, then the lateral force under the tire, Fy, for c = 1 is Fy = Z AP \u03c4y(x, y) dA = Z 0.05 \u22120.05 Z 0.12 \u22120.12 \u03c4yM \u00b3 1\u2212 x 0.05 \u00b4\u00b5 1\u2212 x3 0.053 \u00b6 cos2 \u00b3 y\u03c0 0.24 \u00b4 dy dx = 0.0144\u03c4yM . (3.150) If we calculate the lateral force Fy = 1000N by measuring the lateral acceleration, then the maximum lateral stress is \u03c4yM = Fz 0.014 4 = 69444Pa (3.151) and the lateral stress distribution over the tireprint is \u03c4y(x, y) = 69444 \u00b3 1\u2212 x 0.05 \u00b4\u00b5 1\u2212 x3 0.053 \u00b6 cos2 \u00b3 y\u03c0 0.24 \u00b4 Pa. (3.152) 3. Tire Dynamics 145 3.8 Camber Force Camber angle \u03b3 is the tilting angle of tire about the longitudinal x-axis. Camber angle generates a lateral force Fy called camber trust or camber force. Figure 3.47 illustrates a front view of a cambered tire and the generated camber force Fy. Camber angle is assumed positive \u03b3 > 0, when it is in the positive direction of the x-axis, measured from the z-axis to the tire. A positive camber angle generates a camber force along the \u2212y-axis. The camber force is proportional to \u03b3 at low camber angles, and depends directly on the wheel load Fz. Therefore, Fy = Fy j\u0302 (3.153) Fy = \u2212C\u03b3 \u03b3 (3.154) where C\u03b3 is called the camber stiffness of tire. C\u03b3 = lim \u03b3\u21920 \u2202 (\u2212Fy) \u2202\u03b3 (3.155) In presence of both, camber \u03b3 and sideslip \u03b1, the overall lateral force Fy on a tire is a superposition of the corner force and camber trust. Fy = \u2212C\u03b3 \u03b3 \u2212 C\u03b1 \u03b1 (3.156) Proof. When a wheel is under a constant load and then a camber angle is applied on the rim, the tire will deflect laterally such that it is longer in 146 3. Tire Dynamics the cambered side and shorter in the other side. Figure 3.48 compares the tireprint of a straight and a cambered tire, turning slowly on a flat road. As the wheel turns forward, undeflected treads enter the tireprint region and deflect laterally as well as longitudinally. However, because of the shape of the tireprint, the treads entering the tireprint closer to the cambered side, have more time to be stretched laterally. Because the developed lateral stress is proportional to the lateral stretch, the nonuniform tread stretching generates an asymmetric stress distribution and more lateral stress will be developed on the cambered side. The result of the nonuniform lateral stress distribution over the tireprint of a cambered tire produces the camber trust Fy in the cambered direction. Fy = Fy j\u0302 (3.157) Fy = Z AP \u03c4y dA (3.158) The camber trust is proportional to the camber angle for small angles. Fy = \u2212C\u03b3 \u03b3 (3.159) The camber trust Fy shifts a distance ax\u03b3 forward when the cambered tire turns on the road. The resultant moment Mz = Mz k\u0302 (3.160) Mz = Fy ax\u03b3 (3.161) 3. Tire Dynamics 147 is called camber torque, and the distance ax\u03b3 is called camber trail. Camber trail is usually very small and hence, the camber torque can be ignored in linear analysis of vehicle dynamics. Because the tireprint of a cambered tire deforms to be longer in the cambered side, the resultant vertical force Fz Fz = Z AP \u03c3z dA (3.162) that supports the wheel load, shifts laterally to a distance ay\u03b3 from the center of the tireprint. ay\u03b3 = 1 Fz Z AP y \u03c3z dAp (3.163) The distance ay\u03b3 is called the camber arm, and the resultant moment Mx 148 3. Tire Dynamics is called the camber moment. Mx = Mx k\u0302 (3.164) Mx = \u2212Fz ay\u03b3 (3.165) The camber moment tends to turn the tire about the x-axis and make the tire-plane align with the z-axis. The camber arm ay\u03b3 is proportional to the camber angle \u03b3 for small angles. ay\u03b3 = Cy\u03b3 \u03b3 (3.166) Figure 3.49 shows the camber force Fy for different camber angle \u03b3 at a constant tire load Fz = 4500N. Radial tires generate lower camber force due to their higher flexibility. It is better to illustrate the effect of Fz graphically to visualize the camber force. Figure 3.50 depicts the variation of camber force Fy as a function of normal load Fz at different camber angles for a sample radial tire. If we apply a slip angle \u03b1 to a turning cambered tire, the tireprint will distort similar to the shape in Figure 3.51 and the path of treads become more complicated. The resultant lateral force would be at a distance ax\u03b3 and ay\u03b3 from the center of the tireprint. Both distances ax\u03b3 and ay\u03b3 are functions of angles \u03b1 and \u03b3. Camber force due to \u03b3, along with the corner force due to \u03b1, give the total lateral force applied on a tire. Therefore, the lateral force can be calculated as Fy = \u2212C\u03b1 \u03b1\u2212 C\u03b3 \u03b3 (3.167) 3. Tire Dynamics 149 150 3. Tire Dynamics that is acceptable for \u03b3 . 10 deg and \u03b1 . 5 deg. Presence of both camber angle \u03b3 and slip angle \u03b1 makes the situation interesting because the total lateral force can be positive or negative. Figure 3.52 illustrates an example of lateral force as a function of \u03b3 and \u03b1 at a constant load Fz = 4000N. Similar to lateral force, the aligning moment Mz can be approximated as a combination of the slip and camber angle effects Mz = CM\u03b1 \u03b1+ CM\u03b3 \u03b3. (3.168) For a radial tire, CM\u03b1 \u2248 0.013Nm/deg and CM\u03b3 \u2248 0.0003Nm/deg, while for a non-radial tire, CM\u03b1 \u2248 0.01Nm/deg and CM\u03b3 \u2248 0.001Nm/deg. Example 114 Banked road. Consider a vehicle moving on a road with a transversal slope \u03b2, while its tires remain vertical. There is a downhill component of weight, F1 = mg sin\u03b2, that pulls the vehicle down. There is also an uphill camber force due to camber \u03b3 \u2248 \u03b2 of tires with respect to the road F2 = C\u03b3 \u03b3. The resultant lateral force Fy = C\u03b3 \u03b3 \u2212 mg sin\u03b2 depends on camber stiffness C\u03b3 and determines if the vehicle goes uphill or downhill. Since the camber stiffness C\u03b3 is higher for non-radial tires, it is more possible for a radial tire to go downhill and a non-radial uphill. The effects of cambering are particularly important for motorcycles that produce a large part of the cornering force by cambering. For cars and trucks, the cambering angles are much smaller and in many applications their effect can be negligible. However, some suspensions are designed to make the wheels camber when the axle load varies. 3. Tire Dynamics 151 Example 115 Camber importance and tireprint model. Cambering of a tire creates a lateral force, even though there is no sideslip. The effects of cambering are particularly important for motorcycles that produce a large part of the lateral force by camber. The following equations are presented to model the lateral deviation of a cambered tireprint from the straight tireprint, and expressing the lateral stress \u03c4y due to camber y = \u2212 sin \u03b3 \u00b3q R2g \u2212 x2 \u2212 q R2g \u2212 a2 \u00b4 (3.169) \u03c4y = \u2212\u03b3k \u00a1 a2 \u2212 x2 \u00a2 (3.170) where k is chosen such that the average camber defection is correct in the tireprint Z a \u2212a \u03c4y dx = Z a \u2212a y dx. (3.171) Therefore, k = 3 sin \u03b3 4a3\u03b3 \u00b5 \u2212a q R2g \u2212 a2 +R2g sin \u22121 a Rg \u00b6 (3.172) \u2248 3 4 Rg q R2g \u2212 a2 a2 (3.173) and \u03c4y = \u2212 3 4 \u03b3 Rg q R2g \u2212 a2 a2 \u00a1 a2 \u2212 x2 \u00a2 . (3.174) 3.9 Tire Force Tires may be considered as a force generator with two major outputs: forward force Fx, lateral force Fy, and three minor outputs: aligning moment Mz, roll moment Mx, and pitch moment My. The input of the force generator is the tire load Fz, sideslip \u03b1, longitudinal slip s, and the camber angle \u03b3. Fx = Fx (Fz, \u03b1, s, \u03b3) (3.175) Fy = Fy (Fz, \u03b1, s, \u03b3) (3.176) Mx = Mx (Fz, \u03b1, s, \u03b3) (3.177) My = My (Fz, \u03b1, s, \u03b3) (3.178) Mz = Mz (Fz, \u03b1, s, \u03b3) (3.179) Ignoring the rolling resistance and aerodynamic force, and when the tire is under a load Fz plus only one more of the inputs \u03b1, s, or \u03b3, the major 152 3. Tire Dynamics output forces can be approximated by a set of linear equations Fx = \u03bcx (s) Fz (3.180) \u03bcx (s) = Cs s Fy = \u2212C\u03b1 \u03b1 (3.181) Fy = \u2212C\u03b3 \u03b3 (3.182) where, Cs is the longitudinal slip coefficient, C\u03b1 is the lateral stiffness, and C\u03b3 is the camber stiffness. When the tire has a combination of tire inputs, the tire forces are called tire combined force. The most important tire combined force is the shear force because of longitudinal and sideslips. However, as long as the angles and slips are within the linear range of tire behavior, a superposition can be utilized to estimate the output forces. Driving and braking forces change the lateral force Fy generated at any sideslip angle \u03b1. This is because the longitudinal force pulls the tireprint in the direction of the driving or braking force and hence, the length of lateral displacement of the tireprint will also change. Figure 3.53 illustrates how a sideslip \u03b1 affects the longitudinal force ratio Fx/Fz as a function of slip ratio s. Figure 3.54 illustrates the effect of sideslip \u03b1 on the lateral force ratio Fy/Fz as a function of slip ratio s. Figure 3.55 and 3.56 illustrate the same force ratios as Figures 3.53 and 3.54 when the slip ratio s is a parameter. Proof. Consider a turning tire under a sideslip angle \u03b1. The tire develops a lateral force Fy = \u2212C\u03b1 \u03b1. Applying a driving or braking force on this tire will reduce the lateral force while developing a longitudinal force Fx = \u03bcx (s) Fz. Experimental data shows that the reduction in lateral force in presence of a slip ratio s is similar to Figure 3.54. Now assume the sideslip \u03b1 is reduced to zero. Reduction \u03b1 will increase the longitudinal force while decreasing the lateral force. Increasing the longitudinal force is experimentally similar to Figure 3.55. A turning tire under a slip ratio s develops a longitudinal force Fx = \u03bcx (s) Fz. Applying a sideslip angle \u03b1 will reduce the longitudinal force while developing a lateral force. Experimental data shows that the reduction in longitudinal force in presence of a sideslip \u03b1 is similar to Figure 3.53. Now assume the slip ratio s and hence, the driving or breaking force is reduced to zero. Reduction s will increase the lateral force while decreasing the longitudinal force. Increasing the lateral force is similar to Figure 3.54. Example 116 Pacejka model. An approximate equation is presented to describe force Equations (3.175) 3. Tire Dynamics 153 154 3. Tire Dynamics 3. Tire Dynamics 155 or (3.176). This equation is called the Pacejka model. F = A sin \u00a9 B tan\u22121 \u00a3 Cx\u2212D \u00a1 Cx\u2212 tan\u22121 (Cx) \u00a2\u00a4\u00aa (3.183) A = \u03bcFz (3.184) C = C\u03b1 AB (3.185) B,D = shape factors (3.186) The Pacejka model is substantially empirical. However, when the parameters A, B, C, D, C1, and C2 are determined for a tire, the equation expresses the tire behavior well enough. Figure 3.57 illustrates how the parameters can be determined from a test force-slip experimental result. Example 117 Friction ellipse. When the tire is under both longitudinal and sideslips, the tire is under combined slip. The shear force on the tireprint of a tire under a combined slip can approximately be found using a friction ellipse model.\u00b5 Fy FyM \u00b62 + \u00b5 Fx FxM \u00b62 = 1 (3.187) A friction ellipse is shown in Figure 3.58. Proof. The shear force Fshear, applied on the tire at tireprint, parallel to the ground surface, has two components: the longitudinal force Fx and the lateral force Fy. Fshear = Fx \u0131\u0302+ Fy j\u0302 (3.188) Fx = Cs sFz (3.189) Fy = \u2212C\u03b1 \u03b1 (3.190) 156 3. Tire Dynamics These forces cannot exceed their maximum values FyM and FxM . FyM = \u03bcy Fz FxM = \u03bcx Fz The tire shown in Figure 3.58 is moving along the velocity vector v at a sideslip angle \u03b1. The x-axis indicates the tire-plane. When there is no sideslip, the maximum longitudinal force is FxM = \u03bcx Fz = \u2212\u2192 OA. Now, if a sideslip angle \u03b1 is applied, a lateral force Fy = \u2212\u2212\u2192 OE is generated, and the longitudinal force reduces to Fx = \u2212\u2212\u2192 OB. The maximum lateral force would be FyM = \u03bcy Fz = \u2212\u2212\u2192 OD when there is no longitudinal slip. In presence of the longitudinal and lateral forces, we may assume that the tip point of the maximum shear force vector is on the following friction ellipse: \u00b5 Fy FyM \u00b62 + \u00b5 Fx FxM \u00b62 = 1 (3.191) When \u03bcx = \u03bcy = \u03bc, the friction ellipse would be a circle and Fshear = \u03bcFz. (3.192) 3. Tire Dynamics 157 Example 118 Wide tires. A wide tire has a shorter tireprint than a narrow tire. Assuming the same vehicle and same tire pressure, the area of tireprint would be equal in both tires. The shorter tireprint at the same sideslip has more of its length stuck to the road than longer tireprint. So, a wider tireprint generates more lateral force than a narrower tireprint for the same tire load and sideslip. Generally speaking, tire performance and maximum force capability decrease with increasing speed in both wide and narrow tires. Example 119 sin tire forces model. A few decades ago, a series of applied sine functions were developed based on experimental data to model tire forces. The sine functions, which are explained below, may be used to model tire forces, especially for computer purposes, effectively. The lateral force of a tire is Fy = A sin \u00a9 B tan\u22121 (C\u03a6) \u00aa (3.193) \u03a6 = (1\u2212E) (\u03b1+ \u03b4)\u03bcFz (3.194) C = C\u03b1 AB (3.195) C\u03b1 = C1 sin \u00b5 2 tan\u22121 Fz C2 \u00b6 (3.196) A,B = Shape factors (3.197) C1 = Maximum cornering stiffness (3.198) C2 = Tire load at maximum cornerin stiffness (3.199) 3.10 Summary We attach a coordinate frame (oxyz) to the tire at the center of the tireprint, called the tire frame. The x-axis is along the intersection line of the tire-plane and the ground. The z-axis is perpendicular to the ground, and the y-axis makes the coordinate system right-hand. We show the tire orientation using two angles: camber angle \u03b3 and sideslip angle \u03b1. The camber angle is the angle between the tire-plane and the vertical plane measured about the x-axis, and the sideslip angle \u03b1 is the angle between the velocity vector v and the x-axis measured about the z-axis. A vertically loaded wheel turning on a flat surface has an effective radius Rw, called rolling radius Rw = vx \u03c9w (3.200) where vx is the forward velocity, and \u03c9w is the angular velocity of the 158 3. Tire Dynamics wheel. The effective radius Rw is approximately equal to Rw \u2248 Rg \u2212 Rg \u2212Rh 3 (3.201) and is a number between the unloaded or geometric radius Rg and the loaded height Rh. Rh < Rw < Rg (3.202) A turning tire on the ground generates a longitudinal force called rolling resistance. The force is opposite to the direction of motion and is proportional to the normal force on the tireprint. Fr = \u03bcr Fz (3.203) The parameter \u03bcr is called the rolling friction coefficient and is a function of tire mechanical properties, speed, wear, temperature, load, size, driving and braking forces, and road condition. The tire force in the x-direction is a combination of the longitudinal force Fx and the roll resistance Fr. The longitudinal force is Fx = \u03bcx (s) Fz (3.204) where s is the longitudinal slip ratio of the tire s = Rg\u03c9w vx \u2212 1 (3.205) \u03bcx (s) = Cs s s << 1 (3.206) The wheel force in the tire y-direction, Fy, is a combination of the lateral force and the tire roll resistance Fr. The lateral force is Fy = \u2212C\u03b3 \u03b3 \u2212 C\u03b1 \u03b1 (3.207) where \u2212C\u03b3\u03b3 is called the camber trust and C\u03b1\u03b1 is called the sideslip force. 3. Tire Dynamics 159 3.11 Key Symbols a \u2261 x\u0308 acceleration a, b semiaxes of AP ax\u03b1 ax\u03b3 camber trail ay\u03b3 camber arm AP tireprint area c1, c2, c3, c4 coefficients of the function Fx = Fx (s) C0, C1, C2 coefficients of the polynomial function Fr = Fr (vx) Cs longitudinal slip coefficient Csx , Csy longitudinal and lateral slip coefficients C\u03b1 sideslip coefficient C\u03b3 camber stiffness d distance of tire travel dF no slip tire travel dA actual tire travel D tire diameter E Young modulus f function fk spring force Fr Fr rolling resistance force Fx longitudinal force, forward force Fy lateral force FyM pneumatic trail Fz normal force, vertical force, wheel load g g gravitational acceleration k stiffness k1, k2, k3 nonlinear tire stiffness coefficients keq equivalent stiffness ks slope of Fx (s) versus s at s = 0 kx tire stiffness in the x-direction ky tire stiffness in the y-direction kz tire stiffness in the z-direction K radial and non-radial tires parameter in \u03bcr = \u03bcr (p, vx) m mass Mr Mr rolling resistance moment Mx, Mx roll moment, bank moment, tilting torque, My pitch moment, rolling resistance torque Mz yaw moment, aligning moment, self aligning moment, bore torque n exponent for shape and stress distribution of AP n1 number of tire rotations p tire inflation pressure P rolling resistance power 160 3. Tire Dynamics r radial position of tire periphery r = \u03c9/\u03c9n frequency ratio Rg geometric radius Rh loaded height Rw rolling radius s longitudinal slip sy lateral slip T wheel torque v \u2261 x\u0307, v velocity x, y, z, x displacement x, y, z coordinate axes 4x tire deflection in the x-direction, rolling resistance arm 4y tire deflection in the y-direction 4z tire deflection in the z-direction z\u0307 tire deflection rate in the z-direction \u03b1 sideslip angle \u03b1M maximum sideslip angle \u03b2 transversal slope \u03b3 camber angle \u03b4 deflection 4x tire deflection in the x-direction, rolling resistance arm 4y tire deflection in the y-direction 4z tire deflection in the z-direction \u03b8 tire angular rotation \u03bc0, \u03bc1 nonlinear rolling friction coefficient \u03bcr rolling friction coefficient \u03bcx (s) longitudinal friction coefficient \u03bcdp friction coefficient driving peak value \u03bcds friction coefficient steady-state value \u03c3zM maximum normal stress \u03c3z(x, y) normal stress over the tireprint \u03c3zm normal stress mean value \u03c4x(x, y), \u03c4y(x, y) shear stresses over the tireprint \u03c4xM , \u03c4yM maximum shear stresses \u03d5 contact angle, angular length of AP \u03c9eq equivalent tire angular velocity \u03c9w angular velocity of a wheel \u03c9w actual tire angular velocity 3. Tire Dynamics 161"
        ]
    },
    {
        "image_filename": "designv10_0_0000233_978-1-4614-0460-6-Figure6.35-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000233_978-1-4614-0460-6-Figure6.35-1.png",
        "caption": "Fig. 6.35 Force balance to determine m11",
        "texts": [
            "19 gives the stiffness term: fk \u00bc ka2 l2 \u00fe mg l : (6.20) result that we obtained from the free body diagram approach; see Eq. 6.17. Let\u2019s conclude this section (and chapter) by determining the mass, damping, and stiffness matrices for the automobile suspension model shown in Fig. 6.22. We will begin with the 2 2 mass matrix for the two degree of freedom system. We determine m11 from the force, fm11 , required to apply a unit acceleration to x1 while holding x2 stationary. The force balance is shown in Fig. 6.35, where the motionless x2 is represented as a pivot for the rigid, massless bar. The accelerations at the two masses are both \u20acx1 2= \u00bc 1 2= because they are half the distance from the pivot relative to x1. Summing the moments (clockwise moments are taken to be positive) about the x2 pivot gives: X M \u00bc fm11 2l 3 \u00fe m1 \u20acx1 2 l 3 \u00fe m2 \u20acx1 2 l 3 \u00bc fm11 2l 3 \u00fe m1 1 2 l 3 \u00fe m2 1 2 l 3 \u00bc 0: (6.21) fm mg a O mg \u03b8 l x mg sin(q ) = mgq = mg k x l a The forces are displayed in Fig. 6.36. The force summation is: X f \u00bc fm11 \u00fe fm12 m1 \u20acx1 2 \u00fe m2 \u20acx1 2 \u00bc fm11 \u00fe fm12 m1 1 2 \u00fe m2 1 2 \u00bc 0: (6"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001514_j.engfailanal.2015.06.004-Figure12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001514_j.engfailanal.2015.06.004-Figure12-1.png",
        "caption": "Fig. 12. Cracked tooth model [44].",
        "texts": [
            " [43] proposed an improved mesh stiffness model for the cracked spur gear pair where two parabolic curves are used to simulate the crack propagation path and limiting line, respectively [42] (see Fig. 10). Based on the crack propagation paths determined by FRANC, Pandya and Parey [7] presented an improved TVMS calculation method, which adopts straight lines with different slopes rather than only one straight to simulate the predicted crack path (see Fig. 11), and compared the difference of the TVMS under two crack propagation paths: a straight line and multiple straight lines. Based on the a curved crack propagation path obtained using FRANC (see Fig. 12), Zhao et al. [44] proposed a potential energy method to calculate the TVMS of a cracked gear pair. Considering the combined effects of tooth crack and the plastic inclination, Shao and Chen [45] developed a tooth plastic inclination model for a spur gear pair by considering the cracked tooth as a variable cross-section cantilever beam (see Fig. 13). Considering the effects of the gear tooth root crack which propagates into the gear-body, Zhou et al. [46] presented a revised mathematical model based on the assumption of a linear propagation path (see Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002628_j.autcon.2020.103078-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002628_j.autcon.2020.103078-Figure2-1.png",
        "caption": "Fig. 2. hTetro rotation of Z shape.",
        "texts": [
            " The cost function for translation is determined by calculating the Euclidean distance between the start location (xj,yj) and the end location (xi,yi) of each block, which is described in Eq. (1). Rotation and transformation assume that the position of the center of block B of the robot is fixed. Under this assumption, the cost function for a rotation of angle \u03c0/2 (90\u00b0) is given by Eq. (2). This equation represents the length of the arc in which each block has to travel to achieve this change. An example of this is shown in Fig. 2. The blocks move according to the dotted line. The overall cost as per the given equation would be + +l l l( 2 )\u03c0 2 , where l is the length of a block. \u2211= \u2212 + \u2212C x x y y( ) ( )translation i j A B C D i j i j , \u03f5 , , , 2 2 (1) \u2211= \u00d7 \u2212 + \u2212C pi x x y y 2 ( ) ( )rotation i A C D i B i B \u03f5 , , 2 2 (2) = \u23a7 \u23a8 \u23aa \u23a9\u23aa C a C if a O Z L T J S I C if a clockwise anticlockwise C if a up down left right ( ) \u03f5 { , , , , , , } \u03f5 { , } \u03f5 { , , , } total t transformation t rotation t translation t (3) Transformation costs are dependent on the current shape and the desired shape"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure14.63-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure14.63-1.png",
        "caption": "Figure 14.63 If we take into account the axial deformation, the horizontal support reactions are smaller and the line of force no longer coincides with the (bent) member axis: bending is generated.",
        "texts": [
            " If the deformation by normal forces is ignored (as it was in the cable), it is possible to show that no bending occurs under the given load, and that the normal forces in the frame are equal to the forces in the two-force members in Figure 14.62a. In Figure 14.62c, the frame has been isolated and all the forces acting on it are shown. The line of force coincides everywhere with the bent member axis and there is no bending anywhere. In reality, there is always some axial deformation due to normal forces. As such, the horizontal support reactions are somewhat smaller and, because the vertical support reactions remain equal, the line of force no longer coincides with the bent member axis (see Figure 14.63). Axial deformation therefore induces bending in the two-hinged frame. Since statically indeterminate structures are more sensitive to settling and temperature, statically determinate structures are generally preferable, because the force distribution is more manageable. In Figure 14.64, the statically indeterminate two-hinged frame has been changed into a statically determined three-hinged frame. With the given load, the line of force coincides everywhere with the bent member axis and there is no bending anywhere"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002208_j.ymssp.2019.05.014-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002208_j.ymssp.2019.05.014-Figure3-1.png",
        "caption": "Fig. 3. Schematic of spur gears with web structure: (a) proposed model; (b) three-dimensional finite element model.",
        "texts": [
            " Based on the static transmission error xs obtained from the deformation compatibility equation, the mesh stiffness at the meshing position can be calculated by: k \u00bc F xs min e1; e2 . . . ei . . . en\u00f0 \u00de \u00f012\u00de In this part, the plane elements are used to simulate the spur gears with complex foundation. The face width and the web thickness are mainly controlled by the planar element thickness Le (see Appendix A). In the proposed method, for the spur gears with web structure, the element thickness of the elements associated with the web structure need to be defined as the web thickness. As shown in Fig. 3a, the elements associated with the web structure are red and the elements thickness is web thickness. In order to verify the proposed method, the mesh stiffness of spur gears with web foundation obtained from the proposed method is compared with those obtained from the three-dimensional finite element method (see Fig. 3b). In this way, the spur gears with complex foundation are simulated by the plane elements and the influence of foundation parameters (including the web thickness, rim thickness and slot parameters) on the mesh stiffness will be discussed below. The real crack growth paths have great significance to the research about the meshing characteristics of cracked gears. This part will introduce the crack propagation method using the fracture mechanics theory. It should be noted that the crack propagation angle and the crack propagation increment have great influence on the crack propagation paths"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002794_j.rcim.2021.102165-Figure16-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002794_j.rcim.2021.102165-Figure16-1.png",
        "caption": "Fig. 16. The comparison of the trajectories with measurement noise considered.",
        "texts": [
            " The medium olive and pale maroon ones are generated after L-M and DE calibration respectively. The comparison of the compensated trajectories, generated with no noise considered, with the actual trajectory and theoretical trajectory is shown in Fig. 15. After calibration with L-M and DE algorithms, the deviations have dropped to about 0.06 mm and 0.001 mm, the declining contents of deviations after LM-DEH calibration is almost the same as the decrease after DE calibration. The compensated trajectories shown in Fig. 16 are generated with measurement noise considered. In this case, the deviation dropped to 0.38 mm after L-M calibration, and both the deviations after DE and LM-DEH calibration have decreased by about 47.4% compared with that after L- M calibration. It can be seen in the partial enlarged images that the deviation between the compensated trajectory and theoretical trajectory has significantly decreased compared with the actual trajectory. In G. Luo et al. Robotics and Computer-Integrated Manufacturing 71 (2021) 102165 contrast with trajectory generated after L-M calibration, trajectories generated after LM-DEH and DE calibration are closer to the theoretical trajectory"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001571_s10845-012-0682-1-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001571_s10845-012-0682-1-Figure4-1.png",
        "caption": "Fig. 4 A schematic diagram of a multilayer neural network",
        "texts": [
            " In this work, a neural network was adopted to characterize the complex relationship between input variables and the bead geometry. The structure of the neural network was a multilayer feedforward network which is normally trained with the back-propagation error algorithm. The multilayer feedforward network has the capability of approximating any nonlinear processes. A schematic representation of the structure of the multilayer feedforward network that performs the mapping between the four input process variables and two output responses is shown in Fig. 4. The network was composed of an input layer, a hidden layer and an output layer. Neurons in the hidden layers are computational elements accomplishing nonlinear mapping between process variables and bead geometry. The training data of the neural work were the first 25 inputoutput pairs shown in Table 2. Before training the network, the data should be normalized to ensure that each welding parameter has the identical effect on the network. The linear normalization was employed as it can transform a data range to another without distortion"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000336_bf02783061-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000336_bf02783061-Figure4-1.png",
        "caption": "Fig. 4. The scheme of functioning of a potentiometric immunosensor with",
        "texts": [
            " In principle, this type of laccase-based sensor can be also used to assay lignins and some phenols in waste waters, as well as phenolic compounds in wine. POTENTIOMETRIC IMMUNOSENBOR BASED ON LACCASE The capacity of laccase to catalyze the reactions of mediator-less electroreduction of oxygen and the possibility of conducting the laccaseassisted immunoassays underlied the elaboration of a new type of sensor Applied Biochernistry and Biotechnology Vol. 49, 1994 272 Yaropolov et al. immobilized laccase. (84,85). The mode of action of this sensor is shown in Fig. 4. The test antigen is crosslinked to the electrode of carbon material. Addition of laccase-antibodies conjugate solution into the reaction unit leads to binding the conjugate to antigen on the electrode surface. This results in a rapid growth of the potential on the electrode owing to the mediator-less catalysis of oxygen electroreduction by laccase. A preliminary addition of the test antigen into the reaction unit decelerates the growth of electrode potential because of the competition for binding free and immobilized antigens"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002769_j.cirpj.2021.04.010-Figure63-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002769_j.cirpj.2021.04.010-Figure63-1.png",
        "caption": "Fig. 63. Fabrication of lattice structures using aluminium alloy (a) CAD model, (b) actual [67].",
        "texts": [
            " The buy-to-fly ratio was about 92.12%, which indicates high material utilization. Fabrication of multi-directional parts through GMAW-AW GMAW based additive manufacturing was also employed to fabricate multi-directional parts and lattice structures by tilting the GMAW torch at different angles. A hollow pipe with multiple branches (Fig. 62) was manufactured using YHJ507M wire [44]. Here dimensional and angle error was within 1 mm and 0.5 . A lattice structure was fabricated in Ref. [67] using ER2319 aluminium alloy wire (Fig. 63). Geometrical accuracy of lattice structure directly depends upon strut diameter and angle, as the interconnection struts formed it. Strut diameter was depends upon ig. 56. Relationship between volume ratio of YS308L and chemical composition at he interface [49]. 427 S. Pattanayak and S.K. Sahoo CIRP Journal of Manufacturing Science and Technology 33 (2021) 398\u2013442 droplet diameter, number of droplets transferred onto the substrate, welding current, voltage, and pulse on time. Strut angle depends upon the torch angle and layer offset"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure2.18-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure2.18-1.png",
        "caption": "Figure 2.18 (a) The components of the resultant R that the cables exert on the wreckage and (b) the angles \u03b1x , \u03b1y , \u03b1z that this force makes with the coordinate axes.",
        "texts": [
            " Question: Find the resultant force being exerted by the cables on the wreckage. 34 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Solution: First, we have to resolve the forces F1 and F2 into their components in the same way as described in the previous section: Fx = Fdx/d , etc. The calculation is shown in Table 2.1 (d in m and F in kN). The components of the resultant force R on the wreckage are therefore Rx = \u2211 Fx = \u22129.68 kN, Ry = \u2211 Fy = +9.26 kN, Rz = \u2211 Fz = \u22123.33 kN, so that R = \u221a R2 x + R2 y + R2 z = \u221a (\u22129.68 kN)2 + (9.26 kN)2 + (\u22123.33 kN)2 = 13.80 kN. Figure 2.18a shows the components of the resultant R as they act on the wreckage. Figure 2.18b shows the angles that the resultant makes with the coordinate axes. The angles are calculated as follows: cos \u03b1x = Rx R = \u22129.68 kN 13.80 kN = \u22120.701 \u21d2 \u03b1x = 134.5\u25e6, 2 Statics of a Particle 35 cos \u03b1y = Ry R = +9.26 kN 13.80 kN = +0.671 \u21d2 \u03b1y = 47.8\u25e6, cos \u03b1z = Rz R = \u22123.33 kN 13.80 kN = \u22120.241 \u21d2 \u03b1z = 104.0\u25e6. According to Newton\u2019s first law, if the resultant of all the forces on a particle is zero, it will remain at rest if it was at rest originally. This means that the particle is in equilibrium. If the particle is to be in equilibrium, then \u2211 F = 0"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002151_j.engfailanal.2014.11.015-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002151_j.engfailanal.2014.11.015-Figure1-1.png",
        "caption": "Fig. 1. Transmission paths [8].",
        "texts": [
            " The line of force Fspn(t) is the internal common tangent of the base circles of the sun gear and the n-th planet gear. While the line of force Frpn(t) is the internal common tangent of the base circles of the ring gear and the n-th planet gear. Since dynamic forces Fspn(t) and Frpn(t) are not in the same direction, it is not proper to add weighted Fspn(t) and weighted Frpn(t) together as two scalars. Feng and Zuo [8] investigated possible transmission paths of vibration signals in a planetary gearbox. In Fig. 1, three transmission paths are illustrated from its origin to the transducer. According to their studies, the transducer perceived signal arriving along paths 2 and 3 will have negligible amplitude. Therefore, only the first transmission path was considered in their studies, and the transmission path was modeled by a Hanning function with a time duration of Tc. All the previous studies modeled the effect of transmission path using a Hanning function [7,8,11]. They all assumed that as planet n approached the transducer location, its influence would increase, reaching its maximum when planet n was the closest to the transducer location, then, its influence would decrease to zero as the planet went away from the transducer",
            " Time-varying mesh stiffness is one of the main sources of vibration in a gear transmission system [27]. In Ref. [12], the gear mesh stiffness was approximated by a square waveform. The square waveform reflects only the effect of the change in the tooth contact number, but ignores the effect of the change in the tooth contact position [28]. In addition, the physical damping was ignored in their model. In this study, more accurate physical parameters (mesh stiffness and physical damping) will be adopted in the simulation of vibration signals. Fig. 1 shows a two-dimensional lumped-parameter model which is used in this study to simulate the vibration signals of a planetary gear set, which consists of one sun gear (s), one ring gear (r), one carrier (c) and N planet gears (p). Each component has three degrees of freedom: transverse motions in the x-axis and y-axis, and rotation. The rotation coordinates hi, i = s, r, c, p1, . . ., pN are the angular displacement. The sun gear, ring gear and carrier translations xj, yj, j = s, r, c and planet translations xpn, ypn, n = 1, "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003900_i2007-10196-1-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003900_i2007-10196-1-Figure2-1.png",
        "caption": "Fig. 2. (a\u2013c) Schematic illustrations of molecular alignment configurations: uniaxial planar alignment (a), twisted nematic (TN) alignment (b) and splayed alignment (c). (d) Model of a photomechanical actuator with bending radius r under incident light with intensity I0. The top left corner of the flap is chosen as the origin.",
        "texts": [
            " For crosslinking, a radical photoinitiator (Ciba Irgacure 819) was used. The polymer obtained after photocuring will be denoted by AnP, where n indicates the weight percentage of A3MA in the monomeric mixture. The mixtures of monomers showed a nematic-to-isotropic phase transition between 68 and 75\u25e6C corresponding to 10% and 0% A3MA, respectively. In the nematic phase, these monomeric mixtures self-assemble into ordered systems dictated by surface conditions. Three examples of these assemblies are shown in Figure 2a\u2013c. Polymerization in the nematic phase results in a densely crosslinked, glassy polymer with a high degree of molecular order. Actuators with a uniaxial planar or twisted nematic molecular ordering were made following the approach presented in [6] and [10]. Due to the preference of the monomeric mixture to align homeotropically to the air interface, a splayed molecular ordering can be obtained by photopolymerizing the monomeric mixture on a single substrate with rubbed polyimide (Nissan Sunever 7511)",
            " Therefore, the modeling approach is adopted and expanded to account for different concentrations of azobenzene and changes in the director orientation through the thickness of the film. Theoretical [12] and experimental [6] evidence in the literature suggest that when the light intensity is low with respect to the dye loading, the deformation scales linearly with the light intensity. For sake of simplicity, the absorption of the azo-dye in the photostationary UV-illuminated state is assumed to be isotropic. The coordinate system is chosen as depicted in Figure 2d. The attenuation of the light through the thickness of the film is described by I(x) = I0e (\u2212x\u03d5/d) , (1) where I(x) is the intensity of the light at thickness x, I0 is the intensity of the light entering the film at x = 0, \u03d5 is the weight fraction of azobenzene monomer and d is the attenuation length per fraction azobenzene. In analogy to a mechanical compliance which relates stresses to imposed strains, the relation between absorbed light quanta and resulting strain will be called here photocompliance, with unit cm2/W",
            " However, the effect for the projections on the y-axis is effectively the same and therefore the model treats TN and splay alignments as the same system. In the case of the splay or TN alignments, the modulus in the plane of the film varies with depth and thus has an influence on the bending of the length axis of the film. For the splay and TN system, we approach this modulus profile with a simple quadratic profile, Ey(x) = cos2(\u03b8)E\u2016 + sin2(\u03b8)E\u22a5 . (5) The deformation (in the y-direction) of a cross-sectional element that bends upon illumination as shown in Figure 2d, relative to the straight, non-illuminated state, can be described by \u03b5bending(x) = x/r + c, (6) where r is the radius of curvature and c is a component describing a homogeneous contraction or expansion over the thickness of the film resulting in a pure in-plane deformation. The light-induced deformation prescribes a new \u201cnatural\u201d state and thus the resultant internal stresses are given by \u03c3y(x) = Ey(\u03b5bending \u2212 \u03b5light,y). (7) The deformed state can now be calculated analytically, solving for the bending radius r and homogeneous strain c by meeting the demand that the sum of forces and sum of moments over the cross-section is zero: \u222b h x=0 \u03c3y(x)dx = 0, \u222b h x=0 x\u03c3y(x)dx = 0"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure6.27-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure6.27-1.png",
        "caption": "FIGURE 6.27. Fifteen instant center of rotations for a 6-link mechanism.",
        "texts": [
            " Because two links have only one instant center, we have Iij = Iji. (6.218) The four instant centers of rotation for a four-bar linkage, a slider-crank, and an inverted slider-crank mechanisms are shown in Figure 6.26. The instant center of rotation for two links that slide on each other is at infinity, 350 6. Applied Mechanisms on a line normal to the common tangent. So, I14 in Figure 6.26(b) is on a line perpendicular to the ground, and I34 in Figure 6.26(c) is on perpendicular to the link 3. Figure 6.27 depicts the 15 instant centers for a six-link mechanism. Example 244 Application of instant center of rotation in vehicles. Figure 6.28 illustrates a double A-arm suspension and its equivalent kinematic model. The wheel will be fastened to the coupler link AB, witch connects the upper A-arm BN to the lower A-arm AN . The A-arms are connected to the body with two revolute joints at N and M . The body of the vehicle acts as the ground link for the suspension mechanism, which is a four-bar linkage"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure3.27-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure3.27-1.png",
        "caption": "FIGURE 3.27. Illustration of circumferential waves in a rolling tire at its critical speed.",
        "texts": [
            "74) and experimental data for a radial tire. Generally speaking, the rolling friction coefficient of radial tires show to be less than non-radials. Figure 3.26 illustrates a sample comparison. Equation (3.74) is applied when the speed is below the tire\u2019s critical speed. Critical speed is the speed at which standing circumferential waves appear and the rolling friction increases rapidly. The wavelength of the standing waves are close to the length of the tireprint. Above the critical speed, overheating happens and tire fails very soon. Figure 3.27 illustrates the circumferential waves in a rolling tire at its critical speed. Example 87 Rolling resistance force and vehicle velocity. For computer simulation purposes, a fourth degree equation is presented to evaluate the rolling resistance force Fr Fr = C0 + C1 vx + C2 v 4 x. (3.79) The coefficients Ci are dependent on the tire characteristics, however, the following values can be used for a typical raided passenger car tire: C0 = 9.91\u00d7 10\u22123 C1 = 1.95\u00d7 10\u22125 (3.80) C2 = 1.76\u00d7 10\u22129 120 3. Tire Dynamics Example 88 Road pavement and rolling resistance"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000015_0020717031000099029-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000015_0020717031000099029-Figure4-1.png",
        "caption": "Figure 4. Fifth-order differentiation.",
        "texts": [
            " The attained accuracies are 5:8 10 12, 1:4 10 8, 1:0 10 5 and 0.0031 for the signal tracking, the first, second and third derivatives respectively. The derivative tracking deviations changed to 8:3 10 16, 1:8 10 11, 1:2 10 7 and 0.00036 respectively after was reduced to 10 4. That corresponds to Theorem 7. Fifth-order differentiator. The attained accuracies are 1:1 10 16, 1:29 10 12, 7:87 10 10, 5:3 10 7, 2:0 10 4 and 0.014 for tracking the signal, the first, second, third, fourth and fifth derivatives respectively with \u00bc 10 4 (figure 4(a)). There is no significant improvement with further reduction of . The author wanted to demonstrate the 10th-order differentiation, but found that differentiation of the order exceeding 5 is unlikely to be performed with the standard software. Further calculations are to be carried out with precision higher than the standard long double precision (128 bits per number). D ow nl oa de d by [ U ni ve rs ity L ib ra ry U tr ec ht ] at 0 7: 07 1 3 N ov em be r 20 12 Sensitivity to noises. The main problem of the differentiation is certainly its well-known sensitivity to noises",
            " For comparison, if the successive first-order differentiation were used, the respective maximal error would be at least \"\u00f02 5\u00de \u00bc 0:649 and some additional conditions on the input signal would be required. Taking 10% as a border, achieve that the direct successive differentiation does not give reliable results starting with the order 3, while the proposed differentiator may be used up to the order 5. With the noise magnitude 0.01 and the noise frequency about 1000 the 5th-order differentiator produces estimation errors 0.000 42, 0.0088, 0.076, 0.20, 0.34 and 0.52 for signal (23) and its five derivatives respectively (figure 4(b)). The differentiator performance does not significantly depend on the noise frequency. The author found that the second differentiation scheme (15) provides for slightly better accuracies. 7.2. Output-feedback control simulation Consider a simple kinematic model of car control (Murray and Sastry 1993) _x \u00bc v cos\u2019; _y \u00bc v sin\u2019 _\u2019 \u00bc \u00f0v=l\u00de tan _ \u00bc u where x and y are Cartesian coordinates of the rear-axle middle point, \u2019 is the orientation angle, v is the longitudinal velocity, l is the length between the two axles and is the steering angle (figure 5)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure16.10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure16.10-1.png",
        "caption": "Figure 16.10 (a) Hinged beam with (b) the influence line for the shear force directly to the right of C and (c) the influence line for the shear force at S1.",
        "texts": [
            "9b has a maximum positive value when load F is at S1: MD = +2Fa. The most negative fixed-end moment occurs when the load is at A, the end of the overhang: MD = \u22124Fa. The zeros in the influence lines allow us to check: the fixed-end moment is always zero when the load is placed at one of the supports B, C or D. 1 Note: the unit rotation is applied at D and not at S2! 754 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Example 2 \u2013 Influence line for shear forces The beam from Example 1 is again shown in Figure 16.10a. Figure 16.10b shows the influence line for the shear force V right C directly to the right of C. Figure 16.10c shows the influence line for the shear force VS1 at hinge S1. The influence lines are found by introducing a slide joint directly to the right of C, and at S1, respectively, and there applying a unit displacement such that the shear force performs negative work. The deformed mechanism is then the influence line we are looking for. The (assumed) positive direction of the shear force is shown separately in the figures. The mechanisms are not shown separately. However, a \u2460 shows where in the mechanism a unit displacement was applied. In the influence line for V right C (Figure 16.10b) the paths through S1C and CS2 are parallel. After all, in the mechanism, segments S1C and CS2 can displace only with respect to one another, and cannot rotate with respect to one another. In the influence line for VS1 (Figure 16.10c) the paths through ABS1 and S1C are not parallel, as in accordance with the mechanism the segments ABS1 and S1C can rotate with respect to one another due to the hinge at S1. Example 3 \u2013 Various influence lines Figures 16.11b to 16.11e show various influence lines for the hinged beam in Figure 16.11a. The positive direction of the support reaction Bv at B is shown in Figure 16.11a. For the bending moment and the shear force, the positive directions are related to the xz coordinate system in Figure 16"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure4.18-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure4.18-1.png",
        "caption": "Figure 4.18 A steel roller support.",
        "texts": [
            "16, with the roller track parallel to the x axis, the following applies for the motion at A: ux;A = unknown (free motion), uy;A = 0 (prescribed motion), \u03d5z;A = unknown (free motion). ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM 4 Structures 123 The following applies for the forces in A in the coordinate system given: Fx;A = 0 (prescribed force), Fy;A = unknown (free force), Tz;A = 0 (prescribed force). The roller support therefore has two degrees of freedom and generates one support reaction. Note the parallel with a bar support! Figure 4.18 shows a steel roller support used in older bridge structures. Due to the continuous sideways movement, the roller can end up askew after a while. To prevent this happening, the roller is provided with a tooth structure on its sides (comparable to a cogwheel). In order to prevent displacement in the z direction, a groove is sometimes cut into the roller that fits over an open ridge in the rail and bearing pedestal. This example of a steel roller support provides a good picture of how it works. Roller supports can be made of materials other than steel, but then as sliding supports"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001374_jsen.2017.2726011-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001374_jsen.2017.2726011-Figure1-1.png",
        "caption": "Fig. 1: Experimental setup of multistage centrifugal blower",
        "texts": [
            "html for more information. Finally, Section V presents the conclusion and future work. An experiment was performed using the multistage centrifugal blower fault diagnostic unit at Guangdong Provincial Key Laboratory, which diagnoses petrochemical equipment faults within a large petrochemical rotating machinery fault diagnosis test platform. The unit combines an 11kW fivestage centrifugal blower with transmission, a torque sensor, an inverter motor, a standard plate, and various failure bearings parts, as shown in Fig. 1. The vibration signal is extracted from the acceleration sensor mounted on the bearing housing, and the fault vibration signal is collected by the EMT490 data acquisition device. The experimental parameters were as follows: speed of 1000 r/min, sampling frequency of 1000 Hz, and 1024 sampling points. In this experiment, the vibration signals of four bearing states (normal bearing, bearing with cracked outer ring , bearing with cracked inner ring, lacking ball bearing) were sampled separately, these are given in Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003445_027836498900800301-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003445_027836498900800301-Figure8-1.png",
        "caption": "Fig. 8. Mechanical transducer : a linear potentiometer.",
        "texts": [
            " Some other novel mechanical devices detect the sound generated by movement against the sensor surface. Such a sensor developed at the University of Karlsruhe, West Germany, uses a microphone containing a piezocrystal. Microphones are mounted on opposing jaws of a sensory gripper, so the signals generated by the sensor are the result of actions within the gripping area. Noise from the gripper motor is filtered out by a rubber material covering the crystal. Slip can also be detected by this sensor (Dillman 1982). Figure 8 illustrates a mechanically based sensor. Many of the devices detailed in other sections could technically be described as mechanical sensors (for example, Sato, Heginbotham, and Pugh 1986), but they have been included elsewhere to illustrate an alternative design within a transduction class. 2.2.6. Optical Transduction Methods The development of optical fiber technology and solid-state cameras has led to some interesting new tactile sensor designs. The capability for high-spatialresolution images, freedom from electrical interference, and ease of separation of sensor from processing electronics are some of the attractions of incorporating optical transduction methods into tactile sensors"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002097_978-1-4939-3017-3-Figure1.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002097_978-1-4939-3017-3-Figure1.1-1.png",
        "caption": "Figure 1.1.1. Kinematic chains.",
        "texts": [
            " A set of generalized coordinates is minimal in the sense that no set of fewer variables suffices to determine the locations of all points on the mechanism. The number of variables in a set of generalized coordinates for a mechanical system is called the number of degrees of freedom of the system. 1.1.1 Example (A Simple Kinematic Chain). Simple ideas along this line, which will be generalized to provide the foundation of most of the models studied in this book, may be illustrated using the simple kinematic chain shown in Figure 1.1.1. Here there are drawn two copies of the same mechanism. This mechanism consists of planar rigid bodies connected by massless rods, and the joints are free to rotate in a fixed plane. In the first, the motion of a typical point P is described in terms of coordinate variables (\u03b81, \u03b82), where \u03b82 is the relative angle between the two links in the chain. In Figure 1.1.1 (b), the motion of the typical point P is described in terms of coordinate variables (\u03d51, \u03d52), which are the (absolute) angles of the links with respect to the vertical direction. Other choices of coordinate variables are, of course, possible. In any case, the coordinate variables serve the purpose of describing the location of typical points of the mechanism with respect to a privileged coordinate frame, which we may refer to as an inertial frame. A thorough axiomatic discussion of inertial frames is beyond the scope of this book, but roughly speaking, these are frames that are \u201cnonaccelerating relative to the distant stars"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.5-1.png",
        "caption": "FIGURE 7.5. The required space for a turning two-axle vehicle.",
        "texts": [
            " The track w in the kinematic condition (7.1) refers to the front track wf . The rear track has no effect on the kinematic condition of a front-wheel-steering vehicle. The rear track wr of a FWS vehicle can be zero with the same kinematic steering condition (7.1). Example 259 Space requirement. The kinematic steering condition can be used to calculate the space requirement of a vehicle during a turn. Consider the front wheels of a two-axle vehicle, steered according to the Ackerman geometry as shown in Figure 7.5. The outer point of the front of the vehicle will run on the maximum radius RMax, whereas a point on the inner side of the vehicle at the location of the rear axle will run on the minimum radius Rmin. The front outer point has an overhang distance g from the front axle. The maximum radius RMax is RMax = q (Rmin + w)2 + (l + g)2. (7.16) 384 7. Steering Dynamics Therefore, the required space for turning is a ring with a width 4R, which is a function of the vehicle\u2019s geometry. 4R = RMax \u2212Rmin = q (Rmin + w)2 + (l + g)2 \u2212Rmin (7"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure4.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure4.2-1.png",
        "caption": "Fig. 4.2. Actuator in the i-th joint of robot (the remaining joints are fixed)",
        "texts": [
            "4) 124 4 Control of Robots Now, instead of Eq. (4.3) we may write: (4.5) where A; is matrix of dimensions 2 x 2, h; and}; are vectors of dimensions 2 x 1 given by: A.=[O 1 \u00b0 (4.6) The actuator is driving the i-th joint while all other joints are in some fixed positions qj = q? ,j = 1, 2, ... , n,j #- i. The i-th actuator is driving the mechanical part of the robot around the i-th joint. In the given fixed positions of the joints q?, j > i, the mechanical part of the robot has a constant moment of inertia around the i-th joint Hii(qf) (Fig. 4.2). The actuator practically drives the set of links which all together have a moment of inertia around the i-th joint H;;(qf). These links also produce gravitational moment around the axis of the i-th joint; this moment depends on the fixed positions of all joints (j> i) and the instant (variable) angle of the i-th joint, i.e. Gi(q?, qJ Thus, the moment which the mechanism produces around the shaft of the i-th motor might be written as: (4.7) If we introduce the dynamic model of the mechanism rotation around the i-th joint in the model of actuator (Eq"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003947_robot.2004.1308858-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003947_robot.2004.1308858-Figure2-1.png",
        "caption": "Fig. 2. Location of the ZRAM point, denoted by A, for four different ground geometries. The ZRAM point is located at the intersection of the ground surface and a laterally shifted GRF such that it passes through the CoM, ensuring H c = 0.",
        "texts": [
            " This also causes the GRF Line of action to penetrate the ground at a different point, and this point might not Lie within the convex hull of the foot support area. If the GRF were to act through this shifted point (point A in Fig. Ib), while maintaining its original direction, HG would reduce to zero. We name point A the ZRAM point @ro gate of change of Angular Momentum). The actual position of the ZL&l point will clearly depend on the geometry of the ground as schematically depicted for four different Situations in Fig. 2. ZRAM point possesses several advantages as a stability measure for biped robots. It is important to not lose sight of the fact that it is H G that contains stability information of the robot. ZRAM is derived from H G , and one may perhaps derive other such criteria. The robot controller may be used to directly control H G or one of the derived quantities. Angular momentum rate change is physically central to rotational instability and intuitively more transparent to the phenomena of tipping and tumbling"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003418_physreve.65.041720-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003418_physreve.65.041720-Figure5-1.png",
        "caption": "FIG. 5. Schematic depiction of the potential energy profile of a photosensitive moiety versus configuration showing the relative stabilities of the cis and trans isomers.",
        "texts": [
            " However, the parameters are rather arbitrary and appear to be quite unrelated to the corresponding spontaneous thermal expansion curves. It will shortly be seen that a different, universal theoretical dependence on temperature and irradiation time describes the data 0-5 with a much greater degree of fidelity. Accordingly, the solid lines in Fig. 4 serve only as a rough guide to the eye. The internal energy of rodlike trans, and the kinked cis mesogenic moieties is adequately described as a two-level potential well, with the former configuration energetically more stable than the latter ~Fig. 5!. In the experiments conducted, there are three possible ways of moving between states, each characterized by a separate time constant. ~i! Thermal excitations up, from trans to cis, with the characteristic time given by t tc5t0eU/kT, with t0 the bare attempt rate. ~ii! Thermally driven relaxation down, from cis to trans, with the time constant tct5t0eD/kT. ~iii! uv isomerization, occurring with the rate 1/tuv[h} radiation intensity. One expects that the barrier height U is much greater than the metastable state depth D , and hence one can neglect the rate of spontaneous, thermally driven trans to cis transitions, although certainly not its reverse"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure11.10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure11.10-1.png",
        "caption": "FIGURE 11.10 Example of two cell volumes for inductive coupling for Lmxy.",
        "texts": [
            " The magnetic inductances couple only to magnetization cells. However, they contribute to the vector potential for all cells in the system. Expanding the cross-product in the numerator under the integral in (11.50), we get M(r\u2032) \u00d7 (r \u2212 r\u2032) = x\u0302 [My (z \u2212 z\u2032) \u2212 Mz (y \u2212 y\u2032)] \u2212 y\u0302 [Mx (z \u2212 z\u2032) \u2212 Mz (x \u2212 x\u2032)] + z\u0302 [Mx (y \u2212 y\u2032) \u2212 My (x \u2212 x\u2032)]. (11.52) Using (11.52) and substituting it into (11.50) and by averaging as we did for the partial inductance in (5.18), a magnetic inductive coupling matrix Lm can be defined for the example given in Fig. 11.10. 298 PEEC MODELS FOR MAGNETIC MATERIAL The contribution of My to x is Lmxy km = \ud835\udf070 4\ud835\udf0b 1 x \u222bx,k \u222b \u2032y,m (z \u2212 z\u2032) \u2223 r \u2212 r\u2032 \u22233 d \u2032 y dx, (11.53) where we need to find each of the magnetic inductive couplings for all three magnetization directions. The contribution of Mz to x is Lmxz km = \ud835\udf070 4\ud835\udf0b 1 x \u222bx,k \u222b \u2032 z,m (y \u2212 y\u2032) \u2223 r \u2212 r\u2032 \u22233 d \u2032 z dx. (11.54) The contribution of Mx to y is Lmyx km = \ud835\udf070 4\ud835\udf0b 1 y \u222by,k \u222b \u2032 x,m (z \u2212 z\u2032) \u2223 r \u2212 r\u2032 \u22233 d \u2032 x dy. (11.55) The contribution of Mz to y is Lmyz km = \ud835\udf070 4\ud835\udf0b 1 y \u222by,k \u222b \u2032 z,m (x \u2212 x\u2032) \u2223 r \u2212 r\u2032 \u22233 d \u2032 z dy"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000613_978-94-015-9261-1-Figure8.12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000613_978-94-015-9261-1-Figure8.12-1.png",
        "caption": "Figure 8.12 . Mobile robot (a) in absolute coordinate system Y and (b) wheel placement in robot-fixed coordinate system Z",
        "texts": [
            " The design procedure is divided into i) [orce-iorqu e control associated with the solution to th e rigid body motion problem by using the resultant forces acting on the robot platform: ii) inverse kinematic-dynamic problem consisting in defining the relevant wheel actions. The approach enables llS to overcome known design difficulties caus ed by a speci al allocation of the wheels, the kinematic structure redundancy and complexity of the desired route . Basic model In the beginning we consider the robot platform as a rigid body in th e absolute coord inate system Y . The platform position in Y is characterized by th e pair (y , T(O')), where y = COI(Yl ' Y2) E 1R2 is the vector of th e mass-center position and 0' is the angle of th e platform orientation (Figure 8.12 (a)) . The robot platform dynamics in Cartesian space is described the equations Y V, mV = F n, J n = M , (8.87) (8.88) l The section is wr itt.en by l.V . Miroshnik and A.V . Lyamin 450 CHAPTER 8 where V E 1R2 is the vector of the absolute translational velocities, F E 1R2 is the vector of the (external) active forces, n is the angular velocity, M is the resultant torque , m and J are mass-inertia parameters. The orthogonal matrix T(o:) = I co~o: sin o I = IT[(O:) IE 50(2) - sm 0: cos 0: Ti (0:) is formed by the unit vectors Tl (0:), T2( 0:) and satisfies the differential equa tion (see Property 8",
            "1) where T(o:) = n E T(a), (8.89) E = I~1 ~ I E 80(2). The forces acting on the mobile robot are produced by a wheeled system. In the robot coordinate system Z we introduce the vector V' E 1R2 of translational speeds which can be found as V' = T(a) V (8.90) and the vector of the of resultant forces F' connected with the external forces as T IF = T (o) F. (8.91) Here we consider a multidrive vehicle, equipped with a two steering/driving wheel modules placed (in the robot coordinate system Z, see Figure 8.12 CONTROL OF MECHANICAL SYSTEMS 451 (b )) at the points Cj(zi, z~), j = 1,2. The current angles \u00dfj of the wheel rotation s are associated with the or thogonal matri ces formed by the uni t vectors Tl (\u00dfj) and T2(\u00dfj ) and satisfying the differential equa tions (8.92 ) In th e wheel-fixed coordinat e syst ems, the vectors of the forces acting in the wheels ar e presented (under the assumpt ion th at rolling friction can be neglected) by the variables uj and j j , where uj are th e driving actions and j j are the lateral friction forces"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000206_978-0-387-74315-8-Figure3-1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000206_978-0-387-74315-8-Figure3-1-1.png",
        "caption": "Figure 3-1: Spatial momentum of a rigid body",
        "texts": [
            " The discussion on inertias is concluded with a brief review of alternative representations of inertia. The ease with which inertias can be manipulated is one of the key features of spatial notation. The remaining sections deal with spatial vector analysis. A scalar product is defined for spatial vectors, which leads to the definition of the spatial transpose operator; then various techniques are described for resolving vectors and projecting them into sub-spaces using the sca.lar product. The chapter concludes with a. summary of spatial vector algebra.. 38 Consider the rigid body shown in Figure 3-1, which has a mass of m, position of centre of mass at P, and a rotational inertia of 1* about the centre of mass. The body has a spatial velocity of ~O = [ wT vOT f, and the linear -+ velocity of the centre of mass is v p = va + ro x w. The linear momentum of the rigid body is a line vector with magnitude and direction given by p = m v p and line of action passing through the body's centre of mass. It can be expressed as a spatial vector Pa = [pT (OPX p)TJT. The vector -+ OPX p is the body's moment of momentum about the origin"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure4.23-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure4.23-1.png",
        "caption": "Fig. 4.23: A scheme for loop closure of a four-bar linkage mechanism.",
        "texts": [
            " Although a grasping efficiency is based on the statics aspects of the gripper system, great importance can also be addressed to the motion behavior during the phases of approaching a suitable grasp configuration, since a grasp strongly depends on the finger position and orientation through position of points A and B and angles \u03c81 and \u03c82 , as it can be observed from Fig. 4.17. Among the several methods that are available for kinematic analysis of mechanisms, in the case of gripping mechanisms the technique of closure equations can be considered convenient even for design purposes, since it permits the deduction of closed-form expressions for the kinematics of a mechanism. Referring to the case of a four-bar linkage with the kinematic parameters shown in Fig. 4.23, the method is outlined as reviewing the fundamentals and deducing useful formulation for the analysis of gripping mechanisms. Figure 4.23 shows a general fourbar linkage in which l1 denotes the frame length, l2 and l4 are respectively the input and the output length links, and l3 is the length of the coupler on which the finger is installed. The loop closure can be determined by looking at the vectors that can be identified on the links themselves, as shown in Fig. 4.23. These vectors can identify a closed polygonal circuit that can be formulated as 04321 =+++ llll (4.6.1) Fundamentals of the Mechanics of Robotic Manipulation 267 as the sum of the vectors of the closed circuit. Indeed, one can formulate many closure equations, such as Eq. (4.6.1) as many circuits can be identified in the kinematic chain of a mechanism. But then, independent circuits should also be identified in order to write a suitable number of independent equations that will be equal to the number of unknowns in order to have a solvable system"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000494_1.1756141-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000494_1.1756141-Figure3-1.png",
        "caption": "Fig. 3 An example from \u20203\u2021 of a highly organized buckling pattern for a film\u00d5substrate system. As depicted, the film is forced to buckle into a mold with a square pattern, after which the mold is removed.",
        "texts": [
            " The herringbone buckle pattern appears to be the preferred mode whenever there exists a sufficiently large patch of smooth substrate and when the system has been cooled well below the onset of buckling. There are irregularities to the herringbone pattern, including local distortion most likely due to imperfections in either the film or substrate. Another example is shown in Fig. 2 where the substrate is pre-patterned with a single circular depression of several millimeters diameter at its center. The depression edge determines the orientation of the buckles in its vicinity, but away from the edge the herringbone pattern emerges. The example in Fig. 3, which is taken from @3#, shows the highly ordered mode that forms when a mold with a square Contributed by the Applied Mechanics Division of THE AMERICAN SOCIETY OF MECHANICAL ENGINEERS for publication in the ASME JOURNAL OF APPLIED MECHANICS. Manuscript received by the Applied Mechanics Division, May 14, 2003; final revision October 30, 2003. Editor: R. M. McMeeking. Discussion on the paper should be addressed to the Editor, Prof. Robert M. McMeeking, Journal of Applied Mechanics, Department of Mechanical and Environmental Engineering, University of California\u2013Santa Barbara, Santa Barbara, CA 93106-5070, and will be accepted until four months after final publication in the paper itself in the ASME JOURNAL OF APPLIED MECHANICS",
            " @1,2# were not able to observe the evolution of the buckling patterns as their specimens were cooled from the film deposition temperature, and thus at this time it is not possible to give an experimental description of how the herringbone mode evolves. A word of caution is in order about predicting mode patterns based on minimum energy states. The means by which deformations evolve to the minimum energy state is by no means obvious. Mechanics is replete with problems whose minimum energy states are not easily assessable. The pattern formed by forcing a film to buckle into a mold in Fig. 3 is just such an example. Once the finite amplitude mode has formed and the mold removed, it appears that the mode is locked in place and does not undergo changes towards a lower energy state unless further disturbed. To our knowledge, the nonlinear mechanics governing such behavior in buckled films has not been studied. In the case of the minimum energy herringbone mode, experimental observation confirms its existence, even though it has not been established how it evolves. This work has been supported in part by Grant NSF DMR 0213805 and in part by the Division of Engineering and Applied Sciences, Harvard University"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure6.22-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure6.22-1.png",
        "caption": "Fig. 6.22 Articulated-Body Inertia: IA 2 represents the relationship between the acceleration and the applied force on the link2. The dynamic effects of all mechanisms (gray part) farther than link2 from the body are taken into consideration.",
        "texts": [
            "44) Moreover, the joint torque u2 is a projection of this force and moment u2 = sT2 [ f2 \u03c4 2 ] . (6.45) Substituting (6.43) and (6.44) into this equation and we can solve it for the joint acceleration q\u03082 = u2 \u2212 sT2 (I S 2 (\u03be\u03071 + s\u03072q\u03072) + \u03be2 \u00d7 IS 2 \u03be2) sT2 I S 2 s2 . (6.46) By this equation, we can immediately calculate the joint acceleration q\u03082 for the given joint torque u2 without dealing with the inertia matrix that appeared in (6.39) of the previous section. However, we can use (6.46) only if link2 is the outermost link. If we have extra links starting from link2 as in Fig. 6.22, (6.44) will no longer be valid. 204 6 Dynamic Simulation One way to use (6.46) would be to introduce a new inertia tensor IA 2 which holds all effects of the extra links farther than link2, which satisfies [ f2 \u03c4 2 ] = IA 2 \u03be\u03072 + b2. (6.47) IA 2 is called articulated-body inertia. An articulated-body inertia matrix represents the relationship between the acceleration and the force acting on the link of interest. b2 is the bias force, which includes the Coriolis, centrifugal and gravity forces and the joint torques acting on the mechanism farther than the link2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure4.34-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure4.34-1.png",
        "caption": "Fig. 4.34: A force control scheme for a pneumatic actuator of grippers by using a flow proportional electrovalve.",
        "texts": [
            "33 the pressure control in chamber 1 is obtained by directly using a pressure proportional electrovalve 3/2 (three ways/two positions), which can ensure the required pressure when a suitable electric analogue signal Vrif is imposed by the user or is sent by a control unit. Usually Vrif is computed as the V value in the Fig.4.32 in a preliminary step by considering the static equilibrium of the grasp, but even by suitable start-up calibration. Fundamentals of the Mechanics of Robotic Manipulation 289 In Fig. 4.34 the pressure control in chamber 1 is obtained indirectly by using a flow proportional electrovalve 3/2, which can vary the mass flow into chamber 1 when a suitable Vrif control signal is received by the electrovalve. The electropneumatic circuit of Fig. 4.35 refers to a special solution, which has been proposed and analyzed at the Laboratory of Robotics and Mechatronics, in Cassino. It mainly consists of controlling the air pressure in chamber 1 by using two digital electrovalves 2/2, V1 and V2, which are connected to chamber 1 at the supply and Chapter 4 Fundamentals of the Mechanics of Grasp290 discharge pipelines, respectively, in order to obtain an equivalent electrovalve 3/2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure12.58-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure12.58-1.png",
        "caption": "FIGURE 12.58. Spring connected rectilinear oscillator.",
        "texts": [],
        "surrounding_texts": [
            "Example 436 No mechanical harmonically forced vibration. In mechanics, there is no way to apply a periodic force on an object without attaching a mechanical device and applying a displacement. Hence, the forced vibrating system shown in Figure 12.15 has no practical application in mechanics. However, it is possible to make m from a ferromagnetic material to apply an alternative or periodic magnetic force. Example 437 F Orthogonality of functions sin\u03c9t and cos\u03c9t. Two functions f(t) and g(t) are orthogonal in [a, b] ifZ b a f(t) g(t) dt = 0. (12.110) The functions sin\u03c9t and cos\u03c9t are orthogonal in a period T = [0, 2\u03c0/\u03c9].Z 2\u03c0/\u03c9 0 sin\u03c9t cos\u03c9t dt = 0 (12.111) Example 438 F Beating in linear systems. Consider a displacement x(t) that is produced by two harmonic forces f1 and f2. f1 = F1 cos\u03c91t (12.112) f2 = F2 cos\u03c92t (12.113) Assume that the steady-state response to f1 is x1(t) = X1 cos (\u03c91t+ \u03c61) (12.114) 12. Applied Vibrations 755 and response to f2 is x2(t) = X2 cos (\u03c92t+ \u03c62) (12.115) then, because of the linearity of the system, the response to f1 + f2 would be x(t) = x1(t) + x2(t). x(t) = x1(t) + x2(t) = X1 cos (\u03c91t+ \u03c61) +X2 cos (\u03c92t+ \u03c62) (12.116) It is convenient to express x(t) in an alternative method x(t) = 1 2 (X1 +X2) (cos (\u03c91t+ \u03c61) + cos (\u03c92t+ \u03c62)) + 1 2 (X1 \u2212X2) (cos (\u03c91t+ \u03c61)\u2212 cos (\u03c92t+ \u03c62)) (12.117) and convert the sums to a product. x(t) = (X1 +X2) cos \u00b5 \u03c91 + \u03c92 2 t\u2212 \u03c61 + \u03c62 2 \u00b6 \u00d7 cos \u00b5 \u03c91 \u2212 \u03c92 2 t\u2212 \u03c61 \u2212 \u03c62 2 \u00b6 \u2212 (X1 \u2212X2) sin \u00b5 \u03c91 + \u03c92 2 t\u2212 \u03c61 + \u03c62 2 \u00b6 \u00d7 sin \u00b5 \u03c91 \u2212 \u03c92 2 t\u2212 \u03c61 \u2212 \u03c62 2 \u00b6 (12.118) This equation may be expressed better as x(t) = (X1 +X2) cos (\u21261t\u2212 \u03a61) cos (\u21262t\u2212 \u03a62) \u2212 (X1 \u2212X2) sin (\u21261t\u2212 \u03a61) sin (\u21262t\u2212 \u03a62) (12.119) if we use the following notations. \u21261 = \u03c91 + \u03c92 2 (12.120) \u21262 = \u03c91 \u2212 \u03c92 2 (12.121) \u03a61 = \u03c61 + \u03c62 2 (12.122) \u03a62 = \u03c61 \u2212 \u03c62 2 . (12.123) Figure 12.22 illustrates a sample plot of x(t) for \u03c91 = 10 \u03c92 = 12 \u03c61 = \u03c0 4 \u03c62 = \u03c0 6 (12.124) 756 12. Applied Vibrations The displacement x(t) indicates an oscillation between X1+X2 and X1\u2212 X2, with the higher frequency \u21261 inside an envelope that oscillates, at the lower frequency \u21262. This behavior is called beating. When X1 = X2 = X then x(t) = 2X cos (\u21261t\u2212 \u03a61) cos (\u21262t\u2212 \u03a62) (12.125) which becomes zero at every half period T = 2\u03c0/\u21262. 12.3.2 Base Excitation Figure 12.23 illustrates a one-DOF base excited vibrating system with a mass m supported by a spring k and a damper c. Base excited system is a good model for vehicle suspension system or any equipment the is mounted on a vibrating base. The absolute motion of m with respect to its equilibrium position is measured by the coordinate x. A sinusoidal excitation motion y = Y sin\u03c9t (12.126) is applied to the base of the suspension and makes the system vibrate. The equation of motion for the system can be expressed by either one of the following equations for the absolute displacement x m x\u0308+ c x\u0307+ kx = cY \u03c9 cos\u03c9t+ kY sin\u03c9t (12.127) x\u0308+ 2\u03be\u03c9n x\u0307+ \u03c92n x = 2\u03be\u03c9n \u03c9Y cos\u03c9t+ \u03c92n Y sin\u03c9t (12.128) or either one of the following equations for the relative displacement z. mz\u0308 + c z\u0307 + kz = m\u03c92 Y sin\u03c9t (12.129) z\u0308 + 2\u03be\u03c9n z\u0307 + \u03c92n z = \u03c92 Y sin\u03c9t (12.130) z = x\u2212 y (12.131) 12. Applied Vibrations 757 The equations of motion generate the following absolute and relative frequency responses. x = A2 sin\u03c9t+B2 cos\u03c9t (12.132) = X sin (\u03c9t\u2212 \u03d5x) (12.133) z = A3 sin\u03c9t+B3 cos\u03c9t (12.134) = Z sin (\u03c9t\u2212 \u03d5z) (12.135) The frequency response of x has an amplitude X, and the frequency response of z has an amplitude Z X Y = q 1 + (2\u03ber)2q (1\u2212 r2)2 + (2\u03ber)2 (12.136) Z Y = r2q (1\u2212 r2)2 + (2\u03ber)2 (12.137) with the following phases \u03d5x and \u03d5z for x and z. \u03d5x = tan\u22121 2\u03ber3 1\u2212 r2 + (2\u03ber) 2 (12.138) \u03d5z = tan\u22121 2\u03ber 1\u2212 r2 (12.139) The phase \u03d5x indicates the angular lag of the response x with respect to the excitation y. The frequency responses for X, Z, and \u03d5x as a function of r and \u03be are plotted in Figures 12.24, 12.25, and 12.26. Proof. Newton\u2019s method and the free body diagram of the system, as shown in Figure 12.27, generate the equation of motion mx\u0308 = \u2212c (x\u0307\u2212 y\u0307)\u2212 k (x\u2212 y) (12.140) 758 12. Applied Vibrations 12. Applied Vibrations 759 which, after substituting (12.126), makes the equation of motion (12.127). Equation (12.127) can be transformed to (12.128) by dividing over m and using the definition (12.71)-(12.73) for natural frequency and damping ratio. A practical response for a base excited system is the relative displacement z = x\u2212 y. (12.141) Relative displacement is important because for every mechanical device mounted on a suspension such as vehicle body, we need to control the maximum or minimum distance between the base and the device. Taking derivatives from (12.141) z\u0308 = x\u0308\u2212 y\u0308 (12.142) 760 12. Applied Vibrations and substituting in (12.140) m (z\u0308 + y\u0308) = \u2212c z\u0307 \u2212 kz (12.143) can be transformed to Equations (12.129) and (12.130). The steady-state solution of Equation (12.127) can be (12.132), or (12.133). To find the amplitude and phase of the response, we substitute the solution (12.132) in the equation of motion. \u2212m\u03c92 (A2 sin\u03c9t+B2 cos\u03c9t) +c\u03c9 (A2 cos\u03c9t\u2212B2 sin\u03c9t) +k (A2 sin\u03c9t+B2 cos\u03c9t) = cY \u03c9 cos\u03c9t+ kY sin\u03c9t (12.144) The coefficients of the functions sin\u03c9t and cos\u03c9t must balance on both sides of the equation. kA2 \u2212mA2\u03c9 2 \u2212 cB2\u03c9 = Y k (12.145) kB2 \u2212m\u03c92B2 + c\u03c9A2 = Y c\u03c9 (12.146) Therefore, we find two algebraic equations to calculate A2 and B2.\u2219 k \u2212m\u03c92 \u2212c\u03c9 c\u03c9 k \u2212m\u03c92 \u00b8 \u2219 A2 B2 \u00b8 = \u2219 Y k Y c\u03c9 \u00b8 (12.147) Solving for the coefficients A2 and B2\u2219 A2 B2 \u00b8 = \u2219 k \u2212m\u03c92 \u2212c\u03c9 c\u03c9 k \u2212m\u03c92 \u00b8\u22121 \u2219 Y k Y c\u03c9 \u00b8 = \u23a1\u23a2\u23a2\u23a2\u23a2\u23a3 k \u00a1 k \u2212m\u03c92 \u00a2 + c2\u03c92 (k \u2212m\u03c92) 2 + c2\u03c92 Y c\u03c9 \u00a1 k \u2212m\u03c92 \u00a2 \u2212 ck\u03c9 (k \u2212m\u03c92)2 + c2\u03c92 Y \u23a4\u23a5\u23a5\u23a5\u23a5\u23a6 (12.148) provides the steady-state solution (12.132). The amplitude X and phase \u03d5x can be found by X = q A22 +B2 2 (12.149) tan\u03d5x = \u2212B2 A2 (12.150) which, after substituting A2 and B2 from (12.148), results in the following solutions: X = \u221a k2 + c2\u03c92q (k \u2212m\u03c92) 2 + c2\u03c92 Y (12.151) tan\u03d5x = \u2212cm\u03c93 k (k \u2212m\u03c92) + c2\u03c92 (12.152) 12. Applied Vibrations 761 A more practical expressions for X and \u03d5x are Equations (12.136) and (12.138), which can be found by employing r and \u03be. To find the relative displacement frequency response (12.137), we substitute Equation (12.134) in (12.129). \u2212m\u03c92 (A3 sin\u03c9t+B3 cos\u03c9t) +c\u03c9 (A3 cos\u03c9t\u2212B3 sin\u03c9t) +k (A3 sin\u03c9t+B3 cos\u03c9t) = m\u03c92 Y sin\u03c9t (12.153) Balancing the coefficients of the functions sin\u03c9t and cos\u03c9t kA2 \u2212mA2\u03c9 2 \u2212 cB2\u03c9 = m\u03c92 Y (12.154) kB2 \u2212m\u03c92B2 + c\u03c9A2 = 0 (12.155) provides two algebraic equations to find A3 and B3.\u2219 k \u2212m\u03c92 \u2212c\u03c9 c\u03c9 k \u2212m\u03c92 \u00b8 \u2219 A3 B3 \u00b8 = \u2219 m\u03c92 Y 0 \u00b8 (12.156) Solving for the coefficients A3 and B3\u2219 A3 B3 \u00b8 = \u2219 k \u2212m\u03c92 \u2212c\u03c9 c\u03c9 k \u2212m\u03c92 \u00b8\u22121 \u2219 m\u03c92 Y 0 \u00b8 = \u23a1\u23a2\u23a2\u23a2\u23a3 m\u03c92 \u00a1 k \u2212m\u03c92 \u00a2 (k \u2212m\u03c92)2 + c2\u03c92 Y \u2212 mc\u03c93 (k \u2212m\u03c92) 2 + c2\u03c92 Y \u23a4\u23a5\u23a5\u23a5\u23a6 (12.157) provides the steady-state solution (12.134). The amplitude Z and phase \u03d5z can be found by Z = q A23 +B2 3 (12.158) tan\u03d5z = \u2212B3 A3 (12.159) which, after substituting A3 and B3 from (12.157), results in the following solutions. Z = m\u03c92q (k \u2212m\u03c92)2 + c2\u03c92 Y (12.160) tan\u03d5z = c\u03c9 k \u2212m\u03c92 (12.161) A more practical expression for Z and \u03d5z are Equations (12.137) and (12.139). 762 12. Applied Vibrations Example 439 A base excited system. Consider a mass-spring-damper system with m = 2kg k = 100000N/m c = 100N s/m. (12.162) If a harmonic base excitation y y = 0.002 sin 350t (12.163) is applied on the system, then the absolute and relative steady-state amplitude of vibrations of the mass, X and Z would be X = Y q 1 + (2\u03ber) 2q (1\u2212 r2) 2 + (2\u03ber) 2 = 1.9573\u00d7 10\u22123m (12.164) Z = Y r2q (1\u2212 r2)2 + (2\u03ber)2 = 9.589\u00d7 10\u22124m (12.165) because \u03c9n = r k m = 223.61 rad/ s \u2248 35.6Hz (12.166) \u03be = c 2 \u221a km = 0.1118 (12.167) r = \u03c9 \u03c9n = 1.5652. (12.168) The phases \u03d5x and \u03d5z for x and z are \u03d5x = tan\u22121 2\u03ber3 1\u2212 r2 + (2\u03ber) 2 = 0.489 rad \u2248 28.02 deg (12.169) \u03d5z = tan\u22121 2\u03ber 1\u2212 r2 = 1.8585 rad \u2248 106.48 deg (12.170) Therefore, the steady-state vibrations of the mass m can be expressed by the following functions. x = 1.9573\u00d7 10\u22123 sin (350t\u2212 0.489) (12.171) z = 9.589\u00d7 10\u22124 sin (350t\u2212 1.8585) (12.172) Example 440 Comparison between frequency responses. A comparison shows that Equation (12.137) is equal to Equation (12.98), and therefore the relative frequency response Z Y for a base excited system, 12. Applied Vibrations 763 is the same as acceleration frequency response X\u0308 F/m for a forces excited system. Also a graph for \u03d5z would be the same as Figure 12.17. Comparing Equations (12.136) and (12.103) indicates that the amplitude frequency response of a base excited system, X Y is the same as the transmitted force frequency response of a harmonically force excited system FT F . However, the phase of these two responses are different. Example 441 Absolute velocity and acceleration of a base excited system. Having the position frequency response of a base excited system x = A2 sin\u03c9t+B2 cos\u03c9t = X sin (\u03c9t\u2212 \u03d5x) (12.173) we are able to calculate the velocity and acceleration frequency responses. x\u0307 = A2\u03c9 cos\u03c9t\u2212B2\u03c9 sin\u03c9t = X\u03c9 cos (\u03c9t\u2212 \u03d5x) = X\u0307 cos (\u03c9t\u2212 \u03d5x) (12.174) x\u0308 = \u2212A2\u03c92 sin\u03c9t\u2212B2\u03c9 2 cos\u03c9t = \u2212X\u03c92 sin (\u03c9t\u2212 \u03d5x) = X\u0308 sin (\u03c9t\u2212 \u03d5x) (12.175) The amplitude of velocity and acceleration frequency responses, X\u0307, X\u0308 are X\u0307 = \u03c9 \u221a k2 + c2\u03c92q (k \u2212m\u03c92) 2 + c2\u03c92 Y (12.176) X\u0308 = \u03c92 \u221a k2 + c2\u03c92q (k \u2212m\u03c92)2 + c2\u03c92 Y (12.177) which can be written as X\u0307 \u03c9nY = r q 1 + (2\u03ber)2q (1\u2212 r2)2 + (2\u03ber)2 (12.178) X\u0308 \u03c92nY = r2 q 1 + (2\u03ber)2q (1\u2212 r2)2 + (2\u03ber)2 . (12.179) The velocity and acceleration frequency responses (12.178) and (12.179) are plotted in Figures 12.28 and 12.29. 764 12. Applied Vibrations 12. Applied Vibrations 765 There is a point in both figures, called the switching point or node, at which the behavior of X\u0307 and X\u0308 as a function of \u03be switches. Before the node, X\u0307 and X\u0308 increase by increasing \u03be, while they decrease after the node. To find the node, we may find the intersection between frequency response curves for \u03be = 0 and \u03be = \u221e. We apply this method to the acceleration frequency response. lim \u03be\u21920 X\u0308 \u03c92nY = \u00b1 r2 (1\u2212 r2) (12.180) lim \u03be\u2192\u221e X\u0308 \u03c92nY = \u00b1r2 (12.181) Therefore, the frequency ratio r at the intersection of these two limits is the solution of the equation r2 \u00a1 r2 \u2212 2 \u00a2 = 0. (12.182) The nodal frequency response is then equal to r = \u221a 2. (12.183) The value of acceleration frequency response at the node is not a function of \u03be. lim r\u2192 \u221a 2 X\u0308 \u03c92nY = 2 p 8\u03be2 + 1p 8\u03be2 + 1 = 2 (12.184) Applying the same method for the velocity frequency response results in the same nodal frequency ratio r = \u221a 2 2 . However, the value of the frequency response at the node is different. lim r\u2192 \u221a 2 X\u0307 \u03c9nY = \u221a 2 p 8\u03be2 + 1p 8\u03be2 + 1 = \u221a 2 (12.185) Example 442 Relative velocity and acceleration of a base excited system. We may use the relative displacement frequency response of a base excited system z = A3 sin\u03c9t+B3 cos\u03c9t = Z sin (\u03c9t\u2212 \u03d5z) (12.186) and calculate the relative velocity and acceleration frequency responses. z\u0307 = A3\u03c9 cos\u03c9t\u2212B3\u03c9 sin\u03c9t = Z\u03c9 cos (\u03c9t\u2212 \u03d5z) = Z\u0307 cos (\u03c9t\u2212 \u03d5z) (12.187) 766 12. Applied Vibrations z\u0308 = \u2212A3\u03c92 sin\u03c9t\u2212B3\u03c9 2 cos\u03c9t = \u2212Z\u03c92 sin (\u03c9t\u2212 \u03d5z) = Z\u0308 sin (\u03c9t\u2212 \u03d5z) (12.188) The amplitude of velocity and acceleration frequency responses, Z\u0307, Z\u0308 Z\u0307 = m\u03c93q (k \u2212m\u03c92) 2 + c2\u03c92 Y (12.189) Z\u0308 = m\u03c94q (k \u2212m\u03c92) 2 + c2\u03c92 Y (12.190) can be written as Z\u0307 \u03c9nY = r3q (1\u2212 r2) 2 + (2\u03ber) 2 (12.191) Z\u0308 \u03c92nY = r4q (1\u2212 r2) 2 + (2\u03ber) 2 . (12.192) Example 443 Transmitted force to the base of a base excited system. The transmitted force fT to the ground by a base excited system, such as is shown in Figure 12.23, is equal to the sum of forces in the spring and damper. fT = fk + fc = k (x\u2212 y) + c (x\u0307\u2212 y\u0307) (12.193) which based on the equation of motion (12.140) is also equal to fT = \u2212mx\u0308. (12.194) Substituting x\u0308 from (12.175) and (12.179) shows that the frequency response of the transmitted force can be written as FT kY = \u03c92 \u221a k2 + c2\u03c92q (k \u2212m\u03c92)2 + c2\u03c92 (12.195) = r2 q 1 + (2\u03ber)2q (1\u2212 r2)2 + (2\u03ber)2 . (12.196) The frequency response of FT kY is the same as is shown in Figure 12.29. 12. Applied Vibrations 767 Example 444 F Line of maxima in X/Y . The peak value of the absolute displacement frequency response X/Y happens at different r depending on \u03be. To find this relationship, we take a derivative of X/Y , given in Equation (12.136), with respect to r and solve the equation. d dr X Y = 2r \u00a1 1\u2212 r2 \u2212 2r4\u03be2 \u00a2 p 1 + 4r2\u03be2 \u00b3 (1\u2212 r2) 2 + (2\u03ber) 2 \u00b4 3 2 = 0 (12.197) Let\u2019s indicate the peak amplitude by Xmax and the associated frequency by rmax. The value of r2max is r2max = 1 4\u03be2 \u00b5 \u22121\u00b1 q 1 + 8\u03be2 \u00b6 (12.198) which is only a function of \u03be. Substituting the positive sign of (12.198) in (12.136) determines the peak amplitude Xmax. Xmax Y = 2 \u221a 2\u03be2 4 p 8\u03be2 + 1q 8\u03be2 + \u00a1 8\u03be4 \u2212 4\u03be2 \u2212 1 \u00a2p 8\u03be2 + 1 + 1 (12.199) Figure 12.30 shows Xmax and rmax as a function of \u03be. Example 445 F Line of maxima in Z/Y . The peak value of the relative displacement frequency response Z/Y happens at r > 1 depending on \u03be. To find this relationship, we take a derivative 768 12. Applied Vibrations of Z/Y , given in Equation (12.137), with respect to r and solve the equation. d dr Z Y = 2r \u00a1 1\u2212 r2 \u2212 2r4\u03be2 \u00a2\u00b3 (1\u2212 r2) 2 + (2\u03ber) 2 \u00b4 3 2 = 0 (12.200) Let\u2019s indicate the peak amplitude by Zmax and the associated frequency by rmax. The value of r2max is r2max = 1p 1\u2212 2\u03be2 (12.201) which has a real value for \u03be < \u221a 2 2 . (12.202) Substituting (12.201) in (12.137) determines the peak amplitude Zmax. Zmax Y = 1 2\u03be p 1\u2212 2\u03be2 (12.203) As an example, the maximum amplitude of a system with m = 2kg k = 100000N/m c = 100N s/m \u03c9n = 223.61 rad/ s \u03be = 0.1118 Y = 0.002m (12.204) is Zmax = Y 2\u03be p 1\u2212 2\u03be2 = 9. 058 5\u00d7 10\u22123m (12.205) that occurs at rmax = 1 4 p 1\u2212 2\u03be2 = 1.0063. (12.206) 12.3.3 Eccentric Excitation Figure 12.31 illustrates a one-DOF eccentric excited vibrating system with a mass m supported by a suspension made of a spring k and a damper c. There is an unbalance mass me at a distance e that is rotating with an angular velocity \u03c9. An eccentric excited vibrating system is a good model for vibration analysis of the engine of a vehicle, or any rotary motor that is mounted on a stationary base with a flexible suspension. 12. Applied Vibrations 769 The absolute motion of m with respect to its equilibrium position is measured by the coordinate x. When the lateral motion of m is protected, a harmonic excitation force fx = mee\u03c9 2 sin\u03c9t (12.207) is applied on m and makes the system vibrate. The distance e is called the eccentricity and me is called the eccentric mass. The equation of motion for the system can be expressed by mx\u0308+ c x\u0307+ kx = mee\u03c9 2 sin\u03c9t (12.208) or x\u0308+ 2\u03be\u03c9n x\u0307+ \u03c92n x = \u03b5e\u03c92 sin\u03c9t (12.209) \u03b5 = me m . (12.210) The absolute displacement responses of the system is x = A4 sin\u03c9t+B4 cos\u03c9t (12.211) = X sin (\u03c9t\u2212 \u03d5e) (12.212) which has an amplitude X, and phases \u03d5e X e\u03b5 = r2q (1\u2212 r2)2 + (2\u03ber)2 (12.213) \u03d5e = tan \u22121 2\u03ber 1\u2212 r2 (12.214) Phase \u03d5e indicates the angular lag of the response x with respect to the excitationmee\u03c9 2 sin\u03c9t. The frequency responses forX and \u03d5e as a function of r and \u03be are plotted in Figures 12.32 and 12.33. 770 12. Applied Vibrations 12. Applied Vibrations 771 Proof. Employing the free body diagram of the system, as shown in Figure 12.34, and applying Newton\u2019s method in the x-direction generate the equation of motion mx\u0308 = \u2212c x\u0307\u2212 kx+mee\u03c9 2 sin\u03c9t. (12.215) Equation (12.208) can be transformed to (12.209) by dividing over m and using the following definitions for natural frequency, damping ratio, and frequency ratio. \u03c9n = r k m (12.216) \u03be = c 2 \u221a km (12.217) r = \u03c9 \u03c9n (12.218) The parameter \u03b5 = me m is called the mass ratio and indicates the ratio between the eccentric mass me and the total mass m. The steady-state solution of Equations (12.208) can be (12.211), or (12.212). To find the amplitude and phase of the response, we substitute the solution (12.211) in the equation of motion. \u2212m\u03c92 (A4 sin\u03c9t+B4 cos\u03c9t) +c\u03c9 (A4 cos\u03c9t\u2212B4 sin\u03c9t) +k (A4 sin\u03c9t+B4 cos\u03c9t) = mee\u03c9 2 sin\u03c9t (12.219) The coefficients of the functions sin\u03c9t and cos\u03c9t must balance on both sides of the equation. kA4 \u2212mA4\u03c9 2 \u2212 cB4\u03c9 = mee\u03c9 2 (12.220) kB4 \u2212m\u03c92B4 + c\u03c9A4 = 0 (12.221) 772 12. Applied Vibrations Therefore, we find two algebraic equations to calculate A4 and B4.\u2219 k \u2212 \u03c92m \u2212c\u03c9 c\u03c9 k \u2212 \u03c92m \u00b8 \u2219 A4 B4 \u00b8 = \u2219 e\u03c92me 0 \u00b8 (12.222) Solving for the coefficients A4 and B4\u2219 A4 B4 \u00b8 = \u2219 k \u2212 \u03c92m \u2212c\u03c9 c\u03c9 k \u2212 \u03c92m \u00b8\u22121 \u2219 e\u03c92me 0 \u00b8 = \u23a1\u23a2\u23a2\u23a2\u23a3 k \u2212m\u03c92 \u2212 \u03c92me (k \u2212 \u03c92m)2 + c2\u03c92 e\u03c92me \u2212c\u03c9 (k \u2212 \u03c92m) 2 + c2\u03c92 e\u03c92me \u23a4\u23a5\u23a5\u23a5\u23a6 (12.223) provides the steady-state solution (12.211). The amplitude X and phase \u03d5e can be found by X = q A24 +B2 4 (12.224) tan\u03d5e = \u2212B4 A4 (12.225) which, after substituting A4 and B4 from (12.223), results in the following solutions. X = \u03c92emeq (k \u2212m\u03c92) 2 + c2\u03c92 (12.226) tan\u03d5e = c\u03c9 k \u2212m\u03c92 (12.227) A more practical expression forX and \u03d5e is Equation (12.213) and (12.214), which can be found by employing r and \u03be. Example 446 An eccentric excited system. Consider an engine with a mass m m = 110 kg (12.228) that is supported by four engine mounts, each with the following equivalent stiffness and damping. k = 100000N/m (12.229) c = 1000N s/m. (12.230) The engine is running at \u03c9 = 5000 rpm \u2248 523.60 rad/ s \u2248 83.333Hz (12.231) 12. Applied Vibrations 773 with the following eccentric parameters. me = 0.001 kg (12.232) e = 0.12m (12.233) The natural frequency \u03c9n, damping ratio \u03be, and mass ratio \u03b5 of the system, and frequency ratio r are \u03c9n = r k m = r 400000 110 = 60.302 rad/ s \u2248 9.6Hz (12.234) \u03be = c 2 \u221a km = 0.30151 (12.235) \u03b5 = me m = 0.001 110 = 9.0909\u00d7 10\u22126 (12.236) r = \u03c9 \u03c9n = 523.60 60.302 = 8.683 (12.237) The engine\u2019s amplitude of vibration is X = r2e\u03b5q (1\u2212 r2)2 + (2\u03ber)2 = 1.1028\u00d7 10\u22126m. (12.238) However, if the speed of the engine is at the natural frequency of the system, \u03c9 = 576.0 rpm \u2248 60.302 rad/ s \u2248 9.6Hz (12.239) then the amplitude of the engine\u2019s vibration increases to X = r2e\u03b5q (1\u2212 r2) 2 + (2\u03ber) 2 = 1.8091\u00d7 10\u22126m. (12.240) Example 447 Eccentric exciting systems. All rotating machines such as engines, turbines, generators, and turning machines can have imperfections in their rotating components or have irregular mass distribution, which creates dynamic imbalances. When the unbalanced components rotate, an eccentric load applies to the structure. The load can be decomposed into two perpendicular harmonic forces in the plane of rotation in lateral and normal directions of the suspension. If the lateral force component is balanced by a reaction, the normal component provides a harmonically variable force with an amplitude depending on the eccentricity mee. Unbalanced rotating machines are a common source of vibration excitation. 774 12. Applied Vibrations Example 448 Absolute velocity and acceleration of an eccentric excited system. Using the position frequency response of an eccentric excited system x = A4 sin\u03c9t+B4 cos\u03c9t = X sin (\u03c9t\u2212 \u03d5e) (12.241) we can find the velocity and acceleration frequency responses. x\u0307 = A4\u03c9 cos\u03c9t\u2212B4\u03c9 sin\u03c9t = X\u03c9 cos (\u03c9t\u2212 \u03d5e) = X\u0307 cos (\u03c9t\u2212 \u03d5e) (12.242) x\u0308 = \u2212A4\u03c92 sin\u03c9t\u2212B4\u03c9 2 cos\u03c9t = \u2212X\u03c92 sin (\u03c9t\u2212 \u03d5e) = X\u0308 sin (\u03c9t\u2212 \u03d5e) (12.243) The amplitude of velocity and acceleration frequency responses, X\u0307, X\u0308 are X\u0307 e\u03b5 = \u03c93emeq (k \u2212m\u03c92) 2 + c2\u03c92 (12.244) X\u0308 e\u03b5 = \u03c94emeq (k \u2212m\u03c92)2 + c2\u03c92 (12.245) which can be written as X\u0307 e\u03b5\u03c9n = r3q (1\u2212 r2)2 + (2\u03ber)2 (12.246) X\u0308 e\u03b5\u03c92n = r4q (1\u2212 r2) 2 + (2\u03ber) 2 . (12.247) Example 449 Transmitted force to the base of an eccentric excited system. The transmitted force fT = FT sin (\u03c9t\u2212 \u03d5T ) (12.248) to the ground by an eccentric excited system is equal to the sum of forces in the spring and damper. fT = fk + fc = kx+ cx\u0307 (12.249) 12. Applied Vibrations 775 Substituting x and x\u0307 from (12.211) shows that fT = (kA4 \u2212 c\u03c9B4) sin t\u03c9 + (kB4 + c\u03c9A4) cos t\u03c9 (12.250) therefore the amplitude of the transmitted force is FT = q (kA4 \u2212 c\u03c9B4) 2 + (kB4 + c\u03c9A4) 2 = e\u03c92me s c2\u03c92 + k2 (k \u2212m\u03c92) 2 + c2\u03c92 . (12.251) The frequency response of the transmitted force can be simplified to the following applied equation. FT e\u03c92me = q 1 + (2\u03ber) 2q (1\u2212 r2) 2 + (2\u03ber) 2 (12.252) 12.3.4 F Eccentric Base Excitation Figure 12.35 illustrates a one-DOF eccentric base excited vibrating system with a massm suspended by a spring k and a damper c on a base with mass mb. The base has an unbalance mass me at a distance e that is rotating with angular velocity \u03c9. The eccentric base excited system is a good model for vibration analysis of different equipment that are attached to the engine of a vehicle, or any equipment mounted on a rotary motor. Using the relative motion of m with respect to the base z = x\u2212 y (12.253) we may develop the equation of motion as mmb mb +m z\u0308 + c z\u0307 + kz = mme mb +m e\u03c92 sin\u03c9t (12.254) 776 12. Applied Vibrations or z\u0308 + 2\u03be\u03c9n z\u0307 + \u03c92n z = \u03b5e\u03c92 sin\u03c9t (12.255) \u03b5 = me mb . (12.256) The relative displacement response of the system is z = A5 sin\u03c9t+B5 cos\u03c9t (12.257) = Z sin (\u03c9t\u2212 \u03d5b) (12.258) which has an amplitude Z and phases \u03d5b. Z e\u03b5 = r2q (1\u2212 r2) 2 + (2\u03ber) 2 (12.259) \u03d5b = tan\u22121 2\u03ber 1\u2212 r2 (12.260) The frequency responses for Z, and \u03d5b as a function of r and \u03be are plotted in Figures 12.36 and 12.37. Proof. The free body diagram shown in Figure 12.38, along with Newton\u2019s method in the x-direction, may be used to find the equation of motion. mx\u0308 = \u2212c (x\u0307\u2212 y\u0307)\u2212 k (x\u2212 y) (12.261) mby\u0308 = c (x\u0307\u2212 y\u0307) + k (x\u2212 y)\u2212mee\u03c9 2 sin\u03c9t (12.262) 12. Applied Vibrations 777 778 12. Applied Vibrations Using z = x\u2212 y, and z\u0308 = x\u0308\u2212 y\u0308 (12.263) we may combine Equations (12.261) and (12.262) to find the equation of relative motion. mmb mb +m z\u0308 + c z\u0307 + kz = mme mb +m e\u03c92 sin\u03c9t (12.264) Equation (12.264) can be transformed to (12.255) if we divide it by mmb mb+m and use the following definitions: \u03be = c 2 r k mmb mb +m (12.265) \u03c9n = r k mb +m mmb . (12.266) The parameter \u03b5 = me mb is called the mass ratio and indicates the ratio between the eccentric mass me and the total base mass mb. The steady-state solution of Equation (12.255) can be (12.257). To find the amplitude and phase of the response, we substitute the solution in the equation of motion. \u2212\u03c92 (A5 sin\u03c9t+B5 cos\u03c9t) +2\u03be\u03c9n\u03c9 (A5 cos\u03c9t\u2212B5 sin\u03c9t) +\u03c92n (A5 sin\u03c9t+B5 cos\u03c9t) = \u03b5e\u03c92 sin\u03c9t (12.267) The coefficients of the functions sin\u03c9t and cos\u03c9t must balance on both sides of the equation. \u03c92nA5 \u2212 \u03c92A5 \u2212 2\u03be\u03c9\u03c9nB5 = \u03b5\u03c92e (12.268) 2\u03beA5\u03c9\u03c9n \u2212B5\u03c9 2 +B5\u03c9 2 n = 0 (12.269) Therefore, we find two algebraic equations to calculate A5 and B5.\u2219 \u03c92n \u2212 \u03c92 \u22122\u03be\u03c9\u03c9n 2\u03be\u03c9\u03c9n \u03c92n \u2212 \u03c92 \u00b8 \u2219 A5 B5 \u00b8 = \u2219 \u03b5\u03c92e 0 \u00b8 (12.270) Solving for the coefficients A5 and B5\u2219 A5 B5 \u00b8 = \u2219 \u03c92n \u2212 \u03c92 \u22122\u03be\u03c9\u03c9n 2\u03be\u03c9\u03c9n \u03c92n \u2212 \u03c92 \u00b8\u22121 \u2219 \u03b5\u03c92e 0 \u00b8 = \u23a1\u23a2\u23a2\u23a2\u23a3 \u03c92n \u2212 \u03c92 (\u03c92n \u2212 \u03c92) 2 + (2\u03be\u03c9\u03c9n) 2 \u03b5\u03c9 2e \u22122\u03be\u03c9\u03c9n (\u03c92n \u2212 \u03c92) 2 + (2\u03be\u03c9\u03c9n) 2 \u03b5\u03c9 2e \u23a4\u23a5\u23a5\u23a5\u23a6 (12.271) 12. Applied Vibrations 779 provides the steady-state solution (12.255). The amplitude Z and phase \u03d5b can be found by X = q A25 +B2 5 (12.272) tan\u03d5b = \u2212B5 A5 (12.273) which, after substituting A5 and B5 from (12.271), results in the following solutions: Z = \u03c92e\u03b5q (\u03c92n \u2212 \u03c92)2 + (2\u03be\u03c9\u03c9n) 2 (12.274) tan\u03d5b = 2\u03be\u03c9\u03c9n \u03c92n \u2212 \u03c92 (12.275) Equations (12.274) and (12.275) can be simplified to more practical expressions (12.259) and (12.260) by employing r = \u03c9 \u03c9n . Example 450 F A base eccentric excited system. Consider an engine with a mass mb mb = 110kg (12.276) and an air intake device with a mass m = 2kg (12.277) that is mounted on the engine using an elastic mounts, with the following equivalent stiffness and damping. k = 10000N/m (12.278) c = 100N s/m. (12.279) The engine is running at \u03c9 = 576.0 rpm \u2248 60.302 rad/ s \u2248 9.6Hz (12.280) with the following eccentric parameters. me = 0.001 kg (12.281) e = 0.12m (12.282) The natural frequency \u03c9n, damping ratio \u03be, and mass ratio \u03b5 of the system, 780 12. Applied Vibrations and frequency ratio r are \u03c9n = r k mb +m mmb = 100 rad/ s \u2248 15.9Hz (12.283) \u03be = c 2 r k mmb mb +m = 0.49995 (12.284) \u03b5 = me mb = 9.0909\u00d7 10\u22126 (12.285) r = \u03c9 \u03c9n = 0.60302. (12.286) The relative amplitude of the device\u2019s vibration is Z = e\u03b5r2q (1\u2212 r2)2 + (2\u03ber)2 = 4.525\u00d7 10\u22127m. (12.287) Example 451 F Absolute displacement of the upper mass in an eccentric base excited system. Equation (12.261) x\u0308 = \u2212 c m (x\u0307\u2212 y\u0307)\u2212 k m (x\u2212 y) = \u2212 c m z\u0307 \u2212 k m z (12.288) along with the solution (12.257) may be used to calculate the displacement frequency response of the upper mass m in the eccentric base excited system shown in Figure 12.35. Assuming a steady-state displacement x = A6 sin\u03c9t+B6 cos\u03c9t (12.289) = X sin (\u03c9t\u2212 \u03d5bx) (12.290) we have \u2212\u03c92 (A6 sin\u03c9t+B6 cos\u03c9t) = \u2212 c m z\u0307 \u2212 k m z = \u2212 c m \u03c9 (A5 cos\u03c9t\u2212B5 sin\u03c9t) \u2212 k m (A5 sin\u03c9t+B5 cos\u03c9t) = \u00b5 c m \u03c9B5 \u2212 k m A5 \u00b6 sin t\u03c9 + \u00b5 \u2212 k m B5 \u2212 c m \u03c9A5 \u00b6 cos t\u03c9 (12.291) 12. Applied Vibrations 781 and therefore, \u2212\u03c92A6 = c m \u03c9B5 \u2212 k m A5 (12.292) \u2212\u03c92B6 = \u2212 k m B5 \u2212 c m \u03c9A5. (12.293) Substituting A5 and B5 from (12.271) and using X = q A26 +B2 6 (12.294) tan\u03d5bx = \u2212B6 A6 (12.295) shows that A6 = \u2212 2c\u03be\u03c92\u03c9n + k \u00a1 \u03c92n \u2212 \u03c92 \u00a2 (\u03c92n \u2212 \u03c92) 2 + (2\u03be\u03c9\u03c9n) 2 1 m \u03b5e (12.296) B6 = \u2212c \u00a1 \u03c92n \u2212 \u03c92 \u00a2 + 2k\u03be\u03c9n (\u03c92n \u2212 \u03c92) 2 + (2\u03be\u03c9\u03c9n) 2 1 m \u03b5\u03c9e (12.297) the amplitude X of steady-state vibration of the upper mass in an eccentric base excited system is X = \u221a c2\u03c92 + k2q (\u03c92n \u2212 \u03c92)2 + (2\u03be\u03c9\u03c9n) 2 \u03b5 m e. (12.298) 12.3.5 F Classification for the Frequency Responses of One-DOF Forced Vibration Systems A harmonically excited one-DOF systems can be one of the four systems shown in Figure 12.39 as a concise version of Figure 12.14. The dimensionless amplitude of different applied steady-state responses of these systems is equal to one of the following equations (12.299)-(12.306), and the phase of the motion is equal to one of the equations (12.307)-(12.310). S0 = 1q (1\u2212 r2)2 + (2\u03ber)2 (12.299) S1 = rq (1\u2212 r2) 2 + (2\u03ber) 2 (12.300) S2 = r2q (1\u2212 r2) 2 + (2\u03ber) 2 (12.301) S3 = r3q (1\u2212 r2) 2 + (2\u03ber) 2 (12.302) S4 = r4q (1\u2212 r2) 2 + (2\u03ber) 2 (12.303) G0 = q 1 + (2\u03ber) 2q (1\u2212 r2) 2 + (2\u03ber) 2 (12.304) G1 = r q 1 + (2\u03ber) 2q (1\u2212 r2) 2 + (2\u03ber) 2 (12.305) G2 = r2 q 1 + (2\u03ber) 2q (1\u2212 r2) 2 + (2\u03ber) 2 (12.306) \u03a60 = tan\u22121 2\u03ber 1\u2212 r2 (12.307) \u03a61 = tan\u22121 1\u2212 r2 \u22122\u03ber (12.308) \u03a62 = tan\u22121 \u22122\u03ber 1\u2212 r2 (12.309) \u03a63 = tan\u22121 2\u03ber3 (1\u2212 r2) 2 + (2\u03ber) 2 (12.310) The function S0 and G0 are the main parts in all the amplitude frequency responses. To have a sense about the behavior of different responses, we use 12. Applied Vibrations 783 plot them as a function of r and using \u03be as a parameter. Mass m, stiffness k, and damper c of the system are fixed and hence, the excitation frequency \u03c9 is the only variable. We combine m, k, c, \u03c9 and define two parameters r and \u03be to express frequency responses by two variable functions. To develop a clear classification let\u2019s indicate the frequency responses related to the systems shown in Figure 12.39 by adding a subscript and express their different responses as follow: 1\u2212 For a base excitation system, we usually use the frequency responses of the relative and absolute kinematics ZB, Z\u0307B, Z\u0308B , XB, X\u0307B , X\u0308B, along with the transmitted force frequency response FTB . 2\u2212 For an eccentric excitation system, we usually use the frequency responses of the absolute kinematicsXE , X\u0307E , X\u0308E , along with the transmitted force frequency response FTE . 3\u2212 For an eccentric base excitation system, we usually use the frequency responses of the relative and absolute kinematics ZR, Z\u0307R, Z\u0308R, XR, X\u0307R, X\u0308R, YR, Y\u0307R, Y\u0308R, along with the transmitted force frequency response FTR . 4\u2212 For a forced excitation system, we usually use the frequency responses of the absolute kinematics XF , X\u0307F , X\u0308F , along with the transmitted force frequency response FTF The frequency response of different features of the four systems in Figure 12.39 may be summarized and labeled as follows: S0 = XF F/k (12.311) S1 = X\u0307F F/ \u221a km (12.312) S2 = X\u0308F F/m = ZB Y = XE e\u03b5E = ZR e\u03b5R (12.313) S3 = Z\u0307B \u03c9nY = X\u0307E e\u03b5E\u03c9n = Z\u0307R e\u03b5R\u03c9n (12.314) S4 = Z\u0308B \u03c92nY = X\u0308E e\u03b5E\u03c92n = Z\u0308R e\u03b5R\u03c92n (12.315) G0 = FTF F = XB Y (12.316) G1 = X\u0307B \u03c9nY (12.317) G2 = X\u0308B \u03c92nY = FTB kY = FTE e\u03c92nme = FTR e\u03c92nme \u00b3 1 + mb m \u00b4 (12.318) 784 12. Applied Vibrations Figures A.1-A.8 in Appendix A visualize the frequency responses used in analysis and designing of the systems. However, the exact value of the responses should be found from the associated equations. Proof. The equations of motion for a harmonically forced vibrating one DOF system is always equal to mq\u0308 + cq\u0307 + kq = f (q, q\u0307, t) (12.319) where, the variable q is a general coordinate to show the absolute displacement x, or relative displacement z = x\u2212 y. The forcing term f (x, x\u0307, t) is a harmonic function which in the general case can be a combination of sin\u03c9t and cos\u03c9t, where \u03c9 is the excitation frequency. f (q, q\u0307, t) = a sin\u03c9t+ b cos\u03c9t. (12.320) Depending on the system and the frequency response we are looking for, the coefficients a and b are zero, constant, or proportional to \u03c9, \u03c92, \u03c93, \u03c94, \u00b7 \u00b7 \u00b7 , \u03c9n. To cover every practical harmonically forced vibrating systems, let\u2019s assume a = a0 + a1\u03c9 + a2\u03c9 2 (12.321) b = b0 + b1\u03c9 + b2\u03c9 2. (12.322) We usually divide the equation of motion (12.319) by m to express that with \u03be and \u03c9n q\u0308 + 2\u03be\u03c9nq\u0307 + \u03c92nq = \u00a1 A0 +A1\u03c9 +A2\u03c9 2 \u00a2 sin\u03c9t + \u00a1 B0 +B1\u03c9 +B2\u03c9 2 \u00a2 cos\u03c9t (12.323) where, A0 +A1\u03c9 +A2\u03c9 2 = 1 m \u00a1 a0 + a1\u03c9 + a2\u03c9 2 \u00a2 (12.324) B0 +B1\u03c9 +B2\u03c9 2 = 1 m \u00a1 b0 + b1\u03c9 + b2\u03c9 2 \u00a2 . (12.325) The solution of the equation of motion would be a harmonic response with unknown coefficients. q = A sin\u03c9t+B cos\u03c9t (12.326) = Q sin (\u03c9t\u2212 \u03d5) (12.327) To find the steady-state amplitude of the response Q Q = p A2 +B2 (12.328) \u03d5 = tan\u22121 \u2212B A (12.329) 12. Applied Vibrations 785 we should substitute the solution in the equation of motion. \u2212\u03c92 (A sin\u03c9t+B cos\u03c9t) + 2\u03be\u03c9n\u03c9 (A cos\u03c9t\u2212B sin\u03c9t) +\u03c92n (A sin\u03c9t+B cos\u03c9t) = \u00a1 A0 +A1\u03c9 +A2\u03c9 2 \u00a2 sin\u03c9t+ \u00a1 B0 +B1\u03c9 +B2\u03c9 2 \u00a2 cos\u03c9t (12.330) The coefficients of the functions sin\u03c9t and cos\u03c9t must balance on both sides of the equation. \u03c92nA\u2212 \u03c92A\u2212 2\u03be\u03c9\u03c9nB = A0 +A1\u03c9 +A2\u03c9 2 (12.331) 2\u03beA\u03c9\u03c9n \u2212B\u03c92 +B\u03c92n = B0 +B1\u03c9 +B2\u03c9 2 (12.332) Therefore, we find two algebraic equations to calculate A and B.\u2219 \u03c92n \u2212 \u03c92 \u22122\u03be\u03c9\u03c9n 2\u03be\u03c9\u03c9n \u03c92n \u2212 \u03c92 \u00b8 \u2219 A B \u00b8 = \u2219 A0 +A1\u03c9 +A2\u03c9 2 B0 +B1\u03c9 +B2\u03c9 2 \u00b8 (12.333) Solving for the coefficients A and B\u2219 A B \u00b8 = \u2219 \u03c92n \u2212 \u03c92 \u22122\u03be\u03c9\u03c9n 2\u03be\u03c9\u03c9n \u03c92n \u2212 \u03c92 \u00b8\u22121 \u2219 A0 +A1\u03c9 +A2\u03c9 2 B0 +B1\u03c9 +B2\u03c9 2 \u00b8 = \u23a1\u23a2\u23a2\u23a2\u23a3 Z1 (1\u2212 r2)2 + (2\u03ber)2 Z2 (1\u2212 r2)2 + (2\u03ber)2 \u23a4\u23a5\u23a5\u23a5\u23a6 (12.334) Z1 = 2\u03ber 1 \u03c92n \u00a1 B2\u03c9 2 +B1\u03c9 +B0 \u00a2 + 1 \u03c92n \u00a1 1\u2212 r2 \u00a2 \u00a1 A2\u03c9 2 +A1\u03c9 +A0 \u00a2 (12.335) Z2 = 1 \u03c92n \u00a1 1\u2212 r2 \u00a2 \u00a1 B2\u03c9 2 +B1\u03c9 +B0 \u00a2 \u22122\u03ber 1 \u03c92n \u00a1 A2\u03c9 2 +A1\u03c9 +A0 \u00a2 (12.336) provides the steady-state solution amplitude Q and phase \u03d5 Q = p A2 +B2 (12.337) tan\u03d5 = \u2212B A . (12.338) We are able to reproduce any of the steady-state responses Si and Gi by setting the coefficients A0, A1, A2, B0, B1, and B2 properly. 786 12. Applied Vibrations Example 452 F Base excited frequency responses. A one DOF base excited vibrating system is shown in Figure 12.23. The equation of relative motion z = x \u2212 y with a harmonic excitation y = Y sin\u03c9t is z\u0308 + 2\u03be\u03c9n z\u0307 + \u03c92n z = \u03c92 Y sin\u03c9t. (12.339) This equation can be found from Equation (12.323) if A0 = 0 A1 = 0 A2 = Y B0 = 0 B1 = 0 B2 = 0. (12.340) So, the frequency response of the system would be Z = Q = p A2 +B2 = r2q (1\u2212 r2) 2 + (2\u03ber) 2 Y (12.341) because, \u2219 A B \u00b8 = \u23a1\u23a2\u23a2\u23a2\u23a3 Z1 (1\u2212 r2) 2 + (2\u03ber) 2 Z2 (1\u2212 r2) 2 + (2\u03ber) 2 \u23a4\u23a5\u23a5\u23a5\u23a6 (12.342) Z1 = r2 \u00a1 1\u2212 r2 \u00a2 Y (12.343) Z2 = 2\u03ber3Y. (12.344) 12.4 Time Response of Vibrating Systems Linear vibrating systems have a general equation of motion as the set of differential equations, [m] x\u0308+ [c] x\u0307+ [k]x = F (12.345) with the following initial conditions. x(0) = x0 (12.346) x\u0307(0) = x\u03070 (12.347) 12. Applied Vibrations 787 The time response of the system is the solution x = x(t), t > 0 for a set of coupled ordinary differential equations. Such a problem is called an initial-value problem. Consider a one-DOF vibrating system mx\u0308+ cx\u0307+ kx = f (x, x\u0307, t) (12.348) with the initial conditions x(0) = x0 (12.349) x\u0307(0) = x\u03070. (12.350) The coefficients m, c, k are assumed constant, although, they may be functions of time in more general problems. The solution of such a problem, x = x(t), t > 0, is unique. The order of an equation is the highest number of derivatives. In mechanical vibrations of lumped models, we work with a set of second-order differential equations. If x1(t), x2(t), \u00b7 \u00b7 \u00b7 , xn(t), are solutions of an n-order equation, then its general solution is x(t) = a1x1(t) + a2x2(t) + \u00b7 \u00b7 \u00b7+ anxn(t). (12.351) When f = 0, the equation is called homogeneous, mx\u0308+ cx\u0307+ kx = 0 (12.352) otherwise it is non-homogeneous. The solution of the non-homogeneous equation (12.348) is equal to x(t) = xh(t) + xp(t) (12.353) where, xh(t) is the homogeneous solution, and xp(t) is the particular solution. In mechanical vibration, the homogeneous equation is called free vibration and its solution is called free vibration response. The non-homogeneous equation is called forced vibration and its solution is called forced vibration response. An exponential function x = e\u03bbt (12.354) satisfies every homogeneous linear differential equation. Therefore, the homogeneous response of the second order equation (12.352) is xh(t) = a1e \u03bb1t + a2e \u03bb2t (12.355) where the constants a1 and a2 depend on the initial conditions. The parameters \u03bb1 and \u03bb2 are called characteristic parameters or eigenvalues of the system The eigenvalues are the solution of an algebraic equation, called a characteristic equation, which is the result of substituting the solution 788 12. Applied Vibrations (12.354) in Equation (12.352). The characteristic equation is the condition to make the solution (12.354) satisfy the equation of motion (12.352). A general particular solution of a forced equation is hard to find, however, we know that the forcing function f = f(t) is a combination of the following functions: 1\u2212 a constant, such as f = a 2\u2212 a polynomial in t, such as f = a0 + a1t+ a2t 2 + \u00b7 \u00b7 \u00b7+ ant n 3\u2212 an exponential function, such as f = eat 4\u2212 a harmonic function, such as f = F1 sin at+ F2 cos at if the particular solution xp(t) has the same form as a forcing term. 1\u2212 xp(t) =a constant, such as xp(t) = C 2\u2212 xp(t) =a polynomial of the same degree, such as xp(t) = C0 + C1t+ C2t 2 + \u00b7 \u00b7 \u00b7+ Cnt n 3\u2212 xp(t) =an exponential function, such as xp(t) = Ceat 4\u2212 xp(t) =a harmonic function, such as xp(t) = A sin at+B cos at. If the system is force free, or the forcing term disappears after a while, the solution of the equation is called time response or transient response. The initial conditions are important in transient response. When the system has some damping, the effect of initial conditions disappears after a while, in both transient and forced vibration responses, and a steady-state response remains. If the forcing term is harmonic, then the steady-state solution is called frequency response. Example 453 A homogeneous solution of a second-order linear equation. Consider a system with the following equation of motion: x\u0308+ x\u0307\u2212 2x = 0 (12.356) x0 = 1 (12.357) x\u03070 = 7 (12.358) To find the solution, we substitute an exponential solution x = e\u03bbt in the equation of motion and find the characteristic equation. \u03bb2 + \u03bb\u2212 2 = 0 (12.359) The eigenvalues are \u03bb1,2 = 1,\u22122 (12.360) and therefore, the solution is x = a1e t + a2e \u22122t. (12.361) Taking a derivative x\u0307 = a1e t \u2212 2a2e\u22122t (12.362) 12. Applied Vibrations 789 and employing the initial conditions 1 = a1 + a2 (12.363) 7 = a1 \u2212 2a2 (12.364) provides the constants a1, a2, and the solution x = x(t). a1 = 3 (12.365) a2 = \u22122 (12.366) x = 3et \u2212 2e\u22122t (12.367) Example 454 Natural frequency. Consider a free mass-spring system such as the one shown in Figure 12.40. The system is undamped and free of excitation forces, so its equation of motion is mx\u0308+ kx = 0. (12.368) To find the solution, let\u2019s try a harmonic solution with an unknown frequency. x = A sin\u2126t+B cos\u2126t (12.369) Substituting (12.369) in (12.368) provides \u2212\u21262m (A sin\u2126t+B cos\u2126t) + k (A sin\u2126t+B cos\u2126t) = 0 (12.370) which can be collected as\u00a1 Bk \u2212Bm\u21262 \u00a2 cos\u2126t+ \u00a1 Ak \u2212Am\u21262 \u00a2 sin\u2126t = 0. (12.371) The coefficients of sin\u2126t and cos\u2126t must be zero, and hence, \u2126 = r k m (12.372) x = A sin r k m t+B cos r k m t. (12.373) 790 12. Applied Vibrations The frequency \u2126 = p k/m is the frequency of vibration of a free and undamped mass-spring system. It is called natural frequency and is shown by a special character \u03c9n. \u03c9n = r k m (12.374) A system has as many natural frequencies as its degrees of freedom. Example 455 Free vibration of a single-DOF system. The simplest free vibration equation of motion is mx\u0308+ c x\u0307+ kx = 0 (12.375) which is equivalent to x\u0308+ 2\u03be\u03c9n x\u0307+ \u03c92nx = 0. (12.376) The response of a system to free vibration is called transient response and depends solely on the initial conditions x0 = x(0) and x\u03070 = x\u0307(0). To determine the solution of the linear equation (12.375), we may search for a solution in an exponential form. x = Ae\u03bbt (12.377) Substituting (12.377) in (12.376) provides the characteristic equation \u03bb2 + 2\u03be\u03c9n\u03bb+ \u03c92n = 0 (12.378) to find the eigenvalues \u03bb1,2. \u03bb1,2 = \u2212\u03be\u03c9n \u00b1 \u03c9n q \u03be2 \u2212 1 (12.379) Therefore, the general solution for Equation (12.376) is x = A1 e \u03bb1t +A2 e \u03bb2t = A1 e \u2212\u03be\u03c9n+\u03c9n \u221a \u03be2\u22121 t +A2 e \u2212\u03be\u03c9n\u2212\u03c9n \u221a \u03be2\u22121 t = e\u2212\u03be\u03c9nt \u00a1 A1 e i\u03c9d t +A2 e \u2212i\u03c9d t\u00a2 (12.380) \u03c9d = \u03c9n q 1\u2212 \u03be2 (12.381) where \u03c9d is called damped natural frequency. By using the Euler equation ei\u03b1 = cos\u03b1+ i sin\u03b1 (12.382) we may modify solution (12.380) to the following forms: x = e\u2212\u03be\u03c9nt (B1 sin\u03c9dt+B2 cos\u03c9dt) (12.383) x = Be\u2212\u03be\u03c9nt sin (\u03c9dt+ \u03c6) (12.384) 12. Applied Vibrations 791 where B1 = i (A1 \u2212A2) (12.385) B2 = A1 +A2 (12.386) B = q B2 1 +B2 2 (12.387) \u03c6 = tan\u22121 B2 B1 . (12.388) Because the displacement x is a real physical quantity, coefficients B1 and B2 in Equation (12.383) must also be real. This requires that A1 and A2 be complex conjugates. The motion described by Equation (12.384) consists of a harmonic motion of frequency \u03c9d = \u03c9n p 1\u2212 \u03be2 and a decreasing amplitude Be\u2212\u03be\u03c9nt. Example 456 Under-damped, critically-damped, and over-damped systems. The time response of a damped one-DOF system is given by Equation (12.380). The solution can be transformed to Equation (12.383) as long as \u03be < 1. The value of a damping ratio controls the type of time response of a oneDOF system. Depending on the value of damping, there are three major solution categories: 1\u2212 under-damped, 2\u2212 critically-damped, and 3\u2212 over-damped. An under damped system is when \u03be < 1. For such a system, the char- acteristic parameters (12.379) are a complex conjugate \u03bb1,2 = \u2212\u03be\u03c9n \u00b1 i\u03c9n q 1\u2212 \u03be2 (12.389) and therefore, the general solution (12.380) x = A1 e \u03bb1t +A2 e \u03bb2t (12.390) can be transformed to (12.383) x = e\u2212\u03be\u03c9nt (B1 sin\u03c9dt+B2 cos\u03c9dt) . (12.391) An under-damped system has an oscillatory time response with a decaying amplitude as shown in Figure 12.41 for \u03be = 0.15, \u03c9n = 20\u03c0 rad, x0 = 1, and x\u03070 = 0. The exponential function e\u00b1\u03be\u03c9nt is an envelope for the curve of response. A critically damped system is when \u03be = 1. For such a system, the characteristic parameters (12.379) are equal. \u03bb = \u03bb1,2 = \u2212\u03c9n (12.392) 792 12. Applied Vibrations When the characteristic values are equal, the time response of the system is x = A1 e \u03bbt +A2 t e \u03bbt (12.393) which is equal to x = e\u2212\u03be\u03c9nt (A1 +A2t) . (12.394) Figure 12.42 shows a critically-damped response for \u03be = 1, \u03c9n = 10\u03c0 rad, x0 = 1, and x\u03070 = 0. An over damped system is when \u03be > 1. The characteristic parameters (12.379) for an over-damped system are two real numbers \u03bb1,2 = \u2212\u03be\u03c9n \u00b1 \u03c9n q \u03be2 \u2212 1 (12.395) and therefore, the exponential solution cannot be converted to harmonic 12. Applied Vibrations 793 functions. x = A1 e \u03bb1t +A2 e \u03bb2t (12.396) Starting from any set of initial conditions, the time response of an overdamped system goes to zero exponentially. Figure 12.43 shows an overdamped response for \u03be = 2, \u03c9n = 10\u03c0 rad, x0 = 1, and x\u03070 = 0. Example 457 Free vibration and initial conditions. Consider a one DOF mass-spring-damper in a free vibration. The general motion of the system, given in Equation (12.383), is x = e\u2212\u03be\u03c9nt (B1 sin\u03c9dt+B2 cos\u03c9dt) . (12.397) If the initial conditions of the system are x(0) = x0 (12.398) x\u0307(0) = x\u03070 (12.399) then, x0 = B2 (12.400) x\u03070 = \u2212\u03be\u03c9nB2 +B1\u03c9d (12.401) and hence, B1 = x\u03070 + \u03be\u03c9nx0 \u03c9d (12.402) B2 = x0. (12.403) Substituting B1 and B2 in solution (12.397) generates the general solution for free vibration of a single-DOF system. x = e\u2212\u03be\u03c9nt \u00b5 x\u03070 + \u03be\u03c9nx0 \u03c9d sin\u03c9dt+ x0 cos\u03c9dt \u00b6 (12.404) 794 12. Applied Vibrations The solution can also be written as x = e\u2212\u03be\u03c9nt \u00b5 x0 \u00b5 cos\u03c9dt+ \u03be \u03c9d \u03c9n sin\u03c9dt \u00b6 + x\u03070 \u03c9d sin\u03c9dt \u00b6 . (12.405) If the initial conditions of the system are substituted in solution (12.384) x = Be\u2212\u03be\u03c9nt sin (\u03c9dt+ \u03c6) (12.406) then, x0 = B sin\u03c6 (12.407) x\u03070 = \u2212B\u03be\u03c9n sin\u03c6+B\u03c9d cos\u03c6. (12.408) To solve for B and \u03c6, we may write B = x0 sin\u03c6 (12.409) tan\u03c6 = \u03c9dx0 x\u03070 + \u03be\u03c9nx0 (12.410) and therefore, B = 1 \u03c9d q (\u03c9dx0) 2 + (x\u03070 + \u03be\u03c9nx0) 2 . (12.411) Now the solution (12.406) becomes x = e\u2212\u03be\u03c9nt \u03c9d q (\u03c9dx0) 2 + (x\u03070 + \u03be\u03c9nx0) 2 \u00d7 sin \u00b5 \u03c9dt+ tan \u22121 \u03c9dx0 x\u03070 + \u03be\u03c9nx0 \u00b6 . (12.412) Example 458 Free vibration, initial conditions, and critically damping. If the system is critically damped, then the time response to free vibrations is x = e\u2212\u03be\u03c9nt (A1 +A2t) . (12.413) Using the initial conditions, x(0) = x0, x\u0307(0) = x\u03070, we can find the coefficients A1 and A2 as A1 = x0 (12.414) A2 = x\u03070 + \u03be\u03c9nx0 (12.415) and therefore the general critically-damped response is x = e\u2212\u03be\u03c9nt (x0 + (x\u03070 + \u03be\u03c9nx0) t) . (12.416) 12. Applied Vibrations 795 Example 459 Free vibration, initial conditions, and over damping. If the system is over-damped, then the characteristic parameters \u03bb1,2 are real and the time response to free vibrations is a real exponential function. x = A1 e \u03bb1t +A2 e \u03bb2t (12.417) Using the initial conditions, x(0) = x0, x\u0307(0) = x\u03070, x0 = A1 +A2 (12.418) x\u03070 = \u03bb1A1 + \u03bb2A2 (12.419) we can find the coefficients A1 and A2 as A1 = x\u03070 \u2212 \u03bb2x0 \u03bb1 \u2212 \u03bb2 (12.420) A2 = \u03bb1x0 \u2212 x\u03070 \u03bb1 \u2212 \u03bb2 . (12.421) Hence, the general over-damped response is x = x\u03070 \u2212 \u03bb2x0 \u03bb1 \u2212 \u03bb2 e\u03bb1t + \u03bb1x0 \u2212 x\u03070 \u03bb1 \u2212 \u03bb2 e\u03bb2t. (12.422) Example 460 Work done by a harmonic force. The work done by a harmonic force f(t) = F sin (\u03c9t+ \u03d5) (12.423) acting on a body with a harmonic displacement x(t) = X sin (\u03c9t) (12.424) during one period T = 2\u03c0 \u03c9 (12.425) is equal to W = Z 2\u03c0/\u03c9 0 f(t)dx = Z 2\u03c0/\u03c9 0 f(t) dx dt dt = FX\u03c9 Z 2\u03c0/\u03c9 0 sin (\u03c9t+ \u03d5) cos (\u03c9t) dt = FX Z 2\u03c0 0 sin (\u03c9t+ \u03d5) cos (\u03c9t) d (\u03c9t) = FX Z 2\u03c0 0 \u00a1 sin\u03d5 cos2 \u03c9t+ cos\u03d5 sin\u03c9t cos\u03c9t \u00a2 d (\u03c9t) = \u03c0FX sin\u03d5. (12.426) 796 12. Applied Vibrations The work W is a function of the phase \u03d5 between f and x. When \u03d5 = \u03c0 2 then the work is maximum WMax = \u03c0F0X0 (12.427) and when \u03d5 = 0, the work is minimum. Wmin = 0 (12.428) Example 461 F Response to a step input. Step input is an standard and the most important transient excitation by which we examine and compare vibrating systems. Consider a linear second order system with the following equation of motion. x\u0308+ 2\u03be\u03c9nx\u0307+ \u03c92nx = f (t) (12.429) \u03be < 1 (12.430) A step input, is a sudden change of the forcing function f (t) from zero to a constant and steady value. If the value is unity then, f (t) = \u00bd 1N/ kg t > 0 0 t \u2264 0 . (12.431) This excitation is called unit step input, and the response of the system is called the unit step response. Linearity of the equation of motion guarantees that the response to a non-unit step input is proportional to the unit step response. Consider a force function as f (t) = \u00bd F0N/ kg t > 0 0 t \u2264 0 . (12.432) The general solution of Equation (12.429) along with (12.432) is equal to sum of the homogeneous and particular solutions, x = xh + xp. The homogeneous solution was given by Equation (12.380) in Example 455. The particular solution would be constant xp = C because the input is constant f(t) = F0. Substituting xp = C in Equation (12.429) provides C = F0 \u03c92n . (12.433) Therefore, the general solution of Equation (12.429) is x = xh + xp = F0 \u03c92n + e\u2212\u03be\u03c9nt (A cos\u03c9dt+B sin\u03c9dt) t \u2265 0 (12.434) \u03c9d = \u03c9n q 1\u2212 \u03be2. (12.435) 12. Applied Vibrations 797 The zero initial conditions are the best to explore the natural behavior of the system. Applying a zero initial condition x(0) = 0 (12.436) x\u0307(0) = 0 (12.437) provides two equations for A and B F0 \u03c92n +A = 0 (12.438) \u03be\u03c9nA+ \u03c9dB = 0 (12.439) with the following solutions. A = \u2212F0 \u03c92n (12.440) B = \u2212 \u03beF0 \u03c9d\u03c9n (12.441) Therefore, the step response is x = F0 \u03c92n \u00b5 1\u2212 e\u2212\u03be\u03c9nt \u00b5 cos\u03c9dt+ \u03be\u03c9n \u03c9d sin\u03c9dt \u00b6\u00b6 . (12.442) Figure 12.44 depicts a step input for the following numerical values. \u03be = 0.3 \u03c9n = 1 F0 = 1 (12.443) 798 12. Applied Vibrations There are some characteristics for a step response: rise time tr, peak time tP , peak value xP , overshoot S = xP \u2212 F0 \u03c92n , and settling time ts. Rise time tr is the first time that the response x(t) reaches the value of the step input F0 \u03c92n . tr = 2 \u03c9d tan\u22121 \u03be + 1p 1\u2212 \u03be2 (12.444) Rise time may also be defines as the inverse of the largest slope of the step response, or as the time it takes to pass from 10% to 90% of the steady-state value. Peak time tP is the first time that the response x(t) reaches its maximum value. tP = \u03c0 \u03c9d (12.445) Peak value xP is the value of x(t) when t = tP . xP = F0 \u03c92n \u00b3 1 + e \u2212\u03be\u03c9n \u03c0 \u03c9d \u00b4 = F0 \u03c92n \u00b3 1 + e \u2212\u03be \u03c0\u221a 1\u2212\u03be2 \u00b4 (12.446) Overshoot S indicates how much the response x(t) exceeds the step input. S = xP \u2212 F0 \u03c92n = F0 \u03c92n e \u2212\u03be \u03c0\u221a 1\u2212\u03be2 (12.447) Settling time ts is, by definition, four times of the time constant of the exponential function e\u2212\u03be\u03c9nt. ts = 4 \u03be\u03c9n (12.448) Settling time may also be defines as the required time that the step response x(t) needs to settles within a \u00b1p% window of the step input. The value p = 2 is commonly used. ts \u2248 ln \u00b3 p p 1\u2212 \u03be2 \u00b4 \u03be\u03c9n (12.449) For the given data in (12.443) we find the following characteristic values. tr = 1.966 tP = 3.2933 xP = 1.3723 S = 0.3723 ts = 13.333 (12.450) 12. Applied Vibrations 799 12.5 Vibration Application and Measurement The measurable vibration parameters, such as period T and amplitude X, may be used to identify mechanical characteristics of the vibrating system. In most vibration measurement and test methods, a transient or harmonically steady-state vibration will be examined. Using time and kinematic measurement devices, we measure amplitude and period of response, and use the analytic equations to find the required data. Example 462 Damping ratio determination. Damping ratio of an under-damped one-DOF system can be found by \u03be = 1r 4 (n\u2212 1)2 \u03c02 + ln2 x1 xn ln x1 xn (12.451) \u2248 1 2 (n\u2212 1)\u03c0 ln x1 xn (12.452) which is based on a plot of x = x (t) and peak amplitudes xi. To show this equation, consider the free vibration of an under-damped one-DOF system with the following equation of motion: x\u0308+ 2\u03be\u03c9n x\u0307+ \u03c92nx = 0. (12.453) The time response of the system is given in Equation (12.383) as x = X e\u2212\u03be\u03c9nt cos (\u03c9dt+ \u03c6) (12.454) where the constants X and \u03c6 are dependent on initial conditions. 800 12. Applied Vibrations Figure 12.45 illustrates a sample of the x-response. The peak amplitudes xi are x1 = e\u2212\u03be\u03c9nt1 (X cos (\u03c9dt1 + \u03c6)) (12.455) x2 = e\u2212\u03be\u03c9nt2 (X cos (\u03c9dt2 + \u03c6)) (12.456) ... xn = e\u2212\u03be\u03c9ntn (X cos (\u03c9dtn + \u03c6)) . (12.457) The ratio of the first two peaks is x1 x2 = e\u2212\u03be\u03c9n(t1\u2212t2) cos (\u03c9dt1 + \u03c6) cos (\u03c9dt2 + \u03c6) (12.458) Because the time difference between t1 and t2 is the period of oscillation Td = t2 \u2212 t1 = 2\u03c0 \u03c9d = 2\u03c0 \u03c9n p 1\u2212 \u03be2 (12.459) we may simplify Equation (12.458) to x1 x2 = e\u03be\u03c9nTd cos (\u03c9dt1 + \u03c6) cos (\u03c9d (t1 + Td) + \u03c6) = e\u03be\u03c9nTd cos (\u03c9dt1 + \u03c6) cos (\u03c9dt1 + 2\u03c0 + \u03c6) = e\u03be\u03c9nTd . (12.460) This equation shows that, ln x1 x2 = \u03be\u03c9nTd = 2\u03c0\u03bep 1\u2212 \u03be2 (12.461) which can be used to evaluate the damping ratio \u03be. \u03be \u2248 1r 4\u03c02 + ln2 x1 x2 ln x1 x2 (12.462) For a better evaluation we may measure the ratio between x1 and any other xn, and use the following equation: \u03be \u2248 1r 4 (n\u2212 1)2 \u03c02 + ln2 x1 xn ln x1 xn (12.463) 12. Applied Vibrations 801 If \u03be << 1, then p 1\u2212 \u03be2 \u2248 1, and we may evaluate \u03be from (12.461) with a simpler equation. \u03be \u2248 1 2 (n\u2212 1)\u03c0 ln x1 xn (12.464) Example 463 Natural frequency determination. Natural frequency of a mass-spring-damper system can be found by measuring the static deflection of the system. Consider a one-DOF system shown in Figure 12.46(a) that barely touches the ground. Assume that the spring has no tension or compression. When the system rests on the ground as shown in Figure 12.46(b), the spring is compressed by a static deflection \u03b4s = mg/k because of gravity. We may determine the natural frequency of the system by measuring \u03b4s \u03c9n = r g \u03b4s (12.465) because \u03b4s = mg k = g \u03c92n . (12.466) Example 464 Moments of inertia determination. Mass moments of inertia are important characteristics of a vehicle that affect its dynamic behavior. The main moments of inertia Ix, Iy, and Iz can be calculated by an experiment. Figure 12.47 illustrates an oscillating platform hung from point A. Assume the platform has a mass M and a moment of inertia I0 about the pivot point A. Ignoring the mass of cables, we can write the Euler equation about point A X My = I0 \u03b8\u0308 = \u2212Mgh1 sin \u03b8 (12.467) 802 12. Applied Vibrations and derive the equation of motion. I0 \u03b8\u0308 +Mgh1 sin \u03b8 = 0 (12.468) If the angle of oscillation \u03b8 is very small, then sin \u03b8 \u2248 \u03b8 and therefore, Equation (12.468) reduces to a linear equation \u03b8\u0308 + \u03c92n \u03b8 = 0 (12.469) \u03c9n = r Mgh1 I0 (12.470) where \u03c9n is the natural frequency of the oscillation. \u03c9n can be assumed as the frequency of small oscillation about the point A when the platform is set free after a small deviation from equilibrium position. The natural period of oscillation Tn = 2\u03c0/\u03c9n is what we can measure, and therefore, the moment of inertia I0 is equal to I0 = 1 4\u03c02 Mgh1T 2 n . (12.471) The natural period Tn may be measured by an average period of a few cycles, or more accurately, by an accelerometer. Now consider the swing shown in Figure 12.48. A car with mass m at C is on the platform such that C is exactly above the mass center of the platform. Because the location of the mass center C is known, the distance between C and the fulcrum A is also known as h2. 12. Applied Vibrations 803 To find the car\u2019s pitch mass moment of inertia Iy about C, we apply the Euler equation about point A, when the oscillator is deviated from the equilibrium condition. X My = IA \u03b8\u0308 (12.472) \u2212Mgh1 sin \u03b8 \u2212mgh2 sin \u03b8 = I0 + Iy +mh22 (12.473) Assuming very small oscillation, we may use sin \u03b8 \u2248 \u03b8 and then Equation (12.473) reduces to a linear oscillator \u03b8\u0308 + \u03c92n \u03b8 = 0 (12.474) \u03c9n = s (Mh1 +mh2) g I0 + Iy +mh22 . (12.475) Therefore, the pitch moment of inertia Iy can be calculated by measuring the natural period of oscillation Tn = 2\u03c0/\u03c9n from the following equation. Iy = 1 4\u03c02 (Mh1 +mh2) gT 2 n \u2212 I0 \u2212mh22. (12.476) To determine the roll moment of inertia, we may put the car on the platform as shown in Figure 12.49. Having Ix and Iy we may put the car on the platform, at an angle \u03b1, to find its moment of inertia about the axis passing through C and parallel to the swing axis. Then, the product moment of inertia Ixy can be calculated by transformation calculus. 804 12. Applied Vibrations Example 465 Sample data. Tables 12.1 indicates an example of data for the mass center position, moment of inertia, and geometry of street cars, that are close to a MercedesBenz A-Class. 12.6 F Vibration Optimization Theory The first goal in vibration optimization is to reduce the vibration amplitude of a primary mass to zero, when the system is under a forced vibration. There are two principal methods for decreasing the vibration amplitude of a primary mass: vibration absorber, and vibration isolator. When the suspension of a primary system is not easy to change, we add another vibrating system, known as the vibration absorber or secondary 12. Applied Vibrations 805 system, to absorb the vibrations of the primary system. The vibration absorber increases the DOF of the system, and is an applied method for vibration reduction in frequency domain. It can work very well in a few specific frequencies, and may be designed to work well in a range of frequencies. Consider a mass m1 supported by a suspension made of only a spring k1, as shown in Figure 12.50. There is a harmonic force f = F sin\u03c9t applied on m1. We add a secondary system (m2, c2, k2) to the primary mass m1 and make a two-DOF vibrating system. Such a system is sometimes called Frahm absorber, or Frahm damper. It is possible to design the suspension of the secondary system (c2, k2) to reduce the amplitude of vibration m1 to zero at any specific excitation frequency \u03c9. However, if the excitation frequency is variable, we can adjust k2 at the optimal value k F 2 , kF2 = m1m2 (m1 +m2) 2 k1 (12.477) and select c2 within the range 2m2\u03c91\u03be F 1 < c2 < 2m2\u03c91\u03be F 2 to minimize the amplitude of m1 over the whole frequency range. The 806 12. Applied Vibrations optimal \u03beF1 and \u03beF2 are the positive values of \u03beF1 = s \u2212B \u2212 \u221a B2 \u2212 4AC 2A (12.478) \u03beF2 = s \u2212B + \u221a B2 \u2212 4AC 2A (12.479) where A = 16Z8 \u2212 4r2 (4Z4 + 8Z5) (12.480) B = 4Z9 \u2212 4Z6r2 \u2212 Z7 (4Z4 + 8Z5) + 4Z3Z8 (12.481) C = Z3Z9 \u2212 Z6Z7. (12.482) and Z3 = 2 \u00a1 r2 \u2212 \u03b12 \u00a2 (12.483) Z4 = \u00a3 r2 (1 + \u03b5)\u2212 1 \u00a42 (12.484) Z5 = r2 (1 + \u03b5) \u00a3 r2 (1 + \u03b5)\u2212 1 \u00a4 (12.485) Z6 = 2 \u00a3 \u03b5\u03b12r2 \u2212 \u00a1 r2 \u2212 \u03b12 \u00a2 \u00a1 r2 \u2212 1 \u00a2\u00a4 \u00d7 \u00a3 \u03b5\u03b12 \u2212 \u00a1 r2 \u2212 \u03b12 \u00a2 \u2212 \u00a1 r2 \u2212 1 \u00a2\u00a4 (12.486) Z7 = \u00a1 r2 \u2212 \u03b12 \u00a22 (12.487) Z8 = r2 \u00a3 r2 (1 + \u03b5)\u2212 1 \u00a42 (12.488) Z9 = \u00a3 \u03b5\u03b12r2 \u2212 \u00a1 r2 \u2212 1 \u00a2 \u00a1 r2 \u2212 \u03b12 \u00a2\u00a42 . (12.489) Proof. The equations of motion for the system shown in Figure 12.50 are: m1x\u03081 + c2 (x\u03071 \u2212 x\u03072) + k1x1 + k2 (x1 \u2212 x2) = F sin\u03c9t (12.490) m2x\u03082 \u2212 c2 (x\u03071 \u2212 x\u03072)\u2212 k2 (x1 \u2212 x2) = 0. (12.491) To find the frequency response of the system, we substitute the following solutions in the equations of motion: x1 = A1 cos\u03c9t+B1 sin\u03c9t (12.492) x2 = A2 cos\u03c9t+B2 sin\u03c9t (12.493) Assuming a steady-state condition, we find the following set of equations for A1, B1, A2, B2\u23a1\u23a2\u23a2\u23a3 a11 c2\u03c9 \u2212k2 \u2212c2\u03c9 \u2212c2\u03c9 a22 c2\u03c9 \u2212k2 \u2212k2 \u2212c2\u03c9 a33 c2\u03c9 c2\u03c9 \u2212k2 \u2212c2\u03c9 a44 \u23a4\u23a5\u23a5\u23a6 \u23a1\u23a2\u23a2\u23a3 A1 B1 A2 B2 \u23a4\u23a5\u23a5\u23a6 = \u23a1\u23a2\u23a2\u23a3 0 F 0 0 \u23a4\u23a5\u23a5\u23a6 (12.494) 12. Applied Vibrations 807 where, a11 = a22 = k1 + k2 \u2212m1\u03c9 2 (12.495) a33 = a44 = k2 \u2212m2\u03c9 2. (12.496) The steady-state amplitude X1 for vibration of the primary mass m1 is found by X1 = q A21 +B2 1 (12.497) and is equal to \u00b5 X1 F \u00b62 = \u00a1 k2 \u2212 \u03c92m2 \u00a22 + \u03c92c22 Z21 + \u03c92c22Z 2 2 (12.498) where, Z1 = \u00a1 k1 \u2212 \u03c92m1 \u00a2 \u00a1 k2 \u2212 \u03c92m2 \u00a2 \u2212 \u03c92m2k2 (12.499) Z2 = k1 \u2212 \u03c92m1 \u2212 \u03c92m2. (12.500) Introducing the parameters \u03b5 = m2 m1 (12.501) \u03c91 = r k1 m1 (12.502) \u03c92 = r k2 m2 (12.503) \u03b1 = \u03c92 \u03c91 (12.504) r = \u03c9 \u03c91 (12.505) \u03be = c2 2m2\u03c91 (12.506) \u03bc = X1 F/k1 (12.507) we may rearrange the frequency response (12.498) to the following equation. \u03bc2 = 4\u03be2r2 + \u00a1 r2 \u2212 \u03b12 \u00a22 4\u03be2r2 [r2 (1 + \u03b5)\u2212 1]2 + [\u03b5\u03b12r2 \u2212 (r2 \u2212 1) (r2 \u2212 \u03b12)] 2 (12.508) The parameter \u03b5 is themass ratio betweenm2 and them1, \u03c91 is the angular natural frequency of the main system, \u03c92 is the angular natural frequency of the vibration absorber system, \u03b1 is the natural frequency ratio, r is the excitation frequency ratio, \u03be is the damping ratio, and \u03bc is the amplitude ratio between dynamic amplitude X1 and the static deflection F/k1. 808 12. Applied Vibrations Figure 12.51 illustrates the behavior of frequency response \u03bc for \u03b5 = 0.1 (12.509) \u03b1 = 1 (12.510) and \u03be = 0 (12.511) \u03be = 0.2 (12.512) \u03be = 0.3 (12.513) \u03be = \u221e. (12.514) All the curves pass through two nodes P andQ, independent of the damping ratio \u03be. To find the parameters that control the position of the nodes, we find the intersection points of the curves for \u03be = 0 and \u03be = \u221e. Setting \u03be = 0 and \u03be =\u221e results in the following equations: \u03bc2 = \u00a1 r2 \u2212 \u03b12 \u00a22 [\u03b5\u03b12r2 \u2212 (r2 \u2212 1) (r2 \u2212 \u03b12)] 2 (12.515) \u03bc2 = 1 [r2 (1 + \u03b5)\u2212 1]2 (12.516) When \u03be = 0, the system is an undamped, linear two-DOF system with two natural frequencies. The vibration amplitude of the system approaches infinity \u03bc\u2192\u221e when the excitation frequency approaches either of the natural frequencies. When \u03be =\u221e, there would be no relative motion between 12. Applied Vibrations 809 m1 and m2. The system reduces to an undamped, linear one-DOF system with one natural frequency \u03c9n = r k1 m1 +m2 (12.517) or rn = 1\u221a 1 + \u03b5 . (12.518) The vibration amplitude of the system approaches infinity \u03bc \u2192 \u221e when the excitation frequency approaches the natural frequency \u03c9 \u2192 \u03c9ni or r \u2192 1/ (1 + \u03b5). Using Equations (12.515) and (12.516), we find\u00a1 r2 \u2212 \u03b12 \u00a22 (\u03b5\u03b12r2 \u2212 (r2 \u2212 1) (r2 \u2212 \u03b12))2 = 1 [r2 (1 + \u03b5)\u2212 1]2 (12.519) which can be simplified to \u03b5\u03b12r2 \u2212 \u00a1 r2 \u2212 1 \u00a2 \u00a1 r2 \u2212 \u03b12 \u00a2 = \u00b1 \u00a1 r2 \u2212 \u03b12 \u00a2 \u00a3 r2 (1 + \u03b5)\u2212 1 \u00a4 . (12.520) The negative sign is equivalent to r4\u03b5 = 0 which indicates that there is a common point at r = 0. The plus sign produces a quadratic equation for r2 (2 + \u03b5) r4 \u2212 r2 \u00a1 2 + 2\u03b12 (1 + \u03b5) \u00a2 + 2\u03b12 = 0 (12.521) with two positive solutions r1 and r2 corresponding to nodes P and Q. r21,2 = 1 \u03b5+ 2 \u00b3 \u03b12 \u00b1 p (\u03b52 + 2\u03b5+ 1)\u03b14 \u2212 2\u03b12 + 1 + \u03b12\u03b5+ 1 \u00b4 (12.522) r1 < rn < r2 (12.523) Because the frequency response curves always pass through P and Q, the optimal situation would be when the nodes P and Q have equal height. \u03bc (P ) = \u03bc (Q) (12.524) Because the value of \u03bc2 at P and Q are independent of \u03be, we may substitute r1 and r2 in Equation (12.516) for \u03bc corresponding to \u03be =\u221e. However, \u03bc from Equation (12.516) \u03bc = 1 [r2 (1 + \u03b5)\u2212 1] (12.525) 810 12. Applied Vibrations produces a positive number for r < rn and a negative number for r > rn. Therefore, \u03bc (r1) = \u2212\u03bc (r2) (12.526) generates the equality 1 1\u2212 r21 (1 + \u03b5) = \u22121 1\u2212 r22 (1 + \u03b5) (12.527) which can be simplified to r21 + r22 = 2 1 + \u03b5 . (12.528) The sum of the roots from Equation (12.521) is r21 + r22 = 2 + 2\u03b12 (1 + \u03b5) 1 + \u03b5 (12.529) and therefore, 2 1 + \u03b5 = \u00a3 2 + 2\u03b12 (1 + \u03b5) \u00a4 1 + \u03b5 (12.530) which provides \u03b1 = 1 1 + \u03b5 . (12.531) Equation (12.531) is the required condition to make the height of the nodes P and Q equal, and hence, provides the optimal value of \u03b1. Having an optimal value for \u03b1 is equivalent to designing the optimal stiffness k2 for the secondary suspension, because, \u03b1 = \u03c92 \u03c91 = r m1 m2 r k2 k1 (12.532) and Equation (12.531) simplifies to \u03b1 = m1 m1 +m2 (12.533) to provide the following condition for optimal kF2 : kF2 = k1 m1m2 (m1 +m2) 2 (12.534) To determine the optimal damping ratio \u03be, we force \u03bc to have its maximum at P or Q. Having \u03bcMax at P guarantees that \u03bc (r1) is the highest value in a frequency domain around r1, and having \u03bcMax at Q, guarantees 12. Applied Vibrations 811 that \u03bc (r2) is the highest value in a frequency domain around r2. The position of \u03bcMax is controlled by \u03be, so we may determine two optimal \u03be at which \u03bcMax is at \u03bc (r1) and \u03bc (r2). An example of this situation is shown in Figure 12.52. Using the optimal \u03b1 from (12.531), the nodal frequencies are r21,2 = 1 1 + \u03b5 \u00b5 1\u00b1 r \u03b5 2 + \u03b5 \u00b6 . (12.535) To set the partial derivative \u2202\u03bc2/\u2202r2 equal to zero at the nodal frequencies \u2202\u03bc2 \u2202r2 \u00af\u0304\u0304\u0304 r21 = 0 (12.536) \u2202\u03bc2 \u2202r2 \u00af\u0304\u0304\u0304 r22 = 0 (12.537) we write \u03bc2 by numerator N (r) divided by denominator D (r) \u03bc2 = N (r) D (r) (12.538) which helps to find the derivative easier. \u2202\u03bc2 \u2202r2 = 1 D2 \u00b5 D \u2202N \u2202r2 \u2212N \u2202D \u2202r2 \u00b6 = 1 D \u00b5 \u2202N \u2202r2 \u2212 N D \u2202D \u2202r2 \u00b6 (12.539) 812 12. Applied Vibrations Differentiating gives \u2202N \u2202r2 = \u2202N \u2202r2 = 4\u03be2 + Z3 (12.540) \u2202D \u2202r2 = 4\u03be2Z4 + 8\u03be 2Z5 + Z6. (12.541) Equations (12.540) and (12.541), along with (12.535), must be substituted in (12.539) to be solved for \u03be. After substitution, the equation \u2202\u03bc2/\u2202r2 = 0 would be \u2202N \u2202r2 \u2212 N D \u2202D \u2202r2 = \u00a1 4\u03be2 + Z3 \u00a2 \u00a1 4\u03be2Z8 + Z9 \u00a2 \u2212 \u00a1 4\u03be2r2 + Z7 \u00a2 \u00a1 4\u03be2Z4 + 8\u03be 2Z5 + Z6 \u00a2 = 0 (12.542) because of N D = 4\u03be2r2 + Z7 4\u03be2Z8 + Z9 . (12.543) Equation (12.542) is a quadratic for \u03be2\u00a1 16Z8 \u2212 4r2 (4Z4 + 8Z5) \u00a2 \u03be4 + \u00a1 4Z9 \u2212 4Z6r2 \u2212 Z7 (4Z4 + 8Z5) + 4Z3Z8 \u00a2 \u03be2 +(Z3Z9 \u2212 Z6Z7) = A \u00a1 \u03be2 \u00a22 +B\u03be2 + C = 0 (12.544) with the solution \u03be2 = \u2212B \u00b1 \u221a B2 \u2212 4AC 2A . (12.545) The positive value of \u03be from (12.545) for r = r1 and r = r2 provides the limiting values for \u03beF1 and \u03beF2 . Figure 12.52 shows the behavior of \u03bc for optimal \u03b1 and \u03be = 0, \u03beF1 , \u03be F 2 ,\u221e. Example 466 F Optimal spring and damper for \u03b5 = 0.1. Consider a Frahm vibration absorber with \u03b5 = m2 m1 = 0.1. (12.546) We adjust the optimal frequency ratio \u03b1 form Equation (12.531). \u03b1F = 1 1 + \u03b5 \u2248 0.9091 (12.547) and find the nodal frequencies r21,2 from (12.535). r21,2 = 1 1 + \u03b5 \u00b5 1\u00b1 r \u03b5 2 + \u03b5 \u00b6 = 0.71071, 1.1075. (12.548) 12. Applied Vibrations 813 Now, we set r = r1 = \u221a 0.71071 \u2248 0.843 and evaluate the parameters Z3 to Z9 from (12.483)-(12.489) Z3 = \u22120.231470544 Z4 = 0.0476190476 Z5 = \u22120.1705988426 Z6 = 0.0246326501 Z7 = 0.01339465321 Z8 = 0.03384338136 Z9 = 0.0006378406298 (12.549) and the coefficients A, B, and C from (12.480)-(12.482) A = 3.879887219 B = \u22120.08308086729 C = \u22120.0004775871233 (12.550) to find the first optimal damping ratio \u03be1. \u03beF1 = 0.1616320694 (12.551) Using r = r2 = \u221a 1.1075 \u2248 1.05236 we find the following numbers Z3 = 0.562049056 Z4 = 0.04761904752 Z5 = 0.2658369375 Z6 = \u22120.375123324 Z7 = 0.07897478534 Z8 = 0.05273670508 Z9 = 0.003760704084 (12.552) A = \u22129.421012739 B = 0.1167823931 C = 0.005076228579 (12.553) to find the second optimal damping ratio \u03be1. \u03beF2 = 0.1738496023 (12.554) Therefore, the optimal \u03b1 is \u03b1F = 0.9091, and the optimal \u03be is between 0.1616320694 < \u03beF < 0.1738496023. 814 12. Applied Vibrations Example 467 F The vibration absorber is most effective when r = \u03b1 = 1. When \u03be = 0, then \u03bc = 0 at r = 1, which shows the amplitude of the primary mass reduces to zero if the natural frequency of the primary and secondary systems are equal to the excitation frequency r = \u03b1 = 1. Example 468 F The optimal nodal amplitude. Substituting the optimal \u03b1 from (12.531) in Equation (12.521), r4 \u2212 2 2 + \u03b5 r2 + 2 (2 + \u03b5) (1 + \u03b5) 2 = 0 (12.555) provides the following nodal frequencies: r21,2 = 1 1 + \u03b5 \u00b5 1\u00b1 r \u03b5 2 + \u03b5 \u00b6 (12.556) Applying r1,2 in Equation (12.525) shows that the common nodal amplitude \u03bc (r1,2) is \u03bc = r 2 + \u03b5 \u03b5 . (12.557) Example 469 F Optimal \u03b1 and mass ratio \u03b5. The optimal value of the natural frequency ratio, \u03b1, is only a function of mass ratio \u03b5, as determined in Equation (12.531). Figure 12.53 depicts the behavior of \u03b1 as a function of \u03b5. The value of optimal \u03b1, and hence, the value of optimal k2, decreases by increasing \u03b5 = m2/m1. Therefore, a smaller mass for the vibration absorber needs a stiffer spring. 12. Applied Vibrations 815 Example 470 F Nodal frequencies r1,2 and mass ratio \u03b5. As shown in Equation (12.535), the nodal frequencies r1,2 for optimal \u03b1 (12.531), are only a function of the mass ratio \u03b5. r21,2 = 1 1 + \u03b5 \u00b5 1\u00b1 r \u03b5 2 + \u03b5 \u00b6 (12.558) Figure 12.54 illustrates the behavior of r1,2 as a function of \u03b5. When, \u03b5\u2192 0, the vibration absorber m2 vanishes, and hence, the system becomes a one-DOF primary oscillator. Such a system has only one natural frequency rn = 1 as given by Equation (12.518). It is the frequency that r1,2 will approach by vanishing m2. The nodal frequencies r1,2 are always on both sides of the singe-DOF natural frequency rn r1 < rn < r2 (12.559) while all of them are decreasing functions of the mass ratio \u03b5. Example 471 F Natural frequencies for extreme values of damping. By setting \u03be = 0 for \u03b5 = 0.1, we find \u03bc = \u00af\u0304\u0304\u0304 \u00af r2 \u2212 1 0.1r2 \u2212 (r2 \u2212 1)2 \u00af\u0304\u0304\u0304 \u00af (12.560) and by setting \u03be =\u221e, we find \u03bc = \u00af\u0304\u0304\u0304 1 1.1r2 \u2212 1 \u00af\u0304\u0304\u0304 . (12.561) Having \u03be = 0 is equivalent to no damping. When there is no damping, \u03bc approaches infinity at the real roots of its denominator, rn1 and rn2 , 816 12. Applied Vibrations which are the natural frequencies of the system. As an example, the natural frequencies rn1 and rn2 for \u03b5 = 0.1, are 0.1r2 \u2212 \u00a1 r2 \u2212 1 \u00a22 = 0 (12.562) rn1 = 0.854 31 (12.563) rn2 = 1.170 5. (12.564) Having \u03be = \u221e is equivalent to a rigid connection between m1 and m2. The system would have only one DOF and therefore, \u03bc approaches infinity at the only roots of the denominator, rn 1.1r2 \u2212 1 = 0 (12.565) rn = 0.953 (12.566) where, rn is always between rn1 and rn2 . rn1 < rn < rn2 (12.567) 12.7 Summary Generally speaking, vibration is a harmful and unwanted phenomenon. Vibration is important when a non-vibrating system is connected to a vibrating system. To minimize the effects of vibration, we connect the systems by a damping elastic isolator. For simplicity, we model the isolator by a spring and damper parallel to each other. Such an isolator is called suspension. Vibration can be physically expressed as a result of energy conversion. It can mathematically be expressed by solutions of a set of differential equations. If the system is linear, then its equations of motion can always be arranged in the following matrix form: [M ] x\u0307+ [c] x\u0307+ [k]x = F (x, x\u0307, t) (12.568) Vibration can be separated into free vibrations, when F = 0, and forced vibrations, when F 6= 0. However, in applied vibrations, we usually separate the solution of the equations of motion into transient and steady-state. Transient response is the solution of the equations of motion when F = 0 or F is active for a short period of time. Because most industrial machines are equipped with a rotating motor, periodic and harmonic excitation is very common. Frequency response is the steady-state solution of equations of motion when the system is harmonically excited. In frequency analysis we seek the steady-state response of the system, after the effect of initial conditions dies out. Frequency response of mechanical systems, such as vehicles, is dominated by the natural frequencies of the system and by excitation frequencies. The 12. Applied Vibrations 817 amplitude of vibration increases when an excitation frequency approaches one of the natural frequencies of the system. Frequency domains around the natural frequencies are called the resonance zone. The amplitude of vibration in resonance zones can be reduced by introducing damping. One-DOF, harmonically excited systems may be classified as base excitation, eccentric excitation, eccentric base excitation, and forced excitation. Every frequency response these systems can be expressed by one of the functions Si, Gi, and \u03a6i, each with a specific characteristic. We usually use a graphical illustration to see the frequency response of the system as a function of frequency ratio r = \u03c9/\u03c9n and damping ratio \u03be = c/ \u221a 4km. 818 12. Applied Vibrations 12.8 Key Symbols a \u2261 x\u0308 acceleration a1 distance from mass center to front axle a2 distance from mass center to rear axle [a] , [A] coefficient matrix A,B,C unknown coefficients for frequency responses b1 distance from mass center to left wheel b2 distance from mass center to right wheel c damping cF optimum damping ceq equivalent damping cij element of row i and column j of a damping matrix [c] damping matrix D denominator e eccentricity arm e exponential function E mechanical energy E Young modulus of elasticity f = 1/T cyclic frequency [ Hz] f, F force fc damper force feq equivalent force fk spring force fm required force to move a mass m F amplitude of a harmonic force f = F sin\u03c9t F0 constant force Ft tension force FT transmitted force g gravitational acceleration G0, G1, G2 amplitude frequency response I area moment of inertia for beams I mass moment of inertia for vehicles I identity matrix k stiffness kF optimum stiffness keq equivalent stiffness kij element of row i and column j of a stiffness matrix kR antiroll bar torsional stiffness [k] stiffness matrix K kinetic energy l length m mass mb device mass 12. Applied Vibrations 819 me eccentric mass mij element of row i and column j of a mass matrix ms sprung mass ms mass of spring mu unsprung mass [m] mass matrix M mass of platform N numerator p momentum Q general amplitude r = \u03c9/\u03c9n frequency ratio r, R radius r1, r2 frequency ratio at nodes rn = \u03c9n/\u03c91 dimensionless natural frequency S overshoot S quadrature S0, S1, \u00b7 \u00b7 \u00b7 , S4 amplitude frequency response t time tp peak time tr rise time ts settling time T period Tn natural period v \u2261 x\u0307, v velocity V potential energy w track of a car wf front track of a car wr rear track of a car x, y, z, x displacement x0 initial displacement xh homogeneous solution xp particular solution xP peak displacement x\u03070 initial velocity X,Y,Z, amplitude Zi, i = 1, 2, \u00b7 \u00b7 \u00b7 short notation parameters \u03b1 = \u03c92/\u03c91 natural frequency ratio \u03b4 deflection \u03b4s static deflection \u03b5 mass ratio \u03b8 angular motion \u0398 amplitude of angular vibration \u03bb eigenvalue \u03bc amplitude frequency response 820 12. Applied Vibrations \u03d5 phase angle \u03a60,\u03a61, \u00b7 \u00b7 \u00b7 ,\u03a63 phase frequency response \u03c9 = 2\u03c0f angular frequency [ rad/ s] \u03c9n natural frequency \u03be damping ratio \u03beF optimum damping ratio Subscriptions d driver f front M maximum r rear s sprung mass u unsprung mass 12. Applied Vibrations 821 Exercises 1. Natural frequency and damping ratio. A one DOF mass-spring-damper has m = 1kg, k = 1000N/m and c = 100N s/m. Determine the natural frequency, and damping ratio of the system. 2. Equivalent spring. Determine the equivalent spring for the vibrating system that is shown in Figure 12.55. xk1 m k2 E,I,l 822 12. Applied Vibrations a k b m\u03b8 12. Applied Vibrations 823 m k1 c x y k2 z 824 12. Applied Vibrations 10. Base excited system and spring stiffness. A base excited m-k-c system has m = 200 kg c = 1000N s/m. Determine the stiffness of the spring, k, such that the steady-state amplitude of m is lass than 0.07m when the base is excited as y = 0.05 sin 2\u03c0t at the natural frequency of the system. 11. Base excited system and absolute acceleration. Assume a base excited m-k-c system is vibrating at the node of its absolute acceleration frequency response. If the base is excited according to y = 0.05 sin 2\u03c0t determine \u03c9n, X\u0308, X. 12. Eccentric excitation and transmitted force. An engine with massm = 175 kg and eccentricitymee = 0.4\u00d70.1 kgm is turning at \u03c9e = 4000 rpm. (a) Determine the steady-state amplitude of its vibration, if there are four engine mounts, each with k = 10000N/m and c = 100N s/m. (b) Determine the transmitted force to the base. 13. F Eccentric base excitation and absolute displacement. An eccentric base excited system has m = 3kg, mb = 175 kg, mee = 0.4 \u00d7 0.1 kgm, and \u03c9 = 4000 rpm. If Z/(e\u03b5) = 2 at r = 1, calculate X and Y . 14. Characteristic values and free vibrations. An m-k-c system has m = 250 kg k = 8000N/m c = 1000N s/m. Determine the characteristic values of the system and its free vibration response, for zero initial conditions. 12. Applied Vibrations 825 15. F Response to the step input. Consider an m-k-c system with m = 250 kg k = 8000N/m c = 1000N s/m. Determine the step input parameters, tr, tP , xP , S, and ts for 2% window. 16. Damping ratio determination. Consider a vibrating system that after n = 100 times oscillation, the peak amplitude drops by 2%. Determine the exact and approximate values of \u03be. 17. The car lateral moment of inertia. Consider a car with the following characteristics: b1 746mm b2 740mm mass 1245 kg a1 1100mm a2 1323mm h 580mm Ix 335 kgm2 Iy 1095 kgm2 Determine the period of oscillation when the car is on a solid steel platform with dimension 2000mm\u00d7 3800mm\u00d7 35mm, (a) laterally (b) longitudinally. 18. F Optimal vibration absorber. Consider a primary system with m1 = 250 kg and k = 8000N/m. (a) Determine the best suspension for the secondary system with m2 = 1kg to act as a vibration absorber. (b) Determine the natural frequencies of the two system for the optimized vibration absorber. (c) Determine the nodal frequencies and amplitudes of the primary system. 826 12. Applied Vibrations 19. F Frequency response. Prove the following equations: G2 = FTB kY G2 = FTE e\u03c92nme G2 = FTR e\u03c92nme \u00b3 1 + mb m \u00b4 13 Vehicle Vibrations Vehicles are multiple-DOF systems as the one that is shown in Figure 13.1. The vibration behavior of a vehicle, which is called ride or ride comfort, is highly dependent on the natural frequencies and mode shapes of the vehicle. In this chapter, we review and examine the applied methods of determining the equations of motion, natural frequencies, and mode shapes of different models of vehicles. 13.1 Lagrange Method and Dissipation Function Lagrange equation, d dt \u00b5 \u2202K \u2202q\u0307r \u00b6 \u2212 \u2202K \u2202qr = Fr r = 1, 2, \u00b7 \u00b7 \u00b7n (13.1) or, d dt \u00b5 \u2202L \u2202q\u0307r \u00b6 \u2212 \u2202L \u2202qr = Qr r = 1, 2, \u00b7 \u00b7 \u00b7n (13.2) as introduced in Equations (9.243) and (9.298), can both be applied to find the equations of motion for a vibrating system. However, for small and linear vibrations, we may use a simpler and more practical Lagrange 828 13. Vehicle Vibrations equation such as d dt \u00b5 \u2202K \u2202q\u0307r \u00b6 \u2212 \u2202K \u2202qr + \u2202D \u2202q\u0307r + \u2202V \u2202qr = fr r = 1, 2, \u00b7 \u00b7 \u00b7n (13.3) where K is the kinetic energy, V is the potential energy, and D is the dissipation function of the system K = 1 2 x\u0307T [m] x\u0307 = 1 2 nX i=1 nX j=1 x\u0307imij x\u0307j (13.4) V = 1 2 xT [k]x = 1 2 nX i=1 nX j=1 xikijxj (13.5) D = 1 2 x\u0307T [c] x\u0307 = 1 2 nX i=1 nX j=1 x\u0307icij x\u0307j (13.6) and fr is the applied force on the mass mr. Proof. Consider a one-DOF mass-spring-damper vibrating system. When viscous damping is the only type of damping in the system, we may employ a function known as Rayleigh dissipation function D = 1 2 cx\u03072 (13.7) to find the damping force fc by differentiation. fc = \u2212 \u2202D \u2202x\u0307 . (13.8) Remembering the elastic force fk can be found from a potential energy V fk = \u2212 \u2202V \u2202x then, the generalized force F can be separated to F = fc + fk + f = \u2212\u2202D \u2202x\u0307 \u2212 \u2202V \u2202x + f (13.9) 13. Vehicle Vibrations 829 where f is the non-conservative applied force on mass m. Substituting (13.9) in (13.1) d dt \u00b5 \u2202K \u2202x\u0307 \u00b6 \u2212 \u2202K \u2202x = \u2212\u2202D \u2202x\u0307 \u2212 \u2202V \u2202x + f (13.10) gives us the Lagrange equation for a viscous damped vibrating system. d dt \u00b5 \u2202K \u2202x\u0307 \u00b6 \u2212 \u2202K \u2202x + \u2202D \u2202x\u0307 + \u2202V \u2202x = f (13.11) When the vibrating system has n DOF, then the kinetic energy K, potential energy V , and dissipating function D are as (13.4)-(13.6). Applying the Lagrange equation to the n-DOF system would result n second-order differential equations (13.3). Example 472 A one-DOF forced mass-spring-damper system. Figure 13.2 illustrates a single DOF mass-spring-damper system with an external force f applied on the mass m. The kinetic and potential energies of the system, when it is in motion, are K = 1 2 mx\u03072 (13.12) V = 1 2 kx2 (13.13) and its dissipation function is D = 1 2 cx\u03072. (13.14) 830 13. Vehicle Vibrations Substituting (13.12)-(13.14) in Lagrange equation (13.3), generates the following equation of motion: d dt (mx\u0307) + cx\u0307+ kx = f, (13.15) because \u2202K \u2202x\u0307 = mx\u0307 (13.16) \u2202K \u2202x = 0 (13.17) \u2202D \u2202x\u0307 = cx\u0307 (13.18) \u2202V \u2202x = kx. (13.19) Example 473 An undamped three-DOF system. Figure 13.3 illustrates an undamped three-DOF linear vibrating system. The kinetic and potential energies of the system are: K = 1 2 m1x\u0307 2 1 + 1 2 m2x\u0307 2 2 + 1 2 m3x\u0307 2 3 (13.20) V = 1 2 k1x 2 1 + 1 2 k2 (x1 \u2212 x2) 2 + 1 2 k3 (x2 \u2212 x3) 2 + 1 2 k4x 2 3 (13.21) Because there is no damping in the system, we may find the Lagrangean L L = K \u2212 V (13.22) and use Equation (13.2) with Qr = 0 \u2202L \u2202x1 = \u2212k1x1 \u2212 k2 (x1 \u2212 x2) (13.23) \u2202L \u2202x2 = k2 (x1 \u2212 x2)\u2212 k3 (x2 \u2212 x3) (13.24) \u2202L \u2202x3 = k3 (x2 \u2212 x3)\u2212 k4x3 (13.25) 13. Vehicle Vibrations 831 \u2202L \u2202x\u03071 = m1x\u03071 (13.26) \u2202L \u2202x\u03072 = m2x\u03072 (13.27) \u2202L \u2202x\u03073 = m3x\u03073 (13.28) to find the equations of motion: m1x\u03081 + k1x1 + k2 (x1 \u2212 x2) = 0 (13.29) m2x\u03082 \u2212 k2 (x1 \u2212 x2) + k3 (x2 \u2212 x3) = 0 (13.30) m3x\u03083 \u2212 k3 (x2 \u2212 x3) + k4x3 = 0 (13.31) These equations can be rewritten in matrix form for simpler calculation. \u23a1\u23a3 m1 0 0 0 m2 0 0 0 m3 \u23a4\u23a6\u23a1\u23a3 x\u03081 x\u03082 x\u03083 \u23a4\u23a6 + \u23a1\u23a3 k1 + k2 \u2212k2 0 \u2212k2 k2 + k3 \u2212k3 0 \u2212k3 k3 + k4 \u23a4\u23a6\u23a1\u23a3 x1 x2 x3 \u23a4\u23a6 = 0 (13.32) Example 474 An eccentric excited one-DOF system. An eccentric excited one-DOF system is shown in Figure 12.31 with mass m supported by a suspension made up of a spring k and a damper c. There is also a mass me at a distance e that is rotating with an angular velocity \u03c9. We may find the equation of motion by applying the Lagrange method. The kinetic energy of the system is K = 1 2 (m\u2212me) x\u0307 2 + 1 2 me (x\u0307+ e\u03c9 cos\u03c9t) 2 + 1 2 me (\u2212e\u03c9 sin\u03c9t)2 (13.33) because the velocity of the main vibrating mass m\u2212me is x\u0307, and the velocity of the eccentric mass me has two components x\u0307+ e\u03c9 cos\u03c9t and \u2212e\u03c9 sin\u03c9t. The potential energy and dissipation function of the system are: V = 1 2 kx2 (13.34) D = 1 2 cx\u03072. (13.35) 832 13. Vehicle Vibrations Applying the Lagrange equation (13.3), \u2202K \u2202x\u0307 = mx\u0307+mee\u03c9 cos\u03c9t (13.36) d dt \u00b5 \u2202K \u2202x\u0307 \u00b6 = mx\u0308\u2212mee\u03c9 2 sin\u03c9t (13.37) \u2202D \u2202x\u0307 = cx\u0307 (13.38) \u2202V \u2202x = kx (13.39) provides the equation of motion mx\u0308+ c x\u0307+ kx = mee\u03c9 2 sin\u03c9t (13.40) that is the same as Equation (12.208). Example 475 An eccentric base excited vibrating system. Figure 12.35 illustrates a one DOF eccentric base excited vibrating system. A mass m is mounted on an eccentric excited base by a spring k and a damper c. The base has a mass mb with an attached unbalance mass me at a distance e. The mass me is rotating with an angular velocity \u03c9. We may derive the equation of motion of the system by applying Lagrange method. The required functions are: K = 1 2 mx\u03072 + 1 2 (mb \u2212me) y\u0307 2 + 1 2 me (y\u0307 \u2212 e\u03c9 cos\u03c9t) 2 + 1 2 me (e\u03c9 sin\u03c9t) 2 (13.41) V = 1 2 k (x\u2212 y) 2 (13.42) D = 1 2 c (x\u0307\u2212 y\u0307)2 . (13.43) Applying the Lagrange method (13.3), provides the equations mx\u0308+ c (x\u0307\u2212 y\u0307) + k (x\u2212 y) = 0 (13.44) mby\u0308 +mee\u03c9 2 sin\u03c9t\u2212 c (x\u0307\u2212 y\u0307)\u2212 k (x\u2212 y) = 0 (13.45) because \u2202K \u2202x\u0307 = mx\u0307 (13.46) d dt \u00b5 \u2202K \u2202x\u0307 \u00b6 = mx\u0308 (13.47) \u2202D \u2202x\u0307 = c (x\u0307\u2212 y\u0307) (13.48) \u2202V \u2202x = k (x\u2212 y) (13.49) 13. Vehicle Vibrations 833 \u2202K \u2202y\u0307 = mby\u0307 \u2212mee\u03c9 cos\u03c9t (13.50) d dt \u00b5 \u2202K \u2202y\u0307 \u00b6 = mby\u0308 +mee\u03c9 2 sin\u03c9t (13.51) \u2202D \u2202y\u0307 = \u2212c (x\u0307\u2212 y\u0307) (13.52) \u2202V \u2202y = \u2212k (x\u2212 y) . (13.53) Using z = x \u2212 y, we may combine Equations (13.44) and (13.45) to find the equation of relative motion mmb mb +m z\u0308 + c z\u0307 + kz = mme mb +m e\u03c92 sin\u03c9t (13.54) that is equal to z\u0308 + 2\u03be\u03c9n z\u0307 + \u03c92n z = \u03b5e\u03c92 sin\u03c9t (13.55) \u03b5 = me mb . (13.56) Example 476 F A rolling disc in a circular path. Figure 13.4 illustrates a uniform disc with mass m and radius r. The disc is rolling without slip in a circular path with radius R. The disc may have a free oscillation around \u03b8 = 0. When the oscillation is very small, we may substitute the oscillating disc with an equivalent mass-spring system. To find the equation of motion, we employ the Lagrange method. The energies of the system are K = 1 2 mv2C + 1 2 Ic\u03c9 2 = 1 2 m(R\u2212 r)2\u03b8\u0307 2 + 1 2 \u00b5 1 2 mr2 \u00b6\u00b3 \u03d5\u0307\u2212 \u03b8\u0307 \u00b42 (13.57) V = \u2212mg(R\u2212 r) cos \u03b8. (13.58) When there is no slip, there is a constraint between \u03b8 and \u03d5 R\u03b8 = r\u03d5 (13.59) 834 13. Vehicle Vibrations which can be used to eliminate \u03d5 from K. K = 3 4 m (R\u2212 r) 2 \u03b8\u0307 2 (13.60) Based on the following partial derivatives: d dt \u00b5 \u2202L \u2202\u03b8\u0307 \u00b6 = 3 2 m (R\u2212 r)2 \u03b8\u0308 (13.61) \u2202L \u2202\u03b8 = \u2212mg(R\u2212 r) sin \u03b8 (13.62) we find the equation of motion for the oscillating disc. 3 2 (R\u2212 r) \u03b8\u0308 + g sin \u03b8 = 0 (13.63) When \u03b8 is very small, this equation is equivalent to a mass-spring system with meq = 3 (R\u2212 r) and keq = 2g. Example 477 F A double pendulum. Figure 13.5 illustrates a double pendulum made by a series of two pendulums. There are two massless rods with lengths l1 and l2, and two point masses m1 and m2. The variables \u03b81 and \u03b82 can be used as the generalized coordinates to express the system configuration. To calculate the Lagrangean of the system and find the equations of motion, we start by defining the global position of the masses. 13. Vehicle Vibrations 835 x1 = l1 sin \u03b81 (13.64) y1 = \u2212l1 cos \u03b81 (13.65) x2 = l1 sin \u03b81 + l2 sin \u03b82 (13.66) y2 = \u2212l1 cos \u03b81 \u2212 l2 cos \u03b82 (13.67) Time derivatives of the coordinates are x\u03071 = l1\u03b8\u03071 cos \u03b81 (13.68) y\u03071 = l1\u03b8\u03071 sin \u03b81 (13.69) x\u03072 = l1\u03b8\u03071 cos \u03b81 + l2\u03b8\u03072 cos \u03b82 (13.70) y\u03072 = l1\u03b8\u03071 sin \u03b81 + l2\u03b8\u03072 sin \u03b82 (13.71) and therefore, the squares of the masses\u2019 velocities are v21 = x\u030721 + y\u030721 = l21 \u03b8\u0307 2 1 (13.72) v22 = x\u030722 + y\u030722 = l21 \u03b8\u0307 2 1 + l22 \u03b8\u0307 2 2 + 2l1l2\u03b8\u03071\u03b8\u03072 cos (\u03b81 \u2212 \u03b82) . (13.73) The kinetic energy of the pendulum is then equal to K = 1 2 m1v 2 1 + 1 2 m2v 2 2 = 1 2 m1l 2 1 \u03b8\u0307 2 1 + 1 2 m2 \u00b3 l21 \u03b8\u0307 2 1 + l22 \u03b8\u0307 2 2 + 2l1l2\u03b8\u03071\u03b8\u03072 cos (\u03b81 \u2212 \u03b82) \u00b4 . (13.74) The potential energy of the pendulum is equal to sum of the potentials of each mass. V = m1gy1 +m2gy2 = \u2212m1gl1 cos \u03b81 \u2212m2g (l1 cos \u03b81 + l2 cos \u03b82) (13.75) The kinetic and potential energies make the following Lagrangean: L = K \u2212 V = 1 2 m1l 2 1 \u03b8\u0307 2 1 + 1 2 m2 \u00b3 l21 \u03b8\u0307 2 1 + l22 \u03b8\u0307 2 2 + 2l1l2\u03b8\u03071\u03b8\u03072 cos (\u03b81 \u2212 \u03b82) \u00b4 +m1gl1 cos \u03b81 +m2g (l1 cos \u03b81 + l2 cos \u03b82) (13.76) Employing Lagrange method (13.2) we find the following equations of motion: d dt \u00b5 \u2202L \u2202\u03b8\u03071 \u00b6 \u2212 \u2202L \u2202\u03b81 = (m1 +m2) l 2 1 \u03b8\u03081 +m2l1l2\u03b8\u03082 cos (\u03b81 \u2212 \u03b82) \u2212m2l1l2\u03b8\u0307 2 2 sin (\u03b81 \u2212 \u03b82) + (m1 +m2) l1g sin \u03b81 = 0 (13.77) 836 13. Vehicle Vibrations d dt \u00b5 \u2202L \u2202\u03b8\u03072 \u00b6 \u2212 \u2202L \u2202\u03b82 = m2l 2 2 \u03b8\u03082 +m2l1l2\u03b8\u03081 cos (\u03b81 \u2212 \u03b82) +m2l1l2\u03b8\u0307 2 1 sin (\u03b81 \u2212 \u03b82) +m2l2g sin \u03b82 = 0 (13.78) Example 478 F Chain pendulum. Consider an n-chain-pendulum as shown in Figure 13.6. Each pendulum has a massless length li with a concentrated point mass mi, and a generalized angular coordinate \u03b8i measured from the vertical direction. The xi and yi components of the mass mi are xi = iX j=1 lj sin \u03b8j (13.79) yi = \u2212 iX j=1 lj cos \u03b8j . (13.80) We find their time derivatives x\u0307i = iX j=1 lj \u03b8\u0307j cos \u03b8j (13.81) y\u0307i = iX j=1 lj \u03b8\u0307j sin \u03b8j (13.82) 13. Vehicle Vibrations 837 and the square of x\u0307i and y\u0307i x\u03072i = \u239b\u239d iX j=1 lj \u03b8\u0307j cos \u03b8j \u239e\u23a0\u00c3 iX k=1 lk\u03b8\u0307k cos \u03b8k ! = nX j=1 nX k=1 ljlk\u03b8\u0307j \u03b8\u0307k cos \u03b8j cos \u03b8k (13.83) y\u03072i = \u239b\u239d iX j=1 lj \u03b8\u0307j sin \u03b8j \u239e\u23a0\u00c3 iX k=1 lk\u03b8\u0307k sin \u03b8k ! = iX j=1 iX k=1 ljlk\u03b8\u0307j \u03b8\u0307k sin \u03b8j sin \u03b8k (13.84) to calculate the velocity vi of the mass mi. v2i = x\u03072i + y\u03072i = iX j=1 iX k=1 ljlk\u03b8\u0307j \u03b8\u0307k (cos \u03b8j cos \u03b8k + sin \u03b8j sin \u03b8k) = iX j=1 iX k=1 ljlk\u03b8\u0307j \u03b8\u0307k cos (\u03b8j \u2212 \u03b8k) = iX r=1 l2r \u03b8\u0307 2 r + 2 iX j=1 iX k=j+1 ljlk\u03b8\u0307j \u03b8\u0307k cos (\u03b8j \u2212 \u03b8k) (13.85) Now, we may calculate the kinetic energy, K, of the chain. K = 1 2 nX i=1 miv 2 i = 1 2 nX i=1 mi \u239b\u239d iX r=1 l2r \u03b8\u0307 2 r + 2 iX j=1 iX k=j+1 ljlk\u03b8\u0307j \u03b8\u0307k cos (\u03b8j \u2212 \u03b8k) \u239e\u23a0 = 1 2 nX i=1 iX r=1 mil 2 r \u03b8\u0307 2 r + nX i=1 iX j=1 iX k=j+1 miljlk\u03b8\u0307j \u03b8\u0307k cos (\u03b8j \u2212 \u03b8k) (13.86) The potential energy of the ith pendulum is related to mi Vi = migyi = \u2212mig iX j=1 lj cos \u03b8j (13.87) 838 13. Vehicle Vibrations and therefore, the potential energy of the chain is V = nX i=1 migyi = \u2212 nX i=1 iX j=1 miglj cos \u03b8j (13.88) To find the equations of motion for the chain, we may use the Lagrangean L L = K \u2212 V (13.89) and apply the Lagrange equation d dt \u00b5 \u2202L \u2202q\u0307s \u00b6 \u2212 \u2202L \u2202qs = 0 s = 1, 2, \u00b7 \u00b7 \u00b7n (13.90) or d dt \u00b5 \u2202K \u2202q\u0307s \u00b6 \u2212 \u2202K \u2202qs + \u2202V \u2202qs = 0 s = 1, 2, \u00b7 \u00b7 \u00b7n. (13.91) 13.2 F Quadratures If [m] is an n\u00d7n square matrix and x is an n\u00d7 1 vector, then S is a scalar function called quadrature and is defined by S = xT [m]x. (13.92) The derivative of the quadrature S with respect to the vector x is \u2202S \u2202x = \u00b3 [m] + [m] T \u00b4 x. (13.93) Kinetic energy K, potential energy V , and dissipation function D, are quadratures K = 1 2 x\u0307T [m] x\u0307 (13.94) V = 1 2 xT [k]x (13.95) D = 1 2 x\u0307T [c] x\u0307 (13.96) and therefore, \u2202K \u2202x\u0307 = 1 2 \u00b3 [m] + [m] T \u00b4 x\u0307 (13.97) \u2202V \u2202x = 1 2 \u00b3 [k] + [k] T \u00b4 x (13.98) \u2202D \u2202x\u0307 = 1 2 \u00b3 [c] + [c]T \u00b4 x\u0307. (13.99) 13. Vehicle Vibrations 839 Employing quadrature derivatives and the Lagrange method, d dt \u2202K \u2202x\u0307 + \u2202K \u2202x + \u2202D \u2202x\u0307 + \u2202V \u2202x = F (13.100) the equation of motion for a linear n degree-of-freedom vibrating system becomes [m] x\u0308+ [c] x\u0307+ [k]x = F (13.101) where, [m], [c], [k] are symmetric matrices. [m] = 1 2 \u00b3 [m] + [m] T \u00b4 (13.102) [c] = 1 2 \u00b3 [c] + [c] T \u00b4 (13.103) [k] = 1 2 \u00b3 [k] + [k]T \u00b4 (13.104) Quadratures are also called Hermitian form. Proof. Let\u2019s define a general asymmetric quadrature as S = xT [a]y = X i X j xiaijyj . (13.105) If the quadrature is symmetric, then x = y and S = xT [a]x = X i X j xiaijxj . (13.106) The vectors x and y may be functions of n generalized coordinates qi and time t. x = x (q1, q2, \u00b7 \u00b7 \u00b7 , qn, t) (13.107) y = y (q1, q2, \u00b7 \u00b7 \u00b7 , qn, t) (13.108) q = \u00a3 q1 q2 \u00b7 \u00b7 \u00b7 qn \u00a4T (13.109) The derivative of x with respect to q is a square matrix \u2202x \u2202q = \u23a1\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3 \u2202x1 \u2202q1 \u2202x2 \u2202q1 \u00b7 \u00b7 \u00b7 \u2202xn \u2202q1 \u2202x1 \u2202q2 \u2202x2 \u2202q2 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u2202x1 \u2202qn \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u2202xn \u2202qn \u23a4\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6 (13.110) 840 13. Vehicle Vibrations that can also be expressed by \u2202x \u2202q = \u23a1\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3 \u2202x \u2202q1 \u2202x \u2202q2 \u00b7 \u00b7 \u00b7 \u2202x \u2202qn \u23a4\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6 (13.111) or \u2202x \u2202q = \u2219 \u2202x1 \u2202q \u2202x2 \u2202q \u00b7 \u00b7 \u00b7 \u2202xn \u2202q \u00b8 . (13.112) Now a derivative of S with respect to an element of qk is \u2202S \u2202qk = \u2202 \u2202qk X i X j xiaijyj = X i X j \u2202xi \u2202qk aijyj + X i X j xiaij \u2202yj \u2202qk = X j X i \u2202xi \u2202qk aijyj + X i X j \u2202yj \u2202qk aijxi = X j X i \u2202xi \u2202qk aijyj + X j X i \u2202yi \u2202qk ajixj (13.113) and hence, the derivative of S with respect to q is \u2202S \u2202q = \u2202x \u2202q [a]y+ \u2202y \u2202q [a] T x. (13.114) If S is a symmetric quadrature then, \u2202S \u2202q = \u2202 \u2202q \u00a1 xT [a]x \u00a2 = \u2202x \u2202q [a]x+ \u2202x \u2202q [a] T x. (13.115) and if q = x, then the derivative of a symmetric S with respect to x is \u2202S \u2202x = \u2202 \u2202x \u00a1 xT [a]x \u00a2 = \u2202x \u2202x [a]x+ \u2202x \u2202x [a]T x = [a]x+ [a] T x = \u00b3 [a] + [a]T \u00b4 x. (13.116) 13. Vehicle Vibrations 841 If [a] is a symmetric matrix, then [a] + [a]T = 2 [a] (13.117) however, if [a] is not a symmetric matrix, then [a] = [a]+[a]T is a symmetric matrix because aij = aij + aji = aji + aij = aji (13.118) and therefore, [a] = [a] T . (13.119) Kinetic energy K, potential energy V , and dissipation function D can be expressed by quadratures. K = 1 2 x\u0307T [m] x\u0307 (13.120) V = 1 2 xT [k]x (13.121) D = 1 2 x\u0307T [c] x\u0307 (13.122) Substituting K, V , and D in the Lagrange equation provides the equations of motion: F = d dt \u2202K \u2202x\u0307 + \u2202K \u2202x + \u2202D \u2202x\u0307 + \u2202V \u2202x = 1 2 d dt \u2202 \u2202x\u0307 \u00a1 x\u0307T [m] x\u0307 \u00a2 + 1 2 \u2202 \u2202x\u0307 \u00a1 x\u0307T [c] x\u0307 \u00a2 + 1 2 \u2202 \u2202x \u00a1 xT [k]x \u00a2 = 1 2 \u2219 d dt \u00b3\u00b3 [m] + [m] T \u00b4 x\u0307 \u00b4 + \u00b3 [c] + [c] T \u00b4 x\u0307+ \u00b3 [k] + [k] T \u00b4 x \u00b8 = 1 2 \u00b3 [m] + [m]T \u00b4 x\u0308+ 1 2 \u00b3 [c] + [c]T \u00b4 x\u0307+ 1 2 \u00b3 [k] + [k]T \u00b4 x = [m] x\u0308+ [c] x\u0307+ [k]x (13.123) where [m] = 1 2 \u00b3 [m] + [m]T \u00b4 (13.124) [c] = 1 2 \u00b3 [k] + [k] T \u00b4 (13.125) [k] = 1 2 \u00b3 [c] + [c] T \u00b4 . (13.126) From now on, we assume that every equation of motion is found from the Lagrange method to have symmetric coefficient matrices. Hence, we show the equations of motion as, [m] x\u0308+ [c] x\u0307+ [k]x = F (13.127) 842 13. Vehicle Vibrations and use [m], [c], [k] as a substitute for [m], [c], [k] [m] \u2261 [m] (13.128) [c] \u2261 [c] (13.129) [k] \u2261 [k] . (13.130) Example 479 F A quarter car model with driver\u2019s chair. Figure 13.7 illustrates a quarter car model plus a driver, which is modeled by a mass md over a linear cushion above the sprung mass ms. Assuming y = 0 (13.131) we can find the free vibration equations of motion by the Lagrange method and quadrature derivative. 13. Vehicle Vibrations 843 The kinetic energy K of the system can be expressed by K = 1 2 mux\u0307 2 u + 1 2 msx\u0307 2 s + 1 2 mdx\u0307 2 d = 1 2 \u00a3 x\u0307u x\u0307s x\u0307d \u00a4\u23a1\u23a3 mu 0 0 0 ms 0 0 0 md \u23a4\u23a6\u23a1\u23a3 x\u0307u x\u0307s x\u0307d \u23a4\u23a6 = 1 2 x\u0307T [m] x\u0307 (13.132) and the potential energy V can be expressed as V = 1 2 ku (xu) 2 + 1 2 ks (xs \u2212 xu) 2 + 1 2 kd (xd \u2212 xs) 2 = 1 2 \u00a3 xu xs xd \u00a4\u23a1\u23a3 ku + ks \u2212ks 0 \u2212ks ks + kd \u2212kd 0 \u2212kd kd \u23a4\u23a6\u23a1\u23a3 xu xs xd \u23a4\u23a6 = 1 2 xT [k]x. (13.133) Similarly, the dissipation function D can be expressed by D = 1 2 ku (x\u0307u) 2 + 1 2 ks (x\u0307s \u2212 x\u0307u) 2 + 1 2 kd (x\u0307d \u2212 x\u0307s) 2 = 1 2 \u00a3 x\u0307u x\u0307s x\u0307d \u00a4\u23a1\u23a3 cu + cs \u2212cs 0 \u2212cs cs + cd \u2212cd 0 \u2212cd cd \u23a4\u23a6\u23a1\u23a3 x\u0307u x\u0307s x\u0307d \u23a4\u23a6 = 1 2 x\u0307T [c] x\u0307. (13.134) Employing the quadrature derivative method, we may find derivatives of K, V , and D with respect to their variable vectors as follow: \u2202K \u2202x\u0307 = 1 2 \u00b3 [m] + [m]T \u00b4 x\u0307 = 1 2 \u00b3 [k] + [k] T \u00b4\u23a1\u23a3 x\u0307u x\u0307s x\u0307d \u23a4\u23a6 = \u23a1\u23a3 mu 0 0 0 ms 0 0 0 md \u23a4\u23a6\u23a1\u23a3 x\u0307u x\u0307s x\u0307d \u23a4\u23a6 (13.135) 844 13. Vehicle Vibrations \u2202V \u2202x = 1 2 \u00b3 [k] + [k] T \u00b4 x = 1 2 \u00b3 [k] + [k] T \u00b4\u23a1\u23a3 xu xs xd \u23a4\u23a6 = \u23a1\u23a3 ku + ks \u2212ks 0 \u2212ks ks + kd \u2212kd 0 \u2212kd kd \u23a4\u23a6\u23a1\u23a3 xu xs xd \u23a4\u23a6 (13.136) \u2202D \u2202x\u0307 = 1 2 \u00b3 [c] + [c]T \u00b4 x\u0307 = 1 2 \u00b3 [c] + [c] T \u00b4\u23a1\u23a3 x\u0307u x\u0307s x\u0307d \u23a4\u23a6 = \u23a1\u23a3 cu + cs \u2212cs 0 \u2212cs cs + cd \u2212cd 0 \u2212cd cd \u23a4\u23a6\u23a1\u23a3 x\u0307u x\u0307s x\u0307d \u23a4\u23a6 (13.137) Therefore, we find the system\u2019s free vibration equations of motion. [m] x\u0308+ [c] x\u0307+ [k]x = 0 (13.138) \u23a1\u23a3 mu 0 0 0 ms 0 0 0 md \u23a4\u23a6\u23a1\u23a3 x\u0308u x\u0308s x\u0308d \u23a4\u23a6+ \u23a1\u23a3 cu + cs \u2212cs 0 \u2212cs cs + cd \u2212cd 0 \u2212cd cd \u23a4\u23a6\u23a1\u23a3 x\u0307u x\u0307s x\u0307d \u23a4\u23a6 + \u23a1\u23a3 ku + ks \u2212ks 0 \u2212ks ks + kd \u2212kd 0 \u2212kd kd \u23a4\u23a6\u23a1\u23a3 xu xs xd \u23a4\u23a6 = 0 (13.139) Example 480 F Different [m], [c], and [k] arrangements. Mass, damping, and stiffness matrices [m], [c], [k] for a vibrating system may be arranged in different forms with the same overall kinetic energy K, potential energy V , and dissipation function D. As an example, the potential energy V for the quarter car model that is shown in Figure 13.7 may be expressed by different [k]. V = 1 2 ku (xu) 2 + 1 2 ks (xs \u2212 xu) 2 + 1 2 kd (xd \u2212 xs) 2 (13.140) V = 1 2 xT \u23a1\u23a3 ku + ks \u2212ks 0 \u2212ks ks + kd \u2212kd 0 \u2212kd kd \u23a4\u23a6x (13.141) 13. Vehicle Vibrations 845 V = 1 2 xT \u23a1\u23a3 ku + ks \u22122ks 0 0 ks + kd \u22122kd 0 0 kd \u23a4\u23a6x (13.142) V = 1 2 xT \u23a1\u23a3 ku + ks 0 0 \u22122ks ks + kd 0 0 \u22122kd kd \u23a4\u23a6x (13.143) The matrices [m], [c], and [k], in K, D, and V , may not be symmetric however, the matrices [m], [c], and [k] in \u2202K/\u2202x\u0307, \u2202D/\u2202x\u0307, \u2202V/\u2202x are symmetric. When a matrix [a] is diagonal, it is symmetric and [a] = [a] . (13.144) A diagonal matrix cannot be written in different forms. The matrix [m] in Example 479 is diagonal and hence, K has only one form (13.132). Example 481 F Positive definite matrix. A matrix [a] is called positive definite if xT [a]x > 0 for all x 6= 0. A matrix [a] is called positive semidefinite if xT [a]x \u2265 0 for all x. Kinetic energy is positive definite and it means we cannot have K = 0 unless x\u0307 = 0. Potential energy is positive semidefinite and it means we have V \u2265 0 as long as x > 0, however, it is possible to have a especial x0 > 0 at which V = 0. 13.3 Natural Frequencies and Mode Shapes Unforced and undamped vibrations of a system is a basic response of the system which expresses its natural behavior. We call a system with no damping and no external excitation, a free system. A free system is governed by the following set of differential equations. [m] x\u0308+ [k]x = 0 (13.145) The response of the free system is harmonic x = nX i=1 ui (Ai sin\u03c9it+Bi cos\u03c9it) i = 1, 2, 3, \u00b7 \u00b7 \u00b7n = nX i=1 Ciui sin (\u03c9it\u2212 \u03d5i) i = 1, 2, 3, \u00b7 \u00b7 \u00b7n (13.146) where, \u03c9i are the natural frequencies and ui are the mode shapes of the system. 846 13. Vehicle Vibrations The natural frequencies \u03c9i are solutions of the characteristic equation of the system det \u00a3 [k]\u2212 \u03c92 [m] \u00a4 = 0 (13.147) and the mode shapes ui, corresponding to \u03c9i, are solutions of the following equation. \u00a3 [k]\u2212 \u03c92i [m] \u00a4 ui = 0 (13.148) The unknown coefficients Ai and Bi, or Ci and \u03d5i, must be determined from the initial conditions. Proof. By eliminating the force and damping terms from the general equations of motion [m] x\u0308+ [c] x\u0307+ [k]x = F (13.149) we find the equations for free systems. [m] x\u0308+ [k]x = 0 (13.150) Let\u2019s search for a possible solution of the following form x = u q(t) (13.151) xi = ui q(t) i = 1, 2, 3, \u00b7 \u00b7 \u00b7n. (13.152) This solution implies that the amplitude ratio of two coordinates during motion does not depend on time. Substituting (13.151) in Equation (13.150) [m]u q\u0308(t) + [k]u q(t) = 0 (13.153) and separating the time dependent terms, ends up with the following equation. \u2212 q\u0308(t) q(t) = [[m]u] \u22121 [[k]u] = Pn j=1 kijujPn j=1mijuj i = 1, 2, 3, \u00b7 \u00b7 \u00b7n (13.154) Because the right hand side of this equation is time independent and the left hand side is independent of the index i, both sides must be equal to a constant. Let\u2019s assume the constant be a positive number \u03c92. Hence, Equation (13.154) can be separated into two equations q\u0308(t) + \u03c92q(t) = 0 (13.155) and \u00a3 [k]\u2212 \u03c92 [m] \u00a4 u = 0 (13.156) or nX j=1 \u00a1 kij \u2212 \u03c92mij \u00a2 uj = 0 i = 1, 2, 3, \u00b7 \u00b7 \u00b7n. (13.157) 13. Vehicle Vibrations 847 The solution of (13.155) is q(t) = sin\u03c9t+ cos\u03c9t = sin (\u03c9t\u2212 \u03d5) (13.158) which shows that all the coordinates of the system, xi, have harmonic motion with identical frequency \u03c9 and identical phase angle \u03d5. The frequency \u03c9 can be determined from Equation (13.156) which is a set of homogeneous equations for the unknown u. The set of equations (13.156) has a solution u = 0, which is the rest position of the system and shows no motion. This solution is called trivial solution and is unimportant. To have a nontrivial solution, the determinant of the coefficient matrix must be zero. det \u00a3 [k]\u2212 \u03c92 [m] \u00a4 = 0 (13.159) Determining the constant \u03c9, such that the set of equations (13.156) provide a nontrivial solution, is called eigenvalue problem. Expanding the determinant (13.159) provides an algebraic equation that is called the characteristic equation. The characteristic equation is an nth order equation in \u03c92, and provides n natural frequencies \u03c9i. The natural frequencies \u03c9i can be set in the following order. \u03c91 \u2264 \u03c92 \u2264 \u03c93 \u2264 \u00b7 \u00b7 \u00b7 \u2264 \u03c9n (13.160) Having n values for \u03c9 indicates that the solution (13.158) is possible with n different frequencies \u03c9i, i = 1, 2, 3, \u00b7 \u00b7 \u00b7n. We may multiply the Equation (13.150) by [m]\u22121 x\u0308+ [m]\u22121 [k]x = 0 (13.161) and find the the characteristic equation (13.159) as det [[A]\u2212 \u03bbI] = 0 (13.162) where [A] = [m] \u22121 [k] . (13.163) So, determination of the natural frequencies \u03c9i would be equivalent to determining the eigenvalues \u03bbi of the matrix [A] = [m] \u22121 [k]. \u03bbi = \u03c92i (13.164) Determining the vectors ui to satisfy Equation (13.156) is called the eigenvector problem. To determine ui, we may solve Equation (13.156) for every \u03c9i \u00a3 [k]\u2212 \u03c92i [m] \u00a4 ui = 0 (13.165) and find n different ui. In vibrations and vehicle dynamics, the eigenvector ui corresponding to the eigenvalue \u03c9i is called the mode shape. 848 13. Vehicle Vibrations Alternatively, we may find the eigenvectors of matrix [A] = [m]\u22121 [k] [[A]\u2212 \u03bbiI]ui = 0 (13.166) instead of finding the mode shapes from ((13.165)). Equations (13.165) are homogeneous so, if ui is a solution, then aui is also a solution. Hence, the eigenvectors are not unique and may be expressed with any length. However, the ratio of any two elements of an eigenvector is unique and therefore, ui has a unique shape. If one of the elements of ui is assigned, the remaining n\u2212 1 elements are uniquely determined. The shape of an eigenvector indicates the relative amplitudes of the coordinates of the system in vibration. Because the length of an eigenvector is not uniquely defined, there are many options to express ui. The most common expressions are: 1\u2212 normalization, 2\u2212 normal form, 3\u2212 high-unit, 4\u2212 first-unit, 5\u2212 last-unit. In the normalization expression, we may adjust the length of ui such that uTi [m]ui = 1 (13.167) or uTi [k]ui = 1 (13.168) and call ui a normal mode with respect to [m] or [k] respectively. In the normal form expression, we adjust ui such that its length has a unity value. In the high-unit expression, we adjust the length of ui such that the largest element has a unity value. In the first-unit expression, we adjust the length of ui such that the first element has a unity value. In the last-unit expression, we adjust the length of ui such that the last element has a unity value. Example 482 Eigenvalues and eigenvectors of a 2\u00d7 2 matrix. Consider a 2\u00d7 2 matrix is given as [A] = \u2219 5 3 3 6 \u00b8 . (13.169) To find the eigenvalues \u03bbi of [A], we find the characteristic equation of the matrix by subtracting an unknown \u03bb from the main diagonal, and taking 13. Vehicle Vibrations 849 the determinant. det [[A]\u2212 \u03bbI] = det \u2219\u2219 5 3 3 6 \u00b8 \u2212 \u03bb \u2219 1 0 0 1 \u00b8\u00b8 = det \u2219 5\u2212 \u03bb 3 3 6\u2212 \u03bb \u00b8 = \u03bb2 \u2212 11\u03bb+ 21 (13.170) The solution of the characteristic equation (13.170) are \u03bb1 = 8.5414 (13.171) \u03bb2 = 2.4586. (13.172) To find the corresponding eigenvectors u1 and u2 we solve the following equations. [[A]\u2212 \u03bb1I]u1 = 0 (13.173) [[A]\u2212 \u03bb2I]u2 = 0 (13.174) Let\u2019s denote the eigenvectors by u1 = \u2219 u11 u12 \u00b8 (13.175) u2 = \u2219 u21 u22 \u00b8 (13.176) therefore, [[A]\u2212 \u03bb1I]u1 = \u2219\u2219 5 3 3 6 \u00b8 \u2212 8.5414 \u2219 1 0 0 1 \u00b8\u00b8 \u2219 u11 u12 \u00b8 = \u2219 3u12 \u2212 3.5414u11 3u11 \u2212 2.5414u12 \u00b8 = 0 (13.177) [[A]\u2212 \u03bb2I]u2 = \u2219\u2219 5 3 3 6 \u00b8 \u2212 2.4586 \u2219 1 0 0 1 \u00b8\u00b8 \u2219 u21 u22 \u00b8 = \u2219 2.5414u21 + 3u22 3u21 + 3.5414u22 \u00b8 = 0. (13.178) Assigning last-unit eigenvectors u12 = 1 (13.179) u22 = 1 (13.180) provides u1 = \u2219 \u22121.180 5 1.0 \u00b8 (13.181) u2 = \u2219 0.84713 1.0 \u00b8 . (13.182) 850 13. Vehicle Vibrations Example 483 F Unique ratio of the eigenvectors\u2019 elements. To show an example that the ratio of the elements of eigenvectors is unique, we examine the eigenvectors u1 and u2 in Example 482. u1 = \u2219 3u12 \u2212 3.5414u11 3u11 \u2212 2.5414u12 \u00b8 (13.183) u2 = \u2219 2.5414u21 + 3u22 3u21 + 3.5414u22 \u00b8 (13.184) The ratio u11/u12 may be found from the first row of u1 in (13.183) u11 u12 = 3 3.5414 = 0.84712 (13.185) or from the second row u11 u12 = 2.5414 3 = 0.84713 (13.186) to examine their equality. The ratio u21/u22 may also be found from the first or second row of u2 in (13.184) to check their equality. u21 u22 = \u2212 3 2.5414 = \u22123.5414 3 = \u22121.1805 (13.187) Example 484 F Characteristics of free systems. Free systems have two characteristics: 1\u2212natural frequencies, and 2\u2212mode shapes. An n DOF vibrating system will have n natural frequencies \u03c9i and n mode shapes ui. The natural frequencies \u03c9i are cores for the system\u2019s resonance zones, and the eigenvectors ui show the relative vibration of different coordinates of the system at the resonance \u03c9i. The highest element of each mode shape ui, indicates the coordinate or the component of the system which is most willing to vibrate at \u03c9i. Example 485 Importance of free systems. The response of free systems is the core for all other responses of the vibrating system. When there is damping, then the response of the system is bounded by the free undamped solution. When there is a forcing function, the natural frequencies of the free response indicate the resonance zones at which the amplitude of the response may go to infinity if an excitation frequency of the force function matches. Example 486 F Sign of the separation constant \u03c92. Both, left and right sides of Equation (13.154) must be equal to a constant. The sign of the constant is dictated by physical considerations. A free and undamped vibrating system is conservative and as a constant mechanical energy, so the amplitude of vibration must remain finite when t\u2192\u221e. 13. Vehicle Vibrations 851 If the constant is positive, then the response is harmonic with a constant amplitude, however, if the constant is negative, the response is hyperbolic with an exponentially increasing amplitude. Example 487 F Quarter car natural frequencies and mode shapes. Figure 13.8 illustrates a quarter car model which is made of two solid masses ms and mu denoted as sprung and unsprung masses, respectively. The sprung mass ms represents 1/4 of the body of the vehicle, and the unsprung mass mu represents one wheel of the vehicle. A spring of stiffness ks, and a shock absorber with viscous damping coefficient cs support the sprung mass. The unsprung mass mu is in direct contact with the ground through a spring ku, and a damper cu representing the tire stiffness and damping. The governing differential equations of motion for the quarter car model are ms x\u0308s = \u2212ks (xs \u2212 xu)\u2212 cs (x\u0307s \u2212 x\u0307u) (13.188) mu x\u0308u = ks (xs \u2212 xu) + cs (x\u0307s \u2212 x\u0307u) \u2212ku (xu \u2212 y)\u2212 cu (x\u0307u \u2212 y\u0307) (13.189) which can be expressed in a matrix form [M ] x\u0307+ [c] x\u0307+ [k]x = F (13.190)\u2219 ms 0 0 mu \u00b8 \u2219 x\u0308s x\u0308u \u00b8 + \u2219 cs \u2212cs \u2212cs cs + cu \u00b8 \u2219 x\u0307s x\u0307u \u00b8 +\u2219 ks \u2212ks \u2212ks ks + ku \u00b8 \u2219 xs xu \u00b8 = \u2219 0 kuy + cuy\u0307 \u00b8 . (13.191) 852 13. Vehicle Vibrations To find the natural frequencies and mode shapes of the quarter car model, we have to drop the damping and forcing terms and analyze the following set of equations.\u2219 ms 0 0 mu \u00b8 \u2219 x\u0308s x\u0308u \u00b8 + \u2219 ks \u2212ks \u2212ks ks + ku \u00b8 \u2219 xs xu \u00b8 = 0 (13.192) Consider a vehicle with the following characteristics. ms = 375kg mu = 75kg ku = 193000N/m ks = 35000N/m. (13.193) The equations of motion for this vehicle are\u2219 375 0 0 75 \u00b8 \u2219 x\u0308s x\u0308u \u00b8 + \u2219 35000 \u221235000 \u221235000 2.28\u00d7 105 \u00b8 \u2219 xs xu \u00b8 = 0. (13.194) The natural frequencies of the vehicle can be found by solving its characteristic equation. det \u00a3 [k]\u2212 \u03c92 [m] \u00a4 = det \u2219 35000 \u221235000 \u221235000 2.28\u00d7 105 \u00b8 \u2212 \u03c92 \u2219 375 0 0 75 \u00b8 = det \u2219 35 000\u2212 375\u03c92 \u221235 000 \u221235 000 2.28\u00d7 105 \u2212 75\u03c92 \u00b8 = 28125\u03c94 \u2212 8.8125\u00d7 107\u03c92 + 6.755\u00d7 109 (13.195) \u03c91 = 8.8671 rad/ s \u2248 1.41Hz (13.196) \u03c92 = 55.269 rad/ s \u2248 8.79Hz (13.197) To find the corresponding mode shapes, we use Equation (13.165).\u00a3 [k]\u2212 \u03c921 [m] \u00a4 u1 = \u2219\u2219 35000 \u221235000 \u221235000 2.28\u00d7 105 \u00b8 \u2212 3054.7 \u2219 375 0 0 75 \u00b8\u00b8 \u2219 u11 u12 \u00b8 = \u2219 \u22121.1105\u00d7 106u11 \u2212 35000u12 \u221235000u11 \u2212 1102.5u12 \u00b8 = 0 (13.198) \u00a3 [k]\u2212 \u03c922 [m] \u00a4 u2 = \u2219\u2219 35000 \u221235000 \u221235000 2.28\u00d7 105 \u00b8 \u2212 78.625 \u2219 375 0 0 75 \u00b8\u00b8 \u2219 u21 u22 \u00b8 = \u2219 5515.6u21 \u2212 35000u22 2.221\u00d7 105u22 \u2212 35000u21 \u00b8 = 0 (13.199) 13. Vehicle Vibrations 853 Searching for the first-unit expression of u1 and u2 provides the following mode shapes. u1 = \u2219 1 \u22123.1729\u00d7 10\u22123 \u00b8 (13.200) u2 = \u2219 1 0.157 58 \u00b8 (13.201) Therefore, the free vibrations of the quarter car is x = nX i=1 ui (Ai sin\u03c9it+Bi cos\u03c9it) i = 1, 2 (13.202)\u2219 xs xu \u00b8 = \u2219 1 \u22123.1729\u00d7 10\u22123 \u00b8 (A1 sin 8.8671t+B1 cos 8.8671t) + \u2219 1 0.157 58 \u00b8 (A2 sin 55.269t+B2 cos 55.269t) (13.203) 13.4 Bicycle Car and Body Pitch Mode Quarter car model is excellent to examine and optimize the body bounce mode of vibrations. However, we may expand the vibrating model of a vehicle to include pitch and other modes of vibrations as well. Figure 13.9 illustrates a bicycle vibrating model of a vehicle. This model includes the body bounce x, body pitch \u03b8, wheels hop x1 and x2 and independent road excitations y1 and y2. The equations of motion for the bicycle vibrating model of a vehicle are 854 13. Vehicle Vibrations as follow. mx\u0308+ c1 \u00b3 x\u0307\u2212 x\u03071 \u2212 a1\u03b8\u0307 \u00b4 + c2 \u00b3 x\u0307\u2212 x\u03072 + a2\u03b8\u0307 \u00b4 +k1 (x\u2212 x1 \u2212 a1\u03b8) + k2 (x\u2212 x2 + a2\u03b8) = 0 (13.204) Iz \u03b8\u0308 \u2212 a1c1 \u00b3 x\u0307\u2212 x\u03071 \u2212 a1\u03b8\u0307 \u00b4 + a2c2 \u00b3 x\u0307\u2212 x\u03072 + a2\u03b8\u0307 \u00b4 \u2212a1k1 (x\u2212 x1 \u2212 a1\u03b8) + a2k2 (x\u2212 x2 + a2\u03b8) = 0 (13.205) m1x\u03081 \u2212 c1 \u00b3 x\u0307\u2212 x\u03071 \u2212 a1\u03b8\u0307 \u00b4 + kt1 (x1 \u2212 y1) \u2212k1 (x\u2212 x1 \u2212 a1\u03b8) = 0 (13.206) m2x\u03082 \u2212 c2 \u00b3 x\u0307\u2212 x\u03072 + a2\u03b8\u0307 \u00b4 + kt2 (x2 \u2212 y2) \u2212k2 (x\u2212 x2 + a2\u03b8) = 0 (13.207) As a reminder, the definition of the employed parameters are indicated in Table 13.1. Proof. Figure 13.10 shows a better vibrating model of the system. The body of the vehicle is assumed to be a rigid bar. This bar has a mass m, which is half of the total body mass, and a lateral mass moment of inertia Iy, which is half of the total body mass moment of inertia. The front and real wheels have a mass m1 and m2 respectively. The tires stiffness are indicated by different parameters kt1 and kt2 . It is because the rear tires are usually stiffer than the fronts, although in a simpler model we may assume kt1 = kt2 . Damping of tires are much smaller than the damping of shock absorbers so, we may ignore the tire damping for simpler calculation. To find the equations of motion for the bicycle vibrating model, we use 13. Vehicle Vibrations 855 the Lagrange method. The kinetic and potential energies of the system are K = 1 2 mx\u03072 + 1 2 m1x\u0307 2 1 + 1 2 m2x\u0307 2 2 + 1 2 Iz \u03b8\u0307 2 (13.208) V = 1 2 kt1 (x1 \u2212 y1) 2 + 1 2 kt2 (x2 \u2212 y2) 2 + 1 2 k1 (x\u2212 x1 \u2212 a1\u03b8) 2 + 1 2 k2 (x\u2212 x2 + a2\u03b8) (13.209) and the dissipation function is D = 1 2 c1 \u00b3 x\u0307\u2212 x\u03071 \u2212 a1\u03b8\u0307 \u00b42 + 1 2 c2 \u00b3 x\u0307\u2212 x\u03072 + a2\u03b8\u0307 \u00b4 . (13.210) Applying Lagrange method d dt \u00b5 \u2202K \u2202q\u0307r \u00b6 \u2212 \u2202K \u2202qr + \u2202D \u2202q\u0307r + \u2202V \u2202qr = fr r = 1, 2, \u00b7 \u00b7 \u00b7 4 (13.211) provides the following equations of motion (13.204)-(13.207). These set of equations may be rearranged in a matrix form [m] x\u0308+ [c] x\u0307+ [k]x = F (13.212) where, x = \u23a1\u23a2\u23a2\u23a3 x \u03b8 x1 x2 \u23a4\u23a5\u23a5\u23a6 (13.213) [m] = \u23a1\u23a2\u23a2\u23a3 m 0 0 0 0 Iz 0 0 0 0 m1 0 0 0 0 m2 \u23a4\u23a5\u23a5\u23a6 (13.214) 856 13. Vehicle Vibrations [c] = \u23a1\u23a2\u23a2\u23a3 c1 + c2 a2c2 \u2212 a1c1 \u2212c1 \u2212c2 a2c2 \u2212 a1c1 c1a 2 1 + c2a 2 2 a1c1 \u2212a2c2 \u2212c1 a1c1 c1 0 \u2212c2 \u2212a2c2 0 c2 \u23a4\u23a5\u23a5\u23a6 (13.215) [k] = \u23a1\u23a2\u23a2\u23a3 k1 + k2 a2k2 \u2212 a1k1 \u2212k1 \u2212k2 a2k2 \u2212 a1k1 k1a 2 1 + k2a 2 2 a1k1 \u2212a2k2 \u2212k1 a1k1 k1 + kt1 0 \u2212k2 \u2212a2k2 0 k2 + kt2 \u23a4\u23a5\u23a5\u23a6 (13.216) F = \u23a1\u23a2\u23a2\u23a3 0 0 y1kt1 y2kt2 \u23a4\u23a5\u23a5\u23a6 . (13.217) Example 488 Natural frequencies and mode shapes of a bicycle car model. Consider a vehicle with a heavy solid axle in the rear and independent suspensions in front. the vehicle has the following characteristics. m = 840 2 kg m1 = 53kg m2 = 152 2 kg Iy = 1100 kgm2 (13.218) a1 = 1.4m a2 = 1.47m (13.219) k1 = 10000N/m k2 = 13000N/m kt1 = kt2 = 200000N/m (13.220) The natural frequencies of this vehicle can be found by using the undamped and free vibration equations of motion. The characteristic equation of the system is det \u00a3 [k]\u2212 \u03c92 [m] \u00a4 = 8609\u00d7 109\u03c98 \u2212 1.2747\u00d7 1013\u03c96 +2.1708\u00d7 1016\u03c94 \u2212 1.676\u00d7 1018\u03c92 + 2.9848\u00d7 1019 (13.221) 13. Vehicle Vibrations 857 because [m] = \u23a1\u23a2\u23a2\u23a3 420 0 0 0 0 1100 0 0 0 0 53 0 0 0 0 76 \u23a4\u23a5\u23a5\u23a6 (13.222) [k] = \u23a1\u23a2\u23a2\u23a3 23000 5110 \u221210000 \u221213000 5110 47692 14000 \u221219110 \u221210000 14000 2.1\u00d7 105 0 \u221213000 \u221219110 0 2.13\u00d7 105 \u23a4\u23a5\u23a5\u23a6 . (13.223) To find the natural frequencies we may solve the characteristic equation (13.221) or search for eigenvalues of [A] = [m]\u22121 [k]. [A] = [m]\u22121 [k] = \u23a1\u23a2\u23a2\u23a3 54.762 12.167 \u221223.810 \u221230.952 4.6455 43.356 12.727 \u221217.373 \u2212188.68 264.15 3962.3 0 \u2212171.05 \u2212251.45 0 2802.6 \u23a4\u23a5\u23a5\u23a6 (13.224) The eigenvalues of [A] are \u03bb1 = 37.657 \u03bb2 = 54.943 \u03bb3 = 2806.1 \u03bb4 = 3964.3 (13.225) therefore, the natural frequencies of the bicycle car model are \u03c91 = p \u03bb1 = 6.1365 rad/ s \u2248 0.97665Hz \u03c92 = p \u03bb2 = 7. 412 4 rad/ s \u2248 1.1797Hz \u03c93 = p \u03bb3 = 52.973 rad/ s \u2248 8.4309Hz \u03c94 = p \u03bb4 = 62.963 rad/ s \u2248 10.021Hz. (13.226) The normal form of the mode shapes of the system are u1 = \u23a1\u23a2\u23a2\u23a3 0.61258 \u22120.7854 8.2312\u00d7 10\u22122 \u22123.353\u00d7 10\u22122 \u23a4\u23a5\u23a5\u23a6 (13.227) u2 = \u23a1\u23a2\u23a2\u23a3 0.95459 0.28415 2.6886\u00d7 10\u22122 0.08543 \u23a4\u23a5\u23a5\u23a6 (13.228) 858 13. Vehicle Vibrations u3 = \u23a1\u23a2\u23a2\u23a3 1.1273\u00d7 10\u22122 6.3085\u00d7 10\u22123 3.9841\u00d7 10\u22124 \u22120.99992 \u23a4\u23a5\u23a5\u23a6 (13.229) u4 = \u23a1\u23a2\u23a2\u23a3 6.0815\u00d7 10\u22123 \u22123.2378\u00d7 10\u22123 \u22120.99998 \u22121.9464\u00d7 10\u22124 \u23a4\u23a5\u23a5\u23a6 (13.230) The biggest element of the fourth mode shape u4 belongs to x1. It shows that in the fourth mode of vibrations at \u03c94 \u2248 10.021Hz the front wheel will have the largest amplitude, while the amplitude of the other components are \u0398 = u42 u43 = \u22123.2378\u00d7 10\u22123 \u22120.99998 X1 = 3.2379\u00d7 10\u22123X1 (13.231) X = u41 u43 = 6.0815\u00d7 10\u22123 \u22120.99998 X1 = \u22126.0816\u00d7 10\u22123X1 (13.232) X2 = u44 u43 = \u22123.2378\u00d7 10\u22123 \u22120.99998 X1 = 3.2379\u00d7 10\u22123X1. (13.233) Example 489 Comparison of the mode shapes of a bicycle car model. In Example 488, the biggest element of the first mode shape u1 belongs to \u03b8, the biggest element of the second mode shape u2 belongs to x, and the biggest element of the first mode shape u3 belongs to x2. Similar to the fourth mode shape u4, we can find the relative amplitude of different coordinates at each mode. Consider a car starts to move on a bumpy road at a very small acceleration. By increasing the speed, the first resonance occurs at \u03c91 \u2248 0.97665Hz, at which the pitch vibration is the most observable vibration. The second resonance occurs at \u03c92 \u2248 1.1797Hz when the bounce vibration of the body is the most observable vibration. The third and fourth resonances at \u03c93 \u2248 8.4309Hz and \u03c94 \u2248 10.021Hz are related to rear and front wheels respectively. When the excitation frequency of a multiple DOF system increases, we will see that observable vibration moves from a coordinate to the others in the order of natural frequencies and associated mode shapes. When the excitation frequency is exactly at a natural frequency, the relative amplitudes of vibration are exactly similar to the associated mode shape. If the excitation frequency is not on a natural frequency, then vibration of the system is a combination of all modes of vibration. However, the weight factor of the closer modes are higher. 13.5 Half Car and Body Roll Mode To examine and optimize the roll vibration of a vehicle we may use a half car vibrating model. Figure 13.11 illustrates a half car model of a vehicle. 13. Vehicle Vibrations 859 This model includes the body bounce x, body roll \u03d5, wheels hop x1 and x2 and independent road excitations y1 and y2. The equations of motion for the half car vibrating model of a vehicle are as follow. mx\u0308+ c (x\u0307\u2212 x\u03071 + b1\u03d5\u0307) + c (x\u0307\u2212 x\u03072 \u2212 b2\u03d5\u0307) +k (x\u2212 x1 + b1\u03d5) + k (x\u2212 x2 \u2212 b2\u03d5) = 0 (13.234) Ix\u03d5\u0308+ b1c (x\u0307\u2212 x\u03071 + b1\u03d5\u0307)\u2212 b2c (x\u0307\u2212 x\u03072 \u2212 b2\u03d5\u0307) +b1k (x\u2212 x1 + b1\u03d5)\u2212 b2k (x\u2212 x2 \u2212 b2\u03d5) + kR\u03d5 = 0 (13.235) m1x\u03081 \u2212 c (x\u0307\u2212 x\u03071 + b1\u03d5\u0307) + kt (x1 \u2212 y1) \u2212k (x\u2212 x1 + b1\u03d5) = 0 (13.236) m2x\u03082 \u2212 c (x\u0307\u2212 x\u03072 \u2212 b2\u03d5\u0307) + kt (x2 \u2212 y2) \u2212k (x\u2212 x2 \u2212 b2\u03d5) = 0 (13.237) The half car model may be different for the front half and rear half due to different suspensions and mass distribution. Furthermore, different antiroll bars with different torsional stiffness may be used in the front and rear halves. Proof. Figure 13.12 shows a better vibrating model of the system. The body of the vehicle is assumed to be a rigid bar. This bar has a mass m, which is the front or rear half of the total body mass, and a longitudinal mass moment of inertia Ix, which is half of the total body mass moment of inertia. The left and right wheels have a mass m1 and m2 respectively, although they are usually equal. The tires stiffness are indicated by kt. Damping of tires are much smaller than the damping of shock absorbers so, we may ignore the tire damping for simpler calculation. The suspension of the car has stiffness k and damping c for the left and right wheels. It 860 13. Vehicle Vibrations is common to make the suspension of the left and right wheels mirror. So, their stiffness and damping are equal. However, the half car model has different k, c, and kt for front or rear. The vehicle may also have an antiroll bar with a torsional stiffness kR in front and or rear. Using a simple model, the antiroll bar provides a torque MR proportional to the roll angle \u03d5. MR = \u2212kR\u03d5 (13.238) However, a better model of the antiroll bar effect is MR = \u2212kR \u00b5 \u03d5\u2212 x1 \u2212 x2 w \u00b6 . (13.239) To find the equations of motion for the half car vibrating model, we use the Lagrange method. The kinetic and potential energies of the system are K = 1 2 mx\u03072 + 1 2 m1x\u0307 2 1 + 1 2 m2x\u0307 2 2 + 1 2 Ix\u03d5\u0307 2 (13.240) V = 1 2 kt (x1 \u2212 y1) 2 + 1 2 kt (x2 \u2212 y2) 2 + 1 2 kR\u03d5 2 + 1 2 k (x\u2212 x1 + b1\u03d5) 2 + 1 2 k (x\u2212 x2 \u2212 b2\u03d5) (13.241) and the dissipation function is D = 1 2 c (x\u0307\u2212 x\u03071 + b1\u03d5\u0307) 2 + 1 2 c (x\u0307\u2212 x\u03072 \u2212 b2\u03d5\u0307) . (13.242) Applying the Lagrange method d dt \u00b5 \u2202K \u2202q\u0307r \u00b6 \u2212 \u2202K \u2202qr + \u2202D \u2202q\u0307r + \u2202V \u2202qr = fr r = 1, 2, \u00b7 \u00b7 \u00b7 4 (13.243) 13. Vehicle Vibrations 861 provides the following equations of motion (13.234)-(13.237). The set of equations of motion may be rearranged in a matrix form [m] x\u0308+ [c] x\u0307+ [k]x = F (13.244) where, x = \u23a1\u23a2\u23a2\u23a3 x \u03d5 x1 x2 \u23a4\u23a5\u23a5\u23a6 (13.245) [m] = \u23a1\u23a2\u23a2\u23a3 m 0 0 0 0 Ix 0 0 0 0 m1 0 0 0 0 m2 \u23a4\u23a5\u23a5\u23a6 (13.246) [c] = \u23a1\u23a2\u23a2\u23a3 2c cb1 \u2212 cb2 \u2212c \u2212c cb1 \u2212 cb2 cb21 + cb22 \u2212cb1 cb2 \u2212c \u2212cb1 c 0 \u2212c cb2 0 c \u23a4\u23a5\u23a5\u23a6 (13.247) [k] = \u23a1\u23a2\u23a2\u23a3 2k kb1 \u2212 kb2 \u2212k \u2212k kb1 \u2212 kb2 kb21 + kb22 + kR \u2212kb1 kb2 \u2212k \u2212kb1 k + kt 0 \u2212k kb2 0 k + kt \u23a4\u23a5\u23a5\u23a6 (13.248) F = \u23a1\u23a2\u23a2\u23a3 0 0 y1kt y2kt \u23a4\u23a5\u23a5\u23a6 . (13.249) Example 490 Natural frequencies and mode shapes of a half car model. Consider a vehicle with the following characteristics. m = 840 2 kg m1 = 53 kg m2 = 53 kg Ix = 820 kgm2 (13.250) b1 = 0.7m b2 = 0.75m (13.251) k = 10000N/m kt = kt = 200000N/m kR = 25000Nm/ rad (13.252) 862 13. Vehicle Vibrations The natural frequency of this vehicle is found by using the undamped and free vibration equations of motion. [m] x\u0308+ [k]x = 0 (13.253) The characteristic equation of the system is det \u00a3 [k]\u2212 \u03c92 [m] \u00a4 = 6742\u00d7 108\u03c98 \u2212 8.192 0\u00d7 1012\u03c96 +1.9363\u00d7 1016\u03c94 \u2212 8.3728\u00d7 1018\u03c92 +3.2287\u00d7 1020 (13.254) because [m] = \u23a1\u23a2\u23a2\u23a3 420 0 0 0 0 820 0 0 0 0 53 0 0 0 0 53 \u23a4\u23a5\u23a5\u23a6 (13.255) [k] = \u23a1\u23a2\u23a2\u23a3 2.1\u00d7 105 \u2212500 \u221210000 \u221210000 \u2212500 35525 \u22127000 7500 \u221210000 \u22127000 2.1\u00d7 105 0 \u221210000 7500 0 2.1\u00d7 105 \u23a4\u23a5\u23a5\u23a6 . (13.256) To find the natural frequencies we may solve the characteristic equation (13.221) or search for eigenvalues of [A] = [m]\u22121 [k]. [A] = [m] \u22121 [k] = \u23a1\u23a2\u23a2\u23a3 500.0 \u22121.1905 \u221223.810 \u221223.810 \u22120.609 76 43.323 \u22128.5366 9.1463 \u2212188.68 \u2212132.08 3962.3 0 \u2212188.68 141.51 0 3962.3 \u23a4\u23a5\u23a5\u23a6 (13.257) The eigenvalues of [A] are \u03bb1 = 42.702 \u03bb2 = 497.32 \u03bb3 = 3962.9 \u03bb4 = 3964.9 (13.258) therefore, the natural frequencies of the half car model are \u03c91 = p \u03bb1 = 6.5347 rad/ s \u2248 1.04Hz \u03c92 = p \u03bb2 = 22.301 rad/ s \u2248 3.5493Hz \u03c93 = p \u03bb3 = 62.952 rad/ s \u2248 10.019Hz \u03c94 = p \u03bb4 = 62.967 rad/ s \u2248 10.022Hz. (13.259) 13. Vehicle Vibrations 863 The normal form of the mode shapes of the system are u1 = \u23a1\u23a2\u23a2\u23a3 2.4875\u00d7 10\u22123 0.99878 3.3776\u00d7 10\u22122 \u22123.5939\u00d7 10\u22122 \u23a4\u23a5\u23a5\u23a6 (13.260) u2 = \u23a1\u23a2\u23a2\u23a3 0.99705 \u22121.264\u00d7 10\u22123 5.4246\u00d7 10\u22122 5.4346\u00d7 10\u22122 \u23a4\u23a5\u23a5\u23a6 (13.261) u3 = \u23a1\u23a2\u23a2\u23a3 1.0488\u00d7 10\u22124 3.1886\u00d7 10\u22123 \u22120.71477 0.69935 \u23a4\u23a5\u23a5\u23a6 (13.262) u4 = \u23a1\u23a2\u23a2\u23a3 9.7172\u00d7 10\u22123 \u22121.462\u00d7 10\u22124 \u22120.69932 \u22120.71474 \u23a4\u23a5\u23a5\u23a6 (13.263) Example 491 Comparison of the mode shapes of a half car model. In example 490, the biggest element of the first mode shape u1 belongs to \u03d5, the biggest element of the second mode shape u2 belongs to x, the biggest element of the third mode shape u3 belongs to x2, and the biggest element of the fourth mode shape u4 belongs to x1. Consider a car that starts to move on a bumpy road at a very small acceleration. By increasing the speed, the first resonance occurs at \u03c91 \u2248 1.04Hz, at which the roll vibration is the most observable vibration. The second resonance occurs at \u03c92 \u2248 3.5493Hz when the bounce vibration of the body is the most observable vibration. The third and fourth resonances at \u03c93 \u2248 10.019Hz and \u03c94 \u2248 10.022Hz are related to right and left wheels respectively. Example 492 Antiroll bar affects only the roll mode. If in example 490, we eliminate the antiroll bar by setting kR = 0, the natural frequencies and mode shapes of the half car model would be as follow. \u03bb1 = 12.221 \u03bb2 = 497.41 \u03bb3 = 3962.9 \u03bb4 = 3964.9 (13.264) 864 13. Vehicle Vibrations \u03c91 = p \u03bb1 = 3.4957 rad/ s \u2248 0.556 36Hz \u03c92 = p \u03bb2 = 22.301 rad/ s \u2248 3.5493Hz \u03c93 = p \u03bb3 = 62. 967 rad/ s \u2248 10.022Hz \u03c94 = p \u03bb4 = 62. 967 rad/ s \u2248 10.022Hz. (13.265) u1 = \u23a1\u23a2\u23a2\u23a3 2.332 2\u00d7 10\u22123 0.998 80 3.350 9\u00d7 10\u22122 \u22120.035 67 \u23a4\u23a5\u23a5\u23a6 (13.266) u2 = \u23a1\u23a2\u23a2\u23a3 0.997 05 \u22121.184 6\u00d7 10\u22123 5.424 9\u00d7 10\u22122 5.434 2\u00d7 10\u22122 \u23a4\u23a5\u23a5\u23a6 (13.267) u3 = \u23a1\u23a2\u23a2\u23a3 1.038 2\u00d7 10\u22124 3.164\u00d7 10\u22123 \u22120.714 69 0.699 43 \u23a4\u23a5\u23a5\u23a6 (13.268) u4 = \u23a1\u23a2\u23a2\u23a3 9.717 2\u00d7 10\u22123 \u22121.447 2\u00d7 10\u22124 \u22120.699 4 \u22120.714 66 \u23a4\u23a5\u23a5\u23a6 (13.269) Comparing these results with the results in example 490, shows the antiroll bar affects only the roll mode of vibration. A half car model needs a proper antiroll bar to increase the roll natural frequency. It is recommended to have the roll mode as close as possible to the body bounce natural frequency to have narrow resonance zone around the body bounce. Avoiding a narrow resonance zone would be simpler. 13.6 Full Car Vibrating Model A general vibrating model of a vehicle is called the full car model. Such a model, that is shown in Figure 13.13, includes the body bounce x, body roll \u03d5, body pitch \u03b8, wheels hop x1, x2, x3, and x4 and independent road excitations y1, y2, y3, and y4. A full car vibrating model has seven DOF with the following equations 13. Vehicle Vibrations 865 of motion. mx\u0308+ cf \u00b3 x\u0307\u2212 x\u03071 + b1\u03d5\u0307\u2212 a1\u03b8\u0307 \u00b4 + cf \u00b3 x\u0307\u2212 x\u03072 \u2212 b2\u03d5\u0307\u2212 a1\u03b8\u0307 \u00b4 +cr \u00b3 x\u0307\u2212 x\u03073 \u2212 b1\u03d5\u0307+ a2\u03b8\u0307 \u00b4 + cr \u00b3 x\u0307\u2212 x\u03074 + b2\u03d5\u0307+ a2\u03b8\u0307 \u00b4 +kf (x\u2212 x1 + b1\u03d5\u2212 a1\u03b8) + kf (x\u2212 x2 \u2212 b2\u03d5\u2212 a1\u03b8) +kr (x\u2212 x3 \u2212 b1\u03d5+ a2\u03b8) + kr (x\u2212 x4 + b2\u03d5+ a2\u03b8) = 0 (13.270) Ix\u03d5\u0308+ b1cf \u00b3 x\u0307\u2212 x\u03071 + b1\u03d5\u0307\u2212 a1\u03b8\u0307 \u00b4 \u2212 b2cf \u00b3 x\u0307\u2212 x\u03072 \u2212 b2\u03d5\u0307\u2212 a1\u03b8\u0307 \u00b4 \u2212b1cr \u00b3 x\u0307\u2212 x\u03073 \u2212 b1\u03d5\u0307+ a2\u03b8\u0307 \u00b4 + b2cr \u00b3 x\u0307\u2212 x\u03074 + b2\u03d5\u0307+ a2\u03b8\u0307 \u00b4 +b1kf (x\u2212 x1 + b1\u03d5\u2212 a1\u03b8)\u2212 b2kf (x\u2212 x2 \u2212 b2\u03d5\u2212 a1\u03b8) \u2212b1kr (x\u2212 x3 \u2212 b1\u03d5+ a2\u03b8) + b2kr (x\u2212 x4 + b2\u03d5+ a2\u03b8) +kR \u00b5 \u03d5\u2212 x1 \u2212 x2 w \u00b6 = 0 (13.271) Iy \u03b8\u0308 \u2212 a1cf \u00b3 x\u0307\u2212 x\u03071 + b1\u03d5\u0307\u2212 a1\u03b8\u0307 \u00b4 \u2212 a1cf \u00b3 x\u0307\u2212 x\u03072 \u2212 b2\u03d5\u0307\u2212 a1\u03b8\u0307 \u00b4 +a2cr \u00b3 x\u0307\u2212 x\u03073 \u2212 b1\u03d5\u0307+ a2\u03b8\u0307 \u00b4 + a2cr \u00b3 x\u0307\u2212 x\u03074 + b2\u03d5\u0307+ a2\u03b8\u0307 \u00b4 \u2212a1kf (x\u2212 x1 + b1\u03d5\u2212 a1\u03b8)\u2212 a1kf (x\u2212 x2 \u2212 b2\u03d5\u2212 a1\u03b8) +a2kr (x\u2212 x3 \u2212 b1\u03d5+ a2\u03b8) + a2kr (x\u2212 x4 + b2\u03d5+ a2\u03b8) = 0 (13.272) 866 13. Vehicle Vibrations mf x\u03081 \u2212 cf \u00b3 x\u0307\u2212 x\u03071 + b1\u03d5\u0307\u2212 a1\u03b8\u0307 \u00b4 \u2212 kf (x\u2212 x1 + b1\u03d5\u2212 a1\u03b8) \u2212kR 1 w \u00b5 \u03d5\u2212 x1 \u2212 x2 w \u00b6 + ktf (x1 \u2212 y1) = 0 (13.273) mf x\u03082 \u2212 cf \u00b3 x\u0307\u2212 x\u03072 \u2212 b2\u03d5\u0307\u2212 a1\u03b8\u0307 \u00b4 \u2212 kf (x\u2212 x2 \u2212 b2\u03d5\u2212 a1\u03b8) +kR 1 w \u00b5 \u03d5\u2212 x1 \u2212 x2 w \u00b6 + ktf (x2 \u2212 y2) = 0 (13.274) mrx\u03083 \u2212 cr \u00b3 x\u0307\u2212 x\u03073 \u2212 b1\u03d5\u0307+ a2\u03b8\u0307 \u00b4 \u2212kr (x\u2212 x3 \u2212 b1\u03d5+ a2\u03b8) + ktr (x3 \u2212 y3) = 0 (13.275) mrx\u03084 \u2212 cr \u00b3 x\u0307\u2212 x\u03074 + b2\u03d5\u0307+ a2\u03b8\u0307 \u00b4 \u2212kr (x\u2212 x4 + b2\u03d5+ a2\u03b8) + ktr (x4 \u2212 y4) = 0 (13.276) Proof. Figure 13.14 shows a better vibrating model of the system. The body of the vehicle is assumed to be a rigid slab. This slab has a mass m, which is the total body mass, a longitudinal mass moment of inertia Ix, and a lateral mass moment of inertia Iy. the moments of inertia are only the body mass moments of inertia not the vehicle\u2019s mass moments of inertia. The wheels have a mass m1, m2, m3, and m4 respectively. However, it is common to have m1 = m2 = mf (13.277) m3 = m4 = mr. (13.278) The front and rear tires stiffness are indicated by ktf and ktr respectively. Because the damping of tires are much smaller than the damping of shock absorbers, we may ignore the tires\u2019 damping for simpler calculation. The suspension of the car has stiffness kf and damping cf in the front and stiffness kr and damping cr in the rear. It is common to make the suspension of the left and right wheels mirror. So, their stiffness and damping are equal. The vehicle may also have an antiroll bar in front and in the back, with a torsional stiffness kRf and kRr . Using a simple model, the antiroll bar provides a torque MR proportional to the roll angle \u03d5. MR = \u2212 \u00a1 kRf + kRr \u00a2 \u03d5 = \u2212kR\u03d5 (13.279) 13. Vehicle Vibrations 867 However, a better model of the antiroll bar reaction is MR = \u2212kRf \u00b5 \u03d5\u2212 x1 \u2212 x2 wf \u00b6 \u2212 kRf \u00b5 \u03d5\u2212 x4 \u2212 x3 wr \u00b6 . (13.280) Most cars only have an antiroll bar in front. For these cars, the moment of the antiroll bar simplifies to MR = \u2212kR \u00b5 \u03d5\u2212 x1 \u2212 x2 w \u00b6 (13.281) if we use wf \u2261 w = b1 + b2 (13.282) kRf \u2261 kR. (13.283) To find the equations of motion for the full car vibrating model, we use the Lagrange method. The kinetic and potential energies of the system are K = 1 2 mx\u03072 + 1 2 Ix\u03d5\u0307 2 + 1 2 Iy \u03b8\u0307 2 + 1 2 mf \u00a1 x\u030721 + x\u030722 \u00a2 + 1 2 mr \u00a1 x\u030723 + x\u030724 \u00a2 (13.284) 868 13. Vehicle Vibrations V = 1 2 kf (x\u2212 x1 + b1\u03d5\u2212 a1\u03b8) 2 + 1 2 kf (x\u2212 x2 \u2212 b2\u03d5\u2212 a1\u03b8) 2 + 1 2 kr (x\u2212 x3 \u2212 b1\u03d5+ a2\u03b8) 2 + 1 2 kr (x\u2212 x4 + b2\u03d5+ a2\u03b8) 2 + 1 2 kR \u00b5 \u03d5\u2212 x1 \u2212 x2 w \u00b62 + 1 2 ktf (x1 \u2212 y1) 2 + 1 2 ktf (x2 \u2212 y2) 2 + 1 2 ktr (x3 \u2212 y3) 2 + 1 2 ktr (x4 \u2212 y4) 2 (13.285) and the dissipation function is D = 1 2 cf \u00b3 x\u0307\u2212 x\u03071 + b1\u03d5\u0307\u2212 a1\u03b8\u0307 \u00b42 + 1 2 cf \u00b3 x\u0307\u2212 x\u03072 \u2212 b2\u03d5\u0307\u2212 a1\u03b8\u0307 \u00b42 + 1 2 cr \u00b3 x\u0307\u2212 x\u03073 \u2212 b1\u03d5\u0307+ a2\u03b8\u0307 \u00b42 + 1 2 cr \u00b3 x\u0307\u2212 x\u03074 + b2\u03d5\u0307+ a2\u03b8\u0307 \u00b42 (13.286) Applying Lagrange method d dt \u00b5 \u2202K \u2202q\u0307r \u00b6 \u2212 \u2202K \u2202qr + \u2202D \u2202q\u0307r + \u2202V \u2202qr = fr r = 1, 2, \u00b7 \u00b7 \u00b7 7 (13.287) provides the following equations of motion (13.270)-(13.276). The set of equations of motion may be rearranged in a matrix form [m] x\u0308+ [c] x\u0307+ [k]x = F (13.288) where, x = \u23a1\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3 x \u03d5 \u03b8 x1 x2 x3 x4 \u23a4\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6 (13.289) [m] = \u23a1\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3 m 0 0 0 0 0 0 0 Ix 0 0 0 0 0 0 0 Iy 0 0 0 0 0 0 0 mf 0 0 0 0 0 0 0 mf 0 0 0 0 0 0 0 mr 0 0 0 0 0 0 0 mr \u23a4\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6 (13.290) 13. Vehicle Vibrations 869 [c] = \u23a1\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3 c11 c12 c13 \u2212cf \u2212cf \u2212cr \u2212cr c21 c22 c23 \u2212b1cf b2cf b1cr \u2212b2cr c31 c32 c33 a1cf a1cf \u2212a2cr \u2212a2cr \u2212cf \u2212b1cf a1cf cf 0 0 0 \u2212cf b2cf a1cf 0 cf 0 0 \u2212cr b1cr \u2212a2cr 0 0 cr 0 \u2212cr \u2212b2cr \u2212a2cr 0 0 0 cr \u23a4\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6 (13.291) c11 = 2cf + 2cr c21 = c12 = b1cf \u2212 b2cf \u2212 b1cr + b2cr c31 = c13 = 2a2cr \u2212 2a1cf c22 = b21cf + b22cf + b21cr + b22cr c32 = c23 = a1b2cf \u2212 a1b1cf \u2212 a2b1cr + a2b2cr c33 = 2cfa 2 1 + 2cra 2 2 (13.292) [k] = \u23a1\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3 k11 k12 k13 \u2212kf \u2212kf \u2212kr \u2212kr k21 k22 k23 k24 k25 b1kr \u2212b2kr k31 k32 k33 a1kf a1kf \u2212a2kr \u2212a2kr \u2212kf k42 a1kf k44 \u2212kR w2 0 0 \u2212kf k52 a1kf \u2212kR w2 k55 0 0 \u2212kr b1kr \u2212a2kr 0 0 kr + ktr 0 \u2212kr \u2212b2kr \u2212a2kr 0 0 0 kr + ktr \u23a4\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6 (13.293) k11 = 2kf + 2kr k21 = k12 = b1kf \u2212 b2kf \u2212 b1kr + b2kr k31 = k13 = 2a2kr \u2212 2a1kf k22 = kR + b21kf + b22kf + b21kr + b22kr k32 = k23 = a1b2kf \u2212 a1b1kf \u2212 a2b1kr + a2b2kr k42 = k24 = \u2212b1kf \u2212 1 w kR k52 = k25 = b2kf + 1 w kR k33 = 2kfa 2 1 + 2kra 2 2 k44 = kf + ktf + 1 w2 kR k55 = kf + ktf + 1 w2 kR (13.294) 870 13. Vehicle Vibrations F = \u23a1\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3 0 0 0 y1ktf y2ktf y3ktr y4ktr \u23a4\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6 . (13.295) Example 493 Natural frequencies and mode shapes of a full car model. Consider a vehicle with a the following characteristics. m = 840 kg mf = 53 kg mr = 76 kg Ix = 820 kgm2 Iy = 1100 kgm2 (13.296) a1 = 1.4m a2 = 1.47m b1 = 0.7m b2 = 0.75m (13.297) kf = 10000N/m kr = 13000N/m ktf = ktr = 200000N/m kR = 25000Nm/ rad (13.298) Using the matrix [A] = [m]\u22121 [k] and solving the associated eigenvalue and eigenvector problems, we find the following natural frequencies, and mode shapes for the full car model. \u03c91 = 1.11274Hz \u03c92 = 1.15405Hz \u03c93 = 1.46412Hz \u03c94 = 8.42729Hz \u03c95 = 8.43346Hz \u03c96 = 10.0219Hz \u03c97 = 10.5779Hz. (13.299) 13. Vehicle Vibrations 871 u1 = \u23a1\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3 1 0.0871235 \u22120.257739 0.0747553 0.0562862 0.0347648 0.0426351 \u23a4\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6 (13.300) u2 = \u23a1\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3 \u22120.0927632 1 \u22120.0222029 0.101837 \u22120.110205 \u22120.0514524 0.0389078 \u23a4\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6 (13.301) u3 = \u23a1\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3 .331817 0.0611334 1 \u22120.0453749 \u22120.0585134 0.110767 0.116324 \u23a4\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6 (13.302) u4 = \u23a1\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3 \u22120.0000730868 0.00826940 \u22120.0000799890 0.00233206 \u22120.00241445 1 \u2212.986988 \u23a4\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6 (13.303) u5 = \u23a1\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3 \u22120.0111901 \u22120.000380648 \u22120.0126628 0.000965144 0.00118508 0.978914 1 \u23a4\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6 (13.304) u6 = \u23a1\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3 \u22120.00606295e\u2212 2 0.000156931e\u2212 3 0.00655247e\u2212 2 1 0.99966 \u22120.000528339e\u2212 3 \u22120.000566133e\u2212 3 \u23a4\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6 (13.305) 872 13. Vehicle Vibrations u7 = \u23a1\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3 \u22120.000000327006 0.0137130 0.00000560144 \u22120.999745 1 0.00103530 \u22120.00107039 \u23a4\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6 (13.306) A visual illustration of the mode shapes are shown in Figures 13.15 to 13.21. The biggest element of the mode shapes u1 to u7 are x, \u03d5, \u03b8, x3, x4, x1, x2 respectively. These figures depict the relative amplitude of each coordinate of the full car model at a resonance frequency. The natural frequencies of a full car can be separated in two classes. The first class is the natural frequencies of the body: body bounce, body roll, and body pitch. Body related natural frequencies are always around 1Hz. The 13. Vehicle Vibrations 873 874 13. Vehicle Vibrations 13. Vehicle Vibrations 875 second class is the natural frequencies of the wheels bounce. Wheel related natural frequencies are always around 10Hz. In this example, we assumed the car has independent suspension in front and rear. So, each wheel has only vertical displacement. In case of a solid axle, the left and right wheels make a rigid body with a roll and bounce motion. The energies and hence, the equations of motion should be revised accordingly to show the bounce and roll of the solid axle. 13.7 Summary Vehicles are connected multi-body dynamic systems and hence, their vibrating model has multiple DOF system. The vibrating behavior of multiple DOF systems are very much dependent to their natural frequencies and mode shapes. These characteristics can be determined by solving an eigenvalue and an eigenvector problems. The most practical vibrating model of vehicles, starting from the simplest to more complex, are the one-eight car, quarter car, bicycle car, half car, and full car models. Having symmetric mass, stiffness, and damping matrices of multiple DOF system simplifies the calculation of the eigenvalue and eigenvector problems. To have symmetric coefficient matrices, we may define the kinetic energy, potential energy, and dissipation function of the system by quadratures and derive the equations of motion by applying the Lagrange method. 876 13. Vehicle Vibrations 13.8 Key Symbols a, x\u0308 acceleration a1 distance from mass center to front axle a2 distance from mass center to rear axle [a] , [A] coefficient matrix [A] = [m]\u22121 [k] coefficient matrix of characteristic equation b1 distance from mass center to left wheel b2 distance from mass center to right wheel c damping ceq equivalent damping cij element of row i and column j of [c] [c] damping matrix [c] symmetric damping matrix C mass center D dissipation function e eccentricity arm E mechanical energy f, F harmonic force f = 1 T cyclic frequency [ Hz] fc damper force fk spring force F amplitude of a harmonic force f Fr, Qr generalized force g gravitational acceleration I mass moment of inertia I identity matrix k stiffness keq equivalent stiffness kij element of row i and column j of [k] kR antiroll bar torsional stiffness [k] stiffness matrix [k] symmetric stiffness matrix K kinetic energy l length l wheelbase L Lagrangean m mass me eccentric mass mij element of row i and column j of [m] ms sprung mass mu unsprung mass [m] mass matrix [m] symmetric mass matrix 13. Vehicle Vibrations 877 n number of DOF p momentum qi, Qi generalized force r = \u03c9 \u03c9n frequency ratio r,R radius S quadrature t time T period uij jth element of the ith mode shape u mode shape, eigenvector ui ith eigenvector v,v, x\u0307, x\u0307 velocity V potential energy w track x absolute displacement X steady-state amplitude of x y base excitation displacement Y steady-state amplitude of y z relative displacement Z steady-state amplitude of z Zi short notation parameter \u03b4 deflection \u03be = c 2 \u221a km damping ratio \u03bb eigenvalue \u03bbi ith eigenvalue \u03c9 = 2\u03c0f angular frequency [ rad/ s] \u03c9n natural frequency \u03c9i ith natural frequency Subscript d driver f front r rear s sprung u unsprung 878 13. Vehicle Vibrations Exercises 1. Equation of motion of a multiple DOF system. Figure 13.22 illustrates a two DOF vibrating system. (a) Determine K, V , and D functions. (b) Determine the equations of motion using the Lagrange method. (c) F Rewrite K, V , and D in quadrature form. (d) Determine the natural frequencies and mode shapes of the system. 2. Absolute and relative coordinates. Figure 13.23 illustrates two similar double pendulums. We express the motion of the left one using absolute coordinates \u03b81 and \u03b82, and express the motion of the right one with absolute coordinate \u03b81 and relative coordinate \u03b82. (a) Determine the equation of motion of the absolute coordinate double pendulum. (b) Determine the equation of motion of the relative coordinate double pendulum. (c) Compare their mass and stiffness matrices. 13. Vehicle Vibrations 879 3. One-eight car model. Consider a one-eight car model as a base excited one DOF system. Determine its natural \u03c9n and damped natural frequencies \u03c9d if m = 1245 kg k = 60000N/m c = 2400N s/m. 4. Quarter car model. Consider a quarter car model. Determine its natural frequencies and mode shapes if ms = 1085/4 kg mu = 40 kg ks = 10000N/m ku = 150000N/m cs = 800N s/m. 5. Bicycle car model. 880 13. Vehicle Vibrations Consider a bicycle car model with the following characteristics: m = 1085/2 kg m1 = 40 kg m2 = 40 kg Iy = 1100 kgm2 a1 = 1.4m a2 = 1.47m k1 = 10000N/m kt1 = kt2 = 150000N/m Determine its natural frequencies and mode shapes for (a) k2 = 8000N/m (b) k2 = 10000N/m (c) k2 = 12000N/m. (d) Compare the natural frequencies for different k1/k2 and express the effect of increasing stiffness ratio on the pitch mode. 6. Half car model. Consider a bicycle car model with the following characteristics: m = 1085/2 kg m1 = 40 kg m2 = 40 kg Ix = 820 kgm2 b1 = 0.7m b2 = 0.75m k1 = 10000N/m kt1 = kt2 = 150000N/m Determine its natural frequencies and mode shapes for (a) kR = 0 (b) kR = 10000Nm/ rad (c) kR = 50000Nm/ rad. 13. Vehicle Vibrations 881 (d) Compare the natural frequencies for different kR and express the effect of increasing roll stiffness on the roll mode. (e) Determine kR such that the roll natural frequency be equal to the bounce natural frequency and determine the mode shapes of the half car for that kR. 7. Full car model. Consider a full car model with the following characteristics: m = 1085 kg mf = 40 kg mr = 40 kg Ix = 820 kgm2 Iy = 1100 kgm2 (13.307) a1 = 1.4m a2 = 1.47m b1 = 0.7m b2 = 0.75m (13.308) kf = 10000N/m kr = 10000N/m ktf = ktr = 150000N/m kR = 20000Nm/ rad (13.309) (a) Determine its natural frequencies and mode shapes. (b) Change kR such that the roll mode and pitch modes have the same frequency. (c) Determine the mode shapes of the car for that kR. 14"
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        "image_filename": "designv10_0_0003780_cdc.1993.325686-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003780_cdc.1993.325686-Figure5-1.png",
        "caption": "Figure 5: the general 2-trailer system is not flat.",
        "texts": [
            " , &P+\"(t)) are well defined smooth functions. This paper addresses the standard n-trailer system considered by many authors (see, e.g., [22, 24, 17, 21, 161 and the references herein) and the general n-trailer system that, as far as we know, has never been considered in details. The gcneral n-trai1e.r system is more realistic and differs from the standard one by the fact that trailer i + 1 is hitched to trailer i - 1 not directly at the center of its rear axle, but at a certain distance of this point (see, e.g., figure 5). Section 2 completes [19] whetc parking simulations were presented. The standard n-trailer system is shown to be flat. The Cartesian coordinates of the last trailer give the linearizing output. The natural parameterization of the curve followed by the last trailer and the Fdnet formulas are inuoduced. They lead to a simple geometric construction underlying proposition 2 and provide a global and consmctive solution of the motion planning problem. In section 3, the general 1-trailer system is considered",
            " As in [19], a simple steering program can be directly deduccd from such developments. Thc MATLAB simulations of figure 4 illustrate the interest of flatness combined with such gcomclric constructions. 4 Concluding remarks The concept of flatness, which has been illustrated by thesc nonholonomic systems, may be utilized in many industrial applications, such as the crane [lo], aircraft control [13] and chemical reactors [20, 181. Neverthelcss, all systems are not flat. Using the flatness characterization given in [15], one can prove that the general 2-trailer systcm of figure 5 and the plate-ball system considcrcd in [12] are not kit: their defccts [7, 41 are equal to one. These two nonllat systems are closcly related CO a class of nonlinear second order Mongc cquations studicd in [Ill. The multi-stecring trailer syslcms considered in (1, 26, 231 are also flat: Lhc llat output is then obtaincc1 by adding to the Cartesian coortlinatcs ol' the last trailer, the angles of thc trailers that are directly steered. This generalization is quitc natural in view of thc gcomctric: construction of ligurc 2. Notice finally that, if we add to thc general 2-trailer systcm of figure 5 a new control that steers directly the last trailer, thc system becomes flat: the linearizing output is then formcd of the point P with the angle of the last trailer. This fact cxplains probabily why multi-steering lrailcr systems are cncouiitcrcd B begin end the geometric construction egin Figurc 4: examplc of backward motions for the general 1- uailcr systcm (8 ) with 1 = 1 m, U = 1.5 m, 6 = 2.5 m and T = 15 S. in practice. Acknowledgment The MATLAB programs used for the simulations of figure 4 can be obtained upon request and via electronic mail from the first author (email: rouchon@ cas "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003657_02783649922066475-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003657_02783649922066475-Figure8-1.png",
        "caption": "Fig. 8. The tether stabilizes the robot and equalizes leg loads on steep terrain.",
        "texts": [
            " In the case of the translation drive system, a worst case of propulsion up a 30\u25e6 slope without any assistance from the tether system was assumed. This could occur, for instance, if the robot were climbing a small incline at the crater floor and could not rely on propulsive assistance from the tethering system. Specifications for the turning and translation actuators are given in Table 4. Dante II\u2019s tethering system enables it to traverse steep and nearly vertical terrain by reducing tractive forces at the feet and providing stability. As shown in Figure 8, tension from the tether cable is used to assist the robot in climbing by lifting much of the weight directly with the cable, instead of through transverse shearing forces at the feet (Krishna, Bares, and Mutschler 1997). Without assistance from the tether, the maximum transverse load possible at any foot is limited by either the tractive characteristics of the terrain material, or the transverse in-plane leg strength. On loose material such as that found on many volcanic slopes, a tetherless walking robot would thus be limited to mild inclines. Tether tension also provides stability on steep slopes (also shown in Fig. 8) where the robot would tip forward without the tether force. Note that the tether force also reduces the transverse in-plane foot forces (H1 and H2) which lowers the need for leg structure and thus lowers weight. Force components of the tether tension can also have destabilizing effects, which commonly occur when the robot turns its longitudinal axis away from the departing path of the cable, thus creating a lateral \u201crestoring\u201d force at the tether exit point. (A lateral force on the at East Tennessee State University on June 6, 2015ijr"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001231_s2301385014300017-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001231_s2301385014300017-Figure8-1.png",
        "caption": "Fig. 8. A FCS design example (developed by NUS UAV Research Team on 2005).",
        "texts": [
            " For onboard FCS construction, the elements are generally COTS products. I/O (input/output) compatibility is a primary concern for components selection, and serial communication is widely adopted. The reason for such prevalence is that most researchers sought for a prompt U n. S ys . 2 01 4. 02 :1 75 -1 99 . D ow nl oa de d fr om w w w .w or ld sc ie nt if ic .c om by F L IN D E R S U N IV E R SI T Y L IB R A R Y o n 02 /0 4/ 15 . F or p er so na l u se o nl y. but reliable solution for FCS development. A representative example is shown in Fig. 8. . Onboard software is generally developed based on realtime operating system (RTOS). Three main options include: QNX [129], RTLinux [130], and VxWorks [131]. In terms of software architecture, hierarchical and modular design was commonly adopted. Key tasks such as data acquisition and flight control execution are programmed as individual modules and assigned with different priorities. A benchmark work is presented in [45], which was developed for Stanford's DragonFly UAVs. . Aerial platforms developed for RC purposes are widely adopted as the baseline with minimum modification"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002591_j.ymssp.2020.107014-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002591_j.ymssp.2020.107014-Figure1-1.png",
        "caption": "Fig. 1. Experimental setup of an LM positioner control system.",
        "texts": [
            " Section 2 gives the structure of the LM positioner experimental setup and the plant model of the LM. In Section 3, an adaptive disturbance observer is introduced for disturbance estimation. The design of the robust recursive sliding mode controller is discussed in Section 4, where the stability analysis and parameter selections are also elaborated. Section 5 presents the experiments on the LM positioner and provides the results and comparisons to verify the performance of the proposed method. Finally, conclusion is given in Section 6. Fig. 1 shows the LM positioner experimental setup (by Baldor Electric Company). The positioning stage with a travel range of 500 mm is directly driven by an LM. An equipped position encoder (by Renishaw PLC) with a resolution of 1 lm is used to obtain the position information of the stage. The power amplifier converts the control input signals into current commands to drive the LM. In the setup, since the bandwidth of the power amplifier is 400 Hz, which is much higher than that of the LM dynamics, its model is simply treated as a constant gain"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure3.40-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure3.40-1.png",
        "caption": "Figure 3.40 (a) Forces are exerted on the block at three points of which the lines of action pass through a single point, so that there is moment equilibrium; (b) since all the forces exerted on the block form a closed force polygon there is also force equilibrium.",
        "texts": [
            " Solution: The three unknown forces are determined using the three equilibrium equations: \u2211 Fx = Ah + (4 kN) = 0,\u2211 Fy = Av + Bv \u2212 (6 kN) = 0,\u2211 Tz|A = +Bv \u00d7 (6 m) \u2212 (6 kN)(8 m) \u2212 (4 kN)(3 m) = 0. The first equation gives Ah = \u22124 kN, the third gives Bv = 10 kN, and the second equation gives Av = \u22124 kN. Av can also be found directly from the moment equilibrium about B: \u2211 Tz|B = \u2212Av \u00d7 (6 m) \u2212 (6 kN)(2 m) \u2212 (4 kN)(3 m) = 0 \u21d2 Av = \u22124 kN. The fact that Ah and Av are negative means that they act in a direction opposite to the directions given in Figure 3.39. In Figure 3.40a, the forces are depicted as they act on the block in reality. The block is subject to forces at three points: \u2022 the resultant of the two forces at C, with line of action c, \u2022 the force Bv at B, with line of action b, and \u2022 the resultant of the forces Ah and Av at A, with line of action a. Graphical check of the moment equilibrium (see Figure 3.40a): For a body subject to three forces, the lines of action of the three forces have to pass through a single point. This condition is met. Graphical check of the force equilibrium (see Figure 3.40b): There is force equilibrium if all the forces acting on the block form a closed force polygon. This is the case. 80 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM 3. A body subject to several forces of which the lines of action all pass through a single point (Figure 3.41a) or are all parallel (Figure 3.41b). If for all the forces on a body the lines of action intersect at a single point, the moment equilibrium of the body is assured (see Figure 3.41a). The force equilibrium needs further investigation"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure10.11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure10.11-1.png",
        "caption": "FIGURE 10.11. A two-wheel model for a vehicle moving with no roll.",
        "texts": [
            " According to (3.131), the existence of a sideslip angle is sufficient to generate a lateral force Fy, which is proportional to \u03b1 when the angle is small. Fy = \u2212C\u03b1 \u03b1 (10.107) 10.3.3 Two-wheel Model and Body Force Components Figure 10.10 illustrates the forces in the xy-plane acting at the tireprints of a front-wheel-steering four-wheel vehicle. When we ignore the roll motion of the vehicle, the xy-plane remains parallel to the road\u2019s XY -plane, and we may use a two-wheel model for the vehicle. Figure 10.11 illustrates a two-wheel model for a vehicle with no roll motion. The two-wheel model is also called a bicycle model, although a two-wheel model does not act similar to a bicycle. The force system applied on a two-wheel vehicle, in which only the front wheel is steerable, is Fx = Fxf cos \u03b4 + Fxr \u2212 Fyf sin \u03b4 (10.108) Fy = Fyf cos \u03b4 + Fyr + Fxf sin \u03b4 (10.109) Mz = a1Fyf \u2212 a2Fyr (10.110) where, \u00a1 Fxf , Fxr \u00a2 and \u00a1 Fyf , Fyr \u00a2 are the planar forces on the tireprint of the front and rear wheels. The force system may be approximated by the following equations, if the steer angle \u03b4 is assumed small: Fx \u2248 Fxf + Fxr (10"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001409_j.engfailanal.2014.05.018-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001409_j.engfailanal.2014.05.018-Figure5-1.png",
        "caption": "Fig. 5. Schematic of crack propagation for method 1: (a) q 6 q1max, (b) q > q1max.",
        "texts": [
            " ters of spur gear pairs. meters Gear pair 1 Gear pair 2 Pinion Gear Pinion Gear ber of teeth N 22 22 62 62 g\u2019s modulus E (GPa) 206 206 206 206 on\u2019s ratio v 0.3 0.3 0.3 0.3 ule m (mm) 3 3 3 3 ndum coefficient ha * 1 1 1 1 learance coefficient c* 0.25 0.25 0.25 0.25 h width L (mm) 20 20 20 20 sure angle a ( ) 20 20 20 20 bore radius rint (mm) 11.7 11.7 35.7 35.7 In method 1 the crack path is assumed to be a straight line. The crack starts at the root of the driven gear and then propagates, as is shown in Fig. 5. The geometrical parameters of the crack (q, t, w) are presented in Fig. 5, in which q denotes the crack depth, t the crack propagation direction, w the crack initial position (in this paper w = 35 ). Assuming that the crack extends along the whole tooth width with a uniform crack depth distribution, the tooth with this crack is still considered as a cantilevered beam, the curve of the tooth profile remains perfect and the foundation stiffness is not affected. In this case, based on Eq. (9), the Hertzian contact stiffness will still be a constant since the work surface of the tooth has no defect and the width of the effective work surface will always be a constant L",
            " Therefore, for the gear with a cracked tooth, Hertzian, fillet foundation and axial compressive stiffness can still be calculated according to Eqs. (9), (10), and (14), respectively. However, the bending and shear stiffnesses will change due to the influence of the crack. Therefore, based on Eq. (7) single-tooth-pair mesh stiffness for a cracked gear pair can be calculated as: k \u00bc 1 1 kh \u00fe 1 kb1 \u00fe 1 ks1 \u00fe 1 ka1 \u00fe 1 kf 1 \u00fe 1 kb crack \u00fe 1 ks crack \u00fe 1 ka2 \u00fe 1 kf 2 ; \u00f017\u00de where kb_crack and ks_crack denote bending and shear stiffnesses when a crack is introduced. When the tooth crack is present, and q \u00bc q1 6 q1 max (see Fig. 5a), the effective area moment of inertia and area of the cross section at the position of y can be calculated as: I \u00bc 1 12 \u00f0hq \u00fe x\u00de3L; yP 6 y 6 yG 1 12 \u00f02x\u00de3L \u00bc 2 3 x3L; y < yP or y > yG ( ; \u00f018\u00de A \u00bc \u00f0hq \u00fe x\u00deL; yP 6 y 6 yG 2xL; y < yP or y > yG ; \u00f019\u00de where hq is the distance from the root of the crack to the central line of the tooth, which corresponds to point G on the tooth profile, and hq can be calculated by: hq \u00bc xQ q1 sin t: \u00f020\u00de When the crack propagates to the central line of the tooth, q1 reaches its maximum value q1max = xQ/sint. Then the crack will change direction to q2, which is assumed to be exactly symmetric with q1, and in this stage q = q1max + q2 (see Fig. 5b). When the crack propagates along q2, the effective area moment of inertia and area of the cross section at the position of y can be calculated as: I \u00bc 2 3 x3L; y < ymax 1 12 x \u00f0y ymax\u00dehq yP ymax h i3 L; ymax 6 y < yP 1 12 \u00f0x hq\u00de3L; yP 6 y 6 yG 0; y > yG 8>>>< >>>: ; \u00f021\u00de A \u00bc 2xL; y < ymax x \u00f0y ymax\u00dehq yP ymax h i L; ymax 6 y < yP \u00f0x hq\u00deL; yP 6 y 6 yG 0; y > yG 8>>>< >>: ; \u00f022\u00de where ymax denotes the leftmost position of crack propagation (shown in Fig. 5b), and hq = q2sint. Based on Eqs. (15) and (16), the bending and shear stiffnesses of the cracked gears can be calculated. The schematic of the calculation process for method 1 is presented in Fig. 6. In this paper, the parameters of the studied spur gear pair are shown in Table 2. In order to analyze the influences of crack depth and crack propagation direction on TVMS, 10 crack cases are considered, as are shown in Table 3. The crack depth is expressed as a percentage which is defined as the ratio of actual crack depth with theoretical through-tooth crack size",
            " But the stiffness in Fig. 7 increases slightly with the increase of crack propagation direction angle t, this is because a crack with a pretty large t has little effect on the rigidity of the transition part. So when the transition part of the gear tooth is considered, not only the TVMS of the healthy gear pair will change (see Fig. 4), but the TVMS of the cracked gear pair will vary with the change of crack propagation direction angles. Instead of resorting to a straight line (see straight line PG in Fig. 5) as a limiting line for reducing the tooth thickness in method 1, a parabolic curve between the crack tip and the tooth profile tip is chosen as the limiting line in method 2 (see Fig. 8), which is proposed by Mohammed et al. [1] and the crack propagation path is assumed to be the same as that in method 1. Similarly, once kb_crack and ks_crack are determined, the single-tooth-pair mesh stiffness for cracked gears can be obtained by Eq. (17). Effective area moment of inertia and area of the cross section at the position of y in method 2 can be calculated as follows: When q \u00bc q1 6 q1 max (see Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure4.29-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure4.29-1.png",
        "caption": "Figure 4.29 (a) A truss with by definition solely hinged joints and (b) a frame with by definition exclusively rigid joints.",
        "texts": [
            " Although the structure now consists of only line elements, the transfer of forces is mostly unchanged and occurs through the same planes as in Figure 4.28a. These examples illustrate that spatial structures can be composed of planar structures that consist of line elements. It is therefore certainly worth the effort of further investigating these types of planar structures. 128 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Planar trusses and frames are planar structures that are loaded in their plane (see Figure 4.29). The difference between a truss and a frame is determined by the nature of the joints in the connections. \u2022 in a truss, the bars are joined together by hinges at all the connections;1 \u2022 in frames, all the joints are fixed and entirely stiff. The truss in Figure 4.29a appeared in the bridge in Figure 4.27b. The open circles, which represent the hinged joints, are generally omitted as in a truss all the joints are by definition hinged. The structure in Figure 4.29b is a frame. You will recognise part of the building in Figure 4.28b here, with the vertical floor loading and the horizontal wind loading. Sometimes the stiffnesses of the joints are accentuated by thickenings in the connections, but generally they are omitted. If there are also hinged joints in a frame, they have to be clearly depicted by means of open circles. This is the case in Figure 4.30, which could represent a building made of concrete, on which a steel floor was placed at a later stage"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure6.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure6.2-1.png",
        "caption": "Fig. 6.2. Palletizing task",
        "texts": [
            " Teaching by showing methods are very popular because of their simplicity, but they also have significant disadvantages. First of all, the operator's capa bility of visual alignment is restricted, so that the guiding has to be combined with other methods whenever precise positioning and, especially, precise orienting of the gripper are requested. Second, the trajectory often includes a number of points whose space relationships are known in advance. Apparently, there are many more suitable ways of formulating such known relationships than to \"show\" all of them to the robot. As an example, Fig. 6.2 shows a palletizing problem: parts conveyed by a moving belt should be placed by the manipulator in a pallet with 5 x 5 beds. For the sake of simplicity, suppose that the conveyor stops whenever a part comes in the fixed position PART where it will be picked by the manipulator. In order to ensure safe manipulation, the gripper first has to pass through the approaching position PI and then to move to the gripping position H. After closing its fingers, the gripper carrying the part should lift to the position PI, move to position P2 behind the bed BED[i,j], put the part down, open the fingers again and finally retreat to the position P2",
            " Its programming by pure teaching by showing requires guiding the robot hand through a total of 150 positions. Obviously, it is desirable to reduce the number of configurations that have to be showed to the robot. Another drawback of pure teaching by guiding is that the actual ma nipulator configurations must not deviate from the preplanned configurations. However, in a more realistic situation, some working positions are not known in advance and are partially or completely determined by some sensors during the robot operation. In the example from Fig. 6.2, the programmer would not be able to know the actual trajectory if the picking position of the parts were determined by a camera. In such a case, it is necessary to have tools that are more suitable for representing the positions of the objects and their interrelations. This purpose is served by the external coordinates: position ~ of an arbitrary rigid body is described by connecting an auxiliary coordinate frame to it, and by defining a position of that auxiliary frame to some reference frame. A relative position of one object in relation to another is described in a similar manner. For example, BASE in Fig. 6.2 can represent the position of a frame connected to the manipulator base in relation to the reference frame, while EFFECTOR can represent the position of a frame connected to the manipulator effector with respect to the frame connected to the manipulator base. The camera determines the relative position PART of the part visual centre with respect to its own 6.2 Describing the Motion 185 reference frame CAMERA, etc. Relative positions between the frames are usually defined in terms of cartesian coordinates and Euler's angles, by the equivalent translation and rotation vectors, or via homogeneous trans formations. During the manipulator operation, some relative positions can change and others stay fixed. It is often useful to formulate the unknown position of a particular object by composing the position of some other object which is determined by sensors, and the fixed relative position between the objects. For example, it can be assumed that position H in Fig. 6.2 of the effector grasping the part, relative to the visual centre of the part, should always remain fixed. If we denote the composing operator by \"+\", the absolute gripping position may be expressed as CAMERA + PART + H. If, on the other hand, we express the absolute position of the effector as BASE + EFFECTO R, then the following kinematic equation must hold at the grasping moment: BASE + EFFECTOR = CAMERA + PART + H In a similar manner we may introduce other operations. If X describes a position of some object in relation to the reference frame, we may denote a position of the reference frame relative to that object by - X"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.40-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.40-1.png",
        "caption": "FIGURE 7.40. Illustration of a car that is moving on a road at the point that O is the center of curvature.",
        "texts": [
            " In a negative steering, the rear wheels are steered in the opposite direction as the front wheels to turn in a significantly smaller radius, while in positive steering, the rear wheels are steered in the same direction as the front wheels to increase the lateral force. When the 4WS system is passive, there is a constant proportional ratio 422 7. Steering Dynamics between the front and rear steer angles which is equivalent to have a constant cs. A passive steering may be applied in vehicles to compensate some vehicle tendencies. As an example, in a FWS system, the rear wheels tend to steer slightly to the outside of a turn. Such tendency can reduce stability. Example 291 F Autodriver. Consider a car that is moving on a road, as shown in Figure 7.40. Point O indicates the center of curvature of the road at the car\u2019s position. Center of curvature of the road is supposed to be the turning center of the car at the instant of consideration. There is a global coordinate frame G attached to the ground, and a vehicle coordinate frame B attached to the car at its mass center C. The z and Z axes are parallel and the angle \u03c8 indicates the angle between X and x axes. If (XO, YO) are the coordinates of O in the global coordinate frame G then, 7. Steering Dynamics 423 the coordinates of C in B would be BrO = Rz,\u03c8 GrC + d\u23a1\u23a3 x y 0 \u23a4\u23a6 = \u23a1\u23a3 cos\u03c8 sin\u03c8 0 \u2212 sin\u03c8 cos\u03c8 0 0 0 1 \u23a4\u23a6\u23a1\u23a3 XC YC 0 \u23a4\u23a6+ \u23a1\u23a3 X Y 0 \u23a4\u23a6 = \u23a1\u23a3 X cos\u03c8 + Y sin\u03c8 Y cos\u03c8 \u2212X sin\u03c8 0 \u23a4\u23a6+ \u23a1\u23a3 X Y 0 \u23a4\u23a6 "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure4.12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure4.12-1.png",
        "caption": "FIGURE 4.12 Two conductor example for the capacitance sensitivity.",
        "texts": [
            " MESHING RELATED ACCURACY PROBLEMS FOR PEEC MODEL 79 We only consider in this text the capacitance aspect of the meshing accuracy problem. However, the meshing sensitivity issue presented here can also be related to the meshing for inductance cells in Chapter 5. The PEEC solution approach for capacitances is given in Section 4.3. Also, the construction of meshes is presented in Chapter 8. Here, we consider important issues regarding the influence of the mesh subdivision on the solution accuracy. Since we consider sensitivities only, all dimensions in Fig. 4.12 can be viewed as normalized units. It is clear that the accuracy of the computed capacitance values depend on the quality of the mesh. An important question is how fine the mesh subdivisions have to be for a sufficiently accurate answer without using too many cells or subdivisions. The sensitivity in (4.34) evaluates the quality of a mesh Sk\ud835\udcc1(N) = |||| |Ck\ud835\udcc1(2N) \u2212 Ck\ud835\udcc1(N)| Ck\ud835\udcc1(2N) |||| , (4.34) where N refers to the number of cells in (4.23) and k and \ud835\udcc1 are the two different conductors. Here, Ck\ud835\udcc1(2N) means the capacitance calculated using 2N mesh cells while Ck\ud835\udcc1(N) refers to the capacitance calculated using N cells. It is clear that the accuracy of an off-diagonal capacitance values is usually more sensitive to the cell division. Therefore, we use an off-diagonal coupling capacitance value for this measure. Practical experience with this sensitivity measure has shown that Sk\ud835\udcc1(N) \u2264 0.02 indicates that the number of cells N leads to a sufficiently accurate answer. The sensitivities are given for the mutual capacitance C12 for the two-conductor geometry in Fig. 4.12 for different conductor thicknesses T and conductor-to-conductor spacing S. The sensitivities in Table 4.2 are for 20 and 50 cells, respectively. We can observe that the sensitivity with respect to the number of cells is smaller for thin conductors. Of course, the sensitivity also decreases with larger conductor-to-conductor spacings S. In a problem with many conductors, the sensitivity of conductors with a small mutual capacitance tends to be less accurate. In this section, we consider a very important meshing issue pertaining to the location of the cells on conductors"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000538_acs.chemrev.8b00172-Figure25-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000538_acs.chemrev.8b00172-Figure25-1.png",
        "caption": "Figure 25. (1) Dynamic confinement effect and discriminative amplification on nanoporous electrodes. Once a reactant molecule comes into a nanopore, the probability of interaction between the nanoporous electrode surface and the molecule is augmented (b) compared with that on a flat electrode (a). In the case of a diffusion-controlled reaction, most of reactant molecules (red) undergo faradaic reaction at the outskirts of nanopores where the product molecules (blue) are created (c). On the other hand, the reactant molecules undergoing a sluggish electrokinetic process can enter deep inside the nanopores. A more sluggish reaction leads to longer penetration depth. Therefore, the amperometric responses of sluggish faradaic reactions are selectively amplified on nanoporous electrodes (d). Reproduced with permission from ref 315. Copyright 2012 The Royal Society of Chemistry. (2) Schematic cross-sectional view of Pt films without nanoporous structure with roughness factor (Rf) < 10 (a and e), 3DnpPt films with 10 < Rf < 40 (b and f), 40 < Rf < 300 (c and g), and extremely high Rf (d and h). Blue and red regions indicate the surface area effectively available for faradaic reaction. Reproduced with permission from ref 316. Copyright 2010 Elsevier.",
        "texts": [
            " The nature of the redox reaction determines whether nanoporous electrode structures are beneficial. For example, redox markers such as potassium ferrocyanide and ruthenium hexamine chloride exhibit diffusion-controlled behavior due to fast electron transfer rate constants. When the reaction of these redox markers occurs at the nanoporous electrode, an increased surface area bears no advantage as the molecules do not have enough time to access the inside of the nanocavities and tend to react at the rim of the pores (Figure 25-1c). On the other hand, redox molecules with slow reaction rates and a non-diffusion-controlled system, such as glucose and methylene blue, can realize the benefit of enhanced surface areas as they can enter deeper inside the nanocavities prior to the electrochemical reaction taking place (Figure 25-1d). These observations are also supported by Park and colleagues\u2019 investigation. They studied the relationship between the ratio of the real surface area to apparent area, termed roughness factor (Rf), and the participated electrochemical reactions for analytes with fast and slow electrokinetic processes as displayed in Figure 25-2. Thin-film nanoporous electrodes (low Rf as indicated by Figure 25-2a,b) promote the participation of electrochemical reactions of O2 and H2O2, while in the thick-film electrodes (high Rf as indicated by Figure 25-2c,d) the reaction takes place only on the top surfaces. On the other hand, the reaction of glucose, with slow reaction kinetics, can occur better when the thickness of the nanoporous electrode increases (Figure 25-2f\u2212h). Furthermore, this behavior enabled the selectivity for nonenzymatic detection of glucose realized by nanostructure electrodes as reported by Park et al.317 In 2015, Daggumati et al. comprehensively studied the effect of nanoporous gold (np-Au) morphology on the performance of DNA sensors using methylene blue as a redox marker.318 That study has shown that the annealed nanoporous gold electrodes which are obtained after thermal treatment of the nanoporous gold film (Figure 26-1) enhance the accessibility of DNA probes, DNA targets, and methylene blue, resulting in higher immobilized probe density, greater hybridization events, and ultimately better sensing performance over unannealed and flat gold electrodes (Figure 26-2)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000961_j.euromechsol.2008.07.007-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000961_j.euromechsol.2008.07.007-Figure3-1.png",
        "caption": "Fig. 3. Influence of backup ratio (A) and initial crack angle (B) on crack propagation directions (1\u2014through the tooth; 2\u2014through the rim).",
        "texts": [
            " (8) A tooth root crack typically results from insufficient rim thickness in the design, improperly processed material containing inclusions where cracks can start, severe operating conditions such as overload or misalignment or operation near the resonant frequency of a gear structure (McFadden, 1985). According to Lewicki (1996), the crack propagation depends on the backup ratio defined as rim thickness divided by tooth height. For high backup ratios the analysis predicted cracks that would propagate through the teeth (Fig. 3A direction 1) and through the rim for low backup ratios (Fig. 3A direction 2). The initial crack angle \u03b1c accounts also for the propagation. For low \u03b1c , the propagation is through the rim even for high backup ratios (Fig. 3B). Lewicki (2001) noted that crack propagation paths tend to be smooth, continuous, and rather straight with only a slight curvature. The proposed modelling of gear tooth cracks presented in this paper (Fig. 4) assumes a straight line shape for the crack which is introduced in the tooth root circle. This crack affects the whole width W of the tooth and is defined by its inclination angle \u03b1c and depth pc . This modelling induces a progressive reduction of the actual tooth thickness along the width of the tooth as shown in Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002769_j.cirpj.2021.04.010-Figure51-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002769_j.cirpj.2021.04.010-Figure51-1.png",
        "caption": "Fig. 51. Schematic arrangement of air-jet coupled experimental setup [102].",
        "texts": [
            " When the number of layers increases, the interlayer temperature (temperature difference before and during deposition) increases, so conductive heat flux becomes less. Therefore conduction heat loss becomes lower, 3 S. Pattanayak and S.K. Sahoo CIRP Journal of Manufacturing Science and Technology 33 (2021) 398\u2013442 which enhances the melt pool size, metal over-flow, and nonuniformity in bead geometry [74,102,109,126]. So to overcome these issues, an additional air jet is directed towards the deposited layers (Fig. 51) to promote the convective heat transfer mode. So the previously deposited layers become suitably solidified before depositing a new layer to produce better form quality. In Ref. [102], the air was supplied at 0.6 MPa through a 3 mm diameter nozzle, which makes an angle 45 with the GMAW torch. Here finite element modelling (FEM) was developed to calculate the interlayer cooling time for air jet impingement cooling and no cooling condition. It was reported that a constant interlayer temperature (Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000369_s12555-009-0311-8-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000369_s12555-009-0311-8-Figure2-1.png",
        "caption": "Fig. 2. A quadrotor helicopter configuration with rollpitch-yaw Euler angles [ ].\u03c6 \u03b8 \u03c8, ,",
        "texts": [
            " In Section 3, a feedback linearization controller is described. Section 4 presents an adaptive sliding mode controller and error analysis for the controller. Simulation results are given in Section 5, and Section 6 contains concluding remarks. 2. QUADROTOR HELICOPTER MODEL Quadrotor helicopters we consider have four fixedpitch-angle blades whereas classic helicopters have variable-pitch-angle blades. The control of a quadrotor helicopter is performed by varying the speed of each rotor. A concept of the quadrotor helicopter is shown in Fig. 2. Each rotor produces a lift force and moment. The two pairs of rotors, i.e., rotors (1,3) and rotors (2,4), rotate in opposite directions so as to cancel the moment produced by the other pair. To make a roll angle (\u03c6 ), along the x -axis of the body frame, one can increase the angular velocity of rotor (2) and decrease the angular velocity of rotor (4) while keeping the whole thrust constant. Likewise, the angular velocity of rotor (3) is increased and the angular velocity of rotor (1) is decreased to produce a pitch angle (\u03b8 ), along the y -axis of the body frame",
            " In order to perform yawing motion (\u03c8 ), along the z -axis of the body frame, the speed of rotors (1,3) are increased and the speed of rotors (2,4) are decreased. The quadrotor helicopter is assumed to be symmetric with respect to the x and y axes so that the center of gravity is located at the center of the quadrotor. Each rotor is located at the end of bars, whose length from the center to rotor is l. The rotors generate thrust force Feedback Linearization vs. Adaptive Sliding Mode Control for a Quadrotor Helicopter 421 ( 1 2 3 4)= , , , i F i which are perpendicular to x - y plane as shown in Fig. 2. i J is the moment of inertia with respect to each axis and \u03c1 is the force-to-moment scaling factor. Then the equations of motion of the quadrotor without consideration of air drag can be presented as below. 4 3 3 1 1 ( ) ( ( ) ) , =     = + \u2212     \u2211 i r i x y F Re g z g e m z (1) 2 4 1 1 3 2 1 2 3 4 3 ( ) , ( ) , ( ) . l F F J l F F J F F F F J \u03c6 \u03b8 \u03c8 \u03c1 = \u2212 / = \u2212 + / = \u2212 + \u2212 / (2) [ ], ,x y z represent the position of the quadrotor in the inertial frame, and the attitude state variables in the body frame [ ]\u03c6 \u03b8 \u03c8, , represent roll, pitch and yaw angles, respectively"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003701_j.snb.2004.04.084-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003701_j.snb.2004.04.084-Figure1-1.png",
        "caption": "Fig. 1. Glutamate biosensor\u2013counter electrode system fitting a single cell culture well.",
        "texts": [
            " Several reviews have recently been published covering the field of enzyme-based amperometric sensors with biomedical applications [47\u201349]. Our group has recently started to develop biosensor-based platforms for monitoring glutamate in neuron cultures. One of the approaches, carried out in collaboration with Institute of Microtechnology, Neuchatel\u2013Switzerland, consists in fabricating electrode assemblies on the bottom of cell culture plate with up to 24 wells (see single well equipped with one glutamate sensor in Fig. 1). A sensing chemistry based on glutamate oxidase, a redox polymer bearing Os complexes, and a cross-linker (described elsewhere [50]) was used to modify the working electrodes in order to selectively detect glutamate. When neurons (cultured in commercial culture plate inserts) are placed in the well, glutamate release and uptake cycles could be monitored (see Fig. 2). Based on pre-calibration experiments, the current peak obtained following chemical stimulation corresponds to a glutamate concentration of 3 M"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002794_j.rcim.2021.102165-Figure14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002794_j.rcim.2021.102165-Figure14-1.png",
        "caption": "Fig. 14. The deviation between the actual trajectory and the theoretical trajectory.",
        "texts": [
            " The improvement of the robot\u2019s absolute positioning accuracy can make the contour of the blade after machining within the tolerance range, as shown by the red dashed line in Fig. 13, but the contour shape may have a large deviation from the theoretical contour. Dynamic calibration and compensation methods can further improve the path accuracy so that the contour shape of the blade after grinding and the theoretical profile approximately coincide. Poor absolute positioning accuracy will cause the phenomenon of grinding misplacement. Fig. 14 shows an example of misplacement of grinding when the blade edge is machined by a robot. The area enclosed by black boundary lines is the area to be machined. Blue line is the theoretical trajectory generated through offline programming, and the red line is the actual grinding trajectory of the robot, calculated with pre-set parameter errors. It can be seen from the figure that there is an average deviation about 1.4547 mm between the actual trajectory and the theoretical trajectory, causing that a certain of the non-machining area is processed, while part of the area should be machined remains"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001017_robot.1986.1087552-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001017_robot.1986.1087552-Figure7-1.png",
        "caption": "Figure 7. New parameters for a l i nk wi th more than t w o j o i n t s",
        "texts": [
            " j o i n t (i) connects the l ink ( a ( i ) ) and l i nk ( i) , where a ( i ) is the number of the l i n k a n t e c e d e n t t o l i n k (i) when coming from the base frame (i) is de f ined f ixed wi th r e spec t t o l i n k (i) , and Z is the a x i s o f j o i n t (i) . -i I n t h e case of a l i nk wi th two j o i n t s t h e frame a t the e n d o f t h i s l i n k , s a y j , w i l l b e d e f i n e d w i t h r e s p e c t t o the frame a t the begin ing of the l i n k , say i = a ( j ) , e x a c t l y a s i n t h e case of open-loop robot descr ibed previous ly , i.e by the a id o f 4 parameters ( a . d O.,r ). X . is t h e common perpendicu- l a r on Z and, Z . I n the case o f l i nks w i t h more than two j o i n t s (Fig. 7) , w e def ine the d i f fe ren t f rames as fol lows - f i n d t h e common p e r p e n d i c u l a r s t o Z_. and each of the succeeding axis ,on the same l i n k ti), Z j where i = a ( j ) ( j = k. ,R,m ,... ). - l e t cne of these common perpendiculars be the Zi a x i s , it is p r e f e r e d t o t a k e X . as that corresponding t o t h e common perpendicular of the joint on which is a r t i cu la t ed the longes t b ranch , say k , J ' j ' 1. j T. i-j -1 1177 - t he o the r pe rpend icu la r s w i l l be denoted by X : X!', ... Thus some other auxi l ia r f rames R ! ( 0 ! X ' Y ! Z , ) w i l l be de f ined f ixed wi th r e spec t t o l i nk (i) . -1 The mat r ix T w i l l be def ined by the use of 4 parameters ( %,%,rk,!k) as i n Eq. ( 7 ) . The other succeeding frames R . ( 3 = k,m, -. w i l l be def ined in genera l by the'following parameters (Fig. 7 ) : -I' f -1 J'...l t i k y . angle between X . and X! about Z , 7 -1 --I -1 E , d i s t a n c e between 0 , and 01 7 a . angle between Z , and Z . about X! 7 -1 -3 d , d i s t a n c e between O ! and 2 , 7 8, angle between X! and. X . about 2 , -3 7 -1 -3 -3 r . distance between 0 and X! 3 I n t h i s c a s e T . w i l l be def ined by : i -1 i j T . = T. , T . 7 1 3 i i ' I where IT.$ = Rot(Z,y,) Trans(Z,E , I 3 3 cosy , - s h y . # (d 3 1 = s i n y . cosy , # # 3 7 # # I E . # er 6 1 - - - - - - - - - 3 and T , is i d e n t i c a l t o T "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-FigureD.10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-FigureD.10-1.png",
        "caption": "FIGURE D.10 Relevant geometry for the double-line integrals.",
        "texts": [
            "41) where r and r\u2032 denote the source and observation points on 1 and 2, respectively. By using the Gauss and Stokes theorems, the double surface integral of (D.40) is transformed into the sum of 16 line integrals Pp12 = \u2212 1 4\ud835\udf0b\ud835\udf16  \u2032 4\u2211 i=1 4\u2211 j=1 \u222bl \u2032 i \u222blj R u\u0302ju\u0302 \u2032 i dlj dl \u2032 i = \u2212 1 4\ud835\udf0b\ud835\udf16  \u2032 4\u2211 i=1 4\u2211 j=1 Pij, (D.42) where u\u0302 and u\u0302\u2032 vectors are orthogonal to the circumference of the quadrilateral surfaces as shown in Fig. D.9. The results for the evaluation of the partial potential coefficients is given in Section 7.3.1. The integrals are performed over the lines shown in Fig. D.10. This section illustrates how analytic integration techniques can be helpful in the evaluation of some of the nonorthogonal partial coefficients of potential. It is surprising that the integrals for the nonsingular kernel R are also very complex. These integrals with the kernel R also have applications for Taylor expansion of a retardation term. This is considered in Section 5.8 for nonorthogonal filaments. Integrals with an R-kernel will result, which are considered in Section 7.3.1 such as the two filaments shown in Fig. D.10. Result for the integrals is given in Section 7.3.2. REFERENCES 421 packages. IEEE Journal Solid-State Circuits, 4(4):289\u2013290, August 1973. 2. A. Muesing and J. W. Kolar. Efficient partial element calculation and the extension to cylindrical elements for the PEEC method. In 2008 11th Workshop on Control and Modeling for Power Electronics, COMPEL, Zurich, August 2008. 3. A. E. Ruehli and P. A. Brennan. Efficient capacitance calculations for three-dimensional multiconductor systems. IEEE Transactions on Microwave Theory and Techniques, 21(2):76\u201382, February 1973"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002408_b978-0-08-100433-3.00002-6-Figure2.5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002408_b978-0-08-100433-3.00002-6-Figure2.5-1.png",
        "caption": "Figure 2.5 Schematic illustration of columnar g grains formed by unidirectional solidification, g00 precipitate arrays and the melt pool/layer structure in Inconel 718 processed by selective laser melting [69].",
        "texts": [
            " The columnar structures are the elongated grains that grow across the layers in the direction roughly parallel to the build direction because of the epitaxial grain growth at the partially remelting grains on the previously solidified layers and directional solidification [68]. The dominant {100} texture in the direction of grain growth is developed in SLMprocessed Inconel alloy [61] as the result of the preferential {100} growth direction of cubic crystals [60], which results in the columnar structure shown in Fig. 2.5. In the SLM-processed Ti6Al4V, the average length of prior b grains is in the order of millimetres. This shows that prior b grains grow through tens of layers along a direction tilted at an angle about 20 degree from the build direction (z-axis) [5,60,70]. The inclined angle is attributed to the rotation of the scan direction in each layer. The acicular martensite a0 laths, characterized by high vanadium content and a high dislocation density, are primarily orientated at inclinations of about 45 and 90 degree to the build direction [60,70]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure3.4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure3.4-1.png",
        "caption": "Fig. 3.4. Powder salient pole stators for small single-sided AFPM motors. Courtesy of Mii Technologies, LLC , Lebanon, NH, U.S.A.",
        "texts": [
            " The magnitude of these stresses depends upon the difference in the temperature of the winding and core, difference in coefficients of thermal expansion of both materials and slot fill factor. This problem is more important in powder cores than in laminated cores since the tensile stress of powder cores is at least 25 times lower and their modulus of elasticity is less than 100 GPa (versus 200 GPa for steel laminations). Slotted stators for small disc-type motors fabricated from SMC powders are shown in Fig. 3.3 [160, 265]. SMC powder salient-pole stators for small single-sided AFPM motors manufactured by Mii Technologies, LLC , Lebanon, NH, U.S.A. are shown in Fig. 3.4. The three-phase stator has 9 poles. A single SMC powder salient pole manufactured by Ho\u0308gana\u0308s is shown in Fig. 3.5. 88 3 Materials and Fabrication Magnetic circuits of rotors consist of PMs and mild steel backing rings or discs. Since the air gap is somewhat larger that that in similar RFPM counterparts, high energy density PMs should be used. Normally, surface magnets are glued to smooth backing rings or rings with cavities of the same shape as magnets without any additional mechanical protection against normal attractive forces"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure6.43-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure6.43-1.png",
        "caption": "FIGURE 6.43. A kinematic model for a universal joint.",
        "texts": [
            " Although the driver and driven shafts make a complete revolution at the same time, the velocity ratio is not constant throughout the revolution. A separate illustration of the input, output, and the cross links are shown in Figure 6.42. The angular velocity of the cross-link may be shown by 1\u03c93 = 1\u03c92 + 1 2\u03c93 = 1\u03c94 + 1 4\u03c93 (6.289) 366 6. Applied Mechanisms where, 1\u03c92 is the angular velocity of the driver yoke about the x2-axis and 12\u03c93 is the angular velocity of the cross-link about the axis AB relative to the drive yoke expressed in the ground coordinate frame. Figure 6.43 shows that the unit vectors j\u03022 and j\u03023 are along the arms of the cross link, and the unit vectors \u0131\u03022 and \u0131\u03024 are along the shafts. Having the angular velocity vectors, 1\u03c92 = \u23a1\u23a3 \u03c921 \u0131\u03021 j\u03021 k\u03021 \u23a4\u23a6 = \u23a1\u23a3 \u03c921 0 0 \u23a4\u23a6 (6.290) 1\u03c94 = \u23a1\u23a3 \u03c941 \u0131\u03024 j\u03024 k\u03024 \u23a4\u23a6 = \u23a1\u23a3 \u03c941 0 0 \u23a4\u23a6 (6.291) 2\u03c93 = \u23a1\u23a3 \u0131\u03022 \u03c932 j\u03022 k\u03022 \u23a4\u23a6 (6.292) 3 2\u03c93 = \u23a1\u23a3 \u03c932 \u0131\u03023 j\u03023 k\u03023 \u23a4\u23a6 (6.293) 4\u03c93 = \u23a1\u23a3 \u0131\u03024 \u03c934 j\u03024 k\u03024 \u23a4\u23a6 (6.294) 3 4\u03c93 = \u23a1\u23a3 \u0131\u03023 \u03c934 j\u03023 k\u03023 \u23a4\u23a6 (6.295) 6. Applied Mechanisms 367 we can simplify Equation (6.289) to \u03c932 \u0131\u03023 + \u03c921 \u0131\u03022 = \u03c941 \u0131\u03024 + \u03c934 j\u03023"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure3.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure3.1-1.png",
        "caption": "Fig. 3.1. Stator core segment formed from lamination strip: 1 \u2014 lamination strip, 2 \u2014 groove, 3 \u2014 folding, 4 \u2014 compressed segment, 5 \u2014 finished segment.",
        "texts": [
            " In addition, this manufacturing process allows for making skewed slots to minimize the cogging torque and the effect of slot harmonics. Each stator core has skewed slots in opposite directions. It is recommended that a wave stator winding should be made to obtain shorter end connections and more space for the shaft. An odd 86 3 Materials and Fabrication number of slots, e.g. 25 instead of 24 can help to reduce the cogging torque (VU\u0301ES Brno). Another technique is to form the stator core using trapezoidal segments [257]. Each segment corresponds to one slot pitch (Fig. 3.1). The lamination strip of constant width is folded at distances proportional to the radius. To make folding easy, the strip has transverse grooves on opposite sides of the 3.1 Stator Cores 87 alternative steps. The zigzag laminated segment is finally compressed and fixed using a tape or thermosetting, as shown in Fig. 3.1 [257]. Fabrication of Soft Magnetic Powder Stator Cores The laminated cores of axial flux machines are much more difficult to fabricate than those of radial flux machines. SMC powders simplify the manufacturing process of stator cores with complicated shapes, in general, 3D cores. Mass production of AFPM machines is much more cost effective if soft magnetic powder composites are used as materials for stator cores. Using SMC powders the stator core of an AFPM machine can be made as a slotted core, slotless cylindrical core and salient pole core with one coil per pole"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure4.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure4.2-1.png",
        "caption": "Fig. 4.2. Longitudinal section of the double-sided AFPM brushless servo motor shown in Fig. 4.1. Courtesy of Mavilor , Barcelona, Spain.",
        "texts": [],
        "surrounding_texts": [
            "Iron-cored AFPM machines are distinguished in two ways from coreless (aircored) AFPM machines, namely: (i) iron-cored machines have core losses while coreless machines do not and (ii) the per unit values of the synchronous reactances of iron-cored machines are much higher than those of coreless machines. The core losses are functions of amongst other things the frequency and density of the magnetic flux in the ferromagnetic core \u2013 see eqn (2.55). The frequency of the flux variation in the core is in turn determined by the speed 126 4 AFPM Machines With Iron Cores and number of pole pairs. As AFPM brushless machines tend to have a high number of poles (minimum 2p = 6), the speeds of iron-cored AFPM machines are limited to keep the frequency, and hence the core losses of the machine, within limits; unless the machines are designed to have low flux densities in ferromagnetic cores. Flux frequencies are normally kept below, say, 100 Hz for laminated ferromagnetic cores. If higher frequencies are required, then silicon steel laminations with a thickness of less than 0.2 mm, amorphous alloy ribbons or SMC powders must be used. The higher synchronous inductance of iron-cored AFPM machines affects negatively the voltage regulation of the machine in generator mode, which might be considered as a disadvantage. However, for solid-state converter-fed applications the higher synchronous inductance of iron-cored AFPM motors is an important advantage as it helps reduce the current-ripple due to converter switching. Thus the relatively low inductance of coreless AFPM machines can be a significant disadvantage in switched solid state converter applications."
        ]
    },
    {
        "image_filename": "designv10_0_0001031_j.jsv.2008.03.038-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001031_j.jsv.2008.03.038-Figure5-1.png",
        "caption": "Fig. 5. The cracked tooth model for case 3.",
        "texts": [
            " We will consider the crack path to be a straight line as shown in the right side of Fig. 2. The crack starts at the root of the pinion and then proceeds as shown in Fig. 2. Further referring to Figs. 3\u20136, the intersection angle, u, between the crack and the central line of the tooth is set at a constant 45 . The crack length, q1, grows from zero with an increment size of Dq1 \u00bc 0:1 mm until the crack reaches the tooth\u2019s central line. At that point, q1, reaches its maximum value of 3.9mm. After that, the crack then changes direction to q2 (see Fig. 5), which is assumed to be exactly symmetric around the tooth\u2019s central line. Theoretically, the maximum length of q2 should be the ARTICLE IN PRESS S. Wu et al. / Journal of Sound and Vibration 317 (2008) 608\u2013624 611 same as q1, however, the tooth is expected to suffer sudden breakage before the crack runs through the whole tooth. Thus the maximum length of q2 is assumed to be 60% of q1max and the increment size, Dq2, is also 0.1mm. In later reference of the crack growth, we will use a relative length"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure9.35-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure9.35-1.png",
        "caption": "Fig. 9.35. Longitudinal section of VentrAssistTM hydronamically levitated centrifugal blood pump: 1 - impeller, 2 - PM, 3 - body coil, 4 - body yoke, 5 - cover coil, 6 - cover yoke, 7 - inlet, 8 - outlet.",
        "texts": [],
        "surrounding_texts": [
            "When electricity was new, people had high hopes that it had curative powers. For example, electropathy (electrodes between patient\u2019s hands and ailing body part), very popular from 1850 to 1900, promised to cure most diseases and conditions, including mental illness. The 21st century biomedical engineering community has resurrected magnetic fields to treat depression, e.g. magnetic seizure therapy (high frequency, powerful electromagnets) or transcranial magnetic stimulation (strong pulse magnetic fields). At present, many medical devices use small permanent magnet (PM) electric motors as, for example, high-quality pumps, centrifuges, infusion pumps, hemodialysis machines, precision handpieces and implantable devices (ventricular assist devices, pacemakers, defibrillators, nerve stimulators, etc). Since the reliability of medical products is critical, the electric motor is considered a precision component rather than a commodity device [203]. This section focuses on very small PM brushless motors for implantable devices, in particular, motors for rotary blood pumps [102]. A left ventricular assist device (LVAD) is an electromechanical pump implanted inside the body and intended to assist a weak heart that cannot 314 9 Applications efficiently pump blood on its own. It is used by end-stage heart failure patients who are unable to receive a heart transplant due to donor availability, eligibility, or other factors. Motor driven pumps implanted in the human body must be free of shaft seals. This problem can be solved by embedding PMs in the pump rotor placed in a special enclosure and driven directly by the stator magnetic field. In this case the nonmagnetic air gap is large and high energy PMs are required. Electromagnetic pumps for LVADs can be classified into three categories: \u2022 1st generation (1G), i.e. electromagnetic pulse pumps; \u2022 2nd generation (2G), i.e. electromagnetic rotary pumps; \u2022 3rd generation (3G), i.e. electromagnetic pumps with magnetic or hydro- dynamic bearings. Electromagnetic 1G pumps were driven by electromagnets, linear oscillating motors or linear short-stroke actuators. The pump, integrated with a linear actuator, is heavy, large and noisy. DuraHeartR 3G LVAD developed by Terumo Corporation, Tokyo, Japan, combines centrifugal pump with magnetic levitation technologies (Figs 9.33 and 9.34) [102]. Magnetic levitation allows the impeller to be suspended within the blood chamber by electromagnets and position sensors. The three-phase, 8-pole, axial flux PM brushless motor with slotless stator resembles a floppy disc drive spindle motor (Fig. 9.33). NdFeB PMs are integrated with the impeller. The output power of the motor is 4.5 W, speed 2000 rpm and torque 0.0215 Nm [226]. 9.11 Ventricular Assist Devices 315 An axial flux slotless motor integrated with a centrifugal blood pump is shown in Figs. 9.35 and 9.36. The so-called VentrAssistTM manufactured now by Australian company Ventracor is a new cardiac LVAD, which has only one moving part - a hydrodynamically suspended impeller integrated with a PM rotor. The hydrodynamic forces act on tapered edges of the four blades. The stator of the brushless electric motor is of slotless type and has only upper and lower coils. The three coil winding and four pole rotor use the second harmonic of the magnetic field wave to produce the torque. To provide redundancy, body coils and cover coils are connected in parallel so that the motor still can run even if one coil is damaged. The housing and impeller shell are made of titanium alloy Ti-6Al-4V. Vacodym 510 HR NdFeB PMs are embedded into impeller. To reduce the reluctance for the magnetic flux, laminated silicon steel return paths (yokes) are designed. The 2D FEM simulation of the magnetic field distribution is shown in Figs 9.37 and 9.38 [102]. The measured performance characteristics are shown in Figs 9.39a and 9.39b [226]. For output power between 3 and 7 W and speed between 2000 and 2500 rpm, the efficiency is from 45 to 48% (Fig. 9.39a). At 3 W and 2250 rpm the winding losses are 1.7 W, eddy current loses in titanium 316 9 Applications 1.0 W and losses in laminated yokes are 0.7 W [226]. At load torque 0.03 Nm the fundamental phase current is 0.72 A (Fig. 9.39b). The device weighs 298 g and measures 60 mm in diameter, making it suitable for both children and adults. 9.12 Axial Flux Machines with Superconducting Field Excitation System 317"
        ]
    },
    {
        "image_filename": "designv10_0_0000029_j.actamat.2010.02.004-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000029_j.actamat.2010.02.004-Figure2-1.png",
        "caption": "Fig. 2. Three different scanning strategies, namely identical scanned layers using zigzag (left) or unidirectional scan vectors (centre) and the cross-hatching technique using the zigzag scan vectors (right). The different views that will be considered are indicated in the picture on the left.",
        "texts": [
            " A layer thickness of 30 lm was chosen as a compromise between a practical building time and a good surface roughness, caused by the staircase effect. The previously optimized, but not yet published, parameters for processing of Ti\u20136Al\u20134V alloy in continuous laser mode are a laser power of 42 W, a scanning velocity v of 200 mm s 1 and a hatch spacing h (i.e. the distance between two neighbouring scan vectors) of 75 lm. In this research, the influence of the scanning velocity v, hatching space h and scanning strategy (see Fig. 2) on the properties of bulk material produced by SLM is investigated. Rectangular samples with a width, length and height of 5, 10 and 5 mm, respectively, were produced. The three types of scanning strategy that were used are illustrated in Fig. 2. Two types of scan vectors were used, namely the alternating and the unidirectional scan vector. The scanning strategy for the successive layers can be chosen to be identical, or one can rotate the scanning direction, e.g. through 90 for the cross-hatching strategy. Table 1 gives an overview of the parameters used for the various parts. Sample A was scanned with the optimal parameters (see previous paragraph) and will therefore be named the reference sample hereafter. Most samples were built using the alternating scan vector without a rotation of the scanning direction between different layers, except samples D and H",
            " sample C, the angle is 90 and the pores are aligned vertically. The Vickers hardness values for samples B and C are 426 \u00b1 34 and 381 \u00b1 29 Hv, respectively (see Table 2). These values show a decrease in hardness when the hatch spacing is increased. In the following experiments, rather than varying energy input, different scanning strategies were applied to determine the correlation between the growth direction of the elongated grains and the scanning direction. The different scanning strategies are illustrated in Fig. 2. For sample D, unidirectional scan vectors are applied and the scanning pattern is identical for all layers. In the top view (Fig. 7a), the herringbone pattern is lost due to ith a unidirectional scan vector: (a) top view; (b) side view; (c) front view; , scanned with the use of alternating hatch directions: (e) top view; (f) side the unidirectional scan vectors. In the side view (Fig. 7b), the elongated grains are parallel to each other and are tilted 19 away from the building direction. The layers are scanned from right to left, and from this experiment we can therefore conclude that the grains grow towards the melt pool"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000255_j.compositesb.2019.107496-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000255_j.compositesb.2019.107496-Figure8-1.png",
        "caption": "Fig. 8. (A) Cattail structures [60]; (B) Cross-beam and bionic box bumper model [28]; (C) Bionic cross-beam and box bumper model [28]",
        "texts": [
            " [33] simplified the bionic element to a thin wall and experimentally studied such elements in on three new BMTs mimicking the cross-section and nodes of bamboo: bionic tubes of variable thickness, bionic tubes with rib, and bionic tubes with a bamboo cross-section, as shown in Fig. 6D. Using different designs of the vascular structures, a bionic honeycomb tubular nested structure inspired by bamboo was manufactured by a wire electrical discharge machining technique, as shown in Fig. 6E [56]. All results demonstrated that biomimetic methods could improve energy absorption performance. Cattail plants are found in streams, lakes, marshes, rivers, and other shallow water areas, as shown in Fig. 8A. The stems and leaves have a distinct twisted chiral morphology to adapt to their environment. Zhao et al. [60] found that cattail leaves had evolved multi-cell structures and superior mechanical properties against wind loading, as shown in Fig. 8A. Inspired by cattail structures, Xu et al. [28] proposed a bionic bumper with a cross-beam mimicking the internal rib structure of cattail (Fig. 8C, (b)) and a crash box imitating bamboo structures (Fig. 8B (c)). Simulation results indicated that the bionic design enhanced the SEA of the bumper. Moreover, the bionic cross-beam and bionic box of the bionic bumper had a significant effect on the crashworthiness of the structure. The crush deformation and total weight of the bionic cross-beam and box were reduced by 33.33% and 44.44%, respectively. 2.2.4 Palm The trabeculae structure in the beetle forewing is also an excellent structure for inspiration, as shown in Fig. 10. By observing the internal structures of the ladybeetle and Japanese beetle forewings, Xiang and Du [63] proposed two new BMTs"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002387_tcyb.2017.2681683-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002387_tcyb.2017.2681683-Figure2-1.png",
        "caption": "Fig. 2. Two inverted pendulums on cars with stochastic disturbance and nonlinear function faults.",
        "texts": [
            " Note that the results in [27] and [45] have been used to control the two inverted pendulums systems (76) and (77), while it was under the fault-free condition, and it also did not consider the stochastic disturbance term. In fact, stochastic disturbance and nonlinear function faults often exist in the practical systems, and make the systems unstable. Therefore, in this paper, we will apply the proposed method to control a two inverted pendulums system connected by a moving spring mounted on two cars with stochastic disturbance and nonlinear function faults (see Fig. 2), The dynamics equations are described as S1 : \u23a7 \u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23a8 \u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23a9 dx1,1 = x1,2dt dx1,2 = [[ g cl \u2212 ka(t)(a(t)\u2212cl) cml2 ] x1,1 + u1 cml2 + ka(t)(a(t)\u2212cl) cml2 x2,1 + \u03b2\u03041x2 1,2 + k(a(t)\u2212cl) cml2 (Y1 \u2212 Y2)+ \u03b21(t \u2212 T0) \u00d7 p1(x1, u1) ] dt + g1,2(x1)dw1 y1 = x1,1 (78) S2 : \u23a7 \u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23a8 \u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23a9 dx2,1 = x2,2dt dx2,2 = [[ g cl \u2212 ka(t)(a(t)\u2212cl) cml2 ] x2,1 + u1 cml2 + ka(t)(a(t)\u2212cl) cml2 x1,1 + \u03b2\u03042x2 2,2 + k(a(t)\u2212cl) cml2 (Y2 \u2212 Y1)+ \u03b22(t \u2212 T0) \u00d7 p2(x2, u2) ] dt + g2,2(x2)dw2 y2 = x2,1 (79) where g1,2(x1) = cos(x1,2) and g2,2(x2) = sin(x2,2)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000275_s11661-010-0397-x-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000275_s11661-010-0397-x-Figure1-1.png",
        "caption": "Fig. 1\u2014Key components of the Arcam S12: electron gun column (10-7 mbar) (A), build chamber (2 9 10-3 mbar) (B), powder hoppers (C), rake arm (D), component (E), SS base plate (F), build tank (G), z-axis assembly (H), and coordinate system. CAD data courtesy of Arcam AB.",
        "texts": [
            "[2] discussed the use of EBM for medical applications and described the Ti-6Al-4V microstructure as having a uniform acicular a-phase microstructure (with b along the phase boundaries). In this work, a combination of optical microscopy, scanning electron microscopy (SEM), and electron backscatter diffraction (EBSD) have been used to determine microstructural evolution and the effect of build temperature upon mechanical properties. A numerical model has been used to determine its suitability to describe EBM and predict solidification conditions. All components produced for evaluation were built using an Arcam S12 EBM machine (Figure 1) at the University of Sheffield. Prealloyed Ti-6Al-4V powder, produced by the plasma rotating electrode process, and supplied by Arcam AB was used; the powder size distribution was quoted as 45 to 100 lm. The nominal composition (in wt pct) of as-supplied, virgin powder was 6.1Al, 3.95V, 0.05Fe, 0.17O, 0.008N, 0.011C, and 0.0014H. Recycling of nonmelted powder is standard practice and is achieved by using a powder recovery system (PRS). All powder used during the production of an EBM component is passed through the PRS system in which Ti-6Al-4V powder is used as a blasting medium, to break up sintered blocks of powder recovered from previous processing"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003459_0278364905058363-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003459_0278364905058363-Figure1-1.png",
        "caption": "Fig. 1. Zero moment point. The ZMP is where the ground reaction force acts whereas the CM point is where inertia and the force of gravity originate.",
        "texts": [
            " (5) can be used to reconstruct the ZMP trajectory. Alternatively, at a particular instant in time, eq. (5) can be employed as a constraint equation for deciding joint accelerations consistent with a desired ZMP position, as discussed by Kondak and Hommel (2003). Finally, the ZMP as a function of the CM position, net CM force ( F = M aCM), and net moment about the CM can be expressed as xZMP = xCM \u2212 Fx Fz + Mg zCM \u2212 \u03c4y ( rCM) Fz + Mg (6) and yZMP = yCM \u2212 Fy Fz + Mg zCM + \u03c4x ( rCM) Fz + Mg . As emphasized in Figure 1, the most important notion of the ZMP quantity, applicable for both single and multileg ground support phases, is that it resolves the ground reaction force distribution to a single point. However, one needs to be careful to use this point in an appropriate manner. Most notably, both the vertical component of moment and the CM work performed by the ground reaction force cannot be computed solely on the bases of the ZMP trajectory and the resulting ground reaction force vector. For example, the resultant horizontal ground reaction force could be zero while the vertical component of moment and/or the work performed by the ground reaction force could be non-zero"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure3.42-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure3.42-1.png",
        "caption": "Fig. 3.42: A statistical interpretation of accuracy and repeatability errors.",
        "texts": [
            " The accuracy error can be interpreted as a measure of the difference between a desired position and the mean position of the statistical distribution of the measures. The repeatability error can be related to the dispersion size of the repeated measures and it can be represented with a sphere of radius erS inside which the measures are contained. Fundamentals of Mechanics of Robotic Manipulation 179 The smaller the sphere, the better repeatability the manipulator shows. This corresponds to having a thicker distribution curve in Fig. 3.42 so that there is a high probability that with one measure only one can obtain a measure within the repeatability error bands. Dynamic characteristics are related to motion capability of a robot ensuring repeatability and accuracy performances after a manipulator motion. Thus, dynamic characteristics are generally described by the values of: - maximum velocities; - maximum accelerations; - cycle times; - payload data. Maximum velocities and accelerations are indicated with respect to a single joint or a motion axis"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure9.16-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure9.16-1.png",
        "caption": "Fig. 9.16. Large power double disc AFPM brushless motor. 1 \u2014 PMs, 2 \u2014 stator assembly, 3 \u2014 housing, 4 \u2014 shock snubber, 5 \u2014 shock mount, 6 \u2014 rotor shaft, 7 \u2014 rotor disc clamp, 8 \u2014 shaft seal assembly, 9 \u2014 bearing retainer, 10 \u2014 stator segment, 11 \u2014 centre frame housing, 12 \u2014 spacer housing, 13 \u2014 rotor disc, 14 \u2014 bearing assembly, 15 \u2014 rotor seal runner, 16 \u2014 rotor seal assembly. Courtesy of Kaman Aerospace, EDC, Hudson, MA, U.S.A.",
        "texts": [
            " Stators of large AFPM brushless motors with disc type rotors usually have three basic parts [48]: \u2022 aluminum cold plate; \u2022 bolted ferromagnetic core; \u2022 polyphase winding. 296 9 Applications The cold plate is a part of the frame and transfers heat from the stator to the heat exchange surface. The slots are machined into a laminated core wound of steel ribbon in a continuous spiral in the circumferential direction. The copper winding, frequently a Litz wire, is placed in slots and then impregnated with a potting compound. The construction of a double disc AFPM motor developed by Kaman Aerospace, EDC, Hudson, MA, U.S.A. is shown in Fig. 9.16 [48]. Specifications of large axial flux motors manufactured by Kaman are given in Table 9.4. 9.3 Ship Propulsion 297 An electric propulsion system for submarines requires high output power, high efficiency, very salient and compact motors [64, 195]. Disc-type brushless motors can meet these requirements and run for over 100 000 h without a failure, cooled only by ambient sea water. These motors are virtually silent and operate with minimum vibration level. The output power at rated operating conditions can exceed 2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001999_j.matdes.2017.05.091-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001999_j.matdes.2017.05.091-Figure2-1.png",
        "caption": "Fig. 2. Extraction of the two subvolumes from the bracket.",
        "texts": [
            " Considering the component, two subvolumes having different positions and orientations were extracted from one of the brackets. The part denoted as B1 has a volume of 1297 mm3, while B2 has a volume of 1333 mm3. To obtain results that are directly comparable to those of the specimens, B1 is horizontally oriented, while B2 is almost vertical (inclination of 790). As done for the specimens, CT was performed positioning the axis of symmetry in the vertical direction (z). Their position and shape are depicted in Fig. 2. A series of 6 HCF and 13 LCF tests was performed to define the material resistance to both elastic and plastic conditions. The results of this campaign in terms of number of cycles to failure and material response are presented elsewhere [37,38]. The objective of the present analysis is to understand if a fatigue failure can be foreseen by performing a CT scan before the test. In fact, being able to predict from which defect of the thousands detected the failure will origin, would allow to determine the fatigue limit of the material in the Kitagawa diagram",
            " To avoid all kinds of related issues, the bracket subvolumes were extracted with a material orientation very close to the ones of the specimens, so that the data could directly be compared. Sometimes, down-facing surfaces of non-machined parts showed diffused subsurface porosity [14]. The influence of these skin effects are not the topic of the present analysis and will be discussed in future investigations, therefore only the internal part of the subvolumes were considered. In particular, B1 has a horizontal axis of symmetry, while B2 is almost vertical (see Fig. 2). The results of the CT scan performed are summarized in Table 10. A first consideration is that the main variables are in line with those observed in the specimens (see Section 3): the defect density is 9.6 def/ mm3 for B1 and 22.7 def/mm3 for B2 and the largest defects detected are close to the maxima measured. If the density in B2 seems very high, it is worth noticing that the number of exceedances over u is close to the lowest detected in the samples, with a particular similarity with sample V03U"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure3.39-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure3.39-1.png",
        "caption": "Fig. 3.39: A scheme of an RP manipulator with concentrated masses.",
        "texts": [
            " When k differs from m, Hikm gives an expression for the Coriolis force that is generated by the velocity in motion of joint k and felt at joint i. The entry Gi represents the action of the gravity field on the mass of the link. It is of note that the Lagrangian formulation gives the equations of motion only and does not permit evaluation of other reaction force and torque at the joints. But the Lagrangian formulation also considerably reduces the computational efforts as compared with the Newton\u2013Euler algorithm. Figure 3.39 shows a scheme for a telescopic RP manipulator. The manipulator is represented in the plane of the motion and the Lagrangian coordinates can be chosen as the joint angle \u03b81 and the stroke a1 of the prismatic pair. H\u2013D notation is shown in Fig. 3.39 and the mass distribution has been described by the point masses mR and mL that are the masses for the link associated with revolute and prismatic joints, respectively. In Chapter 3: Fundamentals of the Mechanics of Robots168 particular, mR is the mass of the links including the fixed part of the prismatic joint mass and it is moved by the revolute joint. The mass mL is the mass that is moved by the prismatic pair. The kinetic energy associated with the motion of mR is given by 2 1 2 1aRm 2 1 RT \u03b8= (3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001467_j.mechmachtheory.2010.05.001-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001467_j.mechmachtheory.2010.05.001-Figure3-1.png",
        "caption": "Fig. 3. A PSS leg.",
        "texts": [
            " Since f12 andw12 are ectors, then we get jf12\u00b7v12jmax = jf12\u00b7 w12 \u00d7 a\u00f0 \u00dejmax = a \u00f012\u00de Thus, Eq. (11) can be rewritten as \u03bb12 = jf12\u00b7v12j jf12\u00b7v12jmax = jf12\u00b7v12j a = jcos\u03c6j \u00f013\u00de One may see from Eq. (13) that the input transmission index of the RSS leg is independent of the frame o\u2212xyz and is equal to the absolute value of the cosine of the angle between the velocity of the joint connecting to the input link and the vector along the coupler link. Moreover, this result also fits for some similar legs, such as the RUS (U represents the universal joint) leg and the planar RRR leg. Fig. 3 shows a PSS (P standing for the prismatic joint) leg that contains one actuating P joint and two passive S joints S3 and S4. The transmission wrench of the PSS leg is a pure force along the coupler link. Since the input twist is a translation, the pitch of the input twist screw is infinite.With respect to the coordinate system o\u2212xyz, the unit input twist screw $I can be expressed as where $I = 0;v34\u00f0 \u00de \u00f014\u00de According to Eq. (8), the input transmission index of the PSS leg can thus be obtained as \u03bb34 = jf34\u22c5v34j jf34\u22c5v34jmax = jf34\u22c5v34j = jcos\u03c8j \u00f015\u00de is also free from the coordinate system o\u2212xyz"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000428_j.addma.2019.100877-Figure30-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000428_j.addma.2019.100877-Figure30-1.png",
        "caption": "Fig. 30. (a\u2013c) Finite element predicted residual stress distribution within DLD Inconel 718 build, (d\u2013e) residual stress components obtained from neutron diffraction analysis of SLM Inconel 718 build (reproduced from [32,55]).",
        "texts": [
            " During AM, the pre-deposited layer restrains the free expansion of the newly deposited layer and leads to compressive plastic deformation of the newly deposited layer at a higher temperature. As the new layer cools down, its contraction is restrained by the pre-deposited layer and results in a residual tensile stress within the new layer [35,95]. Various techniques have been employed for measuring the distribution of residual stress in AM Inconel 718 builds, namely X-ray and neutron diffractions [84], layer removal techniques [50] and Vickers indentation [23,35,112,116]. The latter is a simple and fast way and is most often used in the studies [23,35,112,116]. Fig. 30 presents the finite element predicted and experimentally measured residual stress components for Inconel 718 builds made by DLD and SLM. The resulting residual stresses in the AM builds significantly depend on the type and parameters of the AM process. The residual stresses of the EBM parts are much smaller than those for laser-based AM processes [130,133,134]. Although the electron beam has significantly higher power and scanning rate, the employment of high pre-heating temperature in EBM (of the order of 0"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002586_j.addma.2021.102017-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002586_j.addma.2021.102017-Figure1-1.png",
        "caption": "Fig. 1. Four different commercial atomization methods [25]. Schematic illustrations of (a) gas atomization, (b) water atomization (c) plasma atomization with a snapshot of a pre-alloyed wire getting atomized at the apex of plasma torches and (d) the plasma rotating electrode process (PREP) which is a centrifugal atomization process.",
        "texts": [
            " It provides a strong control over the production rate and desired physical characteristics of the resulting powders when compared to other powder production methods such as mechanical or chemical methods. Atomization accounts for a large fraction of manufactured metal powders. In 2015, over 60% of all metal powders produced in North America were produced by atomization [24]. All atomization methods use a source of high energy to melt raw material into a molten stream within an adjusted environment to solidify in a controlled manner [25]. Different methods use different raw materials. Fig. 1 shows four different commercial atomization methods for metal powders used in additive manufacturing. For instance, gas and water atomization use elemental or pre-alloyed feedstock (see Fig. 1a and b). On the other hand, plasma atomization uses a pre-alloyed wire as its raw material (see Fig. 1c). The energy source used to melt the raw material can be an induction coil, a plasma arc or plasma torches. The molten stream will be struck with a high-velocity jet of gas, plasma or water to pulverize it into a cloud of fine particles. Gas Atomization (GA) is a traditional manufacturing process for obtaining close-to-spherical powders, which was first patented in 1872 by Marriott of Huddersfield who used steam as the atomizing gas of the molten material [26\u201328]. Hall [29] filed a patent in 1919, where he constructed a nozzle such that the molten material could be suctioned vertically from its tip by the force of the steam jet",
            " On the other hand, gas atomization is a well-established and mature technology. The process is economical, fast and can produce a broad range of PSDs that can be tailored to different applications. Fig. 2a, b and c show scanning electron microscope (SEM) images of gas-atomized Ti-6Al-4V, nickel alloy 718 and Al-Si10-Mg powders. As seen in these images, the degree of particle sphericity is relatively low, and a high number of satellite particles can be observed. Water atomization technology is similar to the gas atomization process as shown in Fig. 1b. However, a high-pressure water stream is used as the atomizing fluid to produce spattered droplets. Since the heat transfer and momentum is much higher in water atomization (compared to gas atomization where the atomization jet has a much lower mass flow rate and heat capacity), it can increase the shape irregularity and the porosity in the powder. Water atomization benefits from a relatively low production cost and high productivity rate (~ 1 ton/min). It is the most widely used method for the production of ferrous metal powders, but it has also been used to produce nickel alloys or reactive materials like copper and copper alloys [24]",
            " Plasma atomization (PA) was developed in 1996 by Entezarian et al. [40] to produce highly spherical powders of reactive metals with an average particle size of 40 \u00b5m. The method was patented by Pegasus Refractory Materials and Hydro-Que\u0301bec in Canada in 1998 [41]. This method was similar to the PREP method which was developed by Nuclear Metals Inc. in 1980. It enables a one-step operation, combining the P. Moghimian et al. Additive Manufacturing 43 (2021) 102017 melting and atomization steps using a pre-alloyed wire as the raw material, as shown in Fig. 1c. Therefore, the risk of ceramic contamination from the melting of a reactive metal in a ceramic crucible was eliminated, ensuring a high purity in the final powder product [42]. In this method, the molten material is superheated using an extremely hot (up to 11,000 K) argon plasma produced by converging plasma torches. This inhibits the molten metal particles from solidifying rapidly into irregular shapes [25,43]. The molten droplets spend sufficient time in the superheated state, allowing them to adopt an equilibrium shape driven by the surface tension forces: ideal spheres [42,43]",
            " The latter three were historically those benefiting from the high-quality powder produced by atomization before metal additive manufacturing became widely adopted. The benefits of plasma atomized powder in coating, HIP and MIM resulted respectively in an up to 10% more efficiency and a lower product shrinkage. It has been reported that in the MIM process, a good flowability of plasma atomized powder can lead to less required additives for the powder to be processed [43]. In the plasma rotating electrode process (PREP) which is a centrifugal atomization process, shown in Fig. 1d, feedstock material comes in the form of an electrode bar which will be rotated at approximately 15,000 rpm while being melted by a plasma arc. Centrifugal forces eject the molten material from the bar and the particles solidify before they hit the chamber wall. This method was invented as a response to the high consumption volume of gas per unit weight of atomized powder and a generally very low yield of fine powder particles by traditional gas atomization [46]. In the PREP process, this yield is dependent on the centrifugal force thus rotation speed"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure3.16-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure3.16-1.png",
        "caption": "Fig. 3.16. Constraints of D.C. motor",
        "texts": [
            " If we ignore the rotor acceleration effects and friction term, the motor output torque is (3.55) where CM is the torque constant. Combining Eq. (3.54) and (3.55) we obtain If the rotation speed is expressed in terms of r.p.m. (nm= 60 tim) then 2n (3.56) (3.57) Let us introduce the constraint of the maximal input voltage umax \u2022 Then, from Eqs. (3.56) and (3.57) it follows (3.58) Umax CM \u2022 60 UmaX \u2022 hId d where P:= IS the stall torque and n:=-2 .-C IS t e no- oa spee. Rr n E This constraint of maximal input voltage can be represented by a straight line in the pm_nm plane ((1) in Fig. 3.16). We use this constraint in the quadrants I and III of the pm_nm plane. For the quadrants II and IV we introduce the constraint 80 3 Dynamics and Dynamic Analysis of Manipulation Robots of maximal rotor current in order to keep this current smaller than the stall current value: Iii ~ iM = umax/Rr. This constraint is represented by a horizontal line (2) in the pm - nm plane. Finally, we introduce the constraint of maxima in the speed allowed n max ((3) in Fig. 3.16); for instance, nmax=n~. If the viscous friction is not ignored then the no-load speed becomes w~ = Umax I ( CE + R~:e ), where Be is the viscous friction coefficient. This modi fied constraint is represented by dotted line in Fig. 3.16. One example of the constraint is shown in Fig. 3.15. Straight lines represent the constraint P:::ax - nm. It can therefore be concluded that the pm - nm dia grams spread when the working speed increases, i.e. the execution time T decreases. For T= 5s and T=4s the diagrams are wholly within the permissible domain which means that the actuators chosen can produce the manipulator work at this speed. For T=3s the diagram extends beyond the permissible domain i.e. the constraint is violated, so therefore the motor cannot produce the manipulator work at that speed"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure9.20-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure9.20-1.png",
        "caption": "Fig. 9.20. Oil beam pump: 1 - horse head, 2 - walking beam, 3 - counterweights, 4 - Pitmans arm, 5 -cranks, 6 - transmission, 7 - sheaves and belts, 8 - electric motor or other prime mover, 9 - Samson beams, 10 - casing, 11- flow line, 12 - polishing rod.",
        "texts": [
            "4) have been successfully tested to 60g triaxial loading, making them the ideal choice for rough drilling environments; \u2022 AFPM synchronous motors that allow a high level of precision in speed and torque control not available with other motors; \u2022 a modular design that allows drilling to continue at reduced power with one motor; \u2022 direct connection to the rigs a.c. bus and interface with the existing power supply with minimal degradation to the rig power; \u2022 the entire ECI system can be transported in three standard 6-m sea containers. Specifications of the ECI drilling system are given in Table 9.7. An oil beam pump (Fig. 9.20) consists of an overground electromechanical drive and underground reciprocating piston pump. An electric motor drives one end of a heavy beam via pair of cranks with counterweights and transmission. The other end of the beam is equipped with the so called horse head with attached vertical cable. The cable is connected to a polished rod of a pump at the bottom of the well. The cranks move the beam up and down and 9.6 Oil Beam Pumps 303 actuates the pump. Typically, a beam pump delivers 5 to 40 liters of a crude oil-water mixture per each stroke"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure3.5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure3.5-1.png",
        "caption": "Fig. 3.5: An example of modeling an industrial robot: a) the mechanical design; b) the corresponding kinematic functional scheme.",
        "texts": [
            " A kinematic model of a manipulator can be named as \u2018functional\u2019 when its scheme refers to kinematic parameters only, but also permits understanding the motion capability of the manipulator architecture. A kinematic functional model can be determined from the mechanical design of a robot through the following step-by-step procedure: - identification of the type of joints; - identification of the position of each joint axes; - identification of the geometry of the links; - drawing of a scheme for the kinematic chain. Figure 3.5 is an illustrative example for a kinematic functional scheme of a given mechanical design of a robot. It is worthy of note that in such a scheme the link is represented in a way that makes clear the relative motion because of the joints. The kinematic functional model can be completed as a kinematic model by identifying the geometrical sizes and kinematic parameters. These parameters determine the relative position and orientation of a link with respect to a neighborhood one as a function of the Fundamentals of Mechanics of Robotic Manipulation 77 variable coordinates of the joints that are connected through the link"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000961_j.euromechsol.2008.07.007-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000961_j.euromechsol.2008.07.007-Figure4-1.png",
        "caption": "Fig. 4. Schematic graph of a crack.",
        "texts": [
            " For high backup ratios the analysis predicted cracks that would propagate through the teeth (Fig. 3A direction 1) and through the rim for low backup ratios (Fig. 3A direction 2). The initial crack angle \u03b1c accounts also for the propagation. For low \u03b1c , the propagation is through the rim even for high backup ratios (Fig. 3B). Lewicki (2001) noted that crack propagation paths tend to be smooth, continuous, and rather straight with only a slight curvature. The proposed modelling of gear tooth cracks presented in this paper (Fig. 4) assumes a straight line shape for the crack which is introduced in the tooth root circle. This crack affects the whole width W of the tooth and is defined by its inclination angle \u03b1c and depth pc . This modelling induces a progressive reduction of the actual tooth thickness along the width of the tooth as shown in Fig. 4. t1, t2, t3, . . . , etc. define the new thickness of the tooth in the crack zone. The gearmesh stiffness of the gear pair is calculated according to Eq. (8) by taking into account the thickness reduction which will affect the rigidity of the tooth. Table 1 Values of the coefficients of Eq. (4) Ai Bi Ci Di Ei F i L\u2217(h f i , \u03b8 f ) \u22125.574 \u00d7 10\u22125 \u22121.9986 \u00d7 10\u22123 \u22122.3015 \u00d7 10\u22124 4.7702 \u00d7 10\u22123 0.0271 6.8045 M\u2217(h f i , \u03b8 f ) 60.111 \u00d7 10\u22125 28.100 \u00d7 10\u22123 \u221283.431 \u00d7 10\u22124 \u22129.9256 \u00d7 10\u22123 0.1624 0.9086 P\u2217(h f i , \u03b8 f ) \u221250"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003785_j.jsv.2009.09.022-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003785_j.jsv.2009.09.022-Figure3-1.png",
        "caption": "Fig. 3. Schematic illustration of a planetary gear set with four planets and an accelerometer mounted on the stationary ring gear.",
        "texts": [
            " The predicted Fspi(t) and Frpi(t) with or without the other gear errors can be used as input in this earlier formulation [4] to predict the resultant spectra at a fixed point for the rotating carrier case. Referring to Ref. [4] for the details, a brief summary of this formulation will be provided here for completeness purposes. Consider the same planetary gear set with N planets positioned at angles ci (iA[1,N]) and with rotating carrier c and the sun gear s (fixed: ring gear r). For a complete revolution of the carrier, a transducer positioned on the housing shown in Fig. 3 experiences the disturbances from all 2N planet meshes in sequence. As the force transmission path between the meshes of planet pi and the fixed transducer location varies in time as the carrier rotates, the influence of each planet on the transducer was limited in Ref. [4] to a duration of Tc/N, where Tc=2p/oc is the rotational period of the carrier. This gradually increasing and then decaying dominance of the forces of planet pi within a Tc/N time period, was approximated by weighting function in the form of a Hanning window w(t)=0",
            " Two almost symmetric sidebands are evident at both sides of Hm at orders of Hm7Hp/c where Hp/c=Zr/Zp=4.69. Dynamic mesh force spectra for the meshes of planets p2 to p4 are also very similar to those shown in Fig. 4 for the meshes of the first planet. With these gear mesh forces, Fspi(t) and Frpi(t) (iA[1,N]), predicted with the sidebands associated with Ep1, the procedure outlined in Section 2.3 is applied next to predict the acceleration spectra that would be measured on the stationary ring gear as illustrated in Fig. 3. Fig. 5 shows the predicted acceleration spectrum A(o) for this case, given Ss=0 and Sr=1 in Eq. (17). Here, some of the energy of the gear mesh harmonic order of H=Hm=140 is distributed to two sidebands at H=Hm7N=14074 as a direct result of the amplitude modulation caused by the rotating carrier [4]. As the run-out (or eccentricity) vector rotates with the planet gear, it has a different phase angle each time the planet enters its windowed region. This causes the force sidebands at orders Hm7Hp/c of Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure3.4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure3.4-1.png",
        "caption": "Fig. 3.4. Determination of the \"home\" position",
        "texts": [
            " The task of determining the \"home\" position is now reduced to determining KP = (QP, RP, EP)\u00b7 The sets Ri and Ei in the case of a simple kinematic chain have the following form: where ei and ei + 1 stand for vectors eii and ei, i + 1 . On the basis of the assumption that KP-l is known, it follows that ep and -rp-l. i are also known. Further, let us note the set of three orthogonal unit vectors in points Zi-l,i: ep, liP and ep x liP, where liP = - ort(ep x (-rP-l,i x ep)), whose components are known in the reference coordinate system (Fig. 3.4). It should be noticed that ort(\u00b7) denotes the unit vector of ('). -+ -+ -+-+ On the other hand, let us note the set of vectors, ei, iii> and ei x iii, where ai = - ort(~i x (iii x ~J), whose components are known with respect to the local coordinate system Qi' Under the condition qi = 0, these two sets of vectors coincide. Since QP = [liPl liP2 liP3] represents the transformation matrix of the i-th link Ci joint coordinate system into the reference system, it follows that and this completely determines the matrix Qp: Qo [-0 -0 -0 -0] [-:: -:: -:: -:: ]T i = ei ai ei x ai ei ai ei x ai (3",
            " Consequently, for forming all nine equations of qij it is necessary to know the values of the projections of three non-colinear, non-zero vectors in both coordinate systems. In that case the system determinant of linear qij is different from zero, which enables calculation to be made of the transformation matrix Q (see Eqs. (3.1H3.4\u00bb. Let us return to our task of determining the transformation matrices of the robot mechanism. First consider Figs. 3.4 and A.3.t.t. The resting two vectors which are not mutual and related to the unit vectors of the joint axes ei' colinear, are ai and bi = (ei x a-\\.). From Fig. 3.4 is evident that only for the case qi = 0 does the identity ai = ai hold when it is possible to perform the transformation of the i-th segment coordinate system into the fixed system. The vector trihedron -+ -+ -+ -:=. {ei' ai' bJ and {ei, ai' bJ then conduct to coincide in space, so there exist only three non-colinear vectors, the position of which is known in both coordinate systems. The system of Eq. (3.1) can be represented in an expanded form as: q~l ei,l +q~2ei, 2 +q~3ei, 3 =ei,x 0- 0- 0-q12 ei, 1 +q22 ei, 2 + q23 ei, 3 =ei,y qg 1 ei, 1 + qg 2 ei, 2 + qg 3 ei, 3 = ei , z q~l ai,l +q~2ai,2 +q~3ai, 3 =ai, x (A"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003965_978-94-009-1718-7_40-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003965_978-94-009-1718-7_40-Figure1-1.png",
        "caption": "Figure 1. A three-DOF parallel Manipulator",
        "texts": [
            " This manipulator also has a fairly large translational workspace and a closed-form direct kinematics solution. In what follows, we first describe the construction of the mech anism. The sufficient conditions for achieving a pure translational motion of the platform are developed. Then, the inverse and direct kinematics are solved in closed forms. Finally, the workspace and singular conditions of the manipulator are discussed. 2. Mechanism Description A schematic of the parallel manipulator is shown in Fig. 1, where the fixed base is labeled as link 0 and the moving platform is labeled as link 1. Three identical limbs connect the moving platform to the fixed base by universal joints at points Bi and A, i = 1,2, and 3, respectively. Each limb consists of an upper member and a lower member (links 2 and 5 for the first limb, 3 and 6 for the second limb, and 4 and 7 for the third limb) that are connected together by a prismatic joint, Pi. The base connected axes of the universal joints, Ai, i = 1,2, and 3, are co-planar",
            " The three limbs are preferably, but not necessarily, separated by 120 degrees at the points of connection with the moving and base platforms. Ball screws or hydraulic jacks can be used to vary the lengths of the prismatic joints and, therefore, to control the position of the moving platform. Let F denote the degrees of freedom of a mechanism, n the number of links, j the number of joints, Ii the degrees of freedom associated with the ith joint, and>' = 6 the motion parameter. Then, the mobility equation can be written as: F = >'(n - j -1) + L k (1) For the manipulator shown in Fig. 1, we have n = 8, j = 9, and Ii = 2 for the universal joints and 1 for the prismatic joints. Applying Eq. (1) to the manipulator, yields F = 6(8 - 9 -1) + 6 x 2 + 3 x 1 = 3. (2) Hence, the manipulator is a three-DOF mechanism. Since the limbs are connected to the moving platform and the fixed base by universal joints, no bending moments will transmit to the limbs. That is the force acting on each limb is directed along the longitudinal axis of the limb and the only moment acting on each limb is a twisting moment. Hence, these limbs can be made of hollow cylindrical rods to produce a light weight, high stiffness, and high speed manipulator. To facilitate the analysis, two coordinate systems ~ and E are attached to the fixed base and moving platform, respectively, as shown in Fig. 1. Figure 2 shows the geometry of limb i, where WI denotes the based attached axis of the lower universal joint, W5 denotes the moving platform attached axis of the upper universal joint, W2 and W4 denote the axes of the universal joints that are respectively attached to lower and upper members of the limb, UI is a unit vector pointing from 0 to Ai, U2 is a unit vector normal to the plane defined by WI and W2, U4 is a unit vector pointing from Ai to Bi, U5 is a unit vector normal to the plane defined by W4 and W5, and U6 a unit vector pointing from Q to Bi"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000908_j.jmatprotec.2013.01.020-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000908_j.jmatprotec.2013.01.020-Figure3-1.png",
        "caption": "Fig. 3. CAD geometry of the cantilev",
        "texts": [
            " These characteristics are well suited for support structure in metallic AM process for saving material and energy, while reducing build time and cost. This is study is aimed to further continue the preliminary work by Hussein et al. (2011) on lattice support structures. To the best of author\u2019s knowledge, the design and manufacturing of lattices as an external support structure is a new and novel approach in metal AM processes and offers new possibilities in building support structure for complex metallic parts. 3. Experimental procedure 3.1. Design process Fig. 3 shows the cantilever part used for testing the lattice support structure. The large overhang of the cantilever was chosen for studying the effectiveness of the lattice support structure and to collect sufficient data on cantilever deformation. Schoen Gyroid and Schwartz diamond lattice types with cell sizes ranging from 3 mm to 5 mm and volume fractions from 8 to 15% have been generated with the lattice support is then exported as a single STL (standard tessellation language) file format to the SLM machine for manufacturing"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001491_0278364919842269-Figure18-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001491_0278364919842269-Figure18-1.png",
        "caption": "Fig. 18. Still frames convey the simulated soft robot motion. Starting from zero gauge pressure, additional fluid is added to the chamber, resulting in a bending motion.",
        "texts": [
            " The robots began in the straight configuration and increased either n or V at a constant rate. The first simulation increases n to 2\u00d7 10 4 mol, and the second simulation increases the volume of the chamber linearly up to 2:65\u00d7 10 6 m3. These two final configurations are equivalent at steady state. For both simulations the changes occur over a period of 2 seconds, and the dynamic response without any additional actuation is simulated for another 2.5 seconds. The results of the simulation are shown in Figure 18. The real-time ratio was in the range of 26\u201332 for the incompressible fluid and about 28 with air as the fluid (a single-chamber robot is a simple case). The transverse tip response is shown in Figure 19. The limit cycle amplitude is roughly twice as large for the compressible air compared with an incompressible fluid. This indicates that choosing a fluid with greater compressibility results in larger vibrations. In addition, we note that the shear and extension strains turn out to be significant for this robot; at the final steady-state solution the elongation strain average over arclength is 0"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002652_j.jmapro.2019.12.009-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002652_j.jmapro.2019.12.009-Figure1-1.png",
        "caption": "Fig. 1. Schematic diagram of the experimental apparatus of laser cladding.",
        "texts": [
            " The optimal cladding parameters including scanning strategy, Z-axis increment, optimal overlapping rate and multi-layer accumulation pattern are investigated systematically to reduce defects inside the remanufactured tooth and improve the surface quality. Finally, the mechanical properties of the remanufactured tooth are tested and compared with the original one. The laser cladding system used in this research is composed of a KUKA robot arm, a YLR-500W laser generation machine, a water cooler, and a powder delivery machine, as shown in Fig. 1. The coaxial feed nozzle is placed at a distance of 14 mm from the substrate to keep the focus of the powder beam floating near the substrate surface. The powder material is H13 steel. The powder, with a size ranging from 75 \u03bcm to 100 \u03bcm and spherical shape, is produced by plasma rotation electrode process. The powder feeder is a rotary disc feeder in which powder mass flow can be controlled by the rotation speed of a metering disc. The broken tooth 3D model is constructed using a reverse engineering system, which contains a 3D scanner (OKIO-B, Beijing TenYoun 3D Technology Co"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure6.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure6.1-1.png",
        "caption": "Fig. 6.1. Disc type coreless PM brushless machine: 1 \u2014 coreless stator (armature) winding, 2 \u2014 PMs, 3 \u2014 twin rotor, 4 \u2014 shaft, 5 \u2014 bearing, 6 \u2014 frame.",
        "texts": [
            " Factors limiting the single disc design are listed in Section 2.1.6. A reasonable solution for larger torques are double or triple-disc machines. The disc-type PM brushless machines without stator and rotor cores were first manufactured commercially in the late 1990s for servo mechanisms and industrial electromechanical drives [150], solar powered electrical vehicles [225] as well as micromotors for computer peripherals and vibration motors for mobile phones [93]. The AFPM brushless machine without any ferromagnetic core is shown in Fig. 6.1. The machine consists of a twin rotor (3) with rare earth PMs (2) and nonmagnetic supporting structure. The steel-free stator (armature) winding (1) is located between two identical parts of the rotor. The stator polyphase winding fixed to the frame (6) is assembled as \u201cflower petals\u201d (Fig. 3.16) [P83]. Multi-turn coils are arranged in overlapping layers around the shaftaxis of the machine. The whole winding is then embedded in a high mechanical integrity plastic or resin. Since the topology shown in Fig 6.1 does not use any ferromagnetic core with slots, the machine is free of cogging (detent) torque 194 6 AFPM Machines Without Stator and Rotor Cores 6.3 Air Gap Magnetic Flux Density 195 and core losses. The only eddy current losses are losses in the stator winding conductors and metallic parts (if they exist) that reinforce the stator coreless winding. The coreless machine designed as a segmental (modular) machine is shown in Fig. 6.2. The output shaft power can easily be adjusted to desired level by adding more modules"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure2.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure2.3-1.png",
        "caption": "Fig. 2.3. Prismatic joint",
        "texts": [
            " We will be concerned with the manipu lator structure, link, kinematic pair, kinematic chain, joint coordinates, world coordinates, direct and inverse kinematic problems and redundancy. Let us consider the manipulator shown in Fig. 2.1. It consists of n rigid bodies interconnected by joints. The joints are either revolute or prismatic 20 2 Manipulator Kinematic Model (sliding). Revolute joints enable rotational motion of one link with respect to the other (Figure 2.2). In prismatic joints the motion is translational (Figure 2.3). The mechanical structure of the mechanism depends on the type of the joints that are included in the given robot and the disposition of the joints. For example, the robot shown in Fig. 2.1 has six revolute degrees of freedom, with the second and the third always being in parallel (anthropomor phic manipulator). 2.2 Definitions 21 Different schematic representations have been introduced in order to de scribe manipulator configurations simply. Figure 2.4{aHd) shows several types of robots. Here, revolute joints are denoted by cylinders, while prismatic joints are represented by parallelepipeds"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure14.27-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure14.27-1.png",
        "caption": "Figure 14.27 The origin of the xz coordinate system chosen at the point where dz/dx = 0.",
        "texts": [
            " Calculating the terms on both sides of the equals sign gives H cosh u \u00b7 du dx = \u2212g cosh u so that H du dx = \u2212g. By integrating this first degree equation in u we find u = \u2212gx H + C1. Substitution in (26) gives dz dx = sinh u = sinh ( \u2212gx H + C1 ) . (27) Integrating with respect to x gives z = \u2212H g cosh ( \u2212gx H + C1 ) + C2. (28) The integration constants C1 and C2 follow from the boundary conditions. 658 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM If we choose the origin of the coordinate system at the point in the cable where dz/dx = 0, the boundary conditions are (see Figure 14.27) x = 0; z = 0, x = 0; dz dx = 0. With these boundary conditions, (27) and (28) give C1 = 0 and C2 = H/g and the cable shape is1 z = \u2212H g ( cosh gx H \u2212 1 ) . (29) The slope of the cable is dz dx = \u2212 sinh gx H . (30) The vertical component of the cable force is V = H dz dx = \u2212H sinh gx H . (31) The tensile force in the cable is N = \u221a H 2 + V 2 = H \u221a 1 + sinh2 gx H = H cosh gx H . (32a) 1 Hereafter, we use the properties cosh(\u2212u) = cosh(+u) and sinh(\u2212u) = \u2212 sinh(+u). 14 Cables, Lines of Force and Structural Shapes 659 According to (29) H cosh gx H = H \u2212 gz so that the tensile force can also be written as N = H \u2212 gz"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure12.40-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure12.40-1.png",
        "caption": "FIGURE 12.40. A mass-spring, single degree-of-freedom vibrating system.",
        "texts": [
            "358) To find the solution, we substitute an exponential solution x = e\u03bbt in the equation of motion and find the characteristic equation. \u03bb2 + \u03bb\u2212 2 = 0 (12.359) The eigenvalues are \u03bb1,2 = 1,\u22122 (12.360) and therefore, the solution is x = a1e t + a2e \u22122t. (12.361) Taking a derivative x\u0307 = a1e t \u2212 2a2e\u22122t (12.362) 12. Applied Vibrations 789 and employing the initial conditions 1 = a1 + a2 (12.363) 7 = a1 \u2212 2a2 (12.364) provides the constants a1, a2, and the solution x = x(t). a1 = 3 (12.365) a2 = \u22122 (12.366) x = 3et \u2212 2e\u22122t (12.367) Example 454 Natural frequency. Consider a free mass-spring system such as the one shown in Figure 12.40. The system is undamped and free of excitation forces, so its equation of motion is mx\u0308+ kx = 0. (12.368) To find the solution, let\u2019s try a harmonic solution with an unknown frequency. x = A sin\u2126t+B cos\u2126t (12.369) Substituting (12.369) in (12.368) provides \u2212\u21262m (A sin\u2126t+B cos\u2126t) + k (A sin\u2126t+B cos\u2126t) = 0 (12.370) which can be collected as\u00a1 Bk \u2212Bm\u21262 \u00a2 cos\u2126t+ \u00a1 Ak \u2212Am\u21262 \u00a2 sin\u2126t = 0. (12.371) The coefficients of sin\u2126t and cos\u2126t must be zero, and hence, \u2126 = r k m (12.372) x = A sin r k m t+B cos r k m t"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.18-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.18-1.png",
        "caption": "FIGURE 7.18. A six-wheel vehicle with one steerable axle in front.",
        "texts": [
            " When a multi-axle vehicle has only one steerable axle, slip-free rotation is impossible for the non-steering wheels. The kinematic length or wheelbase of the vehicle is not clear, and it is not possible to define an Ackerman condition. Strong wear occurs for the tires, especially at low speeds and large steer angles. Hence, such a combination is not recommended. However, in case of a long three-axle vehicle with two nonsteerable axles close to each other, an approximated analysis is possible for low-speed steering. Figure 7.18 illustrates a six-wheel vehicle with only one steerable axle in front. We design the steering mechanism such that the center of rotation O is on a lateral line, called the midline, between the couple rear axles. The kinematic length of the vehicle, l, is the distance between the front axle and the midline. For this design we have cot \u03b4o \u2212 cot \u03b4i = w l (7.46) 7. Steering Dynamics 397 and R1 = l cot \u03b4o \u2212 w 2 = l cot \u03b4i + w 2 . (7.47) The center of the front axle and the mass center of the vehicle are turning about O by radii Rf and R"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002360_j.ijheatmasstransfer.2019.05.003-Figure10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002360_j.ijheatmasstransfer.2019.05.003-Figure10-1.png",
        "caption": "Fig. 10. (Left) Temperature contour at t = 1910 ms for the first layer along with the mode between the melt pool and the solidified region. (Right) laser scan pattern along with its s D (100 mm, 350 mm) are investigation points which are set at the middle of each scan tr",
        "texts": [
            " stored thermal data, are extracted and the former is fed to the next powder-laying calculation model where it serves as a solid wall with a prescribed restitution coefficient. The latter is used as the initial thermal and fluid condition for the next CFD (fusion) model. This procedure is repeated for three layers, based on Fig. 9. The process parameters and the laser specifications used in this study are given in Table 3. The base metal is nickle alloy IN718. The thermal field at t = 1910 ms along with the applied coordinate system for the computational domain are shown in Fig. 10. The domain size is 600 lm 400 lm 150 lm and the hatch spacing is 100 lm. The temperature field during the scanning of the first layer along with the melt region are both shown in Fig. 11 for different times. According to Fig. 11(a), the maximum temperature zone is formed very close to the centre of the moving laser beam. At 400 ls, most of the powder and base-plate are still at their initial temperature of 300 K. Due to the relatively high scanning velocity of the laser and the presence of air between the powders, the flow (propagation) of the heat in the transverse direction is much slower"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000554_978-3-540-85629-0-Figure1.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000554_978-3-540-85629-0-Figure1.3-1.png",
        "caption": "Fig. 1.3. Sketch of an automatic machine for tea packaging (courtesy IMA).",
        "texts": [
            " \u2022 Approximation: the curve does not pass exactly through the points, but there is an error that may be assigned by specifying a prescribed tolerance, Fig. 1.2(b). The latter approach can be useful when, especially in multi-dimensional trajectories, a reduction of the speed/acceleration values along the curve is desirable, at the expense of a lower precision. 1.2 One-dimensional Trajectories Nowadays the design of high speed automatic machines, whose actuation systems is mainly based on electric drives, generally involves the use of several actuators distributed in the machine and of relatively simple mechanisms, see for example Fig. 1.3, where the sketch of a packaging machine is reported. About twenty motion axes are present in a machine of this type. The socalled electronic cams and electronic gears are employed for the generation 2 At least two (three) segments are necessary for a typical periodic motion com- posed by a rise and a return phase (and, in case, by a dwell phase). 4 1 Trajectory Planning of motion where needed, in place of a single or few actuators and complex kinematic chains. In this manner, more flexible machines can be obtained, able to cope with the different production needs required from the market, [2]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure1.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure1.1-1.png",
        "caption": "Fig. 1.1. Electro-magnetic motor with disc rotor according to N. Tesla\u2019s patent No. 405 858, 1889 [P2].",
        "texts": [
            " Davenport (1837) claimed the first patent [P1] for a radial flux machine, conventional radial flux machines have been widely accepted as the mainstream configuration for electrical machines [33, 53]. The first primitive working prototype of an axial flux machine ever recorded was M. Faraday\u2019s disc (1831) - see Numerical Example 1.2. The disc type construction of electrical machines also appears in N. Tesla\u2019s patents, e.g. U.S. patent No. 405 858 [P2] entitled Electro-Magnetic Motor and published in 1889 (Fig. 1.1). The reasons for shelving the axial flux machine were multi-fold and may be summarised as follows: \u2022 strong axial (normal) magnetic attraction force between the stator and rotor; \u2022 fabrication difficulties, such as cutting slots in laminated cores and other methods of making slotted stator cores; \u2022 high costs involved in manufacturing the laminated stator cores; \u2022 difficulties in assembling the machine and keeping the uniform air gap. Although, the first PM excitation system was applied to electrical machines as early as the 1830s, the poor quality of hard magnetic materials soon discouraged their use"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure10.14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure10.14-1.png",
        "caption": "FIGURE 10.14 Images for g11 where observation and source points are in layer 1.",
        "texts": [
            " The strength of the images requires some thought in that the reflections must be considered carefully. Because there are multi-reflections due to the existence of two parallel dielectric boundaries, the coefficients \ud835\udefci in (10.52) have to be determined by multiple applications of the \ud835\udefc\u2019s in (10.53). A summary of diagrams in Fig. 10.12 is very helpful for the determination of the strength as well as the location of the images. We start with g11 where both the source and the observation points are located in region 1 shown in Fig. 10.14. g11(r, r\u2032) = 1 4\ud835\udf0b\ud835\udf001 [ 1 [\ud835\udf0c2 + (z \u2212 z\u2032)2]1\u22152 + \ud835\udefc1 [\ud835\udf0c2 + (z + z\u2032 \u2212 2d)2]1\u22152 + (1 \u2212 \ud835\udefc2 1) \u221e\u2211 k=0 (\u2212\ud835\udefc1)k(\ud835\udefc2)k+1 [\ud835\udf0c2 + (z + z\u2032 + 2 k d)2]1\u22152 ] (10.53) Each of the situations lead to as new formulation. Also, g33 can be derived from g11 by a change of variables. 268 PEEC MODELS FOR DIELECTRICS For the second Green\u2019s function, the source is still located in layer 1 while the observation point is in the middle layer 2, as shown in Fig. 10.15. By a change of variables we can construct G23 from this result since the observation point is staying in layer 2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.22-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.22-1.png",
        "caption": "FIGURE 8.22. A multi-link suspension mechanism.",
        "texts": [
            " Consider a double A-arm suspension mechanism. The coil spring may be between the lower arm and the chassis, as shown in Figure 8.18. It is also possible to install the spring between the upper arm and the chassis, or between the upper and lower arms. In either case, the lower or the upper arm, which supports the spring, is made stronger and the other arm acts as a connecting arm. Example 310 Multi-link suspension mechanism. When the two side bars of an A-arm are attached to each other with a joint, as shown in Figure 8.22, then the double A-arm is called a multi-link mechanism. A multi-link mechanism is a six-bar mechanism that may have a better coupler motion than a double A-arm mechanism. However, multi- 466 8. Suspension Mechanisms 8. Suspension Mechanisms 467 link suspensions are more expensive, less reliable, and more complicated compare to a double A-arm four-bar linkage. There are vehicles with more than six-link suspension with possibly better kinematic performance. Example 311 Swing arm suspension. An independent suspension may be as simple as a triangle shown in Figure 8"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure3.15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure3.15-1.png",
        "caption": "Figure 3.15 The moment of the force F about point A is equal to the moment of the couple that one has to add to the force if one shifts it parallel to its line of action through A.",
        "texts": [
            "14, the moment of force F with respect to A is seen as positive as F causes a rotation about A in the positive direction of rotation in the xy plane: Tz|A = +Fa = +(10 kN)(4 m) = +40 kNm. The same force F causes the body to rotate about B in the negative direction of rotation. The moment of F about B is therefore negative: Tz|B = \u2212Fb = \u2212(10 kN)(5 m) = \u221250 kNm. The moment of the force F about a point C located on its line of action, is zero: Tz|C = 0. For a force, in contrast to a couple, one has to specify the point about which the moment is being calculated. Here, this is done by including the point in question, after a vertical line, in the expression for the moment. Figure 3.15 shows a single force F acting at point B. Now introduce two equal and opposite forces F1 = F and F2 = F acting at point A. Since F1 and F2 together form an equilibrium system, the single force F at B is statically equivalent to the three forces F at B and F1 and F2 at A. F at B and F2 = F at A together form a couple with moment Fa. The force F = 7 kN at B is therefore statically equivalent with a force F = 7 kN at B and a couple with moment Fa = 21 kNm. Conclusion: The moment of a force F about a point A is equal to the moment of the couple one has to add when moving the force parallel to its a line of action to A"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000369_s12555-009-0311-8-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000369_s12555-009-0311-8-Figure1-1.png",
        "caption": "Fig. 1. Quadrotor helicopter on a landing pad under consideration.",
        "texts": [
            " INTRODUCTION Unmanned aerial vehicles (UAVs) are being used more often for military and civilian purposes such as traffic monitoring, patrolling for forest fires, surveillance, and rescue, in which risks to pilots are often high. Rotorcraft have an evident advantage over fixed-wing aircraft for various applications because of their vertical landing/take-off capability and payload. Among the rotorcraft, quadrotor helicopters can usually afford a larger payload than conventional helicopters due to four rotors as shown in Fig. 1. Moreover, small quadrotor helicopters possess a great maneuverability and are potentially simpler to manufacture. For these advantages, quadrotor helicopters have received much interest in UAV research. The quadrotor we consider is an underactuated system with six outputs and four inputs, and the states are highly coupled. To deal with this system, many modeling approaches have been presented [1,2] and various control methods proposed [3-17]. First of all, several backstepping controllers have been developed"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001975_tro.2013.2281564-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001975_tro.2013.2281564-Figure2-1.png",
        "caption": "Fig. 2. (a) Segment coordinates and frames on two sequential disks. The coordinate frames are illustrated on the disks\u2019 surfaces for clarity\u2014their origins actually align with the disks\u2019 centers of mass. (b) Composition of the resulting segment curvature by its two orthogonal curvatures \u03b2i and \u03b3i . (c) Illustration showing the bending plane angle \u03d5i .",
        "texts": [
            " A vector qi,lcl of these three variables defines the generalized coordinates for a given subsegment, as in (5), shown below, and the collection of these vectors for an N -subsegment continuum robot results in the robot\u2019s generalized coordinates q, shown below in (6): qi,lcl = [\u03b2i, \u03b3i, \u03b5i ] T (5) q = [ q T 1,lcl , q T 2,lcl , . . . , q T N ,lcl ] T . (6) To simplify the analysis, three intermediate variables are defined in (7), shown below: 1) the subsegment curvature magnitude ki ; 2) the bending plane angle \u03d5i ; and 3) the subsegment bending angle \u03b8i , where L0 is the subsegment length. The atan2 function is a four quadrant mapping of the two quadrant atan(\u03b3i /\u03b2i) function ki = \u221a \u03b22 i + \u03b32 i , \u03d5i = atan2(\u03b3i, \u03b2i) \u03b8i = kiL0 . (7) An illustration of these parameters is shown in Fig. 2. Each disk has a local coordinate system coincident with its center of mass (Fig. 2 illustrates these frames on the surfaces for clarity). For disk i, the local coordinate system is xi yi zi . The global frame at the robot\u2019s base is denoted by x0 y0 z0 . Based on geometric analysis [4], the local position vector pi,lcl of the disk i center of mass relative to the previous frame i\u20131 is calculated in (8), shown below. For this and other expressions, special consideration must be made for cases in which ki = 0. Analytically, the expressions all become zero divided by zero. However, as limk\u21920 , the coordinates asymptotically approach pi,lcl = [ 0, 0, L0 ] T "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000639_tie.2014.2364982-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000639_tie.2014.2364982-Figure1-1.png",
        "caption": "Fig. 1. Schematic of a quadrotor UAV.",
        "texts": [
            " To our best knowledge, few previous work have demonstrated real-time experimental results for the I&I approach on quadrotor UAV system. This paper is organized as follows. The dynamic model of the quadrotor and the control objective is described in Section II. Section III details the control development and in Section IV the asymptotic tracking results are proven. Realtime experimental results are presented in Section V. Finally, some conclusion remarks are included in Section VI. II. PROBLEM FORMULATION The schematic of a quadrotor UAV is shown in Fig. 1. The thrusts generated by the four rotors are denoted by fi (t) , i = 1, 2, 3, 4 respectively. Let I = {xI , yI , zI} represents a right hand inertia frame with zI being the vertical direction towards the sky. The body fixed frame, denoted by B = {xB , yB , zB} , is located at the center of gravity of the aircraft. The Euclidean position and Euler angle of the UAV with respect to the frame {I} are represented by \u03be (t) = [x (t) , y (t) , z (t)] T \u2208 R3 and \u03b7 (t) = [\u03d5 (t) , \u03b8 (t) , \u03c8 (t)] T \u2208 R3. The detailed dynamic model of the quadrotor can be found in previous works [21], [22]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000448_s11740-009-0192-y-Figure9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000448_s11740-009-0192-y-Figure9-1.png",
        "caption": "Fig. 9 Cantilever specimen on the base plate",
        "texts": [
            " In general, short scanning vectors, which occur within an island scanning strategy, cause higher temperature gradients than long scanning vectors, e.g. within a x-scanning-type. Concerning the thermo-mechanical effects, an alternating annealing of single islands leads to lower residual stresses than a non-recurring heat influence, which is typical for a conventional scanning strategy. The accordant specimen are manufactured with a powder layer thickness of 50 lm, a laser beam power of 200 W and a scanning velocity of 450 mm/s (cp. Fig. 9). Concerning the residual stresses of the cantilever, a differentiation between longitudinal (x-direction), normal respectively bending stresses (y-direction) and transversal (z-direction) stresses is carried out. Longitudinal stresses in combination with normal stresses are responsible for a bending of the cantilevers in the positive z-direction [34]. The stresses are evaluated in 12 measuring points, which are arranged along the x- (horizontally, 8 points) and the z-coordinate-direction (vertically, 4 points) in the center of the cantilever"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure11.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure11.1-1.png",
        "caption": "FIGURE 11.1 Simple example bar of magnetic material.",
        "texts": [
            " An interesting aspect is that we can solve this class of problems by using a circuit analog. To compare the corresponding statement for the Circuit Oriented Electromagnetic Modeling Using the PEEC Techniques, First Edition. Albert E. Ruehli, Giulio Antonini, and Lijun Jiang. \u00a9 2017 John Wiley & Sons, Inc. Published 2017 by John Wiley & Sons, Inc. 286 PEEC MODELS FOR MAGNETIC MATERIAL electrical and magnetic circuits, we display the electrical circuit equation on the left and the equivalent magnetic one on the right side in the derivation. We first show in Fig. 11.1 part of a magnetic circuit to define the geometrical parameters such as the cross section  and the length \ud835\udcc1. However, we should remark that the reluctance approach does not work well for open-loop problems such as the bar shown in Fig. 11.1. Hence, we assume that it is a portion of a loop without large gaps. Corresponding to the electrical potential, we define an artificial magnetic potential \u03a6m with the integral form E = \u2212\u2207\u03a6, H = \u2212\u2207\u03a6m (11.1) \u03a6 \ud835\udefc,\ud835\udefd = \u222b \ud835\udefd \ud835\udefc E \u22c5 d\ud835\udcc1, \u03a6m,\ud835\udefc,\ud835\udefd = \u222b \ud835\udefd \ud835\udefc H \u22c5 d\ud835\udcc1. (11.2) \ud835\udefc and \ud835\udefd are the line integration starting position and the ending position respectively. The relationship between the current density and electrical fields is in terms of the conductivity \ud835\udf0e for conductors that corresponds to the permeability \ud835\udf07 in the magnetic material"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure7.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure7.3-1.png",
        "caption": "Figure 7.3 Stresses normal to the sides of a triangular volume element.",
        "texts": [
            " In that case, one does not have to draw the entire stress distribution, but a single arrow1 is sufficient. The condition that no shear stresses can act in the material implies that the stresses on the boundary of the volume element have to be of the same magnitude: p1 = p2 = p3. 1 Note: the arrows here cannot be interpreted as forces. on its boundaries. 7 Gas Pressure and Hydrostatic Pressure 247 To demonstrate this, a small triangular part has been isolated from the material parallel to the xy plane in Figure 7.3. The oblique side has an area A. The area of the vertical side is therefore A cos \u03b1, while that of the horizontal side is A sin \u03b1. The triangular part is so small that, for all the stresses on the boundary, it can be assumed that they are uniformly distributed. Assume that a compressive stress p is acting on the oblique side. This stress acts normal to the side as there is no shear stress. In Figure 7.4, the forces (force = stress \u00d7 area) on the edges of the triangular part are shown. The lines of action of the forces pass through a single point"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure3.27-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure3.27-1.png",
        "caption": "Fig. 3.27 Relationship between Robot\u2019s Motion and ZMP (a) Explanation using Inertial Force (b) Force acting on Robot (c) Explanation using the Gravity Compensation Force and Acceleration Force",
        "texts": [
            "81) where their elements are 98 3 ZMP and Dynamics \u23a1 \u23a3 P\u0307x P\u0307y P\u0307z \u23a4 \u23a6 = \u23a1 \u23a3 Mx\u0308 My\u0308 Mz\u0308 \u23a4 \u23a6 (3.82) \u23a1 \u23a3 L\u0307x L\u0307y L\u0307z \u23a4 \u23a6 = \u23a1 \u23a3 M(yz\u0308 \u2212 zy\u0308) M(zx\u0308\u2212 xz\u0308) M(xy\u0308 \u2212 yx\u0308) \u23a4 \u23a6 . (3.83) Substituting the above equation to (3.73) and (3.74), the ZMP can be given by px = x\u2212 (z \u2212 pz)x\u0308 z\u0308 + g (3.84) py = y \u2212 (z \u2212 pz)y\u0308 z\u0308 + g . (3.85) We will use (3.84) in Chapter 4 to generate the biped gait pattern. By intuitively explaining the relationship between the robot\u2019s motion and the ZMP, figures of the point-mass model like Fig. 3.27(a) is often introduced. Here, \u2212Mx\u0308 denotes the virtual force called the inertial force expressing the reaction generated by the acceleration of an object[8]. Fig. 3.27(a) shows that the inertial force and the gravity force balance with the ground reaction force. Here, the forces acting on the robot is apparently the gravity and the ground reaction forces as shown in 3.27(b). Since it is difficult to show the balance of forces, we introduced the inertial force. However, we can explain without introducing the inertial force. In Fig.3.27(c), opposing the gravity force, the center of mass goes up due to the effect of the ground reaction force. At the same time, the center of mass is accelerated due to the effect of it [23]. In this case, the ground reaction force is decomposed into the gravity compensation and the acceleration forces. Of course, the right results can be expected by introducing both explanations. The discussions like \u201cDepending on the motion of the robot, may the ZMP leave the support polygon?\u201d often occurs. Of course, the conclusion is that \u201cthe ZMP never exists outside the support polygon\u201d [37, 88]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure9.17-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure9.17-1.png",
        "caption": "FIGURE 9.17. A turning wheel moving up a step.",
        "texts": [
            " Applied Dynamics where IC is the mass moment of inertia for the wheel about its center. Therefore, 1 2 mv2 + 1 2 IC\u03c9 2 = mg (R+ r) (1\u2212 cos \u03b8) (9.346) and substituting (9.340) and (9.341) provides\u00b5 1 + IC mr2 \u00b6 (R+ r) g cos \u03b8 = 2g (R+ r) (1\u2212 cos \u03b8) (9.347) and therefore, the separation angle is \u03b8 = cos\u22121 2mr2 IC + 3mr2 . (9.348) Let\u2019s examine the equation for a disc wheel with IC = 1 2 mr2. (9.349) and find the separation angle. \u03b8 = cos\u22121 4 7 (9.350) \u2248 0.96 rad \u2248 55.15 deg Example 373 Turning wheel over a step. Figure 9.17 illustrates a wheel of radius R turning with speed v to go over a step with height H < R. We may use the principle of energy conservation and find the speed of the wheel after getting across the step. Employing the conservation of energy, 9. Applied Dynamics 569 we have EA = EB (9.351) KA + VA = KB + VB (9.352) 1 2 mv21 + 1 2 IC\u03c9 2 1 + 0 = 1 2 mv22 + 1 2 IC\u03c9 2 2 +mgH (9.353)\u00b5 m+ IC R2 \u00b6 v21 = \u00b5 m+ IC R2 \u00b6 v22 + 2mgH (9.354) and therefore, v2 = vuutv21 \u2212 2gH 1 + IC mR2 . (9.355) The condition for having a real v2 is v1 > vuut 2gH 1 + IC mR2 "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.63-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.63-1.png",
        "caption": "FIGURE 8.63.",
        "texts": [],
        "surrounding_texts": [
            "where, a1 is the longitudinal distance between C and the front axle, b1 is the lateral distance between C and the tireprint of the tire 1, and h is the height of C from the ground level. If P is a point in the tire frame at T1rP T1rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 . (8.62) then its coordinates in the body frame are BrP = BRT1 T1rP + BdT1 = \u23a1\u23a3 a1 + xP cos \u03b41 \u2212 yP sin \u03b41 yP cos \u03b41 \u2212 b1 + xP sin \u03b41 zP \u2212 h \u23a4\u23a6 . (8.63) The rotation matrix BRT1 is a result of steering about the z1-axis. BRT1 = \u23a1\u23a3 cos \u03b41 \u2212 sin \u03b41 0 sin \u03b41 cos \u03b41 0 0 0 1 \u23a4\u23a6 (8.64) Employing Equation (8.28), we may examine a wheel point P at W rP W rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 (8.65) 8. Suspension Mechanisms 497 and find the body coordinates of the point BrP = BRT1 T1rP + BdT1 = BRT1 \u00a1 T1RW W rP + T1dW \u00a2 + BdT1 = BRT1 T1RW W rP + BRT1 T1dW + BdT1 = BRW W rP + BRT1 T1dW + BdT1 (8.66) BrP = \u23a1\u23a3 a1 + xP cos \u03b41 \u2212 yP cos \u03b3 sin \u03b41 + (Rw + zP ) sin \u03b3 sin \u03b41 xP sin \u03b41 \u2212 b1 + yP cos \u03b3 cos \u03b41 \u2212 (Rw + zP ) cos \u03b41 sin \u03b3 (Rw + zP ) cos \u03b3 + yP sin \u03b3 \u2212 h \u23a4\u23a6 (8.67) where, BRW = BRT1 T1RW = \u23a1\u23a3 cos \u03b41 \u2212 cos \u03b3 sin \u03b41 sin \u03b3 sin \u03b41 sin \u03b41 cos \u03b3 cos \u03b41 \u2212 cos \u03b41 sin \u03b3 0 sin \u03b3 cos \u03b3 \u23a4\u23a6 (8.68) T1dW = \u23a1\u23a3 0 \u2212Rw sin \u03b3 Rw cos \u03b3 \u23a4\u23a6 . (8.69) Example 334 F Wheel-body to vehicle transformation. The wheel-body coordinate frames are always parallel to the vehicle frame. The origin of the wheel-body coordinate frame of the wheel number 1 is at BdW1 = \u23a1\u23a3 a1 \u2212b1 \u2212h+Rw \u23a4\u23a6 . (8.70) Hence the transformation between the two frames is only a displacement. Br = BIW1 W1r+ BdW1 (8.71) 8.6 F Caster Theory The steer axis may have any angle and any location with respect to the wheel-body coordinate frame. The wheel-body frame C (xc, yc, zc) is a frame at the center of the wheel at its rest position, parallel to the vehicle coordinate frame. The frame C does not follow any motion of the wheel. The steer axis is the kingpin axis of rotation. Figure 8.51 illustrates the front and side views of a wheel and its steering axis. The steering axis has angle \u03d5 with (yc, zc) plane, and angle \u03b8 with (xc, zc) plane. The angles \u03d5 and \u03b8 are measured about the yc and xc axes, 498 8. Suspension Mechanisms respectively. The angle \u03d5 is the caster angle of the wheel, while the angle \u03b8 is the lean angle. The steering axis of the wheel, as shown in Figure 8.51, is at a positive caster and lean angles. The steering axis intersect the ground plane at a point that has coordinates (sa, sb,\u2212Rw) in the wheelbody coordinate frame. If we indicate the steering axis by the unit vector u\u0302, then the components of u\u0302 are functions of the caster and lean angles. C u\u0302 = \u23a1\u23a3 u1 u2 u3 \u23a4\u23a6 = 1p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 \u23a1\u23a3 cos \u03b8 sin\u03d5 \u2212 cos\u03d5 sin \u03b8 cos \u03b8 cos\u03d5 \u23a4\u23a6 (8.72) The position vector of the point that u\u0302 intersects the ground plane, is called the location vector s that in the wheel-body frame has the following coordinates: Cs = \u23a1\u23a3 sa sb \u2212Rw \u23a4\u23a6 (8.73) We express the rotation of the wheel about the steering axis u\u0302 by a zero pitch screw motion s\u030c. CTW = C s\u030cW (0, \u03b4, u\u0302, s) = \u2219 CRW Cs\u2212 CRW Cs 0 1 \u00b8 = \u2219 CRW CdW 0 1 \u00b8 (8.74) Proof. The steering axis is at the intersection of the caster plane \u03c0C and the lean plane \u03c0L, both expressed in the wheel-body coordinate frame. The 8. Suspension Mechanisms 499 two planes can be indicated by their normal unit vectors n\u03021 and n\u03022. C n\u03021 = \u23a1\u23a3 0 cos \u03b8 sin \u03b8 \u23a4\u23a6 (8.75) C n\u03022 = \u23a1\u23a3 \u2212 cos\u03d50 sin\u03d5 \u23a4\u23a6 (8.76) The unit vector u\u0302 on the intersection of the caster and lean planes can be found by u\u0302 = n\u03021 \u00d7 n\u03022 |n\u03021 \u00d7 n\u03022| (8.77) where, n\u03021 \u00d7 n\u03022 = \u23a1\u23a3 cos \u03b8 sin\u03d5 \u2212 cos\u03d5 sin \u03b8 cos \u03b8 cos\u03d5 \u23a4\u23a6 (8.78) |n\u03021 \u00d7 n\u03022| = q cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 (8.79) and therefore, C u\u0302 = \u23a1\u23a3 u1 u2 u3 \u23a4\u23a6 = \u23a1\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3 cos \u03b8 sin\u03d5p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 \u2212 cos\u03d5 sin \u03b8p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 cos \u03b8 cos\u03d5p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 \u23a4\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6 . (8.80) Steering axis does not follow any motion of the wheel except the wheel hop in the z-direction. We assume that the steering axis is a fixed line with respect to the vehicle, and the steer angle \u03b4 is the rotation angle about u\u0302. The intersection point of the steering axis and the ground plane defines the location vector s. Cs = \u23a1\u23a3 sa sb \u2212Rw \u23a4\u23a6 (8.81) The components sa and sb are called the forward and lateral locations respectively. Using the axis-angle rotation (u\u0302, \u03b4), and the location vector s, we can define the steering process as a screw motion s\u030c with zero pitch. Employing Equations (5.473)-(5.477), we find the transformation screw for wheel frame W to wheel-body frame C. CTW = C s\u030cW (0, \u03b4, u\u0302, s) (8.82) = \u2219 CRW Cs\u2212 CRW Cs 0 1 \u00b8 = \u2219 CRW Cd 0 1 \u00b8 500 8. Suspension Mechanisms CRW = I cos \u03b4 + u\u0302u\u0302T vers \u03b4 + u\u0303 sin \u03b4 (8.83) CdW = \u00a1\u00a1 I\u2212 u\u0302u\u0302T \u00a2 vers \u03b4 \u2212 u\u0303 sin \u03b4 \u00a2 Cs. (8.84) u\u0303 = \u23a1\u23a3 0 \u2212u3 u2 u3 0 \u2212u1 \u2212u2 u1 0 \u23a4\u23a6 (8.85) vers \u03b4 = 1\u2212 cos \u03b4 (8.86) Direct substitution shows that CRW and CdW are: CRW = \u23a1\u23a3 u21 vers \u03b4 + c\u03b4 u1u2 vers \u03b4 \u2212 u3s\u03b4 u1u3 vers \u03b4 + u2s\u03b4 u1u2 vers \u03b4 + u3s\u03b4 u22 vers \u03b4 + c\u03b4 u2u3 vers \u03b4 \u2212 u1s\u03b4 u1u3 vers \u03b4 \u2212 u2s\u03b4 u2u3 vers \u03b4 + u1s\u03b4 u23 vers \u03b4 + c\u03b4 \u23a4\u23a6 (8.87) CdW = \u23a1\u23a3 (s1 \u2212 u1 (s3u3 + s2u2 + s1u1)) vers \u03b4 + (s2u3 \u2212 s3u2) sin \u03b4 (s2 \u2212 u2 (s3u3 + s2u2 + s1u1)) vers \u03b4 + (s3u1 \u2212 s1u3) sin \u03b4 (s3 \u2212 u3 (s3u3 + s2u2 + s1u1)) vers \u03b4 + (s1u2 \u2212 s2u1) sin \u03b4 \u23a4\u23a6 (8.88) The vector CdW indicates the position of the wheel center with respect to the wheel-body frame. The matrix CTW is the homogeneous transformation from wheel frame W to wheel-body frame C, when the wheel is steered by the angle \u03b4 about the steering axis u\u0302. Example 335 F Zero steer angle. To examine the screw transformation, we check the zero steering. Substituting \u03b4 = 0 simplifies the rotation matrix CRW and the position vector CdW to I and 0 CRW = \u23a1\u23a3 1 0 0 0 1 0 0 0 1 \u23a4\u23a6 (8.89) CdW= \u23a1\u23a3 0 0 0 \u23a4\u23a6 (8.90) indicating that at zero steering, the wheel frameW and wheel-body frame C are coincident. Example 336 F Steer angle transformation for zero lean and caster. Consider a wheel with a steer axis coincident with zw. Such a wheel has no lean or caster angle. When the wheel is steered by the angle \u03b4, we can find the coordinates of a wheel point P in the wheel-body coordinate frame using transformation method. Figure 8.52 illustrates a 3D view, and Figure 8.53 a top view of such a wheel. 8. Suspension Mechanisms 501 502 8. Suspension Mechanisms Assume W rP = [xw, yw, zw] T is the position vector of a wheel point, then its position vector in the wheel-body coordinate frame C is CrP = CRW W rP = Rz,\u03b4 W rP = \u23a1\u23a3 cos \u03b4 \u2212 sin \u03b4 0 sin \u03b4 cos \u03b4 0 0 0 1 \u23a4\u23a6\u23a1\u23a3 xw yw zw \u23a4\u23a6 = \u23a1\u23a3 xw cos \u03b4 \u2212 yw sin \u03b4 yw cos \u03b4 + xw sin \u03b4 zw \u23a4\u23a6 . (8.91) We assumed that the wheel-body coordinate is installed at the center of the wheel and is parallel to the vehicle coordinate frame. Therefore, the transformation from the frame W to the frame C is a rotation \u03b4 about the wheel-body z-axis. There would be no camber angle when the lean and caster angles are zero and steer axis is on the zw-axis. Example 337 F Zero lean, zero lateral location. The case of zero lean, \u03b8 = 0, and zero lateral location, sb = 0, is important in caster dynamics of bicycle model. The screw transformation for this case will be simplified to C u\u0302 = \u23a1\u23a3 u1 u2 u3 \u23a4\u23a6 = \u23a1\u23a3 sin\u03d5 0 cos\u03d5 \u23a4\u23a6 (8.92) Cs = \u23a1\u23a3 sa 0 \u2212Rw \u23a4\u23a6 (8.93) CRW = \u23a1\u23a3 sin2 \u03d5 vers \u03b4 + cos \u03b4 \u2212 cos\u03d5 sin \u03b4 sin\u03d5 cos\u03d5 vers \u03b4 cos\u03d5 sin \u03b4 cos \u03b4 \u2212 sin\u03d5 sin \u03b4 sin\u03d5 cos\u03d5 vers \u03b4 sin\u03d5 sin \u03b4 cos2 \u03d5 vers \u03b4 + cos \u03b4 \u23a4\u23a6 (8.94) Cd = \u23a1\u23a2\u23a3 cos\u03d5 (sa cos\u03d5+Rw sin\u03d5) vers \u03b4 \u2212 (sa cos\u03d5+Rw sin\u03d5) sin \u03b4 \u22121 2 (Rw \u2212Rw cos 2\u03d5+ sa sin 2\u03d5) vers \u03b4 \u23a4\u23a5\u23a6 . (8.95) Example 338 F Position of the tireprint. The center of tireprint in the wheel coordinate frame is at rT W rT = \u23a1\u23a3 0 0 \u2212Rw \u23a4\u23a6 . (8.96) 8. Suspension Mechanisms 503 If we assume the width of the tire is zero and the wheel is steered, the center of tireprint would be at CrT = CTW W rT = \u23a1\u23a3 xT yT zT \u23a4\u23a6 (8.97) where xT = \u00a1 1\u2212 u21 \u00a2 (1\u2212 cos \u03b4) sa + (u3 sin \u03b4 \u2212 u1u2 (1\u2212 cos \u03b4)) sb (8.98) yT = \u2212 (u3 sin \u03b4 + u1u2 (1\u2212 cos \u03b4)) sa + \u00a1 1\u2212 u22 \u00a2 (1\u2212 cos \u03b4) sb (8.99) zT = (u2 sin \u03b4 \u2212 u1u3 (1\u2212 cos \u03b4)) sa \u2212 (u1 sin \u03b4 + u2u3 (1\u2212 cos \u03b4)) sb \u2212Rw (8.100) or xT = sb \u00c3 cos \u03b8 cos\u03d5 sin \u03b4p cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 + 1 4 sin 2\u03b8 sin 2\u03d5 (1\u2212 cos \u03b4) cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 ! +sa \u00b5 1\u2212 cos2 \u03b8 sin2 \u03d5 cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 \u00b6 (1\u2212 cos \u03b4) (8.101) yT = \u2212sa \u00c3 cos \u03b8 cos\u03d5 sin \u03b4p cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 \u2212 1 4 sin 2\u03b8 sin 2\u03d5 (1\u2212 cos \u03b4) cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 ! +sb \u00b5 1\u2212 cos2 \u03d5 sin2 \u03b8 cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 \u00b6 (1\u2212 cos \u03b4) (8.102) zT = \u2212Rw \u2212 sb cos \u03b8 sin\u03d5+ sa cos\u03d5 sin \u03b8p cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 sin \u03b4 + 1 2 sb cos 2 \u03d5 sin 2\u03b8 \u2212 sa cos 2 \u03b8 sin 2\u03d5 cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 (1\u2212 cos \u03b4) (8.103) Example 339 F Wheel center drop. The zT coordinate in (8.100) or (8.103) indicates the amount that the center of the tireprint will move in the vertical direction with respect to the wheel-body frame when the wheel is steering. If the steer angle is zero, \u03b4 = 0, then zT is at zT = \u2212Rw. (8.104) Because the center of tireprint must be on the ground, H = \u2212Rw \u2212 zT indicated the height that the center of the wheel will drop during steering. H = \u2212Rw \u2212 zT (8.105) = sb cos \u03b8 sin\u03d5+ sa cos\u03d5 sin \u03b8p cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 sin \u03b4 \u22121 2 sb cos 2 \u03d5 sin 2\u03b8 \u2212 sa cos 2 \u03b8 sin 2\u03d5 cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 (1\u2212 cos \u03b4) 504 8. Suspension Mechanisms The zT coordinate of the tireprint may be simplifie for different designs: 1\u2212 If the lean angle is zero, \u03b8 = 0, then zT is at zT = \u2212Rw \u2212 1 2 sa sin 2\u03d5 (1\u2212 cos \u03b4)\u2212 sb sin\u03d5 sin \u03b4. (8.106) 2\u2212 If the lean angle and lateral location are zero, \u03b8 = 0, sb = 0, then zT is at zT = \u2212Rw \u2212 1 2 sa sin 2\u03d5 (1\u2212 cos \u03b4) . (8.107) In this case, the wheel center drop may be expressed by a dimensionless equation. H sa = 1 2 sin 2\u03d5 (1\u2212 cos \u03b4) (8.108) Figure 8.54 illustrates H/sa for the caster angle \u03d5 = 5deg, 0 deg, \u22125 deg, \u221210 deg, \u221215 deg, \u221220 deg, and the steer angle \u03b4 in the range \u221210 deg < \u03b4 < 10 deg. In street cars, we set the steering axis with a positive longitudinal location sa > 0, and a few degrees negative caster angle \u03d5 < 0. In this case the wheel center drops as is shown in the figure. 3\u2212 If the caster angle is zero, \u03d5 = 0, then zT is at zT = \u2212Rw + 1 2 sb sin 2\u03b8 (1\u2212 cos \u03b4)\u2212 sa sin \u03b8 sin \u03b4. (8.109) 4\u2212 If the caster angle and lateral location are zero, \u03d5 = 0, sb = 0, then zT is at zT = \u2212Rw \u2212 sa sin \u03b8 sin \u03b4. (8.110) 8. Suspension Mechanisms 505 In this case, the wheel center drop may be expressed by a dimensionless equation. H sa = \u2212 sin \u03b8 sin \u03b4 (8.111) Figure 8.55 illustrates H/sa for the lean angle \u03b8 = 5deg, 0, \u22125 deg, \u221210 deg, \u221215 deg, \u221220 deg and the steer angle \u03b4 in the range \u221210 deg < \u03b4 < 10 deg. The steering axis of street cars is usually set with a positive longitudinal location sa > 0, and a few degrees positive lean angle \u03b8 > 0. In this case the wheel center lowers when the wheel number 1 turns to the right, and elevates when the wheel turns to the left. Comparison of Figures 8.54 and 8.55 shows that the lean angle has much more affect on the wheel center drop than the caster angle. 5\u2212 If the lateral location is zero, sb = 0, then zT is at zT = \u2212Rw \u2212 sa cos\u03d5 sin \u03b8p cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 sin \u03b4 \u22121 2 sa cos2 \u03b8 sin 2\u03d5 cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 (1\u2212 cos \u03b4) (8.112) and the wheel center drop,H, may be expressed by a dimensionless equation. H sa = \u22121 2 cos2 \u03b8 sin2 \u03d5 (1\u2212 cos \u03b4) cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 \u2212 cos\u03d5 sin \u03b8 sin \u03b4p cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 (8.113) Example 340 F Position of the wheel center. As given in Equation (8.88), the wheel center is at CdW with respect to 506 8. Suspension Mechanisms the wheel-body frame. CdW = \u23a1\u23a3 xW yW zW \u23a4\u23a6 (8.114) Substituting for u\u0302 and s from (8.72) and (8.73) in (8.88) provides the coordinates of the wheel center in the wheel-body frame as xW = (sa \u2212 u1 (\u2212Rwu3 + sbu2 + sau1)) (1\u2212 cos \u03b4) + (sbu3 +Rwu2) sin \u03b4 (8.115) yW = (sb \u2212 u2 (\u2212Rwu3 + sbu2 + sau1)) (1\u2212 cos \u03b4) \u2212 (Rwu1 + sau3) sin \u03b4 (8.116) zW = (\u2212Rw \u2212 u3 (\u2212Rwu3 + sbu2 + sau1)) (1\u2212 cos \u03b4) + (sau2 \u2212 sbu1) sin \u03b4 (8.117) or xW = sa (1\u2212 cos \u03b4) + \u00b5 1 2 Rw sin 2\u03d5\u2212 sa sin 2 \u03d5 \u00b6 cos2 \u03b8 + 1 4 sb sin 2\u03b8 sin 2\u03d5 cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 (1\u2212 cos \u03b4) + (sb cos \u03b8 \u2212Rw sin \u03b8)p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 cos\u03d5 sin \u03b4 (8.118) yW = sb (1\u2212 cos \u03b4) \u2212 1 2 \u00a1 Rw sin 2\u03b8 + sb sin 2 \u03b8 \u00a2 cos2 \u03d5\u2212 1 4 sa sin 2\u03b8 sin 2\u03d5 cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 (1\u2212 cos \u03b4) \u2212 Rw sin\u03d5+ sa cos\u03d5p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 cos \u03b8 sin \u03b4 (8.119) zW = \u2212Rw (1\u2212 cos \u03b4) + \u00b5 Rw cos 2 \u03b8 + 1 2 sb sin 2\u03b8 \u00b6 cos2 \u03d5\u2212 1 2 sa cos 2 \u03b8 sin 2\u03d5 cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 (1\u2212 cos \u03b4) \u2212sa cos\u03d5 sin \u03b8 + sb cos \u03b8 sin\u03d5p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 sin \u03b4 (8.120) The zW coordinate indicates how the center of the wheel will move in the vertical direction with respect to the wheel-body frame, when the wheel is steering. It shows that zW = 0, as long as \u03b4 = 0. 8. Suspension Mechanisms 507 The zW coordinate of the wheel center may be simplified for different designs: 1\u2212 If the lean angle is zero, \u03b8 = 0, then zW is at zW = \u2212Rw \u00a1 1\u2212 cos2 \u03d5 \u00a2 (1\u2212 cos \u03b4)\u2212 sb sin\u03d5 sin \u03b4 \u22121 2 sa sin 2\u03d5 (1\u2212 cos \u03b4) . (8.121) 2\u2212 If the lean angle and lateral location are zero, \u03b8 = 0, sb = 0, then zW is at zW = \u2212Rw \u00a1 1\u2212 cos2 \u03d5 \u00a2 (1\u2212 cos \u03b4)\u2212 1 2 sa sin 2\u03d5 (1\u2212 cos \u03b4) . (8.122) 3\u2212 If the caster angle is zero, \u03d5 = 0, then zW is at zW = \u2212Rw \u00a1 1\u2212 cos2 \u03b8 \u00a2 (1\u2212 cos \u03b4)\u2212 sa sin \u03b8 sin \u03b4 + 1 2 sb sin 2\u03b8 (1\u2212 cos \u03b4) . (8.123) 4\u2212 If the caster angle and lateral location are zero, \u03d5 = 0, sb = 0, then zW is at zW = \u2212Rw \u00a1 1\u2212 cos2 \u03b8 \u00a2 (1\u2212 cos \u03b4)\u2212 sa sin \u03b8 sin \u03b4. (8.124) 5\u2212 If the lateral location is zero, sb = 0, then zT is at zW = \u2212Rw (1\u2212 cos \u03b4)\u2212 sa cos\u03d5 sin \u03b8p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 sin \u03b4 + Rw cos 2 \u03b8 cos2 \u03d5\u2212 1 2 sa cos 2 \u03b8 sin 2\u03d5 cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 (1\u2212 cos \u03b4) (8.125) In each case of the above designs, the height of the wheel center with respect to the ground level can be found by adding H to zW . The equations for calculating H are found in Example 340. Example 341 F Camber theory. Having a non-zero lean and caster angles causes a camber angle \u03b3 for a steered wheel. To find the camber angle of an steered wheel, we may determine the angle between the camber line and the vertical direction zc. The camber line is the line connecting the wheel center and the center of tireprint. The coordinates of the center of tireprint (xT , yT , zT ) are given in Equations (8.101)-(8.103), and the coordinates of the wheel center (xW , yW , zW ) are given in Equations (8.118)-(8.120). The line connecting (xT , yT , zT ) to (xW , yW , zW ) may be indicated by the unit vector l\u0302c l\u0302c = (xW \u2212 xT ) I\u0302 + (yW \u2212 yT ) J\u0302 + (zW \u2212 zT ) K\u0302q (xW \u2212 xT ) 2 + (yW \u2212 yT ) 2 + (zW \u2212 zT ) 2 (8.126) 508 8. Suspension Mechanisms in which I\u0302 , J\u0302 , K\u0302, are the unit vectors of the wheel-body coordinate frame C. The camber angle is the angle between l\u0302c and K\u0302, which can be found by the inner vector product. \u03b3 = cos\u22121 \u00b3 l\u0302c \u00b7 K\u0302 \u00b4 = cos\u22121 (zW \u2212 zT )q (xW \u2212 xT ) 2 + (yW \u2212 yT ) 2 + (zW \u2212 zT ) 2 (8.127) As an special case, let us determine the camber angle when the lean angle and lateral location are zero, \u03b8 = 0, sb = 0. In this case, we have xT = sa \u00a1 1\u2212 sin2 \u03d5 \u00a2 (cos \u03b4 \u2212 1) (8.128) yT = \u2212sa cos\u03d5 sin \u03b4 (8.129) zT = zT = \u2212Rw \u2212 1 2 sa sin 2\u03d5 (1\u2212 cos \u03b4) (8.130) xW = \u00b5 sa + 1 2 Rw sin 2\u03d5\u2212 sa sin 2 \u03d5 \u00b6 (1\u2212 cos \u03b4) (8.131) yW = sb (1\u2212 cos \u03b4)\u2212Rw sin\u03d5+ sa cos\u03d5 sin \u03b4 (8.132) zW = \u00b5 Rw \u00a1 cos2 \u03d5\u2212 1 \u00a2 \u2212 1 2 sa sin 2\u03d5 \u00b6 (1\u2212 cos \u03b4) . (8.133) 8.7 Summary There are two general types of suspensions: dependent, in which the left and right wheels on an axle are rigidly connected, and independent, in which the left and right wheels are disconnected. Solid axle is the most common dependent suspension, while McPherson and double A-arm are the most common independent suspensions. The roll axis is the instantaneous line about which the body of a vehicle rolls. Roll axis is found by connecting the roll center of the front and rear suspensions of the vehicle. The instant center of rotation of a wheel with respect to the body is called suspension roll center. So, to find the roll center of the front or rear half of a car, we should determine the suspension roll centers, and find the intersection of the lines connecting the suspension roll centers to the center of their associated tireprints. Three coordinate frames are employed to express the orientation of a tire and wheel with respect to the vehicle: the wheel frame W , wheel-body frame C, and tire frame T . A wheel coordinate frame W (xw, yw, zw) is attached to the center of a wheel. It follows every translation and rotation of the wheel except the spin. Hence, the xw and zw axes are always in the tire-plane, while the yw-axis is always along the spin axis. When the wheel 8. Suspension Mechanisms 509 is straight and the W frame is parallel to the vehicle coordinate frame, we attach a wheel-body coordinate frame C (xc, yc, zc) at the center of the wheel parallel to the vehicle coordinate axes. The wheel-body frame C is motionless with respect to the vehicle coordinate and does not follow any motion of the wheel. The tire coordinate frame T (xt, yt, zt) is set at the center of the tireprint. The zt-axis is always perpendicular to the ground. The xt-axis is along the intersection line of the tire-plane and the ground. The tire frame does not follows the spin and camber rotations of the tire however, it follows the steer angle rotation about the zc-axis. We define the orientation and position of a steering axis by the caster angle \u03d5, lean angle \u03b8, and the intersection point of the axis with the ground surface at (sa, sb) with respect to the center of tireprint. Because of these parameters, a steered wheel will camber and generates a lateral force. This is called the caster theory. The camber angle \u03b3 of a steered wheel for \u03b8 = 0, and sb = 0 is: \u03b3 = cos\u22121 \u00b3 l\u0302c \u00b7 K\u0302 \u00b4 = cos\u22121 (zW \u2212 zT )q (xW \u2212 xT ) 2 + (yW \u2212 yT ) 2 + (zW \u2212 zT ) 2 (8.134) where xT = sa \u00a1 1\u2212 sin2 \u03d5 \u00a2 (cos \u03b4 \u2212 1) (8.135) yT = \u2212sa cos\u03d5 sin \u03b4 (8.136) zT = zT = \u2212Rw \u2212 1 2 sa sin 2\u03d5 (1\u2212 cos \u03b4) (8.137) xW = \u00b5 sa + 1 2 Rw sin 2\u03d5\u2212 sa sin 2 \u03d5 \u00b6 (1\u2212 cos \u03b4) (8.138) yW = sb (1\u2212 cos \u03b4)\u2212Rw sin\u03d5+ sa cos\u03d5 sin \u03b4 (8.139) zW = \u00b5 Rw \u00a1 cos2 \u03d5\u2212 1 \u00a2 \u2212 1 2 sa sin 2\u03d5 \u00b6 (1\u2212 cos \u03b4) . (8.140) 510 8. Suspension Mechanisms 8.8 Key Symbols a, b, c, d lengths of the links of a four-bar linkage ai distance of the axle number i from the mass center A,B, \u00b7 \u00b7 \u00b7 coefficients in equation for calculating \u03b83 b1 distance of left wheels from mass center B (x, y, z) vehicle coordinate frame C mass center C coupler point C (xc, yc, zc) wheel-body coordinate frame C T dW C expression of the position of W with respect to T e, \u03b1 polar coordinates of a coupler point g overhang h = z \u2212 z0 vertical displacement of the wheel center H wheel center drop Iij instant center of rotation between link i and link j IijImn a line connecting Iij and Imn I\u0302 , J\u0302 , K\u0302 unit vectors of the wheel-body frame C I identity matrix J1, J2, \u00b7 \u00b7 \u00b7 length function for calculating \u03b83 l\u0302c unit vector on the line (xT , yT , zT ) to (xW , yW , zW ) ms sprung mass mu unsprung mass n\u03021 normal unit vectors to \u03c0L n\u03022 normal unit vectors to \u03c0C P point q, p, f parameters for calculating couple point coordinate r position vector Rw tire radius TRW rotation matrix to go from W frame to T frame s position vector of the steer axis sa forward location of the steer axis sb lateral location of the steer axis s\u030cW (0, \u03b4, u\u0302, s) zero pitch screw about the steer axis T (xt, yt, zt) tire coordinate system TTW homogeneous transformation to go from W to T u\u0302 steer axis unit vector u\u0303 skew symmetrix matrix associated to u\u0302 uC position vector of the coupler point u\u0302z unit vector in the z-direction vx forward speed x, y suspension coordinate frame xC , yC coordinate of a couple point (xT , yT , zT ) wheel-body coordinates of the origin of T frame 8. Suspension Mechanisms 511 (xW , yW , zW ) wheel-body coordinates of the origin of W frame vers \u03b4 1\u2212 cos \u03b4 W (xwywzw) wheel coordinate system z vertical position of the wheel center z0 initial vertical position of the wheel center \u03b1 angle of a coupler point with upper A-arm \u03b3 camber angle \u03b4 steer angle \u03b5 = ms/mu sprung to unsprung mass ratio \u03b8 lean angle \u03b80 angle between the ground link and the z-direction \u03b8i angular position of link number i \u03b82 angular position of the upper A-arm \u03b83 angular position of the coupler link \u03b84 angular position of link lower A-arm \u03b8i0 initial angular position of \u03b8i \u03c0C caster plane \u03c0L lean plane \u03c5 trust angle \u03d5 caster angle \u03c9 angular velocity 512 8. Suspension Mechanisms Exercises 1. Roll center. Determine the roll ceneter of the kinematic models of vehicles shown in Figures 8.56 to 8.59. 2 1 Body 4 6 3 5 8 7 8. Suspension Mechanisms 513 514 8. Suspension Mechanisms 8. Suspension Mechanisms 515 4. F Position of the roll center and mass ceneter. Figure 8.66 illustrates the wheels and mass center C of a vehicle. Design a double A-arm suspension such that the roll center of the C 516 8. Suspension Mechanisms c d a 2\u03b8 3\u03b8 4\u03b8 A B M N x y 0\u03b8 \u03b1 e z Cb 8. Suspension Mechanisms 517 Determine CTW for \u03d5 = 8deg, \u03b8 = 12deg, and the location vector Cs Cs = \u23a1\u23a3 3.8 cm 1.8 cm \u2212Rw \u23a4\u23a6 . (a) The vehicle uses a tire 235/35ZR19. (b) The vehicle uses a tire P215/65R15 96H. 10. F Wheel drop. Find the coordinates of the tireprint for \u03d5 = 10deg \u03b8 = 10deg Cs = \u23a1\u23a3 3.8 cm 1.8 cm 38 cm \u23a4\u23a6 if \u03b4 = 18deg. How much is the wheel drop H. 11. F Wheel drop and steer angle. Draw a plot to show the wheel drop H at different steer angle \u03b4 for the given data in Exercise 10. 12. F Camber and steering. Draw a plot to show the camber angle \u03b3 at different steer angle \u03b4 for the following characteristics: \u03d5 = 10deg \u03b8 = 0deg Cs = \u23a1\u23a3 3.8 cm 0 cm 38 cm \u23a4\u23a6 Part III Vehicle Dynamics 9"
        ]
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure13.76-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure13.76-1.png",
        "caption": "Figure 13.76 (a) Normal force diagram.",
        "texts": [
            " The maximum bending moment in DE is slightly to the left of the middle G of ED, and will be only marginally larger than MG. From the area under the V diagram we find Mmax = 1 2 \u00d7 21.5 \u00d7 21.5 21.5+24.5 \u00d7 5 \u2212 12.5 = 12.62 kNm ( ). The statically indeterminate structure in Figure 13.77 has a hinged joint at S. All other joints are rigid. Dimensions and loads are given in the figure. The shear forces directly next to joint C are given: V BC C = 2.5 \u221a 2 kN, V CS C = 10 kN. 13 Calculating M, V and N Diagrams 605 Figure 13.77 Statically indeterminate frame with two given shear forces directly next to joint C. Figure 13.76 (b) Shear force diagram and (c) bending moment diagram. 606 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM The directions follow from the deformation symbols given in the figure. Questions: a. Determine the degree of static indeterminacy of the structure. b. Draw the force polygon for the force equilibrium of joint C. c. Determine the support reactions at A and D. d. Determine the M , V and N diagrams for the entire structure, with the deformation symbols. Solution (units kN and m): a. The two parts ABCS and DS provide e = 2 \u00d7 3 = 6 equilibrium equations (see Figure 13"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000138_6.2007-6461-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000138_6.2007-6461-Figure4-1.png",
        "caption": "Figure 4. Free body diagram of a quadrotor helicopter. The roll, pitch and yaw angles (\u03c6, \u03b8, and \u03c8, respectively) are controlled by differential thrust. Differential thrust between opposite motors provides roll and pitch torques. Differential thrust between the two pairs of counter-rotating motors provides yaw torque. Position control, with respect to the North-East-Down (NED) coordinate frame is accomplished by controlling the magnitude and direction of the total thrust. A drag force, Db, also acts on the vehicle, opposite the velocity direction, eV.",
        "texts": [
            " Numerous additional sensors have been tested on the STARMAC platform, including the Videre Systems stereo vision camera,35 the Hokuyo-URG laser range finder36 and the Tracker DTS digital avalanche beacon and receiver.37 The system is designed to function as an independent sensing and computing platform, to enable the vehicles to perform multi-agent missions, such as cooperative search and rescue.1 IV. Vehicle Dynamics The derivation of the nonlinear dynamics is performed in North-East-Down (NED) inertial and body fixed coordinates. Let {eN, eE, eD} denote unit vectors along the respective inertial axes, and {xB,yB, zB} denote unit vectors along the respective body axes, as defined in Figure 4. Euler angles of the body axes are {\u03c6, \u03b8, \u03c8} with respect to the eN, eE and eD axes, respectively, and are referred to as roll, pitch and yaw. The current velocity direction unit vector is ev, in inertial coordinates, and defines coordinates relative to the c.g. referred to as longitudinal, lateral and vertical. The rotor plane does not necessarily align with the xB, yB plane, so for the ith rotor let {xR,i,yR,i, zR,i} denote unit vectors aligned with the plane of the rotor and oriented with respect to the lateral, longitudinal, and vertical directions as shown in Figure 5",
            " The roll, pitch and yaw angles are controlled by differential thrust. Differential thrust between opposite motors provides roll and pitch torques. Differential thrust between the two pairs of counter-rotating motors provides yaw torque. To decouple the control, motors 1 and 3 rotate in the opposite direction of rotors 2 and 4. The vehicle body drag force is defined as Db, vehicle mass is m, acceleration due to gravity is g, and the inertia matrix is Ib \u2208 R3\u00d73. A free body diagram is depicted in Figure 4, with a depiction of the rotor forces and moments in Figure 5. The total force, F, can be summed as, F = \u2212Dbev +mgeD + 4\u2211 i=1 (\u2212TiRRi,Ib zR,i) (8) where RRi,I is the rotation matrix from the plane of rotor i to inertial coordinatesa. Similarly, the total moment, M, is, M = 4\u2211 i=1 (Mi + ri \u00d7 (\u2212TiRRi,BzR,i)) (9) aThe notation RA,B shall refer to rotation matrices from coordinate system A to B throughout. 8 of 20 American Institute of Aeronautics and Astronautics where RRi,B is the rotation matrix from the plane of rotor i to body coordinates"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003558_978-1-4020-2249-4-Figure10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003558_978-1-4020-2249-4-Figure10-1.png",
        "caption": "Figure 10. Subfamily ROQo Jl!. IJO J Figure II.Subfamily QO J l!. HO J RO",
        "texts": [
            " it has only two IKS), with a hole (Fig. 3). All other types are quaternary (Le. with four IKS). The second type is a quaternary manipulator (like the remaining three types) with four cusps on the internal boundary. There are three instances of type 2 according to the number of nodes, as will be shown in section 3 .3. The first one is shown in Fig. 4 with two nodes. The second one was shown in Fig. 2 with no node. The last one is such that d3~c4 Le. with the two singular lines 8.3=\u00b1arccos(-d3/ c4) . It will be shown in Fig. 10 (section 3.3) . -2 0 0.5 1 1.5 2 fJ..[rad) p d:= 1. </,,=0.7. d,=O.). /\":=0.2 \u00b72 0 f/..[rad) 2 3 d:= I. dl = 1.5. d,=0.7. /\":=0.5 p Figure 3. Type 1 manipulator. Figure 4 . Type 2 manipulator. Transition between type 1 and type 2 is a manipulator having a pair of points with four equal IKS, where two nodes and one cusp 223 coincide. Proof of this result can be obtained as a straightforward consequence of transitions in quartics root patterns (Baili, 2003) . Deriving the condition for the inverse kinematic polynomial to have four equal roots yields the equation of the separating surface (Baili, 2003) ( d2 22 d2 2) d - ",
            "2, a third topology exists in this domain, where the internal boundary exhibits a '2-tail fish'. This workspace topology, which we call WT4 , features two nodes like in Fig. 4, but these nodes do not play the same role. They coincide with two isolated singular points, which are associated with the two singularity lines defined by B.3=\u00b1arccos(-d3/ c4) (the operation point lies on the second joint axis and the inverse kinematics admits infinitely many solutions). Also, the nodes do not bound a hole like in Fig. 4 but a region with four IKS (Fig. 10). Transition between WT3 and WT4 is a workspace topology such that the upper and lower segments of the internal boundary meet tangentially (Fig. 11). As shown in (Baili, 2003), this transition is the occurrence of the additional singularity d3 + c3c4 = O. This transition is defined by: (11) Domains 3 and 5: The internal boundary has either 2 cusps (domain 3) or 0 cusp (domain 5). This boundary is either fully inside the external boundary (like in Figs 5 and 7), or it can cross the external boundary, yielding two nodes like in Fig",
            "J6)+ 6(a sincp + scoscp). For the mechanism in Fig. 8 we find F = 6 + 3- 6 x 1 = 3 and for the Do decahedral Zig-Zag Linkage (Fig.9): F = I./; -61 = (20x6+30)-6x31 = -46. 359 For thkosahedral Zig-Zag Linkage we use the Flat Ring Articula tion. With the {a,s,d,t,{3 = arccos[(-3+,j5)/~(20-8,j5) -n/2 - 31, 71\u00b0}and h(cp) = acoscp-ssincp the edge length ofthe icosahedron can be calculated: L( cp) \"\" 2[(1 I 2) + h( cp )cos( n I 5)]/[ cos{3 sine n I 5)] + 6(a sin cp + s cos cp). The mobility of the mechanism shown in Fig.10 (Flat Ring Articulation and 5 Nuremberg scissors) is found by F = ~/; - 6/ .. (10+ 5)- 6x 1 = 9. For the Ikosahedral Zig-Zag Linkage (Figs. 11,12) we obtain with ~/; = j+30= 12xIO+30 and / = 31 from F \"\" ~/; - 6/ = -36, which indicates overconstraintness. This linkage is movable with one degree of freedom which follows from the fact that for any value of the position angle cp the 30 Nuremberg scissors can be built up with the 12 x 5 = 60 gussets. Agraval, S.K., Yim, M.,Suh, W. (2001), Polyhedral Single Degree of Freedom Ex panding Structures, Proceedings of IEEE, Seoul,Korea",
            " Then the sequence PP RO becomes PPRo, which is equivalent to a planar pair G. We obtain a new family ROflO jG employing the equivalencies of G. In a limb of the family ROflO JG the H pair axis passes through the fixed point 0 and is parallel to the plane of G. The limbs of the family ROflO JG, are deducted by replacing G by equivalent generators of planar gliding motion (figure 9). 379 5.3 Family ROCo Jp HD J and family Cop HD JRO In a limb of the family RORopP EP J, the sequence ROE. may become (RO l!J equivalent to QO J and, we obtain the subfamily ROQO J E. EP J (figure 10 left) and its special subfamily ROEP JpP EP J (figure 10 right) derived from RO(EP I\u00a3)E.EP J(figure 10 midle). By the same way, a limb of the type ROPPEP JRo, may be (ROE.) E.EP JRo, which leads to the subfamily QO JE.EP JRo (figure 11). The paper describes new special SPM limbs with no inactive pair nor passive motion. In the new special 5-dof limbs, two generators of two dependent displacement Lie subgroups can be identified and the passive motion that results of the dependency is eliminated. Furthermore particular geometrical arrangements yield several subfamilies of the limb main families. Herve, J"
        ],
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    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure11.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure11.2-1.png",
        "caption": "Figure 11.2 (b) to (d) Over a distance x in the x direction the section forces increase by amounts N , V and M respectively. (e) The section forces on the right-hand sectional plane are then N + N , V + V and M + M .",
        "texts": [
            " Since no misunderstanding is possible, we will omit the index z in Mz and Vz to simplify the writing. 432 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM The differential equations for equilibrium are derived from the equilibrium of a small member segment with a length x that approaches zero ( x \u2192 0). We assume that the displacements due to the deformation of the member are negligible small. Therefore the equilibrium, including that of a small member element, can be related to the non-deformed geometry (see Figure 11.1). In Figure 11.2a, a small segment with length x has been isolated from a member and greatly magnified. The member segment is subjected to qx and qz. The loads act on the member axis (this has not been drawn as such for qz for the sake of clarity). If length x of the member segment is sufficiently small ( x \u2192 0), the distributed loads qx and qz can be considered uniformly distributed. The (unknown) section forces on the left and right-hand sectional planes are shown in their positive direction. The section forces are a function of x, the location of the cross-section, and are generally different in both sectional planes. In Figure 11.2a, it is assumed that the forces on the left-hand section are N , V and M . If the section forces increase over a distance x in the x direction by amounts N , V and M , respectively (see Figures 11.2b to 11.2d), the forces on the right-hand sectional plane are then N + N , V + V and M + M (see Figure 11.2e). From the force equilibrium of the small member segment it follows that \u2211 Fx = \u2212N + (N + N) + qx x = 0, (a) \u2211 Fz = \u2212V + (V + V ) + qz x = 0. (b) From the moment equilibrium it follows that (we have selected the moment 11 Mathematical Description of the Relationship between Section Forces and Loading 433 sum about point A on the right-hand sectional plane) \u2211 Ty |A = \u2212M \u2212 V x + (M + M) + qz x \u00b7 1 2 x = 0. (c) With the three equilibrium conditions (a) to (c) this gives N + qx x = 0, V + qz x = 0, M \u2212 V x + 1 2qz( x)2 = 0. After dividing by x we find N x + qx = 0, V x + qz = 0, M x \u2212 V = \u2212 1 2qz x. N/ x is the increase in the normal force per length in the x direction (see Figure 11.2b). In the limit x \u2192 0 this is known as the derivative from N with respect to x and is written dN/dx: lim x\u21920 N x = dN dx . In the same way: lim x\u21920 V x = dV dx , 434 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM lim x\u21920 M x = dM dx . The three equations for the equilibrium of an elementary member segment with length x in the limit x \u2192 0 are dN dx + qx = 0, dV dx + qz = 0, dM dx \u2212 V = 0. In the last equation, with x \u2192 0, the term 1 2qz x has disappeared. This is justified since the contribution by qz in the equation for the moment equilibrium is one order smaller than the contribution of the other terms"
        ],
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    {
        "image_filename": "designv10_0_0003628_91.705510-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003628_91.705510-Figure2-1.png",
        "caption": "Fig. 2. Membership functions for input s.",
        "texts": [],
        "surrounding_texts": [
            "In this section, we follow the development established in [15] and show that a particular fuzzy controller is an extension of an SMC with a boundary layer [13]\u2013[15]. Suppose the fuzzy controller in this article is constructed from the following IF-THEN rules: If is NB, then is BIGGER If is NM, then is BIG If is ZR, then is MEDIUM If is PM, then is SMALL If is PB, then is SMALLER or equivalently If is , then is where NB is negative big, NM is negative medium, ZR is zero, PM is positive medium, and PB is positive big. NB, NM, , SMALL, SMALLER are labels of fuzzy sets and their corresponding membership functions are depicted in Figs. 2 and 3, respectively. Let and be the input and output space of the fuzzy rules. For any arbitrary fuzzy set in , each rule can determine a fuzzy set in . Use the sup\u2013min compositional rule of inference (11) It can be further simplified by supposing as a fuzzy singleton, i.e., only at it\u2019s support , otherwise, . Then (11) becomes (12) and the deduced membership function of the consequences of all rules is (13) the crisp output is obtained by the center-of-area defuzzifier (14) Fig. 4 is the result of the defuzzified output for a fuzzy input , and has the following form: (15) where if if if if if if (16) IV. IDEAS BEHIND THE DECOUPLED CONTROL The sliding-mode controller design can be applied to the system represented by the canonical form (1). But for such nonlinear models as a cart-pole system, the system dynamic representation is not in the canonical form exactly. Rather, it has a form shown below (17) where is the state vector, , , and , are nonlinear functions, is the control input, and , are external disturbances. The disturbances are assumed to be bounded as , . Define one sliding surface as (18) another sliding surface as (19) then according to the sliding-mode theory presented in the previous section, we can choose a control law as (20) where or (21) where Obviously, if is used to control the system, only state and along the hyperplane will approach to zero asymptotically. If is used to control the system, only state and along the hyperplane will slide to zero asymptotically. In other words, the sliding-mode controller can only control either pole or cart for a system such as (17). Although a nonlinear system described by (17) can be transformed into a canonical form if there exists a diffeomorphism [20] that transforms the system into the canonical form as (1), the mathematical manipulation is too complicated and tedious to work with. To deal with the problem just mentioned, we develop a controller called decoupled fuzzy sliding-mode controller, which can control both pole and cart around zero without transforming (17) into the canonical form. Without loss of generality, it is possible to consider nonlinear systems of the form given by (17). The control objective is to force the system state to the origin equilibrium point. More specifically, a control objective is considered: to determine a feedback control (based on the decoupled fuzzy sliding-mode controller) such that the closed-loop system is globally stable in the sense that all the state variables are uniformly bounded and converge to its equilibrium asymptotically. The control objective of a sliding-mode controller is to let become zero the sliding surface in (2), i.e., , . If the definition of the sliding surface is changed, the control objective changes as well and the control force calculated from (9) will be different accordingly. The main idea of the decoupled sliding-mode controller proposed in this paper can be structured as follows. \u2022 Decouple the whole system into two subsystems A and B. System A contains , and its corresponding sliding surface . System B contains , and its corresponding sliding surface . By doing so, the main control objective is to keep states of system A move toward the surface and converge to asymptotically and a subtarget is to keep states of system B moving toward the surface and sliding to asymptotically. \u2022 Because the main target is to stabilize system A, it is reasonable to consider the information from system B as secondary and this secondary information must be reflected through a mechanism to the primary. An intermediate variable , which represents secondary information, is incorporated into and, therefore, the sliding surface is modified to take the form . This modified reflects the fact that the main target is changed from , to , where is a function of . Notice that the subtarget of system B is embedded to the main target through the variable . Both primary and secondary targets can be controlled simultaneously. \u2022 represents a decaying oscillation signal whose value is assigned through (27) and its interpretation is described in what follows. Having those ideas in mind, now is rewritten as (22) Where is a value transferred from , it has a value proportional to and has the range proper to . Equation (22) means the control objective of is changed from , to , . is still defined as (23) Let the control input be the sliding-mode controller of (20) (24) where (25) Because the sliding-mode controller is used to control the whole system, the boundedness of can be guaranteed by letting (26) where is the upper bound of . Equation (26) means the maximum absolute value of will be limited. Summarizing what we have mentioned above, can be defined as (27) where is the boundary layer of to smooth , transfers to the proper range of , and the definition of function is the same as (10). Notice that is a decaying oscillation signal because is a factor less than one. The control sequences satisfy the following: when , then there is a in (22) which causes (24) to generate a control that reduces . As decreases, decreases, too. When , then , then , then and the control objective can be achieved. To recap briefly, by introducing an intermediate variable in , we decouple the system into two levels. And by changing the output of properly, we gradually reduce and to zero in the same time. The structure of two-level decoupled control is shown in Fig. 5."
        ]
    },
    {
        "image_filename": "designv10_0_0000554_978-3-540-85629-0-Figure8.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000554_978-3-540-85629-0-Figure8.6-1.png",
        "caption": "Fig. 8.6. Change of the orientation.",
        "texts": [
            "2 Orientation of the Tool The orientation R of the end effector6 may be expressed by means of a rotation matrix composed by three orthogonal unit vectors (called in robotics normal, slide, and approach) defining the orientation of the tool with respect to the base world frame: R = [n, s, a]. Therefore, in order to specify the orientation of the end effector in different points of the trajectory, a desired rotation matrix Rk must be specified at each point pk and a proper interpolating technique must be used among them, see Fig. 8.6. 8.2.1 Case of independent position and orientation In defining a trajectory, it is in general not convenient to adopt rotation matrices for the interpolation of the orientation among given poses. As a matter of fact, by interpolating the three unit vectors n, s,a, from the initial value corresponding to the rotation matrix R0 to the final one R1, it is not possible to guarantee that the orthonormality conditions7 hold in every instant. 6 An end effector is the (generic) tool carried by a robotic arm or the operating tool of a multi-axis machine"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000459_1521-4109(200108)13:12<983::aid-elan983>3.0.co;2-#-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000459_1521-4109(200108)13:12<983::aid-elan983>3.0.co;2-#-Figure4-1.png",
        "caption": "Fig. 4. A needle-type glucose biosensor for subcutaneous monitoring (reproduced with permission [14]).",
        "texts": [
            " Problems with biocompatibility have proved to be the major barriers to the development of reliable implantable devices. Most glucose biosensors lack the biocompatibility necessary for a prolonged and reliable operation in whole blood. Alternative sensing sites, particularly the subcutaneous tissue, have thus received growing attention. While the above issues represent a major challenge, significant progress has been made towards the continuous monitoring of glucose. Most of the recent attention has been given to the development of subcutaneously implantable needle-type electrodes (Fig. 4) [14\u201316]. Such devices are designed to operate for a few days and be replaced by the patient. Success in this direction has reached the level of short-term human implantation; continuously functioning devices, possessing adequate (>1 week) stability, are expected in the near future. Such devices would enable a swift and appropriate corrective action (through a closed-loop insulin delivery system, i.e., an artificial pancreas). Algorithms correcting for the transient difference (time lag) between blood and tissue glucose concentrations have been developed [16]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.25-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.25-1.png",
        "caption": "FIGURE 7.25. A lever arm steering system.",
        "texts": [
            "24 illustrates a sample rack-and-pinion steering system. The rack is either in front or behind the steering axle. The driver\u2019s rotary steering command \u03b4S is transformed by a steering box to translation uR = uR (\u03b4S) 404 7. Steering Dynamics of the racks, and then by the drag links to the wheel steering \u03b4i = \u03b4i (uR), \u03b4o = \u03b4o (uR). The drag link is also called the tie rod. The overall steering ratio depends on the ratio of the steering box and on the kinematics of the steering linkage. Example 274 Lever arm steering system. Figure 7.25 illustrates a steering linkage that sometimes is called a lever arm steering system. Using a lever arm steering system, large steering angles at the wheels are possible. This steering system is used on trucks with large wheel bases and independent wheel suspension at the front axle. The steering box and triangle can also be placed outside of the axle\u2019s center. Example 275 Drag link steering system. It is sometimes better to send the steering command to only one wheel and connect the other one to the first wheel by a drag link, as shown in Figure 7"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001599_iros.2010.5648837-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001599_iros.2010.5648837-Figure7-1.png",
        "caption": "Fig. 7. Lifting a heavy object requires both a task force and task torque to compensate for the gravitational force on the object.",
        "texts": [
            " Figure 5 shows the result of the balance controller when the robot is pushed from behind at the middle of the torso with and without posture control. Without posture control, the torso is underdamped and has a large deviation from upright. This is corrected by torso posture control with Kp = 150 and Kd = 150. The robot is also able to handle much larger pushes with this controller. Photos of the robot during this experiment are shown in Figure 6. This task further demonstrates the capabilities of the DBFC-VMC controller. When lifting a heavy object as shown in Figure 7, DBFC-VMC can be used to generate the compensating joint torques. The VMC task controller can be defined to regulate the position of the object, Flift = Kp ( P object des \u2212 P object ) (29) Now (21) takes the form[ D1 D2 ] F = ( mC\u0308des + Fg + Flift( P object \u2212 C ) \u00d7 Flift ) (30) The mass of the object is not needed for VMC control, but could be implemented in a feedforward manner if known, F\u0302lift = Flift + mobjectg. In addition to the task force, a task torque is required because the gravitational force on the object applies a torque about the COM",
            " This means that any angular momentum added to the system must also be removed quickly. Coordinating this tradeoff is difficult and the subject of future work. In this paper, the desired behavior is to minimize the generation of angular momentum by setting the desired change of angular momentum to zero, L\u0307des = 0. In fact, it is generally useful to dissipate angular momentum by setting L\u0307des = \u2212KpL. The DBFC-VMC controller can accommodate a wide variety of tasks. It also inherits from traditional VMC the ability to define tasks in a model-free way. As shown in Figure 7, a model of the object isn\u2019t needed, only a definition of a controller to grasp it. However, if a model is provided, DBFC-VMC can take advantage of the model to achieve improved control. It can be easily extended to related force control tasks like cooperative manipulation as shown in Figure 15. A slightly modified version of the task in Figure 7 would include grasping forces, which are internal to the system. To grasp the object steadily, both arms could apply an equal and opposite force. This force has no effect on the ground contact forces but does affect the joint torques in the two arms. The lifting example also highlights a possible problem with the DBFC-VMC as presented. If the object is heavy enough to tip the robot over, adjustments such as leaning backwards or moving a foot to reshape the base of support may be required [18]. However, with proper placement of the COM and feet, DBFC is a simple choice for determining full body joint torques"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003568_ac000220v-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003568_ac000220v-Figure2-1.png",
        "caption": "Figure 2. Cyclic voltammograms in 0.5 M KHCO3 containing 1 mM CySH: initial potential, 0.0 V (SCE); sweep rate, 20 mV s-1. (a) BDD electrode (five consecutive sweeps); (b) GC electrode (the numbers indicate the first, second, and third sweeps).",
        "texts": [
            "57 V), whereas that for GC tends to a plateau. A lower peak current was also noticed in the case of GC (for the same electrode area), in agreement with the poor activity for CySH oxidation previously reported for this material.8,10 No cathodic peaks were observed on the reverse scan within the investigated potential range (-1.3 to +1.1 V). This was not surprising, because it was ascertained that CySH oxidation is an electrochemically irreversible process,27 and there are oxidation products in addition to cystine.8 Figure 2 shows voltammograms recorded during consecutive runs (sweep rate, 20 mV s-1) in a 1 mM CySH solution both for BDD (Figure 2a) and for GC (Figure 2b). The solution was mixed between consecutive runs and was allowed to stand for 1 min without mixing before starting each scan. It can be seen that, unlike BDD, CySH oxidation results in a deactivation of the GC electrode, the reaction being suppressed after the first run. This behavior indicates that the GC surface is blocked by reaction products. The variation of the sweep rate for the BDD electrodes showed that peak currents are linearly proportional to the square root of the sweep rate within the range 5-300 mV s-1"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003391_5.663540-Figure26-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003391_5.663540-Figure26-1.png",
        "caption": "Fig. 26. Nondevice daughter cards and extender cards, including cards to add more FPGA logic (lower left), bandwidth (double-sized card in the second row), long-distance communication (third row), and edge bandwidth (two cards at front right). All but the edge bandwidth cards have FPGA\u2019s on the bottom.",
        "texts": [
            " The base plates are constructed such that they can be connected together, forming an arbitrarily large surface for placing daughter cards. In many ways, this approach is similar to both mesh-connected multiprocessors and the approach suggested in [119]. As in most other FPGA systems, there is the potential that Springbok\u2019s simple connection scheme will not be able to accommodate all of the logic or routing assigned to a given location. As opposed to a fixed multi-FPGA system, however, Springbok can insert new \u201cextender\u201d cards between a daughter card and the base plate to deal with these problems (Fig. 26). For example, if the logic assigned to a given FPGA simply will not fit, an extender card with another FPGA can be inserted so that it can handle some of the logic. If too many signals need to be routed along a given link in the mesh structure, an extender card spanning several daughter-card positions can be added, with new routing paths included on the inserted card. For signals that must go long distances in the array, sets of extender cards with ribbon-cable connections can be inserted throughout the array to carry these long-distance wires"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure3.36-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure3.36-1.png",
        "caption": "FIGURE 3.36. Bottom view of a laterally deflected tire.",
        "texts": [
            " When a wheel is under a constant load Fz and then a lateral force is applied on the rim, the tire will deflect laterally as shown in Figure 3.35. The tire acts as a linear spring under small lateral forces Fy = ky\u2206y (3.135) with a lateral stiffness ky. The wheel will start sliding laterally when the lateral force reaches a maximum value FyM . At this point, the lateral force approximately remains constant and is proportional to the vertical load FyM = \u03bcy Fz (3.136) 3. Tire Dynamics 137 where, \u03bcy is the tire friction coefficient in the y-direction. A bottom view of the tireprint of a laterally deflected tire is shown in Figure 3.36. If the laterally deflected tire is turning forward on the road, the tireprint will also flex longitudinally. A bottom view of the tireprint for such a laterally deflected and turning tire is shown in Figure 3.37. Although the tire-plane remains perpendicular to the road, the path of the wheel makes an angle \u03b1 with tire-plane. As the wheel turns forward, undeflected treads enter the tireprint region and deflect laterally as well as longitudinally. When a tread moves toward the end of the tireprint, its lateral deflection increases until it approaches the tailing edge of the tireprint"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000520_iros.2006.282433-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000520_iros.2006.282433-Figure1-1.png",
        "caption": "Fig. 1. Quadrotor helicopter",
        "texts": [
            " The backstepping technique will be used to control the z motion and the yaw angle of the fully-actuated subsystem. The paper is organized as follows: in section II, a dynamic model for a quadrotor helicopter is developed. Based on this nonlinear model, we design in section III a backstepping control law. In section IV, some simulations are carried out to show the performance and stability of the proposed controller. Finally, in section V, our conclusions are presented. II. DYNAMIC MODELING OF A QUADROTOR HELICOPTER The quadrotor helicopter is shown in figure 1. The two pairs of rotors (1, 3) and (2, 4) turn in opposite direction in order to balance the moments and produce yaw motions as needed. On varying the rotor speeds altogether with the same quantity the lift forces will change affecting in this case the altitude z of the system. Yaw angle is obtained by speeding up the clockwise motors or slowing down depending on the desired angle direction. The motion direction according (x, y) axes depends on the sense of tilt angles (pitch and roll) whether they are positive or negative. 1-4244-0259-X/06/$20.00 \u00a92006 IEEE The equations describing the attitude and position of a quadrotor helicopter are basically those of a rotating rigid body with six degrees of freedom [6] [7]. They may be separated into kinematic equations and dynamic equations [8]. Let be two main reference frames (see Fig. 1): \u2022 the earth fixed inertial reference frame Ea : (Oa, \u2212\u2192 ea1 , \u2212\u2192 ea2 , \u2212\u2192 ea3). \u2022 the body fixed reference frame Em : (Om, \u2212\u2192 em1 , \u2212\u2192 em2 , \u2212\u2192 em3 ) regidly attached to the aircraft. Let the vector \u03b6 [x, y, z]T and \u03b7 [\u03c6, \u03b8, \u03c8]T denote respectively the altitude positions and the attitude angles of the quadrotor (frame Em) in the frame Ea relative to a fixed origin Oa. The attitude angles {\u03c6, \u03b8, \u03c8} are respectively called pitch angle (\u2212\u03c0 2 < \u03c6 < \u03c0 2 ), roll angle (\u2212\u03c0 2 < \u03b8 < \u03c0 2 ) and yaw angle (\u2212\u03c0 \u2264 \u03c8 < \u03c0)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure3.5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure3.5-1.png",
        "caption": "FIGURE 3.5. SAE tire coordinate system.",
        "texts": [
            " For a cambered tire, it is not always possible to find or define a center point for the tireprint to be used as the origin of the tire coordinate frame. It is more practical to set the origin of the tire coordinate frame at the center of the intersection line between the tire-plane and the ground. So, the origin of the tire coordinate frame is at the center of the tireprint when the tire is standing upright and stationary on a flat road. Example 76 SAE tire coordinate system. The tire coordinate system adopted by the Society of Automotive Engineers (SAE) is shown in Figure 3.5. The origin of the coordinate system is at the center of the tireprint when the tire is standing stationary. The x-axis is at the intersection of the tire-plane and the ground plane. The z-axis is downward and perpendicular to the tireprint. The y-axis is on the ground plane and goes to the right to make the coordinate frame a righthand frame. The sideslip angle \u03b1 is considered positive if the tire is slipping to the right, and the camber angle \u03b3 is positive when the tire leans to the right. The SAE coordinate system is as good as the coordinate system in Figure 3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure5.30-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure5.30-1.png",
        "caption": "Figure 5.30 Three-hinged frame ASB and rod AB isolated from one another, assuming that a tensile force N acts in rod AB.",
        "texts": [
            "29a, the structure has been isolated from its supports. The support reactions follow from the equilibrium of the structure as a whole. For the directions assumed for Ah, Av and Bv we find 174 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM \u2211 F (ASB) x = 40 \u2212 Ah = 0 \u21d2 Ah = 40 kN, \u2211 F (ASB) y = \u221260 \u2212 60 + Av + Bv = \u221260 \u2212 60 + Av + 70 = 0 \u21d2 Av = 50 kN, \u2211 T (ASB) z |A = \u221240 \u00d7 2 \u2212 60 \u00d7 2 \u221260 \u00d7 6 + Bv \u00d7 8 = 0 \u21d2 Bv = 70 kN. The support reactions are shown in Figure 5.29b. b. To calculate the force in rod AB, it is isolated from ASB in Figure 5.30. We can immediately recognise a two-force member in rod AB: the rod is loaded only by forces at its ends A and B and can therefore be in equilibrium only if these forces are equal and opposite with AB as common line of action. It is assumed that a tensile force N acts in rod AB. The magnitude of N follows from the moment equilibrium about S of one of the parts AS or BS. In Figure 5.31a both parts have been isolated at S. In order to simplify the calculation, N has been resolved into a horizontal component Nh and a vertical component Nv: Nh = 2 5 \u221a 5 N, Nv = 1 5 \u221a 5 N"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000091_s00170-015-7077-3-Figure11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000091_s00170-015-7077-3-Figure11-1.png",
        "caption": "Fig. 11 Schematic of two deposition techniques [81]",
        "texts": [
            " A lot of research strategies have been investigated on thermal stresses analysis of material deposition of multi-pass single-layer structure by altering deposition patterns, deposition sequences and preheating or interpass cooling. The selective deposition approach, which deposits a series of small patches (\u201ctowers\u201d) first and joins them joined together later to form a large patch, has been used to minimise the deformation due to thermal stresses for shape deposition manufacturing [81]. As shown in Fig. 11, these \u201ctowers\u201d are only constrained at the bottom and their large surface area to volume ratio allows them to relax significantly as they cool down. Selective deposition can reduce deformation of the final part by allowing much of the deformation produced during cooling to occur before the deposit is fully constrained. The influence of deposition patterns, such as long raster and spiral, on the deflection of the metal part was been studied by Nickel et al. [82, 83] using a combination of finite element analysis and experiments as shown in Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002537_j.matdes.2020.108818-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002537_j.matdes.2020.108818-Figure3-1.png",
        "caption": "Fig. 3. X-ray diffraction setup.",
        "texts": [
            " Due to attenuation, the penetration depth of x-ray in the diffraction method is minimal, about 5 \u03bcm for Inconel 718; therefore, it can be used only to measure surface RS, which are considered plane stresses [4]. A Bruker D8 instrument with a Co-K\u03b1 radiation source was used to measure the surface RS, installed with a 1 mm collimator. At first, several Bragg's angles (\u03b8) were scanned, and a 2theta angle of 111.4\u00b0 was selected as it gave themost definite diffraction peaks. The x-rays emitter was targeted at the center of the coupon top surface, as shown in Fig. 3, with the x-direction of measurement aligned with the scan tracks on the top surface (in scan direction). The y-direction of measurementwas oriented perpendicular to the scan tracks (in hatch direction). During measurement, the coupon is rotated through 5 Phi angles (\u03c6 = 0\u00b0, 72\u00b0, 144\u00b0, 216\u00b0, 288\u00b0) around its axis, to capturemore diffraction frames from one location. Besides, the coupon is tilted sideways to 4 Psi angles (\u03c8= 10\u00b0, 25\u00b0, 40\u00b0, 55\u00b0). Additionally, the coupon was oscillated within 2\u00b0 at each Psi angle to increase the amount of captured diffraction planes"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002652_j.jmapro.2019.12.009-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002652_j.jmapro.2019.12.009-Figure8-1.png",
        "caption": "Fig. 8. Schematic diagram of the cross section in the ideal overlapping state.",
        "texts": [
            " The processing procedures included coloring original model, defect analysis, removing isolated points, smoothing data, simplifying data, removing features, filling curvature and sharpening processing, as shown in Fig. 7. Finally, the model of the broken tooth was obtained by entities Boolean operation in a CAD software (SolidWorks2015). The model was converted to STL model and its printability was verified by a desktop 3D printer. According to Ref [19], the schematic diagram of the cross section in the ideal overlapping state is shown in Fig. 8. It can be assumed that the shape of the track cross section is an arc. In order to obtain a smooth layer surface, the area of the zone A1 should be equal to that of zone A2 as shown in Fig. 8. Therefore, it can be deduced that the area of the rectangle is equal to that of the track cross section, as given in Eq. (1). \u239c \u239f\u2219 \u2212 = \u239b \u239d + \u239e \u23a0 \u239b \u239d + \u239e \u23a0 \u2212 \u2212H W \u03b7 W H H HW W H W W H H(1 ) 8 2 arcsin 4 4 2 ( 8 2 )i 2 2 2 2 2 (1) Then the ideal optimal overlapping rate \u03b7i was calculated, as shown in Eq. (2) and (3). = \u2212 + \u00d7 + + \u00d7 \u2212arcsin \u03bb\u03b7i 1 ((\u03bb 4) (4 /(\u03bb 4)) 4 \u03bb (4 \u03bb ))/(64\u03bb)2 2 2 2 (2) = W H \u03bb (3) where W is the width, H is the height of the cladding track and \u03bb is the ratio of width to height of the cladding track",
            " Therefore, the optimal overlapping rate used in the repairing work is confirmed to be 20 %. It is found that the value of the optimal overlapping rate is less than that of the ideal overlapping rate. This phenomenon may be attributed to the scattering effect of the former cladding track on the laser and powder. In the ideal overlapping state, the material of the overlapping region should fill the gap between the two adjacent cladding tracks to form a smooth surface of the cladding layer, as shown in Fig. 8. Actually, powder and laser will be scattered when they reach the overlapping region. This will cause heat and mass loss [19]. Then the new formed track will partially grow on the overlapping region and induce the depression between the two adjacent cladding tracks. However, smaller overlapping rate will reduce the heat and mass loss and improve the smoothness of the cladding layer. Therefore, the value of the optimal overlapping rate is less than that of the ideal overlapping rate. Fig. 11 shows the top view of the cladding layer of the four specimens which are made by four different scanning strategies as described in Section 2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure8.37-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure8.37-1.png",
        "caption": "Fig. 8.37. Complex change of working zone",
        "texts": [
            " These robots possess a significant carrying capacity of up to 100 kg, due to the mass of the spot-welding gun with cables of 50-80 kg. They demonstrate high motion speeds of up to 1.6 mls due to fast cycle changes, along with high repeatability and positioning accuracy within the limits of 1 mm. The reach of the \"arm\" is usually 2.5 to 3 m, and they possess practically all six active degrees of freedom: three in the basic configuration and three in the spot-welding gun carrier, when fully orientated in space. That this is really necessary can be seen from Fig. 8.37 which shows a small but complex change of the work zone during the welding process of two adjacent window seams on a car body; two angular gripper positions can be seen from the figure, but bearing in mind that with practically all modern cars the rear side windows are oblique, a third angle change, i.e. degree of freedom, also becomes necessary. With this workprocess, which is usually on a transfer line, an extremely important role is played by synchronization of the work of the whole line, which can move in two modes: usually in pairs weld the points of the subject group in this welding phase"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000846_j.ymssp.2017.05.024-Figure15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000846_j.ymssp.2017.05.024-Figure15-1.png",
        "caption": "Fig. 15. Illustration of transmission path [24].",
        "texts": [
            " [226] compared the performance of two gearbox housing models. The first model was represented by a full FE mesh and the second model was replaced by a reduced model with condensed stiffness and mass matrices. A planetary gear set has several sun-planet and ring-planet gear pairs meshing simultaneously. In addition, carrier rotation may cause the distances between vibration sources and a transducer time varying. Therefore, transmission path of a planetary gearbox is more complicated than that of a fixed-axis gearbox. Fig. 15 shows three possible transmission paths for a sun-planet meshing, (a) meshing point \u2013 planet gear \u2013 housing \u2013 transducer, (b) meshing point \u2013 planet gear \u2013 planet gear bearing \u2013 carrier \u2013 carrier bearing \u2013 housing \u2013 transducer and (c) meshing point \u2013 sun gear \u2013 sun gear shaft \u2013 sun gear shaft bearing \u2013 housing \u2013 transducer. Inalpolat and Kahraman [23] and Feng and Zuo [24] simulated transducer perceived vibration signals of a planetary gearbox based on amplitude modulation and frequency modulation characteristics of vibration signals"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure11.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure11.3-1.png",
        "caption": "FIGURE 11.3 Example bar of magnetic material for reluctance computation.",
        "texts": [
            "18) since the L matrix is symmetrical L12 = L21. While it is clear for the self-inductances, we also see that the mutual coupling inductance is given by L12 = kN1N2. Also, the formula (11.17) is the conventional inductance for the series connected windings. INCLUSION OF PROBLEMS WITH MAGNETIC MATERIALS 289 Since the computation of the reluctance resistance is a key part of the magnetic circuit approach, we give a specific example for the computation of a piece of magnetic material. The flux direction is laminar from \ud835\udefc to \ud835\udefd as shown in Fig. 11.3. It is important to consider the details of computing the magnetic reluctance resistance (11.6), Rm = \u03a6m \u2215 \u03a8, or Rm = \u222b \ud835\udefd \ud835\udefc H \u22c5 d\ud835\udcc1 \u222bB \u22c5 d . (11.19) Finally, the reluctance resistance for the simple geometry in Fig. 11.3 is obtained by approximating (11.19) Rm1 = H1 \ud835\udcc11 B1 1 = \ud835\udcc11 \ud835\udf071 1 . (11.20) The last step is clearly dependent on a sufficiently high permeability \ud835\udf071 since the accuracy of the approximation depends on the condition that \ud835\udf071 \u226b 1. A key application of the magnetic circuit approach is the computation of the inductances for large permeability magnetic cores as we showed above. Further, more complex structures can be treated with a similar approach. The following example shows how the technique can be applied for a realistic case"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002281_j.mechmachtheory.2020.103838-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002281_j.mechmachtheory.2020.103838-Figure3-1.png",
        "caption": "Fig. 3. Modelling a spur gear tooth as non-uniform cantilever beam [21] .",
        "texts": [
            " 2 , deformation of a gear under an external load usually consists of tooth deformation ( K i t ), tooth filletfoundation deformation ( K i f ), tooth deformation due to coupling effects from the neighboring teeth ( K i f jk ) which was concerned in recent years, and the contact deformation of the tooth pair ( K h ). Here, the subscripts, i = p , g, denoting the pinion and the gear, respectively, and j and k are the tooth pair number in mesh. For the gear tooth deformation, it is usually regarded as a non-uniform cantilever beam which could facilitate the analytical calculation of the gear tooth deformation. The model for the tooth deformation calculation is shown in Fig. 3 . The tooth deformation is contributed by the bending, shear, and axial compressive deformations, and their corresponding stiffness are calculated as [20 , 21] , 1 K b = \u222b d 0 (x cos \u03b11 \u2212 h sin \u03b11 ) 2 E I x d x (1) 1 K s = \u222b d 0 1 . 2 cos 2 \u03b11 G A x d x (2) 1 K = \u222b d sin 2 \u03b11 E A d x (3) a 0 x where, the symbols, K b , K s , K a , denote the bending stiffness, shear stiffness, and axial compressive stiffness, respectively; d is the effective tooth length for integration; h represents half of the tooth thickness at the position that the mesh force is applied to; \u03b11 is the angle between the mesh force and the direction perpendicular to the tooth center line; E and G are the Young\u2019s modulus and shear modulus of the gear material, respectively"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure4.5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure4.5-1.png",
        "caption": "Figure 4.5 In a detailed study, a joint should be modelled as a spatial structural element.",
        "texts": [
            " It was stated above that the difference between the four kinds of structural elements is the result of the modelling method, and that this strongly depends on the information sought by the research or calculation. To illustrate, refer to the concrete bar structure and its model in Figure 4.2. The model has been created to investigate the mechanical behaviour of the structure as a whole. The lines in the diagram represent the beams and columns, which have been schematised as line elements. The beams and columns are rigidly joined to one another. In the model, these joints are represented as particle elements (capable of transferring both forces as well as concentrated couples). In Figure 4.5, the circled joint between the beam and outer column has been elaborated. Further investigation shows that there is a complex interplay of forces in the joint; the concrete transfers the compressive forces, and the reinforcement bars transfer the tensile forces. This type of investigation is critical for detailed modelling of the joint. Can the concrete transfer the compressive forces; how much reinforcement is required for transferring the tensile forces, and where should this reinforcement be placed"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure5.19-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure5.19-1.png",
        "caption": "FIGURE 5.19 Coupling pairs for two plane conductor cell pair.",
        "texts": [
            " If both conductors have the same cross sections for the symmetric situation where Lpkm = Lpk\u2032m\u2032 , then we can apply what we call a difference inductance as Ldkm\u2032 = (Lpkm \u2212 Lpkm\u2032 ), (5.41) to simplify the (5.40). The concept of difference inductances is also efficient for other applications such as multiple TLs or other double-layer plane inductances as is considered in the following section. Of course, the cell pairs must be located directly above each other with opposing currents. The general case for coupling Lkm between two cell pairs k and m is shown in Fig. 5.19. The coupling inductance can be computed as Lkm = Lpkm \u2212 Lpkm\u2032 + Lpk\u2032m \u2212 Lpk\u2032m\u2032 = 2 (Lpkm \u2212 Lpkm\u2032 ) = 2 Ldkm\u2032 (5.42) with the difference inductance given in (5.41). The evaluation of all the partial or difference inductances can be very time consuming. We consider simplification and speedup of the evaluation of partial inductances for the dense sections of the MNA circuit matrix. First, we observed that the coupling from section to section decays much faster than the 1\u2215R behavior of the conventional partial inductances where R is the center-to-center distance between the conductor cell pairs shown in Fig. 5.19. 112 INDUCTANCE COMPUTATIONS Hence, the difference inductance-based matrices can be sparsified, while this is not the case for matrices of conventional sparse matrices. We want to apply the simplification of the Ld matrices for the general case shown in Fig. 5.19. The geometry illustrated does not only consider the single TL inductance case where both cell pairs are along the x-axis but also the more general case for plane pairs that is treated in Section 5.7.4. The evaluation of the exact formulas for Lpkm is based on using conventional partial inductance formulas given in Appendix C. The basic inductance formulation for the two coupled cell pairs is given in (5.41). A second, fast approximate evaluation of (5.41) is based on the filament approximation given in (5",
            "45) This leads to the approximate partial inductance using (5.21) for distant conductors that couples between the planes at the two different levels Lpkm\u2032 \u2243 \ud835\udf070 4\ud835\udf0b \ud835\udcc1xk \ud835\udcc1xm Rkm \u221a 1 + d2 z R2 km \u2243 \ud835\udf070 4\ud835\udf0b \ud835\udcc1xk \ud835\udcc1xm ( 1 \u2212 d2 z 2 R2 km ) Rkm , (5.46) where in the last step we approximate the square root with a series expansion under the condition that dz \u226a R2 km. This leads to the approximate section-to-section coupling Lkm = 2 Ldkm = 2 (Lpkm \u2212 Lpkm\u2032 ) = \ud835\udf070 4\ud835\udf0b \ud835\udcc1xk \ud835\udcc1xm d2 z R3 km , (5.47) where the center-to-center distance Rkm in Fig. 5.19 is large compared to the plane-toplane distance dz. DIFFERENCE CELL PAIR INDUCTANCE MODELS 113 We give an example for the reduction of the error with cell distance in the approximate section-to-section inductance Lkm. In this example, the cells are square with \ud835\udcc1x = \ud835\udcc1y = 1 mm since square cells are representing a worst case, which is also important for the plane-pair models in the following section. In this example, the vertical cell spacing is 0.2 mm. The fast in-plane spacing decay is evident from Table 5",
            " Now, the equation derived in the previous section (5.47) Lkm gives the coupling for conducting pairs with opposite currents as is the case in TLs, which is based on the difference inductances. We apply the difference coupling equations to the TL formulation in Section 5.7.1 to study the coupling between TL sections. Very often, we do not take the coupling between length sections into account. For this reason, we study impact of the lack of length inductive coupling between the segments. We cut the TLs into N independent sections as shown in Fig. 5.19. In the comparison, we construct two models, one with couplings between all sections and one where the couplings between the sections are ignored. The error between the two solutions with and without couplings is computed. In summary, the section coupling between sections 1 and 2 is evaluated according to (5.40) as L12 = 2 Ld = 2 (Lp12 \u2212 Lp12\u2032 ), (5.48) where the approximate formula (5.47) is used, while the case of all coupled N sections is the same as the difference inductance for the entire two-wire line"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000148_adma.200904059-Figure11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000148_adma.200904059-Figure11-1.png",
        "caption": "Figure 11 . Schematic drawing of the microfl uidic setup used by Ohm et al dispersed in nonmiscible silicone oil in the T-junction, and the resulting dr the structure of the utilized monomer that was introduced by Thomsen et al. Copyright 2009, Wiley-VCH.",
        "texts": [
            " Another example for mechanical orientation is the drawing of fi bers from a polymeric liquid crystal and a quick crosslinking afterwards, before the obtained orientation is lost. This can be done manually with tweezers, [ 30 , 36 ] or more sophisticated in an electrospinning process. [ 37 , 47 ] These samples are not so well suited for mechanical measurements, but they are easily prepared and highly ordered, and to some extent they mimic natural muscles. A special kind of mechanical orientation is the use of the fl ow fi eld from a fl uid moving inside a confi ned space. In this method, a microfl uidic setup ( Figure 11 ) is utilized to prepare small droplets from a liquid crystalline monomeric material and make them fl ow through a thin capillary, dispersed in a carrier fl uid. [ 58 ] Because of the fl ow velocity profi le inside a tube, shear fl ow is transferred to the droplets. They are then crosslinked by UV irradiation through the tube while they are fl owing to preserve their internal orientation. The advantage of the usual stretching technique is that no precrosslinking is necessary because the material does not have to be mechanically stable to be stretched"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000801_j.finel.2014.04.003-Figure9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000801_j.finel.2014.04.003-Figure9-1.png",
        "caption": "Fig. 9. Substrate heating model (1C, case s1).",
        "texts": [
            " Typically, different values are used in front and behind the melt pool for the longitudinal ellipsoid axis c. However, here the same value is used here since the melt pool is nearly spherical in the LENS process. f is a scaling factor; vw is the heating spot travel speed; and t is time. The values used here are: P\u00bc425 W, \u03b7\u00bc .45, a\u00bc1.5 mm, b\u00bc .9 mm, c\u00bc1.5 mm, and f\u00bc1. Eq. (16) is applied on the entire model and beyond the edges of the ellipsoid (defined at the 5% value [46]) and thus no correction for cut-off is needed. FEA mesh: Fig. 9 shows a symmetric model of the substrate only and Fig. 10 shows a symmetric model of the substrate and wall. The substrate model (Fig. 9) is composed of 8480 hex8 elements and 10,086 nodes. The substrate and wall model contains 14,432 hex8 elements and 19,200 nodes. The mesh density is targeted towards obtaining temperature results for follow on mechanical analyses for computing residual stress and distortion. Finer meshes would be required for solidification studies. Time incrementation: During heating, the time increment is set to 0.0885827 s, which results into a 0.75 mm (quarter of deposition width) travel distance per time increment",
            " Material properties for TI64 are used. The thermal conductivity k and specific heat Cp are listed in Table 3 [47]. Material properties are linearly interpolated between the values listed on the table, and kept constant beyond the minimum and maximum listed values. The density is 4.43 10 6 kg/mm3. The latent heat of fusion 365 kj/kg and is spread over a temperature range from 1600 1C to 1670 1C. In this test, heating of the substrate only with no metal deposition is modeled. A model of the substrate only (Fig. 9, case s1) is used as reference to evaluate the accuracy of models with active elements on the substrate and quiet or inactive elements on the wall (cases sq1 and si1). For the quiet element model (shown in Fig. 10), the conductivity and specific heat scaling factors at the quiet elements are set as sk\u00bc .000001 and sCp \u00bc :01. Temperature results along line AA as illustrated in Figs. 9 and 10 are extracted and plotted in Fig. 11. Convection and radiation are also applied on the interface between active and inactive (or quiet) elements"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure3.33-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure3.33-1.png",
        "caption": "FIGURE 3.33. The molecular binding between the tire and road surfaces.",
        "texts": [
            " Tire Dynamics Example 104 Friction mechanisms. Rubber tires generate friction in three mechanisms: 1\u2212 adhesion, 2\u2212 deformation, and 3\u2212 wear. Fx = Fad + Fde + Fwe. (3.114) Adhesion friction is equivalent to sticking. The rubber resists sliding on the road because adhesion causes it stick to the road surface. Adhesion occurs as a result of molecular binding between the rubber and surfaces. Because the real contact area is much less than the observed contact area, high local pressure make molecular binding, as shown in Figure 3.33. Bound occurs at the points of contact and welds the surfaces together. The adhesion friction is equal to the required force to break these molecular bounds and separate the surfaces. The adhesion is also called cold welding and is attributed to pressure rather than heat. Higher load increases the contact area, makes more bounds, and increases the friction force. So the adhesion friction confirms the friction equation Fx = \u03bcx (s) Fz. (3.115) The main contribution to tire traction force on a dry road is the adhesion friction"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.49-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.49-1.png",
        "caption": "FIGURE 8.49. A cycloid (a), curtate cycloid (b), and prolate cycloid (c).",
        "texts": [
            " Hence, the coordinates of point P in the global frame G, would be GrP = W rP + \u23a1\u23a3 vxt 0 Rw \u23a4\u23a6 = \u23a1\u23a3 Rw (\u03c9t\u2212 sin t\u03c9) 0 Rw (1\u2212 cos t\u03c9) \u23a4\u23a6 . (8.54) The path of motion of point P in the (X,Z)-plane can be found by eliminating t between X and Z coordinates. However, it is easier to expressed the path by using \u03c9t as a parameter. Such a path is called cycloid. In general case, point P can be at any distance from the center of the rim frame. If the point is at a distance d 6= Rw, then its path of motion is called the trochoid. A trochoid is called a curtate cycloid when d < Rw and a prolate cycloid when d > Rw. Figure 8.49(a)-(c) illustrate a cycloid, curtate cycloid, and prolate cycloid respectively. Example 332 F Wheel to wheel-body frame transformation. The homogeneous transformation matrix CTW to go from the wheel frame 494 8. Suspension Mechanisms to the wheel-body frame can be found by combined transformation. CTW = CTT TTW (8.55) = \u23a1\u23a2\u23a2\u23a3 c\u03b4 \u2212s\u03b4 0 0 s\u03b4 c\u03b4 0 0 0 0 1 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 c\u03b3 \u2212s\u03b3 \u2212Rw sin \u03b3 0 s\u03b3 c\u03b3 Rw cos \u03b3 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 = \u23a1\u23a2\u23a2\u23a3 cos \u03b4 \u2212 cos \u03b3 sin \u03b4 sin \u03b3 sin \u03b4 Rw sin \u03b3 sin \u03b4 sin \u03b4 cos \u03b3 cos \u03b4 \u2212 cos \u03b4 sin \u03b3 \u2212Rw cos \u03b4 sin \u03b3 0 sin \u03b3 cos \u03b3 Rw cos \u03b3 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 If rP indicates the position vector of a point P in the wheel coordinate frame, W rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 (8"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002568_j.mechmachtheory.2019.103701-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002568_j.mechmachtheory.2019.103701-Figure6-1.png",
        "caption": "Fig. 6. A schematic diagram of gear sliding-rolling and fluid pressurization phenomenon [94 , 110] .",
        "texts": [
            " Results clearly show that lubrication could effectively dissipate heat and reduce surface pitting compared with dry contact condition. With respect to fatigue crack propagation, after formation of the original, tiny fatigue cracks driven by the contact stress field, the pressure generated by fluid entrainment into the high-pressure contact region causes flow into the cracks, forcing them to open, which is referred as fluid pressurization [110] . This mechanism can explain the fact that the motion direction in the contact zone has decisive impact on crack propagation, which can be schematically seen in Fig. 6 . A similar conclusion can be deduced from Ref. [94] , that is, like macro-pitting evolutions, micropitting cracks propagate mainly opposite the sliding direction on the tooth surface. Hence, the initiated cracks converge toward the pitch line of the driving gear and diverge away in the driven gear. Additionally, according to RCF tests using different lubricants carried out by K\u00fcrten et al. [111] , the hydrogen composition may induce RCF failures in the form of white etching area (WEA) or dark etching region (DER) due to lubricant degradation"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003549_robot.1996.509185-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003549_robot.1996.509185-Figure3-1.png",
        "caption": "Fig. 3: Effect of center-of-mass height.",
        "texts": [
            " The Force-Angle stability measure, cy, is given by the minimum of the two angles, weighted by the magnitude of the force vector (as denoted by lifril) for heaviness sensitivity: (1) Q = 81 ' llfrll Critical tipover stability occurs when 6' goes to zero (and therefore f, coincides with 11 or 12) or, when the magnitude off, goes to zero and even the smallest disturbance may topple the vehicle. Iff , lies outside the cone described by 11 and 12 the angle becomes negat,ive and tipover is in progress. For a vehicle which is capable of adjusting its center-of-mass height, or for a vehicle which carries a variable load, the tipover stability margin should be topheavy sensitive. This is illustrated in Fig. 3 for the Force-Angle stability measure where an increase in c.m. height clearly results in a smaller minimum angle and a reduced measure of tipover stability margin. 3.2 General Form Geometry Of all the vehicle contact points with the ground, it is only necessary to consider those outermost points which form a convex support polygon when projected onto the horizontal plane. These points will simply be referred to as the ground contact points. Let pi represent the location of the ith ground contact point, e"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003669_ivs.2007.4290230-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003669_ivs.2007.4290230-Figure1-1.png",
        "caption": "Fig. 1 The quadrotor and its rotors turning directions.",
        "texts": [
            " used a simple vision system for a quad rotor\u2019s local positioning and orientation in indoor flight [21]. P. Castillo et. al., used a Lagrangian model of the quadrotor and controlled it based on Lyapunov analysis [22 , 23, 24]. Finally, A. Tayebi et. al., proposed a controller which is based upon the compensation of the Coriolis and gyroscopic torques and the use of PD 2 feedback structure [25]. II. DYNAMIC MODELING A quadrotor is an under actuated aircraft with fixed pitch angle four rotors as shown in Figure 1. Modeling a vehicle such as a quadrotor is not an easy task because of its complex structure. Our aim is to develop a model of the vehicle as realistically as possible. A 1-4244-1068-1/07/$25.00 \u00a92007 IEEE. 894 Having four rotors with fixed angles makes quadrotor has four input forces which are basically the thrust provided by each propellers as shown in Figure 2. Forward (backward) motion is maintained by increasing (decreasing) speed of front (rear) rotor speed while decreasing (increasing) rear (front) rotor speed simultaneously which means changing the pitch angle",
            " PD controller for y motions is given by 2 1 1( ) ( )p d d du K K\u03c6 \u03c6 \u03c6 \u03c6= \u2212 + \u2212 Yaw angle (\u03c8 ) and altitude can be controlled by PD controllers given by 4 2 2( ) (p d d du K K )\u03c8 \u03c8 \u03c8 \u03c8= \u2212 + \u2212 3 3 1 ( ) ( ) cos cos p d d dg K z z K z z u \u03b8 \u03c6 + \u2212 + \u2212 = K coefficients which used for ui controllers derived by trial and error for best performance. The controllers have been simulated on MATLAB Simulink model. In this simulation shown in Figure 3, helicopter starts at (5, 4, 3) m position with 30 degrees yaw and zero degree pitch and roll angles. The desired position is origin with zero yaw, pitch and roll angles. Figure 4 shows helicopter\u2019s this movement in x, y and z axes. (14) (15) (16) (17) (18) (19) (20) (21) (22) Each rotor of the quadrotor turns in one direction as it is shown in Fig.1, to reach from 5m altitude to zero, quadrotor shuts down its motors and free falls to reach its target altitude as fast as possible. Hence, x and y motions and yaw, pitch and roll angles stay still up to quadrotor actuates its motors to provide its altitude target. IV. USING VISION TO CONTROL THE QUADROTOR HELICOPTER Vision can be used to identify and to estimate relative position of the objects with respect to a flying vehicle. Vision can also be used to estimate flying vehicle\u2019s position and orientation if the reference feature\u2019s position and orientation are known"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000028_0005-1098(94)90209-7-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000028_0005-1098(94)90209-7-Figure7-1.png",
        "caption": "FIG. 7. Asperity deformation under applied force, presliding displacement.",
        "texts": [
            " Dahl (1968, 1976, 1977), studying experimental observations of friction in small rotations of ball bearings, concluded that for small motions, a junction in static friction behaves like a spring and considered the implications for control\u2022 There is a displacement (presliding displacement) which is an approximately linear function of the applied force, up to a critical force, at which breakaway occurs. The elasticity of asperities is suggested schematically in Fig. 6. When forces are applied, the asperities will deform, as suggested by Fig. 7, but recover when the force is removed, as does a spring. In this regime, the tangential force is governed by: F,(x) = - k , x , (1) where F, is the tangential force, k, is the tangential stiffness of the contact and x is displacement away from the equilibrium position. F, and x refer to the force and displacement in the contact before sliding begins, as indicated in Figs 7 and 8. When the applied force exceeds the required breakaway force, the junctions break (in the boundary layer, if present) and true sliding begins, as suggested in Fig",
            " This discussion of friction has focused on sliding between hard metal parts lubricated by oil or grease. For reasons of machine life and performance, these engineering materials make up many of the machines encountered by controls engineers. When these materials are used, the state of understanding supports a friction model that is comprised of four velocity regimes, two time dependent properties and several mechanism dependent properties. (1) The four velocity regimes. (I) Static Friction: displacement (not velocity) is proportional to force [see Fig. 7 and equation (1)]. (II) Boundary Lubrication: friction is dependent on surface properties and lubricant chemistry. (III) Partial Fluid Lubrication: if static friction is greater than Coulomb friction, friction decreases with increasing velocity. A s u r v e y o f f r i c t i o n a n d c o n t r o l s 1 0 9 7 (IV) Full Fluid Lubrication: friction is a function of velocity, a viscous plus Coulomb friction model may model the friction quite accurately. [Regimes I I - I V in Fig. 5, see also equation (8)",
            " (1990) and references for a brief discussion of two stage mechanisms]. The liabilities of the two stage mechanisms are weight, size and complexity: two actuators and two controllers are required per degree of freedom. By capitalizing on presliding displacement, referred to as microdynamics in the nanotechnology literature, it is possible to achieve two modes of control in a A survey of friction and controls 1121 single mechanism: gross motion in the standard way, and fine motion in the presliding displacement. Referring to Fig. 7 above, presliding displacement is motion that occurs by the deformation of asperities in the sliding interface. In this motion, position is a function of applied force; the junction appears to be a stiff spring rather than a sliding (or rolling) bearing. The result is two markedly different mechanism dynamics: 'macrodynamics,' the ordinary dynamics of the mechanism, and 'microdynamics,' which governs motions that depend upon elastic deformation in the frictional contact. Because the dynamms are drastically different, two different controller structures are required, thus dual mode control"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001256_j.ijmachtools.2021.103729-Figure43-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001256_j.ijmachtools.2021.103729-Figure43-1.png",
        "caption": "Fig. 43. This shows the effect of the PBF build orientation on the subsequent machining. The greatest cutting force is generated when the feed direction is parallel to the build direction, which gives rise to anisotropy at machined faces.",
        "texts": [
            " Studies have shown that the PBF microstructure of nickel-based superalloys has implications for the machining process as well. For example, there are peculiar interactions between build orientation and machining strategy [204]. It has been shown that the surface topography and integrity of LPBF IN625 was affected by the relative orientation of cutting direction to the build direction and scan strategy orientation [205]. Indeed, Patel et al. showed that machining with the feed in the build direction generated the greatest cutting force (as shown in Fig. 43) of the orientations tested [205]. Similarly, another study found that feeding the cutter against the build direction resulted in lower peak forces with larger deviations while feeding along the build direction resulted in higher peak forces with lower deviations [206]. Further, LPBF IN718 with HIP and HT were found to have better minimum specific cutting energy, minimum tool wear and minimum surface roughness during milling than wrought IN718 [207]. The peak milling cutting force was found to be dependent upon the feed direction as well as the layer-wise scan rotation employed in fabricating LPBF IN625 [206]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure11.13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure11.13-1.png",
        "caption": "Figure 11.13 (a) The water-retaining slide modelled as a beam with its (b) shear force diagram and (c) bending moment diagram.",
        "texts": [
            " 11 Mathematical Description of the Relationship between Section Forces and Loading 449 Due to a linear distributed load, the shear force is a quadratic (parabolic) function in x, and the bending moment is a third degree (cubic) function in x. The constants C1 and C2 follow from the boundary conditions that the bending moment at both the top and the base is zero: x = 0 : M = 0, x = h : M = 0. Elaboration of the boundary conditions leads to C1 = 1 6\u03c1gbh2, C2 = 0. The expressions for the shear force and the bending moment are therefore V = 1 6\u03c1gb(h2 \u2212 3x2), (a) M = 1 6\u03c1gb(h2x \u2212 x3). (b) The V diagram is a second degree curve (parabola); the M diagram is a third degree curve (cubic). Both diagrams are shown in Figure 11.13. At A and B tangents to the V and M diagrams are also shown. Note that the tangents to the M diagram intersect at x = 2 3h, the location where the resultant R of the triangular load acts. We will make use of this in the next chapter. The bending moment is extreme when dM dx = V = 0. 450 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM In other words: the bending moment is extreme where the shear force is zero. This location can be found from (a): V = 1 6\u03c1gb(h2 \u2212 3x2) = 0 \u21d2 x = 1 3h \u221a 3. Substituting this value of x in the expression for M gives the maximum bending moment: Mmax = M (x= 1 3 h \u221a 3) = 1 6\u03c1gb { h2 ( 1 3h \u221a 3 ) \u2212 ( 1 3h \u221a 3 )3 } = \u221a 3 27 \u03c1gbh3 = 0"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure9.72-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure9.72-1.png",
        "caption": "Figure 9.72 (a) If three members meet in an unloaded joint of which two are in a direct line with one another, the third member is a zero-force member. The normal forces in continuous members 1 and 3 are equal. (b) The forces acting on isolated joint B.",
        "texts": [
            " i Ni (kN) 9 \u2212F \u221a 2 10 +F \u221a 2 11 +2F 12 +4F 13 0 14 \u22124F 15 0 We can often shorten the calculation that needs to be done by first looking for zero-force members in a truss. Zero-force members are members in which no forces are acting (N = 0) due to the present loading. 364 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM There are three situations of frequent occurrence in which zero-force members can be easily recognised: 1. If only two members meet in an unloaded joint, both are zero-force members (see Figure 9.71). 2. If three members meet in an unloaded joint of which two are in a direct line with one another, then the third is a zero-force member (see Figure 9.72). 3. If two members meet in an unloaded joint and the line of action of the load coincides with one of the members, the other member is a zero-force member (see Figure 9.73). These three rules are the direct consequence of the joint equilibrium, as shown below for each of the cases. Rule 1. Two members meet in unloaded joint A in Figure 9.71. The force in one of the members has a component normal to the direction of the other member. If we write down the equilibrium of joint A in the given (local) xy coordinate system, we find \u2211 Fx = N1 + N2 cos \u03b1 = 0,\u2211 Fy = N2 sin \u03b1 = 0 with the solution (because sin \u03b1 = 0):1 N1 = N2 = 0. Equilibrium is possible only if both member forces are zero. 1 In a kinematically determinate truss, members 1 and 2 cannot be an extension of one another, so that \u03b1 = 0 and \u03b1 = 180\u25e6. 9 Trusses 365 Rule 2. In Figure 9.72, three members meet in joint B, of which members 1 and 3 are in a direct line with one another. The force in member 2 has a component normal to members 1 and 3. There can be equilibrium only if this component is zero, or in other words, if N2 = 0. If we write down the equilibrium of joint B in the given (local) xy coordinate system, we find \u2211 Fx = N1 + N2 cos \u03b1 \u2212 N3 = 0,\u2211 Fy = N2 sin \u03b1 = 0 so that N2 = 0 and N1 = N3. In addition to the fact that member 2 is a zero-force member, the normal forces in the continuous members 1 and 3, which are in a direct line with one another, are equal"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003862_s0143-8166(03)00063-0-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003862_s0143-8166(03)00063-0-Figure1-1.png",
        "caption": "Fig. 1. Schematic of the associated physical domains of the laser cladding process.",
        "texts": [
            " The model also takes into account the power attenuation and Brewster effects to include the changes in absorption factor due to the clad geometry. In the following, the modeling approach is first discussed and then the assumptions for the numerical solution are presented. Finally, the numerical results obtained at different laser pulse setting are compared with experimental data. For a laser cladding process, a moving laser beam with a Gaussian distribution intensity strikes on the substrate at t \u00bc 0 as shown in Fig. 1. Due to additive material, the clad forms on the substrate as shown in the figure. The transient ARTICLE IN PRESS E. Toyserkani et al. / Optics and Lasers in Engineering 41 (2004) 849\u2013867 851 temperature distribution T\u00f0x; y; z; t\u00de is obtained from the three-dimensional heat conduction in the substrate as [11] @\u00f0rcpT\u00de @t \u00fer \u00f0rcpUT\u00de r \u00f0KrT\u00de \u00bc Q; \u00f01\u00de where Q is power generation per unit volume of the substrate \u00f0W=m3\u00de; K is thermal conductivity \u00f0W=m K\u00de; cp is specific heat capacity \u00f0J=kg K\u00de; r is density \u00f0kg=m3\u00de; t is time (s), and U is the travel velocity of workpiece (process speed) (m/s)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000233_978-1-4614-0460-6-Figure1.4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000233_978-1-4614-0460-6-Figure1.4-1.png",
        "caption": "Fig. 1.4 (a) Three degrees of freedom for a particle and (b) six degrees of freedom for a rigid body",
        "texts": [
            " 6 1 Introduction As noted previously, it is often the responsibility of the engineer to model vibrating systems to determine their response to arbitrary inputs or decide how to modify a structure to mitigate the effects of a particular forcing function. Regardless, each system requires a certain number of independent coordinates to adequately describe its motion. These coordinates are the degrees of freedom. For a particle in three-dimensional space, three coordinates are necessary \u2013 one for each of the three translation directions. A rigid body, on the other hand, requires six degrees of freedom to describe its motion: three translations and three rotations. See Fig. 1.4. For an elastic body, or one that possesses mass and has finite stiffness, an infinite number of coordinates is required to fully describe its motion. To demonstrate this, let\u2019s imagine a ruler overhung from the edge of a table and clamped flat; see Fig. 1.5. We can consider each tick mark as a point on the beam. Naturally, the motion of each of these points differs as a function of time in response to the end of the ruler being displaced and released. Therefore, we require a coordinate at each of the tick marks"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000875_tie.2015.2455053-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000875_tie.2015.2455053-Figure1-1.png",
        "caption": "Fig. 1. A single-link robot arm",
        "texts": [
            " IV. SIMULATION RESULTS To show the effectiveness of our developed adaptive neural output feedback control scheme, the simulation results are given in this section. In this simulation, the state observer is designed as (14) and the disturbance observers are designed according to (20), (21), (25) and (26). The adaptation laws of neural network values \u03b8\u0302i are decided by (45), (59) and (73). The adaptive output feedback control scheme is given by (69). Consider a single-link robot system shown in Figure 1 which is described by [54] Mq\u0308 + 0.5mgl sin(q) = \u03c4 y = q (82) 0278-0046 (c) 2015 IEEE. Personal use is permitted, but republication/redistribution requires IEEE permission. See http://www.ieee.org/publications_standards/publications/rights/index.html for more information. where M is the inertia; q is the angle position of the link; q\u0307 is the angle velocity of the link; q\u0308 is the angle acceleration of the link; g = 9.8m/s2 is the acceleration due to gravity; l is the length of the link; m is the mass of the link, and \u03c4 is the control force of the link"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure6.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure6.1-1.png",
        "caption": "FIGURE 6.1 Example conductor bar for resistance computations.",
        "texts": [
            " Further, the fundamental concepts on how the capacitive and inductive cell partial elements are formed have already been introduced in Chapters 4 and 5. As shown in Section 3.6, the weighted residual method (WRM) includes important approximations. Some details are discussed in Chapter 1 and in Section 3.6. The first one is the division or meshing of conductors or other materials into finite blocks, bars, and surface elements for which we preassign fixed local current or charge distributions. An example is a bar shown in Fig. 6.1, where we assign the current to be uniformly in the x-direction. Another example is surface cells for which we assign a uniform charge density. This uniform assignment of the local current density or charge on a cell is part of what is called the basis function. Mathematically, in general a large number of different functions are used to represent the current and voltage distributions. The choices include delta functions, piecewise constant, piecewise linear, and Rao\u2013Wilton\u2013Glisson (RWG) [1] set of basis and/or weighting functions",
            " The PEEC method [4] is based on the fundamental concept that basic circuit elements should be symmetrical such as conventional ones. Mathematically, this corresponds to the Galerkin WRM approach. This is accomplished by using a symmetric discrete approximation for the representation of the currents and charges for each circuit element. The basic method we use to create a finite system from the original Maxwell\u2019s equations can be viewed as WRM approximations. We use the current density for a conductor as an example. It is shown in Fig. 6.1, where a current density is defined uniform for the conductor cross section . Hence, the current I is related to the current density by J = I\u2215, where  = TW is the conductor\u2019s cross-sectional Circuit Oriented Electromagnetic Modeling Using the PEEC Techniques, First Edition. Albert E. Ruehli, Giulio Antonini, and Lijun Jiang. \u00a9 2017 John Wiley & Sons, Inc. Published 2017 by John Wiley & Sons, Inc. 134 BUILDING PEEC MODELS area. In the second case, for a thin or zero thickness conductor T = 0, the surface current is defined as per unit width",
            " In general, we include some unusual cell sizes in our PEEC models, such as the use of very large aspect ratios for the cell dimensions. Further, wherever possible we use the analytical evaluation of the partial elements. As part of this, we try to adhere to rectangular or quadrilateral cell shapes. A specific example of the weighting process is given in (6.1) for the averaging of the fields, or Veb = 1 \u222b \u222b e b \u2212E(r) \u22c5 x\u0302 dx d, (6.1) where  is the cross section in the thickness T and width W direction as shown in Fig. 6.1. We also give an example of the relation of PEEC to a Galerkin formulation. The weighting function averaging for the potential as is done in (6.10) and (6.11). The symmetric circuit elements are obtained with such weighting functions. In the mathematical domain, Galerkin proposed the idea of using the same basis and weighting function to solve both differential and integral equation problems [5] in 1915 using WRM. The PEEC method should be included in the class of the so-called discontinuous Galerkin methods [6], which also are popular today for noncircuit-oriented applications",
            "3) where the first term Einc(r) on the right-hand side is an externally applied electric field at r. The scalar potential \u03a6 and vector potential A contribute to the induced scattering fields. For most problems, applied external source fields Einc(r) are zero. However, we also consider the impact of external fields for PEEC in Chapter 12. For a static case, the vector potential derivative term in (6.3) is zero. We can easily derive an equivalent circuit for the resulting equation E(r) = \u2212\u2207\u03a6(r). (6.4) In Fig. 6.1, the voltage between two points \u201ce\u201d for end and \u201cb\u201d for beginning on the conducting rectangular bar is obtained by integration of (6.4). The two end-face contacts are indicated by the crosshatched areas and the dots are used as an electrical node for the contacts. The integration of (6.4) between e and b in x yields Vbe = \u03a6(b) \u2212 \u03a6(e) = \u222b e b E \u22c5 d\ud835\udcf5 = 1 \ud835\udf0e \u222b e b J \u22c5 x\u0302 dx. (6.5) As part of the circuit and WRM approximation, we place the contact to the entire end surfaces with the same potential. Hence, as shown in Fig. 6.1, the current density is uniformly defined to be Jx = I\u2215(TW) = I\u2215, where  is the cross-sectional area. In terms of the previous discussion on WRM, we a priori assume that the current will flow uniformly in the x-direction. In this unusual case, the exact current flow will be in the direction of the one assigned by the WRM model. For more complex models, current representations will be superimposed. Finally, the result for the resistance is the conventional formula Rbe = Vbe I = \ud835\udcc1 \ud835\udf0eTW = \ud835\udcc1 \ud835\udf0e , (6.6) where \ud835\udcc1 = |xe \u2212 xb|",
            " Besides the fact that these models are already useful by themselves for many applications, they also help to represent a key step toward the general PEEC model in Section 6.3. We start with the general integral equation (3.54), or E(r, t) = \u2212\u2207\u03a6(r, t) \u2212 \ud835\udf15 \ud835\udf15t \ud835\udf07 4\ud835\udf0b\u222b \u2032 J(r\u2032, t\u2032)|r \u2212 r\u2032|d \u2032. (6.8) Taking (6.2) into account but not including the capacitive potential coefficient term, we obtain \u2212\u2207\u03a6(r, t) = J(r, t) \ud835\udf0e + \ud835\udf15 \ud835\udf15t \ud835\udf07 4\ud835\udf0b\u222b \u2032 J(r\u2032, t\u2032)|r \u2212 r\u2032|d \u2032, where the vector r is placed inside or on the surface of the conductor. For simplicity, we again start out with the single bar of Fig. 6.1 as the first geometry. As before, we integrate the electric field along the length of the current path, which is between the contacts b and e, or Vbe = \u03a6(b) \u2212 \u03a6(e) = \u2212\u222b e b \u2207\u03a6(r, t) \u22c5 d\ud835\udcf5 = 1 \ud835\udf0e \u222b e b J(r, t) \u22c5 d\ud835\udcf5 + \ud835\udf07 4\ud835\udf0b \ud835\udf15 \ud835\udf15t\u222b \u2032 \u222b e b 1|r \u2212 r\u2032|J(r\u2032, t\u2032) \u22c5 d\ud835\udcf5 d \u2032. (6.9) We also note that we can simplify the dot product J(r\u2032, t\u2032) \u22c5 d\ud835\udcf5 to Jx dx because both vectors are in the same direction as is evident from Fig. 6.1. Again, we assume that the potential on the contacts is constant. It is equivalent to take the average of the potential over the surface of contacts, which is also very important for the second term Vbe = 1 \ud835\udf0e 1 \u222b \u222b b e Jx(r\u2032, t\u2032) dx d + \ud835\udf07 4\ud835\udf0b \ud835\udf15 \ud835\udf15t 1 \u222b \u222b e b \u222b \u2032 1|r \u2212 r\u2032|Jx(r\u2032, t\u2032) d \u2032 dx d, (6.10) INDUCTANCE\u2013RESISTANCE (Lp,R)PEEC MODELS 137 where  is the surface of the contact. Finally, the uniform current density is represented as J = x\u0302 I\u2215. Hence, the voltage Vbe can be approximated as Vbe = 1 \ud835\udf0e I  \u222b e b dx + \ud835\udf07 4\ud835\udf0b \ud835\udf15 \ud835\udf15t I \u2032\u222b \u2032\u222b 1|r \u2212 r\u2032|d d \u2032 ",
            "11) Using what we learned about partial inductances in Chapter 5, the second part is simply contribution of the partial inductance of the bar. Hence, the circuit equations for (6.11) is rewritten as Vbe = R1 I + Lp11 dI dt , (6.12) where R1 is the partial resistance due to the loss of the conductor bar. Lp11 is the partial self-inductance. Their detailed definitions are R1 = 1 \ud835\udf0e \u222b e b dx (6.13) Lp11 = \ud835\udf07 4\ud835\udf0b 1 \u2032\u222b\u222b \u2032 1|r \u2212 r\u2032|d \u2032 d . (6.14) This represents the (Lp,R)PEEC circuit model for the single bar in Fig. 6.1, which is simply Lp11 in series with the resistor R1. Since the PEEC equivalent circuit does not form a closed loop, the inductance of the model will be an open-loop inductance according to Section 5.4. We refer the reader to Section It is important to use this term so that we are aware of the fact that the circuit is not closed, as is also the case for a partial inductance. Fortunately, PEEC elements have electrical nodes to connect parts together such as building blocks. Some of the parts of a PEEC circuit without capacitive elements consist of partial or open-loop inductances"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure14.17-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure14.17-1.png",
        "caption": "Figure 14.17 The support reactions in A and B.",
        "texts": [
            "14a: Av = 24 kN (\u2191). Check: Figure 14.16 shows the forces acting on the pulley at B. The forces NCB and G, both 20 kN, are known. The force equilibrium can be used to find the support reactions at B: Bh = H = 16 kN (\u2192), Bv = (12 kN) + (20 kN) = 32 kN) (\u2191). Check: The horizontal support reaction at B is equal to H . The vertical support reaction is equal to the support reaction at B of the beam in Figure 14.14a, increased with the vertical cable force G. The support reactions at A and B are shown in Figure 14.17. Figure 14.18 shows a cable subject to a distributed load qz = qz(x). The cable shape is z = z(x). In Figure 14.19, a small cable element of length x has been isolated from the deformed cable and blown up. As x \u2192 0 the distributed load qz on the cable element can be considered to be uniformly distributed. Assume the cable force at the left-hand section is a tensile force N , with a horizontal component H and a vertical component V .1 The cable force at the right-hand section could have changed with respect to magnitude and 1 V is not the transverse force in the cable, but the vertical component of the tensile force N "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure2.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure2.3-1.png",
        "caption": "Fig. 2.3. Double-sided AFPM brushless servo motor with built-in brake and encoder: 1 \u2014 stator winding, 2 \u2014 stator core, 3 \u2014 disc rotor with PMs, 4 \u2014 shaft, 5 \u2014 left frame, 6 \u2014 right frame, 7 \u2014 flange, 8 \u2014 brake shield, 9 \u2014 brake flange, 10 \u2014 electromagnetic brake, 11 \u2014 encoder or resolver. Courtesy of Slovak University of Technology STU , Bratislava and Electrical Research and Testing Institute, Nova\u0301 Dubnica, Slovakia.",
        "texts": [
            " the total air gap is equal to two mechanical clearances plus the thickness of a PM with its relative magnetic permeability close to unity. A double-sided machine with parallel connected stators can operate even if one stator winding is broken. On the other hand, a series connection is preferred because it can provide equal but opposing axial attractive forces. 30 2 Principles of AFPM Machines A practical three-phase, 200 Hz, 3000 rpm, double-sided AFPM brushless motor with built-in brake is shown in Fig. 2.3 [157]. The three-phase winding is Y-connected, while phase windings of the two stators are in series. This motor is used as a flange-mounted servo motor. The ratio Xsd/Xsq \u2248 1.0 so the motor can be analysed as a cylindrical non-salient rotor synchronous machine [128, 155, 157]. 2.1 Magnetic Circuits 31 A double-sided machine with internal ring-shaped stator core has a poly-phase slotless armature winding (toroidal type) wound on the surface of the stator ferromagnetic core [101, 173, 244, 276]. In this machine the ring-shaped stator core is formed either from a continuous steel tape or sintered powders"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001478_j.conengprac.2011.04.005-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001478_j.conengprac.2011.04.005-Figure2-1.png",
        "caption": "Fig. 2. Quadrotor helicopters are controlled by varying thrust at each rot",
        "texts": [
            " Then, in Section 3 a nonlinear dynamic model of the quadrotor helicopter is developed, with focus on the effects of vehicle motion on the forces and moments produced by the vehicle. The design choices made for STARMAC II are presented in Section 4, including propulsion, frame, sensors, communications, and computational payloads. Finally, the control system design is described in Section 5, with flight test results for each control loop culminating in multiple waypoint trajectory tracking demonstrations with sub-meter accuracy. Quadrotor helicopters are controlled by varying the thrust of two sets of counter-rotating rotor pairs, as depicted in Fig. 2. Pitch and roll angles are controlled using moments generated by differential thrust between rotors on opposite sides of the vehicle, and the yaw angle is controlled using the difference in reaction torques between the pitch and roll rotor pairs. Vertical position is controlled with the total thrust of all rotors, and lateral acceleration is controlled through the pitch and roll of the aircraft. The first flight-capable quadrotor designs appeared as early as the 1920s (Leishman, 2000), though no practical versions were built until more recent advances in microprocessor capabilities and in micro-electro-mechanical system (MEMS) inertial sensors have allowed for automatic stability augmentation systems"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003750_(sici)1099-1239(199809)8:11<995::aid-rnc373>3.0.co;2-w-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003750_(sici)1099-1239(199809)8:11<995::aid-rnc373>3.0.co;2-w-Figure4-1.png",
        "caption": "Figure 4. Planar ducted fan engine. Thrust is vectored by moving the flaps at the end of the duct, as shown on the right",
        "texts": [
            " The controls u 1 and u 2 correspond to the torques around the rolling and heading axes. This system is differentially flat using the outputs x 1 and x 2 plus knowledge of time. Given x 1 and x 2 , we can use the first two equations to solve uniquely for x 3 and x 4 . Then given these three variables plus time, we can solve for all other variables in the system by differentiation with respect to time. Example 2 (planar ducted fan) Consider the motion of the planar, vectored thrust vehicle shown in Figure 4. This system consists of a rigid body with body fixed forces and is a simplified model for the Caltech ducted fan described in Reference 39. The ducted fan is mounted on a stand with a counterweight that moves in as the fan moves up. This results in inertial masses m x and m y in the x and y direction, respectively, that change with the ( 1998 John Wiley & Sons, Ltd. Int. J. Robust Nonlinear Control 8, 995\u20141020 (1998) y co-ordinate. We do not take the variation of these inertial masses with y into account but take their value around hover"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure2.14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure2.14-1.png",
        "caption": "Fig. 2.14. Angular velocity of link i",
        "texts": [
            " =[i-1p i-1 p i-1 p JT 1 - 1, n nx ny nz the Jacobian columns are obtained in the form [ \u00b0 i-1 \u00b0 i-1] - a(i-1) 11 Pny + a(i-1) 12 Pnx \u00b0 i-1 \u00b0 j-1 - a(i-1)21 Pny+ a(i-l)22 Pnx \u00b0 i-1 \u00b0 i-1 - a(i-1)31 Pny+ a(i-l)23 Pnx , for revolute joint i [ 0 ] a(i-1) 13 :a(i-1)23 ,forslidingjointi a(i-1) 33 (2.69) where \u00b0a(i_1)jk denotes the element (j,k) of matrix \u00b0Ai _ 1 \u2022 Let us now discuss the evaluation of matrix J OJ E R 3 x n relating between angular velocities of the last link, expressed in the base, reference frame, and joint rates. First, we will determine the angular velocity of link i expressed with respect to the reference frame (Fig. 2.14). This velocity can be expressed as the 42 2 Manipulator Kinematic Model x sum of the angular velocity Wi of the previous link and the relative angular velocity between link i and (i - 1). For a revolute joint i, we have l~i~n (2.70) where Zi _ 1 is the ort of the joint axis. All the points belonging to the link have the same angular velocity. In the case of a sliding joint, there is no relative rotation between the links so that the following relation holds (2.71) Relations (2.70) and (2.71) may be united into a single equation: (2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003881_mssp.2001.1414-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003881_mssp.2001.1414-Figure2-1.png",
        "caption": "Figure 2. Coupling between the torsional and transverse motions of the gears and shafts: r p *base circle radius",
        "texts": [
            "ects and so can be neglected. of input pinion; r g *base circle radius of output gear; k s *shaft transverse sti!ness; k mb *linear translation tooth sti!ness along line of contact; c}c*line of contact. The method of coupling between the torsional and transverse motions of the gears and shafts is described in detail by Du [2] and Howard et al. [3] and is presented here for completeness. The coupling between the torsional and transverse motions of the gears and shafts can be developed as shown in Fig. 2. The linear mesh sti!ness (bending sti!ness) of the gears provides an easy means of coupling the torsional and transverse coordinates of the shaft system together. The \"nite element analysis which has been developed by Sirichai [6] and Wang et al. [7] has derived the detailed torsional mesh sti!ness of gear teeth in mesh. In the \"rst instance, a simple relationship between linear and torsional mesh sti!ness can be derived as follows. The torsional mesh sti!ness can be related to the bending sti"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001025_b809009b-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001025_b809009b-Figure4-1.png",
        "caption": "Fig. 4 Standardized stack cell platform.47",
        "texts": [
            " Correlation of these results to scanning electron micrographs suggested that the entrapment method lacked the expected diffusive pathways and most likely expelled enzyme during growth of the film, a result that was in contrast with a bulk of literature on the entrapment method that had never examined the enzyme kinetics in the absence of mass transfer effects. The importance of geometry, experimental standardization, and quantization the statistical reproducibility In a recent paper, a standardized testing platform for biological fuel cell research was proposed by Svoboda et al. (2008).47 The apparatus, shown in Fig. 4, is a modular stack cell with defined geometry and dimensions, whilst retaining flexibility in configurations for various electrochemical studies, thereby permitting controlled conditions that avoid ambiguities related to experimental parameters that are often critical but not consistently controlled. The controlled parameters included electrolyte concentration and its volume in the cell, separation distance between electrodes and their position (which significantly affects the electrolyte resistance and shape of the electric field and its distribution)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001377_s11661-014-2218-0-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001377_s11661-014-2218-0-Figure1-1.png",
        "caption": "Fig. 1\u2014Schematic diagram of the meander scan strategy: the laser scans the area of each layer with alternating parallel scan vectors. The border of each layer is then scanned twice. The scan rotation is rotated by 67 deg at each new layer deposition.",
        "texts": [
            " Cubic components of 1 cm3 were built using a Renishaw SLM250machine.[26] The SLM250 is equipped with an infra-red fiber modulated pulse laser with a wavelength of 1070 nm and a nominal maximum power of 200 W. In this study, themaximum laser power at the workpiece was 157 W due to losses in the optical train. A series of experiments were performed to establish a processing window that could minimize residual porosity and the optimal processing parameters were reported in detail elsewhere.[27] All the parts were built using a meander scan strategy (Figure 1). Each cross section was initially offset 100 lm from the edge.The area enclosed by the offsetwas scanned using alternating parallel scan vectors. The laser then scanned twice the offset boarder in order to improve the surface roughness of the parts and reduce the number of defects near the external surface of the components. It is noteworthy that the alternating parallel scan vectors were rotated by 67 deg at each new layer deposition as depicted in Figure 1. The cubes were built directly on top of a Ti-6Al-4V build plate pre-heated at 338.15 K (65.15 C) in a protective Ar atmosphere. The microstructure of the as-built components was studied from three orthogonal sections of the specimens, i.e., the frontal, lateral, and horizontal planes (xz-, yzand xy-planes according to ASTM F2921-11). Sample surfaces were prepared by standard metallographic preparation procedures and then etched in Kroll\u2019s solution for optical/SEM. The microstructure of the components was examined in a Reichert-Jung MEF3 optical microscope and a Carl Zeiss (Leo) 1530 VP field emission gun-SEM system",
            " It is not clear to show in the micrograph where each layer deposition has occurred, however it can be estimated that b grains have grown through tens of layers as each powder bed layer on the scan platform was approximately 50 lm. Prior b columnar grain boundaries also emerge from the optical microscopy analysis of the lateral yz-plane (Figure 2(b)). Figure 2(b) shows that the prior b grain growth direction ~g is inclined ~20 deg to the build direction (z-axis) as a result of the adopted rotating scan strategy.[9] Although the scan vectors are hatching each layer in alternating METALLURGICAL AND MATERIALS TRANSACTIONS A direction (Figure 1), it is reasonable to assume that the 67 deg rotation of the scan direction imposed by the scan strategy causes local heat gradients in the y direction (Figure 1). These thermal gradients would explain the inclination of the b grains shown in Figure 2(b). The horizontal plane (xy-plane) reveals the cross sections of the prior columnar b grains (Figure 2(c)) but the correlation between the scan strategy and the morphology of the prior b grains does not appear as clear as reported in other studies[9] probably due to the complexity of the meander scan strategy. Further heat transfer modeling is required to work out the detailed thermal gradient for this particular scan strategy"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003403_978-1-4615-5633-6-Figure6.15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003403_978-1-4615-5633-6-Figure6.15-1.png",
        "caption": "Figure 6.15. Block diagram of the converter controller",
        "texts": [
            "39) dVdi = _1 [I _ I .] dt Cdi d d. (6.40) dIdi 1 dt = La . [-Rc;ldi + Vdi - Ed;] eq. (6.41) Equations (6.37) to (6.41) after linearization can be expressed in the compact form as dXd dt = [Ad]Xd + [Bd]Ud Yd = [Cd]Xd where u~ = [~Edr ~Edi]' Y~ = [~Idr ~ Vdr ~ldi ~ Vdi] 6.3.4 Modelling of converter control (6.42) (6.43) It is assumed that the rectifier controls the current and the inverter regulates the dc voltage or the extinction angle. The current (voltage) controller at the rectifier (inverter) is shown in Fig.6.15. Here the gain K is assumed to be positive. The negative sign associated with the gain K, implies that the delay angle ordered (0' R) at the rectifier or the angle of ad vance (f3 R) ordered at the inverter is decreased whenever the error positive. It is to be noted that actual 0' at the rectifier or f3 at the inverter is also influenced by the change in the phase angle (~Jc) of the ac voltage at the converter bus - with EPC scheme of firing pulse generation. The switch 51 is assumed to be open when IPC firing scheme is assumed",
            " The switch 52 is assumed to be open with pure EPC scheme whereas with 52 closed, the transfer function of the synchronizing circuit is also considered. Typically, SYNC(s) is given by e- STD 5Y NC(s) = T 1 + s s (6.44) (6.45 ) and B Ws is the band wid th of the synchronizing circuit. A typical value of B W. is about 4 Hz. At this frequency, the synchronizing circuit has little impact on TI. TD represents the average transport delay due to the sampling action of the frequency transducer used for synchronization. The inverter control can be used to regulate the extinction angle rather than dc voltage. The block diagram shown in Fig.6.15 also applies in this case except that the controller gain is positive(The angle of advance increases if \"{rei is higher than measured value of the extinction angle). The linearized state equations for the converter controllers can be expressed as dXdc [ [] ( ) t = Adc]Xdc + Bdc Udc 6.46 Ydc = [Cdc]Xdc + [Ddc]Udc (6.47) where 6.3.5 Combining AC network and DC system equations The linearized ac network equations, expressed in D-Q components can be written in the compact form as where XN = [AN]XN + [BNd]UNd + [BNg]UNg YNd = [CNd]XN U~d = [~IrD Y~d = [~v,"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure7.35-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure7.35-1.png",
        "caption": "Figure 7.35 (a) The water pressure on the partition increases linearly with the depth. (b) The water pressure is constant on a small horizontal strip and is equal to \u03c1gz.",
        "texts": [
            " Question: Determine the resultant R of the water pressure on the partition. Solution: The water pressure on the partition varies linearly. At a depth z the water pressure is \u03c1gz, whereby \u03c1 is the density of water, and g is the gravitational 1 Archimedes (287\u2013212 BC), Greek scientist from Syracuse. He addressed issues relating to integral calculus and was one of the founders of statics (equilibrium of solids) and hydrostatics (equilibrium of fluids). the contained space. 7 Gas Pressure and Hydrostatic Pressure 267 field intensity (see Figure 7.35a). For the partition, take a very narrow horizontal strip dz at depth z (see Figure 7.35b). The width of the strip is b(z). The water pressure on this narrow strip is constant and equal to \u03c1gz. This contribution dR of the strip to the resulting water pressure R on the strip is dR = \u03c1gz \u00b7 b(z) \u00b7 dz, (1) whereby b(z) = 2r cos \u03d5, (2) z = zC \u2212 r cos \u03d5, (3) dz = dz d\u03d5 d\u03d5 = r sin \u03d5 d\u03d5. (4) Substitute (2) to (4) into (1) and we find dR = 2\u03c1gr2(zC \u2212 r cos \u03d5)(sin \u03d5)2 d\u03d5. To find the resulting water pressure R, one has to sum up the contributions of all the strips. This is done by integrating between the limits \u03d5 = 0 and \u03d5 = \u03c0 : R = 2\u03c1gr2 \u222b \u03c0 0 (zC \u2212 r cos \u03d5)(sin \u03d5)2 d\u03d5"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure9.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure9.2-1.png",
        "caption": "Figure 9.2 A hinged joint.",
        "texts": [
            " Calculating a plane truss, hereafter referred to as truss, is based on the following assumptions: \u2022 all members are straight; \u2022 all members are connected at hinged joints; \u2022 the load consists of forces that act in the plane of the structure and apply at the joints. 322 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM This implies that all the members in the truss behave as two-force members and can only transfer tensile and compressive forces between the joints (see also Sections 3.2.2 and Figure 3.35). In the past, one tried to realise the connections in the joints as real hinges (see Figure 9.2). These days, all the members are rigidly connected, either directly or via a so-called gusset plate. Figure 9.3 shows two examples of a joint with a gusset plate: one made of steel (a) and the other made of wood (b). It is clear that these joints are not hinged. One can show, however, that whether or not the joints are hinged, this in fact has little impact on the force flow. A condition is, however, that the member axes intersect at the joints \u2013 clearly the case in Figure 9.3 \u2013 and that the load is applied at the joints"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001470_j.actamat.2009.08.027-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001470_j.actamat.2009.08.027-Figure2-1.png",
        "caption": "Fig. 2. Schematic overview of the numerical setup. The powder layer is indicated in light grey, the starting point of the laser source in dark grey.",
        "texts": [
            "5 times the regular cell volume, which will be explained in Section 2.6 on the shrinking algorithm. The stencil for the solution of the enthalpy equation is modified accordingly. The laser source is assumed to be Gaussian, given by: P \u00bc 2 pr2 b exp 2 \u00f0x xS\u00de2 \u00fe \u00f0y yS\u00de 2 r2 b \" # \u00f02\u00de in which P is the power of the stationary laser source, rb the beam radius, and \u00f0xS ; yS\u00de the position of the source. The calculations are performed in a fixed coordinate system, so the source is moving with a velocity v along the x-axis, as shown in Fig. 2. This means that the x-coordinate of the source is given by xS \u00bc xS\u00f00\u00de \u00fe vt and the radiative heat input from the laser in a certain point (x,y) can be obtained through a time integration of Eq. (2): QP \u00bc Z t\u00feDt t Pdt \u00f03\u00de The heat input into a computational cell centered on (x, y) is then given by a space integral of Eq. (3) over the top surface of the cell: QS \u00bc Z x\u00feDx=2 x Dx=2 Z y\u00feDy=2 y Dy=2 QP dxdy \u00f04\u00de In our case, the beam radius rb in the experiments and thus also in the simulations is 26 lm"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002501_j.jsv.2019.115144-Figure14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002501_j.jsv.2019.115144-Figure14-1.png",
        "caption": "Fig. 14 (a) Experimental setup, (b) test bearings, (c) spectrum of healthy test bearing, (d) spectrum of healthy simulation bearing, (e) spectrum of unhealthy test bearing, and (f) spectrum of unhealthy simulation bearing.",
        "texts": [
            " It indicates that the additional excitation zone has some influence on the vibrations of the RBHS, which should be formulated in the dynamic modelling methods for the RBHS considering the faults. (a) (b) (c) An experiment for the studied bearing N306 is conducted based on a BVT-5 test rig. The rotor speed and external load are 1750 r/min and 300 N, respectively. The defect length and width are 0.5 mm and 11.8 mm, respectively. According the method in Ref. [50], the roller passing the outer raceway frequency (RPFO) is 137.6 Hz. The results from the proposed model and experiment are compared in Fig. 14. In Fig. 14, the peak at the RPFO is observed in the simulation and experimental results. The RPFO values for the simulation and experiment are 137.6 Hz and 138 Hz, respectively. The difference between the simulation and experimental results are -0.29%, which can give a validation for the presented model. The difference of the amplitude between the simulation and experimental results may be produced by the damping and vibration of the test rig structures. This investigation presents a new analytical dynamic model for a RBHS"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003389_s0956-5663(01)00126-9-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003389_s0956-5663(01)00126-9-Figure4-1.png",
        "caption": "Fig. 4. Schematic of OPH-modified fiber-optic biosensor.",
        "texts": [
            " Organophosphorus insecticides measured using this technique included parathion, methylparathion, dursban, fensulfothion, crotoxyphos, diazinon, mevinphos, dichlorvos, and coumaphos. The technique was used to measure coumaphos in biodegradation samples of cattle dip wastes and showed a high correlation (r2=0.998) to a HPLC method (Roger et al., 1999). In another fiber-optic biosensor format, the analysis was based on the relationship between the amount of OP hydrolyzed and the amount of chromophoric product formed (quantified by measuring the absorbance at the max of the product) by the enzyme catalyzed hydroysis. The fiber-optic setup (Fig. 4) used in the study consisted of a 75-W xenon arc lamp light source housed in PowerArc lamp housing (b), powered by a constant voltage dc lamp power (a) with a monochromator (c) attached to one arm of the bifurcated fiber-optic bundle, set at a desired cutoff wavelength (400 nm for p-nitrophenol, hydrolysis product of paraoxon and parathion, and 348 nm for chlorferon, hydrolysis product of coumaphos), a 0.5-m bifurcated quartz fiber-optic bundle (d), a second monochromator (e), set at the same wavelength as the first one, attached to the other arm of the bifurcated fiber bundle, a photomultiplier detection system (f), and a strip chart recorder (g)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000637_978-3-642-83006-8-Figure3.23-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000637_978-3-642-83006-8-Figure3.23-1.png",
        "caption": "Fig. 3.23. Nominal trajectories of active internal angles",
        "texts": [],
        "surrounding_texts": [
            "256\n~ + + + -~ only e 4 and e 13 , at the knees e 5 and e 12 , and at the hip joints e 6 and ~ ~ ~ e 1 . At the trunk joint, both d.o.f., represented by e 15 and e 16 are involved in the motion. The rest of them are inactive and locked, and do not change their relative position during the mechanism motion. The hands are crossed on the chest. Such choice of the active legs joints enables their motion in the planes parallel to the frontal plane, only. Thus, the compensating movements are executed by .the trunk: in the frontal plane about e15 , in sagittal about e16 . Fig. 3.22. shows the stick diagram of the nominal biped gait in the sagittal plane. Here 0\ndenotes the nominal position of ZMP under the supporting foot, the other leg is being in swing phase. In Figs. 3.23. and 3.24. are given the nominal trajectories of all the internal active angles and of the corresponding driving torques, respectively.\nEaCh active joint is powered by the corresponding actuator. The actua tors used were DC motors of two types: Moore - Reed PCP 50 for actua tor M1 , TRW GLOBE 102~ 200-8 for actuator M2 . Mathematica~ model of each powered subsystem is given by (3.3.2), with matrix A~ and vectors b i and fi, as well as their distribution per joints and saturation vol-c c tage being given in Table 3.3. The motion was simulated for one half-step period, and for the single support phase, only. Duration time of simulated motion was T = 0.75[s]. The perturbed motion of the system around the nominal trajectory has been simulated, where, trunk angular displacement from the nominal trajectory in the frontal plane, of",
            "257",
            "258\nSimulation was performed for different control schemes applied. These\ncases were:\nwhile k~5 was introduced only at the joint whe~e an initial angular displacement occurred (trunk or ankle) .\nfrom force sensors at mechanism's sole of the supporting foot to the compensating subsystem, selected in advance.\nG m~x 2 an~le joint, i;15 fo~ trunk) def~ned by k i5 . They are In~aJ=3[rad/s J,\nInl 1=5[rad/s J, Inl 1=7[rad/s J. max max\nThe results belonging to one simulated gait case are given as a set of five diagrams. The first diagram shows angular deviation from nominal value for trunk, the second one gives the same for the ankle and hip joints of the supporting leg. Each diagram includes the results for all three values of ni . The next two diagrams present the example (for . 2 max Inl 1=7[rad/s J only) of the control input to the actuator and drimax ving torque deviation for the ankle and hip joints of the supporting leg, as well as for the trunk.\nThe last diagram of each set of results shows the deviation of ZMP from the nominal position.\nThe first set of simulation results describes the case when feedback is defined as in 1), and the trunk displacement for Aq15(O) = O.2[radJ is adopted as a d.isturba,nce. Simulation results are given in Fig. 3.25. Fig. 3.25.a) shows the trunk position deviation from the nominal angle, and Fig. 3.25.b) shows the same for the hip and ankle joint. Fig. 3.25.c) illustra,tes control inputs to actuators, while diagram of driving tor que deviations is ommitted because Api~O (i=4, 6, 15). Fig. 3.25.d) shows the ZMP position deviation from its nominal position. The nomi nal ZMP position in the simulated gait, fixed to the origin of the"
        ]
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.42-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.42-1.png",
        "caption": "FIGURE 7.42. The geometry of the optimal trapezoidal steering mechanism and the estimated design.",
        "texts": [
            " We may set a value for \u03b2, say \u03b2 = 6deg, and evaluate \u03b4Do and \u03b4Ao at n = 100 different values of \u03b4i for a working range such as \u221240 deg \u2264 \u03b4i \u2264 40 deg. Then, we calculate the associated error function e e = vuut 1 100 100X i=1 (\u03b4Do \u2212 \u03b4Ao) 2 (7.152) for the specific \u03b2. Now we conduct our calculation again for new values of \u03b2, such as \u03b2 = 8deg, 9 deg, \u00b7 \u00b7 \u00b7 . Figure 7.41 depicts the function e = e(\u03b2) with a minimum at \u03b2 \u2248 12 deg. The geometry of the optimal trapezoidal steering mechanism is shown in Figure 7.42(a). The two side arms intersect at point G on their extensions. For an optimal mechanism, the intersection of point G is at the outer side of the rear axle. However, it is recommended to put the intersection point at the center of the rear axle and design a near optimal trapezoidal steering mechanism. Using the recommendation, it is possible to eliminate the optimization process and get a good enough design. Such an estimated design is shown in Figure 7.42(b). The angle \u03b2 for the optimal design is \u03b2 = 12.6 deg, and for the estimated design is \u03b2 = 13.9 deg. Example 294 F There is no exact Ackerman mechanism. It is not possible to make a simple steering linkage to work exactly based on the Ackerman steering condition. However, it is possible to optimize various steering linkages for a working range, to work closely to the Ackerman condition, and be exact at a few points. An isosceles trapezoidal linkage is 7. Steering Dynamics 427 not as exact as the Ackerman steering at every arbitrary turning radius, however, it is simple enough to be mass produced, and exact enough work in street cars"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure9.16-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure9.16-1.png",
        "caption": "FIGURE 9.16. A wheel turning, without slip, over a cylindrical hill.",
        "texts": [
            " (9.335) The inner product of equations of motion with r\u0307i and adding the equations nX i=1 mir\u0307i \u00b7 r\u0308i = \u2212 nX i=1 r\u0307i \u00b7\u2207iV (9.336) and then, integrating over time 1 2 nX i=1 mir\u0307i \u00b7 r\u0307i = \u2212 Z nX i=1 ri \u00b7\u2207iV (9.337) shows that K = \u2212 Z nX i=1 \u00b5 \u2202V \u2202xi xi + \u2202V \u2202yi yi + \u2202V \u2202zi zi \u00b6 = \u2212V +E (9.338) 9. Applied Dynamics 567 where E is the constant of integration. E is called the mechanical energy of the system and is equal to kinetic plus potential energies. E = K + V (9.339) Example 372 Falling wheel. Figure 9.16 illustrates a wheel turning, without slip, over a cylindrical hill. We may use the conservation of mechanical energy to find the angle at which the wheel leaves the hill. At the initial instant of time, the wheel is at point A. We assume the initial kinetic and potential, and hence, the mechanical energies are zero. When the wheel is turning over the hill, its angular velocity, \u03c9, is \u03c9 = v r (9.340) where v is the speed at the center of the wheel. At any other point B, the wheel achieves some kinetic energy and loses some potential energy"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001347_j.procir.2015.08.061-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001347_j.procir.2015.08.061-Figure5-1.png",
        "caption": "Fig. 5. Workflow of the Pre-processing.",
        "texts": [
            " Afterwards the powder is measured again without humidity. A high residual moisture has a negative influence on the flow ability and as well as on the metal microstructure of the build part. The maximum acceptable value is 0.025 %. The average detected residual moisture was around 0.017 %. Therefore the residual moisture is suitable for SLM. 4. Production of SLM testing geometries 4.1. Pre-Processing The workflow of the pre-processing for the production of TiAl6V4 testing geometries is shown in Figure 5. The CADmodel is converted into a stereolithography format (STL). After that follows the support generation and the slicing of the part. 4.2. SLM machine and manufacturing parameters The production of TiAl6V4 specimens was carried out using a SLM 250HL machine by the company SLM Solutions GmbH, L\u00fcbeck, Germany. The significant parameters used for the manufacturing of testing geometries are the laser power PL, the scanning velocity vs and the focus diameter ds, which depend on the focus position"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001161_j.actamat.2014.05.039-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001161_j.actamat.2014.05.039-Figure1-1.png",
        "caption": "Fig. 1. FEM model which uses the activating process of pre-set elements. Insert shows the interpolating domain for stochastic analysis.",
        "texts": [
            " The heat loss due to the convection and radiation of the ambient gas is considered with the energy equation as follows: qsurface \u00bc 0:45 qlaser hc\u00f0T surface T ambient gas\u00de re\u00f0T 4 surface T 4 ambient gas\u00de \u00f02\u00de where qsurface is the heat flux on the surface, qlaser is the laser energy, hc is the convection heat transfer coefficient, r is the Stefan\u2013Boltzmann constant, e is the emissivity and Tsurface and Tambient_gas are the temperatures of the surface and ambient gas, respectively. A factor of 0.45 is used for the absorptivity of the laser energy by the substrate, based on a comparison of calculated and experimentally observed fusion zone dimensions. The initial temperature of the powder is set as the liquidus temperature of the powder material as described by Ref. [31]. The configuration of the FEM model is shown in Fig. 1. The properties of the materials used in the FEM simulation and the following stochastic analysis are listed in Table 1 [26]. In the solidification analysis, it was assumed that there is no grain movement in the liquid, and the nucleation rate is [28]: @N @t \u00bc 2lN \u00f0DT \u00de @T @t \u00f01 fs\u00de \u00f03\u00de where N is the number of nuclei, DT is the bulk undercooling, lN is a nucleation parameter,@T @t is the cooling rate and fs is the volume fraction of the solid. In order to simulate the competition between equiaxed and columnar dendrites, nucleation parameter lN in Eq"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001791_j.matdes.2017.08.049-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001791_j.matdes.2017.08.049-Figure3-1.png",
        "caption": "Fig. 3.Neutron diffraction set-up showing (a) the scan directions, (b) the location of the scanned (L) in the samples.",
        "texts": [
            " RS profiles were obtained by means of ND measurements at the E3 beamline of the neutron reactor BER II at Helmholtz-Zentrum f\u00fcr Materialien und Energie, Berlin (HZB). Details about the equipment andmeasurementmethods are available in [36]. Monochromatic radiation with an approximate wavelength of \u03bb = 0.1476 nm was used. ed rectangleswithin the specimen correspond to the SD-BD plane fromwhich the samples Following the Bragg's lawof diffraction, the 2D detectorwasfixed at diffraction angle 2\u03b8 = 85.3\u00b0, corresponding to the Ni-311 peak. Stresses were characterized in the three directions shown in Fig. 3a: longitudinal (L), transversal (T) and normal (N). For comparison, these orientations correspond to the HD, SD, and BD directions, respectively, as shown in Fig. 1. For stress analysis, these directions are assumed to be the principal. Fig. 3b shows the two lines that were scanned along the length of the specimens at the top surface: in the middle (Line 1) and near the edge (Line 2). Ten and nine equi-distant points were measured along the lines 1 and 2, respectively, as shown in Fig. 3b. Diffraction set-up with respect to the orientation of the longitudinal stress component is represented in Fig. 3c. The gauge volume used was 4 mm \u00d7 4 mm \u00d7 2 mm with acquisition times ranging from 5 to 90 min depending on the direction and/or location. Three different specimensweremeasured using [37] to determine the optimal stress-free reference; the raw powder, shavings produced from the AM sample, and a cuboid (5 mm \u00d7 5.5 mm \u00d7 6 mm) EDM cut from the sample. To evaluate the possibility of RS changes induced by sample preparation, all three d0 values were compared and found to be within the range of experimental error",
            " These structures are commonly observed in SLM where the successive re-melting and re-solidifying of the overlappingmelt pools with the previously solidified layers develop columnar grains with a \u3008001\u3009 texture. Although not shown, similar microstructural features were also observed in the columnar grains of the longer hatch length sample 2. Development of these features is reported for the SLM of other metallic materials and in AM processes other than SLM. The formation of a ge line scans shown in Fig. 3b. (a, b) Longitudinal (L), (c, d) transverse (T), and (e, f) normal located at 50 mm. columnar dendritic structure, with extremely small secondary arms, was reported by Cloots et al. [40] for the SLM of superalloy IN738LC. Similar features were also reported by Helmer et al. [41] on selective electron beam melting (SEBM) of Inconel 718 alloy. SEBM is a hotter process as compared to SLM, hence, this comparable observation offers insight into the effect of temperature and temperature gradient on the resulting microstructure in AM parts"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001674_j.mechmachtheory.2012.10.008-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001674_j.mechmachtheory.2012.10.008-Figure7-1.png",
        "caption": "Fig. 7. FE model for the calculation of structural deflection.",
        "texts": [
            " Furthermore, if both gears were considered simultaneously, the model would have to incorporate specific contact elements requiring a certain consistency between the two FE meshes or the definition of functional relationships between the displacements of points located in the vicinity of the contact. In this case, as the local behaviour is defined by an analytical formulation, the FE model only has to provide the global deflections taking into account the bending, compression and shearing of the teeth, the twisting of the wheel body [19] and the influence that a couple of teeth in contact can have on the contact of the adjacent pairs. To carry out this task, two FEmodels have been developed for each gear. The first one has a limited number of teeth depending on the contact ratio (see Fig. 7). Considering gear 1 (the upper one) in Fig. 6, in the extreme case (though unlikely) there could be a contact at the point designated as 3 and simultaneously at point 6. Therefore, it is necessary to establish the relationship between loads and deformations for all the possible contact points as described in subsection 2.2. In the example in Fig. 6 a contact ratio between 1 and 2 is used, and thus a model should be developed that has 4 teeth arranged in the clockwise direction from the tooth for contact 3",
            " As a summary, to be able to consider all the possible contacts for a certain transmission, the number of teeth to be included in the FE model should be that defined by expression (10). Then, a FEmodel is constructed generating a certain number of teeth depending of the contact ratio of the gear pair as defined in the paragraphs above. Besides the teeth profiles, the geometry of the FE model is completed by a circle arc whose radius is the root radius and it is also included the central hole where the shaft is inserted, so that the nodes in this circle are considered as fixed (see Fig. 7). Once the FE model mesh is defined, the loads are applied on the left flank (hereinafter active flank) of the tooth located on the vertical axis in Fig. 7. On each of the nodes in that flank, a unitary force Fi in the direction normal to the tooth profile is applied successively, obtaining the displacement of the nodes in the rest of the flanks both in the right but also in the left one in order to apply the procedure in the case of two tooth flanks contact. These displacements designated as (\u03b2k i, j) represent the flexibility of point j (on the radius Rj), located on the flank k when the unitary force is applied at point i (on the radius Ri) in the active flank (see Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.16-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.16-1.png",
        "caption": "FIGURE 8.16. A solid axle suspension with coil springs.",
        "texts": [
            " A live axle is a casing that contains a differential, and two drive shafts. The drive shafts are connected to the wheel hubs. A live axle can be three to four times heavier than a dead I-beam axle. It is called live axle because of rotating gears and shafts inside the axle. Example 307 Solid axle and coil spring. To decrease the unsprung mass and increase vertical flexibility of solid axle suspensions, it is possible to equip them with coil springs. A sample of a solid axle suspension with coil spring is shown in Figure 8.16. The suspension mechanism is made of four longitudinal bars between the axle and chassis. The springs may have some lateral or longitudinal angle to introduce some lateral or longitudinal compliance. Example 308 De Dion axle. When a solid axle is a dead axle with no driving wheels, the connecting beam between the left and right wheels may have different shapes to do different jobs, usually to give the wheels independent flexibility. We may also modify the shape of a live axle to attach the differential to the chassis and reduce the unsprung mass"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003947_robot.2004.1308858-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003947_robot.2004.1308858-Figure3-1.png",
        "caption": "Fig. 3. General configuration of a biped mbot under interaction forcdtorque from ground and environment. The biped feet are posed on two different planar surfaces and are subjected to individual forcelmoment pairs from each, F ~ / M I (left foot) and F,JM, (right foot). The biped interacts with the environment through individual forcdmoments at the hands, Q,/uI at left hand and QJur at right hand. There can be unexpected interaction forcdtorque Qi/ui active at any arbitrary point on the robot body., COP is not well-defined, nor. are FRI point and GCoM in this case. H o can however successfully determine the state of stability of the biped. An arbitmy inertial coordinate frame is shown situated at 0.",
        "texts": [
            " With the objective of controlling ZMP through linear and angular momenta, ZMP was expressed in terms of the latter quantities. This work was extended to resolved momentum control [23]. We strongly agree with the view that angular momentum can be exploited for general motion planning of legged robots. In this paper our focus is somewhat different, we wish to underline the relationship between angular momentum and biped stability. Iv. ANGULAR MOMENTUM RATE CHANGE FOR A GENERAL BIPED A. The general case The robot (refer to Fig. 3) feet are assumed to be on two different planar support surfaces and subjected to forcelmoments FI/MI (left foot) and F , / M , (right foot). M I and M , are normal to the respective support surfaces which are oriented in a general way in the 3D space. Consequently, each M l and M , bas one non-zero component along the respective surface normals. Robot hands are similarly subjected to a completely general forcelmoment QI/q (left) and Qrju7 (right). Due to the hands' grasping capability, UI and ur are not constrained to be normal to any surface"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure3.33-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure3.33-1.png",
        "caption": "Fig. 3.33: An example of singular configuration of a planar 3R manipulator for which outward radial velocity is impossible and transversal velocity is not a unique function of joint variables.",
        "texts": [
            " Singular configurations can occur at workspace boundary configurations when the manipulator is fully stretched or folded back on itself with the end-effector at its extreme position. In addition, singularities can occur within the workspace when two or more joint axes are lined up. In this last occurrence, the loss of d.o.f.s correspond also to the fact that the motion of the end-effector can also occur along a possible direction but it is not uniquely determined as function of joint motions, as shown in the example of Fig. 3.33. Therefore, investigating the Jacobian for singular configurations gives a procedure for singularity analysis of manipulators. The case of a planar 2R manipulator can be studied by referring to the velocity analysis in the example in 3.3.1. By looking at the expression of 0 v3 and 3 v3 a 2 \u00d7 2 Jacobian matrix can be obtained as the mapping of Eq. (3.4.7) but referring to linear velocity only, in the form 12212211 12212211 0 cacaca sasasa J + \u2212\u2212\u2212 = (3.4.11) and 2211 11 3 aaca 0sa J + = (3.4.12) Fundamentals of Mechanics of Robotic Manipulation 147 If angular velocity is considered, a 3 \u00d7 2 Jacobian matrix can be obtained by including a third row with the entries equal to 1, as the expression of 3\u03c93. But in this case the Jacobian will not be a square matrix that can be useful for determining singular configurations of the manipulator. The determinant of J3 can be computed straightforwardly as 2213 saaJdet = (3.4.13) whose condition for singularity gives 0sin 2 =\u03b8 (3.4.14) which refers to the manipulator in fully stretched and folded back configurations. In these configurations the manipulator has lost one d.o.f., as in the illustrative case of Fig. 3.33. The computation of the Jacobian by using the derivatives of position equations would have given the same results for the 2 \u00d7 2 Jacobian as J0 and J3. But it would not have permitted the computation of the angular velocity entry since there is no orientation vector whose derivative gives the angular velocity vector. One of the most important features of robotic manipulators is the positioning of objects or end-effector in static configurations. In this task a manipulator acts like a structure when the joints are locked in a suitable way"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.11-1.png",
        "caption": "FIGURE 7.11. A rear-wheel-steering vehicle.",
        "texts": [
            " The steering behavior of a vehicle with a differential is relatively stable under changing tire-road conditions. However, the total thrust may be reduced when the traction conditions are different for each drive wheel. Example 262 F Rear-wheel-steering. Rear-wheel-steering is used where high maneuverability is a necessity on a low-speed vehicle, such as forklifts. Rear-wheel-steering is not used on street vehicles because it is unstable at high speeds. The center of rotation for a rear-wheel-steeringe vehicle is always a point on the front axle. Figure 7.11 illustrates a rear-wheel-steering vehicle. The kinematic steering condition (7.1) remains the same for a rear-wheel steering vehicle. cot \u03b4o \u2212 cot \u03b4i = w l (7.30) Example 263 F Alternative kinematic steer angles equation. Consider a rear-wheel-drive vehicle with front steerable wheels as shown in Figure 7.12. Assume that the front and rear tracks of the vehicle are equal and the drive wheels are turning without slip. If we show the angular velocities of the inner and outer drive wheels by \u03c9i and \u03c9o, respectively, the kinematical steer angles of the front wheels can be expressed by \u03b4i = tan\u22121 \u00b5 l w \u00b5 \u03c9o \u03c9i \u2212 1 \u00b6\u00b6 (7"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000763_j.jmbbm.2013.01.021-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000763_j.jmbbm.2013.01.021-Figure1-1.png",
        "caption": "Fig. 1 \u2013 Arrangement of the tensile specimens on the base plate with the angle between the building and longitudinal direction: (a) 01, (b) 451, and (c) 901.",
        "texts": [
            " The laser scan speed and powder stacking thickness were fixed at 50 mm s 1 and 0.05 mm, respectively. The longitudinal direction of the dumbbell specimens was fixed parallel to the building direction for each condition. To investigate the mechanical anisotropy of the SLM build, the dumbbell specimens (n\u00bc3) were fabricated at an angle between the building and longitudinal direction (i.e., the tensile direction) of 01, 451, and 901 under a laser power of 200 W, scan spacing of 0.2 mm, and a stacking thickness of 0.05 mm, as shown in Fig. 1. To investigate the corrosion resistance, plate-shape specimens (15 mm 10 mm 1 mm, n\u00bc5) were prepared using SLM under a laser power of 200 W and 150 W and scan spacing of 0.1 and 0.2 mm, respectively. The narrow side of the rectangle (10 mm) was fixed parallel to the building direction. The laser scan speed and powder stacking thickness were fixed at 50 mm s 1 and 0.05 mm, respectively. Cr\u20136Mo alloy powder (mass%). Fe Ni C O N 0.11 0.02 0.009 0.160 0.021 Dumbbell and plate-shape specimens with the same alloy composition were also prepared by conventional dental casting for comparison"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000403_j.matdes.2013.05.070-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000403_j.matdes.2013.05.070-Figure3-1.png",
        "caption": "Fig. 3. Transient temperature distribution during layer melting (a) at the beginning of first track scan, (b) at the end of the first track scan at time = 0.091 s and (c) at the end of the 5th track scan.",
        "texts": [
            " The equivalent stress or VonMises can be computed as, reqm \u00bc ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 2 \u00bd\u00f0r2 r2\u00de2 \u00fe \u00f0r2 r3\u00de2 \u00fe \u00f0r3 r1\u00de2 r \u00f018\u00de The equivalent stress is related to the equivalent strain by this relationship, r0 \u00bc Ee0 \u00f019\u00de For stress analysis, in addition to density q, the following thermo-structural material properties depending on temperature are required are thermal expansion coefficient a, elastic modulus E, Poisson\u2019s ratio m, and yield strength ry, as listed in Table 2. In SLM, the temperature distribution in the powder bed and consolidated layers changes rapidly with time and space. Fig. 3a shows the temperature at the beginning of the laser scanning, from which, the very high temperature gradients in the vicinity of the laser spot on the powder bed can be clearly seen due to an applied Gaussian heat source. The temperature of the powder particles is elevated rapidly under the action of absorbed energy, causing a molten pool when the temperature exceeds the melting temperature and heat affected zones in the surrounding loose powder. Note that the energy intensity of the source might also be high enough to cause the material to evaporate [16]. The highest predicted temperature corresponding to the molten zone of the powder material is 2600 K for [P = 100 W, V = 100 mm/s] and exceeds the melting temperature of 316L stainless steel (1672 K). However, this maximum temperature at the start of track 1 is reduced at the end of first track scan to 2392 K and so at the end of fifth track to 2225 K as shown in Fig. 3b and c respectively. The drop of maximum temperature can be attributed to the increased conductivity of the previously solidified regions of the track compared to the low thermal conductivity available initially in the powder bed. The thermal field changes as the laser source moves along the track and the melt pool moves along with the laser source. It is further observed that the temperature gradient in the front side of the moving laser beam is much steeper than that in the rear side. The melt pool shape resembles as in comet tail profile (see Fig. 3c). This trend of skewed temperature distribution towards the rear of the laser was also reported in other temperature simulations [13,22]. This can be attributed to the fact that the rapidly cooling molten material has greater conductive properties than the untreated powder in front of the laser. The temperature distribution in the layer is very much affected by the energy density which is a function of laser power, spot size, scanning speed, hatch spacing and scanning strategy. Additionally, the temperature gradient in the layer is similarly influenced by conductivity of the material underneath the deposited layer being a loose powder bed, previously re-melted layer, support structure or a solid substrate"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001533_j.ymssp.2012.06.021-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001533_j.ymssp.2012.06.021-Figure2-1.png",
        "caption": "Fig. 2. A two-stage planetary gearbox test rig: (a) experimental system and (b) simulated model.",
        "texts": [
            " Gears with a chipped tooth and a missing tooth are respectively installed in the test rig and vibration signals are collected under the loaded condition with the value of 13.5 nm and various motor speeds. The diagnosis results validate that the ASR method is able to discover the weak characteristics and diagnose the gear faults effectively. In order to demonstrate the effectiveness of the proposed ASR method in fault diagnosis of planetary gearboxes, a planetary gearbox test rig is established and experiments on it are conducted. The experimental system is shown in Fig. 2(a) and the simulated model of the gearbox test rig in Fig. 2(b). The test rig includes two gearboxes, a 3-hp motor for driving the gearboxes, and a magnetic brake for loading. The motor rotating speed is controlled by a speed controller, which allows the tested gear to operate under various speeds. The load is provided by the magnetic brake connected to the output shaft and the load can be adjusted by a brake controller. As shown in Fig. 2(b), there are two gearboxes in the test rig: a two-stage planetary gearbox and a two-stage fixed-axis gearbox. The present study just concerns the two-stage planetary gearbox. In each of the two stages, an inner sun gear is surrounded by three or four rotating planet gears, and a stationary outer ring gear. The load is transmitted through the sun gear to the planets, which ride on a planetary carrier. The planetary carrier, in turn, transmits torque to the output shaft. In a planetary gearbox, sun gear teeth easily suffer damage because their multiplicity of meshes with the planet gears increases the potential for damage on the sun gear"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure5.28-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure5.28-1.png",
        "caption": "FIGURE 5.28 Two wires in series for computation of retarded partial inductance.",
        "texts": [
            "17) is exact if (xe \u2212 xs) is much larger than the tube diameter d. From the previous section, we get the impedance model for a retarded partial self-inductance as ZR Lp,11 = j\ud835\udf14Lp11 + \ud835\udf14 2 v \ud835\udf070 4\ud835\udf0b (xe \u2212 xs)2, (5.65) where Lp11 is the quasistatic partial inductance computed with the above formula and the resistive part is given by (5.63). The evaluation of the retarded partial mutual inductance between two distant wire segments is more challenging. Here, we give a small example for the simpler case when the two wire segments are lined up as shown in Fig. 5.28. A question is how the large phase factor associated with a mutual inductance is evaluated for the coupling term. The integral (5.64) reduces for this case to LpR 12 = \ud835\udf070 4\ud835\udf0b \u222b e1 x1=0 \u222b e2 x2=b2 e\u2212j\ud835\udefd|x2\u2212x1||x2 \u2212 x1| dx2 dx1. (5.66) PARTIAL INDUCTANCES WITH FREQUENCY DOMAIN RETARDATION 123 We use the following change in the variables where we replace x2 by setting x = x2 \u2212 b2 in (5.66). This leads to the following result: LpRM 12 = \ud835\udf070e\u2212j\ud835\udefdb2 4\ud835\udf0b \u222b e1 x1=0 \u222b (e2\u2212b2) x=0 e\u2212j\ud835\udefd|x\u2212x1||b2 + x \u2212 x1| dx dx1, (5.67) where we take the constant delay factor e\u2212j\ud835\udefdb2 outside of the integral. Finally, if we add and subtract Lp12 e\u2212j\ud835\udefdb2 , the result will be LpR 12 = e\u2212j\ud835\udefdb2 Lp12 \u2212 e\u2212j\ud835\udefdb2 Lp12 + LpRM 12 . (5.68) Different approaches can be used to approximate the second and third part on the right hand side of (5.68) since the delay term under the third term is now smaller such that series expansions can be used. We observe from Fig. 5.28 that the starting points of the two wires are in general much farther apart, leading to a large delay factor e\u2212j\ud835\udefdb2 , which is for most cases much larger than the small ones for the evaluation due to the wires segments with a length of \ud835\udf06\u221520 or smaller. Also, we can view the second term as a correction factor to the conventional solution where the quasistatic solution is multiplied by the delay factor e\u2212j\ud835\udefdb2 Lp12. An example where this approach is used is given in Section 13.5.6. We again refer to the example in Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure2.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure2.6-1.png",
        "caption": "Figure 2.6",
        "texts": [
            " The starting point and the end point of the force polygon determine the direction and magnitude of the resultant R for all forces: the arrow for R runs from starting point A to end point B (see Figure 2.4). The magnitude and direction of R can be measured or calculated using the drawing. If you look closely, you will recognise the force polygon in Figure 2.2. 26 Figure 2.5 shows all the forces in changing order in a force polygon. The order clearly does not influence the end result (the vector addition is associative and commutative, see Section 1.3.7). Example 2 Two forces F1 and F2 are acting on the ring in Figure 2.6. Their directions are shown in the figure. The forces are not shown to scale. Question: Find the magnitude and direction of the resultant force on the ring if F1 = 1000 N, F2 = 750 N. Solution: In Figure 2.7, the forces have been drawn to scale, with 1 cm =\u0302 250 N. Using the parallelogram rule in Figure 2.7a or the force polygon in Figure 2.7b, we can construct the resultant R. Through measuring we find that R has a length of approximately 5.95 cm, so that R \u2248 5.95 \u00d7 250 N = 1488 N. With a protractor, we find that the line of action of R makes an angle \u03b3 of approximately 22"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000028_0005-1098(94)90209-7-Figure11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000028_0005-1098(94)90209-7-Figure11-1.png",
        "caption": "FIG. 11. Partial fluid lubrication, regime III of the Stribeck",
        "texts": [
            " Some is expelled by pressure arising from the load, but viscosity prevents all of the lubricant from escaping and thus a film is formed. The entrainment process is dominated by the interaction of lubricant viscosity, motion speed and contact geometry. The greater viscosity or motion velocity, the thicker the fluid film will be. When the film is not thicker than the height of the asperities, some solid-to-solid contact will result and there will be partial fluid lubrication. When the film is sufficiently thick, separation is complete and the load is fully supported by fluid. Partial fluid lubrication is shown schematically in Fig. 11. The dynamics of partial fluid lubrication can perhaps be understood by analogy with a water skier. At zero velocity the skier is supported buoyantly in the water. Above some critical velocity the skier will be supported dynamically by his motion. Between floating and skiing there is a range of velocities wherein the skier is partially hydrodynamically supported. These velocities are analogous to the regime of partial fluid lubrication. The analogy is imperfect in that the buoyant support is not like solid-to-solid contact; and the dynamic support of the skier is due to fluid inertia as opposed to viscosity, the dominant force in lubrication"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002611_j.ijfatigue.2020.105737-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002611_j.ijfatigue.2020.105737-Figure3-1.png",
        "caption": "Fig. 3. Surface textures of different series obtained by CT scans over the notch region; (a) spec. A3, (b) spec. B13, (c) spec. C1, (d) spec. D1, and (e) spec. E1.",
        "texts": [
            "2 but it is worth summarizing the results in Table 3, where: \u2022 Ra (mean roughness value) and Rz (average maximum height of the profile) were calculated as the average value of all measurements for each series of specimens. \u2022 Maximum values of Rv (maximum depth of valley), Rp (maximum profile peak), and Rt (maximum height of the profile) were measured as the maximum value for each single specimen and then the mean values were calculated for each series of specimens. . An overview of the surface for different orientations can be taken from Fig. 3, which are 3D visualizations of CT scans (the method as applied to additive manufacturing was reviewed in [73]). More details of CT scans of these specimens are reported in detail in [74], where the surface roughness is evaluated in more detail to find \u201ckiller\u201d notches attributed to the roughness. Some of these may be missed by tactile probe due to the complex nature of the additive surface, with undercuts and deep notches. Briefly, CT scan settings of 150 kV and 100 \u03bcA with a voxel size of 15 \u03bcm were used and detailed correlation of failure is made in relation to individual surface features. In the present work, the focus is on the fatigue performance. In Fig. 3 it is seen that a type of stair-case effect due to the low curvature of the specimens in series A is visible and the rough surface in series D with roughness increasing along one side of the curve can be related to the downskin (downward facing) surface (even if roughness parameters in Table 3 are very similar). Series B, C, and E have approximately the same mean surface roughness values but smaller values for valley depths (shallower) for series C with respect to series B and E. The CT scans for series B/C/E do not show significant differences but it is clear that the smoothest surface is found for series B"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure16.12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure16.12-1.png",
        "caption": "Figure 16.12 (a) Hinged beam with set of loads and the influence lines for (b) the bending moment at D and (c) the shear force at D.",
        "texts": [
            " Influence lines are often used for determining the most unfavourable placement of the load, the placement where the load has the most severe effect on the quantity in question. We also provide a number of examples of this. Example 1 \u2013 Set of loads From an influence line, we can read off the influence of a point load with a variable position on a certain quantity (support reaction, section force) at a 756 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM fixed location. We can also say that the influence line gives the variation of a certain quantity due to a movable unit load. The value of the quantity is found at the position of the point load. Figure 16.12 shows a hinged beam with the influence lines for the bending moment and the shear force at D. The positive directions of MD and VD are shown with the influence lines. The hinged beam is loaded in field BC by a set of forces. The length of the beam, and the positions and magnitudes of the forces are shown in the figure. For each of the point loads, the influence line shows the associated influence quantity at the position of the load. The influence quantity for the bending moment (MD/F ) has the dimension of length. Figure 16.12b therefore includes the values in metres. The influence quantity for the shear force (VD/F ) is dimensionless. The influence lines give the following for the bending moment: MD = \u2212 ( 5 4 m ) \u00d7 (15 kN) \u2212 ( 5 6 m) ) \u00d7 (30 kN) \u2212 ( 5 12 m ) \u00d7 (45 kN) = \u221262.5 kNm and for the shear force VD = \u2212 1 4 \u00d7 (15 kN) \u2212 1 6 \u00d7 (30 kN) \u2212 1 12 \u00d7 (45 kN) = \u221212.5 kN. The correctness of these values can be checked by considering the equilibrium equations. Example 2 \u2013 Uniformly distributed load In Figure 16.13a, the beam in the previous example is loaded along its entire length by a uniformly distributed load q = 80 kN/m"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000637_978-3-642-83006-8-Figure2.19-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000637_978-3-642-83006-8-Figure2.19-1.png",
        "caption": "Fig. 2.19. Modelling the two-link foot of suppoting leg",
        "texts": [
            " Modelling the two-linked foot vement of the foot that is being in contact with the ground is accom plished in three stages, as shown in Fig. 2.18. on the toes. Correspondingly, the position of the ZMP is under the heel, under the foot and under the toes. After stage III, the ZMP jumps over to a position under the other foot, at the very moment this foot tou ches the ground. Afterwards, the motion sequences are repeated. modelled as if composed of two links. The first link corresponds to the toes and the second to the rest of the foot (Fig. 2.19). This kind of the foot modelling and displacement of the ZMP is executed in the programme package within the subroutine called SUPP in the fol lowing way. Let us consider the foot which is in the support phase. It always be longs to the 1-st kinematic chain which is represented by the 1-st row in the matrix MS. Let the foot be in phase III when two-link model is 95 required. Then, the link \"in corresponds to the toes and \"i+1\" to the rest of the foot (Fig. 2.19. (a)). The law of the change of the angle of the joint \"i+1\", connecting these two links, is prescribed. When the foot is in the stage I or II, it is modelled as a single link (Fig. 2.19. (b)) according to the following procedure. The data related to the complete foot, are associated with the i-th link, while the \"i+1\"-st link is considered as fictious (its mass, inertia moments and length are zero) which is in Fig. 2.19(b) represented by the dashed line. Now, i+1 the angle q is constant during the phases I and II. The flow chart of this procedure is presented in Fig. 2.20. 96 2.7. The Prescribed Part of Dynamics The prescribed part of dynamics comprises data on the synergy of lower extremities of an anthropomorphic robot and the law of the ZMP displa cement. The data on movements of legs are obtained from biometric mea surements of a normal human gait. These data are referred to the angle trajectories of the hip, knee, and ankle joint as a function of time",
            " T e 1 8 = ( 1 , 0, 0) ct - T e 1 9 = (- 1 ,0 ,0 ) ~ T e 2 0 = (- 1 ,0 ,0 ) ~ '\" '\" 127 sagittal (b) plane 128 second step is to calculate the right-hand sides of the differential equations describing the system's equilibrium in both the sagittal and frontal plane. 129 This procedure enables numerical integration of the system in both phases of the gait. As in the previous examples, an iterative procedu re is to be employed for the synthesis of nominal dynamics. Compensating movements for different duration periods p, and ZI1P law (Fig. 2.23) of the double-support phase are presented in Fig. 2.46-2.47. Example 5. This example is related to the modelling of a two-link foot Fig. 2.19. The mechanism consists of 19 links and 19 revolute jOints (Fig. frontal (2.49a) and sagittal (2.49b) plane, respectively. The right foot is modelled by links 2 and 3, the left one by links 13 and 14. The arms are fi xed, i.e. q16=q18=1.862253 [radj and q17=q19=1.5707 [rad]. The conditi ons q8: = q8_q 4 and q9: = q9+q 4 ensure the pelvis (link 8) is in the horizontal position and the legs are parallel. in the frontal plane. Other input data are: N=3; 11=1,12=8,1 3 =9; K1 =14, K2 =11, K3 =11,NT=19; t;i=o, i=1, \u2022\u2022\u2022 ,19; t;~=1, t;~5=1; value t;~iS1fOriE:{4,13} and 0 for other values of index i; r01 =(O, 0, -O"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure4.33-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure4.33-1.png",
        "caption": "Figure 4.33 Examples of two-hinged frames.",
        "texts": [
            " 4 Structures 129 fa\u00e7ade made of posts and girders (columns and beams), with perpendicular wind loading, can sometimes also be seen as a beam grillage. Calculating the forces and deformations in a beam grillage is in fact a threedimensional problem. For information about the spatial character, refer to Section 3.3.4, examples 1 and 2. Frames are planar, bent beams structures that are loaded in the plane of the structure. Such structures are often used to cover a space (warehouse, sports arena, and so forth). Figure 4.32 shows a number of simple examples of frames. In Figure 4.33, both fixed supports have been replaced by hinged supports, so that the structure is now referred to as a two-hinged frame. If the structure with hinged supports itself consists of two parts joined by a hinge, this is referred to as a three-hinged frame (see Figure 4.34). If the beam structure is not bent but arched, then the structure in Figure 4.35a is called a two-hinged arch, and the structure in Figure 4.35b a three-hinged arch. 130 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM It will be clear that a wide range of planar structures can be constructed using line elements"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001185_j.ymssp.2017.08.002-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001185_j.ymssp.2017.08.002-Figure5-1.png",
        "caption": "Fig. 5. Rolling bearing fault test rig.",
        "texts": [
            " In Case 1, diagnosis for the single fault is performed. In Case 2, diagnosis for the compound faults is carried out. Table 2 The parameters of rolling bearing fault test rig. Parameter description Value Bearing specs NSK Rotating speed 800 rpm Sampling frequency 25.6 kHz Roller number 10 Pressure angle 6.35 mm Contact angle 0 Outer race diameter 27 mm Inner race diameter 14 mm In order to create different kinds of single fault, a custom-built rolling bearing fault test rig is established. As shown in Fig. 5, the test rig mainly consists of motor, shaft, transmission, flange and testing bearing. The accelerometer is placed on the shaft box for measuring the vibration signals of the testing bearing, and the measured vibration data is stored by advanced data acquisition system. More parameters in this experiment are available in Table 2. In this case study, six operating conditions of bearing are created, including normal condition, slightly outer race fault condition, inner race fault condition, ball fault condition, severe outer race fault condition and rotor-bearing unbalance fault condition"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001496_ja9076793-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001496_ja9076793-Figure2-1.png",
        "caption": "Figure 2. (a) Scheme of a droplet in a channel. The presence of HDA at the liquid-air interface gives rise to convective flows. Since more HDA is present in the direction facing the source of acid, the flows and forces are asymmetric. (b) Velocity field based on the theoretical model described in the main text (calculated using the Fluent computational fluid dynamics package from Ansys). (c) Experimental image of the convection rolls visualized using Neutral Red indicator (see also video 3 in the SI). (d) Numerical solution of the concentration profile of HDA in solution. The profile shows asymmetry (more HDA on the \u201cright\u201d side of the droplet facing the source of HCl than on the \u201cleft\u201d side), which explains the motion of the droplet toward the acidic gel.",
        "texts": [
            " The difference in the rate of motion could be due to the difference in partition coefficients of HDA between the phases (mineral oil/KOH solution vs DCM/KOH solution), which can affect the rates of HDA release from the droplets. We emphasize that in all of these experiments, the presence of both HDA and the pH gradient was crucial to the droplet\u2019s \u201cchemotaxis\u201d: when the droplets lacked HDA or when the pH was uniform throughout the maze, no directional motion of the droplets was observed.12 Maze solving by HDA-containing drops can be explained by surface tension effects stemming from the nonuniform distribution of HDA at the liquid-air interface (Figure 2). As HDA diffuses from the droplet, it partially deprotonates in the pH gradient. Importantly, more protonated HDA is found in the direction facing the source of acid (i.e., toward lower pH). This asymmetric distribution translates into a gradient of surface tension that is determined predominantly by the concentration of the protonated HDA at the liquid-air interface (deprotonated HDA is much less surface-active; see ref 13). The surface tension gradients in turn give rise to convective flows,14 which are ultimately responsible for particle motion. In this context, it should be noted that although there are Marangoni-like flows inside of the droplet, their velocities (\u223c0.1 mm s-1) are too small to account for droplet\u2019s translation with the experimentally observed velocities. The above, qualitative description can be formalized in the form of Navier-Stokes (NS) equations describing the fluid dynamics of the system. Consider the model geometry illustrated in Figure 2a, where the HDA droplet is in the middle of a channel and the acidsoaked gel is at one end of the channel. The conservation equation for HDA can be written as v \u00b7\u2207c ) D\u22072c - kBc, where v is the velocity of the liquid, c is the concentration of HDA, D is its diffusion coefficient, and kB is the rate of consumption of HDA by the basic solution. Since the amount of HDA emitted by the droplet is small (\u223c10-3 \u00b5L min-1, estimated by monitoring the droplet size as a function of time), the equation assumes a quasi-steady-state approximation",
            " In the dilute approximation, which is applicable at the micromolar concentrations used here, the relationship between surface tension and HDA concentration is linear:16 \u03b3 ) \u03b30 - bc, where \u03b30 is the surface tension of pure water and b is a constant of proportionality. With the parameter values D ) 10-9 m2 s-1, kB ) 1 s-1, |Jdroplet| ) 10-5 mol m-2 s-1, |Jgel| ) 10-3 mol m-2 s-1, and b ) 0.05 N m2 mol-1 (see ref 17 and also section 3 in the SI), the system of NS equations can be solved numerically. Figure 2b shows the calculated velocity vector field in the liquid, where the flows form two convection rolls, one on each side of the droplet. Such rolls were also observed in the experiments (Figure 2c), and the calculated and experimentally measured fluid velocities were both on the order of centimeters per second. Importantly, the concentration gradients of HDA are steeper in the direction facing the source of acid (to the \u201cright\u201d in Figure 2d), and consequently, the fluid velocities are also higher in this direction. By Bernoulli\u2019s principle [applicable because the Reynolds numbers for the system are Re \u223c O(103)], higher velocities translate into a lower pressure and a net force, Fnet, that originates from the sum of two major contributions integrated over the particle\u2019s surface: (i) pressure forces, Fpress ) \u222bP dy, and (ii) viscous drag forces due to shear stresses, Fvis ) \u222b\u00b5(\u2202Vx/\u2202y) dx. Overall, Fnet is directed toward the acidic gel, and its value is \u223c6 \u00d7 10-9 N"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002400_tcyb.2017.2692767-Figure9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002400_tcyb.2017.2692767-Figure9-1.png",
        "caption": "Fig. 9. Inverted double pendulums connected by a spring and damper.",
        "texts": [
            " Further, the following modified signal proposed in [34]: x\u2032 1,d = \u03c6(t)yd + (1 \u2212 \u03c6(t))x\u2032 1 (69) is introduced for comparison. As displayed in Fig. 8, a better transient response performance is obtained via applying (66), in the sense that faster convergence rate and less overshoot are achieved. Additionally, the control method in [34] requires that the link angular velocity is measurable directly. B. Inverted Double Pendulums To certify the effectiveness of the developed method for the MIMO case, we perform another typical example, i.e., inverted double pendulums, as shown in Fig. 9. The inverted pendulum is a model abstracted from a variety of practical systems, such as rocket launchers, biped robots and two-wheel self-balancing vehicles. Building an accurate mathematical model for inverted pendulums is difficult, due to the high nonlinearity and strongcoupling. In addition, the cost of outfitting velocity sensors for such systems is too high. Hence, a model-free output feedback controller is desired in practical applications. Consider the following inverted double pendulums: \u03b8\u03081 = ( m1gr J1 \u2212 kr2 4J1 ) sin(\u03b81)+ kr 2J1 (l \u2212 b)+ u1 J1 + kr2 4J1 sin(\u03b8\u03072)+ \u03c91 \u03b8\u03082 = ( m2gr J2 \u2212 kr2 4J2 ) sin(\u03b82)+ kr 2J2 (l \u2212 b)+ u2 J2 + kr2 4J2 sin ( \u03b8\u03071 )+ \u03c92 (70) where \u03b81 and \u03b82 denote angular positions; \u03b8\u03071 and \u03b8\u03072 represent angular rates; u1 and u2 are the control torques; m1 and m2 stand for the pendulum masses; J1 and J2 denote the moments of inertia; k is the spring coefficient; r is the pendulum height; l is the spring length; g denotes the gravitational acceleration; \u03c91 and \u03c92 represent external disturbances; and b is the distance between the pendulum hinges"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.34-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.34-1.png",
        "caption": "FIGURE 8.34. The kinematic model of the suspension of a front left wheel: (a) an internal roll center, and (b) an external roll center.",
        "texts": [
            " Suspension Mechanisms 473 find the roll center of the front or rear half of a car, we determine each suspension roll center and then find the intersection of the lines connecting the suspension roll centers to the center of the associated tireprint. Example 317 Roll center of double A-arm suspension. The roll center of an independent suspension such as a double A-arm can be internal or external. The kinematic model of a double A-arm suspension for the front left wheel of a car is illustrated in Figure 8.34. The suspension roll center in Figure 8.34(a) is internal, and in Figure 8.34(b) is external. An internal suspension roll center is toward the vehicle body, while an external suspension roll center goes away from the vehicle body. A suspension roll center may be on, above, or below the road surface, as shown in Figure 8.35(a)-(c) for an external suspension roll center. When the suspension roll center is on the ground, above the ground, or below the ground, the vehicle roll center would be on the ground, below the ground, and above the ground, respectively. Example 318 F Camber variation of double A-arm suspension"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001786_j.mechatronics.2011.02.007-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001786_j.mechatronics.2011.02.007-Figure2-1.png",
        "caption": "Fig. 2. Coordinates and assumption of rotor spin directions.",
        "texts": [
            " Using Assumption 1 and ignoring other secondary effects a standard Newton\u2013Euler formulation [27] gives the dynamics expressed in the body fixed frame J _#\u00fe # \u00f0J#\u00fe hre3\u00de \u00bc T; \u00f01\u00de where J = diag(JX, JY, JZ) is the inertia matrix T 2 R3 is the applied torque from the propellers, # \u00bc \u00bd _/ _h _w T are angular velocities, hr is the angular momentum of the rotors and propellers, e3 = [0 0 1]T defines the axis of rotation of the rotors. Again using Assumption 1 rotor angular momentum is approximately hr \u00bc JrXr \u00bc Jr\u00f0 X1 \u00feX2 X3 \u00feX4\u00de; \u00f02\u00de where Jr is rotor inertia and rotor speeds X1,X2,X3,X4 have directions defined in Fig. 2. In this work, no experiments identify numerical values of the parameters in (1). Rather, we use the mathematical form of (1) to design adaptive controls. As a rational for our approach, we point out that model-based linear controls exhibit stability in a limited region of attraction near the hovering position. Modelbased nonlinear controls would not only require time-consuming system identification for each individual quadrotor, but may be inadequate for aircraft that pick up payloads. A state-space model suitable for control design must include Z or Ze, since vertical movement occurs as a function of rotor angular speeds X = [Xi,X2,X3,X4]T",
            " The proposed method then significantly outperforms e-modification when acquiring a payload, while preventing weight drift and bursting. A.1. Control signals Here we show how simple state-space controls proportional to forces/torques can be transformed into rotor speeds, which are in turn transformed into supply voltages. The first state is Z and the actuating force is proportional to the square of rotor speeds FZ \u00bc bX2 1 \u00fe bX2 2 \u00fe bX2 3 \u00fe bX2 4; \u00f0A:1\u00de where b is a constant thrust coefficient of the propellers. The torques provided by propellers rotating as in Fig. 2 are T \u00bc T/ Th Tw 2 64 3 75 \u00bc Lb X2 4 X2 2 Lb X2 3 X2 1 D X2 1 \u00feX2 3 X2 2 X2 4 2 66664 3 77775; \u00f0A:2\u00de where L is length from rotor to quadrotor center of mass, and D is a constant drag factor. Solving (A.1) and (A.2) for angular speeds gives X2 1 X2 2 X2 3 X2 4 2 66664 3 77775 \u00bc 0:25 0 0:5 0:25 0:25 0:5 0 0:25 0:25 0 0:5 0:25 0:25 0:5 0 0:25 2 6664 3 7775 FZ=b T/=\u00f0Lb\u00de Th=\u00f0Lb\u00de Tw=D 2 6664 3 7775: \u00f0A:3\u00de Thus, rather than design speeds directly we choose to design controls u1 = FZ/b, u2 = T//(Lb), u3 = Th/(Lb), u4 = Tw/D"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure9.15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure9.15-1.png",
        "caption": "FIGURE 9.15 Two volume filament (VFI) model bars separated by distance s.",
        "texts": [
            " The inductive impedance at high frequencies \ud835\udf14L dominates over the resistance and the current flow is determined by the inductances. We observe that the coupled voltages due to the partial mutual inductances are larger toward the center of the conductor. They are represented by the voltage sources Vk in Fig. 9.14. We fundamentally can understand that the high-frequency current will be concentrated toward the periphery of the conductor due to the stronger mutual coupling for the internal conductor filaments. This is a physical interpretation of the skin effect. We use a two-conductor example in Fig. 9.15 to consider the proximity effect. It consists of how the presence of the second conductor will impact the current flow or the partial inductance of the first conductor. 230 SKIN EFFECT MODELING To study the proximity effect, we again use the same VFI 1D current flow model for both conductors where the ends are shorted as shown in Fig. 9.16. This shorted model was also used for including the skin effect in thin conductors in Ref. [44]. The equivalent circuit for this 1D-VFI model is shown in Fig",
            "39) where I1T is the sum of currents of the vector I1T , which contains all the filament currents on conductor 1, and I2T is the sum of currents of the vector I2T , which yields all the filament 232 SKIN EFFECT MODELING currents on conductor 2. Hence, the impedances can easily be obtained by summing up all the currents. Z11 = V1T\u2211N k=1 Ik , Z21 = V1T\u22112N k=N+1 Ik . (9.40) We finally can rewrite this in a form where each of the four impedances Zij consists of a frequency-dependent resistance and inductance, or[ V1T V2T ] = [ R11(\ud835\udf14) + j\ud835\udf14L11(\ud835\udf14) R12(\ud835\udf14) + j\ud835\udf14L12(\ud835\udf14) R21(\ud835\udf14) + j\ud835\udf14L21(\ud835\udf14) R22(\ud835\udf14) + j\ud835\udf14L22(\ud835\udf14) ] [ I1T I2T ] . (9.41) Next, we give an example for the frequency-dependent parameters for two rectangular conductors in Fig. 9.15 for the above derivation. They consist of two bars with \ud835\udcc1 = 1 cm, W = T = 50 \u03bcm, and spacing S = 55 to 100 \u03bcm. The subdivision of the conductors with nine segments along W and T results in a total of 81 filament conductors for each of the two conductors. Hence, the matrix size in (9.32) is 162 \u00d7 162. The choice of the number of subdivisions should be such that the cross-sectional dimensions of each subbar is less than a skin depth at the highest frequency of interest. The results given are for a frequency range of 100 kHz to 100 MHz",
            " However, the GIBC model requires a sufficiently large number of surface cells to form a good closed surface for a reasonably accurate result. This aspect is different for the VFI model that works well with a reduced accuracy even for a moderate number of cells or filaments. 244 SKIN EFFECT MODELING PROBLEMS 9.1 Volume filament approach Using the VFI approach, compute the impedance matrix of two parallel conductors (copper \ud835\udf0e = 5.8 \u00d7 107 S\u2215m), illustrated in Fig. 9.34. Use the subdivisions as shown in Fig. 9.15 and the shorted ends as in Fig. 9.16. Consider the dimensions xs1 = xs2 = 0, xe1 = xe2 = 2 cm, ys1 = 0, ye1 = 0.1 cm, ys2 = ye1 + 0.2 cm, ye2 = ys2 + 0.1 cm, zs1 = zs2 = 0, ze1 = ze2 = 50 \u03bcm. The result will be a frequency-dependent 2\u00d72 impedance matrix. PROBLEMS 245 9.2 Construct a Single-conductor VFI skin-effect model Construct a single-conductor skin-effect model similar to the previous problem. Compute the impedance from end to end of the rectangular conductor (copper \ud835\udf0e = 5.8 \u00d7 107 S\u2215m), illustrated in Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003641_ichr.2007.4813931-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003641_ichr.2007.4813931-Figure2-1.png",
        "caption": "Fig. 2. When only applying ankle torque, the ground reaction force points from the center of pressure (CoP) to the center of mass (CoM), creating zero moment about the CoM.",
        "texts": [
            " The stability of the robot can be determined by the statespace location, (xCoM, x\u0307CoM), of the center of mass. We want to find the limits on the state of the center of mass that can be stopped from leaving the base of support by a saturated ankle torque. The relation between horizontal acceleration of the CoM and the tangential ground reaction force, Fx, is x\u0308CoM = Fx m (1) where Fx is the tangential ground reaction force. If only ankle torque is used, then the ground reaction force points from the CoP to the CoM, as shown in Fig. 2. Then Fx is related to the normal force, Fz , by Fx = Fz zCoM (xCoM \u2212 xCoP) (2) where xCoM and xCoP are the locations of the CoM and CoP, respectively. If we assume the vertical motion of the CoM is negligible, then Fz = mg. Also, if we assume the ankle torque is saturated with the CoP at the edge of the foot at xCoP = \u03b4\u00b1, where \u03b4\u2212 and \u03b4+ are the back and front edges of support region, respectively, then the maximum acceleration is given by x\u0308max CoM = g L (xCoM \u2212 \u03b4\u00b1), (3) where L = zCoM is a constant height"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.38-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.38-1.png",
        "caption": "FIGURE 8.38. A double A-arm suspension mechanism when the wheel is at: (a) a positive displacement, (b) rest position, and (c) a negative displacement.",
        "texts": [
            "24) If the angle \u03b82 at the rest position is at \u03b820 = 121.5 deg (8.25) then the initial angle of the other links are: \u03b830 = 18.36 deg \u03b840 = 107.32 deg (8.26) At the rest position, the coupler point is at: xC0 = \u221222.73 cm yC0 = 9.23 cm z0 = 16.92 cm (8.27) We may calculate h and \u03b3 by varying the parameter \u03b82. Figure 8.37 illustrates h as a function of the camber angle \u03b3. For this suspension mechanism, the wheel gains a positive camber when the wheel moves up, and gains a negative camber when the it moves down. The mechanism is shown in Figure 8.38, when the wheel is at the rest position and has a positive or a negative displacement. 8.4 F Car Tire Relative Angles There are four major wheel alignment parameters that affect vehicle dynamics: toe, camber, caster, and trust angle. 8. Suspension Mechanisms 479 8.4.1 F Toe When a pair of wheels is set so that their leading edges are pointed toward each other, the wheel pair is said to have toe-in. If the leading edges point away from each other, the pair is said to have toe-out. Toe-in and toe-out front wheel configurations of a car are illustrated in Figure 8"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure4.7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure4.7-1.png",
        "caption": "FIGURE 4.7 (a) Parallel plate conductor. (b) Side view and field lines.",
        "texts": [
            " Further, a combination of the different techniques works best for some applications. We also can combine some models based on DE techniques together with partial PEEC model inductances. For this reason, we also consider simple Simple DE models are not well suited for single plate capacitance computations. The IE methods in the previous section yield relatively simple solutions for this type of problem. Problems for which DE method yields good approximate solutions are related to a parallel plate capacitor in Fig. 4.7. Starting from one of Maxwell\u2019s equations (3.1 c), we get \u2207 \u22c5 \ud835\udf160\ud835\udf16rE = q, (4.28) which can also be regarded as the differential form of the Gauss theorem. It is assumed that between the plates there we only have a uniform z component of the electric field. Applying the Gauss theorem to a surface S enclosing one of the plates, we obtain E = Q \ud835\udf160\ud835\udf16rS , (4.29) where Q = \u222b q dS = qS = q \ud835\udcc1 w is the total charge on the plates and we assume that a uniform dielectric with a relative dielectric constant \ud835\udf16r exists in the area between the plates",
            " For the twodimensional case, we assume that the structure is infinitely long in the z direction such as a transmission line structure. Hence, the result will be per unit length capacitances in the z direction. In this model, plates of the local capacitors are always between two nodes. The incomplete equivalent circuit corresponding to the mesh in Fig. 4.8 is shown in Fig. 4.9. A simple application of the equivalent circuit in Fig. 4.9 for a two-dimensional parallel capacitance problem is shown in Fig. 4.7. The upper plate is represented by the connected nodes 7\u20139 and the lower plate is represented by the connected nodes 12\u201314. Hence, the nodes and capacitances such as 7\u20138 are shorted. For the symmetric problem, we can assume that the upper plate is+0.5 V and the lower plate is\u22120.5 V as shown in Fig. 4.10. For clarity, we left all capacitances between the node out. Since we apply 1 V between the plates, we get Q = CppV = Cpp. Hence, we need to evaluate the total charge induced on one of the plates",
            "4 Elements of the short circuit capacitance matrix The off diagonal elements of a correctly computed short circuit capacitance matrix are negative. Please explain why the mutual elements are negative. We assume that all conductors are perfect electrical conductors (PEC) in a free space environment for the model. Explain the relationship between the capacitance model of the structure and component of the capacitance matrix. 4.5 Capacitance modeling Assume there are two congruent square perfect electrical conductor (PEC) plates as is shown in Fig. 4.7a. Each one has again a dimension of 1 \u00d7 1 m. The two plates are separated by a distance d = 0.2 m. Extend your program from problem 4.2 to this two-plate case. Apply the technique from Section 4.2.4 to compute the total capacitance of the two-conductor structure. What does this mean? To earn more points, try to apply some of the other techniques you did learn in this chapter. 1. J. C. Maxwell. A Treatease of Electricity and Magnetism. Dover Publications, New York, 1954. 2. D. K. Reitan and T. J. Higgins"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002629_j.ast.2020.105716-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002629_j.ast.2020.105716-Figure1-1.png",
        "caption": "Fig. 1. Quadrotor UAVs.",
        "texts": [
            " Section 2 describes the dynamic model of the quadrotor, including the fault-free model and fault-tolerant model. Section 3 designs the NTSMC based complete fault-tolerant control system, including the fault-tolerant controller, the fault-free controller, the FDD method and the external disturbances estimator. Section 4 shows the great performance of the proposed flight control method through numerical simulations. Conclusions are presented in Section 5. The dynamical model of the quadrotor UAVs is shown in Fig. 1. Four independent motors on which propellers are fixed are installed at the endpoint of an +-shaped frame. The motors are represented as Mi , i = 1, 2, 3, 4 and corresponding thrust forces generated by them are f i, i = 1, 2, 3, 4. The motors M1 and M3 spin in the clockwise direction with angular velocities \u03c91 and \u03c93. While the angular velocities of motors M2 and M4 are denoted as \u03c92 and \u03c94 in the counter-clockwise direction. The states of the quadrotor UAV can be controlled by changing the angular velocities of four motors. For instance, the quadrotor hovers with certain four equal angular velocities of motors, and the quadrotor rises or descends when increasing or decreasing the angular velocities of motors simultaneously. When f1 = f3, the quadrotor is conducted at the working mode of pitching. And the quadrotor works at the rolling mode when f2 = f4. As shown in Fig. 1, two frames are applied to describe the motion of the quadrotor. The earth-based coordinate system {E}(O , x, y, z) is an inertial frame which is used to express the position [x, y, z]T of the quadrotor. The body-fixed coordinate system {B}(O b, xb, yb, zb) is fixed to the center of mass of the quadrotor. The attitude angles [\u03c6, \u03b8,\u03c8]T between the earth-based coordinate and the body-fixed coordinate describe the orientation of the body frame, where \u03c6, \u03b8 and \u03c8 represent the roll angle, pitch angle and yaw angle, respectively"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure3.43-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure3.43-1.png",
        "caption": "Figure 3.43 The moment of the force F about a point A is defined as the vector product T = r \u00d7 F ; the moment vector T is perpendicular to the plane through r and F .",
        "texts": [
            " Nothing can be said about the sign associated with the direction of the couple until a coordinate system is chosen. So far, we looked at the equilibrium of a body only in the simple case in which all the forces and couples act in one plane. The moment was taken about a point in the same plane. In this section we look at the general three dimensional case. Here we have to define the concept of moment of forces and couples more generally. 3 Statics of a Rigid Body 81 Imagine a force F in space, with point of application B (see Figure 3.43). The moment T of this force about point A is now defined as the vector product (cross product) of the position vector r , from A to B, and the force vector F : T = r \u00d7 F . The vector product of two vectors r and F is a vector with magnitude rF sin \u03b8 and perpendicular to both r and F . Here r and F are the magnitudes of r and F respectively, and \u03b8 is the smaller angle between the vectors r and F when both are drawn outwards from the same point. There are two useful rules for finding the direction of the moment vector T "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002672_j.mechmachtheory.2020.103786-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002672_j.mechmachtheory.2020.103786-Figure2-1.png",
        "caption": "Fig. 2. Schematic of gear teeth: (a) crack propagation path of gear; (b) geometric sketch of gear crack.",
        "texts": [
            " (14) according to the involute properties: h y 2 = R b [ ( \u03b12 + \u03b1) cos \u03b1 \u2212 sin \u03b1] (14) Finally, the total time-varying stiffness k t of a pair can be calculated as follows: k t = 1 1 k h + 1 k b1 + 1 k s1 + 1 k a1 + 1 k f1 + 1 k b2 + 1 k s2 + 1 k a2 + 1 k f2 (15) where subscript 1 represents the pinion, and subscript 2 the gear. Because of excessive working load, inappropriate operating conditions, tooth fatigue, and other reasons, cracks may occur at the gear tooth root. This study assumes that the crack propagation path of the root of the tooth is a straight line. The crack runs through the entire tooth width, distributes uniformly in the direction of the tooth width, and gradually grows along the direction of crack depth. The propagation path is shown in Fig. 2 (a). As shown in Fig. 2 (b), q 1 is the initial crack length that does not cross the tooth centerline, q 2 is the crack length that goes through the tooth centerline with the aggravation of gear failure. When the gear cracks, the Hertzian stiffness and compression stiffness of the gear teeth will not change, but the bending stiffness and shear stiffness will change. After the occurrence of crack in the gear teeth, the cracked part will fail and affect the moment of inertia and area of the cross section of the gear teeth"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003908_978-1-84882-730-1-Figure11.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003908_978-1-84882-730-1-Figure11.1-1.png",
        "caption": "Figure 11.1 General framework of ship path-following",
        "texts": [
            "t/; 271 272 11 Path-following of Underactuated Ships Using Polar Coordinates where all of symbols have been defined in Section 3.3. The bounded time-varying terms, wu.t/; wv.t/; and wr .t/, are the environmental disturbances induced by waves, wind, and ocean currents with j wu.t/j wumax <1, j wv.t/j wvmax < 1, and j wr .t/j wrmax <1. In this chapter, we consider a control objective of designing the surge force u and the yaw moment r to force the underactuated ship (11.1) to follow a specified path \u02dd , see Figure 11.1. In this figure, P is the ship\u2019s center of mass and Pd is a point attached to the virtual ship, which moves along the path with speed of u0. If we are able to steer the ship to closely follow a virtual ship that moves along the path with a desired speed u0, then the control objective is fulfilled, i.e., the ship is in a tube of nonzero diameter centered on the reference path and moves along the specified path at the speed u0. Roughly speaking, the approach is to steer the ship so that it heads toward the virtual ship and eliminates the distance between itself and the virtual ship",
            "1) follow the path \u02dd given by xd D xd .s/; yd D yd .s/; (11.4) where s is the path parameter variable, such that lim t!1ze.t/ ze; lim t!1 j e.t/j e; (11.5) with ze and e being arbitrarily small positive constants. Assumption 11.1. 1. The reference path is regular, i.e., 0 < Rmin @xd @s 2C @yd @s 2 Rmax <1. Remark 11.1. 1. Assumption 11.1 ensures that the path is feasible for the ship to follow. 2. The angle d is not defined at ze D 0 but limze!0 d D s with s being the orientation angle of the virtual ship on the path, see Figure 11.1. Therefore, the fulfillment of the control objective guarantees that the ship closely follows the path in terms of both position and orientation. If the reference path is not regular, then we can often split it into regular pieces and consider each of them separately. This is a case of point-to-point navigation. 3. The path parameter, s, is not the arclength of the path in general. For example, a circle with radius of R centered at the origin can be described as xd D Rcos.s/ and yd DR sin.s/, see [136] for more details"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000416_s0022-460x(88)80365-1-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000416_s0022-460x(88)80365-1-Figure2-1.png",
        "caption": "Figure 2. A general dynamic model for a gear tooth. k. =equivalent stiffness of gear tooth; m. =equivalent mass of gear tooth; w = transmitted load.",
        "texts": [
            " Then Newton's second law of motion was applied to each rigid disk in the equivalent system by using the gear tooth force induced torques and a given output torque. From these equations together with the equations obtained from the\u00b7definition ofangular transmission error and from the force equilibrium condition, the unknown gear tooth forces were determined by simply solving a set of algebraic equations for each time increment. The basic characteristic of the models in this group is that the only compliance considered is due to the gear tooth and that all other elements are assumed to be perfectly rigid. The resulting models are either translational (Figure 2) or torsional (Figure 3). The distinction between translational and torsional models is not made according to the appearance of the model, but according to whether the translational motion of the tooth or the rotational motion of gear is modelled. As can be seen from Figure 3(b) and (c) some torsional models are presented in the form of their translational equivalents, while, in general, in such models the system is idealized as a pair of inertias coupled by a spring which permits relative motion"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000014_revmodphys.85.1143-Figure19-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000014_revmodphys.85.1143-Figure19-1.png",
        "caption": "FIG. 19. Sketch of a rotating spiral defect in an active nematic fluid with vanishing elastic anisotropy K \u00bc 0. The gray arrows mark the director, and the solid lines show its spiral structure. The hydrodynamic velocity field is in the azimuthal direction, and its profile is indicated by the dark arrows. Adapted from Kruse et al., 2004.",
        "texts": [],
        "surrounding_texts": [
            "The nonequilibrium thermodynamic description of active systems is a systematic approach based on symmetries and, in particular, on invariance against time inversion. It is, however, based on a linear expansion of fluxes in terms of forces and can in principle describe only systems close to equilibrium where tends to zero. One of the main applications though is to biological systems that are mostly far from equilibrium systems. There is no systematic extension of the theory to systems far from equilibrium. One must rely either on a microscopic description that generates nonlinear contributions by coarse graining to large length scales and long-time scales or on experimental results that emphasize specific nonlinear aspects, which can then be introduced in theory. Microscopic or mesoscopic descriptions of molecular motors are a good example of the first case and lead to motor forces or velocities that are not linear in (Ju\u0308licher and Prost, 1997; Liverpool et al., 2009). Several experiments suggest that the treadmilling associated with the polymerization and depolymerization of actin in a cell depends on the force applied on the filaments or the local stress in a nonlinear way (Mogilner and Oster, 1999; Prost et al., 2007).5 The force-dependent treadmilling is essential for many cellular processes such as cell migration or cell adhesion and a full description including these effects in the active gel theory has not yet been proposed (Keren et al., 2008). 5In a cell actin filaments polymerize at the so-called barbed end and depolymerize at the so-called pointed end. This constant turnover of active filaments is called treadmilling. Rev. Mod. Phys., Vol. 85, No. 3, July\u2013September 2013 Another intrinsic limitation of the current active gel theory is the assumption of linear rheology and the use of the Maxwell model with a single relaxation time. A large body of experimental work shows that for many types of cells as well as for actomyosin gels there is a broad distribution of relaxation times leading to a complex modulus decreasing as a power law of frequency with a small exponent between 0.1 and 0.25 (Fabry et al., 2001). An ad hoc power-law distribution of relaxation times could be introduced in the active gel hydrodynamic theory but despite some efforts, this type of law is not understood on general grounds (Balland et al., 2006). The nonlinear rheology of actin networks has also been studied in detail, and actin networks in the absence of molecular motors have been found to strain thicken (Gardel et al., 2004). This behavior can at least in part be explained by the inextensibility of the actin filaments. In the presence of myosin motors, the active stress induced by the molecular motors itself stiffens the active gel (Koenderink et al., 2009). The hydrodynamic description of active polar or nematic gels is very close to that of nematic liquid crystals. Nevertheless the existence of an active stress leads to several nonintuitive and spectacular phenomena. The most spectacular is the flow instability described below that leads to spontaneously flowing states. Other spectacular effects are associated with active noise in these systems. In all active systems the noise has a thermal component and a nonthermal active component. The properties of the active noise cannot be inferred from the macroscopic hydrodynamic theory and must be derived in each case from a specific microscopic theory. The study of tracer diffusion in a thin active film, for example (Basu et al., 2012), leads to an anomalous form of the diffusion constant that does not depend on the size of the particle but only on the thickness of the film. We believe that there are still many unusual properties of active gels to be discovered and that in many cases, this will require detailed numerical studies of the active gel hydrodynamic equations such as the one performed by Marenduzzo, Orlandini, and Yeomans (2007). Active gel models have also been used to describe crosslinked motor-filament systems that behave as soft solids at large scales (MacKintosh and Levine, 2008; Levine and MacKintosh, 2009; Yoshinaga et al., 2010; Banerjee and Marchetti, 2011) and also to understand the origin of sarcomeric oscillations, both from a microscopic and from a continuum viewpoint (Gu\u0308nther and Kruse, 2007; Banerjee and Marchetti, 2011). Further, active gel models have been shown to successfully account for the experimentally observed traction stresses exerted by cells and cell sheets on compliant substrates (Banerjee and Marchetti, 2011; Edwards and Schwarz, 2011; Mertz et al., 2012). Finally a large part of the theoretical activity on active gels aims at a quantitative description of biological phenomena at the scale of the cell and of cellular processes involving the cytoskeleton. Some success has already been obtained in discussing lamellipodium motion (Kruse et al., 2006) or the formation of contractile rings during cell division (Salbreux, Prost, and Joanny, 2009). An important step is the connection of the parameters of the hydrodynamic theory with the more microscopic parameters that can be monitored experimentally which requires an explicit coarse-graining of the microscopic theories as discussed in Sec. IV.D. At larger scales one can build a hydrodynamic theory of tissues that shares many features with the active gel theory described here (Ranft et al., 2010)."
        ]
    },
    {
        "image_filename": "designv10_0_0001446_j.oceaneng.2016.06.041-Figure10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001446_j.oceaneng.2016.06.041-Figure10-1.png",
        "caption": "Fig. 10. Rudder configuration of the AUV in the experiment.",
        "texts": [
            " This implies that the sliding mode based control can be used to address the model uncertainties and external disturbances while the input chattering is involved. For practical use, we should carefully select the control parameters and employ some methods to reduce chattering, such as a disturbance observer. In addition, due to the absence of persistent excitation, we cannot guarantee that the estimated value of a parameter converges to the real value. 6. Experimental results In this section, we provide experimental results on an AUV whose attitudes are controlled by four separate rudders with x-type rudders, as shown in Fig. 10. The relationship between the rudders \u03b4 =i, 1, 2, 3, 4i , and the control inputs \u03b4r, \u03b4e and \u03b4d used in this work can be written as \u03b4 \u03b4 \u03b4 \u03b4 \u03b4 \u03b4 \u03b4 \u03b4 \u03b4 \u03b4 \u03b4 \u03b4 \u03b4 \u03b4 \u2212 \u2212 \u2212 \u2212 = \u2212 \u2212 \u2212 \u2212 ( ) \u23a1 \u23a3 \u23a2 \u23a2 \u23a2 \u23a2 \u23a4 \u23a6 \u23a5 \u23a5 \u23a5 \u23a5 \u23a1 \u23a3 \u23a2 \u23a2 \u23a2 \u23a2 \u23a4 \u23a6 \u23a5 \u23a5 \u23a5 \u23a5 \u23a1 \u23a3 \u23a2 \u23a2 \u23a2 \u23a2 \u23a4 \u23a6 \u23a5 \u23a5 \u23a5 \u23a5 \u23a1 \u23a3 \u23a2 \u23a2 \u23a2 \u23a2 \u23a4 \u23a6 \u23a5 \u23a5 \u23a5 \u23a5 \u23a1 \u23a3 \u23a2 \u23a2 \u23a2 \u23a2 \u23a4 \u23a6 \u23a5 \u23a5 \u23a5 \u23a5 \u23a1 \u23a3 \u23a2 \u23a2 \u23a2 \u23a2 \u23a4 \u23a6 \u23a5 \u23a5 \u23a5 \u23a5 1 1 1 1 1 1 1 1 1 1 1 1 , and 1 1 1 1 1 1 1 1 1 1 1 1 48 e r d e r d 1 2 3 4 1 2 3 4 The diameter and the length of the AUV are 200 mm and 2015 mm, respectively. The nominal hydrodynamic parameters of the AUV, which are calculated using a commercial CFD software, are shown in Table 1"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000637_978-3-642-83006-8-Figure2.27-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000637_978-3-642-83006-8-Figure2.27-1.png",
        "caption": "Fig. 2.27. Compensating movements for the single-support gait upon level ground for T=1.5, 5=0.6 andZMPlawsfromFig. 2.22.",
        "texts": [],
        "surrounding_texts": [
            "are shown the reaction forces and driving torques, and in Fig. 2.30. is shown the power at joints. [rad] 0.1 o 9 q 0.1 0.2 105 T=I, S=0.4 S=0.8 T=.5 S=0.6 10 q [rad] 0.3 106 t[s) Fig. 2.29.b. Ankle joint torques for the single-support gait upon level ground for T=1.5, 5=0.6 and different ZMP laws - supporting phase Fig. 2.29.c. Knee joint torques for the single-support gait upon level ground for T=1.5, 5=0.6 and different ZMP laws - supporting phase [Nm) P5 30 10 t[s) 4-------------------~~~~-=~~~~~~ O ~ .75 -50 -100 -150 .~-.-. I! /-...... ~~\"L I z ...... \u00b7-\u00b7\u00b7\u00b7~, .. ::::::.;;;o- 1 i \"/ . . . . I , i i i ; u Fig. 2.29.d. Hip joint torques for the single-support gait upon level ground for T=1.5, S=0.6 and different ZMP laws - supporting phase Fig. 2.29.e. Ankle joint torques for the single-support gait upon [Nm) level ground for T=1.5, S=0.6 and different ZMP laws P 7 - swing phase . 15 10 5 o t[s) 0.75 Fig. 2.29.f. Knee joint torques for the single-support gait upon level ground for T=1.5, S=0.6 and different ZMP laws - swing phase 107 108 [NmJ Ps 20 10 0 Fig. 2.29.g. [NmJ Pg 30 10 0 -50 -100 -150 t[sJ 0.75 Hip joint torques for the single-support gait upon level ground for T=1.5, S=0.6 and different ZMP laws - swing phase Fig. 2.29.h. Trunk joint torques in the sagittal plane for the single-support gait upon level ground for T=l .5, S=0.6 and different ZMP laws [NmJ 10 o -20 -40 -- t[sJ 0.75 .j Fig. 2.29.i. Trunk joint torques in the frontal plane for the single-support gait upon level ground for T=1.5, S=0.6 and different ZMP laws 109 [W) 10 t[s) \\ . ~ I 1.5 -10 l __ \\.j \\1 Fig. 2.30.a. Power at ankle joint for the single-support gait upon level ground for T=1.5, 5=0.6 and different ZMP laws n [W) 20 10 t[s) 0 1.5 -20 . ; -40 \\i Fig. 2.30.b. Power at knee joint for the single-support gait upon level ground for T=1.5, 5=0.6 and different ZMP laws [W) 60 40 Fig. 2.30.c. Power at hip joint for the single-support gait upon level ground for T=1.5, 5=0.6 and different ZMP laws 110 [Wj A H 50 30 10 0 v Fig. 2.30.d. Power at trunk joint in the sagittal plane for the single-support gait upon level ground for T=1.5, 8=0.6 and different Zl\u00b71P laws Fig. 2.30.e. Power at trunk joint in the frontal plane for the single-support gait upon level ground for T=1.5, 8=0.6 and different ZMP laws .5 111 Example 2. In this example we consider the anthropomorphic mechanism shown in Fig. 2.31, which consists of 17 links and 17 revolute joints of the 5-th class. The topplogic,al structure represented by the serief' of .'+\" joints is of the form: MS \u2022 [: 2 3 4 5 6 7 8 9 10 11 1 ~] 2 3 4 5 6 7 13 14 15 0 (2.8.3) 2 3 4 5 6 7 13 16 17 0 The kinematic and dynamic parameters of the mechanism are given in Table 2.2. Fig. 2.32. shows the q3 and q13 d.o.f. for compensation in the frontal (2.32a) and sagittal (Fig. 2.32.b.) plane, respectively, un der the condition link 7 (pelvis) is parallel to the ground and the projection of the legs are parallel in the frontal plane. This condi tion is fulfilled by putting q7+q7_q 3 andq8+q8+q3. As in Example 1, the mechanism's arms are fixed, Le., q14 = q16 = 1.862253 [rad] and q15 = q17 = 1.5707 [rad]. The other data are: N = 31 I1 = 1, I2 = 7, I3 = 81 K1 = 12, K23= 10, K3 = 101 NT=171 E;i = 0, i=1, .\u2022\u2022 ,171 E;~ = 1, E;~3: 11 values E;i is 1m for iE{3, 13} and 0 for other values of index i1 r o1 =(O, 0, -0.0001)~1 e1 = (1, 0, 0) T 1 the initial values of the compensating part of the. system (q3(O), q13(O), q3(O) and q13(O\u00bb1 the repeatibility conditions are: w -1 o o o o o o o o (2.8.4) o -1 o o o Figs. 2.33 - 2.38. show the compensating movements for the different parameters and gait types. Example 3. In this case we consider the realization of an artificial anthropomorphic gait by a mechanism with free arms (Fig. 2.39). The mechanism consists of 19 links and 19 revolute joints of the 5-th class. Its topological structure can be represented by a series of \"+\" joint in the form of the matrix 112 ~4 Mechanical scheme of the antropomorphic mechanism with fixed arms sagittal (b) pl~ne T a b le :. 2 . K in e m a ti c a n d d y n a m ic p a ra m e te rs o f th e m e c h a n is m !-l as s M om en t of i n er ti a [k gm 2 ] D is ta nc e of t he a xe s ce nt re s of j o in ts f ro m th e Li nk Jo in t u n it a xe s [k g] J x J y Jz 1i nk c en tr e [m ] 1 2 3 4 5 6 7 1 0. 0 0. 0 0. 0 0. 0 z T r 1 ,1 = ( 0, 0, 0. 00 01 ) ; 1 1 ,2 = ( 0, 0, -0 .G O G 1) T z T e 1 = ( 1, 0, 0) 2 1. 5 3 C .0 00 06 0. 00 05 5 0. 00 04 5 f 2 ,2 = ( 0, 0, 0. 03 0) T ; f 2 ,3 = ( 0, 0, -O .0 70 )T \"t T e 2 = (0 ,1 ,0 ) 3 0. 0 0. 0 0. 0 0. 0 z T r 3 ,3 = ( 0, 0, 0. O e0 1) ; z T r 3 ,4 = ( 0, 0, -0 .0 00 1) \"t T e 3 = ( 1, 0, 0) -- 4 3. 21 0. 00 39 3 0. 00 39 3 0. 00 03 8 7 T r 4 ,4 = ( 0, 0, 0. 21 0) ; 7 T r 4 ,5 = ( 0, 0, -0 .2 10 ) \"t T e 4 = ( 0, 1, 0) 5 8. 41 0. 01 12 0 0. 01 20 0 0. O C 30 0 z T r 5 ,5 = ( 0, 0, 0. 22 0) ; t 5 ,6 = ( 0, 0, -O .2 20 )T -zT e 5 = ( 0, 1, 0 ) 6 0. 0 0. 0 0. 0 0. 0 \"t T r 6 ,6 = ( 0, 0, 0. 00 01 ) ; -z- T r 6 ,7 = ( 0, 0, -0 .0 0G 1) -zT e 6 = ( 0 , 1, 0) 7 6. 96 0. 00 70 0 0. 00 56 5 0. 00 62 7 \"t T r l ,7 = ( 0, 0. 13 5, 0 .1 ) ; z T r 7 ,8 = ( 0, -0 .1 35 , 0. 1) ; -z- T e 7 = ( 1 , Q , 0) -zT r 7 ,1 3= (0 , C, -0 .0 5) ; 8 0. 0 0. 0 0. 0 0. 0 \"t T r 8 ,8 = ( 0 , 0, -0 .0 00 1) ; z T r 8 ,9 = ( 0, 0, 0. 00 01 ) 7 T e 8 = ( 1 , C, 0) 9 15 .41 0. 01 12 0 0. 01 20 0 0. 00 30 0 f 9 ,9 = ( 0, 0, -0 .2 20 )T ; f 9 ,1 0 = ( 0, 0, C .2 2C )T 7 T e 9 = ( 0 , -1 , 0) , 10 3. 21 0. 00 39 3 0. 00 39 3 0. 00 03 8 ~1 0, 10 = ( 0 , 0, -O .2 10 )T ; z T r 1 o ,1 1 = ( 0 , 0, 0. 21 0) -z' T e 1 0 = ( 0, - 1 , 0) 11 0. 0 0. 0 0. 0 0. 0 -zT \"t T \"t T r 1 1 ,1 1 = ( 0, 0, -0 .0 00 1) ; r 1 1 ,1 2 = ( 0 ,0 , O .O G 01 ) e 1 1 = (0 ,- 1 ,0 ) ~ c.> T a b le 2 .2 . C o n t. ~ ~ 1 2 3 4 5 6 7 12 1. 53 0. 00 00 6 0. 00 05 5 0. 00 04 5 ~ T r 1 2 ,1 2 = ( 0, 0, -0 .0 70 ) ~ T e 1 2 = (1 ,0 ,0 ) 13 30 .8 5 0. 15 14 0 0. 13 70 0 0. 02 83 0 7 T r 1 3 ,1 3 = ( 0, 0, 0. 34 ) ; f 1 3 ,1 4 = ( 0, 0. 2, _0 .6 )T ; 7 T e 13 = ( 0 ,1 , 0) ~ T r 1 3 ,1 6 = ( 0, -0 .2 , -0 .0 6) ~ T -+ ~ T 14 2. 07 0. 00 20 0 0. 00 20 0 0. O C 02 2 r 1 4 ,1 4 = ( 0, 0, -0 .1 54 ) ; f 1 4 ,1 5 = ( 0, 0, 0. 15 4) e 1 4 = (1 ,Q ,0 ) 15 1. 14 0. 00 25 0 0. 00 42 5 0. O O C1 4 = T r 1 5 ,1 5 = ( 0, 0, -0 .1 32 ) ~ T e 1 5 = ( 1 , 0, 0) 16 2. 07 0. 00 20 0 0. 00 20 0 0. 00 02 2 = T r 1 6 ,1 6 = ( 0, 0, -0 .1 54 ) ; = T r 1 6 ,1 7 = ( 0 , 0, 0. 15 4) ~ T e 1 6 = (- 1 ,0 ,0 ) 17 1. 1 4 0. 00 25 0 0. 00 42 5 0. 00 01 4 z T r 1 7 ,1 7 = ( 0, 0, -0 .1 32 ) ~ T e 1 7 = ( -1 , 0, 0) 115 parameters T and S 116 117 118 2 2 2 3 3 3 4 4 4 5 5 5 6 6 6 7 7 7 8 13 13 9 14 17 10 15 13 11 16 19 12 0 J o (2.8.5 ) The kinematic and dynamic parameters for this mechanism are given in Table"
        ]
    },
    {
        "image_filename": "designv10_0_0002194_s11071-016-2760-y-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002194_s11071-016-2760-y-Figure1-1.png",
        "caption": "Fig. 1 Quadrotor configuration frame scheme with the fixed body and the inertia frames",
        "texts": [
            " The quadrotor UAV is a typical underactuated system with six degrees of freedom, including translational motions and rotational motions that are driven using only four actual voltage inputs. These inputs are imposed on rotors to produce the necessary thrust and torque. The dynamic equations that describe the attitude and altitude motions of a quadrotor are basically those of the rotating rigid bodywith six degrees of freedom. For the convenience of an intuitive understanding of the quadrotorUAVconfiguration, the schematic configuration, the illustrations of reference frames, and the force from each of the four rotors are shown in Fig. 1. Ideally, the four rotors are completely the same and are placed at the endpoints of an X-shaped frame to avoid the yawdrift from reactive torques. The rotors are divided into two groups, Pitch group {Rotor 1, Rotor 3} and Roll group {Rotor 2, Rotor 4}. The two rotors in the Pitch group rotate in a counterclockwise direction while the ones in the Roll group rotate in a clockwise direction to generate the lift force and balance the yaw torque. A fixed inertia frame {E} represented by Oxeyeze is established to locate the absolute position of the center of the mass of the quadrotor vehicle"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure13.4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure13.4-1.png",
        "caption": "FIGURE 13.4. A uniform disc, rolling in a circular path.",
        "texts": [
            "47) \u2202D \u2202x\u0307 = c (x\u0307\u2212 y\u0307) (13.48) \u2202V \u2202x = k (x\u2212 y) (13.49) 13. Vehicle Vibrations 833 \u2202K \u2202y\u0307 = mby\u0307 \u2212mee\u03c9 cos\u03c9t (13.50) d dt \u00b5 \u2202K \u2202y\u0307 \u00b6 = mby\u0308 +mee\u03c9 2 sin\u03c9t (13.51) \u2202D \u2202y\u0307 = \u2212c (x\u0307\u2212 y\u0307) (13.52) \u2202V \u2202y = \u2212k (x\u2212 y) . (13.53) Using z = x \u2212 y, we may combine Equations (13.44) and (13.45) to find the equation of relative motion mmb mb +m z\u0308 + c z\u0307 + kz = mme mb +m e\u03c92 sin\u03c9t (13.54) that is equal to z\u0308 + 2\u03be\u03c9n z\u0307 + \u03c92n z = \u03b5e\u03c92 sin\u03c9t (13.55) \u03b5 = me mb . (13.56) Example 476 F A rolling disc in a circular path. Figure 13.4 illustrates a uniform disc with mass m and radius r. The disc is rolling without slip in a circular path with radius R. The disc may have a free oscillation around \u03b8 = 0. When the oscillation is very small, we may substitute the oscillating disc with an equivalent mass-spring system. To find the equation of motion, we employ the Lagrange method. The energies of the system are K = 1 2 mv2C + 1 2 Ic\u03c9 2 = 1 2 m(R\u2212 r)2\u03b8\u0307 2 + 1 2 \u00b5 1 2 mr2 \u00b6\u00b3 \u03d5\u0307\u2212 \u03b8\u0307 \u00b42 (13.57) V = \u2212mg(R\u2212 r) cos \u03b8. (13.58) When there is no slip, there is a constraint between \u03b8 and \u03d5 R\u03b8 = r\u03d5 (13"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure1.9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure1.9-1.png",
        "caption": "Fig. 1.9 a-c. Remotely controlled manipulator. a direct control, b master-slave control, c semi automatic control. SC special purpose computer",
        "texts": [
            " Direct control manipulators are defined as the human operator remotely switching on the actuator of each manipulator joint by pressing the cor responding button. With such a regime teaching of industrial robots is often performed from the command panel. The so-called teleoperators for hazardous work areas are operated analogously. Master- slave manipulators, being in a hazardous area, are controlled re motely by the human operator from a distant safe position by means of a device which is kinematically similar to the manipulator itself (Fig. 1.9b). Therefore motion of each command manipulator joint is transferred to the corresponding joint of the working (executive) mechanism (manipulator) using the tracking system principle. Such manipulators are used in nuclear environment, contami nated atmosphere and other aggressive conditions. Semi-automatic manipulators, compared with the master- slave, on the com mand panel of the operator as a command mechanism, the kinematic of which can be arbitrary to suit small movements of a human hand, use a multi functional joy-stick. The electric signals from the joy-stick are transformed by means of a special-purpose computer (Fig. 1.9c) into the input command signals of the manipulator actuators. Different control algorithms are possible [3]. Robots with supervisory control (Fig. 1.4) are defined as having all the elements of the operations they perform being pre-programmed and generated automatically. The human operator supervising from a distance the robot work being in a hazardous zone sends only individual command signals which according to the individual programmes of the automatic robot work are then 8 1 General Introduction to Robots \u2022 I I , ,"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure4.11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure4.11-1.png",
        "caption": "FIGURE 4.11 Three-dimensional capacitance model for dielectric region. (a) Subdivision of space. (b) Equivalent circuit.",
        "texts": [
            " We can choose any one of the nodes on the lower plate as the zero voltage reference for all the other nodes. If we choose node 13, the symmetry is preserved. Finally, the capacitance is given by Cpp = Q7 + Q8 + Q9. Further, symmetry can be used to simplify the system. The modeling approach is evident from the above explanation and the extension to three dimensions is structurally complex but otherwise straightforward. For this reason, we consider only one local three- dimensional model in more detail. Figure 4.11a shows a capacitance plates for x and y-directions for the node located at the origin. For simplicity only one plate for the z-direction is shown. The plates in the dash-dot lines are the capacitor plates in the x- and z-directions. The solid line plates are for capacitors in the y-direction. Only one dashed line plate is for capacitors in the z-direction. Finally, the equivalent circuit is shown in Fig. 4.11b. The circuit solution of the problem represented by the capacitance model in Fig. 4.11 is evident from Chapter 2. MESHING RELATED ACCURACY PROBLEMS FOR PEEC MODEL 79 We only consider in this text the capacitance aspect of the meshing accuracy problem. However, the meshing sensitivity issue presented here can also be related to the meshing for inductance cells in Chapter 5. The PEEC solution approach for capacitances is given in Section 4.3. Also, the construction of meshes is presented in Chapter 8. Here, we consider important issues regarding the influence of the mesh subdivision on the solution accuracy"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure3.55-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure3.55-1.png",
        "caption": "Figure 3.55 (a) The couples acting on the junction represented by their moment vectors. (b) If there is no resultant couple, the three moment vectors must form a closed polygon.",
        "texts": [
            " There is no resultant force, but there is a resultant couple T of which the moment vector is in the xy plane (see Figure 3.53b). Its magnitude is T = \u221a 752 + 1002 = 125 kNm. The resultant couple acts in a plane perpendicular to the moment vector. 92 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Example 3 Figure 3.54 shows a junction of three coplanar tubes that are rigidly connected at equal angles of 120\u25e6. The tubes are loaded (by torsion) by the couples TA, TB and TC. The resultant couple on the junction is zero. Question: How large are the couples TA and TB if TC = 75 Nm? Solution: In Figure 3.55a, the couples are represented by their moment vectors. The three vectors are in the xy plane, the plane in which the tubes are located. The resultant moment on the junction is zero if the three vectors form a closed polygon, analogous to the closed force polygon for force equilibrium. The equilateral triangle in Figure 3.55b gives TA = TB = TC = 75 Nm. This can of course also be determined analytically. If there is no resultant couple, then \u2211 Tx = \u2212 1 2TA \u2212 1 2TB + TC = \u2212 1 2TA \u2212 1 2TB + (75 Nm) = 0,\u2211 Ty = + 1 2TA \u221a 3 \u2212 1 2TB \u221a 3 = 0. The result of these two equations is again TA = TB = 75 Nm. 3 Statics of a Rigid Body 93 Generalising the equilibrium equations for a rigid body is relatively simple. After all, equilibrium demands that both the resultant force and the resultant moment about an arbitrary point A are zero"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.64-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.64-1.png",
        "caption": "FIGURE 8.64.",
        "texts": [],
        "surrounding_texts": [
            "where, a1 is the longitudinal distance between C and the front axle, b1 is the lateral distance between C and the tireprint of the tire 1, and h is the height of C from the ground level. If P is a point in the tire frame at T1rP T1rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 . (8.62) then its coordinates in the body frame are BrP = BRT1 T1rP + BdT1 = \u23a1\u23a3 a1 + xP cos \u03b41 \u2212 yP sin \u03b41 yP cos \u03b41 \u2212 b1 + xP sin \u03b41 zP \u2212 h \u23a4\u23a6 . (8.63) The rotation matrix BRT1 is a result of steering about the z1-axis. BRT1 = \u23a1\u23a3 cos \u03b41 \u2212 sin \u03b41 0 sin \u03b41 cos \u03b41 0 0 0 1 \u23a4\u23a6 (8.64) Employing Equation (8.28), we may examine a wheel point P at W rP W rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 (8.65) 8. Suspension Mechanisms 497 and find the body coordinates of the point BrP = BRT1 T1rP + BdT1 = BRT1 \u00a1 T1RW W rP + T1dW \u00a2 + BdT1 = BRT1 T1RW W rP + BRT1 T1dW + BdT1 = BRW W rP + BRT1 T1dW + BdT1 (8.66) BrP = \u23a1\u23a3 a1 + xP cos \u03b41 \u2212 yP cos \u03b3 sin \u03b41 + (Rw + zP ) sin \u03b3 sin \u03b41 xP sin \u03b41 \u2212 b1 + yP cos \u03b3 cos \u03b41 \u2212 (Rw + zP ) cos \u03b41 sin \u03b3 (Rw + zP ) cos \u03b3 + yP sin \u03b3 \u2212 h \u23a4\u23a6 (8.67) where, BRW = BRT1 T1RW = \u23a1\u23a3 cos \u03b41 \u2212 cos \u03b3 sin \u03b41 sin \u03b3 sin \u03b41 sin \u03b41 cos \u03b3 cos \u03b41 \u2212 cos \u03b41 sin \u03b3 0 sin \u03b3 cos \u03b3 \u23a4\u23a6 (8.68) T1dW = \u23a1\u23a3 0 \u2212Rw sin \u03b3 Rw cos \u03b3 \u23a4\u23a6 . (8.69) Example 334 F Wheel-body to vehicle transformation. The wheel-body coordinate frames are always parallel to the vehicle frame. The origin of the wheel-body coordinate frame of the wheel number 1 is at BdW1 = \u23a1\u23a3 a1 \u2212b1 \u2212h+Rw \u23a4\u23a6 . (8.70) Hence the transformation between the two frames is only a displacement. Br = BIW1 W1r+ BdW1 (8.71) 8.6 F Caster Theory The steer axis may have any angle and any location with respect to the wheel-body coordinate frame. The wheel-body frame C (xc, yc, zc) is a frame at the center of the wheel at its rest position, parallel to the vehicle coordinate frame. The frame C does not follow any motion of the wheel. The steer axis is the kingpin axis of rotation. Figure 8.51 illustrates the front and side views of a wheel and its steering axis. The steering axis has angle \u03d5 with (yc, zc) plane, and angle \u03b8 with (xc, zc) plane. The angles \u03d5 and \u03b8 are measured about the yc and xc axes, 498 8. Suspension Mechanisms respectively. The angle \u03d5 is the caster angle of the wheel, while the angle \u03b8 is the lean angle. The steering axis of the wheel, as shown in Figure 8.51, is at a positive caster and lean angles. The steering axis intersect the ground plane at a point that has coordinates (sa, sb,\u2212Rw) in the wheelbody coordinate frame. If we indicate the steering axis by the unit vector u\u0302, then the components of u\u0302 are functions of the caster and lean angles. C u\u0302 = \u23a1\u23a3 u1 u2 u3 \u23a4\u23a6 = 1p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 \u23a1\u23a3 cos \u03b8 sin\u03d5 \u2212 cos\u03d5 sin \u03b8 cos \u03b8 cos\u03d5 \u23a4\u23a6 (8.72) The position vector of the point that u\u0302 intersects the ground plane, is called the location vector s that in the wheel-body frame has the following coordinates: Cs = \u23a1\u23a3 sa sb \u2212Rw \u23a4\u23a6 (8.73) We express the rotation of the wheel about the steering axis u\u0302 by a zero pitch screw motion s\u030c. CTW = C s\u030cW (0, \u03b4, u\u0302, s) = \u2219 CRW Cs\u2212 CRW Cs 0 1 \u00b8 = \u2219 CRW CdW 0 1 \u00b8 (8.74) Proof. The steering axis is at the intersection of the caster plane \u03c0C and the lean plane \u03c0L, both expressed in the wheel-body coordinate frame. The 8. Suspension Mechanisms 499 two planes can be indicated by their normal unit vectors n\u03021 and n\u03022. C n\u03021 = \u23a1\u23a3 0 cos \u03b8 sin \u03b8 \u23a4\u23a6 (8.75) C n\u03022 = \u23a1\u23a3 \u2212 cos\u03d50 sin\u03d5 \u23a4\u23a6 (8.76) The unit vector u\u0302 on the intersection of the caster and lean planes can be found by u\u0302 = n\u03021 \u00d7 n\u03022 |n\u03021 \u00d7 n\u03022| (8.77) where, n\u03021 \u00d7 n\u03022 = \u23a1\u23a3 cos \u03b8 sin\u03d5 \u2212 cos\u03d5 sin \u03b8 cos \u03b8 cos\u03d5 \u23a4\u23a6 (8.78) |n\u03021 \u00d7 n\u03022| = q cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 (8.79) and therefore, C u\u0302 = \u23a1\u23a3 u1 u2 u3 \u23a4\u23a6 = \u23a1\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3 cos \u03b8 sin\u03d5p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 \u2212 cos\u03d5 sin \u03b8p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 cos \u03b8 cos\u03d5p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 \u23a4\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6 . (8.80) Steering axis does not follow any motion of the wheel except the wheel hop in the z-direction. We assume that the steering axis is a fixed line with respect to the vehicle, and the steer angle \u03b4 is the rotation angle about u\u0302. The intersection point of the steering axis and the ground plane defines the location vector s. Cs = \u23a1\u23a3 sa sb \u2212Rw \u23a4\u23a6 (8.81) The components sa and sb are called the forward and lateral locations respectively. Using the axis-angle rotation (u\u0302, \u03b4), and the location vector s, we can define the steering process as a screw motion s\u030c with zero pitch. Employing Equations (5.473)-(5.477), we find the transformation screw for wheel frame W to wheel-body frame C. CTW = C s\u030cW (0, \u03b4, u\u0302, s) (8.82) = \u2219 CRW Cs\u2212 CRW Cs 0 1 \u00b8 = \u2219 CRW Cd 0 1 \u00b8 500 8. Suspension Mechanisms CRW = I cos \u03b4 + u\u0302u\u0302T vers \u03b4 + u\u0303 sin \u03b4 (8.83) CdW = \u00a1\u00a1 I\u2212 u\u0302u\u0302T \u00a2 vers \u03b4 \u2212 u\u0303 sin \u03b4 \u00a2 Cs. (8.84) u\u0303 = \u23a1\u23a3 0 \u2212u3 u2 u3 0 \u2212u1 \u2212u2 u1 0 \u23a4\u23a6 (8.85) vers \u03b4 = 1\u2212 cos \u03b4 (8.86) Direct substitution shows that CRW and CdW are: CRW = \u23a1\u23a3 u21 vers \u03b4 + c\u03b4 u1u2 vers \u03b4 \u2212 u3s\u03b4 u1u3 vers \u03b4 + u2s\u03b4 u1u2 vers \u03b4 + u3s\u03b4 u22 vers \u03b4 + c\u03b4 u2u3 vers \u03b4 \u2212 u1s\u03b4 u1u3 vers \u03b4 \u2212 u2s\u03b4 u2u3 vers \u03b4 + u1s\u03b4 u23 vers \u03b4 + c\u03b4 \u23a4\u23a6 (8.87) CdW = \u23a1\u23a3 (s1 \u2212 u1 (s3u3 + s2u2 + s1u1)) vers \u03b4 + (s2u3 \u2212 s3u2) sin \u03b4 (s2 \u2212 u2 (s3u3 + s2u2 + s1u1)) vers \u03b4 + (s3u1 \u2212 s1u3) sin \u03b4 (s3 \u2212 u3 (s3u3 + s2u2 + s1u1)) vers \u03b4 + (s1u2 \u2212 s2u1) sin \u03b4 \u23a4\u23a6 (8.88) The vector CdW indicates the position of the wheel center with respect to the wheel-body frame. The matrix CTW is the homogeneous transformation from wheel frame W to wheel-body frame C, when the wheel is steered by the angle \u03b4 about the steering axis u\u0302. Example 335 F Zero steer angle. To examine the screw transformation, we check the zero steering. Substituting \u03b4 = 0 simplifies the rotation matrix CRW and the position vector CdW to I and 0 CRW = \u23a1\u23a3 1 0 0 0 1 0 0 0 1 \u23a4\u23a6 (8.89) CdW= \u23a1\u23a3 0 0 0 \u23a4\u23a6 (8.90) indicating that at zero steering, the wheel frameW and wheel-body frame C are coincident. Example 336 F Steer angle transformation for zero lean and caster. Consider a wheel with a steer axis coincident with zw. Such a wheel has no lean or caster angle. When the wheel is steered by the angle \u03b4, we can find the coordinates of a wheel point P in the wheel-body coordinate frame using transformation method. Figure 8.52 illustrates a 3D view, and Figure 8.53 a top view of such a wheel. 8. Suspension Mechanisms 501 502 8. Suspension Mechanisms Assume W rP = [xw, yw, zw] T is the position vector of a wheel point, then its position vector in the wheel-body coordinate frame C is CrP = CRW W rP = Rz,\u03b4 W rP = \u23a1\u23a3 cos \u03b4 \u2212 sin \u03b4 0 sin \u03b4 cos \u03b4 0 0 0 1 \u23a4\u23a6\u23a1\u23a3 xw yw zw \u23a4\u23a6 = \u23a1\u23a3 xw cos \u03b4 \u2212 yw sin \u03b4 yw cos \u03b4 + xw sin \u03b4 zw \u23a4\u23a6 . (8.91) We assumed that the wheel-body coordinate is installed at the center of the wheel and is parallel to the vehicle coordinate frame. Therefore, the transformation from the frame W to the frame C is a rotation \u03b4 about the wheel-body z-axis. There would be no camber angle when the lean and caster angles are zero and steer axis is on the zw-axis. Example 337 F Zero lean, zero lateral location. The case of zero lean, \u03b8 = 0, and zero lateral location, sb = 0, is important in caster dynamics of bicycle model. The screw transformation for this case will be simplified to C u\u0302 = \u23a1\u23a3 u1 u2 u3 \u23a4\u23a6 = \u23a1\u23a3 sin\u03d5 0 cos\u03d5 \u23a4\u23a6 (8.92) Cs = \u23a1\u23a3 sa 0 \u2212Rw \u23a4\u23a6 (8.93) CRW = \u23a1\u23a3 sin2 \u03d5 vers \u03b4 + cos \u03b4 \u2212 cos\u03d5 sin \u03b4 sin\u03d5 cos\u03d5 vers \u03b4 cos\u03d5 sin \u03b4 cos \u03b4 \u2212 sin\u03d5 sin \u03b4 sin\u03d5 cos\u03d5 vers \u03b4 sin\u03d5 sin \u03b4 cos2 \u03d5 vers \u03b4 + cos \u03b4 \u23a4\u23a6 (8.94) Cd = \u23a1\u23a2\u23a3 cos\u03d5 (sa cos\u03d5+Rw sin\u03d5) vers \u03b4 \u2212 (sa cos\u03d5+Rw sin\u03d5) sin \u03b4 \u22121 2 (Rw \u2212Rw cos 2\u03d5+ sa sin 2\u03d5) vers \u03b4 \u23a4\u23a5\u23a6 . (8.95) Example 338 F Position of the tireprint. The center of tireprint in the wheel coordinate frame is at rT W rT = \u23a1\u23a3 0 0 \u2212Rw \u23a4\u23a6 . (8.96) 8. Suspension Mechanisms 503 If we assume the width of the tire is zero and the wheel is steered, the center of tireprint would be at CrT = CTW W rT = \u23a1\u23a3 xT yT zT \u23a4\u23a6 (8.97) where xT = \u00a1 1\u2212 u21 \u00a2 (1\u2212 cos \u03b4) sa + (u3 sin \u03b4 \u2212 u1u2 (1\u2212 cos \u03b4)) sb (8.98) yT = \u2212 (u3 sin \u03b4 + u1u2 (1\u2212 cos \u03b4)) sa + \u00a1 1\u2212 u22 \u00a2 (1\u2212 cos \u03b4) sb (8.99) zT = (u2 sin \u03b4 \u2212 u1u3 (1\u2212 cos \u03b4)) sa \u2212 (u1 sin \u03b4 + u2u3 (1\u2212 cos \u03b4)) sb \u2212Rw (8.100) or xT = sb \u00c3 cos \u03b8 cos\u03d5 sin \u03b4p cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 + 1 4 sin 2\u03b8 sin 2\u03d5 (1\u2212 cos \u03b4) cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 ! +sa \u00b5 1\u2212 cos2 \u03b8 sin2 \u03d5 cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 \u00b6 (1\u2212 cos \u03b4) (8.101) yT = \u2212sa \u00c3 cos \u03b8 cos\u03d5 sin \u03b4p cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 \u2212 1 4 sin 2\u03b8 sin 2\u03d5 (1\u2212 cos \u03b4) cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 ! +sb \u00b5 1\u2212 cos2 \u03d5 sin2 \u03b8 cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 \u00b6 (1\u2212 cos \u03b4) (8.102) zT = \u2212Rw \u2212 sb cos \u03b8 sin\u03d5+ sa cos\u03d5 sin \u03b8p cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 sin \u03b4 + 1 2 sb cos 2 \u03d5 sin 2\u03b8 \u2212 sa cos 2 \u03b8 sin 2\u03d5 cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 (1\u2212 cos \u03b4) (8.103) Example 339 F Wheel center drop. The zT coordinate in (8.100) or (8.103) indicates the amount that the center of the tireprint will move in the vertical direction with respect to the wheel-body frame when the wheel is steering. If the steer angle is zero, \u03b4 = 0, then zT is at zT = \u2212Rw. (8.104) Because the center of tireprint must be on the ground, H = \u2212Rw \u2212 zT indicated the height that the center of the wheel will drop during steering. H = \u2212Rw \u2212 zT (8.105) = sb cos \u03b8 sin\u03d5+ sa cos\u03d5 sin \u03b8p cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 sin \u03b4 \u22121 2 sb cos 2 \u03d5 sin 2\u03b8 \u2212 sa cos 2 \u03b8 sin 2\u03d5 cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 (1\u2212 cos \u03b4) 504 8. Suspension Mechanisms The zT coordinate of the tireprint may be simplifie for different designs: 1\u2212 If the lean angle is zero, \u03b8 = 0, then zT is at zT = \u2212Rw \u2212 1 2 sa sin 2\u03d5 (1\u2212 cos \u03b4)\u2212 sb sin\u03d5 sin \u03b4. (8.106) 2\u2212 If the lean angle and lateral location are zero, \u03b8 = 0, sb = 0, then zT is at zT = \u2212Rw \u2212 1 2 sa sin 2\u03d5 (1\u2212 cos \u03b4) . (8.107) In this case, the wheel center drop may be expressed by a dimensionless equation. H sa = 1 2 sin 2\u03d5 (1\u2212 cos \u03b4) (8.108) Figure 8.54 illustrates H/sa for the caster angle \u03d5 = 5deg, 0 deg, \u22125 deg, \u221210 deg, \u221215 deg, \u221220 deg, and the steer angle \u03b4 in the range \u221210 deg < \u03b4 < 10 deg. In street cars, we set the steering axis with a positive longitudinal location sa > 0, and a few degrees negative caster angle \u03d5 < 0. In this case the wheel center drops as is shown in the figure. 3\u2212 If the caster angle is zero, \u03d5 = 0, then zT is at zT = \u2212Rw + 1 2 sb sin 2\u03b8 (1\u2212 cos \u03b4)\u2212 sa sin \u03b8 sin \u03b4. (8.109) 4\u2212 If the caster angle and lateral location are zero, \u03d5 = 0, sb = 0, then zT is at zT = \u2212Rw \u2212 sa sin \u03b8 sin \u03b4. (8.110) 8. Suspension Mechanisms 505 In this case, the wheel center drop may be expressed by a dimensionless equation. H sa = \u2212 sin \u03b8 sin \u03b4 (8.111) Figure 8.55 illustrates H/sa for the lean angle \u03b8 = 5deg, 0, \u22125 deg, \u221210 deg, \u221215 deg, \u221220 deg and the steer angle \u03b4 in the range \u221210 deg < \u03b4 < 10 deg. The steering axis of street cars is usually set with a positive longitudinal location sa > 0, and a few degrees positive lean angle \u03b8 > 0. In this case the wheel center lowers when the wheel number 1 turns to the right, and elevates when the wheel turns to the left. Comparison of Figures 8.54 and 8.55 shows that the lean angle has much more affect on the wheel center drop than the caster angle. 5\u2212 If the lateral location is zero, sb = 0, then zT is at zT = \u2212Rw \u2212 sa cos\u03d5 sin \u03b8p cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 sin \u03b4 \u22121 2 sa cos2 \u03b8 sin 2\u03d5 cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 (1\u2212 cos \u03b4) (8.112) and the wheel center drop,H, may be expressed by a dimensionless equation. H sa = \u22121 2 cos2 \u03b8 sin2 \u03d5 (1\u2212 cos \u03b4) cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 \u2212 cos\u03d5 sin \u03b8 sin \u03b4p cos2 \u03b8 sin2 \u03d5+ cos2 \u03d5 (8.113) Example 340 F Position of the wheel center. As given in Equation (8.88), the wheel center is at CdW with respect to 506 8. Suspension Mechanisms the wheel-body frame. CdW = \u23a1\u23a3 xW yW zW \u23a4\u23a6 (8.114) Substituting for u\u0302 and s from (8.72) and (8.73) in (8.88) provides the coordinates of the wheel center in the wheel-body frame as xW = (sa \u2212 u1 (\u2212Rwu3 + sbu2 + sau1)) (1\u2212 cos \u03b4) + (sbu3 +Rwu2) sin \u03b4 (8.115) yW = (sb \u2212 u2 (\u2212Rwu3 + sbu2 + sau1)) (1\u2212 cos \u03b4) \u2212 (Rwu1 + sau3) sin \u03b4 (8.116) zW = (\u2212Rw \u2212 u3 (\u2212Rwu3 + sbu2 + sau1)) (1\u2212 cos \u03b4) + (sau2 \u2212 sbu1) sin \u03b4 (8.117) or xW = sa (1\u2212 cos \u03b4) + \u00b5 1 2 Rw sin 2\u03d5\u2212 sa sin 2 \u03d5 \u00b6 cos2 \u03b8 + 1 4 sb sin 2\u03b8 sin 2\u03d5 cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 (1\u2212 cos \u03b4) + (sb cos \u03b8 \u2212Rw sin \u03b8)p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 cos\u03d5 sin \u03b4 (8.118) yW = sb (1\u2212 cos \u03b4) \u2212 1 2 \u00a1 Rw sin 2\u03b8 + sb sin 2 \u03b8 \u00a2 cos2 \u03d5\u2212 1 4 sa sin 2\u03b8 sin 2\u03d5 cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 (1\u2212 cos \u03b4) \u2212 Rw sin\u03d5+ sa cos\u03d5p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 cos \u03b8 sin \u03b4 (8.119) zW = \u2212Rw (1\u2212 cos \u03b4) + \u00b5 Rw cos 2 \u03b8 + 1 2 sb sin 2\u03b8 \u00b6 cos2 \u03d5\u2212 1 2 sa cos 2 \u03b8 sin 2\u03d5 cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 (1\u2212 cos \u03b4) \u2212sa cos\u03d5 sin \u03b8 + sb cos \u03b8 sin\u03d5p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 sin \u03b4 (8.120) The zW coordinate indicates how the center of the wheel will move in the vertical direction with respect to the wheel-body frame, when the wheel is steering. It shows that zW = 0, as long as \u03b4 = 0. 8. Suspension Mechanisms 507 The zW coordinate of the wheel center may be simplified for different designs: 1\u2212 If the lean angle is zero, \u03b8 = 0, then zW is at zW = \u2212Rw \u00a1 1\u2212 cos2 \u03d5 \u00a2 (1\u2212 cos \u03b4)\u2212 sb sin\u03d5 sin \u03b4 \u22121 2 sa sin 2\u03d5 (1\u2212 cos \u03b4) . (8.121) 2\u2212 If the lean angle and lateral location are zero, \u03b8 = 0, sb = 0, then zW is at zW = \u2212Rw \u00a1 1\u2212 cos2 \u03d5 \u00a2 (1\u2212 cos \u03b4)\u2212 1 2 sa sin 2\u03d5 (1\u2212 cos \u03b4) . (8.122) 3\u2212 If the caster angle is zero, \u03d5 = 0, then zW is at zW = \u2212Rw \u00a1 1\u2212 cos2 \u03b8 \u00a2 (1\u2212 cos \u03b4)\u2212 sa sin \u03b8 sin \u03b4 + 1 2 sb sin 2\u03b8 (1\u2212 cos \u03b4) . (8.123) 4\u2212 If the caster angle and lateral location are zero, \u03d5 = 0, sb = 0, then zW is at zW = \u2212Rw \u00a1 1\u2212 cos2 \u03b8 \u00a2 (1\u2212 cos \u03b4)\u2212 sa sin \u03b8 sin \u03b4. (8.124) 5\u2212 If the lateral location is zero, sb = 0, then zT is at zW = \u2212Rw (1\u2212 cos \u03b4)\u2212 sa cos\u03d5 sin \u03b8p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 sin \u03b4 + Rw cos 2 \u03b8 cos2 \u03d5\u2212 1 2 sa cos 2 \u03b8 sin 2\u03d5 cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 (1\u2212 cos \u03b4) (8.125) In each case of the above designs, the height of the wheel center with respect to the ground level can be found by adding H to zW . The equations for calculating H are found in Example 340. Example 341 F Camber theory. Having a non-zero lean and caster angles causes a camber angle \u03b3 for a steered wheel. To find the camber angle of an steered wheel, we may determine the angle between the camber line and the vertical direction zc. The camber line is the line connecting the wheel center and the center of tireprint. The coordinates of the center of tireprint (xT , yT , zT ) are given in Equations (8.101)-(8.103), and the coordinates of the wheel center (xW , yW , zW ) are given in Equations (8.118)-(8.120). The line connecting (xT , yT , zT ) to (xW , yW , zW ) may be indicated by the unit vector l\u0302c l\u0302c = (xW \u2212 xT ) I\u0302 + (yW \u2212 yT ) J\u0302 + (zW \u2212 zT ) K\u0302q (xW \u2212 xT ) 2 + (yW \u2212 yT ) 2 + (zW \u2212 zT ) 2 (8.126) 508 8. Suspension Mechanisms in which I\u0302 , J\u0302 , K\u0302, are the unit vectors of the wheel-body coordinate frame C. The camber angle is the angle between l\u0302c and K\u0302, which can be found by the inner vector product. \u03b3 = cos\u22121 \u00b3 l\u0302c \u00b7 K\u0302 \u00b4 = cos\u22121 (zW \u2212 zT )q (xW \u2212 xT ) 2 + (yW \u2212 yT ) 2 + (zW \u2212 zT ) 2 (8.127) As an special case, let us determine the camber angle when the lean angle and lateral location are zero, \u03b8 = 0, sb = 0. In this case, we have xT = sa \u00a1 1\u2212 sin2 \u03d5 \u00a2 (cos \u03b4 \u2212 1) (8.128) yT = \u2212sa cos\u03d5 sin \u03b4 (8.129) zT = zT = \u2212Rw \u2212 1 2 sa sin 2\u03d5 (1\u2212 cos \u03b4) (8.130) xW = \u00b5 sa + 1 2 Rw sin 2\u03d5\u2212 sa sin 2 \u03d5 \u00b6 (1\u2212 cos \u03b4) (8.131) yW = sb (1\u2212 cos \u03b4)\u2212Rw sin\u03d5+ sa cos\u03d5 sin \u03b4 (8.132) zW = \u00b5 Rw \u00a1 cos2 \u03d5\u2212 1 \u00a2 \u2212 1 2 sa sin 2\u03d5 \u00b6 (1\u2212 cos \u03b4) . (8.133) 8.7 Summary There are two general types of suspensions: dependent, in which the left and right wheels on an axle are rigidly connected, and independent, in which the left and right wheels are disconnected. Solid axle is the most common dependent suspension, while McPherson and double A-arm are the most common independent suspensions. The roll axis is the instantaneous line about which the body of a vehicle rolls. Roll axis is found by connecting the roll center of the front and rear suspensions of the vehicle. The instant center of rotation of a wheel with respect to the body is called suspension roll center. So, to find the roll center of the front or rear half of a car, we should determine the suspension roll centers, and find the intersection of the lines connecting the suspension roll centers to the center of their associated tireprints. Three coordinate frames are employed to express the orientation of a tire and wheel with respect to the vehicle: the wheel frame W , wheel-body frame C, and tire frame T . A wheel coordinate frame W (xw, yw, zw) is attached to the center of a wheel. It follows every translation and rotation of the wheel except the spin. Hence, the xw and zw axes are always in the tire-plane, while the yw-axis is always along the spin axis. When the wheel 8. Suspension Mechanisms 509 is straight and the W frame is parallel to the vehicle coordinate frame, we attach a wheel-body coordinate frame C (xc, yc, zc) at the center of the wheel parallel to the vehicle coordinate axes. The wheel-body frame C is motionless with respect to the vehicle coordinate and does not follow any motion of the wheel. The tire coordinate frame T (xt, yt, zt) is set at the center of the tireprint. The zt-axis is always perpendicular to the ground. The xt-axis is along the intersection line of the tire-plane and the ground. The tire frame does not follows the spin and camber rotations of the tire however, it follows the steer angle rotation about the zc-axis. We define the orientation and position of a steering axis by the caster angle \u03d5, lean angle \u03b8, and the intersection point of the axis with the ground surface at (sa, sb) with respect to the center of tireprint. Because of these parameters, a steered wheel will camber and generates a lateral force. This is called the caster theory. The camber angle \u03b3 of a steered wheel for \u03b8 = 0, and sb = 0 is: \u03b3 = cos\u22121 \u00b3 l\u0302c \u00b7 K\u0302 \u00b4 = cos\u22121 (zW \u2212 zT )q (xW \u2212 xT ) 2 + (yW \u2212 yT ) 2 + (zW \u2212 zT ) 2 (8.134) where xT = sa \u00a1 1\u2212 sin2 \u03d5 \u00a2 (cos \u03b4 \u2212 1) (8.135) yT = \u2212sa cos\u03d5 sin \u03b4 (8.136) zT = zT = \u2212Rw \u2212 1 2 sa sin 2\u03d5 (1\u2212 cos \u03b4) (8.137) xW = \u00b5 sa + 1 2 Rw sin 2\u03d5\u2212 sa sin 2 \u03d5 \u00b6 (1\u2212 cos \u03b4) (8.138) yW = sb (1\u2212 cos \u03b4)\u2212Rw sin\u03d5+ sa cos\u03d5 sin \u03b4 (8.139) zW = \u00b5 Rw \u00a1 cos2 \u03d5\u2212 1 \u00a2 \u2212 1 2 sa sin 2\u03d5 \u00b6 (1\u2212 cos \u03b4) . (8.140) 510 8. Suspension Mechanisms 8.8 Key Symbols a, b, c, d lengths of the links of a four-bar linkage ai distance of the axle number i from the mass center A,B, \u00b7 \u00b7 \u00b7 coefficients in equation for calculating \u03b83 b1 distance of left wheels from mass center B (x, y, z) vehicle coordinate frame C mass center C coupler point C (xc, yc, zc) wheel-body coordinate frame C T dW C expression of the position of W with respect to T e, \u03b1 polar coordinates of a coupler point g overhang h = z \u2212 z0 vertical displacement of the wheel center H wheel center drop Iij instant center of rotation between link i and link j IijImn a line connecting Iij and Imn I\u0302 , J\u0302 , K\u0302 unit vectors of the wheel-body frame C I identity matrix J1, J2, \u00b7 \u00b7 \u00b7 length function for calculating \u03b83 l\u0302c unit vector on the line (xT , yT , zT ) to (xW , yW , zW ) ms sprung mass mu unsprung mass n\u03021 normal unit vectors to \u03c0L n\u03022 normal unit vectors to \u03c0C P point q, p, f parameters for calculating couple point coordinate r position vector Rw tire radius TRW rotation matrix to go from W frame to T frame s position vector of the steer axis sa forward location of the steer axis sb lateral location of the steer axis s\u030cW (0, \u03b4, u\u0302, s) zero pitch screw about the steer axis T (xt, yt, zt) tire coordinate system TTW homogeneous transformation to go from W to T u\u0302 steer axis unit vector u\u0303 skew symmetrix matrix associated to u\u0302 uC position vector of the coupler point u\u0302z unit vector in the z-direction vx forward speed x, y suspension coordinate frame xC , yC coordinate of a couple point (xT , yT , zT ) wheel-body coordinates of the origin of T frame 8. Suspension Mechanisms 511 (xW , yW , zW ) wheel-body coordinates of the origin of W frame vers \u03b4 1\u2212 cos \u03b4 W (xwywzw) wheel coordinate system z vertical position of the wheel center z0 initial vertical position of the wheel center \u03b1 angle of a coupler point with upper A-arm \u03b3 camber angle \u03b4 steer angle \u03b5 = ms/mu sprung to unsprung mass ratio \u03b8 lean angle \u03b80 angle between the ground link and the z-direction \u03b8i angular position of link number i \u03b82 angular position of the upper A-arm \u03b83 angular position of the coupler link \u03b84 angular position of link lower A-arm \u03b8i0 initial angular position of \u03b8i \u03c0C caster plane \u03c0L lean plane \u03c5 trust angle \u03d5 caster angle \u03c9 angular velocity 512 8. Suspension Mechanisms Exercises 1. Roll center. Determine the roll ceneter of the kinematic models of vehicles shown in Figures 8.56 to 8.59. 2 1 Body 4 6 3 5 8 7 8. Suspension Mechanisms 513 514 8. Suspension Mechanisms 8. Suspension Mechanisms 515 4. F Position of the roll center and mass ceneter. Figure 8.66 illustrates the wheels and mass center C of a vehicle. Design a double A-arm suspension such that the roll center of the C 516 8. Suspension Mechanisms c d a 2\u03b8 3\u03b8 4\u03b8 A B M N x y 0\u03b8 \u03b1 e z Cb 8. Suspension Mechanisms 517 Determine CTW for \u03d5 = 8deg, \u03b8 = 12deg, and the location vector Cs Cs = \u23a1\u23a3 3.8 cm 1.8 cm \u2212Rw \u23a4\u23a6 . (a) The vehicle uses a tire 235/35ZR19. (b) The vehicle uses a tire P215/65R15 96H. 10. F Wheel drop. Find the coordinates of the tireprint for \u03d5 = 10deg \u03b8 = 10deg Cs = \u23a1\u23a3 3.8 cm 1.8 cm 38 cm \u23a4\u23a6 if \u03b4 = 18deg. How much is the wheel drop H. 11. F Wheel drop and steer angle. Draw a plot to show the wheel drop H at different steer angle \u03b4 for the given data in Exercise 10. 12. F Camber and steering. Draw a plot to show the camber angle \u03b3 at different steer angle \u03b4 for the following characteristics: \u03d5 = 10deg \u03b8 = 0deg Cs = \u23a1\u23a3 3.8 cm 0 cm 38 cm \u23a4\u23a6 Part III Vehicle Dynamics 9"
        ]
    },
    {
        "image_filename": "designv10_0_0003667_60.678979-FigureI-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003667_60.678979-FigureI-1.png",
        "caption": "Fig. I . Elementary salient-pole machine with symmetric placement of windings in the gap and eccentric rotor.",
        "texts": [],
        "surrounding_texts": [
            "157\nmagnetomotive force (MMF), and the resultant air-gap permeance including rotor and stator slotting, saturation and eccentricity were presented. Then, the flux density distribution in the air gap was computed by the product of permeance and the MMF. The harmonic content in the flux density waveform was investigated and concluded that since the harmonic fluxes move relative to the stator, then, they should induce corresponding current harmonics in the stationary stator windings.\nDorrell et al. [4] presented a similar paper on the analysis of air-gap flux, current and vibration signals as a function of both static and dynamic air-gap eccentricity in 3-phase induction motors. They used the same approach, the air-gap permeance approach, as in [3] for calculating the flux density and unbalanced magnetic forces caused by eccentricity; except that they suggested that the dynamic and static eccentricity should both be considered simultaneously and a new theoretical analysis was presented. Also, it was suggested that in addition to monitoring the line current signature, the vibration analysis should be put forward to identify which particular form of eccentricity is dominant.\nVas [ 5 ] even went further and presented a formula for computing the frequency components in the stator currents of an induction machine which is due to air-gap eccentricity. The frequency components are found to be a function of the fundamental stator frequency, number of rotor slots, slip, type of eccentricity and stator MMF time harmonics.\nVerma and Natarajan [6] have studied the changes in the air-gap field as a function of static eccentricity using search coils in the stator core. However, the installation of air-gap search coils is neither practical nor economical for monitoring the condition of the motors that are already in service such as the ones existing in off-shore installations.\nToliyat et al. [7] have also proposed the detection of airgap eccentricity in induction machines by measuring the harmonic content in the machine line currents. However, they proposed a new way for modeling the machine under eccentricity. The winding function approach accounting for all the space harmonics in the machine was used to calculate all the mutual and magnetizing inductances for the induction machine with eccentric rotors.\nThere are fewer publications on the detection of rotor eccentricity using electrical techniques for synchronous machines. Htsui, and Stein [8] have proposed using the shaft voltage signals generated by the shaft flux linkages to detect eccentricities and shorted rotor field in synchronous machines. The magnitude and the thickness of the shaft signal loci reflect static and dynamic eccentricities.\nThe interest in the condition monitoring, on-line and/or off-line, of ac machines has increased tremendously in the last few years because of economic pressures, smaller profit margins and high costs of replacement and spare parts. Therefore, in order to achieve this goal, it is very important to be able to develop simple models for the machines under fault conditions and then analyze the effect of faults on the\nmachines\u2019 behavior. This study is an attempt toward this goal.\nIn this paper, a new linear model, i.e. saturation not included, for synchronous machines by using the geometry and the physical layout of all windings will be developed. This model can be used to simulate the machine under fault conditions such as air-gap eccentricity. The developed model will be used to investigate the effect of dynamic air-gap eccentricity on the stator and rotor magnetizing and mutual inductances of a salient-pole synchronous motor.\nFinally, the stator current signature patterns of the linear model will be examined to find specific harmonic components that are related to dynamic air-gap eccentricity.\nThe winding function approach accounting for all space harmonics in the machine will be used to compute all of the relevant inductances of the stator phases and the rotor field windings of the salient-pole synchronous machine. The inductances will be computed for both cases, when the rotor does and does not experience air-gap eccentricity. In the case of eccentric rotors, the aiir-gap is not uniform. Therefore, a new formulation for computing the inductances has to be developed.\n11. THE MODIFIISD WINDING FUNCTION APPROACH (MWFA)\nAll conventional machines rely upon magnetic fields for the purpose of energy conversion. Windings are arranged on a stationary member (stator) and on the rotating member (rotor) so as to set up a field distribution of magnetic flux density in the space which separates the stator from the rotor (air-gap). Working directly with the electromagnetic fields as an approach to study electrical machines leads to a better understanding of flux, current density, and force distributions in the machine, moreover where hot spots might appear. However, the analysis of machines as a magnetic field problem needs extensive digital computer horsepower. The geometry of electric machines normally leads to complicated boundary conditions even for simplified models. For this reason, it is more convenient and the problem will be much more simplified if coupled magnetic circuits approach is used.\nAn elementary two-pole, salient-pole synchronous machine shown in Figs. 1 with an eccentric rotor will be used for developing the modified winding function approach (MWFA). A single conductor is threaded back and forth in the gap forming the total number of turns for the winding which need not be associated with either the stator or rotor at this point. The stator shape is assumed to be cylindrical . Also, the permeability of the stator and rotor iron cores is assumed to be infinite when compared to the permeability of the air-gap.\nThe stator reference position, the angle cp, of the closed path abcda of Fig. 1 is taken at an arbitrary point along the gap. Points a and d are located on the stator corresponding to angles 0 and cp respectively and points b and c are located on the rotor. In the case of salient-pole synchronous machines,",
            "158\nthe flux lines will take irregular paths in the air-gap but intersect with the stator and rotor at right angles. Therefore, paths ab and cd are defined to lie along the lines of flux even though these flux lines cannot be uniquely defined without using flux plots. However, using Gauss's law for magnetic fields, points b and c are uniquely defined since two flux lines can never originate from the same point if points a and d are fixed on the stator.\nConsider the path abcda of Fig. 1 for an arbitrary O<(p<2n, by Ampere's law\n#H.dl = J J.dS (1) abcda S where S is the surface enclosed by the path abcda. Since all the windings enclosed by the closed path carry the same current i, (1) reduces to the following:\n+H.dl=n(y , ,O) . i (2) abcda where H is the magnetic field intensity and dl is defined to be along the flux lines originating or terminating at two points of the closed path abcda. The function n(y,,O) is called the turns function and represents the number of turns of the winding enclosed by the path abcda. In general, for a rotating coil it is assumed to be a function of 9 and the rotor position angle 6. For a stationary coil it is only a function of q. Turns carrying currents i into the page are considered positive while the turns carrying currents i out of the page are considered negative. In terms of MMF drops in a magnetic circuit, (2) can be written\n(3) Since the iron is considered to be infinitely permeable, the MMF drops Fbc and Fda are negligible and (3) reduces to Fab +- Fbc t- Fcd i- Fda = n(y,,6). i\nFab(0,6)+ F c d ( v , 6 ) = n ( n 6 ) . i (4) Gauss's law for magnetic field can be used to find an expression for the MMF drop at y, = 0, Fab (0,6), which is\narbitrary volume. Taking the surface S to be a cylindrical volume located just inside the stator inner surface, (5) can be written as\n2w L\n0 0 I j h f f ( y , ? O ) . @ ) ( d y , ) = 0 (6)\nwhere L is the axial stack length of the machine, r is the stator inner radius, is the free space permeability and 6 is the angular position of the rotor with respect to stator. Since B does not vary with respect to the axial length, and MMF is the product of flux radial length and the magnetic field intensity, then\nDividing equation (4) by the air-gap function g(y,O), and then integrating from 0 to 2n yields\n2n Fab (0,o) + Fcd (9,e) (9,e) 0 I g(y,, 6 ) dy,=!)yo i d 9 (8)\nSince the second term of the left hand side is zero as found from Gauss's law, (8) will reduce to the following\nwhere < g Substituting (9) in (4) and solving for F c d ( q ) yields > is the average value of the inverse gap function.\n(10) From the last equation, the modified winding function, in general, can be defined as follows\n(1 1) where\nM ( y , , 6 ) = n(y,,O) - < w e ) >\nIf the rotor is not eccentric , i.e. the air-gap is symmetric, and the north and south poles of the saIient-pole machine are identical, then the inverse air-gap function g -'(y,,O) has only even harmonics including a de value. Also, for practical machines, the windings are distributed in such a way that the turns function n(y,,O) has only odd harmonics. Therefore, in the cases where the air-gap has only even harmonics, i.e. noneccentric rotor, (12) reduces to\nwhere < n > is the dc value of the turns function of the winding.\n< M ( O ) > = < n > (13)\ngiven by # B.dS = 0\n111. INDUCTANCE CALCULATIONS ( 5 ) In the previous section, the relative permeability of the\n3 iron was assumed to be infinity, i.e. the MMF drop in the iron was neglected. The MMF distribution of machine windings in the air gap can simply be found by product of the modified where B represents magnetic flux density and the surface integral is carried out over the boundary surface of an",
            "159\nwinding function calculated from (1 1) and the current flowing in the winding.\nAs shown in Fig. 1, windings A and B are located in the air gap and could be associated with either the rotor or the stator. The mutual inductance of winding B due to current i~ flowing in winding A is to be calculated. Winding B is arranged arbitrarily in the air gap and for demonstration is assumed to have two coil sides, 1-1' and 2-2', with different turns distribution in the air gap. The reference angle cp cannot be selected freely and should be the same reference position that has been previously used to calculate the modified winding function M~(q , , , 8 ) .\nThe MMF distribution in the air gap due to current i A can be calculated as follows\n~ A ( q , e ) = ~ A ( q , e ) .iA (14) It is known that the flux in a magnetic circuit is the product of the MMF (F) and the permeance (P) of the flux path. Thus,\n@ = F . P (15) and the permeance is given by\n(16) p=- P A 1\nwhere y is the permeability, A is the cross sectional area and 1 is the length of the magnetic path. The differential flux across the gap through a differential volume of length g(q, 0) and cross-sectional area of (r.L.dq) from the rotor to the stator is\nThe flux linking the coil sides 1-1' of winding B can be calculated using the following integration\nwhere nBl(q,e) is equal to the number of turns of coil sides 1- 1' between the reference angles q1 and 61 of Fig. 1 and zero otherwise. Coil side 1' is the return path for coil side 1. Continuing the process of calculating the flux linking the other coil sides of winding B and in general for any set of coil sides k-k' the flux linkage is\n(19)\nwhere nBk(q , e), FA( q ,e) and g - ' ( q , e) must have the same position reference q. The total flux linking winding B due to current in winding A can be defined as follows\nqi 212\nk = l 0\n4i &A = k = l C @ k - k ' = / 4 J r L c \" B ( ( q , e ) F A ( q , ' ) g - ' ( q , ' ) ~ y ] (20)\nor\nwhere\nis the turns function for ithe winding B assuming that the qr coil sides are connected in series and qr can be defined as follows for the salient-pole synchronous machines:\nstator slots number number of layers X\n\" = phase number 2 for the stator windings and\nq, = number of poles for rotor field windings. winding B due to the current i A in winding A would be The mutual inductance LBA of\nUsing the same process, the magnetizing inductance of winding A can be defined as\nIV. INDUCTANCE CALCULATIONS UNDER DYNAMIlC ECCENTRICITY\nIn general, rotor eccentricity in motors takes two forms, namely static air-gap (eccentricity and dynamic air-gap eccentricity. In the case of static air-gap eccentricity, the position of minimum radial air-gap length is fixed in space and the center of rotation and the center of rotor are the same. For example, static eccentricity can be caused by oval stator cores or by the incorrect positioning of the stator or rotor. However, in the case OS dynamic air-gap eccentricity, the center of rotation and the center of the rotor are not the same and the minimum air-gap rotates with the rotor. So, dynamic eccentricity is both time and space dependent. Dynamic eccentricity can be caused by misalignment of bearings, mechanical resonance at critical speeds, a bent rotor shaft, wear of bearings and so on [3]. In this paper, the inductances of a salient, 3-phase, four-pole synchronous machine whose parameters are given in the Appendix including the effect of dynamic eccentricity will be calculated.\nFig.2 illustrates the stator phase a turns function. The ac component of the phases b and c turns functions are the same as phase a except that they are shifted by 60\" and 120\" mechanical degrees respectively.\n%* Fig. 2. Nsll Stator phase a turns function.\nThe rotor field turns function can be found similarly except that it is not stationary. The rotor position with respect to the previously selected reference angle 'p is presented in Fig. 3 by the mechanical angle 8,."
        ]
    },
    {
        "image_filename": "designv10_0_0003387_jtbi.2001.2279-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003387_jtbi.2001.2279-Figure2-1.png",
        "caption": "FIG. 2. Contour curves showing constant O 2 consumption, above standing still, per unit distance as a function of v and f (ml O 2 (kg body mass~1)m~1). The data used to construct these curves are from Molen et al. (1972). The v}f relationships expected from constrained optimization of this function under three conditions are drawn. The curve passing through point A indicates the optimization to constrained speed (v) in which the curve represents vertically oriented tangents to the cost contours. The curve passing through B indicates optimization to constrained step frequency ( f ) in which the curve represents horizontally oriented tangents to the cost contours. The curve passing through C indicates optimization to constrained step length (d) in which the curve represents tangents of the cost contours and lines emanating from the origin (where each line from the origin indicates a di!erent constant step length).",
        "texts": [
            " So F\"F(x i ). In a given situation an individual should minimize F subject to the constraints imposed. In the standard treadmill experiment we can think of v as constrained and f as unconstrained and measurable. Most of the x i are un- constrained and at their optimal values so we can think of a reduced objective function F\"F (v, f ) that depends only on these directly constrained and easily measurable variables. One could consider as a plausible F the metabolic cost of transport per unit distance. Figure 2 shows contours derived from the metabolic cost surface from one subject in Molen et al. (1972b) (Molen et al. provide detailed data for this individual but report substantial variation between individuals). Each curve indicates the combinations of speed and step frequency (and implicitly step length) that demand equal O 2 consumption per unit distance traveled. Each of these contours depicts a cost of walking that is less metabolically costly than the contour that resides outside it. The set of contours emanate from the point of least cost, generally considered to be at about the preferred walking speed and frequency (Cavagna, 1975; Cavagna & Franzetti, 1986)",
            " The optimization prediction for the v}f curve for treadmill walking can be constructed from the objective function F( f, v) and the constraint function v!C\"0 with the method of Lagrange multipliers (at a constrained optimum the gradient of the cost function is parallel to the gradient of the constraint function). For constrained v, and the other cases we consider below, the optimization result can also be found by a simple geometric construction (that essentially reproduces the Lagrange multiplier method). For every \"xed v, the subject is constrained to a line perpendicular to the speed axis in Fig. 2. On a given vertical line, F is then to be minimized. This will be on a level line (metabolic cost contour) that just grazes the constraint line. Otherwise, there would be a smaller F also consistent with the constraint line. Thus, the minimum F for a constrained v is at the point on the v}f plane where a level line of F is tangent with the line v\"constant. A sample point A is shown in Fig. 2. The v}f curve is then the sequence of points constructed in like manner, as the points where the level lines of F have vertical tangents. The v}f curve generated in this manner indicates the optimization solution for the imposed constraint of speci\"ed speed. Consider now constrained step frequency f, as for walking in time to music, a drumbeat or a metronome. What determines the selection of v and d (and the other x i ) in this situation? Tak- ing this as a constrained optimization problem, we look for points like B of Fig. 2. These are points that minimize F on a given line of constant f. Using the same reasoning as for constrained v above, point B is on a level line of F at a point of horizontal tangency. The derived v}f curve for constrained f experiments is generated as a locus of similarly determined points of horizontal tangency. The last simple situation we consider is that of walking at \"xed step length, such as for walking on regularly spaced stepping stones, colored tiles, or railroad ties. The constant d curve on the v}f plane is a line in which v/f\"constant. A set of possible v}f points for constant step length is represented by a straight line that passes through the origin in Fig. 2, the slope representing a speci\"c constant step length. Again the minimum F on a constant d line will be at the point of tangency of the constraint line with a level line of F, as shown for the typical point C in Fig. 2. The v}f curve predicted for an optimization to constrained d is generated as a locus of similarly constructed points. All three of the curves above pass through, and thus intersect at, the minimum of the objective function F at a point one could call the preferred unconstrained speed and frequency. One could imagine still other constrained experiments that would generate other curves on the v}f plane, but no others seem to correspond to such simple experimental situations. We have experimentally observed three distinct v}f curves for each subject, depending on constraint, that roughly intersect at a common point near the unconstrained values of v and f",
            "ect that we did not measure (avoided by limiting the number of repetitions the subjects performed for each circumstance), is how quickly people adapt their condition to the change in constraint, or whether, in fact, the optimization gets better as people practice in the constrained situations. Although we have shown agreement between the trends predicted by optimization and the choices made by individuals, we have not made a direct comparison between the observed walking v}f behavior of our study subjects and O 2 contours for those same individuals. We reserve for further study whether the variations between subjects observable in Fig. 2 correlates with differences between oxygen consumption surfaces. An individual may seem more commonly to be constrained by the demands of providing a particular walking speed v, but the v}f relationship generated from speed constraint is not really any more fundamental than the other v}f relations from the other constraints. The three conditions we tested have counterparts in common experience. There are other constraints that would fall into the realm of natural behavior such as the carrying of weights, walking bent over to clear overhead obstacles, walking in circles, walking with given heel strike (to control sound genera- tion while hunting, say)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000135_c002690p-Figure10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000135_c002690p-Figure10-1.png",
        "caption": "Fig. 10 Electrochemical decorations of graphene sheets with catalyst nanoparticles. Nucleation sites for Pd at defects at the edges of the intact graphene islands. (Reproduced with permission.29 Copyright 2008 Wiley-VCH.)",
        "texts": [
            ") This journal is c The Royal Society of Chemistry 2010 Chem. Soc. Rev., 2010, 39, 4146\u20134157 | 4153 D ow nl oa de d by U ni ve rs ity o f D el aw ar e on 0 9 Ju ne 2 01 2 Pu bl is he d on 0 9 Ju ly 2 01 0 on h ttp :// pu bs .r sc .o rg | do i:1 0. 10 39 /C 00 26 90 P with a diameter of about 7\u20139 nm can be electrochemically deposited on the defects of graphene sheets due to the preferred nucleation of the palladium at vacancies along the edges of the intact, nanometre-sized graphene islands (see Fig. 10). The electrochemical route of synthesis of catalyst nanoparticles is very attractive because such nanoparticles nucleate at the electroactive sites of carbon nanomaterials. There are a wide variety of applications for such catalyst modified graphenes.30 Spectroelectrochemistry enables us to gain a deeper insight into redox reaction because the spectroscopic method allows identification of participating species.31 Multilayer graphene sheets are transparent in the UV-Vis region even at thicknesses of about 24 nm and possess high conductivity"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure8.29-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure8.29-1.png",
        "caption": "Fig. 8.29. Gripper with \"fingers\" (SIMRIT - FR Germany)",
        "texts": [
            " For these special solutions like the one from the Japanese firm Fanuc are applied for secure gripping of round work pieces of different diameters (see Fig. 8.28). By parallelogram linking of each of the three jaw elements, their movement which runs parallel to the gripped surface under a constant angle is ensured, so that twisting of the workpiece during gripping is avoided which can be important in some cases. There are often unusual solutions for grippers handling fragile, breakable, weak or workpieces with a specially treated (e.g. polished) surface. Thus the gripper illustrated in Fig. 8.29 is executed in the form of two or more (usually three) \"fingers\" made of a special rubber compound which by using solution of an internal plane and external ribbed surface have the property of bending when subdued to internal air prt1ssure and so gripping the workpiece \"softly\". 246 8 Elements, Structures and Application of Industrial Robots For gripping large plane or slightly curved relatively lightweight workpieces, like glass, plastic or sheet metal plates, as well as big box-like products with large flat surfaces (plastic inlays for refrigerators, big boxes, etc"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003543_28.738985-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003543_28.738985-Figure2-1.png",
        "caption": "Fig. 2. The stator and rotor flux linkages in different reference frames.",
        "texts": [
            " The basic idea of the DTC is in the control of the stator flux vector (in amplitude and angular position) with feedback of observed torque and stator flux from measured voltages and currents from motor terminals. (Using modern devices, computations necessary for these two quantities can now be carried out in real time.) The observers can be set up in the stationary reference frame, which leads, eventually, to the elimination of the rotor position sensor. The stator flux linkage vector and rotor flux linkage vector , can be drawn in the rotor flux ( ), stator flux ( ), and stationary ( ) reference frames, as in Fig. 2. The angle between the stator and rotor flux linkages, , is the torque angle. In the steady state, is constant corresponding to load torque, and both stator and rotor flux rotate at the synchronous speed. In transient operation, varies and the stator and rotor flux rotate at different speeds. Since the electrical time constant is normally much smaller than the mechanical time constant, the rotating speed of stator flux with respect to the rotor flux can be easily changed. The stator flux linkage and the torque angle are given by (2) and (3) (2) and (3) With the transformation in (4), the flux linkage and torque in (1) and (2) can be transformed to the stator flux reference frame, as described in [10]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure9.18-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure9.18-1.png",
        "caption": "FIGURE 9.18. A trebuchet at starting position.",
        "texts": [
            "351) KA + VA = KB + VB (9.352) 1 2 mv21 + 1 2 IC\u03c9 2 1 + 0 = 1 2 mv22 + 1 2 IC\u03c9 2 2 +mgH (9.353)\u00b5 m+ IC R2 \u00b6 v21 = \u00b5 m+ IC R2 \u00b6 v22 + 2mgH (9.354) and therefore, v2 = vuutv21 \u2212 2gH 1 + IC mR2 . (9.355) The condition for having a real v2 is v1 > vuut 2gH 1 + IC mR2 . (9.356) The second speed (9.355) and the condition (9.356) for a solid disc are v2 = r v21 \u2212 4 3 Hg (9.357) v1 > r 4 3 Hg (9.358) because we assumed that IC = 1 2 mR2. (9.359) Example 374 Trebuchet. A trebuchet, shown schematically in Figure 9.18, is a shooting weapon of war powered by a falling massive counterweight m1. A beam AB is pivoted to the chassis with two unequal sections a and b. The figure shows a trebuchet at its initial configuration. The origin of a global coordinate frame is set at the pivot point. The counterweight m1 is at (x1, y1) and is hinged at the shorter arm of the beam at a distance c from the end B. The mass of the projectile is m2 and it is at the end of a massless sling with a length l attached to the end of the longer arm of the beam",
            " Applied Dynamics 579 (c) the pivot O has a uniform motion on a circle rO = R cos\u03c9t I\u0302 +R sin\u03c9t J\u0302. 13. Equations of motion from Lagrangean. Consider a physical system with a Lagrangean as L = 1 2 m (ax\u0307+ by\u0307) 2 \u2212 1 2 k (ax+ by) 2 . and find the equations of motion. The coefficients m, k, a, and b are constant. 14. Lagrangean from equation of motion. Find the Lagrangean associated to the following equations of motions: (a) mr2\u03b8\u0308 + k1l1\u03b8 + k2l2\u03b8 +mgl = 0 580 9. Applied Dynamics (b) r\u0308 \u2212 r \u03b8\u0307 2 = 0 r2 \u03b8\u0308 + 2r r\u0307 \u03b8\u0307 = 0 15. Trebuchet. Derive the equations of motion for the trebuchet shown in Figure 9.18. 16. Simplified trebuchet. Three simplified models of a trebuchet are shown in Figures 9.23 to 9.25. Derive and compare their equations of motion. 9. Applied Dynamics 581 10 In this chapter we develop a dynamic model for a rigid vehicle in a planar motion. When the forward, lateral and yaw velocities are important and are enough to examine the behavior of a vehicle, the planar model is applicable. 10.1 Vehicle Coordinate Frame The equations of motion in vehicle dynamics are usually expressed in a set of vehicle coordinate frame B(Cxyz), attached to the vehicle at the mass center C, as shown in Figure 10"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002549_j.matdes.2019.108076-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002549_j.matdes.2019.108076-Figure3-1.png",
        "caption": "Fig. 3. Structures with different optimization design methods.",
        "texts": [
            " By using topological optimization and laser additive manufacturing process, the weight of antenna bracket was reduced from 1.626 kg to 0.936 kg, achieving a weight reduction of 42%. However, compared with the general topology optimized structure, the lattice structure exhibits a combination of high performance features such as high strength, low mass, low thermal conductivity and excellent toughness, sound absorption and vibration damping properties [36], which can meet the design and versatility requirements of ultra-light components [13,14]. Fig. 3 shows the structure and weight designed by three different methods. 54% weight reduction can be achieved using topology optimization design compared to 62% reduction by threedimensional lattice structure. Therefore, the use of lattice structure instead of traditional structure design has becoming one of the key research and development directions of lightweight design and manufacturing of laser additive manufacturing to generate lattice structures. Recently, the research on metal three-dimensional lattice structure mainly focuses on body-centered cubic (BCC) and other topologies derived from it"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001112_bit.260191010-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001112_bit.260191010-Figure1-1.png",
        "caption": "Fig. 1. Scheme of the microbial electrode. 1, Bacteria-collagen membrane; 2, Teflon membrane; 3, cathode (Pt); 4, electrolyte (KOH); 5, anode (Pb); 6, O-ring.",
        "texts": [
            "5 g/liter K2HP04, and 0.5 g/liter FeS04. The cells were washed twice with oxygen-free 0.1M phosphate buffer (pH 7.0, 5\u00b0C). The suspension for preparation of the bacteria-collagen membrane contained 1.8 g collagen fibrils and 0.6 g wet cells. The bacteriacollagen membrane was prepared by casting the suspension on a Teflon plate a t 20\u00b0C. The bacteria-collagen membrane was treated with 0.1% glutarsldehyde solution for 1 min and dried a t 4\u00b0C. The scheme of the microbial electrode is illustrated in Figure 1. The electrode consists of a double membrane of which one layer is the bacteria-collagen membrane (thickness 50 pm) and the other is an oxygen permeable Teflon membrane (thickness 27 pm) , an alkaline electrolyte, a platinum cathode, and a lead anode. The double membrane is in direct contact with the platinum cathode and is tightly secured to the cell with rubber rings. A schematic diagram of the microbial electrode system is illustrated in Figure 2. The microbial electrode was inserted into a sample solution (60 ml), and the sample solution was saturated with dissolved oxygen and stirred magnetically while measurements were taken"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure13.55-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure13.55-1.png",
        "caption": "Figure 13.55 (a) Isolated beam ASC with (b) bending moment diagram, (c) shear force diagram and (d) normal force diagram.",
        "texts": [
            " The magnitude of the forces F AD D and F BD D can be found from the force polygon. In order to interpret them as normal forces N , with the correct signs (positive as tensile force and negative as compressive force), we first have to check whether the forces F AD D and F BD D from the force polygon 592 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM exert tension or compression on joint D (see Figure 13.54b). We find NAD = +F AD D = +20 \u221a 109 kN (= +208.8 kN), NBD = \u2212F BD D = \u221240 \u221a 13 kN (= \u2212144.2 kN). There is therefore tension in AD and CD and compression in BD. b. In Figure 13.55a, the beam ABSC has been isolated. At A and C, in addition to the support reactions, the components of the tensile forces in AD and CD are also active. At B the components of the compressive force in BD are acting. In the figure, the distributed loads in the fields AB, BS and SC have been replaced by their resultants. c. In Figures 13.55b to 13.55d the M , V and N diagrams are shown. The M and V diagrams due to the resultants of the field loads are shown by means of dashed lines. They give the correct values in the field boundaries. Here the dashed M diagram also gives the tangents. The final M diagram shown in Figure 13.55b with a solid line, can be checked using the rise p of the parabolas for both fields: p1 = 1 8 \u00d7 12 \u00d7 82 = 96 kNm, p2 = 1 8 \u00d7 12 \u00d7 62 = 54 kNm. These values of p fit in the M diagram shown. Note that in Figure 13.55c the shear forces at the supports A and C are not equal to the support reactions there. This is caused by the vertical components of the member forces in AD and CD. Also note that the shear force in all fields has the same slope, equal to the distributed load of 12 kN/m. 13 Calculating M, V and N Diagrams 593 There is a compressive force over the entire length of beam ABSC (see Figure 13.55d). At B, a step change in the N diagram occurs due to the horizontal component of the member force in BD. d. The largest bending moment in an absolute sense is the support moment at B: MB = 96 kNm ( ). In addition, there are extreme field moments at E and G three metres from the supports, where the shear force is zero (see Figure 13.55c). The easiest way to find their magnitudes is from the hatched area of the V diagram: ME = MG = 1 2 \u00d7 3 \u00d7 36 = 54 kNm ( ). MG is also equal to the maximum bending moment in the simply supported beam SC with uniformly distributed full load: MG = 1 8 \u00d7 12 \u00d7 62 = 54 kNm ( ). Note that the M diagram has mirror symmetry about B. The mast ABC in Figure 13.56a is supported sideways by a number of bars. Dimensions and load are shown in the figure. Questions: a. Determine the support reactions at A and D"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001972_978-3-319-15171-7-Figure7.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001972_978-3-319-15171-7-Figure7.1-1.png",
        "caption": "Fig. 7.1 Illustration of the control objective",
        "texts": [
            ", Passivity-Based Control and Estimation in Networked Robotics, Communications and Control Engineering, DOI 10.1007/978-3-319-15171-7_7 139 tracking performance for the resulting closed-loop system. The results are verified through experiments using a 2-D manipulator while comparing the present method with the method in [157]. The goal of this section is to design the camera body velocity so that the relative rigid body motion gco is regulated to a desirable configuration gd = (pd , e\u03be\u0302\u03b8d ) \u2208 SE(3) (Fig. 7.1) by using only visual measurement f defined in (6.8) with (6.7). In the following, the reference gd is assumed to be constant, i.e., g\u0307d = 0. We start with reviewing some equations formulated in Chap. 6. The relative rigid body motion gco obeys the differential equation g\u0307co = \u2212V\u0302 b wcgco + gcoV\u0302 b wo, where V b wc is the camera body velocity and V b wo is the object body velocity. The model built in the observer is formulated as \u02d9\u0304g = \u2212V\u0302 b wcg\u0304 \u2212 g\u0304u\u0302e, (7.1) where the notation ue is the observer input to drive the estimated relative rigid body motion g\u0304 to the actual gco"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.58-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.58-1.png",
        "caption": "FIGURE 7.58. Steady state configuration of a car-trailer combination.",
        "texts": [
            " The car is traveling along a circle of radius R > 1, based on a normalized length in which the length of the trailer is 1. In a circular motion with a normalized angular velocity \u03c9 = 1 and period T = 2\u03c0, the position of the car is given by the following time-dependent vector function: r = \u2219 xc(t) yc(t) \u00b8 = \u2219 R cos(t) R sin(t) \u00b8 . (7.230) The initial position of the trailer must lie on a unit circle with a center at r(0) = \u00a3 xc(0) yc(0) \u00a4T . s(0) = \u2219 xt(0) yt(0) \u00b8 = \u2219 xc(0) yc(0) \u00b8 + \u2219 cos \u03b8 sin \u03b8 \u00b8 (7.231) The car-trailer combination approaches a steady-state configuration as shown in Figure 7.58. Substituting r\u0307 = \u2219 \u2212R sin(t) R cos(t) \u00b8 . (7.232) 7. Steering Dynamics 443 444 7. Steering Dynamics and the initial conditions (7.231) in (7.202) will generate two differential equations for trailer position. x\u0307t = R (R cos t\u2212 xt) (xt sin t\u2212 yt cos t) (7.233) y\u0307t = R (R sin t\u2212 yt) (xt sin t\u2212 yt cos t) (7.234) Assuming r(0) = \u00a3 0 0 \u00a4T the steady-state solutions of these equations are xt = c cos(t\u2212 \u03b1) (7.235) yt = c sin(t\u2212 \u03b1) (7.236) where c is the trailer\u2019s radius of rotation, and \u03b1 is the angular position of the trailer behind the car"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000846_j.ymssp.2017.05.024-Figure11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000846_j.ymssp.2017.05.024-Figure11-1.png",
        "caption": "Fig. 11. Simplified models for gear tooth crack path.",
        "texts": [
            " More details about the effect of other factors on crack propagation path is available in Ref. [168]. Belsak and Flasker [169] concluded that gear crack propagation paths were smooth, continuous, and in most cases, rather straight with only a slight curvature. Kramberger et al. [170] indicated that crack mostly initiated at the point of the maximum principal stress in the tensile side of a gear tooth. In dynamic modeling, gear tooth crack paths are usually simplified to a straight line [37,59] or a slight curve [62,171] initiating in the tooth root as shown in Fig. 11. Most research papers assumed tooth crack going through the whole tooth width with a constant crack depth. Chen and Shao [101] first proposed a crack model propagating along both tooth width and crack depth. The crack effect can be simulated using the lumped parameter modeling or the finite element modeling [32]. In the lumped parameter modeling, gear crack fault is mostly reflected in the time-varying mesh stiffness. Tooth crack causes gear mesh stiffness reduction leading to abnormal vibration of gears"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure9.61-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure9.61-1.png",
        "caption": "Figure 9.61 (a) The forces in members 7 and 8 follow from the equilibrium of joint D. (b) The closed force polygon for the equilibrium of joint D. FD;4 and FD;5 are known forces. (c) Joint D with all the forces acting on it. From this figure we can see that N7 and N8 are compressive.",
        "texts": [],
        "surrounding_texts": [
            "In the method of joints, all the joints are isolated, and we investigate the force equilibrium of the individual joints. For the truss in Figure 9.53a, all the joints have been isolated in Figure 9.53b. On the isolated joints are acting \u2022 loads (joints C and D); \u2022 support reactions (joints A and B); \u2022 member forces. Here, the support reactions and member forces are the unknown forces. Since only two equations for the force equilibrium are available per joint, we have to start the calculation at a joint where no more than two forces 352 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM are unknown. These forces are determined from the joint equilibrium, after which we move to the next joint where, again, no more than two forces are unknown. In this way, we pass along each of the joints in the truss. If there are k joints, it is not the intention to first generate all 2k equations for the force equilibrium, and then to solve them together as a system of equations. We will often encounter the problem in which we cannot start with a joint with only two unknowns, as in Figure 9.53. This can be avoided by previously determining the support reactions from the truss as a whole. In Figure 9.53b, we can now start the procedure at one of the joints A or B. The method of joints is mostly used if one wants to find all the member forces in a truss. If you want only to calculate the member force somewhere in the middle of the truss, you will often have to work out the equilibrium for several joints. In that case, the method of sections is faster. Calculating the two unknown forces per joint can be done either analytically or graphically. The graphical approach is preferable; it is not only faster but also gives a better insight in the force flow. The method of joints is illustrated using a number of examples. Example 1 The truss crane in Figure 9.54 is loaded at A by means of a vertical force 4F . Question: Determine all the member forces, with the correct sign for tension and compression. Solution: In this case, we do not have to determine the support reactions as we can start directly at joint A. Here, two forces are unknown: N1 and N2. These forces can be determined both analytically and graphically. 9 Trusses 353 Figure 9.54 A truss crane. We can start the method of joints in A without having to first determine the support reactions. Figure 9.55 (a) The isolated joint A. The unknown forces N1 and N2 exerted by the members 1 and 2 on joint A are shown as tensile forces. (b) The interaction forces between joint A and members 1 and 2 as they really act. Member 1 is a compression member and member 2 is a tension member. Analytical solution for the equilibrium of joint A: In Figure 9.55a, all the forces acting on joint A are shown. In this figure, the member forces are again shown as tensile forces. For a tensile force, N is by convention positive. For the angles \u03b11 and \u03b12 shown in the figure, the equilibrium equations are \u2211 Fx = \u2212N1 cos \u03b11 \u2212 N2 cos \u03b12 = 0,\u2211 Fy = \u2212N1 sin \u03b11 \u2212 N2 sin \u03b12 \u2212 4F = 0. From the slopes of the members 1 and 2 we find sin \u03b11 = cos \u03b11 = 1 2 \u221a 2, sin \u03b12 = 1 5 \u221a 5 and cos \u03b12 = 2 5 \u221a 5. Both equations in N1 and N2 now become \u2212N1 \u00d7 1 2 \u221a 2 \u2212 N2 \u00d7 2 5 \u221a 5 = 0, \u2212N1 \u00d7 1 2 \u221a 2 \u2212 N2 \u00d7 1 5 \u221a 5 = 4F with solution: N1 = \u22128F \u221a 2, N2 = +4F \u221a 5. Member 1 is a compression member and exerts a compressive force on joint A. Member 2 is a tension member. Figure 9.55b shows the forces as they really act on both the joint and on the two members. 354 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Graphical solution for the equilibrium of joint A: FA;1 and FA;2 are the forces that members 1 and 2 exert on joint A. The forces FA;1 and FA;2 have their line of action along the members 1 and 2, but we do not know their magnitudes, nor their directions (see Figure 9.56a). Joint A is in equilibrium if all forces acting on joint A form a closed force polygon. Figure 9.56b shows the closed force polygon for the equilibrium of joint A. From here, we can read off the magnitude of FA;1 and FA;2 (or calculate it): FA;1 = 8F \u221a 2, FA;2 = 4F \u221a 5. From the force polygon, we can also find the directions of FA;1 and FA;2, but we cannot see whether they are tensile or compressive forces. To do so, we first have to draw the forces found as they act on joint A, see Figure 9.56c. Only then we can see that FA;1 is a compressive force, and FA;2 is a tensile force, so that N(1) = \u2212FA;1 = \u22128F \u221a 2, N(2) = +FA;2 = +4F \u221a 5. Note that the forces in the force polygon have not been denoted as N . The force polygon provides information only on the magnitude of the member forces, and not on the sign for tension or compression. The order in which one writes down the forces in a force polygon does not influence the result (vector addition is associative and commutative). Figure 9.57 shows two equivalent force polygons. The first force polygon is created by ranking the various forces acting on joint A in an order that is associated with an anti-clockwise rotation about joint A: 4F \u21d2FA;2 \u21d2FA;1. 9 Trusses 355 The second force polygon arises from ranking the forces in a clockwise order, so that 4F \u21d2 FA;1 \u21d2 FA;2. In Figure 9.58, the order (a) to (h) shows how, per joint, we can consecutively calculate two member forces (and then the support reactions in G and H). The members for which the forces are known are shown in bold. Figure 9.58a shows the initial situation. A is the only joint with two unknown member forces. Once we have calculated these, we get the situation shown in Figure 9.58b. Now B is the only joint with only two unknown member forces. Once these have been determined, we get the situation in Figure 9.58c, and so forth. The order in which the joint equilibrium is determined, with no more than two unknowns per joint, is A \u21d2 B \u21d2 C \u21d2 D \u21d2 E \u21d2 G \u21d2 H. 356 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM For calculating the still unknown member forces, we now use the graphical method. After A, the next joint is B, where we can calculate the member forces. Joint B is subject to the forces FB;1, FB;3 and FB;4, of which FB;1 is known. Earlier, we found that the force in member 1 is a compressive force: N1 = \u22128F \u221a 2. Member 1 therefore exerts a compressive force on joint B of 8F \u221a 2, so that FB;1 = 8F \u221a 2 (see Figure 9.59a). The two unknowns FB;3 and FB;4 can be determined from the closed force polygon for the equilibrium of joint B (see Figure 9.59b): FB;3 = 4F, FB;4 = 4F \u221a 5. In Figure 9.59c, the forces from the force polygon are shown as they act on joint B in reality. Here we see that FB;3 and FB;4 are both compressive forces. Converted into the normal forces in the members 3 and 4, with the correct sign for tension and compression, we therefore get N3 = \u2212FB;3 = \u22124F, N4 = \u2212FB;4 = \u22124F \u221a 5. The following joint with only two unknowns is C. The forces that the members 2 and 3 exert on the joint are known (see Figure 9.60a): FC;2 = 4F \u221a 5, FC;3 = 4F. The unknown forces FC;5 and FC;6 follow from the force polygon in Figure 9.60b: 9 Trusses 357 FC;5 = F \u221a 5, FC;6 = 3F \u221a 5. In Figure 9.60c, all the forces are shown as they act on joint C in reality. Member 5 presses against the joint and is a compression member, member 6 pulls on the joint and is a tension member: N5 = \u2212F \u221a 5, N6 = +3F \u221a 5. In Figures 9.61 to 9.64, the other member forces are calculated using the same method. Table 9.4 provides a summary of all the member forces. 358 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM 9 Trusses 359 360 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM In Figures 9.63 and 9.64 the support reactions in G and H have also been calculated: Gv = 10F, Hh = 0, Hv = 6F. In Figure 9.65, the support reactions are shown with the directions in which they are acting. To check the calculation, we can look at the equilibrium of the truss as a whole: \u2211 Fx = Hh = 0,\u2211 Fy = Gv \u2212 Hv \u2212 4F = 10F \u2212 6F \u2212 4F = 0,\u2211 Tz|H = Gv \u00d7 2a \u2212 4F \u00d7 5a = 10F \u00d7 2a \u2212 4F \u00d7 5a = 0. The truss as a whole meets the equilibrium conditions. Example 2 The truss in Figure 9.66 is loaded at joint E by a vertical force of 120 kN. Question: Calculate the member forces, with the correct sign for tension and compression. Solution: In this truss, we cannot find a joint with only two unknown forces. Before we can start the procedure for the joint equilibrium, we first have to determine the support reactions from the truss as a whole. Then we can start the calculation at joint A or B. 9 Trusses 361 Figure 9.67 The order (a) to (g) shows how, starting at A, we can consecutively determine two member forces per joint. The members for which we know the normal force are shown in bold. In Figure 9.67, the order (a) to (g) shows how, starting at A, we can consecutively determine two member forces per joint. The members for which we know the normal force are shown in bold. We will look at the joints in the following order: A \u21d2 C \u21d2 D \u21d2 E \u21d2 G \u21d2 H \u21d2 K or B. The last two joints K and B both offer an opportunity to check the results: both force polygons have to be closed and give the same force in member 13. Table 9.5 provides a summary of all the member forces. 362 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Example 3 You are given the (Baltimore) truss beam in Figure 9.68. Question: Determine the member forces N1 to N15 using the method of joints. In which order should we handle the joint equilibrium? Solution: After first determining the support reactions from the equilibrium of the truss as a whole, we can determine the forces in members 1 to 6 from the equilibrium of joints A, B and C respectively. In the situation shown in Figure 9.69a we get stuck, as more than two member forces are unknown in both D and E. Since members 8 and 12, and 10 and 13 are in a direct line with one another, we can determine the forces in the members 11 and 9 from the equilibrium of joints H and G. The vertical equilibrium of joint H in Figure 9.70 gives N11 = 2F. We now have the situation as shown in Figure 9.69b. From the equilibrium in G in the direction normal to members 10 and 13 we find next (see Figure 9.70): N9 + 1 2N11 \u221a 2 = N9 + 1 2 \u00d7 2F \u00d7 \u221a 2 = 0 \u21d2 N9 = \u2212F \u221a 2. Now that N9 is known (see Figure 9.69c), we can find the remaining member forces by consecutively elaborating the equilibrium of joints D, E, G, H and L. The order in which we handle the joints is therefore A \u21d2 B \u21d2 C \u21d2 H \u21d2 G \u21d2 D \u21d2 E \u21d2 G \u21d2 H \u21d2 L. 9 Trusses 363 Instead of determining N9 and N11 from the equilibrium of joints H and G, it is far easier to revert to the method of sections. With the section shown in Figure 9.69d across members 12, 13 and 14, we can determine the force in member 14 from \u2211 Tz|K = 0. The other member forces are then found from the equilibrium for the successive joints E, D, G, H and L. In certain cases, it can be useful to switch from one method to the other at the right moment. Table 9.6 provides a summary of member forces N1 to N15. Table 9.6 Member forces Example 3. Mem. no. i Ni (kN) 1 +4F 2 \u22124F \u221a 2 3 +2F 4 +4F 5 \u2212F \u221a 2 6 \u22123F \u221a 2 7 +2F 8 +4F Mem. no. i Ni (kN) 9 \u2212F \u221a 2 10 +F \u221a 2 11 +2F 12 +4F 13 0 14 \u22124F 15 0"
        ]
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure13.23-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure13.23-1.png",
        "caption": "Figure 13.23 Isolated beam ABC with the load resultants for the fields AB and BC.",
        "texts": [
            " The vertical support reaction Av (\u2191) at A is found from the moment equilibrium about G, the intersection of the two-force members BE and CD: \u2211 T|G = (4 \u221a 2 m)(54 \u221a 2 kN) \u2212 (4 m) \u00d7 Av (\u2191) = 0 \u21d2 Av = 108 kN (\u2191). The vertical support reaction Ev (\u2191) in E is then found from the moment equilibrium about C: \u2211 T|C = (4 \u221a 2 m)(54 \u221a 2 kN) \u2212 (6 m)(108 kN) \u2212 (2 m) \u00d7 Ev (\u2191) = 0 so that Ev (\u2191) = \u2212108 kN or in other words Ev = 108 kN (\u2193). Finally, the support reactions at D follow from the horizontal and vertical force equilibrium of the structure: Dh = 54 kN (\u2190), Dv = 54 kN (\u2191). 566 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM This final calculation is left to the reader. b. In Figure 13.23, the beam ABC has been isolated, and all forces acting on it at A, B and C are shown. To draw the M and V diagrams, the distributed loads in fields AB and BC have been replaced by their resultants RAB and RBC. For the triangular load on BC RBC = 1 2 \u00d7 (6 kN/m)(2 \u221a 2 m) = 6 \u221a 2 kN. The trapezoidal load on AB is divided into triangular loads (1) and (2); their resultants can be calculated easier: R(1) = 1 2 \u00d7 (18 kN/m)(4 \u221a 2 m) = 36 \u221a 2 kN, R(2) = 1 2 \u00d7 (6 kN/m)(4 \u221a 2 m) = 12 \u221a 2 kN, RAB = R(1) + R(2) = 48 \u221a 2 kN"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure9.8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure9.8-1.png",
        "caption": "FIGURE 9.8 Section of round wire for internal skin-effect models.",
        "texts": [
            "16) where Lp is the zero thickness surface partial inductance matrix, accounting for the magnetic field coupling occurring among surface currents, Zs is the surface impedance ONE DIMENSIONAL CURRENT FLOW TECHNIQUES 223 accounting for the skin effect, Pd and M are coefficients of potential matrices describing the electric field coupling among surface charges, A\ud835\udcc1 is the connectivity matrix, and Ii(s) represents potential current sources at the nodes. The model in the previous section is efficient for thin sheet-type geometries. Another similar class of problems are thin wires as they often occur in IC interconnect wires and wire bonds. An example of a wire segment is shown in Fig. 9.8. Starting out with (9.3) and (9.4), a cylindrical skin-effect model can be derived provided that the proximity effect is small enough for multiple parallel wires. This is the case if the center-to-center distance between the parallel wires is >2d. The current is strictly in the z-direction with a circular symmetric current density J = Jzz\u0302 and the length of the tube is \u0394z = ze \u2212 zb is sufficiently long for the approximation. Then, the internal field is centered around the dashed line in Fig. 9.8. Hence, the magnetic field in the \ud835\udf19 direction in a cylindrical coordinate system is given by H = H \ud835\udf19 \u0302\ud835\udf53. These assumptions considerably simplify the vector operations. If we take (9.4) and integrate it over the cross section, we get \u222bS (\u2207 \u00d7 H) \u22c5 n\u0302 ds = \u222e\ud835\udcc1 H \u22c5 d\ud835\udcc1 = \u222bS J \u22c5 n\u0302 ds, (9.17) where the second step is based on Stokes\u2019 theorem (3.34). We can get Iz(r, t) = 2\ud835\udf0brH \ud835\udf19 (r, t) (9.18) for 0 \u2264 r \u2264 a in our specific case. Next, we take Maxwell\u2019s equation (3.1a) or (9.3) where we used J = \ud835\udf0eE to get \u2207 \u00d7 J = \u2212\ud835\udf07\ud835\udf0e \ud835\udf15H \ud835\udf15t ",
            "2 Compare the impedance of the previous problem with a skin effect with a model that is constructed with the high-frequency 1D skin-effect model in Section 9.2.2. Hint: Assume that all four side surfaces are represented with such a 1D impedance model. Increase the cross section to ye = 3 mm and ze = 1 mm. Compare the result between the VFI model and the 1D surface impedance model for this larger cross section. 9.4 Skin-effect circuit model for tube Build a five-layer tube equivalent circuit skin-effect model for a conductor as shown in Fig. 9.8 with d = 0.1 mm and \ud835\udcc1 = ze \u2212 zb = 0.5 mm. With one segment along the length use (9.29) for the computation of the impedance along the length as a function of frequency. Compare the result with the conventional result obtained with the Bessel function result, for example, Ref. [22, p. 184]. 246 SKIN EFFECT MODELING 9.5 Generalized impedance boundary condition Derive the generalized impedance boundary condition (GIBC) given in (9.55). 1. H. A. Wheeler. Formulas for the skin effect. Proceedings of the IRE, 30(9):412\u2013424, September 1942"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001362_epe.2007.4417204-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001362_epe.2007.4417204-Figure8-1.png",
        "caption": "Fig. 8. Modified rotor",
        "texts": [],
        "surrounding_texts": [
            "For a three-phase machine (q = 3) and with Nc = 4, we obtain: Ns = 12 and Nr = 10. For a diphasic machine (q = 2) and with Nc = 4, we obtain: Ns = 8 and Nr = 6. We built a three-phase machine. On figures 4 and 5 we can see the stator. On figure 4, there is only the carcass out of aluminium with ferromagnetic sheets. On figure 5, we can see windings of the three phases and excitation circuit.\nAlso, all the active parts are arranged on the static part (stator) which is beneficial to evacuating the copper and iron losses.\nIn Fig. 6, we show that the no-load voltage is almost sinusoidal and that it is possible to modulate their amplitude (dexc is the current density of the wound excitation in A/mm2).\nThis amplitude modulation is useful under driving operation and also under generating operation associated a bridge of diode.\nIn order to eliminate the harmonics components, the iron sheets of the rotor are mounted with a shift angular angle. The shift electrical angle is 7.2\u00b0 to eliminate the 5n harmonics components.",
            "In the following figure, the no-load voltage with the modified rotor is clearly most sinusoidal.\nBy virtue of its passive rotor, this machine displays highly robust qualities. Moreover, it is capable of attaining a good level of performance (continuous thermal specific torque). In association with a three phase voltage bridge converter, this machine can work with a constant maximum power over a theoretically infinite range of speeds in the flux weakening mode [5-7]. In fig. 10, we present the classical associated converter and we specify the experimental measurement.\nThe hybrid excitation allows the modulation of the permanent magnets flux when energy needs are not maximal such as at \u201cat no-load work\u201d. In this structure, the iron losses can be reduced with the flux weakening and with the wound excitation. It\u2019s necessary to use a DC-DC converter to create the current",
            "for the wound excitation. The power of this converter is about 200 W when the converter power with the machine associated with three phase voltage bridge converter is about 3 kW.\nIn fig. 11, we present the assembly experimental test. An induction machine, MAS (3000 rpm \u2013 5 kW), supply with an inverter can be used in motor mode or in generator mode. The DC bus (300 V) is the same for the two converter. The total power (UDC \u00d7 I0T) is equal to the sum of the losses. With the contactors KMS and KMAS and with the coupling A, we can have different solutions to test the prototype, MS.\nTo measure the torque, we feed the machine with sinusoidal currents, and we use a mechanical assembly balances. The current density can vary up to 63 A/mm2 (electronic limit). The thermal torque (permanent working) is obtained with a rated value of 10 A/mm2 and with a current density of the wound excitation of 13 A/mm2.\nWe can see on fig. 12, which the permanent massive torque is about 2.2 Nm/kg and for transient working is can reach about 6 Nm/kg.\nWe have measured the iron losses in alternator mode at no-load (KMAS closed, A closed , KMS open). In Fig. 13, we can see that the modulated excitation can reduce iron losses."
        ]
    },
    {
        "image_filename": "designv10_0_0003578_mech-34238-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003578_mech-34238-Figure3-1.png",
        "caption": "Figure 3. Schematic of a 3\u2013RPS Parallel Manipulator.",
        "texts": [
            " 2a. This singular configuration was discussed by Tsai and Joshi [12]. On the other hand, the last three rows of J represent three moments of constraint, each being perpendicular to the universal joint of a limb. Hence, the manipulator will be under singularity if these three moments either lie on a common plane or are parallel to one another as shown in Fig. 2a and 2b. These conditions were presented by Di Gregorio and Parenti-Castelli [2] and Zlatanov et al. [17]. THE 3\u2013RPS PARALLEL MANIPULATOR Figure 3 shows the 3\u2013RPS parallel manipulator which consists of a moving platform, a fixed base, and three limbs of identical kinematic structure. Each limb connects the fixed base to the moving platform by a revolute joint followed by a prismatic joint and a spherical joint. A linear actuator drives each of the three prismatic joints [9]. The connectivity of each limb is equal to 5. Hence, the instantaneous twist of the moving platform, $p, can be expressed as a linear combination of 5 instantaneous twists: $p \u03b8\u03071 i $\u03021 i d\u03072 i $\u03022 i \u03b8\u03073 i $\u03023 i \u03b8\u03074 i $\u03024 i \u03b8\u03075 i $\u03025 i for i 1 2 3 (16) Copyright  2002 by ASME s of Use: http://www"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000669_j.engappai.2003.09.006-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000669_j.engappai.2003.09.006-Figure2-1.png",
        "caption": "Fig. 2. Experimental test rig.",
        "texts": [
            " Moments of order higher than nine are not considered in the present work to keep the input vector within a reasonable size without sacrificing the accuracy of diagnosis. The roles of different vibration signals and signal preprocessing techniques are investigated. The results show the effectiveness of the extracted features from the acquired and preprocessed signals in diagnosis of the machine condition. The procedure is illustrated using the vibration data of an experimental setup with normal and defective bearings. Fig. 2 shows the schematic diagram of the experimental test rig. The rotor is supported on two ball ARTICLE IN PRESS B. Samanta et al. / Engineering Applications of Artificial Intelligence 16 (2003) 657\u2013665 659 bearings. The rotor was driven by an electrical AC motor through a flexible coupling. Two accelerometers were mounted on the right-hand side (RHS) bearing support, with an angle of 90 to measure vibrations in vertical and horizontal directions (x and y). Separate measurements were obtained for two conditions, one with normal bearings and the other with a fault on the outer race of the RHS bearing"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure6.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure6.6-1.png",
        "caption": "Fig. 6.6 Definition of spatial velocity: vo = v1 \u2212\u03c9\u00d7 p1. If we choose a difference point p2 on the rigid body, vo = v2 \u2212 \u03c9 \u00d7 p2 gives the same value. In a two dimension, when we glue a big grass plate on the rigid body, vo is the speed of grass passing on the origin.",
        "texts": [
            " For the 1 We might have to take care of such rotation to clean up space debris orbiting the earth which potentially could harm future space crafts. 6.2 Spatial Velocity 187 position of reference point p1, the speed of an arbitrary point p on the rigid body is given by v(p) = v1 + \u03c9 \u00d7 (p\u2212 p1). (B) Spatial velocity: The following vector is regarded as the translational speed of a rigid body [30] vo = v1 \u2212 \u03c9 \u00d7 p1. (6.7) With given instantaneous state of a rigid body, we obtain a unique vo despite of the choice of p1 as illustrated in Fig. 6.6. Therefore, we can regard vo as an intrinsic translational speed of a rigid body. In a world frame, the speed of a point p on a rigid body moving at (vo,\u03c9) is given by v(p) = vo + \u03c9 \u00d7 p. (6.8) A six dimensional vector made of (vo,\u03c9) is called a spatial velocity of a rigid body[30, 101]. Usually, translational speed of a rigid body is represented by the way of (A). Indeed, in chapter 2, we have defined translational speed by (A) taking a local frame origin as a reference point. 188 6 Dynamic Simulation On the other hand, by using the spatial velocity of (B), the treatment of acceleration is much simplified, and it finally results Featherstone\u2019s fast dynamic calculation which will be discussed at the end of this section"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001609_j.ijmecsci.2010.05.005-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001609_j.ijmecsci.2010.05.005-Figure3-1.png",
        "caption": "Fig. 3. Schematic diagram of a ball bearing.",
        "texts": [
            " Using this we get K \u00bc 1 1=Ki 1=n \u00fe 1=Ko 1=n \" #n \u00f02\u00de Ki and Ko inner and outer raceways to ball contact stiffness, respectively, which are obtained using Kp \u00bc 2:15 105 X r 1=2\u00f0d \u00de 3=2 \u00f03\u00de where P r is the curvature sum which is calculated using the radii of curvature in a pair of principal planes passing through the point contact. dn is the dimensionless contact deformation obtained using curvature difference (Refer Table 6.1, Essential Concepts of Rolling Bearing Technology , Tedric Harris, Taylor and Francis Group, 2007). The value of K for 6305 bearing is 8.37536 109 N/m3/2. The schematic diagram for determining the radial deflection is shown in Fig. 3. If x and y are the deflections along X- and Y-axis and Cr is the internal radial clearance, the radial deflection at the ith ball, at any angle yi is given by [(x cos yi+y sin yi) Cr]. Substituting in Eq. (1), F \u00bc K \u00f0xcosyi\u00feysinyi\u00de Cr 3=2 \u00f04\u00de Since the Hertzian forces arise only when there is contact deformation, the springs are required to act only in compression. In other words the respective spring force comes into play when the instantaneous spring length is shorter than its unstressed length (the term in the bracket should be positive); otherwise the separation between ball and race takes place and the resulting force is set to zero"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002750_j.jmatprotec.2020.117039-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002750_j.jmatprotec.2020.117039-Figure1-1.png",
        "caption": "Fig. 1. Schematic of hybrid in situ micro-rolled wire + arc additive manufacturing (HRAM).",
        "texts": [
            " Ultimately, considering the size and location of the critical defects, a fatigue life model was proposed using a combination of POT and a modified fatigue crack growth (FCG) program NASA/FLACGRO (NASGRO) model. A hybrid micro-rolling additive manufacturing system recently selfdeveloped by Huazhong University of Science and Technology was adopted to produce the HRAM material. The AM device consists of a customized plate rolling machine, a SAF\u22c5FRO DIGI@WAVE\u2122 280 metal inert gas welder, a 3D computer numerical control machine, and an integrated control system (see Fig. 1). A schematic of the single-pass and single-layer deposition process is also given in Fig. 1. The roller follows the arc torch and cyclically rolls over the material surface according to the expected scanning path. Due to the relatively short distance between the welding gun and roller, the metal specimen can form immediately by hot-rolling-induced deposition. The rolling force can be controlled according to feedback from a force sensor installed between the screw and roller rack. Commercial ER 5087 (Al-Mg4.5 Mn) wire with a diameter of 1.2 mm was used as the filler material, and a 2219-T851 aluminum alloy rolled plate with dimensions of 400 \u00d7 140 \u00d7 40 mm3 was adopted as the substrate"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003947_robot.2004.1308858-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003947_robot.2004.1308858-Figure5-1.png",
        "caption": "Fig. 5. In this figure IJKLM is the support polygon and R acts through P. Since R does not pass tbmugh G. it creates a non-zero moment. The ZRAM point is at A outside the support polygon. If the side JK shifts outward by a distance d it would recapture A, thereby making Hc = 0.",
        "texts": [],
        "surrounding_texts": [
            "robot located at CoM, Q = Czl Q, is the resultant of all the external non-ground forces and a is the acceleration of the CoM. Eq. 1 can be solved for the magnitude and direction of R hut not the location of its line of action. From this equation alone we cannot determine whether R passes through G or not. For that we need to solve the moment equation.\nThe moment equation must be formulated either at the robot CoM or at any inertial reference point. Taking moments about an arbitrary inertial point 0, we have\nP M I + M, + OPi x R, + OP, x R + X u ,\n,=I\n\" n \" H . = Hci + OG; x mea, (3)\n;=I ;=I ;=I\nwhere Hci is the centroidal angular momentum of the ith\nSe!$?t'may be reduced to\nm\nM + O P l ' x R, + OP, x R + o+c OS, x Qi + OG x mg k l\n= H c + OG x ma (4)\nwhere M = M , + M , , u = CP=lu, and H c = EHc,+ 1 GG, x m,a,.\nTa@ng moments about the CoM G, we have the simpler equatlon\nm\nM + G P ! x R, + GP, x Q + U + CGS, x Q~ = H c (5) ;=I\nEq. 5 forms the basis of analysis and control of balance in 3 bipedrobots.\nE. Discussion In a general setting we have to re-evaluate the validity of\nassumptions that are made in special situations. Although a common practice, it is, in general, improper to ignore the gmund reaction moment M in Eq. 5 because it may contribute towards robot tipping. This becomes clear if we imagine a biped standing with two feet on two rock surfaces, none of which is horizontal. The normals to these surfaces are not vertical and footlground frictional moments generated due to these surfaces can act to weaken or strengthen stability.\nConsidered vital to terrestrial locomotion, the gravity term mg does not appear in Eq. '5. Although this is caused by our specific choice of moment center, G, it is instructive to realize that gravity is not an integral pan of rotational instability. In fact one may deliberately set mg = 0 in Eq. 4 to perform stability analysis of a spacewalker inside an orbiting satellite. As the person navigates using both hands and feet while floating in space HG reliably provides the stability information. Similar mechanics is applicable to skyscrape1 window cleaning robot that is suspended from above by a cable to compensate for self-weigbt.\nWe should also point out that the LHS of Eq. 5 is a collection of all the centroidal moments, regardless of their origin. As such, the reaction forcdmoments at the feet are not special and have analogous effects as the reaction forcdmoments at the hands or at any other location of the robot body. This is aligned with the spirit of humanoid robots performing realistic and more useful functions, and especially using bands. In order to incorporate hand interaction forces, traditional definition of ZMP was augmented [24] and imaginary surfaces were constructed [25]. HC may be used unchanged throughout interactions of all types.\nIn a radically different application rotational stability of planar parts is closely studied for automated triage and pans feeding [26]. HG may be used to analyze the turning of these parts on a horizontal treadmill caused by friction, inertia and constraint forces.\nC. Simpler case -free biped walk on level ground Eq. 5 is fairly general except for the support surface beneath individual feet assumed planar. Given specific situations we may relax certain conditions to obtain simpler versions of Eq. 4 and Eq. 5. What follows in the remainder of this section is the exploration of HG and its derived condition, ZRAM for the special case when, U = 0, Qi = 0, left and right feet are posed on the same horizontal plane, and M has a non-zero vertical component that does not contribute to tipover instability. Under these conditions R = R1 + I?+ is the resultant GRF passing through P, and Eq. 4 reduces to\nOP x R + OG x mg = H , = Ho + OG x ma (6)\nCharacterisrics of ZRAM paint: Moments taken at G results in an especially simple result:\nG P X R = H ~ (7)\nIn general, H c = G P x R # 0. But let us suppose that there is a point A on the ground such that G A x R = 0. Point A is called the ZRAM point and is found by projecting robot's CoM along the resultant force [271, [28] (see Fig. 1).\nThe ZRAM point has two characteristics: 1) GAllR and 2) AP x R = Hc. Longer is the distance A P larger is the amount of moment on the robot's CoM and larger is Hc. Conversely, as A gets closer to P, the amount of unbalanced moment at the CoM is also reduced, and finally becqmes zero as the ZRAM point coincides with COP. Note that H a # 0.\nRecall that FRI point is a point on the footlground contact surface where the net ground reaction force would have lo act to keep the foot stationary [Z]. To ensure no foot rotation, the FRI point must remain within the convex hull of the foot support area. Refer to Fig. 4.\nThe distinct advantage of the ZRAM point over FRI point is that the former is not defined on the basis of physical foot rotation and is therefore valid during both the single and double support phases of walking.",
            "V. CONTROL STRATEGIES This section outlines three control strategies that may be\nused to recapture balance. Each strategy attempts to make H c = 0 in a specific way. In this section we will allow interaction forces U and Q. With this relaxed condition Eq. 6 may be re-written as:\nm\nGP x R + GS. x Q< No (81 i=1\nCompare this equation with Eq. 5. Since the interaction forces are beyond direct control of the robot, it can attempt one of three things:\n1) Enlarge suppon polygon such that it encompasses the\n2) Move G with respect to P such that R passes through\n3) Change GRF direction by means of changing the cen-\nZRAM point A.\nG in its new location G\u2019.\ntroidal acceleration a to a\u2018.\nTo en&rge support polygon:\nLet us suppose that IJKLAJ is the current support polygon and that the side J K reauires an outward shift bv an amount d in order to just include-the ZRAM point A (see big. 5) . This can be achieved be re-deploying the foot at a distance d.\nWe can write\nGJ\u2019 = GJ t d(k x e) (9)\nJK I I where e = JR and k is a unit vector perpendicular to the plane of support polygon if it is planar. Otherwise k is a unit vector normal to the plane containing J K and J\u2019K. d can be expressed as:\n- e . [GJ x R t U] e , [ (k x e) x RI d =\nTo move G:\nsatisfies H Q = 0. In other words, Suppose that when G moves to a different position G\u2018 it\nm C P x Rt u+c G\u2019S; x Q, = 0 (11)\nand from Eq. 8 and Eq. 11 we obtain, by setting G P = <=I\nGG\u2019 + G\u2019P and GSj = GG\u2019 + G\u2019Sj,\nG G X ( R + Q ) = & (12)\nEq. 12 is of the standard form A x B = C and can be solved for A. The support stability indicator [29] applies a similar concept for multi-legged robots. To change GRF direction:\nLet us rewrite Eq. 8 by setting R = ma - mg - Q, m\nGP x (ma - mg - Q) t U + 1 GS; x Q, = kc (131\nSuppose that HG = 0 is obtained by changing ma to ma\u2019.\n;=I\nFrom Eq. 13 we get\nm GP x (ma\u2019 - mg - Q) + r + GS, x Q, = o (14)\n,=I",
            "From Eq. 13 and 14 we get,\nGP x ma\u2018 = GP x ma- k c (15)\nWe again find the standard form A x E = C where A = G P , B = ma\u2019 and C = G P x m a - H G . We can subsequently solve for ma\u2018.\nVI. CONCLUSION AND FUTURE WORK We have re-firmed that the rate of change of centroidal angular momentum H G is a useful criterion for the analysis and control of postural balance in biped robots in a very\n191 V. Zatsiorsky, Kinetics of Human Morion. Champaign, IL: Human Kinetics, 2002. [IO] M. VukobratoviC, \u201cHow to control artificial anthropomorphic systems:\u2019 IEEE Tmmnroctiom on Sysremr, Man, ond Cykrnericr, vol. SMC-3. no. 5. pp. 497-507, 1973. I l l ] Q. Li. A. Takanishi, and I. Kata. \"Learning control of compensative vunk motion far biped walking mbot based on zmp stability criterion:\u2019 in f\u201dlenutioM/.Confer=~ on Intelligent Robots and System, vol. 1. July 7-10 1992. pp. 597603. 1121 C.-L. Shih, \u2018The dynamics and conval of a biped walkiing rohat with seven degrees of freedam,*\u2019 ASME Joumol of Dymmic Sysrem, Measurement. nrld Contml, vol. 118. pp. 683690. December 1996. 1131 K. Hirai, M. Hirase. Y. Haikawa and T. Takenaka. \u2018The development of honda humanoid robot,\u201d in IEEE Intemotionol Conference on Robotics and Autonution, 1998. Belgium, pp. 1321-1326. 1141 Q. Huang. K. Yokoi. S. Kajita. K. Kaneko. H. Aria N . Koyachi. and K. Tanie.. \u201dPlanning walking panerns for a biped robot:\u2019 IEEE Tmnsonions on Robotics and Automotion, vol. 17. no. 3. DD. 280-289. ..\ngeneral situation. Loss of balance implies that HG is non-zero.\nderived from H G and is applicable to walking on planar\nmay be used for balance recapture. Athough H G indicates an overall centroidal moment (and hence \u201dangular acceleration\u201d) this does not directly indicate an i d n e n t fall, F~~ that, .one have the bowledge of the angular momentum HG. The d i w i o n s and magnitudes of HG and H G together will determine the rotational behavior of the robot.\nEfficient computation Of the two quantities HG and H G is of vital necessity for a good balance controller. These quantities are functions of the robot\u2019s link geomeuy, mass, and inertia properties, as well as the angular position, velocity\n2001. 1151 T. Sugihara, Y. Nakamura, and H. Inoue. \u201cRealtime humanoid motion\ngeneration through Imp manipulation based on inverted pendulum\nWashington, D.C., 2002. 1161 R. Kuraume. T. Hasegawa, and K. Yoneda, \u2018The sway compensation\ntrajectoly of a biped robot.\u201d in IEEE Internotionol Confernice on\n1171 T. Sugihara and Y. Nakamura, \u201cWhole-body caoperative cog control through zmp manipulation for humanoid robots:\u2019 in Pmc. 2nd Int. S?mp on Adoptive Motion of Adnlols and Mochines, Kyoto, Japan. 2003. 1181 A. Takanishi, ~ . - o . Lim, M. Tsuda, and I. Kato. \u201cRealization of dynamic biped walking stabilized by vunk motion an a sagitally uneven surface.\u2019\u2019 International Confemnce on Intelligent Robots ond System, pp. 323- 330, 1990. 1191 M. Yamada, 1. Furusho. and A. Sana, \u201cDynamic conml of walking robot with kick-action:\u2019 in Pmc. of In!. Conf on Adwncd Robotics. ICAR \u201885, 1985, pp. 405412. 1201 A. San0 and J. FuNsho, \u201cControl of the toque distribution far the BLRG2 biaed robot.\u201d in Pmc. of Int. Cont on Advonced Robotics. ICAR \u201891.\nwe have also introduced the meaSuIe\nsurfaces. Finally we have outlined three control strategies that\na control:\u2019 in IEEElnler~tional conference On Roboticrand Automotion,\nRobotics ~ ~ t ~ ~ t i ~ ~ , 2003, Taipei, Taiwan, pp. 925-931.\nand acceleration. ~ i s a \u2019 1 9 9 1 . m 728-734. . _......__. ~~~ . ~ ~ ~ ~ , . ~ ,y~ough we use a biped robt as a example, our approach is completely general and applicable to multi-legged robots as well. 1211 S. Kagami, F. Kanehim, Y. Tamiya. M. lnaba and H. Inoue. \u201cAutobalancer: An online dynamic balance compensation scheme for humanoid robots,\u201d in Pmc. qih In:. Workrhop on Algorithmic Foundomrims on Robotics. WAFR \u201830, 2wO. pp. SA-79SA-89.\n1221 N. E. Sian, K. Yokoi, S. Kajita, F. Kanehiro, and K. Tanie. \u201cWhole-body teleoperatian of a humanoid robot - a method of integrating operator\u2019s intention and robot\u2019s autonomy:\u2019 in IEEE International Conference on Robotics and Automotion, Taipei. Taiwan, Sept. 14-19 2003, pp. 1613- ACKNOWLEDGMENT\n, Viutha Kallem wishes to thank Honda Research Institute USA, Inc. for summer internship support.\nREFERENCES\nD. KatiC and M. VukabratoviC, \u201cSurvey of intelligent conml techniques for humanoid robots,\u201d Journal of Intelligent and Robotic System, vol. 37, pp. 117-141. 2003. A. Goswami, \u201cPostural stability of biped rnbots and the foot rotation indicator (FRI) point.\u201d Ltemotionol Journal of Robotic3 Reseoreh, vol. 18, no. 6, pp. 523-533, 1999. M. Vukobratovif, A. A. Frank, and D. Juricic, \u201dOn the stability of biped loeomotian:\u2019 IEEE Tmnsocrions on Biomedicol Engineering, vol. BME17, no. 1, pp. 25-36, January 1970. M. Vukobratovif, B. Borovac, D. Surla, and D. Stokic, Scinctific Fundamentals of Robotics 7. Biped Locomotion: Qnamics Stabilio, Conlml and Applicotion. E M . Silva and 1. A. T. Machado, \u201cGoal-oriented biped walking based on fone interaction conml:\u2019 in IEEE Internotional Conference on Robotics ondhtomorion, May 21-26 2001, pp. 41224127. S. 110, Y. Saka. and H. Kawasaki, \u201cA consideration on position of center of ground reaction force in upright pasture:\u2019 in SlCEAnnual Conference, Aueust 2002. Springer-Verlag. New York 1990.\n1619. 1231 S. Kajita, R Kanehim, K. Kaneko. K. Fujiwara, K. Harada K. Yokoi. and\nH. Hirukawa \u201cResolved momentum control Humanoid motion planning bared on the linear and angular momentum:\u2019 in Inremorionol Conference on Intclligenr Robots and System, 2W3, Las Vegas, NV, USA, pp. 1644- 1650.\n1241 K. Harada, S. Kajita K. Kaneko. and H. Hirukawa, \u201cPushing manipulation by humanoids considering two-kindsaf zmps:\u2019 in IEEE 1 n l c r ~ - tional Conference on Robotics and Automotion. September 14-19 2003, Taipei, Taiwan. pp. 1627-1632. I251 S. Kagami, K. Nishiwaki. T. Kitagawa, T. Sugihara. M. Inaba, and H. h u e , \u201cA fast generation method of a dynamically stable humanoid robot mjkectory with enhancedmp constraint,\u201d in P m . 1st IEEE-RAS Int. Conf on Humonoid Robots, Humomids2ooO. September 7-8 2wO. I261 K. M. Lynch and M. T. Mason, \u201cDynamic nonprehensile manipulation: Controllability. planning. and experiments:\u2019 Inlemorional Jounlol of RoboticsReseorch, vol. 18. no. 1. pp. 64-92. 1999. 1271 D. Orin. \u201cInteractive control of a six-legged vehicle with optimization of both stability and energy,\u201d Ph.D. dissemtion, The Ohio State University, 1976. 1281 B. S. Lin and S. M. Sang. \u201cDynamic modeling, stability and energy efficiency of a quadrupedal walking machine:\u2018 in IEEE Inremtionol Conference on Robotics and Automorion, 1993, Atlanta CA, pp. 367- 373. I291 F. Hardanon, \u2019\u2019Stability analysis &d synthesis of statically balanced walking far quadruped robots,\u201d W.D. dissenation, Royal Institute of Stokholm, KTH, 2002. 171 D. T. Greenwood. Principles of Llyamics. New Jersey: Prentice-Hall, 181 1. Peny, Gail AnaIy3iS: N o m 1 ond Pathologic01 Function. Thorfm. Inc., 1965.\nNJ: SLACK Incorporated. 1992."
        ]
    },
    {
        "image_filename": "designv10_0_0003928_tpel.2006.872371-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003928_tpel.2006.872371-Figure4-1.png",
        "caption": "Fig. 4. Finite element field solution with phase overlap under long flux path excitation.",
        "texts": [
            " As can be seen, for a fixed turn-on instant, significantly different peak currents can be achieved by altering the turn-off instant. This point should be taken into account when designing optimal or protective control strategies. In order to efficiently operate an SRM drive, use of substantial overlap between two phases is commonly practiced. Creation of an overlap, once properly tuned, contributes to higher torque productivity and mitigation of torque pulsation. In an 8/6 SRM a maximum overlap of 15 (mechanical) is allowed. Fig. 4 shows the distribution of flux density in the SRM under multiphase excitation. Notably, a part of the back iron, located between two excited stator phases, exhibits a very low flux density. This is due to the fact that fluxes generated by each phase oppose each other within this region. In the remaining 75% of the back iron, on the other hand, fluxes generated by two stator phases are added. This explains the existence of a relatively high flux density in this part of the back iron. This pattern of magnetization has been referred to as \u201clong flux path excitation (LFPE)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001478_j.conengprac.2011.04.005-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001478_j.conengprac.2011.04.005-Figure8-1.png",
        "caption": "Fig. 8. Blade flapping with stiff rotor blades modeled as hinged blades with offset and spring.",
        "texts": [
            " Let {eN,eE,eD} denote unit vectors along the respective inertial axes, and {xB, yB, zB} denote unit vectors along the respective body axes, as shown in Fig. 3. Euler angles to rotate from NED axes to body-fixed axes are the 3\u20132\u20131 sequence fc,y,fg, referred to as yaw, pitch, and roll, respectively. The current velocity direction unit vector is ev, in inertial coordinates. The direction of the projection of ev onto the xB\u2013yB plane defines the direction of elon in the body-fixed longitudinal, lateral, vertical frame, {elon, elat, ever}, as shown in Fig. 8. Due to blade flapping, the rotor plane does not necessarily align with the xB, yB plane, so for the jth rotor let fxRj ,yRj ,zRj g denote unit vectors aligned with the plane of the rotor and oriented with respect to the {elon, elat, ever} frame. Let x be defined as the position vector from the inertial origin to the vehicle c.g., and let xB be defined as the angular velocity of the aircraft in the body frame. The rotors, numbered 1\u20134, are mounted outboard on the xB-, yB-, xB and yB-axes, respectively, with position vectors rj with respect to the c",
            " (14) as lb \u00bc ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1\u00fe 3 2 ef \u00fe kb IbO 2 s \u00f015\u00de Blade flapping causes both longitudinal and lateral thrust forces and moments. For quadrotor helicopters, however, the moments generated by lateral deflections cancel when yaw rates are low relative to airspeed, and generation of unbalanced moments is due entirely to the longitudinal deflection, a1s. The backward tilt of the rotor plane generates longitudinal thrust Tlon (see Fig. 8), Tb,lon \u00bc Tsina1s \u00f016\u00de If the center of gravity (c.g.) is not vertically aligned with the rotor plane, this longitudinal force will generate a moment about the c.g., Mb,lon \u00bc Tb,lonzcg , where zcg is the vertical distance from the rotor plane to the c.g. of the vehicle. For stiff rotors, as are used in most current quadrotor helicopters, the tilt of the blades also generates a moment at the rotor hub Mb,s \u00bc kba1s \u00f017\u00de where kb is the stiffness of the rotor blade in N m/rad. Finally, Mbf\u00bcMb,lon\u00feMb,s is the total longitudinal moment created by blade flapping"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002547_tac.2021.3061646-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002547_tac.2021.3061646-Figure1-1.png",
        "caption": "Fig. 1. Pendulum system.",
        "texts": [
            " Even though not explicit in the statement of Theorem 2, Vn(t, x) solves the following family of HamiltonJacobi-Bellman equations parameterized by \u03b2 \u2208 [2,+\u221e) \u2202Vn \u2202t + \u2202Vn \u2202x F + 1 2 Tr { GT2 \u22022Vn \u2202x2 G2 } \u2212\u03b2 2 \u00b5\u03b4n\u03b1n\u03be 4 n + l(t, x) 2\u03b2 = 0. (105) Remark 8. By choosing k5 = \u22123\u03b4n, S(t, x) = 2\u03b2Vn(t, x), l(t, x) in (95), \u03b3(r) = r4 and R(x) = ( 27 16\u03b22\u03b13 n )1/4 , Theorem 2 solves the prescribed-time inverse optimal mean-square stabilization problem described in Definition 2. V. TWO SIMULATION EXAMPLES In this section, we give two simulation examples to show the effectiveness of the prescribed-time control schemes developed in this paper. Example 2. Consider the pendulum system shown in Fig.1. By Newton\u2019s law of motion, the system is described as [43] m0l\u03b8\u0308 = \u2212m0g sin \u03b8 \u2212 kl\u03b8\u0307 + 1 l T\u0304 , (106) where l denotes the length of the rod, m0 denotes the mass of the bob and g denotes the acceleration due to gravity. Assume the rod is rigid and has zero mass. Let \u03b8 denote the angle subtended by the rod and the vertical axis through the pivot point. The pendulum is free to swing in the vertical plane. The bob of the pendulum moves in a circle of radius l. There is also a frictional force resisting the motion, which we assume to be proportional to the speed of the bob with a coefficient of friction k"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure7.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure7.2-1.png",
        "caption": "Fig. 7.2. Optical encoder",
        "texts": [
            " Shaft encoders are angle positional sensors which, because of the specific construction, convert the angle position directly into digital information. There are several types of encoders (optical, contact, magnetic), but in practice optical encoders are mostly used, having some advantages over others. Optical encoders may be absolute and incremental. An incremental encoder detects the quantum of the change of the angular position, and an absolute one shows the absolute angular position. The encoder consists of the disk (see Figure 7.2) which is connected to the rotating shaft. The disk has concentric ring(s) with alternate transparent and opaque segments. On one side of the disk there are light sources (usually LEDs), and on the other side light detectors are situated. The signal from the detector is led to the electronics for the signal conditioning and then to the control system. The advantage of the incremental encoder over the absolute one is the use of only one pair of light source and detector, while the absolute uses n pairs, where n is its resolution in bits"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003683_1.2787276-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003683_1.2787276-Figure1-1.png",
        "caption": "Fig. 1 A parametrically and externally excited mechanical oscil lator subject to a clearance",
        "texts": [
            " Finally, several cases of measured chaotic motions are also illustrated through time-traces, phase portraits, Fourier spectra, and Poincare plots which form strange attractors. 1 Introduction Certain classes of mechanical systems having clearances at a contact interface which transmits motion or force are also subject to periodically time-varying forcing and parametric excitations. This is especially common in rotating machinery. Such systems give rise to complex dynamic behavior exhibiting a rich spectrum of nonlinear phenomena. The single-degree-offreedom mechanical oscillator shown in Fig. 1 can be considered as a generic model for such systems. The equation of motion for this oscillator is given in dimensionless form as __d2x + 24 dx = d7-2 ~ + w(r)g[x(7-)] f(7-) ( la ) where x ( r ) is the displacement of the unit mass, ~ is the viscous damping ratio, and T represents dimensionless time. The timevarying stiffness w( r ) and external excitation f ( r ) functions are assumed to be periodic in ~- and can therefore be expressed in terms of the Fourier series W(T) = Wo + ~ [WrCOs(rAT) +~r s in ( rAT- ) ] , ( lb) r=l f(7-) =fo + ~ [fr COS (rAT-) +j~ sin (rAT)]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003774_j.1469-185x.1974.tb01299.x-Figure10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003774_j.1469-185x.1974.tb01299.x-Figure10-1.png",
        "caption": "Fig. 10 Fig. II",
        "texts": [
            " At low Reynolds number the force exerted on each element of the cylinder is proportional to the local velocity. The movement of each portion of the cylinder can be described in terms of two components of velocity, one normal (VN) to the cylinder and the other tangential (V,) to it. The coefficients of resistance CN and C, are defined as the dimensional constants which give the relationship between both the normal SFN and tangential SFT force elements and VN and V,, respectively, on an element of cylinder 6s in length (see Fig. 10). Thus Gray & Hancock (1955) obtained the following relationship between the normal and tangential coefficients of resistance, which follows from the discussion in (u) : where (4) c\u2018 = log (2h/r0) - 4 \u2019 Fig. 10. Illustrates the Gray and Hancock theory. An element 6s of the organelle is moving with velocity V with components V, in the normal and V, in the tangential directions. To obtain the direction of the force exerted on the fluid, we double the V, coordinate, while keeping VT the same, then join up the rectangle to obtain the direction as shown. Fig. I I. Comparison between the velocity profile for an inert and self-propelling body. The penetration depth of noticeable fluid motion is S for the self-propelling body"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003440_1.13030-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003440_1.13030-Figure1-1.png",
        "caption": "Fig. 1 Two possible choices for \u03b1o c and \u00b5o c to solve the (\u03c7, \u03b3) control.",
        "texts": [],
        "surrounding_texts": [
            "online approximators are defined in Sec. VII. Section VIII contains a simulation example and discussion of the controller properties."
        ]
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure14.67-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure14.67-1.png",
        "caption": "Figure 14.67 (a) A three-hinged frame whose shape has been somewhat adapted to the shape of the line of pressure. (b) Support reactions. (c) Bending moment diagram.",
        "texts": [
            "66b and 14.66c show the support reactions and the bending moment diagram respectively. The distance b between the moment zeros follows from 1 8qb2 = 36 kNm in which q = 4 kN/m. We find b = \u221a 8 \u00d7 (36 kNm) 4 kN/m = \u221a 72 m2 \u2248 8.5 m. The location of hinge S is then a = \u2212 b 2 \u2248 1.75 m. 14 Cables, Lines of Force and Structural Shapes 685 By moving hinge S we adapt the line of pressure to the shape of the frame. Alternatively, the shape of the frame can also be adapted to the line of pressure. This is shown in Figure 14.67a.1 For the horizontal support reaction H it holds that H = 1 8q 2 p = 1 8 \u00d7 (4 kN/m)(12 m)2 4 m = 18 kN. In Figure 14.67b the support reactions are shown. They are of equal magnitude to the support reactions of the three-hinged frame in Figure 14.65. Figure 14.67c shows the bending moment diagram. The moment distribution is determined below for girder SD. For this reason, in Figure 14.68 a part directly adjacent to hinge S has been isolated. Only the horizontal compressive force H acts at hinge S; there is no vertical force. In the other section, there is a bending moment M , shear force V and normal force N . From the moment equilibrium about this section we find 1 Here we assume that the distributed load is still a force per horizontally measured length. 686 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM M = H \u00b7 1 3x \u2212 qx \u00b7 1 2x = 1 3Hx \u2212 1 2qx2 (a) in which H = 18 kN and q = 4 kN/m. Checking expression (a) for the bending moment at D, with x = 6 m M = MD = 1 3 \u00d7 (18 kN)(6 m) \u2212 1 2 \u00d7 (4 kN/m)(4 m)2 = \u221236 kNm. This minus sign indicates that the bending moment at D acts opposite to the direction shown in Figure 14.68. This is in agreement with the M diagram in Figure 14.67c. The bending moment is zero at E (see Figure 14.67c). With M = ME = 0, it follows from (a) that: x = xE = 2H 3q = 2 \u00d7 (18 kN) 3 \u00d7 (4 kN/m) = 3 m. The field moment in SD is a maximum at G. Here dM/dx = 0. Differentiating expression (a) gives dM dx = 1 3H \u2212 qx = 0 so that x = xG = 1 3H q = 1 3 \u00d7 (18 kN) 4 kN/m = 1.5 m. From here, (a) gives M = MG = 1 3 \u00d7 (18 kN)(1.5 m) \u2212 1 2 \u00d7 (4 kN/m)(1.5 m)2 = 4.5 kNm. 14 Cables, Lines of Force and Structural Shapes 687 Figure 14.70 (a) Arch bridge with upper deck. The horizontal forces in the arch are directly transferred to the foundation"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure9.23-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure9.23-1.png",
        "caption": "FIGURE 9.23. A simplified models of a trebuchet.",
        "texts": [],
        "surrounding_texts": [
            "Dynamics of a rigid vehicle may be considered as the motion of a rigid body with respect to a fixed global coordinate frame. The principles of Newton and Euler equations of motion that describe the translational and rotational motion of the rigid body are reviewed in this chapter. 9.1 Force and Moment In Newtonian dynamics, the forces acting on a system of connected rigid bodied can be divided into internal and external forces. Internal forces are acting between connected bodies, and external forces are acting from outside of the system. An external force can be a contact force, such as the traction force at the tireprint of a driving wheel, or a body force, such as the gravitational force on the vehicle\u2019s body. External forces and moments are called load, and a set of forces and moments acting on a rigid body, such as forces and moments on the vehicle shown in Figure 9.1, is called a force system. The resultant or total force F is the sum of all the external forces acting on a body, and the resultant or 522 9. Applied Dynamics total moment M is the sum of all the moments of the external forces. F = X i Fi (9.1) M = X i Mi (9.2) Consider a force F acting on a point P at rP . The moment of the force about a directional line l passing through the origin is Ml = lu\u0302 \u00b7 (rP \u00d7F) (9.3) where u\u0302 is a unit vector on l. The moment of the force F, about a point Q at rQ is MQ = (rP \u2212 rQ)\u00d7F (9.4) so, the moment of F about the origin is M = rP \u00d7F. (9.5) The moment of a force may also be called torque or moment. The effect of a force system is equivalent to the effect of the resultant force and resultant moment of the force system. Any two force systems are equivalent if their resultant forces and resultant moments are equal. If the resultant force of a force system is zero, the resultant moment of the force system is independent of the origin of the coordinate frame. Such a resultant moment is called couple. When a force system is reduced to a resultant FP andMP with respect to a reference point P , we may change the reference point to another point Q and find the new resultants as FQ = FP (9.6) MQ = MP + (rP \u2212 rQ)\u00d7FP = MP + QrP \u00d7FP . (9.7) The momentum of a moving rigid body is a vector quantity equal to the total mass of the body times the translational velocity of the mass center of the body. p = mv (9.8) The momentum p is also called translational momentum or linear momentum. Consider a rigid body with momentum p. The moment of momentum, L, about a directional line l passing through the origin is Ll = lu\u0302 \u00b7 (rC \u00d7 p) (9.9) 9. Applied Dynamics 523 where u\u0302 is a unit vector indicating the direction of the line, and rC is the position vector of the mass center C. The moment of momentum about the origin is L = rC \u00d7 p. (9.10) The moment of momentum L is also called angular momentum. A bounded vector is a vector fixed at a point in space. A sliding or line vector is a vector free to slide on its line of action. A free vector is a vector that may move to any point as long as it keeps its direction. Force is a sliding vector and couple is a free vector. However, the moment of a force is dependent on the distance between the origin of the coordinate frame and the line of action. The application of a force system is emphasized by Newton\u2019s second and third laws of motion. The second law of motion, also called the Newton\u2019s equation of motion, states that the global rate of change of linear momentum is proportional to the global applied force. GF = Gd dt Gp = Gd dt \u00a1 mGv \u00a2 (9.11) The third law of motion states that the action and reaction forces acting between two bodies are equal and opposite. The second law of motion can be expanded to include rotational motions. Hence, the second law of motion also states that the global rate of change of angular momentum is proportional to the global applied moment. GM = Gd dt GL (9.12) Proof. Differentiating from angular momentum (9.10) shows that Gd dt GL = Gd dt (rC \u00d7 p) = \u00b5 GdrC dt \u00d7 p+ rC \u00d7 Gdp dt \u00b6 = GrC \u00d7 Gdp dt = GrC \u00d7 GF = GM. (9.13) Kinetic energy K of a moving body point P with mass m at a position GrP , and having a velocity GvP , is K = 1 2 mGv2P = 1 2 m \u00b3 Gd\u0307B + BvP + B G\u03c9B \u00d7 BrP \u00b42 . (9.14) 524 9. Applied Dynamics The work done by the applied force GF on m in moving from point 1 to point 2 on a path, indicated by a vector Gr, is 1W2 = Z 2 1 GF \u00b7 dGr. (9.15) However, Z 2 1 GF \u00b7 dGr = m Z 2 1 Gd dt Gv \u00b7 Gvdt = 1 2 m Z 2 1 d dt v2dt = 1 2 m \u00a1 v22 \u2212 v21 \u00a2 = K2 \u2212K1 (9.16) that shows 1W2 is equal to the difference of the kinetic energy between terminal and initial points. 1W2 = K2 \u2212K1 (9.17) Equation (9.17) is called principle of work and energy. Example 342 Position of center of mass. The position of the mass center of a rigid body in a coordinate frame is indicated by BrC and is usually measured in the body coordinate frame. BrC = 1 m Z B Br dm (9.18) \u23a1\u23a3 xC yC zC \u23a4\u23a6 = \u23a1\u23a2\u23a2\u23a3 1 m R B x dm 1 m R B y dm 1 m R B z dm \u23a4\u23a5\u23a5\u23a6 (9.19) Applying the mass center integral on the symmetric and uniform L-section rigid body with \u03c1 = 1 shown in Figure 9.2 provides the position of mass center C of the section. The x position of C is xC = 1 m Z B xdm = 1 A Z B x dA = \u2212b 2 + ab\u2212 a2 4ab+ 2a2 (9.20) and because of symmetry, we have yC = \u2212xC = b2 + ab\u2212 a2 4ab+ 2a2 . (9.21) 9. Applied Dynamics 525 When a = b, the position of C reduces to yC = \u2212xC = 1 2 b. (9.22) Example 343 F Every force system is equivalent to a wrench. The Poinsot theorem states: Every force system is equivalent to a single force, plus a moment parallel to the force. Let F andM be the resultant force and moment of a force system. We decompose the moment into parallel and perpendicular components,Mk andM\u22a5, to the force axis. The force F and the perpendicular moment M\u22a5 can be replaced by a single force F0 parallel to F. Therefore, the force system is reduced to a force F0 and a moment Mk parallel to each other. A force and a moment about the force axis is called a wrench. The Poinsot theorem is similar to the Chasles theorem that states: Every rigid body motion is equivalent to a screw, which is a translation plus a rotation about the axis of translation. Example 344 F Motion of a moving point in a moving body frame. The velocity and acceleration of a moving point P as shown in Figure 5.12 are found in Example 200. GvP = Gd\u0307B + GRB \u00a1 BvP + B G\u03c9B \u00d7 BrP \u00a2 (9.23) GaP = Gd\u0308B + GRB \u00a1 BaP + 2 B G\u03c9B \u00d7 BvP + B G\u03c9\u0307B \u00d7 BrP \u00a2 +GRB \u00a1 B G\u03c9B \u00d7 \u00a1 B G\u03c9B \u00d7 BrP \u00a2\u00a2 (9.24) Therefore, the equation of motion for the point mass P is GF = mGaP = m \u00b3 Gd\u0308B + GRB \u00a1 BaP + 2 B G\u03c9B \u00d7 BvP + B G\u03c9\u0307B \u00d7 BrP \u00a2\u00b4 +m GRB \u00a1 B G\u03c9B \u00d7 \u00a1 B G\u03c9B \u00d7 BrP \u00a2\u00a2 . (9.25) 526 9. Applied Dynamics Example 345 Newton\u2019s equation in a rotating frame. Consider a spherical rigid body, such as Earth, with a fixed point that is rotating with a constant angular velocity. The equation of motion for a moving point vehicle P on the rigid body is found by setting Gd\u0308B = B G\u03c9\u0307B = 0 in the equation of motion of a moving point in a moving body frame (9.25) BF = mBaP +mB G\u03c9B \u00d7 \u00a1 B G\u03c9B \u00d7 BrP \u00a2 + 2mB G\u03c9B \u00d7 B r\u0307P (9.26) 6= mBaP which shows that the Newton\u2019s equation of motion F = ma must be modified for rotating frames. Example 346 Coriolis force. The equation of motion of a moving vehicle point on the surface of the Earth is BF = mBaP +mB G\u03c9B \u00d7 \u00a1 B G\u03c9B \u00d7 BrP \u00a2 + 2mB G\u03c9B \u00d7 BvP (9.27) which can be rearranged to BF\u2212mB G\u03c9B \u00d7 \u00a1 B G\u03c9B \u00d7 BrP \u00a2 \u2212 2mB G\u03c9B \u00d7 BvP = mBaP . (9.28) Equation (9.28) is the equation of motion for an observer in the rotating frame, which in this case is an observer on the Earth. The left-hand side of this equation is called the effective force Feff , Feff = BF\u2212mB G\u03c9B \u00d7 \u00a1 B G\u03c9B \u00d7 BrP \u00a2 \u2212 2mB G\u03c9B \u00d7 BvP (9.29) because it seems that the particle is moving under the influence of this force. The second term is negative of the centrifugal force and pointing outward. The maximum value of this force on the Earth is on the equator r\u03c92 = 6378.388\u00d7 103 \u00d7 \u00b5 2\u03c0 24\u00d7 3600 366.25 365.25 \u00b62 = 3.3917\u00d7 10\u22122m/ s2 (9.30) which is about 0.3% of the acceleration of gravity. If we add the variation of the gravitational acceleration because of a change of radius from R = 6356912m at the pole to R = 6378388m on the equator, then the variation of the acceleration of gravity becomes 0.53%. So, generally speaking, a sportsman such as a pole-vaulter who has practiced in the north pole can show a better record in a competition held on the equator. The third term is called the Coriolis force or Coriolis effect, FC, which is perpendicular to both \u03c9 and BvP . For a mass m moving on the north hemisphere at a latitude \u03b8 towards the equator, we should provide a 9. Applied Dynamics 527 lateral eastward force equal to the Coriolis effect to force the mass, keeping its direction relative to the ground. FC = 2mB G\u03c9B \u00d7 Bvm = 1.4584\u00d7 10\u22124 Bpm cos \u03b8 kgm/ s2 (9.31) The Coriolis effect is the reason why the west side of railways, roads, and rivers wears. The lack of providing the Coriolis force is the reason for turning the direction of winds, projectiles, flood, and falling objects westward. Example 347 Work, force, and kinetic energy in a unidirectional motion. A vehicle with mass m = 1200 kg has an initial kinetic energy K = 6000 J. The mass is under a constant force F = F I\u0302 = 4000I\u0302 and moves from X(0) = 0 to X(tf ) = 1000m at a terminal time tf . The work done by the force during this motion is W = Z r(tf ) r(0) F \u00b7 dr = Z 1000 0 4000 dX = 4\u00d7 106Nm = 4MJ (9.32) The kinetic energy at the terminal time is K(tf ) =W +K(0) = 4006000 J (9.33) which shows that the terminal speed of the mass is v2 = r 2K(tf ) m \u2248 81.7m/ s. (9.34) Example 348 Direct dynamics. When the applied force is time varying and is a known function, then, F(t) = m r\u0308. (9.35) The general solution for the equation of motion can be found by integration. r\u0307(t) = r\u0307(t0) + 1 m Z t t0 F(t)dt (9.36) r(t) = r(t0) + r\u0307(t0)(t\u2212 t0) + 1 m Z t t0 Z t t0 F(t)dt dt (9.37) This kind of problem is called direct or forward dynamics. 528 9. Applied Dynamics 9.2 Rigid Body Translational Dynamics Figure 9.3 depicts a moving body B in a global coordinate frame G. Assume that the body frame is attached at the mass center of the body. Point P indicates an infinitesimal sphere of the body, which has a very small mass dm. The point mass dm is acted on by an infinitesimal force df and has a global velocity GvP . According to Newton\u2019s law of motion we have df = GaP dm. (9.38) However, the equation of motion for the whole body in a global coordinate frame is GF = mGaB (9.39) which can be expressed in the body coordinate frame as BF = mB GaB +m B G\u03c9B \u00d7 BvB (9.40)\u23a1\u23a3 Fx Fy Fz \u23a4\u23a6 = \u23a1\u23a3 max +m (\u03c9yvz \u2212 \u03c9zvy) may \u2212m (\u03c9xvz \u2212 \u03c9zvx) maz +m (\u03c9xvy \u2212 \u03c9yvx) \u23a4\u23a6 . (9.41) In these equations, GaB is the acceleration vector of the body mass center C in the global frame,m is the total mass of the body, and F is the resultant of the external forces acted on the body at C. 9. Applied Dynamics 529 Proof. A body coordinate frame at the mass center is called a central frame. If frame B is a central frame, then the center of mass, C, is defined such that Z B Brdm dm = 0. (9.42) The global position vector of dm is related to its local position vector by Grdm = GdB + GRB Brdm (9.43) where GdB is the global position vector of the central body frame, and therefore, Z B Grdm dm = Z B GdB dm+ GRB Z m Brdm dm = Z B GdB dm = GdB Z B dm = mGdB. (9.44) A time derivative of both sides shows that mGd\u0307B = mGvB = Z B Gr\u0307dm dm = Z B Gvdm dm (9.45) and another derivative is mGv\u0307B = mGaB = Z B Gv\u0307dm dm. (9.46) However, we have df = Gv\u0307P dm and therefore, mGaB = Z B df . (9.47) The integral on the right-hand side accounts for all the forces acting on the body. The internal forces cancel one another out, so the net result is the vector sum of all the externally applied forces, F, and therefore, GF = m GaB = m Gv\u0307B . (9.48) In the body coordinate frame we have BF = BRG GF = m BRG GaB = m B GaB = m BaB +m B G\u03c9B \u00d7 BvB. (9.49) 530 9. Applied Dynamics The expanded form of the Newton\u2019s equation in the body coordinate frame is then equal to BF = m BaB +m B G\u03c9B \u00d7 BvB\u23a1\u23a3 Fx Fy Fz \u23a4\u23a6 = m \u23a1\u23a3 ax ay az \u23a4\u23a6+m \u23a1\u23a3 \u03c9x \u03c9y \u03c9z \u23a4\u23a6\u00d7 \u23a1\u23a3 vx vy vz \u23a4\u23a6 = \u23a1\u23a3 max +m (\u03c9yvz \u2212 \u03c9zvy) may \u2212m (\u03c9xvz \u2212 \u03c9zvx) maz +m (\u03c9xvy \u2212 \u03c9yvx) \u23a4\u23a6 . (9.50) 9.3 Rigid Body Rotational Dynamics The rigid body rotational equation of motion is the Euler equation BM = Gd dt BL = BL\u0307+B G\u03c9B \u00d7 BL = BI B G\u03c9\u0307B + B G\u03c9B \u00d7 \u00a1 BI B G\u03c9B \u00a2 (9.51) where L is the angular momentum BL = BI B G\u03c9B (9.52) and I is the moment of inertia of the rigid body. I = \u23a1\u23a3 Ixx Ixy Ixz Iyx Iyy Iyz Izx Izy Izz \u23a4\u23a6 (9.53) The elements of I are functions of the mass distribution of the rigid body and may be defined by Iij = Z B \u00a1 r2i \u03b4mn \u2212 ximxjn \u00a2 dm , i, j = 1, 2, 3 (9.54) where \u03b4ij is Kronecker\u2019s delta. \u03b4mn = \u00bd 1 if m = n 0 if m 6= n (9.55) The expanded form of the Euler equation (9.51) is Mx = Ixx\u03c9\u0307x + Ixy\u03c9\u0307y + Ixz\u03c9\u0307z \u2212 (Iyy \u2212 Izz)\u03c9y\u03c9z \u2212Iyz \u00a1 \u03c92z \u2212 \u03c92y \u00a2 \u2212 \u03c9x (\u03c9zIxy \u2212 \u03c9yIxz) (9.56) 9. Applied Dynamics 531 My = Iyx\u03c9\u0307x + Iyy\u03c9\u0307y + Iyz\u03c9\u0307z \u2212 (Izz \u2212 Ixx)\u03c9z\u03c9x \u2212Ixz \u00a1 \u03c92x \u2212 \u03c92z \u00a2 \u2212 \u03c9y (\u03c9xIyz \u2212 \u03c9zIxy) (9.57) Mz = Izx\u03c9\u0307x + Izy\u03c9\u0307y + Izz\u03c9\u0307z \u2212 (Ixx \u2212 Iyy)\u03c9x\u03c9y \u2212Ixy \u00a1 \u03c92y \u2212 \u03c92x \u00a2 \u2212 \u03c9z (\u03c9yIxz \u2212 \u03c9xIyz) . (9.58) which can be reduced to M1 = I1\u03c9\u03071 \u2212 (I2 \u2212 I2)\u03c92\u03c93 M2 = I2\u03c9\u03072 \u2212 (I3 \u2212 I1)\u03c93\u03c91 (9.59) M3 = I3\u03c9\u03073 \u2212 (I1 \u2212 I2)\u03c91\u03c92 in a special Cartesian coordinate frame called the principal coordinate frame. The principal coordinate frame is denoted by numbers 123 to indicate the first, second, and third principal axes. The parameters Iij , i 6= j are zero in the principal frame. The body and principal coordinate frame sit at the mass center C. Kinetic energy of a rotating rigid body is K = 1 2 \u00a1 Ixx\u03c9 2 x + Iyy\u03c9 2 y + Izz\u03c9 2 z \u00a2 \u2212Ixy\u03c9x\u03c9y \u2212 Iyz\u03c9y\u03c9z \u2212 Izx\u03c9z\u03c9x (9.60) = 1 2 \u03c9 \u00b7 L (9.61) = 1 2 \u03c9T I \u03c9 (9.62) that in the principal coordinate frame reduces to K = 1 2 \u00a1 I1\u03c9 2 1 + I2\u03c9 2 2 + I3\u03c9 2 3 \u00a2 . (9.63) Proof. Let mi be the mass of the ith particle of a rigid body B, which is made of n particles and let ri = Bri = \u00a3 xi yi zi \u00a4T (9.64) be the Cartesian position vector of mi in a central body fixed coordinate frame Oxyz. Assume that \u03c9 = B G\u03c9B = \u00a3 \u03c9x \u03c9y \u03c9z \u00a4T (9.65) is the angular velocity of the rigid body with respect to the ground, expressed in the body coordinate frame. 532 9. Applied Dynamics The angular momentum of mi is Li = ri \u00d7mir\u0307i = mi [ri \u00d7 (\u03c9 \u00d7 ri)] = mi [(ri \u00b7 ri)\u03c9 \u2212 (ri \u00b7 \u03c9) ri] = mir 2 i\u03c9 \u2212mi (ri \u00b7 \u03c9) ri. (9.66) Hence, the angular momentum of the rigid body would be L = \u03c9 nX i=1 mir 2 i \u2212 nX i=1 mi (ri \u00b7 \u03c9) ri. (9.67) Substitution for ri and \u03c9 gives us L = \u00b3 \u03c9x \u0131\u0302+ \u03c9y j\u0302+ \u03c9z k\u0302 \u00b4 nX i=1 mi \u00a1 x2i + y2i + z2i \u00a2 \u2212 nX i=1 mi (xi\u03c9x + yi\u03c9y + zi\u03c9z) \u00b7 \u00b3 xi\u0131\u0302+ yij\u0302+ zik\u0302 \u00b4 (9.68) and therefore, L = nX i=1 mi \u00a1 x2i + y2i + z2i \u00a2 \u03c9x \u0131\u0302 + nX i=1 mi \u00a1 x2i + y2i + z2i \u00a2 \u03c9y j\u0302 + nX i=1 mi \u00a1 x2i + y2i + z2i \u00a2 \u03c9z k\u0302 \u2212 nX i=1 mi (xi\u03c9x + yi\u03c9y + zi\u03c9z)xi\u0131\u0302 \u2212 nX i=1 mi (xi\u03c9x + yi\u03c9y + zi\u03c9z) yij\u0302 \u2212 nX i=1 mi (xi\u03c9x + yi\u03c9y + zi\u03c9z) zik\u0302 (9.69) or L = nX i=1 mi \u00a3\u00a1 x2i + y2i + z2i \u00a2 \u03c9x \u2212 (xi\u03c9x + yi\u03c9y + zi\u03c9z)xi \u00a4 \u0131\u0302 + nX i=1 mi \u00a3\u00a1 x2i + y2i + z2i \u00a2 \u03c9y \u2212 (xi\u03c9x + yi\u03c9y + zi\u03c9z) yi \u00a4 j\u0302 + nX i=1 mi \u00a3\u00a1 x2i + y2i + z2i \u00a2 \u03c9z \u2212 (xi\u03c9x + yi\u03c9y + zi\u03c9z) zi \u00a4 k\u0302 (9.70) 9. Applied Dynamics 533 which can be rearranged as L = nX i=1 \u00a3 mi \u00a1 y2i + z2i \u00a2\u00a4 \u03c9x\u0131\u0302 + nX i=1 \u00a3 mi \u00a1 z2i + x2i \u00a2\u00a4 \u03c9y j\u0302 + nX i=1 \u00a3 mi \u00a1 x2i + y2i \u00a2\u00a4 \u03c9zk\u0302 \u2212 \u00c3 nX i=1 (mixiyi)\u03c9y + nX i=1 (mixizi)\u03c9z ! \u0131\u0302 \u2212 \u00c3 nX i=1 (miyizi)\u03c9z + nX i=1 (miyixi)\u03c9x ! j\u0302 \u2212 \u00c3 nX i=1 (mizixi)\u03c9x + nX i=1 (miziyi)\u03c9y ! k\u0302. (9.71) By introducing the moment of inertia matrix I with the following elements, Ixx = nX i=1 \u00a3 mi \u00a1 y2i + z2i \u00a2\u00a4 (9.72) Iyy = nX i=1 \u00a3 mi \u00a1 z2i + x2i \u00a2\u00a4 (9.73) Izz = nX i=1 \u00a3 mi \u00a1 x2i + y2i \u00a2\u00a4 (9.74) Ixy = Iyx = \u2212 nX i=1 (mixiyi) (9.75) Iyz = Izy = \u2212 nX i=1 (miyizi) (9.76) Izx = Ixz = \u2212 nX i=1 (mizixi) . (9.77) we may write the angular momentum L in a concise form Lx = Ixx\u03c9x + Ixy\u03c9y + Ixz\u03c9z (9.78) Ly = Iyx\u03c9x + Iyy\u03c9y + Iyz\u03c9z (9.79) Lz = Izx\u03c9x + Izy\u03c9y + Izz\u03c9z (9.80) 534 9. Applied Dynamics or in a matrix form L = I \u00b7 \u03c9 (9.81)\u23a1\u23a3 Lx Ly Lz \u23a4\u23a6 = \u23a1\u23a3 Ixx Ixy Ixz Iyx Iyy Iyz Izx Izy Izz \u23a4\u23a6\u23a1\u23a3 \u03c9x \u03c9y \u03c9z \u23a4\u23a6 . (9.82) For a rigid body that is a continuous solid, the summations must be replaced by integrations over the volume of the body as in Equation (9.54). The Euler equation of motion for a rigid body is BM = Gd dt BL (9.83) where BM is the resultant of the external moments applied on the rigid body. The angular momentum BL is a vector defined in the body coordinate frame. Hence, its time derivative in the global coordinate frame is GdBL dt = BL\u0307+B G\u03c9B \u00d7 BL. (9.84) Therefore, BM = dL dt = L\u0307+ \u03c9 \u00d7 L = I\u03c9\u0307 + \u03c9\u00d7 (I\u03c9) (9.85) or in expanded form BM = (Ixx\u03c9\u0307x + Ixy\u03c9\u0307y + Ixz\u03c9\u0307z) \u0131\u0302 +(Iyx\u03c9\u0307x + Iyy\u03c9\u0307y + Iyz\u03c9\u0307z) j\u0302 +(Izx\u03c9\u0307x + Izy\u03c9\u0307y + Izz\u03c9\u0307z) k\u0302 +\u03c9y (Ixz\u03c9x + Iyz\u03c9y + Izz\u03c9z) \u0131\u0302 \u2212\u03c9z (Ixy\u03c9x + Iyy\u03c9y + Iyz\u03c9z) \u0131\u0302 +\u03c9z (Ixx\u03c9x + Ixy\u03c9y + Ixz\u03c9z) j\u0302 \u2212\u03c9x (Ixz\u03c9x + Iyz\u03c9y + Izz\u03c9z) j\u0302 +\u03c9x (Ixy\u03c9x + Iyy\u03c9y + Iyz\u03c9z) k\u0302 \u2212\u03c9y (Ixx\u03c9x + Ixy\u03c9y + Ixz\u03c9z) k\u0302 (9.86) and therefore, the most general form of the Euler equations of motion for a rigid body in a body frame attached to C are Mx = Ixx\u03c9\u0307x + Ixy\u03c9\u0307y + Ixz\u03c9\u0307z \u2212 (Iyy \u2212 Izz)\u03c9y\u03c9z \u2212Iyz \u00a1 \u03c92z \u2212 \u03c92y \u00a2 \u2212 \u03c9x (\u03c9zIxy \u2212 \u03c9yIxz) (9.87) My = Iyx\u03c9\u0307x + Iyy\u03c9\u0307y + Iyz\u03c9\u0307z \u2212 (Izz \u2212 Ixx)\u03c9z\u03c9x \u2212Ixz \u00a1 \u03c92x \u2212 \u03c92z \u00a2 \u2212 \u03c9y (\u03c9xIyz \u2212 \u03c9zIxy) (9.88) Mz = Izx\u03c9\u0307x + Izy\u03c9\u0307y + Izz\u03c9\u0307z \u2212 (Ixx \u2212 Iyy)\u03c9x\u03c9y \u2212Ixy \u00a1 \u03c92y \u2212 \u03c92x \u00a2 \u2212 \u03c9z (\u03c9yIxz \u2212 \u03c9xIyz) . (9.89) 9. Applied Dynamics 535 Assume that we are able to rotate the body frame about its origin to find an orientation that makes Iij = 0, for i 6= j. In such a coordinate frame, which is called a principal frame, the Euler equations reduce to M1 = I1\u03c9\u03071 \u2212 (I2 \u2212 I2)\u03c92\u03c93 (9.90) M2 = I2\u03c9\u03072 \u2212 (I3 \u2212 I1)\u03c93\u03c91 (9.91) M3 = I3\u03c9\u03073 \u2212 (I1 \u2212 I2)\u03c91\u03c92. (9.92) The kinetic energy of a rigid body may be found by the integral of the kinetic energy of a mass element dm, over the whole body. K = 1 2 Z B v\u03072dm = 1 2 Z B (\u03c9 \u00d7 r) \u00b7 (\u03c9 \u00d7 r) dm = \u03c92x 2 Z B \u00a1 y2 + z2 \u00a2 dm+ \u03c92y 2 Z B \u00a1 z2 + x2 \u00a2 dm+ \u03c92z 2 Z B \u00a1 x2 + y2 \u00a2 dm \u2212\u03c9x\u03c9y Z B xy dm\u2212 \u03c9y\u03c9z Z B yz dm\u2212 \u03c9z\u03c9x Z B zx dm = 1 2 \u00a1 Ixx\u03c9 2 x + Iyy\u03c9 2 y + Izz\u03c9 2 z \u00a2 \u2212Ixy\u03c9x\u03c9y \u2212 Iyz\u03c9y\u03c9z \u2212 Izx\u03c9z\u03c9x (9.93) The kinetic energy can be rearranged to a matrix multiplication form K = 1 2 \u03c9T I \u03c9 (9.94) = 1 2 \u03c9 \u00b7 L. (9.95) When the body frame is principal, the kinetic energy will simplify to K = 1 2 \u00a1 I1\u03c9 2 1 + I2\u03c9 2 2 + I3\u03c9 2 3 \u00a2 . (9.96) Example 349 A tilted disc on a massless shaft. Figure 9.4 illustrates a disc with mass m and radius r, mounted on a massless shaft. The shaft is turning with a constant angular speed \u03c9. The disc is attached to the shaft at an angle \u03b8. Because of \u03b8, the bearings at A and B must support a rotating force. We attach a principal body coordinate frame at the disc center as shown in the figure. The angular velocity vector in the body frame is B G\u03c9B = \u03c9 cos \u03b8 \u0131\u0302+ \u03c9 sin \u03b8 j\u0302 (9.97) 536 9. Applied Dynamics and the mass moment of inertia matrix is BI = \u23a1\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3 mr2 2 0 0 0 mr2 4 0 0 0 mr2 4 \u23a4\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6 . (9.98) Substituting (9.97) and (9.98) in (9.90)-(9.92), with 1 \u2261 x, 2 \u2261 y, 3 \u2261 z, provides that Mx = 0 (9.99) My = 0 (9.100) Mz = mr2 4 \u03c9 cos \u03b8 sin \u03b8. (9.101) Therefore, the bearing reaction forces FA and FB are FA = \u2212FB = \u2212Mz l = \u2212mr2 4l \u03c9 cos \u03b8 sin \u03b8. (9.102) Example 350 Steady rotation of a freely rotating rigid body. The Newton-Euler equations of motion for a rigid body are GF = mGv\u0307 (9.103) BM = I B G\u03c9\u0307B + B G\u03c9B \u00d7 BL. (9.104) 9. Applied Dynamics 537 Consider a situation in which the resultant applied force and moment on the body are zero. GF = BF = 0 (9.105) GM = BM = 0 (9.106) Based on the Newton\u2019s equation, the velocity of the mass center will be constant in the global coordinate frame. However, the Euler equation reduces to \u03c9\u03071 = I2 \u2212 I3 I1 \u03c92\u03c93 (9.107) \u03c9\u03072 = I3 \u2212 I1 I22 \u03c93\u03c91 (9.108) \u03c9\u03073 = I1 \u2212 I2 I3 \u03c91\u03c92 (9.109) that show the angular velocity can be constant if I1 = I2 = I3 (9.110) or if two principal moments of inertia, say I1 and I2, are zero and the third angular velocity, in this case \u03c93, is initially zero, or if the angular velocity vector is initially parallel to a principal axis. Example 351 Angular momentum of a two-link manipulator. A two-link manipulator is shown in Figure 9.5. Link A rotates with angular velocity \u03d5\u0307 about the z-axis of its local coordinate frame. Link B is attached to link A and has angular velocity \u03c8\u0307 with respect to A about the xA-axis. We assume that A and G were coincident at \u03d5 = 0, therefore, the rotation matrix between A and G is GRA = \u23a1\u23a3 cos\u03d5(t) \u2212 sin\u03d5(t) 0 sin\u03d5(t) cos\u03d5(t) 0 0 0 1 \u23a4\u23a6 . (9.111) Frame B is related to frame A by Euler angles \u03d5 = 90deg, \u03b8 = 90deg, and \u03c8, hence, ARB = \u23a1\u23a3 c\u03c0c\u03c8 \u2212 c\u03c0s\u03c0s\u03c8 \u2212c\u03c0s\u03c8 \u2212 c\u03c0c\u03c8s\u03c0 s\u03c0s\u03c0 c\u03c8s\u03c0 + c\u03c0c\u03c0s\u03c8 \u2212s\u03c0s\u03c8 + c\u03c0c\u03c0c\u03c8 \u2212c\u03c0s\u03c0 s\u03c0s\u03c8 s\u03c0c\u03c8 c\u03c0 \u23a4\u23a6 \u23a1\u23a3 \u2212 cos\u03c8 sin\u03c8 0 sin\u03c8 cos\u03c8 0 0 0 \u22121 \u23a4\u23a6 (9.112) 538 9. Applied Dynamics and therefore, GRB = GRA ARB (9.113) = \u23a1\u23a3 \u2212 cos\u03d5 cos\u03c8 \u2212 sin\u03d5 sin\u03c8 cos\u03d5 sin\u03c8 \u2212 cos\u03c8 sin\u03d5 0 cos\u03d5 sin\u03c8 \u2212 cos\u03c8 sin\u03d5 cos\u03d5 cos\u03c8 + sin\u03d5 sin\u03c8 0 0 0 \u22121 \u23a4\u23a6 . The angular velocity of A in G, and B in A are G\u03c9A = \u03d5\u0307K\u0302 (9.114) A\u03c9B = \u03c8\u0307\u0131\u0302A. (9.115) Moment of inertia matrices for the arms A and B can be defined as AIA = \u23a1\u23a3 IA1 0 0 0 IA2 0 0 0 IA3 \u23a4\u23a6 (9.116) BIB = \u23a1\u23a3 IB1 0 0 0 IB2 0 0 0 IB3 \u23a4\u23a6 . (9.117) These moments of inertia must be transformed to the global frame GIA = GRB AIA GRT A (9.118) GIB = GRB BIB GRT B . (9.119) 9. Applied Dynamics 539 The total angular momentum of the manipulator is GL = GLA + GLB (9.120) where GLA = GIA G\u03c9A (9.121) GLB = GIB G\u03c9B = GIB \u00a1 G A\u03c9B + G\u03c9A \u00a2 . (9.122) Example 352 Poinsot\u2019s construction. Consider a freely rotating rigid body with an attached principal coordinate frame. HavingM = 0 provides a motion under constant angular momentum and constant kinetic energy L = I \u03c9 = cte (9.123) K = 1 2 \u03c9T I \u03c9 = cte. (9.124) Because the length of the angular momentum L is constant, the equation L2 = L \u00b7 L = L2x + L2y + L2z = I21\u03c9 2 1 + I22\u03c9 2 2 + I23\u03c9 2 3 (9.125) introduces an ellipsoid in the (\u03c91, \u03c92, \u03c93) coordinate frame, called the momentum ellipsoid. The tip of all possible angular velocity vectors must lie on the surface of the momentum ellipsoid. The kinetic energy also defines an energy ellipsoid in the same coordinate frame so that the tip of the angular velocity vectors must also lie on its surface. K = 1 2 \u00a1 I1\u03c9 2 1 + I2\u03c9 2 2 + I3\u03c9 2 3 \u00a2 (9.126) In other words, the dynamics of moment-free motion of a rigid body requires that the corresponding angular velocity \u03c9(t) satisfy both Equations (9.125) and (9.126) and therefore lie on the intersection of the momentum and energy ellipsoids. For clarity, we may define the ellipsoids in the (Lx, Ly, Lz) coordinate system as L2x + L2y + L2z = L2 (9.127) L2x 2I1K + L2y 2I2K + L2z 2I3K = 1. (9.128) Equation (9.127) is a sphere and Equation (9.128) defines an ellipsoid with\u221a 2IiK as semi-axes. To have a meaningful motion, these two shapes must intersect. The intersection may form a trajectory, as shown in Figure 9.6. 540 9. Applied Dynamics It can be deduced that for a certain value of angular momentum there are maximum and minimum limit values for acceptable kinetic energy. Assuming I1 > I3 > I3 (9.129) the limits of possible kinetic energy are Kmin = L2 2I1 (9.130) Kmax = L2 2I3 (9.131) and the corresponding motions are turning about the axes I1 and I3 respectively. Example 353 F Alternative derivation of Euler equations of motion. Assume that the moment of the small force df is shown by dm and a mass element is shown by dm, then, dm = Grdm \u00d7 df = Grdm \u00d7 Gv\u0307dm dm. (9.132) The global angular momentum dl of dm is equal to dl = Grdm \u00d7 Gvdm dm (9.133) and according to (9.12) we have dm = Gd dt dl (9.134) Grdm \u00d7 df = Gd dt \u00a1 Grdm \u00d7 Gvdm dm \u00a2 . (9.135) 9. Applied Dynamics 541 Integrating over the body results inZ B Grdm \u00d7 df = Z B Gd dt \u00a1 Grdm \u00d7 Gvdm dm \u00a2 = Gd dt Z B \u00a1 Grdm \u00d7 Gvdm dm \u00a2 . (9.136) However, utilizing Grdm = GdB + GRB Brdm (9.137) where GdB is the global position vector of the central body frame, can simplify the left-hand side of the integral toZ B Grdm \u00d7 df = Z B \u00a1 GdB + GRB Brdm \u00a2 \u00d7 df = Z B GdB \u00d7 df + Z B G Brdm \u00d7 df = GdB \u00d7 GF+ GMC (9.138) where MC is the resultant external moment about the body mass center C. The right-hand side of Equation (9.136) is Gd dt Z B \u00a1 Grdm \u00d7 Gvdm dm \u00a2 = Gd dt Z B \u00a1\u00a1 GdB + GRB Brdm \u00a2 \u00d7 Gvdm dm \u00a2 = Gd dt Z B \u00a1 GdB \u00d7 Gvdm \u00a2 dm+ Gd dt Z B \u00a1 G Brdm \u00d7 Gvdm \u00a2 dm = Gd dt \u00b5 GdB \u00d7 Z B Gvdmdm \u00b6 + Gd dt LC = Gd\u0307B \u00d7 Z B Gvdmdm+ GdB \u00d7 Z B Gv\u0307dmdm+ d dt LC . (9.139) We use LC for angular momentum about the body mass center. Because the body frame is at the mass center, we haveZ B Grdm dm = mGdB = mGrC (9.140)Z B Gvdmdm = mGd\u0307B = mGvC (9.141)Z B Gv\u0307dmdm = mGd\u0308B = mGaC (9.142) and therefore, Gd dt Z B \u00a1 Grdm \u00d7 Gvdm dm \u00a2 = GdB \u00d7 GF+ Gd dt GLC . (9.143) 542 9. Applied Dynamics Substituting (9.138) and (9.143) in (9.136) provides the Euler equation of motion in the global frame, indicating that the resultant of externally applied moments about C is equal to the global derivative of angular momentum about C. GMC = Gd dt GLC . (9.144) The Euler equation in the body coordinate can be found by transforming (9.144) BMC = GRT B GMC = GRT B Gd dt LC = Gd dt GRT B LC = Gd dt BLC = BL\u0307C + B G\u03c9B \u00d7 BLC . (9.145) 9.4 Mass Moment of Inertia Matrix In analyzing the motion of rigid bodies, two types of integrals arise that belong to the geometry of the body. The first type defines the center of mass and is important when the translation motion of the body is considered. The second is the moment of inertia that appears when the rotational motion of the body is considered. The moment of inertia is also called centrifugal moments, or deviation moments. Every rigid body has a 3 \u00d7 3 moment of inertia matrix I, which is denoted by I = \u23a1\u23a3 Ixx Ixy Ixz Iyx Iyy Iyz Izx Izy Izz \u23a4\u23a6 . (9.146) The diagonal elements Iij , i = j are called polar moments of inertia Ixx = Ix = Z B \u00a1 y2 + z2 \u00a2 dm (9.147) Iyy = Iy = Z B \u00a1 z2 + x2 \u00a2 dm (9.148) Izz = Iz = Z B \u00a1 x2 + y2 \u00a2 dm (9.149) and the off-diagonal elements Iij , i 6= j are called products of inertia Ixy = Iyx = \u2212 Z B xy dm (9.150) 9. Applied Dynamics 543 Iyz = Izy = \u2212 Z B yz dm (9.151) Izx = Ixz = \u2212 Z B zx dm. (9.152) The elements of I for a rigid body, made of discrete point masses, are defined in Equation (9.54). The elements of I are calculated about a body coordinate frame attached to the mass center C of the body. Therefore, I is a frame-dependent quantity and must be written like BI to show the frame it is computed in. BI = Z B \u23a1\u23a3 y2 + z2 \u2212xy \u2212zx \u2212xy z2 + x2 \u2212yz \u2212zx \u2212yz x2 + y2 \u23a4\u23a6 dm (9.153) = Z B \u00a1 r2I\u2212 r rT \u00a2 dm (9.154) = Z B \u2212r\u0303 r\u0303 dm. (9.155) Moments of inertia can be transformed from a coordinate frame B1 to another coordinate frame B2, both installed at the mass center of the body, according to the rule of the rotated-axes theorem B2I = B2RB1 B1I B2RT B1 . (9.156) Transformation of the moment of inertia from a central frame B1 located at B2rC to another frame B2, which is parallel to B1, is, according to the rule of parallel-axes theorem, B2I = B1I +mr\u0303C r\u0303TC . (9.157) If the local coordinate frame Oxyz is located such that the products of inertia vanish, the local coordinate frame is called the principal coordinate frame and the associated moments of inertia are called principal moments of inertia. Principal axes and principal moments of inertia can be found by solving the following equation for I:\u00af\u0304\u0304\u0304 \u00af\u0304 Ixx \u2212 I Ixy Ixz Iyx Iyy \u2212 I Iyz Izx Izy Izz \u2212 I \u00af\u0304\u0304\u0304 \u00af\u0304 = 0 (9.158) det ([Iij ]\u2212 I [\u03b4ij ]) = 0. (9.159) Since Equation (9.159) is a cubic equation in I, we obtain three eigenvalues I1 = Ix I2 = Iy I3 = Iz (9.160) 544 9. Applied Dynamics that are the principal moments of inertia. Proof. Two coordinate frames with a common origin at the mass center of a rigid body are shown in Figure 9.7. The angular velocity and angular momentum of a rigid body transform from the frame B1 to the frame B2 by vector transformation rule B2\u03c9 = B2RB1 B1\u03c9 (9.161) B2L = B2RB1 B1L. (9.162) However, L and \u03c9 are related according to Equation (9.52) B1L = B1I B1\u03c9 (9.163) and therefore, B2L = B2RB1 B1I B2RT B1 B2\u03c9 = B2I B2\u03c9 (9.164) which shows how to transfer the moment of inertia from the coordinate frame B1 to a rotated frame B2 B2I = B2RB1 B1I B2RT B1 . (9.165) Now consider a central frame B1, shown in Figure 9.8, at B2rC , which rotates about the origin of a fixed frame B2 such that their axes remain parallel. The angular velocity and angular momentum of the rigid body transform from frame B1 to frame B2 by B2\u03c9 = B1\u03c9 (9.166) B2L = B1L+ (rC \u00d7mvC) . (9.167) 9. Applied Dynamics 545 Therefore, B2L = B1L+mB2rC \u00d7 \u00a1 B2\u03c9\u00d7B2rC \u00a2 = B1L+ \u00a1 m B2 r\u0303C B2 r\u0303TC \u00a2 B2\u03c9 = \u00a1 B1I +m B2 r\u0303C B2 r\u0303TC \u00a2 B2\u03c9 (9.168) which shows how to transfer the moment of inertia from frame B1 to a parallel frame B2 B2I = B1I +mr\u0303C r\u0303TC . (9.169) The parallel-axes theorem is also called the Huygens-Steiner theorem. Referring to Equation (9.165) for transformation of the moment of inertia to a rotated frame, we can always find a frame in which B2I is diagonal. In such a frame, we have B2RB1 B1I = B2I B2RB1 (9.170) or \u23a1\u23a3 r11 r12 r13 r21 r22 r23 r31 r32 r33 \u23a4\u23a6\u23a1\u23a3 Ixx Ixy Ixz Iyx Iyy Iyz Izx Izy Izz \u23a4\u23a6 = \u23a1\u23a3 I1 0 0 0 I2 0 0 0 I3 \u23a4\u23a6\u23a1\u23a3 r11 r12 r13 r21 r22 r23 r31 r32 r33 \u23a4\u23a6 (9.171) which shows that I1, I2, and I3 are eigenvalues of B1I. These eigenvalues can be found by solving the following equation for \u03bb:\u00af\u0304\u0304\u0304 \u00af\u0304 Ixx \u2212 \u03bb Ixy Ixz Iyx Iyy \u2212 \u03bb Iyz Izx Izy Izz \u2212 \u03bb \u00af\u0304\u0304\u0304 \u00af\u0304 = 0. (9.172) 546 9. Applied Dynamics The eigenvalues I1, I2, and I3 are principal moments of inertia, and their associated eigenvectors are called principal directions. The coordinate frame made by the eigenvectors is the principal body coordinate frame. In the principal coordinate frame, the rigid body angular momentum is\u23a1\u23a3 L1 L2 L3 \u23a4\u23a6 = \u23a1\u23a3 I1 0 0 0 I2 0 0 0 I3 \u23a4\u23a6\u23a1\u23a3 \u03c91 \u03c92 \u03c93 \u23a4\u23a6 . (9.173) Example 354 Principal moments of inertia. Consider the inertia matrix I I = \u23a1\u23a3 20 \u22122 0 \u22122 30 0 0 0 40 \u23a4\u23a6 . (9.174) We set up the determinant (9.159)\u00af\u0304\u0304\u0304 \u00af\u0304 20\u2212 \u03bb \u22122 0 \u22122 30\u2212 \u03bb 0 0 0 40\u2212 \u03bb \u00af\u0304\u0304\u0304 \u00af\u0304 = 0 (9.175) which leads to the following characteristic equation. (20\u2212 \u03bb) (30\u2212 \u03bb) (40\u2212 \u03bb)\u2212 4 (40\u2212 \u03bb) = 0 (9.176) Three roots of Equation (9.176) are I1 = 30.385, I2 = 19.615, I3 = 40 (9.177) and therefore, the principal moment of inertia matrix is I = \u23a1\u23a3 30.385 0 0 0 19.615 0 0 0 40 \u23a4\u23a6 . (9.178) Example 355 Principal coordinate frame. Consider the inertia matrix I I = \u23a1\u23a3 20 \u22122 0 \u22122 30 0 0 0 40 \u23a4\u23a6 (9.179) the direction of a principal axis xi is established by solving\u23a1\u23a3 Ixx \u2212 Ii Ixy Ixz Iyx Iyy \u2212 Ii Iyz Izx Izy Izz \u2212 Ii \u23a4\u23a6\u23a1\u23a3 cos\u03b1i cos\u03b2i cos \u03b3i \u23a4\u23a6 = \u23a1\u23a3 0 0 0 \u23a4\u23a6 (9.180) 9. Applied Dynamics 547 for direction cosines, which must also satisfy cos2 \u03b1i + cos 2 \u03b2i + cos 2 \u03b3i = 1. (9.181) For the first principal moment of inertia I1 = 30.385 we have\u23a1\u23a3 20\u2212 30.385 \u22122 0 \u22122 30\u2212 30.385 0 0 0 40\u2212 30.385 \u23a4\u23a6\u23a1\u23a3 cos\u03b11 cos\u03b21 cos \u03b31 \u23a4\u23a6 = \u23a1\u23a3 0 0 0 \u23a4\u23a6 (9.182) or \u221210.385 cos\u03b11 \u2212 2 cos\u03b21 + 0 = 0 (9.183) \u22122 cos\u03b11 \u2212 0.385 cos\u03b21 + 0 = 0 (9.184) 0 + 0 + 9.615 cos \u03b31 = 0 (9.185) and we obtain \u03b11 = 79.1 deg (9.186) \u03b21 = 169.1 deg (9.187) \u03b31 = 90.0 deg . (9.188) Using I2 = 19.615 for the second principal axis\u23a1\u23a3 20\u2212 19.62 \u22122 0 \u22122 30\u2212 19.62 0 0 0 40\u2212 19.62 \u23a4\u23a6\u23a1\u23a3 cos\u03b12 cos\u03b22 cos \u03b32 \u23a4\u23a6 = \u23a1\u23a3 0 0 0 \u23a4\u23a6 (9.189) we obtain \u03b12 = 10.9 deg (9.190) \u03b22 = 79.1 deg (9.191) \u03b32 = 90.0 deg . (9.192) The third principal axis is for I3 = 40\u23a1\u23a3 20\u2212 40 \u22122 0 \u22122 30\u2212 40 0 0 0 40\u2212 40 \u23a4\u23a6\u23a1\u23a3 cos\u03b13 cos\u03b23 cos \u03b33 \u23a4\u23a6 = \u23a1\u23a3 0 0 0 \u23a4\u23a6 (9.193) which leads to \u03b13 = 90.0 deg (9.194) \u03b23 = 90.0 deg (9.195) \u03b33 = 0.0 deg . (9.196) 548 9. Applied Dynamics Example 356 Moment of inertia of a rigid rectangular bar. Consider a homogeneous rectangular link with mass m, length l, width w, and height h, as shown in Figure 9.9. The local central coordinate frame is attached to the link at its mass center. The moments of inertia matrix of the link can be found by the integral method. We begin with calculating Ixx Ixx = Z B \u00a1 y2 + z2 \u00a2 dm = Z v \u00a1 y2 + z2 \u00a2 \u03c1dv = m lwh Z v \u00a1 y2 + z2 \u00a2 dv = m lwh Z h/2 \u2212h/2 Z w/2 \u2212w/2 Z l/2 \u2212l/2 \u00a1 y2 + z2 \u00a2 dx dy dz = m 12 \u00a1 w2 + h2 \u00a2 (9.197) which shows Iyy and Izz can be calculated similarly Iyy = m 12 \u00a1 h2 + l2 \u00a2 (9.198) Izz = m 12 \u00a1 l2 + w2 \u00a2 . (9.199) Since the coordinate frame is central, the products of inertia must be zero. To show this, we examine Ixy. Ixy = Iyx = \u2212 Z B xy dm = Z v xy\u03c1dv = m lwh Z h/2 \u2212h/2 Z w/2 \u2212w/2 Z l/2 \u2212l/2 xy dxdy dz = 0 (9.200) 9. Applied Dynamics 549 Therefore, the moment of inertia matrix for the rigid rectangular bar in its central frame is I = \u23a1\u23a3 m 12 \u00a1 w2 + h2 \u00a2 0 0 0 m 12 \u00a1 h2 + l2 \u00a2 0 0 0 m 12 \u00a1 l2 + w2 \u00a2 \u23a4\u23a6 . (9.201) Example 357 Translation of the inertia matrix. The moment of inertia matrix of the rigid body shown in Figure 9.10, in the principal frame B(oxyz) is given in Equation (9.201). The moment of inertia matrix in the non-principal frame B0(ox0y0z0) can be found by applying the parallel-axes transformation formula (9.169). B0 I = BI +m B0 r\u0303C B0 r\u0303TC (9.202) The mass center is at B0 rC = 1 2 \u23a1\u23a3 l w h \u23a4\u23a6 (9.203) and therefore, B0 r\u0303C = 1 2 \u23a1\u23a3 0 \u2212h w h 0 \u2212l \u2212w l 0 \u23a4\u23a6 (9.204) that provides B0 I = \u23a1\u23a3 1 3h 2m+ 1 3mw2 \u221214 lmw \u221214hlm \u221214 lmw 1 3h 2m+ 1 3 l 2m \u221214hmw \u221214hlm \u2212 14hmw 1 3 l 2m+ 1 3mw2 \u23a4\u23a6 . (9.205) 550 9. Applied Dynamics Example 358 Principal rotation matrix. Consider a body inertia matrix as I = \u23a1\u23a3 2/3 \u22121/2 \u22121/2 \u22121/2 5/3 \u22121/4 \u22121/2 \u22121/4 5/3 \u23a4\u23a6 . (9.206) The eigenvalues and eigenvectors of I are I1 = 0.2413 , \u23a1\u23a3 2.351 1 1 \u23a4\u23a6 (9.207) I2 = 1.8421 , \u23a1\u23a3 \u22120.8511 1 \u23a4\u23a6 (9.208) I3 = 1.9167 , \u23a1\u23a3 0 \u22121 1 \u23a4\u23a6 . (9.209) The normalized eigenvector matrix W is equal to the transpose of the required transformation matrix to make the inertia matrix diagonal W = \u23a1\u23a3 | | | w1 w 2 w 3 | | | \u23a4\u23a6 = 2RT 1 = \u23a1\u23a3 0.856 9 \u22120.515 6 0.0 0.364 48 0.605 88 \u22120.707 11 0.364 48 0.605 88 0.707 11 \u23a4\u23a6 . (9.210) We may verify that 2I \u2248 2R1 1I 2RT 1 =WT 1I W = \u23a1\u23a3 0.2413 \u22121\u00d7 10\u22124 0.0 \u22121\u00d7 10\u22124 1.842 1 \u22121\u00d7 10\u221219 0.0 0.0 1.916 7 \u23a4\u23a6 . (9.211) Example 359 F Relative diagonal moments of inertia. Using the definitions for moments of inertia (9.147), (9.148), and (9.149) it is seen that the inertia matrix is symmetric, andZ B \u00a1 x2 + y2 + z2 \u00a2 dm = 1 2 (Ixx + Iyy + Izz) (9.212) and also Ixx + Iyy \u2265 Izz (9.213) Iyy + Izz \u2265 Ixx (9.214) Izz + Ixx \u2265 Iyy. (9.215) 9. Applied Dynamics 551 Noting that (y \u2212 z) 2 \u2265 0 it is evident that \u00a1 y2 + z2 \u00a2 \u2265 2yz and therefore Ixx \u2265 2Iyz (9.216) and similarly Iyy \u2265 2Izx (9.217) Izz \u2265 2Ixy. (9.218) Example 360 F Coefficients of the characteristic equation. The determinant (9.172)\u00af\u0304\u0304\u0304 \u00af\u0304 Ixx \u2212 \u03bb Ixy Ixz Iyx Iyy \u2212 \u03bb Iyz Izx Izy Izz \u2212 \u03bb \u00af\u0304\u0304\u0304 \u00af\u0304 = 0 (9.219) for calculating the principal moments of inertia, leads to a third-degree equation for \u03bb, called the characteristic equation. \u03bb3 \u2212 a1\u03bb 2 + a2\u03bb\u2212 a3 = 0 (9.220) The coefficients of the characteristic equation are called the principal invariants of [I]. The coefficients of the characteristic equation can directly be found from the following equations: a1 = Ixx + Iyy + Izz = tr [I] (9.221) a2 = IxxIyy + IyyIzz + IzzIxx \u2212 I2xy \u2212 I2yz \u2212 I2zx = \u00af\u0304\u0304\u0304 Ixx Ixy Iyx Iyy \u00af\u0304\u0304\u0304 + \u00af\u0304\u0304\u0304 Iyy Iyz Izy Izz \u00af\u0304\u0304\u0304 + \u00af\u0304\u0304\u0304 Ixx Ixz Izx Izz \u00af\u0304\u0304\u0304 = 1 2 \u00a1 a21 \u2212 tr \u00a3 I2 \u00a4\u00a2 (9.222) a3 = IxxIyyIzz + IxyIyzIzx + IzyIyxIxz \u2212 (IxxIyzIzy + IyyIzxIxz + IzzIxyIyx) = IxxIyyIzz + 2IxyIyzIzx \u2212 \u00a1 IxxI 2 yz + IyyI 2 zx + IzzI 2 xy \u00a2 = det [I] (9.223) 552 9. Applied Dynamics Example 361 F The principal moments of inertia are coordinate invariants. The roots of the inertia characteristic equation are the principal moments of inertia. They are all real but not necessarily different. The principal moments of inertia are extreme. That is, the principal moments of inertia determine the smallest and the largest values of Iii. Since the smallest and largest values of Iii do not depend on the choice of the body coordinate frame, the solution of the characteristic equation is not dependent of the coordinate frame. In other words, if I1, I2, and I3 are the principal moments of inertia for B1I, the principal moments of inertia for B2I are also I1, I2, and I3 when B2I = B2RB1 B1I B2RT B1 . We conclude that I1, I2, and I3 are coordinate invariants of the matrix [I], and therefore any quantity that depends on I1, I2, and I3 is also coordinate invariant. The matrix [I] has only three independent invariants and every other invariant can be expressed in terms of I1, I2, and I3. Since I1, I2, and I3 are the solutions of the characteristic equation of [I] given in (9.220), we may write the determinant (9.172) in the form (\u03bb\u2212 I1) (\u03bb\u2212 I2) (\u03bb\u2212 I3) = 0. (9.224) The expanded form of this equation is \u03bb3 \u2212 (I1 + I2 + I3)\u03bb 2 + (I1I2 + I2I3 + I3I1) a2\u03bb\u2212 I1I2I3 = 0. (9.225) By comparing (9.225) and (9.220) we conclude that a1 = Ixx + Iyy + Izz = I1 + I2 + I3 (9.226) a2 = IxxIyy + IyyIzz + IzzIxx \u2212 I2xy \u2212 I2yz \u2212 I2zx = I1I2 + I2I3 + I3I1 (9.227) a3 = IxxIyyIzz + 2IxyIyzIzx \u2212 \u00a1 IxxI 2 yz + IyyI 2 zx + IzzI 2 xy \u00a2 = I1I2I3. (9.228) Being able to express the coefficients a1, a2, and a3 as functions of I1, I2, and I3 determines that the coefficients of the characteristic equation are coordinate-invariant. Example 362 F Short notation for the elements of inertia matrix. Taking advantage of the Kronecker\u2019s delta (5.138) we may write the el- 9. Applied Dynamics 553 ements of the moment of inertia matrix Iij in short notation forms Iij = Z B \u00a1\u00a1 x21 + x22 + x23 \u00a2 \u03b4ij \u2212 xixj \u00a2 dm (9.229) Iij = Z B \u00a1 r2\u03b4ij \u2212 xixj \u00a2 dm (9.230) Iij = Z B \u00c3 3X k=1 xkxk\u03b4ij \u2212 xixj ! dm (9.231) where we utilized the following notations: x1 = x x2 = y x3 = z. (9.232) Example 363 F Moment of inertia with respect to a plane, a line, and a point. The moment of inertia of a system of particles may be defined with respect to a plane, a line, or a point as the sum of the products of the mass of the particles into the square of the perpendicular distance from the particle to the plane, line, or point. For a continuous body, the sum would be definite integral over the volume of the body. The moments of inertia with respect to the xy, yz, and zx-plane are Iz2 = Z B z2dm (9.233) Iy2 = Z B y2dm (9.234) Ix2 = Z B x2dm. (9.235) The moments of inertia with respect to the x, y, and z axes are Ix = Z B \u00a1 y2 + z2 \u00a2 dm (9.236) Iy = Z B \u00a1 z2 + x2 \u00a2 dm (9.237) Iz = Z B \u00a1 x2 + y2 \u00a2 dm (9.238) and therefore, Ix = Iy2 + Iz2 (9.239) Iy = Iz2 + Ix2 (9.240) Iz = Ix2 + Iy2 . (9.241) 554 9. Applied Dynamics The moment of inertia with respect to the origin is Io = Z B \u00a1 x2 + y2 + z2 \u00a2 dm = Ix2 + Iy2 + Iz2 = 1 2 (Ix + Iy + Iz) . (9.242) Because the choice of the coordinate frame is arbitrary, we can say that the moment of inertia with respect to a line is the sum of the moments of inertia with respect to any two mutually orthogonal planes that pass through the line. The moment of inertia with respect to a point has similar meaning for three mutually orthogonal planes intersecting at the point. 9.5 Lagrange\u2019s Form of Newton\u2019s Equations of Motion Newton\u2019s equation of motion can be transformed to d dt \u00b5 \u2202K \u2202q\u0307r \u00b6 \u2212 \u2202K \u2202qr = Fr r = 1, 2, \u00b7 \u00b7 \u00b7n (9.243) where Fr = nX i=1 \u00b5 Fix \u2202fi \u2202q1 + Fiy \u2202gi \u2202q2 + Fiz \u2202hi \u2202qn \u00b6 . (9.244) Equation (9.243) is called the Lagrange equation of motion, where K is the kinetic energy of the n degree-of-freedom (DOF ) system, qr, r = 1, 2, \u00b7 \u00b7 \u00b7 , n are the generalized coordinates of the system, F = \u00a3 Fix Fiy Fiz \u00a4T is the external force acting on the ith particle of the system, and Fr is the generalized force associated to qr. Proof. Let mi be the mass of one of the particles of a system and let (xi, yi, zi) be its Cartesian coordinates in a globally fixed coordinate frame. Assume that the coordinates of every individual particle are functions of another set of coordinates q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, and possibly time t. xi = fi(q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, t) (9.245) yi = gi(q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, t) (9.246) zi = hi(q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, t) (9.247) If Fxi, Fyi, Fzi are components of the total force acting on the particle mi, then the Newton equations of motion for the particle would be Fxi = mix\u0308i (9.248) Fyi = miy\u0308i (9.249) Fzi = miz\u0308i. (9.250) 9. Applied Dynamics 555 We multiply both sides of these equations by \u2202fi \u2202qr \u2202gi \u2202qr \u2202hi \u2202qr respectively, and add them up for all the particles to have nX i=1 mi \u00b5 x\u0308i \u2202fi \u2202qr + y\u0308i \u2202gi \u2202qr + z\u0308i \u2202hi \u2202qr \u00b6 = nX i=1 \u00b5 Fxi \u2202fi \u2202qr + Fyi \u2202gi \u2202qr + Fzi \u2202hi \u2202qr \u00b6 (9.251) where n is the total number of particles. Taking a time derivative of Equation (9.245), x\u0307i = \u2202fi \u2202q1 q\u03071 + \u2202fi \u2202q2 q\u03072 + \u2202fi \u2202q3 q\u03073 + \u00b7 \u00b7 \u00b7+ \u2202fi \u2202qn q\u0307n + \u2202fi \u2202t (9.252) we find \u2202x\u0307i \u2202q\u0307r = \u2202 \u2202q\u0307r \u00b5 \u2202fi \u2202q1 q\u03071 + \u2202fi \u2202q2 q\u03072 + \u00b7 \u00b7 \u00b7+ \u2202fi \u2202qn q\u0307n + \u2202fi \u2202t \u00b6 = \u2202fi \u2202qr . (9.253) and therefore, x\u0308i \u2202fi \u2202qr = x\u0308i \u2202x\u0307i \u2202q\u0307r = d dt \u00b5 x\u0307i \u2202x\u0307i \u2202q\u0307r \u00b6 \u2212 x\u0307i d dt \u00b5 \u2202x\u0307i \u2202q\u0307r \u00b6 . (9.254) However, x\u0307i d dt \u00b5 \u2202x\u0307i \u2202q\u0307r \u00b6 = x\u0307i d dt \u00b5 \u2202fi \u2202qr \u00b6 = x\u0307i \u00b5 \u22022fi \u2202q1\u2202qr q\u03071 + \u00b7 \u00b7 \u00b7+ \u22022fi \u2202qn\u2202qr q\u0307n + \u22022fi \u2202t\u2202qr \u00b6 = x\u0307i \u2202 \u2202qr \u00b5 \u2202fi \u2202q1 q\u03071 + \u2202fi \u2202q2 q\u03072 + \u00b7 \u00b7 \u00b7+ \u2202fi \u2202qn q\u0307n + \u2202fi \u2202t \u00b6 = x\u0307i \u2202x\u0307i \u2202qr (9.255) and we have x\u0308i \u2202x\u0307i \u2202q\u0307r = d dt \u00b5 x\u0307i \u2202x\u0307i \u2202q\u0307r \u00b6 \u2212 x\u0307i \u2202x\u0307i \u2202qr (9.256) 556 9. Applied Dynamics which is equal to x\u0308i x\u0307i q\u0307r = d dt \u2219 \u2202 \u2202q\u0307r \u00b5 1 2 x\u03072i \u00b6\u00b8 \u2212 \u2202 \u2202qr \u00b5 1 2 x\u03072i \u00b6 . (9.257) Now substituting (9.254) and (9.257) in the left-hand side of (9.251) leads to nX i=1 mi \u00b5 x\u0308i \u2202fi \u2202qr + y\u0308i \u2202gi \u2202qr + z\u0308i \u2202hi \u2202qr \u00b6 = nX i=1 mi d dt \u2219 \u2202 \u2202q\u0307r \u00b5 1 2 x\u03072i + 1 2 y\u03072i + 1 2 z\u03072i \u00b6\u00b8 \u2212 nX i=1 mi \u2202 \u2202qr \u00b5 1 2 x\u03072i + 1 2 y\u03072i + 1 2 z\u03072i \u00b6 = 1 2 nX i=1 mi d dt \u2219 \u2202 \u2202q\u0307r \u00a1 x\u03072i + y\u03072i + z\u03072i \u00a2\u00b8 \u22121 2 nX i=1 mi \u2202 \u2202qr \u00a1 x\u03072i + y\u03072i + z\u03072i \u00a2 . (9.258) where 1 2 nX i=1 mi \u00a1 x\u03072i + y\u03072i + z\u03072i \u00a2 = K (9.259) is the kinetic energy of the system. Therefore, the Newton equations of motion (9.248), (9.249), and (9.250) are converted to d dt \u00b5 \u2202K \u2202q\u0307r \u00b6 \u2212 \u2202K \u2202qr = nX i=1 \u00b5 Fxi \u2202fi \u2202qr + Fyi \u2202gi \u2202qr + Fzi \u2202hi \u2202qr \u00b6 . (9.260) Because of (9.245), (9.246), and (9.247), the kinetic energy is a function of q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn and time t. The left-hand side of Equation (9.260) includes the kinetic energy of the whole system and the right-hand side is a generalized force and shows the effect of changing coordinates from xi to qj on the external forces. Let us assume that the coordinate qr alters to qr+ \u03b4qr while the other coordinates q1, q2, q3, \u00b7 \u00b7 \u00b7 , qr\u22121, qr+1, \u00b7 \u00b7 \u00b7 , qn and time t are unaltered. So, the coordinates of mi are changed to xi + \u2202fi \u2202qr \u03b4qr (9.261) yi + \u2202gi \u2202qr \u03b4qr (9.262) zi + \u2202hi \u2202qr \u03b4qr (9.263) 9. Applied Dynamics 557 Such a displacement is called virtual displacement. The work done in this virtual displacement by all forces acting on the particles of the system is \u03b4W = nX i=1 \u00b5 Fxi \u2202fi \u2202qr + Fyi \u2202gi \u2202qr + Fzi \u2202hi \u2202qr \u00b6 \u03b4qr. (9.264) Because the work done by internal forces appears in opposite pairs, only the work done by external forces remains in Equation (9.264). Let\u2019s denote the virtual work by \u03b4W = Fr (q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, t) \u03b4qr. (9.265) Then we have d dt \u00b5 \u2202K \u2202q\u0307r \u00b6 \u2212 \u2202K \u2202qr = Fr (9.266) where Fr = nX i=1 \u00b5 Fxi \u2202fi \u2202qr + Fyi \u2202gi \u2202qr + Fzi \u2202hi \u2202qr \u00b6 . (9.267) Equation (9.266) is the Lagrange form of equations of motion. This equation is true for all values of r from 1 to n. We thus have n second-order ordinary differential equations in which q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn are the dependent variables and t is the independent variable. The coordinates q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn are called generalized coordinates and can be any measurable parameters to provide the configuration of the system. The number of equations and the number of dependent variables are equal, therefore, the equations are theoretically sufficient to determine the motion of all mi. Example 364 Equation of motion for a simple pendulum. A pendulum is shown in Figure 9.11. Using x and y for the Cartesian position of m, and using \u03b8 = q as the generalized coordinate, we have x = f(\u03b8) = l sin \u03b8 (9.268) y = g(\u03b8) = l cos \u03b8 (9.269) K = 1 2 m \u00a1 x\u03072 + y\u03072 \u00a2 = 1 2 ml2\u03b8\u0307 2 (9.270) and therefore, d dt \u00b5 \u2202K \u2202\u03b8\u0307 \u00b6 \u2212 \u2202K \u2202\u03b8 = d dt (ml2\u03b8\u0307) = ml2\u03b8\u0308. (9.271) The external force components, acting on m, are Fx = 0 (9.272) Fy = mg (9.273) 558 9. Applied Dynamics and therefore, F\u03b8 = Fx \u2202f \u2202\u03b8 + Fy \u2202g \u2202\u03b8 = \u2212mgl sin \u03b8. (9.274) Hence, the equation of motion for the pendulum is ml2\u03b8\u0308 = \u2212mgl sin \u03b8. (9.275) Example 365 A pendulum attached to an oscillating mass. Figure 9.12 illustrates a vibrating mass with a hanging pendulum. The pendulum can act as a vibration absorber if designed properly. Starting with coordinate relationships xM = fM = x (9.276) yM = gM = 0 (9.277) xm = fm = x+ l sin \u03b8 (9.278) ym = gm = l cos \u03b8 (9.279) we may find the kinetic energy in terms of the generalized coordinates x and \u03b8. K = 1 2 M \u00a1 x\u03072M + y\u03072M \u00a2 + 1 2 m \u00a1 x\u03072m + y\u03072m \u00a2 = 1 2 Mx\u03072 + 1 2 m \u00b3 x\u03072 + l2\u03b8\u0307 2 + 2lx\u0307\u03b8\u0307 cos \u03b8 \u00b4 (9.280) Then, the left-hand side of the Lagrange equations are d dt \u00b5 \u2202K \u2202x\u0307 \u00b6 \u2212 \u2202K \u2202x = (M +m)x\u0308+ml\u03b8\u0308 cos \u03b8 \u2212ml\u03b8\u0307 2 sin \u03b8 (9.281) d dt \u00b5 \u2202K \u2202\u03b8\u0307 \u00b6 \u2212 \u2202K \u2202\u03b8 = ml2\u03b8\u0308 +mlx\u0308 cos \u03b8. (9.282) 9. Applied Dynamics 559 The external forces acting on M and m are FxM = \u2212kx (9.283) FyM = 0 (9.284) Fxm = 0 (9.285) Fym = mg. (9.286) Therefore, the generalized forces are Fx = FxM \u2202fM \u2202x + FyM \u2202gM \u2202x + Fxm \u2202fm \u2202x + Fym \u2202gm \u2202x = \u2212kx (9.287) F\u03b8 = FxM \u2202fM \u2202\u03b8 + FyM \u2202gM \u2202\u03b8 + Fxm \u2202fm \u2202\u03b8 + Fym \u2202gm \u2202\u03b8 = \u2212mgl sin \u03b8 (9.288) and finally the Lagrange equations of motion are (M +m)x\u0308+ml\u03b8\u0308 cos \u03b8 \u2212ml\u03b8\u0307 2 sin \u03b8 = \u2212kx (9.289) ml2\u03b8\u0308 +mlx\u0308 cos \u03b8 = \u2212mgl sin \u03b8. (9.290) Example 366 Kinetic energy of the Earth. Earth is approximately a rotating rigid body about a fixed axis. The two motions of the Earth are called revolution about the sun, and rotation about an axis approximately fixed in the Earth. The kinetic energy of the Earth due to its rotation is K1 = 1 2 I\u03c921 = 1 2 2 5 \u00a1 5.9742\u00d7 1024 \u00a2\u00b56356912 + 6378388 2 \u00b62\u00b5 2\u03c0 24\u00d7 3600 366.25 365.25 \u00b62 = 2.5762\u00d7 1029 J 560 9. Applied Dynamics and the kinetic energy of the Earth due to its revolution is K2 = 1 2 Mr2\u03c922 = 1 2 \u00a1 5.9742\u00d7 1024 \u00a2 \u00a1 1.49475\u00d7 1011 \u00a22\u00b5 2\u03c0 24\u00d7 3600 1 365.25 \u00b62 = 2.6457\u00d7 1033 J where r is the distance from the sun and \u03c92 is the angular speed about the sun. The total kinetic energy of the Earth is K = K1 +K2. However, the ratio of the revolutionary to rotational kinetic energies is K2 K1 = 2.6457\u00d7 1033 2.5762\u00d7 1029 \u2248 10000. Example 367 F Explicit form of Lagrange equations. Assume the coordinates of every particle are functions of the coordinates q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn but not the time t. The kinetic energy of the system made of n massive particles can be written as K = 1 2 nX i=1 mi \u00a1 x\u03072i + y\u03072i + z\u03072i \u00a2 = 1 2 nX j=1 nX k=1 ajkq\u0307j q\u0307k (9.291) where the coefficients ajk are functions of q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn and ajk = akj . (9.292) The Lagrange equations of motion d dt \u00b5 \u2202K \u2202q\u0307r \u00b6 \u2212 \u2202K \u2202qr = Fr r = 1, 2, \u00b7 \u00b7 \u00b7n (9.293) are then equal to d dt nX m=1 amrq\u0307m \u2212 1 2 nX j=1 nX k=1 ajk \u2202qr q\u0307j q\u0307k = Fr (9.294) or nX m=1 amrq\u0308m + nX k=1 nX n=1 \u0393rk,nq\u0307kq\u0307n = Fr (9.295) where \u0393ij,k is called the Christoffel operator \u0393ij,k = 1 2 \u00b5 \u2202aij \u2202qk + \u2202aik \u2202qj \u2212 \u2202akj \u2202qi \u00b6 . (9.296) 9. Applied Dynamics 561 9.6 Lagrangian Mechanics Assume for some forces F = \u00a3 Fix Fiy Fiz \u00a4T there is a function V , called potential energy, such that the force is derivable from V F = \u2212\u2207V. (9.297) Such a force is called potential or conservative force. Then, the Lagrange equation of motion can be written as d dt \u00b5 \u2202L \u2202q\u0307r \u00b6 \u2212 \u2202L \u2202qr = Qr r = 1, 2, \u00b7 \u00b7 \u00b7n (9.298) where L = K \u2212 V (9.299) is the Lagrangean of the system and Qr is the nonpotential generalized force. Proof. Assume the external forces F = \u00a3 Fxi Fyi Fzi \u00a4T acting on the system are conservative. F = \u2212\u2207V (9.300) The work done by these forces in an arbitrary virtual displacement \u03b4q1, \u03b4q2, \u03b4q3, \u00b7 \u00b7 \u00b7 , \u03b4qn is \u2202W = \u2212\u2202V \u2202q1 \u03b4q1 \u2212 \u2202V \u2202q2 \u03b4q2 \u2212 \u00b7 \u00b7 \u00b7 \u2202V \u2202qn \u03b4qn (9.301) then the Lagrange equation becomes d dt \u00b5 \u2202K \u2202q\u0307r \u00b6 \u2212 \u2202K \u2202qr = \u2212 \u2202V \u2202q1 r = 1, 2, \u00b7 \u00b7 \u00b7n. (9.302) Introducing the Lagrangean function L = K \u2212 V converts the Lagrange equation to d dt \u00b5 \u2202L \u2202q\u0307r \u00b6 \u2212 \u2202L \u2202qr = 0 r = 1, 2, \u00b7 \u00b7 \u00b7n (9.303) for a conservative system. The Lagrangean is also called kinetic potential. If a force is not conservative, then the virtual work done by the force is \u03b4W = nX i=1 \u00b5 Fxi \u2202fi \u2202qr + Fyi \u2202gi \u2202qr + Fzi \u2202hi \u2202qr \u00b6 \u03b4qr = Qr \u03b4qr (9.304) and the equation of motion would be d dt \u00b5 \u2202L \u2202q\u0307r \u00b6 \u2212 \u2202L \u2202qr = Qr r = 1, 2, \u00b7 \u00b7 \u00b7n (9.305) where Qr is the nonpotential generalized force doing work in a virtual displacement of the rth generalized coordinate qr. 562 9. Applied Dynamics Example 368 Spherical pendulum. A pendulum analogy is utilized in modeling of many dynamical problems. Figure 9.13 illustrates a spherical pendulum with mass m and length l. The angles \u03d5 and \u03b8 may be used as describing coordinates of the system. The Cartesian coordinates of the mass as a function of the generalized coordinates are \u23a1\u23a3 X Y Z \u23a4\u23a6 = \u23a1\u23a3 r cos\u03d5 sin \u03b8 r sin \u03b8 sin\u03d5 \u2212r cos \u03b8 \u23a4\u23a6 (9.306) and therefore, the kinetic and potential energies of the pendulum are K = 1 2 m \u00b3 l2\u03b8\u0307 2 + l2\u03d5\u03072 sin2 \u03b8 \u00b4 (9.307) V = \u2212mgl cos \u03b8. (9.308) The kinetic potential function of this system is then equal to L = 1 2 m \u00b3 l2\u03b8\u0307 2 + l2\u03d5\u03072 sin2 \u03b8 \u00b4 +mgl cos \u03b8 (9.309) which leads to the following equations of motion: \u03b8\u0308 \u2212 \u03d5\u03072 sin \u03b8 cos \u03b8 + g l sin \u03b8 = 0 (9.310) \u03d5\u0308 sin2 \u03b8 + 2\u03d5\u0307\u03b8\u0307 sin \u03b8 cos \u03b8 = 0. (9.311) Example 369 Controlled compound pendulum. A massive arm is attached to a ceiling at a pin joint O as illustrated in Figure 9.14. Assume that there is viscous friction in the joint where an ideal motor can apply a torque Q to move the arm. The rotor of an ideal motor has no moment of inertia by assumption. 9. Applied Dynamics 563 The kinetic and potential energies of the manipulator are K = 1 2 I\u03b8\u0307 2 = 1 2 \u00a1 IC +ml2 \u00a2 \u03b8\u0307 2 (9.312) V = \u2212mg cos \u03b8 (9.313) where m is the mass and I is the moment of inertia of the pendulum about O. The Lagrangean of the manipulator is L = K \u2212 V = 1 2 I\u03b8\u0307 2 +mg cos \u03b8 (9.314) and therefore, the equation of motion of the pendulum is M = d dt \u00b5 \u2202L \u2202\u03b8\u0307 \u00b6 \u2212 \u2202L \u2202\u03b8 = I \u03b8\u0308 +mgl sin \u03b8. (9.315) The generalized force M is the contribution of the motor torque Q and the viscous friction torque \u2212c\u03b8\u0307. Hence, the equation of motion of the manipulator is Q = I \u03b8\u0308 + c\u03b8\u0307 +mgl sin \u03b8. (9.316) Example 370 An ideal 2R planar manipulator dynamics. An ideal model of a 2R planar manipulator is illustrated in Figure 9.15. It is called ideal because we assume the links are massless and there is no friction. The masses m1 and m2 are the mass of the second motor to run 564 9. Applied Dynamics the second link and the load at the endpoint. We take the absolute angle \u03b81 and the relative angle \u03b82 as the generalized coordinates to express the configuration of the manipulator. The global positions of m1 and m2 are\u2219 X1 Y2 \u00b8 = \u2219 l1 cos \u03b81 l1 sin \u03b81 \u00b8 (9.317)\u2219 X2 Y2 \u00b8 = \u2219 l1 cos \u03b81 + l2 cos (\u03b81 + \u03b82) l1 sin \u03b81 + l2 sin (\u03b81 + \u03b82) \u00b8 (9.318) and therefore, the global velocity of the masses are\u2219 X\u03071 Y\u03071 \u00b8 = \u2219 \u2212l1\u03b8\u03071 sin \u03b81 l1\u03b8\u03071 cos \u03b81 \u00b8 (9.319) \u2219 X\u03072 Y\u03072 \u00b8 = \u23a1\u23a3 \u2212l1\u03b8\u03071 sin \u03b81 \u2212 l2 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 sin (\u03b81 + \u03b82) l1\u03b8\u03071 cos \u03b81 + l2 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 cos (\u03b81 + \u03b82) \u23a4\u23a6 . (9.320) The kinetic energy of this manipulator is made of kinetic energy of the masses and is equal to K = K1 +K2 = 1 2 m1 \u00b3 X\u03072 1 + Y\u0307 2 1 \u00b4 + 1 2 m2 \u00b3 X\u03072 2 + Y\u0307 2 2 \u00b4 = 1 2 m1l 2 1\u03b8\u0307 2 1 + 1 2 m2 \u00b5 l21\u03b8\u0307 2 1 + l22 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b42 + 2l1l2\u03b8\u03071 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 cos \u03b82 \u00b6 . (9.321) 9. Applied Dynamics 565 The potential energy of the manipulator is V = V1 + V2 = m1gY1 +m2gY2 = m1gl1 sin \u03b81 +m2g (l1 sin \u03b81 + l2 sin (\u03b81 + \u03b82)) . (9.322) The Lagrangean is then obtained from Equations (9.321) and (9.322) L = K \u2212 V (9.323) = 1 2 m1l 2 1\u03b8\u0307 2 1 + 1 2 m2 \u00b5 l21\u03b8\u0307 2 1 + l22 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b42 + 2l1l2\u03b8\u03071 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 cos \u03b82 \u00b6 \u2212 (m1gl1 sin \u03b81 +m2g (l1 sin \u03b81 + l2 sin (\u03b81 + \u03b82))) . which provides the required partial derivatives as follows: \u2202L \u2202\u03b81 = \u2212 (m1 +m2) gl1 cos \u03b81 \u2212m2gl2 cos (\u03b81 + \u03b82) (9.324) \u2202L \u2202\u03b8\u03071 = (m1 +m2) l 2 1\u03b8\u03071 +m2l 2 2 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 +m2l1l2 \u00b3 2\u03b8\u03071 + \u03b8\u03072 \u00b4 cos \u03b82 (9.325) d dt \u00b5 \u2202L \u2202\u03b8\u03071 \u00b6 = (m1 +m2) l 2 1\u03b8\u03081 +m2l 2 2 \u00b3 \u03b8\u03081 + \u03b8\u03082 \u00b4 +m2l1l2 \u00b3 2 \u03b8\u03081 + \u03b8\u03082 \u00b4 cos \u03b82 \u2212m2l1l2\u03b8\u03072 \u00b3 2\u03b8\u03071 + \u03b8\u03072 \u00b4 sin \u03b82 (9.326) \u2202L \u2202\u03b82 = \u2212m2l1l2\u03b8\u03071 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 sin \u03b82 \u2212m2gl2 cos (\u03b81 + \u03b82) (9.327) \u2202L \u2202\u03b8\u03072 = m2l 2 2 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 +m2l1l2\u03b8\u03071 cos \u03b82 (9.328) d dt \u00b5 \u2202L \u2202\u03b8\u03072 \u00b6 = m2l 2 2 \u00b3 \u03b8\u03081 + \u03b8\u03082 \u00b4 +m2l1l2\u03b8\u03081 cos \u03b82 \u2212m2l1l2\u03b8\u03071\u03b8\u03072 sin \u03b82 (9.329) Therefore, the equations of motion for the 2R manipulator are Q1 = d dt \u00b5 \u2202L \u2202\u03b8\u03071 \u00b6 \u2212 \u2202L \u2202\u03b81 = (m1 +m2) l 2 1\u03b8\u03081 +m2l 2 2 \u00b3 \u03b8\u03081 + \u03b8\u03082 \u00b4 +m2l1l2 \u00b3 2 \u03b8\u03081 + \u03b8\u03082 \u00b4 cos \u03b82 \u2212m2l1l2\u03b8\u03072 \u00b3 2\u03b8\u03071 + \u03b8\u03072 \u00b4 sin \u03b82 +(m1 +m2) gl1 cos \u03b81 +m2gl2 cos (\u03b81 + \u03b82) (9.330) 566 9. Applied Dynamics Q2 = d dt \u00b5 \u2202L \u2202\u03b8\u03072 \u00b6 \u2212 \u2202L \u2202\u03b82 = m2l 2 2 \u00b3 \u03b8\u03081 + \u03b8\u03082 \u00b4 +m2l1l2\u03b8\u03081 cos \u03b82 \u2212m2l1l2\u03b8\u03071\u03b8\u03072 sin \u03b82 +m2l1l2\u03b8\u03071 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 sin \u03b82 +m2gl2 cos (\u03b81 + \u03b82) . (9.331) The generalized forces Q1 and Q2 are the required forces to drive the generalized coordinates. In this case, Q1 is the torque at the base motor and Q2 is the torque of the motor at m1. The equations of motion can be rearranged to have a more systematic form Q1 = \u00a1 (m1 +m2) l 2 1 +m2l2 (l2 + 2l1 cos \u03b82) \u00a2 \u03b8\u03081 +m2l2 (l2 + l1 cos \u03b82) \u03b8\u03082 \u22122m2l1l2 sin \u03b82 \u03b8\u03071\u03b8\u03072 \u2212m2l1l2 sin \u03b82 \u03b8\u0307 2 2 +(m1 +m2) gl1 cos \u03b81 +m2gl2 cos (\u03b81 + \u03b82) (9.332) Q2 = m2l2 (l2 + l1 cos \u03b82) \u03b8\u03081 +m2l 2 2\u03b8\u03082 +m2l1l2 sin \u03b82 \u03b8\u0307 2 1 +m2gl2 cos (\u03b81 + \u03b82) . (9.333) Example 371 Mechanical energy. If a system of masses mi are moving in a potential force field Fmi = \u2212\u2207iV (9.334) their Newton equations of motion will be mir\u0308i = \u2212\u2207iV i = 1, 2, \u00b7 \u00b7 \u00b7n. (9.335) The inner product of equations of motion with r\u0307i and adding the equations nX i=1 mir\u0307i \u00b7 r\u0308i = \u2212 nX i=1 r\u0307i \u00b7\u2207iV (9.336) and then, integrating over time 1 2 nX i=1 mir\u0307i \u00b7 r\u0307i = \u2212 Z nX i=1 ri \u00b7\u2207iV (9.337) shows that K = \u2212 Z nX i=1 \u00b5 \u2202V \u2202xi xi + \u2202V \u2202yi yi + \u2202V \u2202zi zi \u00b6 = \u2212V +E (9.338) 9. Applied Dynamics 567 where E is the constant of integration. E is called the mechanical energy of the system and is equal to kinetic plus potential energies. E = K + V (9.339) Example 372 Falling wheel. Figure 9.16 illustrates a wheel turning, without slip, over a cylindrical hill. We may use the conservation of mechanical energy to find the angle at which the wheel leaves the hill. At the initial instant of time, the wheel is at point A. We assume the initial kinetic and potential, and hence, the mechanical energies are zero. When the wheel is turning over the hill, its angular velocity, \u03c9, is \u03c9 = v r (9.340) where v is the speed at the center of the wheel. At any other point B, the wheel achieves some kinetic energy and loses some potential energy. At a certain angle, where the normal component of the weight cannot provide more centripetal force, mg cos \u03b8 = mv2 R+ r . (9.341) the wheel separates from the surface. Employing the conservation of energy, we have EA = EB (9.342) KA + VA = KB + VB. (9.343) The kinetic and potential energy at the separation point B are KB = 1 2 mv2 + 1 2 IC\u03c9 2 (9.344) VB = \u2212mg (R+ r) (1\u2212 cos \u03b8) (9.345) 568 9. Applied Dynamics where IC is the mass moment of inertia for the wheel about its center. Therefore, 1 2 mv2 + 1 2 IC\u03c9 2 = mg (R+ r) (1\u2212 cos \u03b8) (9.346) and substituting (9.340) and (9.341) provides\u00b5 1 + IC mr2 \u00b6 (R+ r) g cos \u03b8 = 2g (R+ r) (1\u2212 cos \u03b8) (9.347) and therefore, the separation angle is \u03b8 = cos\u22121 2mr2 IC + 3mr2 . (9.348) Let\u2019s examine the equation for a disc wheel with IC = 1 2 mr2. (9.349) and find the separation angle. \u03b8 = cos\u22121 4 7 (9.350) \u2248 0.96 rad \u2248 55.15 deg Example 373 Turning wheel over a step. Figure 9.17 illustrates a wheel of radius R turning with speed v to go over a step with height H < R. We may use the principle of energy conservation and find the speed of the wheel after getting across the step. Employing the conservation of energy, 9. Applied Dynamics 569 we have EA = EB (9.351) KA + VA = KB + VB (9.352) 1 2 mv21 + 1 2 IC\u03c9 2 1 + 0 = 1 2 mv22 + 1 2 IC\u03c9 2 2 +mgH (9.353)\u00b5 m+ IC R2 \u00b6 v21 = \u00b5 m+ IC R2 \u00b6 v22 + 2mgH (9.354) and therefore, v2 = vuutv21 \u2212 2gH 1 + IC mR2 . (9.355) The condition for having a real v2 is v1 > vuut 2gH 1 + IC mR2 . (9.356) The second speed (9.355) and the condition (9.356) for a solid disc are v2 = r v21 \u2212 4 3 Hg (9.357) v1 > r 4 3 Hg (9.358) because we assumed that IC = 1 2 mR2. (9.359) Example 374 Trebuchet. A trebuchet, shown schematically in Figure 9.18, is a shooting weapon of war powered by a falling massive counterweight m1. A beam AB is pivoted to the chassis with two unequal sections a and b. The figure shows a trebuchet at its initial configuration. The origin of a global coordinate frame is set at the pivot point. The counterweight m1 is at (x1, y1) and is hinged at the shorter arm of the beam at a distance c from the end B. The mass of the projectile is m2 and it is at the end of a massless sling with a length l attached to the end of the longer arm of the beam. The three independent variable angles \u03b1, \u03b8, and \u03b3 describe the motion of the device. We consider the parameters a, b, c, d, l, m1, and m2 constant, and determine the equations of motion by the Lagrange method. Figure 9.19 illustrates the trebuchet when it is in motion. The position coordinates of masses m1 and m2 are x1 = b sin \u03b8 \u2212 c sin (\u03b8 + \u03b3) (9.360) y1 = \u2212b cos \u03b8 + c cos (\u03b8 + \u03b3) (9.361) 570 9. Applied Dynamics and x2 = \u2212a sin \u03b8 \u2212 l sin (\u2212\u03b8 + \u03b1) (9.362) y2 = \u2212a cos \u03b8 \u2212 l cos (\u2212\u03b8 + \u03b1) . (9.363) Taking a time derivative provides the velocity components x\u03071 = b\u03b8\u0307 cos \u03b8 \u2212 c \u00b3 \u03b8\u0307 + \u03b3\u0307 \u00b4 cos (\u03b8 + \u03b3) (9.364) y\u03071 = b\u03b8\u0307 sin \u03b8 \u2212 c \u00b3 \u03b8\u0307 + \u03b3\u0307 \u00b4 sin (\u03b8 + \u03b3) (9.365) x\u03072 = l (c\u2212 \u03b1\u0307) cos (\u03b1\u2212 \u03b8)\u2212 a\u03b8\u0307 cos (\u03b8) (9.366) y\u03072 = a\u03b8\u0307 sin \u03b8 \u2212 l \u00b3 \u03b8\u0307 \u2212 \u03b1\u0307 \u00b4 sin (\u03b1\u2212 \u03b8) . (9.367) 9. Applied Dynamics 571 which shows that the kinetic energy of the system is K = 1 2 m1v 2 1 + 1 2 m2v 2 2 = 1 2 m1 \u00a1 x\u030721 + y\u030721 \u00a2 + 1 2 m2 \u00a1 x\u030722 + y\u030722 \u00a2 = 1 2 m1 \u00b3\u00a1 b2 + c2 \u00a2 \u03b8\u0307 2 + c2\u03b3\u03072 + 2c2\u03b8\u0307\u03b3\u0307 \u00b4 \u2212m1bc\u03b8\u0307 \u00b3 \u03b8\u0307 + \u03b3\u0307 \u00b4 cos \u03b3 + 1 2 m2 \u00b3\u00a1 a2 + l2 \u00a2 \u03b8\u0307 2 + l2\u03b1\u03072 \u2212 2l2\u03b8\u0307\u03b1\u0307 \u00b4 \u2212m2al\u03b8\u0307 \u00b3 \u03b8\u0307 \u2212 \u03b1\u0307 \u00b4 cos (2\u03b8 \u2212 \u03b1) . (9.368) The potential energy of the system can be calculated by y-position of the masses. V = m1gy1 +m2gy2 = m1g (\u2212b cos \u03b8 + c cos (\u03b8 + \u03b3)) +m2g (\u2212a cos \u03b8 \u2212 l cos (\u2212\u03b8 + \u03b1)) (9.369) Having the energies K and V , we can set up the Lagrangean L. L = K \u2212 V (9.370) Using the Lagrangean, we are able to find the three equations of motion. d dt \u00b5 \u2202L \u2202\u03b8\u0307 \u00b6 \u2212 \u2202L \u2202\u03b8 = 0 (9.371) d dt \u00b5 \u2202L \u2202\u03b1\u0307 \u00b6 \u2212 \u2202L \u2202\u03b1 = 0 (9.372) d dt \u00b5 \u2202L \u2202\u03b3\u0307 \u00b6 \u2212 \u2202L \u2202\u03b3 = 0. (9.373) The trebuchet appeared in 500 to 400 B.C. China and was developed by Persian armies around 300 B.C. It was used by the Arabs against the Romans during 600 to 1200 A.D. The trebuchet is also called the manjaniq, catapults, or onager. The \"Manjaniq\" is the root of the words \"machine\" and \"mechanic\". 9.7 Summary The translational and rotational equations of motion for a rigid body, expressed in the global coordinate frame, are GF = Gd dt Gp (9.374) GM = Gd dt GL (9.375) 572 9. Applied Dynamics where GF and GM indicate the resultant of the external forces and moments applied on the rigid body, measured at the mass center C. The vector Gp is the momentum and GL is the moment of momentum for the rigid body at C p = mv (9.376) L = rC \u00d7 p. (9.377) The expression of the equations of motion in the body coordinate frame are BF = Gp\u0307+ B G\u03c9B \u00d7 Bp = m BaB +m B G\u03c9B \u00d7 BvB (9.378) BM = BL\u0307+B G\u03c9B \u00d7 BL = BI B G\u03c9\u0307B + B G\u03c9B \u00d7 \u00a1 BI B G\u03c9B \u00a2 (9.379) where I is the moment of inertia for the rigid body. I = \u23a1\u23a3 Ixx Ixy Ixz Iyx Iyy Iyz Izx Izy Izz \u23a4\u23a6 . (9.380) 9. Applied Dynamics 573 The elements of I are functions of the mass distribution of the rigid body and are defined by Iij = Z B \u00a1 r2i \u03b4mn \u2212 ximxjn \u00a2 dm , i, j = 1, 2, 3 (9.381) where \u03b4ij is Kronecker\u2019s delta. Every rigid body has a principal body coordinate frame in which the moment of inertia is in the form BI = \u23a1\u23a3 I1 0 0 0 I2 0 0 0 I3 \u23a4\u23a6 . (9.382) The rotational equation of motion in the principal coordinate frame simplifies to M1 = I1\u03c9\u03071 \u2212 (I2 \u2212 I2)\u03c92\u03c93 M2 = I2\u03c9\u03072 \u2212 (I3 \u2212 I1)\u03c93\u03c91 (9.383) M3 = I3\u03c9\u03073 \u2212 (I1 \u2212 I2)\u03c91\u03c92. The equations of motion for a mechanical system having n DOF can also be found by the Lagrange equation d dt \u00b5 \u2202L \u2202q\u0307r \u00b6 \u2212 \u2202L \u2202qr = Qr r = 1, 2, \u00b7 \u00b7 \u00b7n (9.384) L = K \u2212 V (9.385) where L is the Lagrangean of the system, K is the kinetic energy, V is the potential energy, and Qr is the nonpotential generalized force. Qr = nX i=1 \u00b5 Qix \u2202fi \u2202q1 +Qiy \u2202gi \u2202q2 +Qiz \u2202hi \u2202qn \u00b6 (9.386) The parameters qr, r = 1, 2, \u00b7 \u00b7 \u00b7 , n are the generalized coordinates of the system, Q = \u00a3 Qix Qiy Qiz \u00a4T is the external force acting on the ith particle of the system, and Qr is the generalized force associated to qr. When (xi, yi, zi) are Cartesian coordinates in a globally fixed coordinate frame for the particle mi, then its coordinates may be functions of another set of generalized coordinates q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn and possibly time t. xi = fi(q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, t) (9.387) yi = gi(q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, t) (9.388) zi = hi(q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, t) (9.389) 574 9. Applied Dynamics 9.8 Key Symbols a, b, w, h length a acceleration C mass center d position vector of the body coordinate frame df infinitesimal force dm infinitesimal mass dm infinitesimal moment E mechanical energy F force FC Coriolis force g gravitational acceleration H height I moment of inertia matrix I1, I2, I3 principal moment of inertia K kinetic energy l directional line L moment of momentum L = K \u2212 V Lagrangean m mass M moment p momentum P,Q points in rigid body r radius of disc r position vector R radius R rotation matrix t time u\u0302 unit vector to show the directional line v \u2261 x\u0307, v velocity V potential energy w eigenvector W work W eigenvector matrix x, y, z, x displacement \u03b4ij Kronecker\u2019s delta \u0393ij,k Christoffel operator \u03bb eigenvalue \u03d5, \u03b8, \u03c8 Euler angles \u03c9,\u03c9 angular velocity k parallel \u22a5 orthogonal 9. Applied Dynamics 575 Exercises 1. Kinetic energy of a rigid link. Consider a straight and uniform bar as a rigid bar. The bar has a mass m. Show that the kinetic energy of the bar can be expressed as K = 1 6 m (v1 \u00b7 v1 + v1 \u00b7 v2 + v2 \u00b7 v2) where v1 and v2 are the velocity vectors of the endpoints of the bar. 2. Discrete particles. There are three particles m1 = 10kg, m2 = 20 kg, m3 = 30 kg, at r1 = \u23a1\u23a3 1 \u22121 1 \u23a4\u23a6 r1 = \u23a1\u23a3 \u22121\u22123 2 \u23a4\u23a6 r1 = \u23a1\u23a3 2 \u22121 \u22123 \u23a4\u23a6 . Their velocities are v1 = \u23a1\u23a3 2 1 1 \u23a4\u23a6 v1 = \u23a1\u23a3 \u221210 2 \u23a4\u23a6 v1 = \u23a1\u23a3 3 \u22122 \u22121 \u23a4\u23a6 . Find the position and velocity of the system at C. Calculate the system\u2019s momentum and moment of momentum. Calculate the system\u2019s kinetic energy and determine the rotational and translational parts of the kinetic energy. 3. Newton\u2019s equation of motion in the body frame. Show that Newton\u2019s equation of motion in the body frame is\u23a1\u23a3 Fx Fy Fz \u23a4\u23a6 = m \u23a1\u23a3 ax ay az \u23a4\u23a6+ \u23a1\u23a3 0 \u2212\u03c9z \u03c9y \u03c9z 0 \u2212\u03c9x \u2212\u03c9y \u03c9x 0 \u23a4\u23a6\u23a1\u23a3 vx vy vz \u23a4\u23a6 . 4. Work on a curved path. A particle of mass m is moving on a circular path given by GrP = cos \u03b8 I\u0302 + sin \u03b8 J\u0302 + 4 K\u0302. Calculate the work done by a force GF when the particle moves from \u03b8 = 0 to \u03b8 = \u03c0 2 . (a) GF = z2 \u2212 y2 (x+ y) 2 I\u0302 + y2 \u2212 x2 (x+ y) 2 J\u0302 + x2 \u2212 y2 (x+ z) 2 K\u0302 576 9. Applied Dynamics (b) GF = z2 \u2212 y2 (x+ y) 2 I\u0302 + 2y x+ y J\u0302 + x2 \u2212 y2 (x+ z) 2 K\u0302 5. Principal moments of inertia. Find the principal moments of inertia and directions for the following inertia matrices: (a) [I] = \u23a1\u23a3 3 2 2 2 2 0 2 0 4 \u23a4\u23a6 (b) [I] = \u23a1\u23a3 3 2 4 2 0 2 4 2 3 \u23a4\u23a6 (c) [I] = \u23a1\u23a3 100 20 \u221a 3 0 20 \u221a 3 60 0 0 0 10 \u23a4\u23a6 6. Rotated moment of inertia matrix. A principal moment of inertia matrix B2I is given as [I] = \u23a1\u23a3 3 0 0 0 5 0 0 0 4 \u23a4\u23a6 . The principal frame was achieved by rotating the initial body coordinate frame 30 deg about the x-axis, followed by 45 deg about the z-axis. Find the initial moment of inertia matrix B1I. 7. Rotation of moment of inertia matrix. Find the required rotation matrix that transforms the moment of inertia matrix [I] to an diagonal matrix. [I] = \u23a1\u23a3 3 2 2 2 2 0.1 2 0.1 4 \u23a4\u23a6 8. F Cubic equations. The solution of a cubic equation ax3 + bx2 + cx+ d = 0 9. Applied Dynamics 577 where a 6= 0, can be found in a systematic way. Transform the equation to a new form with discriminant 4p3 + q2, y3 + 3py + q = 0 using the transformation x = y \u2212 b 3a , where, p = 3ac\u2212 b2 9a2 q = 2b3 \u2212 9abc+ 27a2d 27a3 . The solutions are then y1 = 3 \u221a \u03b1\u2212 3 p \u03b2 y2 = e 2\u03c0i 3 3 \u221a \u03b1\u2212 e 4\u03c0i 3 3 p \u03b2 y3 = e 4\u03c0i 3 3 \u221a \u03b1\u2212 e 2\u03c0i 3 3 p \u03b2 where, \u03b1 = \u2212q + p q2 + 4p3 2 \u03b2 = \u2212q + p q2 + 4p3 2 . For real values of p and q, if the discriminant is positive, then one root is real, and two roots are complex conjugates. If the discriminant is zero, then there are three real roots, of which at least two are equal. If the discriminant is negative, then there are three unequal real roots. Apply this theory for the characteristic equation of the matrix [I] and show that the principal moments of inertia are real. 9. Kinematics of a moving car on the Earth. The location of a vehicle on the Earth is described by its longitude \u03d5 from a fixed meridian, say, the Greenwich meridian, and its latitude \u03b8 from the equator, as shown in Figure 9.20. We attach a coordinate frame B at the center of the Earth with the x-axis on the equator\u2019s plane and the y-axis pointing to the vehicle. There are also two coordinate frames E and G where E is attached to the Earth and G is the global coordinate frame. Show that the angular velocity of B and the velocity of the vehicle are B G\u03c9B = \u03b8\u0307 \u0131\u0302B + (\u03c9E + \u03d5\u0307) sin \u03b8 j\u0302B + (\u03c9E + \u03d5\u0307) cos \u03b8 k\u0302 B GvP = \u2212r (\u03c9E + \u03d5\u0307) cos \u03b8 \u0131\u0302B + r\u03b8\u0307 k\u0302. Calculate the acceleration of the vehicle. 578 9. Applied Dynamics \u03b8 X Y Z x y z r \u03c9E Z (a) (b) G B Y z x y E \u03d5 \u03b8 y z \u03c9E P P 9. Applied Dynamics 579 (c) the pivot O has a uniform motion on a circle rO = R cos\u03c9t I\u0302 +R sin\u03c9t J\u0302. 13. Equations of motion from Lagrangean. Consider a physical system with a Lagrangean as L = 1 2 m (ax\u0307+ by\u0307) 2 \u2212 1 2 k (ax+ by) 2 . and find the equations of motion. The coefficients m, k, a, and b are constant. 14. Lagrangean from equation of motion. Find the Lagrangean associated to the following equations of motions: (a) mr2\u03b8\u0308 + k1l1\u03b8 + k2l2\u03b8 +mgl = 0 580 9. Applied Dynamics (b) r\u0308 \u2212 r \u03b8\u0307 2 = 0 r2 \u03b8\u0308 + 2r r\u0307 \u03b8\u0307 = 0 15. Trebuchet. Derive the equations of motion for the trebuchet shown in Figure 9.18. 16. Simplified trebuchet. Three simplified models of a trebuchet are shown in Figures 9.23 to 9.25. Derive and compare their equations of motion. 9. Applied Dynamics 581 10"
        ]
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure3.17-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure3.17-1.png",
        "caption": "FIGURE 3.17. Absolute value of a \u03c4y distribution model for n = 1.",
        "texts": [
            " The force distribution on the tireprint is not constant and is influenced by tire structure, load, inflation pressure, and environmental conditions: The tangential stress \u03c4x in the x-direction may be modeled by the following equation. \u03c4x(x, y) = \u2212\u03c4xM \u00b5 x2n+1 a2n+1 \u00b6 sin2 \u00b3x a \u03c0 \u00b4 cos \u00b3 y 2b \u03c0 \u00b4 n \u2208 N (3.32) 3. Tire Dynamics 109 \u03c4x is negative for x > 0 and is positive for x < 0, showing an inward longitudinal stress. Figure 3.16 illustrates the absolute value of a \u03c4x distribution for n = 1. The y-direction tangential stress \u03c4y may be modeled by the equation \u03c4y(x, y) = \u2212\u03c4yM \u00b5 x2n a2n \u2212 1 \u00b6 sin \u00b3y b \u03c0 \u00b4 n \u2208 N (3.33) where \u03c4y is positive for y > 0 and negative for y < 0, showing an outward lateral stress. Figure 3.17 illustrates the absolute value of a \u03c4y distribution for n = 1. 3.4 Effective Radius Consider a vertically loaded wheel that is turning on a flat surface as shown in Figure 3.18. The effective radius of the wheel Rw, which is also called a rolling radius, is defined by Rw = vx \u03c9w (3.34) where, vx is the forward velocity, and \u03c9w is the angular velocity of the wheel. The effective radius Rw is approximately equal to Rw \u2248 Rg \u2212 Rg \u2212Rh 3 (3.35) and is a number between the unloaded or geometric radius Rg and the loaded height Rh"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001805_j.matdes.2019.108091-Figure11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001805_j.matdes.2019.108091-Figure11-1.png",
        "caption": "Figure 11. Comparisons of defect diameter distributions for four shells with equal thickness as control volumes using exponential probability plots, (a) Ti-6AL-4V M290 vertical annealed machined surface, (b) Ti-6AL-4V M290 vertical HIPed machined surface, (c) Ti-6AL-4V AM250 vertical annealed machined surface, (d) 17-4 PH M290 vertical heat treated as-built surface, (e) Ti-6AL-4V AM250 vertical annealed as-built surface.",
        "texts": [
            " This means that characterization of only one of these control volumes will not provide a comprehensive picture of the distribution of defects in the specimens and might cause missing critical information. K-S tests show no significant variability in quadrants for M290 annealed specimens and as expected no significant difference between the distributions after HIPing. To inspect the variability of the defect characteristics distributions through the wall thickness of the specimens, as shown in Figure 4(c), the scanned volume was divided into four shells with thicknesses of ~ 0.3 mm and heights of 10 mm. Exponential probability plots for the defect distribution in these four shells are shown in Figure 11. These plots show very distinct differences between the distributions of defect diameter in different shells for as-built surface specimens. These results were confirmed by the K-S test results from 3D analysis as listed in Table 3. The K-S test results in this table do not show significant differences between the shells of machined surface specimens and the distribution of defects through wall thickness is more uniform after removing the surface defects by machining. The larger defects seem to be located closer to inner and/or outer surfaces"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.12-1.png",
        "caption": "FIGURE 7.12. Kinematic condition of a FWS vehicle using the angular velocity of the inner and outer wheels.",
        "texts": [
            " Rear-wheel-steering is used where high maneuverability is a necessity on a low-speed vehicle, such as forklifts. Rear-wheel-steering is not used on street vehicles because it is unstable at high speeds. The center of rotation for a rear-wheel-steeringe vehicle is always a point on the front axle. Figure 7.11 illustrates a rear-wheel-steering vehicle. The kinematic steering condition (7.1) remains the same for a rear-wheel steering vehicle. cot \u03b4o \u2212 cot \u03b4i = w l (7.30) Example 263 F Alternative kinematic steer angles equation. Consider a rear-wheel-drive vehicle with front steerable wheels as shown in Figure 7.12. Assume that the front and rear tracks of the vehicle are equal and the drive wheels are turning without slip. If we show the angular velocities of the inner and outer drive wheels by \u03c9i and \u03c9o, respectively, the kinematical steer angles of the front wheels can be expressed by \u03b4i = tan\u22121 \u00b5 l w \u00b5 \u03c9o \u03c9i \u2212 1 \u00b6\u00b6 (7.31) \u03b4o = tan\u22121 \u00b5 l w \u00b5 1\u2212 \u03c9i \u03c9o \u00b6\u00b6 . (7.32) 390 7. Steering Dynamics To prove these equations, we may start from the following equation, which is the non-slipping condition for the drive wheels: Rw \u03c9o R1 + w 2 = Rw \u03c9i R1 \u2212 w 2 "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003676_bf00363992-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003676_bf00363992-Figure1-1.png",
        "caption": "Fig. 1. Schematization of lower limbs",
        "texts": [
            " The same recording procedure was used for all the seven examined subjects. To verify the right position of the electrodes, before each experiment the subject was requested to perform suitable contractions so as to activate one of the three muscles at a time. A signal indicating the stride phases was recorded on a fourth trace. For each group of three muscles 10 to 20 strides were recorded. As a reliable repeatibility of the time occurence of signals was obtained in this way, a typical one could be selected for analysis. 2.1, Mathematical Model As shown in Figure 1 the lower limb is an articulated system of links with their own mechanical characteristics. During locomotion it is in dynamical equilibrium under the action of external forces exerted by the rest of the body through the hip joint, by the ground reactions and the internal actions which are the torques performed by the muscles on each joint. By using the mechanics of rigid bodies and limiting the analysis to the plane of progression, the system is characterized by five degrees of freedom (the two coordinates of the hip joint, the angular displacement of the hip, knee, and ankle joints)",
            " Finally, although the present results suggest that proprioceptive signals are important in conjunction with other sensory input for implementing the programs of movements, it should be emphasized that a detailed investigation of the neural mechanism correlated with the resulting biomechanic variables, including the dynamic aspects of movement, is required for a better understanding of the organization of the system controlling locomotion. Appendix The muscular moments about axes orthogonal to the sagittal plane of the hip, knee and ankle are obtained by applying the dynamic equilibrium D'Alambert's Principle on the system illustrated in Figure 1. The resulting differential non-linear equation are: M a = Jfil\" 3 + m f 2 ~ d f cos r/3 + mfZHdf sin ~13 + mf~l I l f l f cos (t/i - t/3) + mill 2l,dJ, sin (t/3 - r/t) - mfi~2lfll cos (q2 -- ~]3) + mlil~ l f l f sin, (q2 + ~13) + rn fO3d} + my9 d f sin t 13 - R~(x,. - x H - I, sin r/1 + Is sin rh) - R x ( Z n - 1~ cos t/1 - 1~ cos t/2) M 2,: = dj]2 + mjC'Hd, cos r/2 -- rG~)d s sin q2 + rGi~ l l,d~ cos, (r/i +q2)-ms i l~ l f l~ sin (t/i +rl2)-ms~2d2s - m s g d , sin ~72 + mfSiH1 s COS r/2 -- rnf~Hl, sin r/2 -~/7~f/I 11tls COS 011 + r/a ) -- mfil 2 Iris, sin (r/1 + r/z) -- myijzl~ + myii3dyl , cos (~/2 + r/3)- mfi l~df ls , sin (r/3 +t/2) -mj "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000091_s00170-015-7077-3-Figure21-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000091_s00170-015-7077-3-Figure21-1.png",
        "caption": "Fig. 21 Intersecting stiffened panels produced byWAAM. a and b carbon steel. c and d aluminium. e titanium-stiffened panel. f Ti thick wall crossover. g Ti residual stress balanced cruciform. h Ti intersections including machining. All test pieces are collected by [26]",
        "texts": [],
        "surrounding_texts": [
            "As shown in Figs. 20, 21 and 22, wire-feed AM are still only applied on simple structures. Real engineering components are more complex. A multi-direction deposition system which enhances the capability of the AM is required. However, the majority of the existing 3D slicing strategies apply only to a subset of possible part geometries. More emphasis should be paid in multi-direction AM to develop robust algorithms capable of automatically slicing any 3D model which satisfy support-less and collision-free layered deposition. Path planning for wire-feed AM technology is more complicated since special concerns should be paid on the residual stress evolution and the dimensional relationship between the weld bead and the part geometry. The investigation of the evolution of residual stresses could guide the path planning in terms of patterns and sequences. For components with complex geometry, dimension-related issue is another obstacle for automatic path planning."
        ]
    },
    {
        "image_filename": "designv10_0_0000492_02783649922066376-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000492_02783649922066376-Figure6-1.png",
        "caption": "Fig. 6. Simple planar examples. The ankle joint in Example 1 is unactuated (a). The FRI point is situated on the line O1G1 (extrapolated) at its penetration point on the ground. In Example 2, the ankle torque is just sufficient to counterbalance the gravity moment, and the system is stationary (b). In this case, as in all other stationary mechanisms, the FRI point coincides with the GCoM and the CoP.",
        "texts": [
            " (20) From above, P and C coincide if ( \u2211 H\u0307Gi + \u2211 PGi \u00d7 miai ) t = 0, which is possible if the robot is stationary or has uniform linear and angular velocities in all the joints. The objective of this section is to elucidate the idea behind the FRI point by means of four simple examples, depicted in Figures 6 and 7. The examples are based on an idealized planar point-mass model of the shank (an inverted pendulum) connected through an \u201cankle\u201d joint to a triangular foot. We consider an unactuated ankle joint, \u03c41 = 0, \u03b8\u03071 6= 0, \u03b8\u03081 6= 0, as shown in Figure 6a. From eq. (6), we have at UNIVERSITY OF BRIGHTON on July 15, 2014ijr.sagepub.comDownloaded from at UNIVERSITY OF BRIGHTON on July 15, 2014ijr.sagepub.comDownloaded from ( FO1 \u00d7 R1 ) t = 0, assuming that m1 \u2248 0. For a frictionless ankle joint, R1 is always directed toward O1G2; in other words, if we simply extend the line O1G2, the point where it penetrates the ground is the position of the FRI point. Two extreme shank configurations beyond which foot rotation occurs are shown as C1 and C2 in the figure",
            " If the shank rotates clockwise, the foot will remain stable until the shank arrives at configuration C2, at which point the foot starts rotating counterclockwise about A. On the other hand, for counterclockwise rotation of the shank, the foot starts rotating clockwise around B once the shank crosses the configuration C1. Although the opposite rotations of the shank and the foot may appear counterintuitive at first, it is better understood by recalling that the forces acting on the two segments at the ankle joint O1 are equal and opposite. Next we consider an actuated system (Fig. 6b) with an ankle torque that precisely compensates for the gravitational moment but does not generate any shank motion; i.e., \u03b8\u03071 = 0 and \u03b8\u03081 = 0. To determine the position of the FRI point of this system, we use \u03c41 = \u2212O1G2 \u00d7 m2g and R1 = \u2212m2g in eq. (6). We get \u2211 FGi \u00d7 mig = 0. This means that F falls on the CG gravity line of the system. This property is valid not only for the foot/shank, but for any stationary mechanism (Shih et al. 1990). In the next example, shown in Figure 7 (left), the shank configuration corresponds to a GCoM position C outside the support polygon"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure2.22-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure2.22-1.png",
        "caption": "Fig. 2.22 Relative Position of Two Links",
        "texts": [
            " We found the implementation with this restriction to be rather error-prone, so we instead adopted the method outlined above. 48 2 Kinematics transform of a single link as shown in Fig. 2.21. We need to set a local coordinate system \u03a3j which has it\u2019s origin on the joint axis. The joint axis vector seen from the parent coordinates is aj and the origin of \u03a3j is bj . The joint angle is qj and the attitude of the link when the joint angle is 0 is, E. The homogeneous transform relative to the parent link is: iT j = [ ea\u0302jqj bj 0 0 0 1 ] . (2.56) Next let us assume there are two links as shown in Fig. 2.22. We will assume that the absolute position and attitude of the parent link pi,Ri is known. Therefore, the homogeneous transform to \u03a3i becomes: T i = [ Ri pi 0 0 0 1 ] . (2.57) From the chain rule of homogeneous transforms \u03a3j is: T j = T i iT j . (2.58) From (2.56), (2.57) and (2.58) the absolute position (pj) and attitude (Rj) of \u03a3j can be calculated as being, pj = pi +Ribj (2.59) Rj = Rie a\u0302jqj (2.60) Using this relationship and the recursive algorithm, the Forward Kinematics can be performed by an extremely simple script shown in Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002443_j.jallcom.2021.159567-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002443_j.jallcom.2021.159567-Figure1-1.png",
        "caption": "Fig. 1. Schematic of (a) LPBF [20] and (b) DED process [38] (Reprinted with permissions from Refs. [20] and [38], respectively).",
        "texts": [
            " Since powder-based AM methods have been used much more widely than the wire-based AM methods, the current study focuses on the former and uses LPBF and DED when referring to powder bed and powder feed AM processes, respectively. Variations of these two basic AM processes are introduced in Table 1. In PBF processes, a thin layer of metallic powder is uniformly spread onto a building platform and is selectively fused using a focused heat source such as a laser or an electron beam [30]. The process goes on by distributing the next powder layer from a preplaced powder mixture over the previous one using a re-coater blade or a roller (Fig. 1a). In fact, an inter-layer metallurgical bonding is to be achieved by partial re-melting of the already fused layer. The manufactured part is usually fixed to the substrate via support, i.e. a lattice-like structure meant for dissipation of heat and part fixation in the powder bed [1]. Fig. 2 depicts solid as well as geometrically complicated porous objects produced by LPBF methods using Ti and Ta elemental powders [31]. In most DED processes, the metallic powders are directly fed through nozzles into a melt pool, already created on the substrate by a heat source (Fig. 1b) [32,33]. While a pre-alloyed powder or a mixture of elemental powders can be used as the feed material, each elemental powder can also be fed via its own nozzle, providing for engineered tailoring of the final as well as the local alloy compositions [21,34,35]. Although less common, some DED processes use wire as the feedstock, which provides for both higher deposition efficiency and deposition rate compared to powder-based DED methods where some loss of the blown powders is inevitable. Wire feed products, however, are rougher on the surface and require more extensive surface finishing [14]",
            " Ta is a \u03b2 stabilizing and biocompatible element and is featured with a good corrosion resistance and acceptable mechanical properties, making it a suitable candidate for biomedical applications. Full dense Ti-Ta parts were successfully produced using LPBF with either spherical [31,62] or irregular shaped [84,126\u2013128] Ta powders. Unmelted Ta powders were seen in all the works [31,62,84,126\u2013128]. With regard to the young modulus, a determining criterion for orthopedic implants, Ti\u201325Ta was reportedly best matched with the human bone elastic properties [62,110,126]. Ti\u2013Ta Scaffolds produced using LPBF method (Fig. 1) not only shows comparable mechanical [31] and biological [126] results with those of the widely used Ti\u20136Al\u20134V, but also have the advantage of being cost-effective and free of toxic elements, and showing enhanced corrosion properties. Shape memory NiTi alloys have recently been of interest to multiple AM research groups [16,28,48,71,80,83,105,106,129\u2013135] as their application is mainly in the biomedical field and include parts which are usually geometrically complicated. Intensive cares should be taken when processing NiTi in order to avoid impurities and inclusions since the SME and SE features of the alloy are highly sensitive to the composition and the secondary phases"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure1.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure1.6-1.png",
        "caption": "Fig. 1.6. Double-sided AFPM brushless machine with three-phase, 9-coil external salient-pole stator and 8-pole internal rotor. 1 \u2014 PM, 2 \u2014 stator backing ferromagnetic disc, 3 \u2014 stator pole, 4 \u2014 stator coil.",
        "texts": [
            " The diverse topologies of AFPM brushless machines may be classified as follows: \u2022 single-sided AFPM machines \u2013 with slotted stator (Fig. 1.4a) \u2013 with slotless stator \u2013 with salient-pole stator \u2022 double-sided AFPM machines \u2013 with internal stator (Fig. 1.4b) \u00b7 with slotted stator \u00b7 with slotless stator \u00b7 with iron core stator \u00b7 with coreless stator (Fig. 1.4d) \u00b7 without both rotor and stator cores \u00b7 with salient pole stator (Fig. 1.5) \u2013 with internal rotor (Fig. 1.4c) \u00b7 with slotted stator \u00b7 with slotless stator \u00b7 with salient pole stator (Fig. 1.6) \u2022 multi-stage (multidisc) AFPM machines (Fig. 1.7) 1.5 Topologies and Geometries 7 The air gap of the slotted armature AFPM machine is relatively small. The mean magnetic flux density in the air gap decreases under each slot opening due to increase in the reluctance. The change in the mean magnetic flux density caused by slot openings corresponds to a fictitious increase in the air gap [121]. The relation between fictitious g\u2032 and physical air gap g is expressed with the aid of Carter coefficient kC > 1, i",
            " The disadvantages include the use of more PM material, lower winding inductances sometimes causing problems for inverter-fed motors and significant eddy current losses in slotless conductors [49]. In the double-sided, salient-pole AFPM brushless machine shown in Fig. 1.5, the stator coils with concentrated parameters are wound on axially laminated poles. To obtain a three-phase self-starting motor, the number of the stator poles should be different from the number of the rotor poles, e.g. 12 stator poles and 8 rotor poles [173, 174, 185]. Fig. 1.6 shows a double-sided AFPM machine with external salient pole stators and internal PM rotor. There are nine stator coils and eight rotor poles for a three-phase AFPM machine. Depending on the application and operating environment, slotless stators may have ferromagnetic cores or be completely coreless. Coreless stator configurations eliminate any ferromagnetic material from the stator (armature) system, thus making the associated eddy current and hysteresis core losses nonexisting. This type of configuration also eliminates axial magnetic attraction forces between the stator and rotor at zero-current state"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure9.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure9.3-1.png",
        "caption": "FIGURE 9.3 PEEC loop with additional skin-effect impedance source.",
        "texts": [
            " This results in an equivalent impedance of the cell of Zs = Rs + j\ud835\udf14Ls, with Rs = \u0394x \ud835\udf0e\ud835\udeff\u0394y , Ls = \u0394x \ud835\udf14\ud835\udf0e\ud835\udeff\u0394y . (9.2) ONE DIMENSIONAL CURRENT FLOW TECHNIQUES 217 Further, due to the small diffusion distance \ud835\udeff into the conductor, we also can assume that the lateral loss coupling between two neighboring conductor surface cells is small. As a consequence, all the skin-effect loss couplings to other surface cells can be ignored. To couple the two models, we simply place the above impedance Zs = Rs + j\ud835\udf14Ls into the PEEC circuit in Fig. 9.3, where the current in the model is I = I\ud835\udcc11. In the following two sections, we consider the class of 1D current flow general surface impedance GSI models. In these models, the mutual inductive coupling between the inside of a conductor and the outside of the surface are approximated. They relate the inductance part to the internal\u2013external inductance concepts that have been used for many years [28, 29]. Also, these issues have been reviewed in a paper dedicated to Clayton Paul [30]. Figure 9.3 shows how the model is embedded in the PEEC model. The internal inductance and loss model is based on a differential equation (DE) formulation. This model is connected in series and is shown as the impedance Zs in Fig. 9.3. In this section, we treat the case where the conductors are thin, whereas thick conductors are considered in the following section. The external inductive model consists of a zero thickness partial inductance Lp11, which is connected in series to the internal DE-based model Zs representing the impedance in Fig. 9.3 for the conductor surfaces [31]. This avoids the inductive couplings from the inside of the VFI skin-effect model. At least for thin conductors, the approach can be less costly than the VFI model, since many couplings are ignored. The solution of the current diffusion problem inside the conductor is fundamentally different from the external PEEC propagation model. For the thin conductor problem, the skin-effect model must represent the EXP current penetration on opposite surfaces. Hence, the mesh cells on both side surfaces need to be lined up and can be coupled",
            "51), a cEFIE for only the equivalent current Jse can be obtained. The interesting result is that we obtain an equivalent scalar impedance Zs for the equivalent electric current as[ t\u0302 \u22c5 1E(r, r\u2032) \u2212 t\u0302 \u22c51E(t\u0302 \u22c5 2E)\u22121 t\u0302 \u22c52E ] Jse(r\u2032) = \u2212t\u0302 \u22c5 Einc(r) (9.52) or [ t\u0302 \u22c5 1E(r, r\u2032) \u2212 Zs(r, r\u2032)] Jse(r\u2032) = \u2212t\u0302 \u22c5 Einc(r) (9.53) which is expanded j\ud835\udf14\ud835\udf071 t\u0302 \u22c5 \u222bS ( I + \u2207\u2207 \ud835\udefd 2 1 ) g1(r, r\u2032) \u22c5 Jse(r\u2032) d \u2032 \u2212 Zs(r, r\u2032)Jse(r\u2032) = \u2212t\u0302 \u22c5 Einc(r), (9.54) where Zs is the GIBC that can be implemented in a PEEC model as shown in Fig. 9.3, Zs(r, r\u2032)Jse(r\u2032) = t\u0302 \u22c51E(t\u0302 \u22c5 2E)\u22121 t\u0302 \u22c52E(r, r\u2032)Jse(r\u2032) = t\u0302 \u22c5 \u222bS d \u2032\u2032\u2032 \u2207g1(r, r\u2032\u2032\u2032) \u00d7 [ t\u0302 \u22c5 j\ud835\udf14\ud835\udf072\u222bS ( I + \u2207\u2032\u2032\u2032\u2207\u2032\u2032\u2032 \ud835\udefd 2 2 ) g2(r\u2032\u2032\u2032, r\u2032\u2032)\u22c5 ]\u22121 d \u2032\u2032 t\u0302 \u22c5 \u222bS \u2207\u2032\u2032 g2(r\u2032\u2032, r\u2032) \u00d7 Jse(r\u2032) d \u2032 . (9.55) It is noted that the operator inverse (t\u0302 \u22c5 2E)\u22121 is evaluated through the multifrontal method [52]. The multifrontal method is a direct method for solving the linear equation inversion. It organizes operations during the factorization of sparse matrices and uses the low rank condition. A tree is employed to handle the dependency between partial factorizations"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001579_joe.2016.2569218-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001579_joe.2016.2569218-Figure1-1.png",
        "caption": "Fig. 1. Geometrical illustration of LOS guidance.",
        "texts": [
            " For an MSV located at (x, y), the along-track error xe and the cross-track error ye can be expressed in a path-tangential reference frame as follows: [ xe ye ] = [ cos \u03b1k \u2212 sin \u03b1k sin \u03b1k cos \u03b1k ]T [ x \u2212 xk (\u03c9) y \u2212 yk (\u03c9) ] . (3) Taking the time derivative of xe , we have x\u0307e = x\u0307 cos \u03b1k + y\u0307 sin \u03b1k \u2212 x\u0307k (\u03c9) cos \u03b1k \u2212 y\u0307k (\u03c9) sin \u03b1k + \u03b1\u0307k [\u2212 (x \u2212 xk (\u03c9)) sin \u03b1k + (y \u2212 yk (\u03c9)) cos \u03b1k ] \ufe38 \ufe37\ufe37 \ufe38 cross-track error ye . (4) Using (2), it follows that: x\u0307e= u cos (\u03c8 \u2212 \u03b1k ) \u2212 \u03c5 sin (\u03c8 \u2212 \u03b1k ) + \u03b1\u0307k ye \u2212 \u03c9\u0307 \u221a x \u20322 k (\u03c9) + y \u20322 k (\u03c9) cos (\u03b1k + \u03c6) = U cos (\u03c8 \u2212 \u03b1k + \u03b2) + \u03b1\u0307k ye \u2212 up (5) where \u03c6 = atan2 (\u2212y\u2032 k (\u03c9) , x\u2032 k (\u03c9)) = \u2212\u03b1k ; U = \u221a u2 + \u03c52 > 0 is the speed of the MSV; \u03b2 = atan2 (\u03c5, u) represents the sideslip angle (see Fig. 1); and up is the speed of the virtual reference point expressed by up = \u03c9\u0307 \u221a x \u20322 k (\u03c9) + y \u20322 k (\u03c9) (6) which is considered as an input to stabilize xe . Similarly, taking the time derivative of ye gives y\u0307e = \u2212 x\u0307 sin\u03b1k + y\u0307 cos \u03b1k + x\u0307k (\u03c9) sin \u03b1k \u2212 y\u0307k (\u03c9) cos \u03b1k \u2212 \u03b1\u0307k [(x \u2212 xk (\u03c9)) cos \u03b1k + (y \u2212 yk (\u03c9)) sin\u03b1k ] \ufe38 \ufe37\ufe37 \ufe38 along-track error xe . (7) Substituting (2) into (7), we obtain y\u0307e= u sin (\u03c8 \u2212 \u03b1k ) + \u03c5 cos (\u03c8 \u2212 \u03b1k ) \u2212 \u03b1\u0307kxe + \u03c9\u0307 \u221a x\u2032 k (\u03c9)2 + y\u2032 k (\u03c9)2 sin (\u03b1k + \u03c6) = U sin (\u03c8 \u2212 \u03b1k + \u03b2) \u2212 \u03b1\u0307kxe "
        ],
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    },
    {
        "image_filename": "designv10_0_0002401_j.ymssp.2017.12.008-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002401_j.ymssp.2017.12.008-Figure2-1.png",
        "caption": "Fig. 2. (a) The engineering drawing of the two-stage planetary gearbox, and (b) the corresponding transmission mechanism chart. Note: In the figure, A denotes the sun gear in the first gear stage, B denotes the planet gear in the first gear stage, C denotes the ring gear, D denotes the sun gear in the second gear stage, E-F denotes the planet-planet mesh pair in the second gear stage, and G denotes the carrier.",
        "texts": [
            " The organization of this paper is as follows: Section 2 presents the studied planetary gearbox; Section 3 describes the fundamentals of AMMFmethod; Section 4 gives the detailed steps of MHPE method and validates the superiority of MHPE using simulation signals; Section 5 introduces the LS for the feature ranking. Section 6 illustrates the procedures of the proposed feature extraction method based on AMMF and MHPE. Section 7 discusses experimental test results obtained using the proposed method. Finally, conclusions are drawn in Section 8. In this paper, we conduct our research on a two-stage compound planetary gearbox as shown in Fig. 1. To further show its special characteristics, Fig. 2 illustrates the engineering drawing and the transmission mechanism of the two-stage planetary gearbox. We can find that the gear set consists of two gear stages, which are connected via a common carrier G. The ring gears are made into one part denoted as C. The first gear stage is a basic planetary gear set with three planets denoted by B. Each planet is in mesh with the sun gear A and the left part of the ring gear C. The second gear stage is a meshedplanet planetary gear set and has three planet-planet mesh pairs E-F"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.45-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.45-1.png",
        "caption": "FIGURE 7.45. The input and output angles of the two 4-bar linkages.",
        "texts": [],
        "surrounding_texts": [
            "Equation (6.1) that is repeated below, provides the angle \u03b84 as a function of \u03b82. \u03b84 = 2 tan \u22121 \u00c3 \u2212B \u00b1 \u221a B2 \u2212 4AC 2A ! (7.162) A = J3 \u2212 J1 + (1\u2212 J2) cos \u03b82 (7.163) B = \u22122 sin \u03b82 (7.164) C = J1 + J3 \u2212 (1 + J2) cos \u03b82 (7.165) 7. Steering Dynamics 431 J1 = d1 a1 (7.166) J2 = d1 c1 (7.167) J3 = a21 \u2212 b21 + c21 + d21 2a1c1 (7.168) J4 = d1 b1 (7.169) J5 = c21 \u2212 d21 \u2212 a21 \u2212 b21 2a1b1 (7.170) The same equation (7.162) can be used to connect the input-output angles of the right four-bar linkage. \u03d54 = 2 tan \u22121 \u00c3 \u2212B \u00b1 \u221a B2 \u2212 4AC 2A ! (7.171) A = J3 \u2212 J1 + (1\u2212 J2) cos\u03d52 (7.172) B = \u22122 sin\u03d52 (7.173) C = J1 + J3 \u2212 (1 + J2) cos\u03d52 (7.174) J1 = d2 a2 (7.175) J2 = d2 c2 (7.176) J3 = a22 \u2212 b22 + c22 + d22 2a2c2 (7.177) J4 = d2 b2 (7.178) J5 = c22 \u2212 d22 \u2212 a22 \u2212 b22 2a2b2 (7.179) Starting with a guess value for x, we are able to calculate the length of the links. Using Equations (7.162) and (7.171), along with (7.160) and (7.161), we calculate \u03b42 for a given value of \u03b41. Let\u2019s start with x = 0, then a1 = 0.37978m b1 = 0.89043m (7.180) c1 = 0.22m a2 = 0.22m b2 = 0.89043m (7.181) c2 = 0.37978m. 432 7. Steering Dynamics Using Equations (7.160) and (7.162), we may calculate the output of the first four-bar linkage, \u03b84, for a range of the left steer angle \u221215 deg < \u03b41 < 15 deg. The following constraint, provides the numerical values for \u03d52 to be used as the input of the right four-bar linkage. \u03d52 = \u03b84 \u2212 2 tan\u22121 x 0.44 (7.182) Then, using Equations (7.171) and (7.162), we can calculate the steer angle \u03b42 for the right wheel. Figure 7.46 depicts the numerical values of the steer angles \u03b42 and \u03b4Ac versus \u03b41. The angle \u03b4Ac is the steer angle of the right wheel based on the Ackerman equation (7.154). Having \u03b42 and \u03b4Ac, we calculate the difference \u2206 \u2206 = \u03b42 \u2212 \u03b4Ac (7.183) for n different values of \u03b41 in the working range angle \u221215 deg < \u03b41 < 15 deg. Based on the n numbers for \u2206, we may find the error e. e = r \u22062 n (7.184) Changing the value of x and recalculating e, results an error function e = e(x). Figure 7.47 illustrates the result of the calculation. It shows that the error is minimum for x = \u22120.824m, which is the best length for the base of the triangle PBC. The behavior of the multi-link steering mechanism for different values of x, is shown in Figure 7.48. The Ackerman condition is also plotted to 7. Steering Dynamics 433 434 7. Steering Dynamics compare with the optimal multi-link mechanism. The optimality of x = \u22120.824m may be more clear in Figure 7.49 that shows the difference \u2206 = \u03b42 \u2212 \u03b4Ac for different values of x. The optimal multi-link steering mechanism along with the length of its links is shown in Figure 7.50. The mechanism and the meaning of negative value for x are shown in Figure 7.51 where the mechanism is in a positive turning position. 7.7 F Trailer-Truck Kinematics Consider a car pulling a one-axle trailer, as shown in Figure 7.52. We may normalize the dimensions such that the length of the trailer is 1. The 7. Steering Dynamics 435 positions of the car at the hinge point and the trailer at the center of its axle are shown by vectors r and s. Assuming r is a given differentiable function of time t, we would like to examine the behavior of the trailer by calculating s, and predict jackknifing. When the car is moving forward, we say the car and trailer are jackknifed if r\u0307 \u00b7 z < 0 (7.185) where z = r\u2212 s. (7.186) A jackknifed configuration is shown in Figure 7.53, while Figure 7.52 is showing an unjackknifed configuration. Mathematically, we want to know if the truck-trailer will jackknife for a 436 7. Steering Dynamics given path of motion r = r(t) and what conditions we must impose on r(t) to prevent jackknifing. The velocity of the trailer can be expressed by s\u0307 = c (r\u2212 s) (7.187) where c = r\u0307 \u00b7 z (7.188) and the unjackknifing condition is c > 0. (7.189) Assume the twice continuously differentiable function r is the path of car motion. If |z| = 1, and r has a radius of curvature R(t) > 1, and r\u0307(0) \u00b7 z(0) > 0 (7.190) then r\u0307(t) \u00b7 z(t) > 0 (7.191) for all t > 0. Therefore, if the car is moving forward and the car-trailer combination is not originally jackknifed, then it will remain unjackknifed. Proof. The normalized trailer length is 1 and is constant, therefore, z is a unit vector |z| = |r\u2212 s| = 1 (7.192) 7. Steering Dynamics 437 and (r\u2212 s) \u00b7 (r\u2212 s) = 1. (7.193) The nonslip wheels of the trailer constrain the vector s such that its velocity vector s\u0307 must be directed along the trailer axis indicated by z. s\u0307 = c (r\u2212 s) = cz (7.194) Differentiating (7.193) yields 2 (r\u0307\u2212 s\u0307) \u00b7 (r\u2212 s) = 0 (7.195) r\u0307 \u00b7 (r\u2212 s) = s\u0307 \u00b7 (r\u2212 s) (7.196) and therefore, r\u0307 \u00b7 (r\u2212 s) = c (r\u2212 s) \u00b7 (r\u2212 s) (7.197) c = r\u0307 \u00b7 (r\u2212 s) = r\u0307 \u00b7 z. (7.198) Having c enables us to write Equation (7.194) as s\u0307 = [r\u0307 \u00b7 (r\u2212 s)] (r\u2212 s) = (r\u0307 \u00b7 z) z. (7.199) There are three situations 1. When c > 0, the velocity vector of the trailer s\u0307 is along the trailer axis z. The trailer follows the car and the system is stable. 2. When c = 0, the velocity vector of the trailer s\u0307 is zero. In this case, the trailer spins about the center of its axle and the system is neutralstable. 3. When c < 0, the velocity vector of the trailer s\u0307 is along the trailer axis \u2212z. The trailer does not follow the car and the system is unstable. Using a Cartesian coordinate expression, we may show the car and trailer position vectors by r = \u2219 xc yc \u00b8 (7.200) s = \u2219 xt yt \u00b8 (7.201) 438 7. Steering Dynamics and therefore, s\u0307 = \u2219 x\u0307t y\u0307t \u00b8 = [r\u0307 \u00b7 (r\u2212 s)] (r\u2212 s) = \u2219 x\u0307c (xc \u2212 xt) 2 + (xc \u2212 xt) (yc \u2212 yt) y\u0307c x\u0307c (xc \u2212 xt) (yc \u2212 yt) + (yc \u2212 yt) 2 y\u0307c \u00b8 (7.202) c = (xc \u2212 xt) x\u0307c + (yc \u2212 yt) y\u0307c = x\u0307cxc + y\u0307cyc \u2212 (x\u0307cxt + y\u0307cyt) . (7.203) Let\u2019s define a function f(t) = r\u0307 \u00b7 z and assume that conclusion (7.191) is wrong while assumption (7.190) is correct. Then there exists a time t1 > 0 such that f(t1) = 0 and f 0(t1) \u2264 0. Using |z| = 1 and r\u0307 6= 0, we have r\u0307(t1) \u00b7 z(t1) = 0 and therefore, r\u0307(t1) is perpendicular to z(t1). The derivative f 0(t) would be f 0(t) = r\u0308 \u00b7 z+ r\u0307 \u00b7 z\u0307 = r\u0308 \u00b7 z+ r\u0307 \u00b7 (r\u0307\u2212 s\u0307) = r\u0308 \u00b7 z+ |r\u0307|2 \u2212 r\u0307 \u00b7 s\u0307 = r\u0308 \u00b7 z+ |r\u0307|2 \u2212 r\u0307 \u00b7 ((r\u0307 \u00b7 z) z) = r\u0308 \u00b7 z+ |r\u0307|2 \u2212 (r\u0307 \u00b7 z)2 = r\u0308 \u00b7 z+ |r\u0307|2 \u2212 f2(t) (7.204) and therefore, f 0(t1) = r\u0308 \u00b7 z+ |r\u0307|2 . (7.205) The acceleration r\u0308 in a normal-tangential coordinate frame (e\u0302n, e\u0302t) is r\u0308 = d |r\u0307| dt e\u0302t + \u03ba |r\u0307|2 e\u0302n (7.206) \u03ba = 1 R (7.207) where e\u0302n and e\u0302t are the unit normal and tangential vectors. e\u0302t is parallel to r\u0307(t1), and e\u0302n is parallel to z(t1). Hence, r\u0308 \u00b7 z = \u00b1\u03ba(t1) |r\u0307(t1)|2 (7.208) and f 0(t1) = |r\u0307(t1)|2 \u00b1 \u03ba(t1) |r\u0307(t1)|2 = [1\u00b1 \u03ba(t1)] |r\u0307(t1)|2 . (7.209) Because \u03ba(t1) = 1/R(t) > 0, we conclude that f 0(t1) > 0, and it is not possible to have f 0(t1) \u2264 0. Example 296 F Straight motion of the car with constant velocity. Consider a car moving forward in a straight line with a constant velocity. We may use a normalization and set the speed of the car as 1 moving in positive x direction starting from x = 0. Using a two-dimensional vector expression we have r = \u2219 xc yc \u00b8 = \u2219 t 0 \u00b8 . (7.210) Because of (7.192), we get z(0) = r(0)\u2212 s(0) = \u2212s(0) (7.211) and therefore, the initial position of the trailer must lie on a unit circle as shown in Figure 7.54. Using two dimensional vectors, we may express z(0) as a function of \u03b8 z(0) = \u2212s(0) = \u2219 xt(0) yt(0) \u00b8 = \u2219 cos \u03b8 sin \u03b8 \u00b8 (7.212) and simplify Equation (7.202) as s\u0307 = \u2219 x\u0307t y\u0307t \u00b8 = \u2219 (t\u2212 xt) 2 \u2212yt (t\u2212 xt) \u00b8 . (7.213) 440 7. Steering Dynamics Equation (7.213) is a set of two coupled first-order ordinary differential equations with the solution s = \u2219 xt yt \u00b8 = \u23a1\u23a2\u23a2\u23a3 t+ e\u22122t \u2212 C1 e\u22122t + C1 C2e \u2212t e\u22122t + C1 \u23a4\u23a5\u23a5\u23a6 . (7.214) Applying the initial conditions (7.212) we find C1 = cos \u03b8 \u2212 1 cos \u03b8 + 1 (7.215) C2 = 2 sin \u03b8 cos \u03b8 + 1 . (7.216) If \u03b8 6= k\u03c0, then the solution depends on time, and when time goes to infinity, the solution leads to the following limits asymptotically: lim t\u2192\u221e xt = t\u2212 1 lim t\u2192\u221e yt = 0 (7.217) When the car is moving with a constant velocity, this solution shows that the trailer will approach the position of straight forward moving, following the car. We may also consider that the car is backing up. In this situation, the solution shows that, except for the unstable initial condition \u03b8 = \u03c0, all solutions ultimately approach the jackknifed position. If \u03b8 = 0, then C1 = 0 C2 = 0 xt = t+ 1 yt = 0 (7.218) and the trailer moves in an unstable configuration. Any deviation from \u03b8 = 0 ends up to change the situation and leads to the stable limiting solution (7.217). If \u03b8 = \u03c0, then C1 = \u221e C2 = \u221e xt = t\u2212 1 yt = 0 (7.219) and the trailer follows the car in an stable configuration. Any deviation from \u03b8 = 0 will disappear after a while. 7. Steering Dynamics 441 Example 297 F Straight car motion with different initial \u03b8. Consider a car moving on an x-axis with constant speed. The car is pulling a trailer, which is initially at \u03b8 such as shown in Figure 7.52. Using a normalized length, we assume the distance between the center of the trailer axle and the hinge is the length of trailer, and is equal to 1. If we show the absolute position of the car at hinge by r = \u00a3 xc yc \u00a4T and the absolute position of the trailer by s = \u00a3 xt yt \u00a4T then the position of the trailer is a function of the car\u2019s motion. When the position of the car is given by a time-dependent vector function r = \u2219 xc(t) yc(t) \u00b8 (7.220) the trailer position can be found by solving two coupled differential equation. x\u0307t = (xc \u2212 xt) 2 x\u0307c + (xc \u2212 xt) (yc \u2212 yt) y\u0307c (7.221) y\u0307t = (xc \u2212 xt) (yc \u2212 yt) x\u0307c + (yc \u2212 yt) 2 y\u0307c (7.222) For a constantly uniform car motion r = \u00a3 t 0 \u00a4T , Equations (7.221) and (7.222) reduce to x\u0307t = (t\u2212 xt) 2 (7.223) y\u0307t = \u2212yt (t\u2212 xt) (7.224) The first equation (7.223) is independent of the second equation (7.224) and can be solved independently. xt = C1 e 2t (t\u2212 1)\u2212 t\u2212 1 C1 e2t \u2212 1 = t+ e\u22122t \u2212 C1 e\u22122t + C1 (7.225) Substituting Equation (7.225) in (7.224) generates the following differential equation: y\u0307t = e\u22122t \u2212 C1 e\u22122t + C1 yt (7.226) with the solution yt = C2e \u2212t e\u22122t + C1 . (7.227) When the trailer starts from s = \u00a3 cos \u03b8 sin \u03b8 \u00a4T , the constants of integral would be equal to Equations (7.215) and (7.216) and therefore, xt = t+ e\u22122t (cos \u03b8 + 1)\u2212 cos \u03b8 + 1 e\u22122t (cos \u03b8 + 1) + cos \u03b8 \u2212 1 (7.228) yt = 2e\u2212t sin \u03b8 e\u22122t (cos \u03b8 + 1) + cos \u03b8 \u2212 1 . (7.229) 442 7. Steering Dynamics Figures 7.55, 7.56, and 7.57 illustrate the behavior of the trailer starting from \u03b8 = 45deg, \u03b8 = 90deg, \u03b8 = 135deg. Example 298 F Circular motion of the car with constant velocity. Consider a car pulling a trailer such as Figure 7.52 shows. The car is traveling along a circle of radius R > 1, based on a normalized length in which the length of the trailer is 1. In a circular motion with a normalized angular velocity \u03c9 = 1 and period T = 2\u03c0, the position of the car is given by the following time-dependent vector function: r = \u2219 xc(t) yc(t) \u00b8 = \u2219 R cos(t) R sin(t) \u00b8 . (7.230) The initial position of the trailer must lie on a unit circle with a center at r(0) = \u00a3 xc(0) yc(0) \u00a4T . s(0) = \u2219 xt(0) yt(0) \u00b8 = \u2219 xc(0) yc(0) \u00b8 + \u2219 cos \u03b8 sin \u03b8 \u00b8 (7.231) The car-trailer combination approaches a steady-state configuration as shown in Figure 7.58. Substituting r\u0307 = \u2219 \u2212R sin(t) R cos(t) \u00b8 . (7.232) 7. Steering Dynamics 443 444 7. Steering Dynamics and the initial conditions (7.231) in (7.202) will generate two differential equations for trailer position. x\u0307t = R (R cos t\u2212 xt) (xt sin t\u2212 yt cos t) (7.233) y\u0307t = R (R sin t\u2212 yt) (xt sin t\u2212 yt cos t) (7.234) Assuming r(0) = \u00a3 0 0 \u00a4T the steady-state solutions of these equations are xt = c cos(t\u2212 \u03b1) (7.235) yt = c sin(t\u2212 \u03b1) (7.236) where c is the trailer\u2019s radius of rotation, and \u03b1 is the angular position of the trailer behind the car. c = p R2 \u2212 1 (7.237) sin\u03b1 = 1 R (7.238) cos\u03b1 = c R (7.239) 7. Steering Dynamics 445 We may check the solution by employing two new variables, u and v, such that u = xt sin t\u2212 yt cos t (7.240) v = xt cos t+ yt sin t (7.241) and u\u0307 = v. (7.242) Using the new variables we find xt = u sin t+ v cos t (7.243) yt = \u2212u cos t+ v sin t (7.244) x\u0307t = Ru(R cos t\u2212 u sin t\u2212 v cos t) (7.245) y\u0307t = Ru(R sin t+ u cos(t)\u2212 v sin t). (7.246) Direct differentiating from (7.240), (7.241), (7.243), and (7.244) shows that u\u0307 = xt cos t+ yt sin t+ x\u0307t (sin t)\u2212 y\u0307t (cos t) (7.247) v\u0307 = \u2212xt sin t+ yt cos t+ x\u0307t cos t+ y\u0307t sin t (7.248) x\u0307t = u\u0307 sin t+ v\u0307 cos t+ u cos t\u2212 v sin t (7.249) y\u0307t = u\u0307 cos t\u2212 v\u0307 sin t\u2212 u sin t\u2212 v cos t (7.250) and therefore, the problem can be expressed in a new set of equations. u\u0307 = v \u2212Ru2 (7.251) v\u0307 = u \u00a1 R2 \u2212Rv \u2212 1 \u00a2 (7.252) At the steady-state condition the time differentials must be zero, and therefore, the steady-state solutions would be the answers to the following algebraic equations: v \u2212Ru2 = 0 (7.253) u \u00a1 R2 \u2212Rv \u2212 1 \u00a2 = 0 (7.254) There are three sets of solutions. {u = 0, v = 0} (7.255) {u = c R , v = c2 R } (7.256) {u = \u2212 c R , v = c2 R } (7.257) 446 7. Steering Dynamics The first solution is associated with s = 0, xt = 0 (7.258) yt = 0 (7.259) which shows that the center of the trailer\u2019s axle remains at the origin and the car is turning on a circle R = 1. This is a stable motion. The second solution is associated with xt = c R sin t+ c2 R cos t (7.260) yt = \u2212 c R cos t+ c2 R sin t (7.261) which are equivalent to (7.235) and (7.236). To examine the stability of the second solution, we may substitute a perturbed solution u = c R + p (7.262) v = c2 R + q (7.263) in the linearized equations of motion (7.251) and (7.252) at the second set of solutions u\u0307 = v \u2212 2cu (7.264) v\u0307 = \u2212cv (7.265) to get two equations for the perturbed functions p and q. p\u0307 = q \u2212 2cp (7.266) q\u0307 = cq (7.267) The set of linear perturbed equations can be set in a matrix form\u2219 p\u0307 q\u0307 \u00b8 = \u2219 \u22122c 1 0 \u2212c \u00b8 \u2219 p q \u00b8 . (7.268) The stability of Equation (7.268) is determined by the eigenvalues \u03bbi of the coefficient matrix, which are \u03bb1 = \u2212c \u03bb2 = \u22122c. Because both eigenvalues \u03bb1 and \u03bb2 are negative, the solution of the perturbed equations symptomatically goes to zero. Therefore, the second set of 7. Steering Dynamics 447 solutions (7.256) is stable and it absorbs any near path that starts close to it. The third solution is associated with xt = \u2212 c R sin t+ c2 R cos t (7.269) yt = c R cos t+ c2 R sin t. (7.270) The linearized equations of motion at the third set of solutions (7.257) are u\u0307 = v + 2cu (7.271) v\u0307 = cv. (7.272) The perturbed equations would then be\u2219 p\u0307 q\u0307 \u00b8 = \u2219 2c 1 0 c \u00b8 \u2219 p q \u00b8 (7.273) with two positive eigenvalues \u03bb1 = c \u03bb2 = 2c Positive eigenvalues show that the solution of the perturbed equations diverges and goes to infinity. Therefore, the third set of solutions (7.257) is unstable and repels any near path that starts close to it. 7.8 Summary Steering is required to guide a vehicle in a desired direction. When a vehicle turns, the wheels closer to the center of rotation are called the inner wheels, and the wheels further from the center of rotation are called the outer wheels. If the speed of a vehicle is very slow, there is a kinematic condition between the inner and outer steerable wheels, called the Ackerman condition. Street cars are four-wheel vehicles and usually have front-wheel-steering. The kinematic condition between the inner and outer steered wheels is cot \u03b4o \u2212 cot \u03b4i = w l (7.274) where \u03b4i is the steer angle of the inner wheel, \u03b4o is the steer angle of the outer wheel, w is the track, and l is the wheelbase of the vehicle. Track w and wheel base l are considered the kinematic width and length of the vehicle. 448 7. Steering Dynamics The mass center of a steered vehicle will turn on a circle with radius R, R = q a22 + l2 cot2 \u03b4 (7.275) where \u03b4 is the cot-average of the inner and outer steer angles. cot \u03b4 = cot \u03b4o + cot \u03b4i 2 . (7.276) The angle \u03b4 is the equivalent steer angle of a bicycle having the same wheelbase l and radius of rotation R. 7. Steering Dynamics 449 7.9 Key Symbols 4WS four-wheel-steering a, b, c, d lengths of the links of a four-bar linkage ai distance of the axle number i from the mass center A,B,C input angle parameters of a four-bar linkage AWS all-wheel-steering b1 distance of the hinge point from rear axle b2 distance of trailer axle from the hinge point c stability index of a trailer motion c trailer\u2019s radius of rotation c1 longitudinal distance of turn center and front axle of a 4WS car c2 longitudinal distance of turn center and rear axle of a 4WS car C mass center C1, C2, \u00b7 \u00b7 \u00b7 constants of integration d arm length in trapezoidal steering mechanism e error e length of the offset arm FWS front-wheel-steering g overhang distance J link parameters of a four-bar linkage l wheelbase n number of increments O center of rotation in a turn r yaw velocity of a turning vehicle r position vector of a car at the hinge R radius of rotation at mass center R1 radius of rotation at the center of the rear axle for FWS R1 horizontal distance of O and the center of axles Rt radius of rotation at the center of the trailer axle Rw radius of the rear wheel RWS rear wheel steering s position vector of a trailer at the axle center t time uR steering rack translation v \u2261 x\u0307, v vehicle velocity vri speed of the inner rear wheel vro speed of the outer rear wheel w track wf front track wr rear track x, y, z, x displacement z = r\u2212 s position vector of a trailer relative to the car 450 7. Steering Dynamics \u03b2 arm angle in trapezoidal steering mechanism \u03b4 cot-average of the inner and outer steer angles \u03b41 = \u03b4fl front left wheel steer angle \u03b42 = \u03b4fr front right wheel steer angle \u03b4Ac steer angle based on Ackerman condition \u03b4fl front left wheel steer angle \u03b4fr front right wheel steer angle \u03b4i inner wheel \u03b4rl rear left wheel steer angle \u03b4rr rear right wheel steer angle \u03b4o outer wheel \u03b4S steer command \u2206 = \u03b42 \u2212 \u03b4Ac steer angle difference \u03b8 angle between trailer and vehicle longitudinal axes \u03c9 angular velocity \u03c9i = \u03c9ri angular velocity of the rear inner wheel \u03c9o = \u03c9ro angular velocity of the rear outer wheel 7. Steering Dynamics 451 Exercises 1. Bicycle model and radius of rotation. Mercedes-Benz GL450TM has the following dimensions. l = 121.1 in wf = 65.0 in wr = 65.1 in R = 39.7 ft Assume a1 = a2 and use an average track to determine the maximum steer angle \u03b4 for a bicycle model of the car. 2. Radius of rotation. Consider a two-axle truck that is offered in different wheelbases. l = 109 in l = 132.5 in l = 150.0 in l = 176.0 in If the front track of the vehicles is w = 70 in and a1 = a2, calculate the radius of rotations if \u03b4 = 30deg. 3. Required space. Consider a two-axle vehicle with the following dimensions. l = 4m w = 1.3m g = 1.2m Determine Rmin, RMax, and \u2206R for \u03b4 = 30deg. 4. Rear wheel steering lift truck. A battery powered lift truck has the following dimensions. l = 55 in w = 30 in Calculate the radius of rotations if \u03b4 = 55deg for a1 = a2. 452 7. Steering Dynamics 5. Wheel angular velocity. Consider a two-axle vehicle with the following dimensions. l = 2.7m w = 1.36m What is the angular velocity ratio of \u03c9o/\u03c9i? 6. A three-axle vehicle. A three-axle vehicle is shown in Figure 7.17. Find the relationship between \u03b42 and \u03b43, and also between \u03b41 and \u03b46. 7. A three-axle truck. Consider a three-axle truck that has only one steerable axle in front. The dimensions of the truck are a1 = 5300mm a2 = 300mm a3 = 1500mm w = 1800mm. Determine maximum steer angles of the front wheels if the truck is supposed to be able to turn with R = 11m. 8. A vehicle with a one-axle trailer. Determine the angle between the trailer and vehicle with the following dimensions. a1 = 1000mm a2 = 1300mm wv = 1500mm b1 = 1200mm b2 = 1800mm wt = 1100mm g = 800mm \u03b4i = 12deg . What is the rotation radius of the trailer Rt, and the vehicle R? Determine minimum radius Rmin, maximum RMax, and difference radius \u2206R? 7. Steering Dynamics 453 9. F Turning radius of a 4WS vehicle. Consider a FWS vehicle with the following dimensions. l = 2300mm wf = 1457mm wr = 1607mm a1 a2 = 38 62 Determine the turning radius of the vehicle for \u03b4fl = 5deg. What should be the steer angles of the front and rear wheels to decrease 10% of the turning radius, if we make the vehicle 4WS? 10. F Coordinates of the turning center. Determine the coordinates of the turning center for the vehicle in Exercise 9 if \u03b4fl = 5deg and c1 = 1300mm. 11. F Different front and rear tracks. Lotus 2-ElevenTM is a RWD sportscar with the following specifications. l = 2300mm wf = 1457mm wr = 1607mm Front tire = 195/50R16 Rear tire = 225/45R17 Fz1 Fz2 = 38 62 Determine the angular velocity ratio of \u03c9o/\u03c9i, R, \u03b4i, and \u03b4o for \u03b4 = 5deg. 12. F Coordinates of turning center. Determine the coordinates of turning center of a 4WS vehicle in terms of outer steer angles \u03b4of and \u03b4or. 13. F Turning radius. Determine the turning radius of a 4WS vehicle in terms of \u03b4r. cot \u03b4r = 1 2 (cot \u03b4ir + cot \u03b4or) 14. F A three-axle car. 454 7. Steering Dynamics Consider a three-axle off-road pick-up car. Assume a1 = 1100mm a2 = 1240mm a3 = 1500mm w = 1457mm and determine \u03b4o, R1, Rf , and R if \u03b4i = 10deg. 15. F Steering mechanism optimization. Find the optimum length x for the multi-link steering mechanism shown in Figure 7.59 to operate as close as possible to the kinematic steering condition. The vehicle has a track w = 2.64m and a wheelbase l = 3.84m, and must be able to turn the front wheels within a working range equal to \u221222 deg \u2264 \u03b4 \u2264 22 deg. 8"
        ]
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure5.38-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure5.38-1.png",
        "caption": "Figure 5.38 The support reactions of the frame as they really act.",
        "texts": [
            " Determine all the forces acting on bar SD. Solution: a. You will recognise a three-hinged frame in the structure. There are two self-contained parts that are connected in a hinge at S and are supported by hinges at A and B (see Figure 5.37b). The structure in Figure 5.37a is therefore also referred to as a shored three-hinged frame. The support reactions can be derived in the standard way for a three-hinged frame (see Section 5.3). The calculation, which will be left to the reader, leads to the support reactions shown in Figure 5.38. b. The shoring bars are loaded only by forces at the end of the bars and therefore act as two-force members. Suppose that a tensile force N(1) acts in the left shoring bar (1) and a tensile force N(2) in the right shoring bar (2). In Figure 5.39, AC and BD have been isolated. The unknown interaction forces at C and D are not shown here. 1 Since the vertical weight causes tension in these bars. 5 Calculating Support Reactions and Interaction Forces 179 It is now possible to deduce N(1) from the moment equilibrium of AC about C: \u2211 T (AC) z |C = +(40 \u221a 2 kN)(2 \u221a 2 m) + N(1) \u00d7 ( \u221a 2 m) = 0 so that N(1) = \u221280 \u221a 2 kN"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure9.14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure9.14-1.png",
        "caption": "Fig. 9.14. GM\u2019s AFPM wheel motor: (a) Section of the AFPM motor showing details of slotted stator, (b) Basic layout of the AFPM motor showing complete stator located between two PM rotor discs, (c) AFPM wheel motor mounted on GM S-10 show truck. Courtesy of General Motors Corporation, U.S.A.",
        "texts": [
            " The two stators are directly attached to 292 9 Applications the vehicle body whilst the PM rotor is free to move in radial directions. It can be observed that in this case the wheel and disc rotor form the unsprung mass, whilst the stators of the motor become sprung mass supported on the chassis [101]. Given the advantages of using wheel motor in electric vehicle drive system such as more onboard space, better vehicle control and modular drive train topology, there is a growing interest in automotive industry to develop wheel motors for future electric vehicles. Figure 9.14 shows the AFPM wheel motor developed at General Motors Advanced Technology Center, Torrance, CA, U.S.A. The developed wheel hub motor has a single stator located between two PM rotor discs. This topology features higher torque density design as it utilizes copper from both surfaces of the stator for torque production [66, 199, 223]. To remove heat from inside stator winding, provision is made of an aluminum ring with internal liquid coolant passages along end winding of the outer stator winding [P149]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001777_j.optlastec.2016.04.009-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001777_j.optlastec.2016.04.009-Figure2-1.png",
        "caption": "Fig. 2. The established three-dimensional finite element model (a) and laser scan strategy (b) during SLM process (Point 1, Point 2 and Point 3 at the center of the 1st scan track, 2nd scan track and 4th scan track, respectively).",
        "texts": [
            " These three coupled heat transfer mechanisms make the thermal behavior of SLM process become extremely complex [20]. 2.2. Basic setup of the finite element model The numerical simulation is carried out using ANSYS multiphysics finite element package to obtain a thorough understanding of the temperature evolution behavior and resultant phenomenon like temperature gradient, metallurgical defects and deformation during SLM process. The established three-dimensional finite model and laser scan strategy during SLM process are shown in Fig. 2. The dimensions of the computational model of TiC/Inconel 718 powder bed are 1.4 mm 0.385 mm 0.1 mmwith two layers. The height of each power layer is 50 mm and, the C45 medium carbon steel block with the dimensions of 1.7 mm 0.7 mm 0.3 mm is taken as the metal substrate beneath the powder layers. The coordinate origin of this physical model lies at the left bottom of the second powder layer as shown in Fig. 2(a) and, its X-axis, Y-axis and Z-axis are located along the laser scan direction, perpendicular to the laser scan direction and from top surface to bottom of the powder layers. Considering the computational precision and simulation efficiency, the ANSYS Solid70 hexahedron element with the fine mesh of 0.0175 mm 0.0175 mm 0.025 mm is utilized in the powder bed, while, a relatively coarse tetrahedron mesh is adopted in the substrate. The three-dimensional simulation model is meshed into 11,745 nodes and 25,532 elements in all",
            " In order to mimic the cooling time involved during the delivery of each powder layer, there is a 10-ms cooling time between the fabrications of neighbor layers. This time is converted by the real time during powder delivery of actual laser processing according to the ratio of the top surface area of the physical model and the area of the real powder delivery during previous experiments. Every calculating step is divided into two smaller sub-steps for improving calculation precision. Each powder layer is processed track by track in a reciprocating raster pattern as shown in Fig. 2(b). Point 1, Point 2 and Point 3 lie at the center of the first scan track, second scan track and fourth scan track, respectively. Path A-B locates on the top surface of the first powder layer along the Y direction at X\u00bc0.7 mm and Z\u00bc0.05 mm. The applied laser processing parameters are listed in Table 1. In selective laser melting, the localized heating of the powder bed by the laser beam results in a heat transfer in the material dominated by conductive heat transfer. The spatial and temporal distribution of the temperature field satisfies the heat conduction equation, which can be expressed as [22]: \u03c1 \u2202 \u2202 = \u2202 \u2202 \u2202 \u2202 + \u2202 \u2202 \u2202 \u2202 + \u2202 \u2202 \u2202 \u2202 + ( ) \u2022\u239b \u239d\u239c \u239e \u23a0\u239f \u239b \u239d\u239c \u239e \u23a0\u239f \u239b \u239d\u239c \u239e \u23a0\u239fC T t x k T x y k T y z k T z Q 1 p where T is the temperature of the powder system, t is the interaction time between laser beam and powder bed, (x, y, z) are the spatial co-ordinates, k is the effective thermal conductivity of powder bed, \u03c1 is the material density, Cp is the specific heat capacity and, Q is the heat generated per volume within the component",
            " The cross-sectional microstructures of the SLM-produced TiC/Inconel 718 parts were characterized using a PMG3 optical microscopy (Olympus Corporation, Japan). The characteristic surface morphologies of the parts were characterized by scanning electron microscopy (SEM; Hitachi model S-4800, Japan) in secondary electron mode at 10 kV. Fig. 3 shows the transient temperature distribution on the top surface and longitudinal view of the molten pool as laser beam reaches Point 1, Point 2 and Point 3, respectively, with v of 100 mm/s and P of 125 W (Fig. 2(b)). At the center of the first track (Point 1, Fig. 2(b)), the isotherm curves on the top surface of the molten pool are similar to a series of ellipses and, the ellipses of fore part are more intensive than those at the back-end of it. Meantime, it is asymmetry of isotherms along the laser scanning direction. However, this phenomenon is not conspicuous enough as shown in temperature contour pots (Fig. 3). The dashed line circle shown in the temperature contour plots presents the melting temperature of Inconel 718 (1300 \u00b0C). The temperature in this dashed line circle is higher compared with the melting point of Inconel 718, which induces a small molten pool within this region",
            " It is responsible for the attendant microcracks in SLMprocessed parts and distortion of the parts. It thus can be concluded that a SLM-fabricated part with high density and outstanding metallurgical bonding ability cannot be obtained under the above mentioned laser processing conditions. Fig. 5 shows the effects of laser processing parameters on the temperature distribution during SLM of TiC/Inconel 718 powder system along the Y-direction at X\u00bc0.7 mm and Z\u00bc0.05 mm (path A-B), and Z-direction at Point 2 (Fig. 2(b)). The slope of the curves shown in the figures presents the temperature gradient of the powder bed, and a steep slope means a relatively high temperature gradient. When the laser power is increased from 75 W to 150 W, the maximum temperature gradient along the path A-B pronouncedly increases from 1.30 104 \u00b0C/mm to 2.60 104 \u00b0C/ mm (Fig. 5(a)). The maximum temperature gradient along the Z-direction also significantly increases from 1.36 104 \u00b0C/mm to 2.62 104 \u00b0C/mm. Furthermore, the temperature trend of each curve along Z-direction is obviously divided into four parts"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002780_j.jmapro.2021.01.008-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002780_j.jmapro.2021.01.008-Figure3-1.png",
        "caption": "Fig. 3. Schematic showing the location in the deposited alloy from which samples were taken for tensile tests.",
        "texts": [
            " The entire arc additive manufacturing process was carried out at the center of the electromagnet pole face, and the uniformity of the magnetic field was ensured before the alloy deposition, i.e. the magnetic field strength was measured at different points to evaluate its uniformity. The measurements showed that the error was within the allowable range and the error tolerance was 10 % (\u00b12 mT). After WAAM deposition of the alloy without and with magnetic field, test specimens for mechanical properties evaluation were machined by wire electric discharge machine (WEDM). The test specimens were taken from identical/similar locations in the respective deposited samples (as shown in Fig. 3) to study their microstructure and mechanical properties. Upon grinding and polishing, microstructure of the deposited samples was examined using optical microscope (OM), scanning electron microscope (SEM) and energy dispersive spectrometer (EDS). The laser confocal microscope (OLS40-CB) was used to quantitatively analyze the size of the microstructure of the deposited parts. The microhardness of the deposited samples was measured by a Vicker\u2019s micro-hardness tester (HXD-1000TM) with 500 g load, applied for a duration of 10 s",
            " Journal of Manufacturing Processes 64 (2021) 10\u201319 indents were performed at each test point to ensure that the results were repeatable. Tensile tests of standard sheet specimens were carried out in a universal tensile testing machine at the loading rate of 1 mm min\u2212 1. The elongation was tested using an extensometer. Three tensile tests were performed for each condition (with magnetic field or without magnetic field) to ensure that the results were repeatable and the representative one would be selected in this study. Fig. 3 shows the dimension of the tensile test specimens (WAAM Inconel 625 samples were shown in Fig. 4). Fig. 5 shows the cross-section of the WAAM deposited Inconel 625 alloy samples deposited without and with magnetic field. It can be seen from the figure that the samples deposited under both the conditions have irregular cross-section due to (i) the high heat generated during the arc additive manufacturing process and (ii) the outward flow of the molten pool. In order to study the effect of magnetic field on the forming ability of Inconel 625 alloy, the verticality and horizontality are defined as a measure of the precision of sample forming"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-FigureC.7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-FigureC.7-1.png",
        "caption": "FIGURE C.7 Two zero thickness conductors at a 90\u2218 angle.",
        "texts": [
            "25) is Lp12 = \ud835\udf070 4\ud835\udf0b 1 (ye1 \u2212 ys1) (ye2 \u2212 ys2) 4\u2211 k=1 4\u2211 m=1 (\u22121)m+k [ b2 m \u2212 Z2 2 ak log (ak + rkm + \ud835\udf16) + a2 k \u2212 Z2 2 bm log (bm + rkm + \ud835\udf16) \u2212 1 6 (b2 m \u2212 2 Z2 + a2 k) rkm \u2212 bm Z aktan\u22121 ( ak bm rkm Z )] , (C.26) where a1 = xs2 \u2212 xe1, a2 = xe2 \u2212 xe1 (C.27) a3 = xe2 \u2212 xs1, a4 = xs2 \u2212 xs1 (C.28) b1 = ys2 \u2212 ye1, b2 = ye2 \u2212 ye1 (C.29) b3 = ye2 \u2212 ys1, b4 = ys2 \u2212 ys1 (C.30) Z = z2 \u2212 z1 + \ud835\udf16, rkm = \u221a a2 k + b2 m + Z2 , (C.31) where we added \ud835\udf16 = 10\u221237 to the equation as a very small number. C.1.6 Lp12 for Two Orthogonal Rectangular Current Sheets As is shown in Fig. C.7, two zero thickness sheets are at an angle of 90\u2218 while the current flow is in the x-direction for both sheets. This formula is again very useful. Of course, the PARTIAL INDUCTANCE FORMULAS FOR ORTHOGONAL GEOMETRIES 391 currents in the sheets as indicated in Fig. C.7 need to be parallel to each other for a nonzero partial inductance. Its original definition is Lp12 = 1 (ye1 \u2212 ys1)(ze2 \u2212 zs2) \ud835\udf070 4\ud835\udf0b \u222b ye1 ys1 \u222b xe1 xs1 \u222b ze2 zs2 \u222b xe2 xs2 1 R1,2 dx2 dz2 dx1 dy1, (C.32) where the final result for Lp12 for the two orthogonal current sheets is Lp12 = \ud835\udf070 4\ud835\udf0b 1 (ye1 \u2212 ys1) (ze2 \u2212 zs2) 4\u2211 k=1 2\u2211 m=1 2\u2211 \ud835\udcc1=1 (\u22121)\ud835\udcc1+m+k+1 [( a2 k 2 \u2212 c2 \ud835\udcc1 6 ) c\ud835\udcc1 log(bm + rkm\ud835\udcc1 + \ud835\udf16) + ( a2 k 2 \u2212 b2 m 6 ) bm log(c\ud835\udcc1 + rkm\ud835\udcc1 + \ud835\udf16) + akbmc\ud835\udcc1 log(ak + rkm\ud835\udcc1 + \ud835\udf16) \u2212 bmc\ud835\udcc1 3 rkm\ud835\udcc1 \u2212 a3 k 6 tan\u22121 ( bmc\ud835\udcc1 ak rkm\ud835\udcc1 ) \u2212 b2 mak 2 tan\u22121 ( akc\ud835\udcc1 bm rkm\ud835\udcc1 ) \u2212 akc2 \ud835\udcc1 2 tan\u22121 ( akbm c\ud835\udcc1 rkm\ud835\udcc1 )] (C"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003841_08905458708905124-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003841_08905458708905124-Figure7-1.png",
        "caption": "Fig. 7 The Stanford robot arm",
        "texts": [
            " Therefore, bodies 2 and 7 become the first parallel group in the acceleration recovery sequence. After computing accelerations for bodies 2 and 7, joint accelerations between these bodies and bodies 3 and 8 can be computed in parallel. Similarly, the third group is bodies 4, 6, and 9 and the fourth group is bodies 5 and 10. Parallel processor implementation of this recursive formulation is considered in Part 111 of this paper, to appear in this journal. BAE AND HAUG VII I . NUMERICAL EXAMPLE OF A ROBOT The Stanford arm manipulator shown in Fig. 7 is used to demonstrate use of the method developed. The mechanism consists of 6 bodies, not counting ground. Bodies 2 , 3 , 5 , 6 , and 7 are connected to their preceding bodies by revolute joints and body 4 is connected to its preceding body by a translational joint. As a result, this mechanism has 6 degrees-of-freedom. Inertia data are listed in Table 1. Inverse dynamic analysis has been performed in Ref. 18. Control torques and forces, defined at each joint in Ref. 18, are slightly modified and used as input forces for dynamic analysis in this paper"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001920_j.optlastec.2018.08.012-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001920_j.optlastec.2018.08.012-Figure3-1.png",
        "caption": "Fig. 3. Geometry and mesh used in the finite element simulation.",
        "texts": [
            " In this study, the adaptive time step algorithm was used with an initial time step of 7 \u03bcs. An adaptive time step allows users to give a reference time step, where the software changes the time step based on the convergence state during simulation. The minimum and maximum values for the time step used in numerical simulations was between 0.2 \u03bcs and 6 \u03bcs. In order to accelerate the simulations, only half of the geometry was built based on symmetry, since the domain is symmetric about the vertical plane containing the laser-beam moving line, as presented in Fig. 3. The symmetry constraint was set on the symmetric plane. In addition, radiative (Eq. (2)) and convective (Eq. (3)) heat losses at the top surface of the powder layer to the ambient air were considered. The convective heat transfer coefficient and the emissivity coefficient were chosen as 15W/mK [37] and 0.5 [38], respectively. The ambient temperature was set to 293 K. The other sides of the domain were specified as a fixed preheating temperature, 353 K. Experiments were carried out on an EOS M 290 LPBF machine"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003403_978-1-4615-5633-6-Figure2.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003403_978-1-4615-5633-6-Figure2.1-1.png",
        "caption": "Figure 2.1. Synchronous machine",
        "texts": [
            " While it is possible to model in detail, the complex structures on the turbine rotors, for SSR studies it is adequate to treat all the rotors as lumped masses interconnected by springs (shaft sections). The torsional (mechanical) system can thus be modelled as a linear, lumped mass-spring-damper network, which is analogous to a lumped, linear R-L-C (electrical)network. Modal analysis of the torsional system is used to compute the modal masses, mode shapes and resonant frequencies. It will be shown that the analysis of the electrical analogue also yields some of these results. The synchronous machine considered is shown in Fig. 2.1. This shows a three phase armature windings (a,b and c) on the stator and four windings on the rotor including the field winding '/'. The amortisseur (or damper) circuits in the salient pole machine or the eddy-current effects in the rotor are represented by a set of coils with constant parameters. Three damper coils, 'h' in the d-axis and g, k in the q-axis are shown in Fig. 2.1. The following assumptions are used in the derivation of the basic equations of the machine. The representation of the saturation will be considered later in this chapter. In what follows, the machine is assumed to have two poles. There is no loss of generality in doing this as the rotor angle () (with respect to a stationary axis) is assumed to be the electrical angle and the equations are invariant with respect to the number of poles. The mechanical angle ()m is related to () by (2.1) where P is the number of poles",
            "22) 24 ANALYSIS OF SUBSYNCHRONOUS RESONANCE IN POWER SYSTEMS where [ Ld 0 [L~31 = 0 Lq o 0 Ld = Laao - Labo + I Laa2 } Lq = Laao - Labo - 2 Laa2 Lo = Laao + 2Labo (Afc:h) 0 0 [ (~) [L~rl = ~ 0 (~) (~) q q 0 0 0 [L:.I = [ (IMaJkd) 0 : [(2Mahkd) 0 0 (~Magkq) 0 (~Makkq) Remarks 1. [L~slt f. [L~rl unless 2 2 2 2 kd = 3' kq = 3 1 (2.23) (2.24) (2.25) (2.26) axis are negative for kq < 0 unless Mag and Mak are both negative. It is to be noted that when the q-axis is lagging the direct axis (in the direction of rotation) as assumed in Fig. 2.1, Mag and Mak are positive. These terms are negative only if q-axis is assumed to be leading the d-axis. Hence, if d-axis is assumed to lead q-axis, it would be convenient to choose positive value of kq. 4. Eq. (2.22) shows that stator coils 'a','b' and 'c' are replaced by fictitious 'd', 'q' and '0' coils from Park's transformation. Out of these, '0' coil (in which zero-sequence current io flows) has no coupling with the rotor coils and may be neglected if io = O. Since the (transformed) mutual inductance terms between d,q coils and the rotor coils are constants, it can be interpreted that d and q coils rotate at the same speed as the rotor",
            "102) d5i 2Hidt + Di(5i - 5io) + Di,i-d5i - 5i-d + Di,i+d5i - 5i+d (2.103) +T;,i-l + Ti,i+l = Tmi - Tei (2.104) (2.105) where T;,i-l is torque in the shaft section connecting mass i and (i-I). It is not difficult to see that the inertia (2H) is analogous to a capacitance, slip analogous to voltage and torque analogous to current. The spring constant in pu (/{ W B) is analogous to the reciprocal of inductance. The per unit damping coefficient (D) is analogous to conductance. For the six mass system shown in Fig.2.1O, the electrical analogue is shown in Fig.2.11. There is no loss of generality in assuming 5io (slip at the operating point) as zero. Actually, at equilibrium point all the slips will be equal (50)' If 50 is not zero, it is equivalent to saying that the voltage of the reference bus in Fig.2.11 is nonzero. This has no effect, particularly when linearized analysis is done. Note that the state variables for the network shown in Fig.2.11 are only 11 given by xt m = [51 52 53 54 55 56 T12 T23 T34 T45 T56]t (2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001156_j.addma.2017.10.011-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001156_j.addma.2017.10.011-Figure3-1.png",
        "caption": "Fig. 3. Possible part failures due to residual stresses.",
        "texts": [
            " Existing issues in selective laser melting process Despite the benefits of flexible manufacturing, SLM still remains complex metallurgical process subjected to various defects and ssues related to process or material changes which significantly ffect resultant build quality of the part. .3.1. Residual stresses The higher irradiation energy used in SLM creates steeper ther- al gradients as compared to SLS which increases residual stress ccumulation during part forming that propagates over multiple uilt layers and may result in severe thermal deformations, delamnation or distortions such as part warping (refer to Fig. 3). Rapid elting and solidification cycles in SLM introduce large thermal uctuations across the solidified material during built-up which nfluence both compressive and tensile conditions. Two main ther- al mechanisms are involved in the build-up of residual stresses uring the SLM process: (1) Temperature Gradient Mechanism TGM) and (2) Cool-down phase [33]. Based on TGM, the preolidified material underneath the melted layer is heated up rapidly pon laser irradiation, which readily expands but is constricted by he cold and rigid portions of the solidified piece"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001545_j.rcim.2015.01.003-Figure12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001545_j.rcim.2015.01.003-Figure12-1.png",
        "caption": "Fig. 12. Geometry 1, solid structure with multiple branches. (a) Geometry is represented by black lines, MAT represented by dotted red lines, and red solid lines stand for branches. (b) Generated trimmed path. (For interpretation of the references to color in this figure legend, the reader is referred to the web version of this article.)",
        "texts": [],
        "surrounding_texts": [
            "Five geometries sliced from CAD models are tested to validate the effectiveness and robustness of the developed algorithms. As shown in Figs. 12\u201316, various types of geometries are tested including solid structures with or without holes, and thin-walled structures. Table 2 provides basic information of the geometries. The MATs and the trimmed paths for all geometries are successfully generated using the developed algorithms. The material efficiency of AM, based on the proposed MATbased path at various step-over distances, is simulated for the different geometries as shown in Fig. 17. Material efficiency of the traditional CNC machining is also provided as a comparison. It is found that, in general, material efficiency decreases with the increasing of step-over distance. This is intuitive as step-over distance represents the resolution of the deposition system. To fill geometry with deposition material, the greater the step-over distance, the more excess material will need to be deposited. The results also show that AM technology has a much higher material efficiency compared to traditional subtractive manufacturing. The effects of the step-over distance on part building time are also discussed. The building time t is determined by the sum of the total path length L and the travel speed, V, of the deposition head. If the build time at the step-over distance of 1 mm, t1, is set to be the characteristic time, the non-dimensional building time T at the step- over distance of d is expressed as T t t L V L V L L/ ( / )/( / ) /d d d1 1 1= = = where Ld represents the total path length at the step-over distance of d; and the travel speed V is assumed to be the same at various stepover distances. The non-dimensional building time as the function of step-over distances are calculated as shown in Fig. 17. With the increasing of the step-over distance, the build time T declines exponentially. It reveals that it is always better to have a small stepover distance for highmaterial efficiency whereas that would be at the expense of productivity. For powder-based AM system, the step-over distance generally ranges from 0.01 mm to 2 mm. Material efficiency in this range decreases only slightly, indicating that the step-over distance does not have a significant effect on the material efficiency of powderbased AM. As inferred above, material efficiency is determined by the resolution of the deposition system in relation to the size of the geometry. Building time changes significantly in this range, therefore, build rate will be an important factor for powder-based AM system. On the other hand, for WAAM technology, the typical step-over distance varies from 2 mm to 12 mm for mild steel materials. Building time in this range as shown in Fig. 17 does not change y of the given geometry. Red lines represent the MAT. Green lines stand for the ath. (b) Trimmed paths. (For interpretation of the references to color in this figure significantly and therefore build rate for WAAM is not the major concern. Material efficiency in WAAM using the proposed MATbased paths is shown in Fig. 18. It is found that for solid structures (Geometry 1 and Geometry 2) material efficiency is relatively constant, with a slight decrease as step-over distance increases. However for thin-walled structures, such as Geometry 3, Geometry 4, and Geometry 5, the variation of material efficiency corresponding to the step-over distance are significant. Although the generally-descending trend of material efficiency can still be found, the variation cannot be predicted since the shapes of various geometries are very different. The optimal step-over distances for Geometry 1 and Geometry 2 are 4 mm, while for Geometry 3 is around 4 mm, for Geometry 4 is near 6 mm and for Geometry 5 is approximately 3 mm. Variations of material efficiency in WAAM for the five tested geometries are shown in Fig. 19. It is clear that while the variation of material efficiency for different step-over distances is minimal for solid structures, it is quite significant for thin-walled structures. For Geometry 5, by choosing the optimal step-over distance, material efficiency could be increased by 2.4 times from 38.63% to 94.15%. This indicates that step-over distance plays an important role on material efficiency when fabricating thin-walled structures using WAAM technology. Experimental results of the proposed MAT-based path generation strategy are conducted using a robotic welding system at the University of Wollongong as shown schematically in Fig. 20. The details of the system can be found in previous publications [24,27]. In this study, a section of Geometry 5 is fabricated using both the proposed MAT-based path patterns and the traditional contour path patterns. Comparisons between the proposed path patterns and the traditional contour path patterns are shown in Fig. 21. For the Fig. 20. Schematic diagram of the proposed path patterns (Fig. 21a), layers are generated by offsetting the deposition head from the medial axis to outside, and the deposited layers are slightly larger than real geometry. For the traditional path patterns (Fig. 21b), deposited layers are consistent to the real geometry boundaries; however, gaps are created since the thicknesses of the walls vary. After surface milling, it can be seen that, while the traditional contour path patterns (Fig. 21d) leaves gaps on the component, the proposed MAT-based path patterns are able to produce gap-free walls. Depending on the thickness of the walls, more material may need to be removed for the parts generated using the proposed MAT-based path patterns, in order to produce the desired geometry. However, for certain applications such as AM components subjected to high mechanical loading, the extra material removal is less of an issue than leaving gaps in the interior of the deposited parts using traditional contour path patterns."
        ]
    },
    {
        "image_filename": "designv10_0_0003953_s0022-5193(84)80179-4-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003953_s0022-5193(84)80179-4-Figure6-1.png",
        "caption": "FIG. 6. Cerebellar motor coordination: transformation of covariant intention to contravariant execution by the neuronal network acting as the Moore-Penrose generalized inverse of the covariant metric tensor of the motor space. The covariant motor intention (i) is converted to contravariant motor execution (e) by the network of the cerebellum. Matrix 3 yields the numerical components of the generalized inverse of the covariant metric tensor of the 0-25-37 degree motor frame. The boxes e\" and i k show that both vectors are assigned to the same invariant displacement (arrowheads), yet the difference is that the contravariant expression executes the movement, while the covariant vectorial version cannot physically generate it.",
        "texts": [
            " (C) C E R E B E L L A R C O O R D I N A T I O N . T R A N S F O R M A T I O N O F C O V A R I A N T M O T O R I N T E N T I O N I N T O C O N T R A V A R I A N T M O T O R E X E C U T I O N T H R O U G H T H E M O O R E - P E N R O S E G E N E R A L I Z E D I N V E R S E O F T H E C O V A R I A N T M E T R I C T E N S O R O F O V E R C O M P L E T E M A N I F O L D S Acting in a metric tensor-like manner in the motor hyperspace, is assumed to be the role of the cerebellum (Pellionisz & Llinas, 1980). Via this transformation, shown in Fig. 6, the covariant motor intention ik is converted by the matrix gk,, implemented by the cerebellar connectivity, into a contravariant-type (physically executable) expression in the motor frame e\". This motor execution vector e\", arriving at the alpha, beta, gamma executor muscles, is capable of generating the required displacement. The cerebellar metric-type transformation is explained in more detail in Fig. 7. Indeed, tensor network theory was developed, in the first place, to interpret the structuro-functional properties of the cerebellar network so that the overall cerebellar function of motor coordination could be precisely explained in a formal mathematical manner",
            " Its inverse can also be uniquely established either by regular inversion or, in the event of singularity, which is the case in overcomplete embedding, by the Moore-Penrose generalized inverse (Albert, 1972; Ben-Israel & Greville, 1980). Such a solution ensures that in both the non-overcomplete and overcomplete case the eigenvectors of the covariant-to-contravariant (and opposite) transformation-matrices are identical. This generalized inverse is calculated from the cosine-matrix of the depicted 0o-250--37 \u00b0 motor frame, as shown in matrix (3) in Fig. 6. The biological importance of the overcompleteness is given by the fact that for a particular musculoskeletal system, the above coordination-paradigm requires only the availability in the CNS of a suitable matrix, implemented by a neuronal net. Such covariant-to-contravariant transformation is an exact basis for a theory of cerebellar coordination. An answer to the question beyond the explanation of coordination, i.e. how such networks are generated by the CNS or in man-made organisms, will be provided elsewhere"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure3.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure3.6-1.png",
        "caption": "FIGURE 3.6. Vertically loaded tire at zero camber.",
        "texts": [
            "3) where the coefficient kx and ky are called tire stiffness in the x and y directions. Proof. The deformation behavior of tires to the applied forces in any three directions x, y, and z are the first important tire characteristics in tire dynamics. Calculating the tire stiffness is generally based on experiment and therefore, they are dependent on the tire\u2019s mechanical properties, as well as environmental characteristics. Consider a vertically loaded tire on a stiff and flat ground as shown in Figure 3.6. The tire will deflect under the load and generate a pressurized contact area to balance the vertical load. Figure 3.7 depicts a sample of experimental stiffness curve in the (Fz,4z) plane. The curve can be expressed by a mathematical function Fz = f (4z) (3.4) however, we may use a linear approximation for the range of the usual application. Fz = \u2202f \u2202 (4z) 4z (3.5) 100 3. Tire Dynamics F1 < F2 < F3 3. Tire Dynamics 101 The coefficient \u2202f \u2202(4z) is the slope of the experimental stiffness curve at zero and is shown by a stiffness coefficient kz kz = tan \u03b8 = lim 4z\u21920 \u2202f \u2202 (4z) "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001598_tpel.2012.2227808-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001598_tpel.2012.2227808-Figure4-1.png",
        "caption": "Fig. 4. Flux in teeth and back iron",
        "texts": [
            " They are commonly classified into phase-to-phase short circuit, phase-to-ground short circuit or inter-turn short circuit [16]. In phase-to-phase short circuit, fuses might burn and machine could stop. In a phase-to-ground short circuit, if machine still continues to operate, a large torque ripple can be found. In an inter-turn short circuit, the faulty winding has a smaller number of effective turns than the other healthy windings, so one can find an asymmetry in machine\u2019s armature current or armature MMF. This signature can be used as an indicator in this paper. Figure 4 shows the path for the coupling of flux when only the armature MMFs are considered. Applying KCL, one gets 1 1 2 2A a b c      (8) where \u03bbA is the flux linkage through tooth A; \u03bba is the flux linkage produced by coil around tooth A; \u03bbb is the flux linkage produced by coil around tooth B and \u03bbc is the flux linkage produced by coil around tooth C. The phasor diagram for this arrangement is shown in Fig 5. It shows that when a phase-to-ground short circuit occurs at phase A, there is still 1/3 flux linkage remaining, which is produced by adjacent armature windings"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003908_978-1-84882-730-1-Figure13.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003908_978-1-84882-730-1-Figure13.1-1.png",
        "caption": "Figure 13.1 General framework of underwater vehicle path-following",
        "texts": [
            "t/ are the environmental disturbance forces or moments acting on the surge, sway, heave, roll, pitch, and yaw axes, respectively. We assume that these disturbances are bounded as follows: j Eu.t/j max Eu <1; j Ev.t/j max Ev <1; j Ew.t/j max Ew <1;\u02c7\u030c Ep.t/ \u02c7\u030c max Ep <1; \u02c7\u030c Eq.t/ \u02c7\u030c max Eq <1; j Er .t/j max Er <1: (13.9) Since the sway and heave control forces are not available in the sway and heave dynamics, the vehicle model (13.1) is underactuated. In this chapter, we consider a control objective of designing the control inputs 1 and 2 to force the underactuated vehicle (13.1) to follow a specified path \u02dd , see Figure 13.1. If we are able to drive the vehicle to follow closely a virtual vessel that moves along the path with a desired speed u0, then the control objective is fulfilled, i.e., the vessel is in a tube of nonzero diameter centered on the reference path and moves along the specified path at the speed u0. Roughly speaking, the approach is to steer the vessel such that it heads toward the virtual one and diminishes the distance between the real and the virtual vessels. 316 13 Path-following of Underactuated Underwater Vehicles In Figure 13.1,A is the center of the real vehicle andAs is a point on the reference path attached to the virtual vehicle. Define the following path-following errors xe D xd x; ye D yd y; ze D zd z; de D q x2e Cy2e Cz2e ; (13.10) where xd ;yd and zd are the coordinates of As . Then the terms ai ; 1 i 3 in Figure 13.1 are obtained from xe; ye and ze by rotating the body frame around the earth-fixed frame OEXEYEZE the roll, pitch, and yaw angles, i.e., 2 4 a1 a2 a3 3 5D J1. 2/ 2 4 xe ye ze 3 5 : (13.11) Expanding (13.11) yields a1 D xeJ 11 1 . 2/CyeJ 21 1 . 2/CzeJ 31 1 . 2/; a2 D xeJ 12 1 . 2/CyeJ 22 1 . 2/CzeJ 32 1 . 2/; a3 D xeJ 13 1 . 2/CyeJ 23 1 . 2/CzeJ 33 1 . 2/; (13.12) where J ij1 . 2/ is the element of J1. 2/ at the i th row and j th column. Therefore the path-following orientation errors are defined by the angles \u02db and \u02c7",
            "2 Coordinate Transformations 317 with de , , \u02db, and \u02c7 being arbitrarily small positive constants. Assumption 13.1. The reference path is regular, i.e. there exist strictly positive constants a3min; a3max; a2min and a2max such that a3min s @xd @s 2 C @yd @s 2 C @zd @s 2 a3max; a2min s @xd @s 2 C @yd @s 2 a2max: (13.14) Remark 13.1. 1. We might refer to the above objective as a path-tracking one. However, we use the term \u201cpath-following\u201d since our approach is to make the real vehicle follow the virtual one, see Figure 13.1. 2. Assumption 13.1 ensures that the path is feasible for the vessel to follow, see Section 13.3. The condition (13.14) implies that the reference trajectory cannot contain a vertical straight line to avoid singularity of J2. 2/ at the pitch angle D \u02d90:5 . 3. If the reference path is not regular, then we can often split it into regular pieces and consider each of them separately. This is the case of point-to-point navigation, which will be addressed in Section 13.6. 4. The path parameter, s, is not the arc length of the path in general. For example, a circle with radius R centered at the origin can be described as xd D Rcos.s/ and yd DR sin.s/. 13.2 Coordinate Transformations From (13.10) and (13.12), we have the position kinematic error dynamics as follows: Pde D 1 de xe @xd @s Cye @yd @s Cze @zd @s Ps a1 de u a2 de v a3 de w: (13.15) For the path-following orientation errors, referring to Figure 13.1 and the control objective stated in the previous section, one can see that the following holds a1 de D cos. /D cos.\u02db/cos.\u02c7/; (13.16) which in turn implies that ( lim t!1\u02db.t/D 0 lim t!1\u02c7.t/D 0 , lim t!1 .t/D 0) lim t!1 a1.t/ de.t/ D 1: (13.17) 318 13 Path-following of Underactuated Underwater Vehicles Hence we can either choose the angles \u02db and \u02c7, or the angle , or the term a1=de as the orientation coordinates for the control design. We now discuss the above options and then choose one that results in a simple control design and enhance feasible initial conditions. Using Angles \u02db and \u02c7. In this case, the path-following orientation errors are defined as follows, see Figure 13.1: \u02db D e\u02db 2 n\u02db.e\u02db/; \u02c7 D e\u02c7 2 n\u02c7 .e\u02c7 /; (13.18) where e\u02db D 8< : 2arctan a3 a1 I .a3;a1/\u00a4 .0;0/; 0I .a3;a1/D .0;0/; e\u02c7 D 8\u0302 \u02c6\u0302\u0302 <\u0302 \u02c6\u0302\u0302\u0302 :\u0302 2arctan 0 B@ a2q a21Ca23 1 CA I a2; q a21Ca23 \u00a4 .0;0/; 0 I a2; q a21Ca23 D .0;0/: (13.19) The functions n\u02db.e\u02db/ and n\u02c7 .e\u02c7 / take values in .0;\u02d91;\u02d92; :::/ such that \u02db and \u02c7 belong to . ; . Hence \u02db and \u02c7 are periodic and piecewise continuous functions with respect to e\u02db and e\u02c7 . The reason for introducing (13.18) is to convert all equilibrium points of \u02db and \u02c7 to the origin",
            "A\u02db\u02c7. //D cos.\u02db/cos. / sin.\u02db/sin. /cos. / sin. /sin. / tan.\u02c7/: (13.25) It can be seen from (13.25) that the matrix A\u02db\u02c7. / is not globally invertible even when D 0 and \u00a4 \u02d90:5 . It is also observed that if we choose the angles \u02db and \u02c7 as the orientation coordinates for control design, there are a number of discontinuous surfaces, see (13.19), (13.20), and (13.21), which make the stability analysis difficult. Using Angle . In this case, the path-following orientation error is defined as follows, see Figure 13.1: D e 2 n .e /; (13.26) 320 13 Path-following of Underactuated Underwater Vehicles where e D 8\u0302 \u02c6\u0302\u0302\u0302 < \u02c6\u0302\u0302\u0302 :\u0302 2arctan 0 B@ q a22Ca23 a1 1 CA I a1; q a22Ca23 \u00a4 .0;0/; 0I a1; q a22Ca23 D .0;0/: (13.27) The interpretation of the above expressions is similar to that of (13.18) and (13.19). It can be seen from (13.27) that e is discontinuous on the following surface: D D .a1;a2;a3/ W q a22Ca23 \u00a4 0;a1 D 0 : (13.28) It can also be seen from (13.26) that is discontinuous on the surface: C D f.a1;a2;a3/ W D g : (13"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001270_jproc.2020.3046112-Figure21-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001270_jproc.2020.3046112-Figure21-1.png",
        "caption": "Fig. 21. Segmented magnet IPM machine: 2-D FEA model (1/4th model) [59].",
        "texts": [
            " 3) Segmented Magnet IPM Machine: The segmented magnet IPM configuration is very similar to the state-ofthe-art configurations using inner rotor topology where each pole has two magnets arranged in V-shape. However, instead of using a single bar in each pocket, each magnet is circumferentially segmented into three or five small magnets. The corner magnets are thicker than the middle magnets to provide higher demagnetization withstand capability. The magnets are displaced from the starting of the cavity, which reduces the effect of the demagnetization field on the edges, as illustrated in Fig. 21. We consider the design of a segmented magnet IPM machine for demonstrating the potentials and challenges PROCEEDINGS OF THE IEEE 15 Authorized licensed use limited to: Univ of Calif Santa Barbara. Downloaded on May 15,2021 at 01:41:56 UTC from IEEE Xplore. Restrictions apply. to be overcome for high-speed HRE-free IPM machines. In this design, a double-layer distributed winding with stranded wire is considered to minimize the torque pulsations and magnet loss. The maximum operating speed and peak power have been considered to be 20 000 rpm and 100 kW, respectively, following the DOE 2025 research goal [1]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000989_s11661-004-0094-8-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000989_s11661-004-0094-8-Figure1-1.png",
        "caption": "Fig. 1\u2014Single line build consisting of 18 layers of Ti-6Al-4V laser deposited onto a substrate of the same material. Laser motion is in the x direction and layers are added in the z direction.",
        "texts": [
            " Typical processing conditions for the LMD of Ti-6Al-4V include a laser power of 11 kW (CO2), laser velocity of 2.54 mm/s, and a mass-deposition rate of 0.936 kg/h.[13] Eighteen layers (individually denoted hereafter as L1 through L18) of Ti-6Al-4V were deposited on a 7-mm-thick Ti-6Al-4V substrate that was previously mill annealed at 700 \u00b0C to 730 \u00b0C for 2 hours. Each deposited layer was about 6-mm thick; however, due to overlap and remelting of the layer due to the addition of a subsequent layer, each ripple seen in Figure 1 is about 3-mm thick. A layer is defined as the distance between cusps created by the intersection of the concave portion of each layer. Samples were removed from the deposit, polished using standard metallographic techniques, and etched such that the macro and microstructures of the indicated y-z plane of the deposit were revealed. A solution of 5 vol pct hydrofluoric acid in deionized water was used as the etchant. Samples were submersed in a watch glass containing the etchant for approximately 10 seconds, rinsed in a beaker of deionized water, and then dried"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure2.20-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure2.20-1.png",
        "caption": "FIGURE 2.20. A three-axle car moving on an inclined road.",
        "texts": [
            " A slight lateral motion of the rear wheels, by any disturbance, develops a yaw motion because of unbalanced lateral forces on the front and rear wheels. The yaw moment turns the vehicle about the z-axis until the rear end leads the front end and the vehicle turns 180 deg. Figure 2.19 illustrates a 180 deg sliding rotation of a rear-wheel-locked car. 74 2. Forward Vehicle Dynamics The lock-up of the front tires does not cause a directional instability, although the car would not be steerable and the driver would lose control. 2.7 F Vehicles With More Than Two Axles If a vehicle has more than two axles, such as the three-axle car shown in Figure 2.20, then the vehicle will be statically indeterminate and the normal forces under the tires cannot be determined by static equilibrium equations. We need to consider the suspensions\u2019 deflection to determine their applied forces. The n normal forces Fzi under the tires can be calculated using the following n algebraic equations. 2 nX i=1 Fzi \u2212mg cos\u03c6 = 0 (2.241) 2 nX i=1 Fzixi + h (a+mg sin\u03c6) = 0 (2.242) Fzi ki \u2212 xi \u2212 x1 xn \u2212 x1 \u00b5 Fzn kn \u2212 Fz1 k1 \u00b6 \u2212 Fz1 k1 = 0 for i = 2, 3, \u00b7 \u00b7 \u00b7 , n\u2212 1 (2.243) where Fxi and Fzi are the longitudinal and normal forces under the tires attached to the axle number i, and xi is the distance of mass center C from the axle number i",
            " Forward Vehicle Dynamics where [X] = \u00a3 Fz1 Fz2 Fz3 \u00b7 \u00b7 \u00b7 Fzn \u00a4T (2.253) [A] = \u23a1\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3 2 2 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 2 2x1 2x2 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 2xn xn \u2212 x2 k1l 1 k2 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 x2 \u2212 x1 knl \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 xn \u2212 xi k1l \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 1 ki \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 xi \u2212 x1 knl \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 xn \u2212 xn\u22121 k1l \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 1 kn\u22121 xn\u22121 \u2212 x1 knl \u23a4\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6 (2.254) l = x1 \u2212 xn (2.255) [B] = \u00a3 mg cos\u03c6 \u2212h (a+mg sin\u03c6) 0 \u00b7 \u00b7 \u00b7 0 \u00a4T . (2.256) Example 71 F Wheel reactions for a three-axle car. Figure 2.20 illustrates a three-axle car moving on an inclined road. We start counting the axles of a multiple-axle vehicle from the front axle as axle-1, and move sequentially to the back as shown in the figure. The set of equations for the three-axle car, as seen in Figure 2.20, is 2Fx1 + 2Fx2 + 2Fx3 \u2212mg sin\u03c6 = ma (2.257) 2Fz1 + 2Fz2 + 2Fz3 \u2212mg cos\u03c6 = 0 (2.258) 2Fz1x1 + 2Fz2x2 + 2Fz3x3 + 2h (Fx1 + Fx2 + Fx3) = 0 (2.259) 1 x2 \u2212 x1 \u00b5 Fz2 k2 \u2212 Fz1 k1 \u00b6 \u2212 1 x3 \u2212 x1 \u00b5 Fz3 k3 \u2212 Fz1 k1 \u00b6 = 0 (2.260) which can be simplified to 2Fz1 + 2Fz2 + 2Fz3 \u2212mg cos\u03c6 = 0 (2.261) 2Fz1x1 + 2Fz2x2 + 2Fz3x3 + hm (a+ g sin\u03c6) = 0 (2.262) (x2k2k3 \u2212 x3k2k3)Fz1 + (x1k1k2 \u2212 x2k1k2)Fz3 \u2212 (x1k1k3 \u2212 x3k1k3)Fz2 = 0. (2.263) The set of equations for wheel loads is linear and may be rearranged in a matrix form [A] [X] = [B] (2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003774_j.1469-185x.1974.tb01299.x-Figure15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003774_j.1469-185x.1974.tb01299.x-Figure15-1.png",
        "caption": "Fig. 15 . Illustration of regular array of cilia. (a) From above and (b) side-on view showing the metachronism. The spacing in the xl-direction is a, while in the %,-direction it is b.",
        "texts": [
            " We need to consider the influence on the velocity profile of (u) the cell surface and (b) the interaction effects of all the other cilia. These considerations have been discussed in Section IV and in more detail by Blake (1972~) . The mathematical model (Blake, 1972~) for the cilia sublayer of micro-organisms can be described briefly as follows. Each cilium is modelled by an elongated body of length L with its base attached to an infinite plane. The bases of the cilia are distributed in a regular array over the infinite plane with spacing a in the xl-direction and b in the x2-direction. This is illustrated in Fig. 15. In protozoa the effective stroke of the cilium can be in any direction, but for the purposes of this model it is taken in the direction of increasing xl , although the beat need not be planar. We can represent the movement of a cilium in Cartesian coordinates by the vector function [(s, t), where s is the distance along the cilium from the base and t is time (Fig. 8a). Thus the mean velocity profile, in terms of components Ul and U2, can be obtained in terms of the following integral equation (Blake, 1972u), where and The bar over the integral sign indicates a time average, x3 is the vertical coordinate, &: the vertical coordinate of the cilium above the cell surface, Fl and F2 are the force I I 2 J"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003908_978-1-84882-730-1-Figure14.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003908_978-1-84882-730-1-Figure14.2-1.png",
        "caption": "Figure 14.2 Robot parameters",
        "texts": [
            " Explanations of the above motion tasks are similar to those of ocean vehicles, see Section 3.2. 14.1.2 Modeling and Control Properties 14.1.2.1 Modeling We consider a unicycle-type mobile robot, which under the assumption of no wheel slips has the following dynamics [142]: 14.1 Mobile Robots 343 P D J . /!; (14.1) M P! D C . P /! D!C ; where D \u0152x y T denotes the position .x;y/, the coordinates of the middle point, P0, between the left and right driving wheels, and heading angle of the robot coordinated in the earth-fixed frame OXY , see Figure 14.2, ! D \u0152!1 !2 T , with !1 and !2 being the angular velocities of the wheels of the robot, D \u0152 1 2 T , with 1 and 2 being the control torques applied to the wheels of the robot. The rotation matrix J . /, mass matrix M , Coriolis matrix C . P /, and damping matrix D in (14.1) are given by J . /D r 2 2 4 cos. / cos. / sin. / sin. / b 1 b 1 3 5 ; M D m11 m12 m12 m11 ; C . P /D 0 c P c P 0 ; D D d11 0 0 d22 ; (14.2) with c D 1 2b r2mca; m11 D 1 4b2 r2.mb2CI /CIw ; m12 D 1 4b2 r2.mb2 I /; mDmc C2mw ; I Dmca 2C2mwb 2CIc C2Im; (14.3) where mc and mw are the masses of the body and wheel with a motor; Ic ;Iw , and Im are the moments of inertia of the body about the vertical axis through Pc (center of mass), the wheel with the rotor of a motor about the wheel axis, and the wheel with the rotor of a motor about the wheel diameter, respectively; r , a, and b are defined in Figure 14.2; the nonnegative constants d11 and d22 are the damping coefficients. If these damping coefficients are zero, we have an undamped case. On the other hand, if the damping coefficients are positive, we have a damped case. We take the physical parameters from [142]: b D 0:75 m, a D 0:3 m, r D 0:15 m, mc D 30 kg, mw D 1 kg, Ic D 15:625 kgm2, Iw D 0:005 kgm2, Im D 0:0025 kgm2, d11 D d22 D 5 kgs 1 for numerical simulations. For convenience, we convert the wheel angular velocities .!1;!2/ of the robot to its linear, v, and angular, w, velocities by: $ D B 1"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003413_ac0010882-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003413_ac0010882-Figure4-1.png",
        "caption": "Figure 4. Cyclic voltammograms of 1 \u00b5M DA at a, an overoxidized bare CFME, and b, an OPPD-modified CFME: medium, pH 7.4 phosphate buffer; a1 and b1, first scan; a2 and b2, second scan; and b3, third scan. Scan rate, 100 mV/s; accumulation conditions, 10 s at -0.2 V.",
        "texts": [
            " These observations indicate that not only the extent of porosity and amount of negative charge but also some other factors (e.g., analyte and/or oxidation product adsorption/desorption processes, van der Waals interactions, ion-pairing, etc.) may play a particular role in governing the electrochemical behavior of the OPPD film. Because the heterogeneous mechanisms of the oxidation of catechols and ascorbic acid at carbon and some coated electrodes have been reported in detail,32,38,40,41 our attention was mainly focused on the particular electrochemical behavior of DA and AA at the OPPD-modified CFME. Figure 4 displays cyclic voltammograms of DA in pH 7.4 phosphate buffer at an overoxidized bare (39) Komura, T.; Funahasi, Y.; Yamaguti, T.; Takahasi, K. J. Electroanal. Chem. 1998, 446, 113-123. (40) Deakin, M. R.; Kovach, P. M.; Stutts, K. J.; Wightman, R. M. Anal. Chem. 1986, 58, 1474-1480. (41) Malem, F.; Mandler, D. Anal. Chem. 1993, 65, 37-41. Analytical Chemistry, Vol. 73, No. 6, March 15, 2001 1199 CFME (curves a1, a2) and an OPPD-CFME (b1-b3), demonstrating a two-electron redox reaction, with anodic and cathodic peaks (at the OPPD-CFME) at 0",
            "; Clarke, J. Langmuir 1994, 10, 3675-3683. (44) Maeda, H.; Yamauchi, Y.; Yoshida, M.; Ohmori, H. Anal. Sci. 1995, 11, 947-952. (45) Nuwer, M. J.; Osteryoung, J. Anal. Chem. 1989, 61, 1954-1959. 1200 Analytical Chemistry, Vol. 73, No. 6, March 15, 2001 throughout this work, except in some characterization experiments, due to the excellent peak resolution, well-shaped peaks, and negligible interference from dissolved oxygen. Because the ability of the OPPD film to accumulate DA at physiological pH (Figure 4) was considered to be an important analytical factor, the preconcentration parameters were carefully optimized. The response to DA greatly increased when accumulation potentials were altered from the positive to negative side but remained constant when applying preconcentration potentials in the range of -0.1 to -0.5 V. In the presence of AA, -0.2 V was selected as an optimum preconcentration potential to enable recording of the AA signal, whereas in the absence of AA, the accumulation potential can be set to -0"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001593_j.mechmachtheory.2015.11.017-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001593_j.mechmachtheory.2015.11.017-Figure8-1.png",
        "caption": "Fig. 8. Schematic of tooth profile modification.",
        "texts": [
            " The solution process of the mesh stiffness for the meshing teeth is similar to that of the fillet-foundation stiffness in Section 2.1. More details about the modeling process using FE method refer to Refs. [1,7,8,39]. The tooth profile modification curve has to be defined before simulation. In this paper, linear tip relief is adopted. The main design for this kind of profile modification is to determine the length of profile modification La and the amount of profile modification Ca, as shown in Fig. 8. The equation of TPM curve is generally defined as [42]: C \u00bc Ca \u03c4\u2010\u03c4s \u03c4e\u2212\u03c4s ; \u00f016\u00de where \u03c4s and \u03c4e denote angular displacements of the starting point and the ending point of modification, respectively; \u03c4 denotes angular displacement of arbitrary point at the involute curve; and C represents the amount of material removal at the angular displacement \u03c4. It is worth noting that other TPM parameter \u0394LT is adopted instead of La by Fernandez del Rincon et al. [29]. In order to compare the results obtained from the proposed method with the results of Fernandez del Rincon et al"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure3.57-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure3.57-1.png",
        "caption": "Fig. 3.57: Kinematic chain and design parameters of CaPaMan.",
        "texts": [
            " The above-mentioned characteristics can also have a practical interpretation for assembling applications that can be characterized through suitable values of workspace, precision error, velocity performance as given by values of workspace volume, Chapter 3: Fundamentals of the Mechanics of Robots202 compliant displacements, and velocity mapping. This example illustrates the fundamental concepts for analysis and design of parallel manipulators as applied to CaPaMan (Cassino Parallel Manipulator), a specific 3-d.o.f.s parallel manipulator that has been developed at LARM Laboratory of Robotics and Mechatronics in Cassino. Fundamentals of Mechanics of Robotic Manipulation 203 The kinematic chain of CaPaMan is illustrated in Fig. 3.57 and a built prototype that is available at LARM is reported in Fig. 3.58. Figure 3.57 shows the kinematic chain of the parallel mechanism, which is composed of a movable plate MP that is connected to a fixed plate FP by means of three-leg mechanisms. Each leg mechanism is composed of an articulated parallelogram AP whose coupler carries a prismatic joint SJ, a connecting bar CB which transmits the motion from AP to MP through SJ, and a spherical joint BJ which is installed on MP. Thus, CB may translate along the prismatic guide of SJ keeping its vertical posture while the BJ allows the MP to rotate in the space",
            " Finally, the size of MP and FP are given by rp and rf, respectively, where H is the center point of MP, O is the center point of FP, Hk is the center point of the k-th BJ, and Ok is the middle point of the frame link ak. Platform MP is driven by the three leg mechanisms through the corresponding articulation points H1, H2, H3. A motor on the input crankshaft can actuate each leg mechanism so that the device is a 3-d.o.f.s spatial mechanism. In addition, the kinematic architecture has been useful to propose an alternative lowcost solution for the leg design, as shown in Fig. 3.59. The versions CaPaMan 2 in Fig. 3.59 a) and CaPaMan 3 in Fig. 3.59 b) have been obtained from the kinematic design of CaPaMan in Fig. 3.57 by substituting the prismatic joint with a suitable revolute joint and the four-bar linkage with a slider-crank mechanism, respectively. Chapter 3: Fundamentals of the Mechanics of Robots204 In order to describe the motion of MP with respect to FP, a world frame OXYZ has been assumed as fixed to FP and a moving frame HXPYPZP has been fixed to MP, as shown in Fig. 3.57. Particularly, OXYZ has been fixed with Z axis orthogonal to the FP plane, X axis as coincident with the line joining O to O1 , and Y axis to give a Cartesian frame. The moving frame HXPYPZP has been fixed in an analogous way to the movable plane MP with ZP orthogonal to the MP plane, XP axis as coincident to the line joining H to H1 and YP to give a Cartesian frame. A prototype of Fig. 3.57 has been built as in Fig. 3.58 at LARM, the Laboratory of Robotics and Mechatronics in Cassino, with the dimensional size of Table 3.1. The linkage geometry of the parallelograms with commercial prismatic joints on the couplers can be conveniently exploited for an easy modular mechanical design and assembly, which may allow easy changes in the leg dimensions. The kinematic feasibility of CaPaMan has been studied and characterized in terms of an analytical formulation for direct and inverse kinematics, which has been obtained as closed-form expressions because of the geometry of the parallel assembly and the parallelograms in the legs"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure3.74-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure3.74-1.png",
        "caption": "Fig. 3.74: An equivalent model for stiffness evaluation of the CaPaMan leg mechanism.",
        "texts": [
            " A simple way to build an equivalent scheme is to use the basic concept of stiffness composition for serial and parallel spring systems. These well-known rules give the equivalent stiffness for a series of springs as the sum of the stiffness parameters of components and for a parallel system of springs as the sum of the inverse of the stiffness parameters of components. Thus, from the leg mechanism scheme of Fig. 3.73 with torsion stiffness parameters kTb and kTd that are considered in both the frame joints, one can easily obtain the equivalent scheme of Fig. 3.74 by using the above-mentioned rules. In particular one can compute a stiffness coefficient of a CaPaMan leg mechanism from Eq. (3.9.86) in the form t Leqp zyKR \u0394\u0394= (3.9.101) with the so-called \u2018equivalent spring matrix\u2019 that is defined as k0 0k K z y Leq = (3.9.102) with b k kcos k 1 k 1 k Ty c2 db y ++\u03b1+= (3.9.103) b k ksin k 1 k 1 k Tz h2 db z ++\u03b1+= in which 2 TdTb Ty sin k 1 k 1 k \u03b1+= (3.9.104) 2 TdTb Ty cos k 1 k 1 k \u03b1+= Chapter 3: Fundamentals of the Mechanics of Robots236 Furthermore, by again using the composition rules for parallel spring systems\u2019 stiffness coefficients for CaPaMan can be computed for a 3D Cartesian scheme as k sin k sin k sin k y3 3 y2 2 y1 1 xsCaPaMan \u03b4 +\u03b4+\u03b4=\u2212 k cos k cos k cos k y3 3 y2 2 y1 1 ysCaPaMan \u03b4 +\u03b4+\u03b4=\u2212 (3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000609_j.procir.2018.05.039-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000609_j.procir.2018.05.039-Figure2-1.png",
        "caption": "Fig. 2. Typical residual stress in build direction of as-SLM part w.o. the substrate: (a) cross-sectional residual stress contours measured using neutron diffraction [22], (b) predicted residual stress [29].",
        "texts": [
            " [26] utilized stress-free lattice spacing determined from stress relief treated DED samples and the neutron diffraction measured residual stress field obtained very good agreement with numerical prediction. Residual stress field of metal AM parts could depend on scanning strategies, dwell time and many other process parameters that have huge impact on thermal history. Wu et al. [22] studied the surface-level residual stress of SLM processed stainless steel 316 L-shaped bar (off substrate) in as-build state by digital image correlation method and neutron diffraction. Residual stress near the center of part tends to be compressive and tensile near surfaces as shown in Fig. 2a. Effects of parameters including scanning strategy, laser power, scanning speed, and build orientation on residual stress has been systematically investigated. A smaller scan island size and increased length energy density would result in a reduced residual stress field. Kruth et al. [27] studied the effects of scanning strategies on distortion and concluded that the island scanning strategy (material deposited in a \u201cchessboard\u201d pattern) would cause less distortion than other scanning strategy. Lu et al",
            " It was found that the island size of 2\u00d72 mm2 resulted the smallest residual stress, but severe cracking was also observed in the sample built using this island size. An island size of 5\u00d75 mm2 was considered the optimal strategy due to higher density, better mechanical properties, and relatively lower residual stresses. Mercelis et al. [11] also concluded that high tensile residual stress in build direction of a cutoff part in asbuild state is located just below the top surface followed by a compressive zone in the middle and a tensile zone at the bottom surface. And this trend was also captured by simulation work [29] as shown in Fig. 2b that a large compressive residual stress was predicted in the core of the part and experiment work of DED processed Waspaloy by Moat et al. [30]. X-ray diffraction residual stress measurement of SLM processed stainless steel and Ti6Al4V small-sized samples conducted by Yadroitsev et al. [31] has shown that residual stress in scanning direction is more tensile than in the perpendicular direction and reaches its maximum at the buildsubstrate interface. Kruth et al. [15] assessed the residual stress of the SLM processed part using the bridge curvature method"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure6.9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure6.9-1.png",
        "caption": "Fig. 6.9. Exploded view of the AFPM brushless motor with film coil coreless stator winding and twin PM rotor. Courtesy of EmBest , Soeul, South Korea.",
        "texts": [
            " CSIRO Telecommunications and Industrial Physics, Lindfield, NSW, Australia manufactures in-wheel coreless AFPM brushless motors for solar powered vehicles [225]. Solar-powered vehicles compete in the World Solar Challenge solar car race organized every two years in Australia between Darwin and Adelaide which has become a well-known international event. CSIRO provides in-wheel AFPM brushless motors both with surface PMs glued to steel discs and with PMs arranged in Halbach array. The motor structure, PM disc and stator winding are shown in Fig. 6.8 [225]. Fig. 6.9 shows a coreless brushless motor with film coil stator winding and twin PM rotor manufactured by EmBest , Soeul, South Korea. The coreles stator has a foil winding at both sides. The 8-pole PM rotor is designed as a twin external rotor. A low speed, three-phase, Y-connected AFPM brushless motor rated at 10 kW, 750 rpm, 28.5 A has been designed and investigated. The rotor does not have any ferromagnetic core and consists of trapezoidal coils embedded in a high mechanical integrity resin. Average quality sintered NdFeB PMs with Br = 1"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure2.7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure2.7-1.png",
        "caption": "FIGURE 2.7. An accelerating car on a level pavement.",
        "texts": [
            "77) At the ultimate angle \u03c6 = \u03c6M , all wheels will begin to slide simultaneously and therefore, Fx1 = \u03bcx1Fz1 (2.78) Fx2 = \u03bcx2Fz2 . (2.79) The equilibrium equations show that 2\u03bcx1Fz1 + 2\u03bcx2Fz2 \u2212mg sin\u03c6M = 0 (2.80) 2Fz1 + 2Fz2 \u2212mg cos\u03c6M = 0 (2.81) \u22122Fz1a1 + 2Fz2a2 \u2212 \u00a1 2\u03bcx1Fz1 + 2\u03bcx2Fz2 \u00a2 h = 0. (2.82) 50 2. Forward Vehicle Dynamics Assuming \u03bcx1 = \u03bcx2 = \u03bcx (2.83) will provide Fz1 = 1 2 mg a2 l cos\u03c6M \u2212 1 2 mg h l sin\u03c6M (2.84) Fz2 = 1 2 mg a1 l cos\u03c6M + 1 2 mg h l sin\u03c6M (2.85) tan\u03c6M = \u03bcx. (2.86) 2.3 Accelerating Car on a Level Road When a car is speeding with acceleration a on a level road as shown in Figure 2.7, the vertical forces under the front and rear wheels are Fz1 = 1 2 mg a2 l \u2212 1 2 mg h l a g (2.87) Fz2 = 1 2 mg a1 l + 1 2 mg h l a g . (2.88) The first terms, 12mg a2l and 1 2mg a1l , are called static parts, and the second terms \u00b11 2mg hl a g are called dynamic parts of the normal forces. Proof. The vehicle is considered as a rigid body that moves along a horizontal road. The force at the tireprint of each tire may be decomposed to a normal and a longitudinal force. The equations of motion for the accelerating car come from Newton\u2019s equation in x-direction and two static 2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001832_adfm.201504699-Figure9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001832_adfm.201504699-Figure9-1.png",
        "caption": "Figure 9. Actuation based on magnetic force derived torque: a) Ferrogel occilated in a magnetic fi eld with a ruler behind to indicate the amplitude, which was around 3 mm. Adapted with permission. [ 132 ] 1998, Elsevier. b) A piece of polymer magnet was attached to a copper beam to perform bending when turinging on the magnetic fi eld generated by the coil beneath. The left image describing the numerical analysis result. Adapted with permission. [ 133 ] Copyright 1999, Institute of Electrical and Electronics Engineers. c) With the presence of a changing magnetic fi eld, a piece of magnetic aerogel bends upward and then downward to absorb the water droplet. Adapted with permission. [ 135 ] Copyright 2010, Nature Publishing Group.",
        "texts": [
            " The ferrogel was fi xed on one side to create a fulcrum and was subjected to a strong magnetic fi eld gradient to create a mechanical torque that was not quantifi ed. The deformation of the ferrogel could be enhanced either by increasing the magnetic fi eld gradient or by reducing the strength of the network by lowering the amount of cross-linker (glutaraldehyde). The strong torque experienced here has led to further characterization of these ferrogels as actuators including investigation of their kinetic motion and cyclic behavior ( Figure 9 a). [ 132 ] The dynamic response of ferrogels was also characterized by switching a magnetic fi eld gradient on and off at a range of frequencies and amplitudes. Ferrogels were able to respond to switching fi eld gradients at frequencies up to 40 Hz with good resilience to cyclic testing. To overcome the poor mechanical strength and weak actuation forces of ferrogels, fully dense magnetic polymer composites have been investigated. Magnetic polymers and papers exhibit signifi cant strength and durability. Lagorce et al. [ 133 ] synthesized a magnetic polymer composite of magnetic ceramic ferrite imbedded into an epoxy resin. This magnetic polymer composite was shaped into a disk shape attaching to a copper cantilever and actuated with external magnetic fi elds (Figure 9 b). In addition, magnetic paper has been synthesized that responds to external magnetic fi elds. Liang et al. [ 134 ] developed a stiff magnetic paper created from graphene sheets and magnetic nanoparticles. The resultant composite exhibited both the excellent electrical conductivity of graphene and the magnetic response of the magnetic nanoparticles. This magnetic paper had suffi cient fl exibility to be used as a switch in an electrical circuit that could be actuated by external magnetic fi elds over 1 million of cyclic tests. Olsson et al. [ 135 ] developed another magnetic paper from compacted, lightweight porous magnetic aerogels. Here, freeze dried bacterial cellulose nanofi brils were employed as templates to grow cobalt ferrite nanoparticles from thermal precipitation. [ 136,137 ] The resultant composite aerogels had high fl exibility and could be actuated by magnetic fi elds (Figure 9 c). Like the other magnetic papers, this aerogel demonstrated high endurance when exposed to repeated and large deformations. Magnetic papers offer a robust actuator compared to ferrogels that can be cyclically loaded and withstand reasonable mechanical stresses. Material scientists are teaming with biologists to focus on the interface of materials and biology in the burgeoning fi eld of biomaterials. Biomaterials have shown promise in a wide variety of applications including delivery of drugs or biofactors, monitoring or directing cellular response and fate, and creating sensitive diagnostic systems"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000417_s11661-012-1444-6-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000417_s11661-012-1444-6-Figure2-1.png",
        "caption": "Fig. 2\u2014Schematic diagram showing the extraction of tensile and fatigue specimens from (a) Build 1 and (b) Build 2. In labeling the specimens, the first category indicates the orientation: V indicates the specimen was machined parallel to the build direction and H indicates perpendicular. The second category indicates the type of test: T for tensile and F for fatigue. M indicates the specimens used for microstructural investigation and Blank are those specimens not used in this study.",
        "texts": [
            " Tensile specimens according to the requirements of ASTM E8-04 were extracted parallel and perpendicular to the build direction. The tensile specimens had a dogbone shape with a gauge length of 75 mm and a 12.5 9 2.5 mm cross section, as shown in Figure 1. Three control specimens were extracted from a separate forged Ti-6Al-4V bar (MIL-T 9047) and tested to provide a baseline comparison. Eighteen tensile specimens were tested\u2014twelve in the vertical and six in horizontal direction. They were extracted from the walls using the arrangement shown in Figure 2. Tensile tests were carried out with an Instron 5500R Electromechanical (Instron, Coronation Road, High Wycombe, Bucks) machine with a 100 kN load cell. The cross head displacement speed was 0.5 mm/min. An Instron 2620- 604 dynamic strain gauge extensometer was attached to each specimen while testing to measure 0.2 pct yield stress and ultimate tensile stress accurately. The extensometer had a gage length of 25 mm. Twenty-four fatigue specimens were tested\u2014fourteen in the vertical direction and ten in the horizontal direction, which were extracted from the walls as shown in the Figure 2. The specimens were tested according to EN6072. Five specimens machined from a forged Ti-6Al-4V bar (MIL-T 9047)were also tested as a baseline comparison. The fatigue specimens had a dog-bone shape with a gauge length of 40 mm and a 12.7 9 2.5 mm cross section, as indicated in Figure 1. The fatigue specimens were ground and had a roughness, Ra = 0.4 lm. All the fatigue specimens were tested at a maximum stress of 600 MPa, a load ratio R = Pmin/Pmax = 0.1 with a sinusoidal waveform at 30 Hz. The fatigue endurance METALLURGICAL AND MATERIALS TRANSACTIONS A VOLUME 44A, FEBRUARY 2013\u2014969 limit was defined when a sample reached 107 cycles without failure",
            " The material then cooled rapidly from the b phase transforming to a fine a/a\u00a2 structure which was not subjected to further thermal cycling below the b transus (Figure 4(b)); the upper last 5 deposited layers of the wall were completely transformed during deposition of the final layer (Figure 5) and therefore contained no banding (see Figure 4(a)). In contrast, the material in the layers below this underwent additional cycles of reheating below the b 972\u2014VOLUME 44A, FEBRUARY 2013 METALLURGICAL AND MATERIALS TRANSACTIONS A transus temperature, in which the time the temperature reached was sufficiently high to significantly coarsen the a phase. The continuous grain-boundary a layer found along some of the prior b grain boundaries (Figure 2(c)) in the bottom part of the build is also evidence of the effects of this repeated heat treatment. The white bands thus correspond to the region where the peak temperature was just below the b transus, which leads to a coarsened a plate structure. The tensile test results are summarized in Figure 6. Baseline specimens from a forged Ti-6Al-4V bar, which had a duplex microstructure, are also shown for comparison purposes. No systematic correlation was found between the position of a sample within the wall and its tensile test results",
            " Sixteen of the as-deposited specimens did not fail after ten million (107) cycles when the test was stopped. In total, twenty-one samples had a fatigue life well above three million cycles. However, there were still three samples where failure occurred below one million cycles, as indicated by the arrows in the 974\u2014VOLUME 44A, FEBRUARY 2013 METALLURGICAL AND MATERIALS TRANSACTIONS A Figure 7. With the limited dataset tested, no statistically valid relationship could be demonstrated between a specimen\u2019s orientation and location (see Figure 2) and its fatigue life. However, of the few specimens that failed below one million cycles, two were orientated horizontally within the wall and the vertical sample that failed prematurely was machined from close to the wall end. Comparing the fatigue test results of the WAAM Ti-6Al-4V samples to the baseline bar specimens shows that the average high-cycle fatigue resistance of WAAM specimens, at this load level, was significantly better than that of the baseline bar specimens. This difference can be attributed to intrinsic microstructural differences between the baseline bar and the deposited material, which will be discussed further below"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure6.28-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure6.28-1.png",
        "caption": "Figure 6.28 (a) The dimensions of the square footing. (b) On the bottom of the footing, the earth pressure has to provide an equilibrium with the force of 56 kN from the column and the footing\u2019s dead weight of 4.6 kN. The resultant of the earth pressure is therefore 60.6 kN.",
        "texts": [
            " For the cross-sectional dimensions of the column in Figure 6.27a, the dead weight per length is (0.2 m)(0.4 m)(24 kN/m3) = 1.92 kN/m. This is a uniformly distributed load acting in the direction of the column axis (see Figure 6.27b). The total dead weight of the column is (5 m)(1.92 kN/m) = 9.6 kN. At its base, the column has to be kept in equilibrium by a force of (46.4 kN) + (5 m)(1.92 kN/m) = 56 kN. 6 Loads 233 An equally large, opposite force is acting on the footing of the column. If the footing is square, and has the dimensions given in Figure 6.28a, the dead weight of the footing is (0.8 m)(0.8 m)(0.3 m)(24 kN/m3) = 4.6 kN. The earth pressure on the bottom of the footing has to be in equilibrium with the force of 56 kN from the column, and the footing\u2019s dead weight of 4.6 kN (see Figure 6.28b) (56 kN) + (4.6 kN) = 60.6 kN. If the earth pressure is uniformly distributed, it equals p = 60.6 \u00d7 103 N (800 mm)(800 mm) = 0.095 N/mm2. In general, the earth pressure is not uniformly distributed. The value given for p is then referred to as the average earth pressure. How the load exerted by the footing is transferred further into the ground, is a problem addressed by the special field of soil mechanics. In reality, the earth pressure on the footing in Figure 6.29 consists of a very large number of very small forces provided by the grains of soil"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000233_978-1-4614-0460-6-Figure6.36-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000233_978-1-4614-0460-6-Figure6.36-1.png",
        "caption": "Fig. 6.36 Force balance to determine m12",
        "texts": [
            " The force balance is shown in Fig. 6.35, where the motionless x2 is represented as a pivot for the rigid, massless bar. The accelerations at the two masses are both \u20acx1 2= \u00bc 1 2= because they are half the distance from the pivot relative to x1. Summing the moments (clockwise moments are taken to be positive) about the x2 pivot gives: X M \u00bc fm11 2l 3 \u00fe m1 \u20acx1 2 l 3 \u00fe m2 \u20acx1 2 l 3 \u00bc fm11 2l 3 \u00fe m1 1 2 l 3 \u00fe m2 1 2 l 3 \u00bc 0: (6.21) fm mg a O mg \u03b8 l x mg sin(q ) = mgq = mg k x l a The forces are displayed in Fig. 6.36. The force summation is: X f \u00bc fm11 \u00fe fm12 m1 \u20acx1 2 \u00fe m2 \u20acx1 2 \u00bc fm11 \u00fe fm12 m1 1 2 \u00fe m2 1 2 \u00bc 0: (6.23) Solving Eq. 6.23 for fm12 gives: fm12 \u00bc fm11 \u00fe m1 2 m2 2 : (6.24) Substituting for fm11 from Eq. 6.22 gives the final expression for fm12 : fm12 \u00bc m1 \u00fe m2 4 \u00fe m1 2 m2 2 \u00bc m1 3m2 4 : (6.25) We will next find m22 by applying unit acceleration to x2 while holding x1 stationary. The corresponding forces are shown in Fig. 6.37. The moment summation about the x1 pivot is: X M \u00bc fm22 2l 3 m1 \u20acx1 2 l 3 m2 3\u20acx1 2 3l 3 \u00bc fm22 2l 3 m1 1 2 l 3 m2 3 2 3l 3 \u00bc 0: (6",
            "^2); realX1_F1 = realQ1_R1 + realQ2_R2; imagX1_F1 = imagQ1_R1 + imagQ2_R2; freq = omega/2/pi; figure(1) subplot(211) plot(freq, realX1_F1, 'k') set(gca,'FontSize', 14) axis([200 800 -2e-6 2e-6]) ylabel('Real({X_1}/{F_1}) (m/N)') grid subplot(212) plot(freq, imagX1_F1, 'k') set(gca,'FontSize', 14) axis([200 800 -3e-6 5e-7]) xlabel('Frequency (Hz)') ylabel('Imag({X_1}/{F_1}) (m/N)') grid Plot your modal fit together with the measured FRF and comment on their agreement. 5. Figures P6.5a through P6.5e show direct, X1/F1, and cross FRFs, X2/F1 through X5/F1, measured on a fixed-free beam. They were measured at the beam\u2019s free end and in 20 mm increments toward its base; see Fig. 6.36. Determine the mode shape associated with the 200 Hz natural frequency. Exercises 239 0 100 200 300 400 \u20135 \u20134 \u20133 \u20132 \u20131 0 x 10\u20136 Frequency (Hz) Im ag (X 1/ F 1) ( m /N ) Fig. P6.5a Direct FRF X1/F1 x 10\u20136 0 100 200 300 400 \u20133.5 \u20133 \u20132.5 \u20132 \u20131.5 \u20131 \u20130.5 0 Frequency (Hz) Im ag (X 2/ F 1) ( m /N ) Fig. P6.5b Cross FRF X2/F1 240 6 Model Development by Modal Analysis x 10\u20136 0 100 200 300 400 \u20132 \u20131.5 \u20131 \u20130.5 0 Frequency (Hz) Im ag (X 3/ F 1) ( m /N ) Fig. P6.5c Cross FRF X3/F1 x 10\u20136 0 100 200 300 400 \u20131 \u20130"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure9.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure9.1-1.png",
        "caption": "FIGURE 9.1. The force system of a vehicle is the applied forces and moments at the tireprints.",
        "texts": [
            "1 Force and Moment In Newtonian dynamics, the forces acting on a system of connected rigid bodied can be divided into internal and external forces. Internal forces are acting between connected bodies, and external forces are acting from outside of the system. An external force can be a contact force, such as the traction force at the tireprint of a driving wheel, or a body force, such as the gravitational force on the vehicle\u2019s body. External forces and moments are called load, and a set of forces and moments acting on a rigid body, such as forces and moments on the vehicle shown in Figure 9.1, is called a force system. The resultant or total force F is the sum of all the external forces acting on a body, and the resultant or 522 9. Applied Dynamics total moment M is the sum of all the moments of the external forces. F = X i Fi (9.1) M = X i Mi (9.2) Consider a force F acting on a point P at rP . The moment of the force about a directional line l passing through the origin is Ml = lu\u0302 \u00b7 (rP \u00d7F) (9.3) where u\u0302 is a unit vector on l. The moment of the force F, about a point Q at rQ is MQ = (rP \u2212 rQ)\u00d7F (9"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure2.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure2.2-1.png",
        "caption": "Fig. 2.2 Origin of World Coordinates defined directly below robot in initial position. ph is the location of the end of the hands in World Coordinates.",
        "texts": [
            "1 illustrates the Humanoid Robot HRP-2, which was developed during the Humanoid Robotics Project [65]. It has a height of 154 cm, and its weight including the batteries is 58 kg. It can walk for about one hour using the batteries alone. The HRP-2 has a total of 30 joints which can be controlled independently. Fig 2.1(b) shows the names of the Links and the position of their Local Coordinates. Our first task is to clearly define the position of each part of the robot. To do this we define a special point directly below the robot in its initial position1 as shown in Fig 2.2. Taking this point as the origin we define a fixed coordinate system whose x axis faces forward, y axis to the left and z axis faces up. We will name these coordinates \u03a3W and hereafter will use this to describe our robots and their motion. Coordinates such as this one are called World Coordinates. By using a common coordinate system to define the robot itself and surrounding objects we can check a safe landing of a foot on the ground, successful hand grasping of an object, undesired collisions with the environment, and so on. 1 Actually it is the intersection between a perpendicular line which goes through the origin of the the Local Coordinates of the Waist Link and the floor. S. Kajita et al., Introduction to Humanoid Robotics, 19 Springer Tracts in Advanced Robotics 101, DOI: 10.1007/978-3-642-54536-8_2, c\u00a9 Springer-Verlag Berlin Heidelberg 2014 20 2 Kinematics Positions defined using World Coordinates are called Absolute Positions. For example, we describe the absolute position of the left hand tip in Fig. 2.2 by the following 3D vectors ph = \u23a1 \u23a3 phx phy phz \u23a4 \u23a6 . We will also use the terms, Absolute Attitude and Absolute Velocity to show they are defined in the World Coordinates. Lets take a look at how the hand tip position ph changes by the rotation of the robot\u2019s shoulder. From Fig. 2.3(a) we can see that the absolute position of the left shoulder is defined by vector pa, and vector r shows the position of the hand end relative to the shoulder. As we can see from this figure, we have ph = pa + r. 2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure4.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure4.2-1.png",
        "caption": "Fig. 4.2 2D inverted pendulum: The simplest model for a human or a walking robot. It consists of the center of mass (CoM) and massless telescopic leg. We measure the inclination of the pendulum \u03b8 from vertical, positive for counter clockwise rotation.",
        "texts": [
            " First, we assume that all the mass of the robot is concentrated at its center of mass (CoM). Second, we assume that the robot has massless legs, whose tips contact the ground at single rotating joints. At last, we only consider the forward/backward and the up/down motions of the robot, neglecting lateral motion. In other words, we assume the robot motion is constrained to the sagittal plane defined by the axis of walking direction and vertical axis. With these assumptions, we model a robot as a 2D inverted pendulum as shown in Fig. 4.2. 108 4 Biped Walking The inputs of the pendulum are the torque \u03c4 at the pivot and the kick force f at the prismatic joint along the leg. The dynamics of the pendulum are described as a couple of differential equations1 as follows r2\u03b8\u0308 + 2rr\u0307\u03b8\u0307 \u2212 gr sin \u03b8 = \u03c4/M r\u0308 \u2212 r\u03b8\u03072 + g cos \u03b8 = f/M. We can simulate the behavior of the inverted pendulum by integrating them numerically with given input torque. One of the important limitations is we cannot use big torque \u03c4 since the feet of biped robot is very small",
            "10 shows the time profile of the position and the velocity of the CoM. The position graph takes origin at each support point, therefore, the plot jumps at the moment of support exchange (b, d) and the amount of the jump indicates the step length. The velocity graph depicts the dynamic change of CoM speed and the peaks at support exchange. Although the method explained so far was limited to a walk on level ground, we can use the same method for walking on uneven terrain with a small modification. Let us explain this. Returning to the inverted pendulum model of Fig. 4.2, we consider the case that the center of mass moves on a sloped line as illustrated in Fig. 4.11 described by z = kx+ zc (4.20) where k is the inclination of the line, zc is the z intersection of the line. We call the line which the CoM moves along as constraint line. Let us calculate the kick force f to realize such motion. First, we decompose the kick force f into the horizontal part fx and the vertical part fz fx = f sin \u03b8 = (x/r)f (4.21) fz = f cos \u03b8 = (z/r)f. (4.22) 4.2 Two Dimensional Walking Pattern Generation 117 To let the CoM move along the constraint line, the sum of the kick force and the gravity force must be parallel with the constraint line"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002200_j.matpr.2020.02.635-Figure17-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002200_j.matpr.2020.02.635-Figure17-1.png",
        "caption": "Fig. 17. (a) Process schematic diagram o",
        "texts": [
            " It is currently a very popular method for fabrication of metal parts. In SLM, a laser is fitted on the top with the set of lenses focused on the powdered material for the solidification of the layer. Once a layer is sintered the build platform goes down and the recoated arm makes a new layer on the top of the bed, this whole process will repeat until the whole part is not produced f SLM, (b) Products of SLM [68,69]. on process in additive manufacturing: An overview, Materials Today: Pro- [64,65]. The minimum thickness of powder material layer is 0.020 mm (Fig. 17). Recently, high power laser with the development of fibre optics are also added in SLM to process different type of metallic materials such as tungsten, copper & aluminium. Therefore, various types of metallic materials can be sintered on SLM. This enabled many research opportunities are developed in SLM for composite and ceramic materials. This methods has the advantages to produce high precision and high quality products [66,67]. Direct Metal Laser Sintering (DMLS): Direct metal laser sintering (DMLS) was first developed by EOS firm, Germany in 1995"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000252_j.automatica.2006.10.008-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000252_j.automatica.2006.10.008-Figure3-1.png",
        "caption": "Fig. 3. Pendulum and performance of controllers (25), (26), (27) (b, f) and (25), (26), (28) (c, d, e).",
        "texts": [
            " In the presence of small noises and sampling intervals the resulting motion will take place in a small vicinity of the 2-sliding mode = \u0307 = 0. Thus, if this motion does not leave , the homogeneous dynamics is still in charge, and the statements of Theorems 1\u20134 are true. A number of new controllers from the previous section are demonstrated here. Consider an academic example of a variable-length pendulum with motions restricted to some vertical plane. A load of a known mass m moves without friction along the pendulum rod (Fig. 3a). Its distance from O equals R(t) and is not measured. An engine transmits a torque u, which is considered as control. The task is to track some function xc given in real time by the angular coordinate x of the rod. The system is described by the equation x\u0308 = \u22122 R\u0307 R x\u0307 \u2212 g 1 R sin x + 1 mR2 u, (24) where m = 1 and g = 9.81 is the gravitational constant. Let 0 < Rm R RM, R\u0308, R\u0308, x\u0307c and x\u0308c be bounded, = x \u2212 xc be available. Following are the \u201cunknown\u201d functions R and xc considered in the simulation: R = 0",
            " Note that it is the first time when such a twisting-type convergence is demonstrated for this controller. The phase trajectories in the plane , \u0307 and the first 0.1 s of the differentiator convergence are shown in Figs.3b,c respectively, the corresponding accuracies being | | = |x \u2212 xc| 4.2 \u00d7 10\u22125, |\u0307| = |x\u0307 \u2212 x\u0307c| 2.7 \u00d7 10\u22122. After the sampling step was reduced from 10\u22124 to 10\u22125 the resulting accuracies changed to |x \u2212 xc| 4.8 \u00d7 10\u22127, |\u0307| = |x\u0307 \u2212 x\u0307c| 3.1 \u00d7 10\u22123 which corresponds to Theorem 4. The differentiator convergence is demonstrated in Fig. 3f. Consider the quasi-continuous controller u = \u221210 \u0307 + | |1/2 sign |\u0307| + | |1/2 . (28) The trajectory and 2-sliding tracking performance in the absence of noises are shown in Figs. 3d and c, respectively, the corresponding accuracies being | | = |x \u2212 xc| 5.4 \u00d7 10\u22126, |x\u0307 \u2212 x\u0307c| 1.0 \u00d7 10\u22122 with = 0.0001. Control (28) is demonstrated in Fig. 3e. It is seen from the graph that the control remains continuous until the entrance into the 2-sliding mode = \u0307 = 0. The tracking results of (25), (26), (28) and the differentiator performance in the presence of a noise with the magnitude 0.01 are demonstrated in Figs. 4c,d, respectively, the tracking accuracy being | | = |x \u2212 xc| 0.036. The noise was a periodic non-smooth function with non-zero average. The performance does not significantly change, when the frequency of the noise varies from 10 to 100 000",
            " Without changing the control values, (33) is constructed so that the overflow be avoided during the computer simulation (or practical implementation). In the absence of noises the controller provides in finite time for keeping + \u0307 \u2261 0. As a result the asymptotically stable 3-sliding mode =\u0307=\u0308=0 is established. Note that since (19) is quasi-continuous, the applied controller (29)\u2013(34) produces the control u, whose derivative u\u0307 remains continuous until the entrance into the 2-sliding mode =\u0307=0. The graph of u\u0307 is very similar to Fig. 3e and is omitted. The graphs of 3-sliding deviations , \u0307, \u0308 and the control u are demonstrated in Figs. 5a,c, respectively. The 2-sliding convergence in the plane , \u0307 is demonstrated in Fig. 5b. It is seen from Fig. 5d that + \u0307 \u2261 0 is kept in 2-sliding mode. Due to this equality the accuracies achieved at t = 10 are of the same order: | | = |x \u2212 xc| 4.0\u00d710\u22124, |\u0307|=|x\u0307\u2212 x\u0307c| 4.0\u00d710\u22124, |\u0308|=|x\u0308\u2212 x\u0308c| 4.0 \u00d7 10\u22124. 2-Sliding homogeneity and contractivity are shown to provide for all needed features of 2-sliding-mode controllers (Theorems 1\u20134)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000166_978-1-4471-4396-3_11-Figure11.13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000166_978-1-4471-4396-3_11-Figure11.13-1.png",
        "caption": "Fig. 11.13 Top and front views of a conventional aircraft in Example 11.5",
        "texts": [
            "1 gives a summary of the four robustness modifications that were introduced in this chapter. Table 11.2 presents an overview of the continuous Projection Operator, which acts on a pair of n-dimensional vectors y and y. The next example illustrates key design points in application of the projectionbased MRAC to lateral-directional dynamics of an aircraft. Example 11.5 Aircraft Lateral-Directional Dynamics and Control Lateral-directional motion of a conventional aircraft is controlled by vertical tail panels (rudders) and wing-mounted surfaces (ailerons). Figure 11.13 shows a sketch. The rudder dr\u00f0 \u00de is the primary control device for turning the aircraft, thus regulating the vehicle yaw rate r and the sideslip angle b . Moving ailerons differentially (i.e., left aileron trailing edge down and right aileron trailing edge up, da) will force the aircraft to roll (right wing down), changing (increasing) its roll rate p, and thus the bank angle \u2019, with some induced coupling into the yaw and sideslip dynamics. For small angles, the aircraft lateral-directional dynamics can be approximated by a linear time-invariant system in the form whereg \u00bc 32:174 is the acceleration due to gravity (ft/s2),V is the trimmed airspeed (positive constant, ft/s), and the system matrices Ap; Bp are comprised of the vehicle aerodynamic stability and control derivatives"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure1.13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure1.13-1.png",
        "caption": "Fig. 1.13. Faraday\u2019s disc: a nonmagnetic conductive disc rotating in a stationary magnetic field according to Numerical example 1.2 .",
        "texts": [
            "21 rad/s Critical speed of rotation of the system according to Dunkerley equation ncr = \u2126cr 2\u03c0 = 614.21 2\u03c0 = 97.75 rev/s = 5865.3 rpm Critical speed of rotation of the system according to Rayleigh\u2019s method - eqn (1.8) ncr = 1 2\u03c0 \u221a 9.81(5.372\u00d7 2.356\u00d7 10\u22129 + 13.519\u00d7 1.135\u00d7 10\u22127 + 2.178\u00d7 2.59\u00d7 10\u22125) 5.372\u00d7 (2.356\u00d7 10\u22129)2 + 13.519\u00d7 (1.135\u00d7 10\u22127)2 + 2.178\u00d7 (2.59\u00d7 10\u22125)2 = 99.3 rev/s = 5958.15 rpm The results according to Dunkerley and Rayleigh are similar. 22 1 Introduction Numerical Example 1.2 A copper disc with its dimensions as shown in Fig. 1.13 rotates with the speed of 12 000 rpm between U-shaped laminated pole shoes of a PM. A sliding contact consisting of two brushes is used to collect the electric current generated by this primitive homopolar generator: one brush slides on the external diameter Dout = 0.232 m and the second brush is located directly below one of the poles at the distance of 0.5Din = 0.03 m from the axis of the disc. The remanent magnetic flux density of the NdFeB PM is Br = 1.25 T, coercivity is Hc = 950 000 A/m and height 2hM = 0"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000148_adma.200904059-Figure24-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000148_adma.200904059-Figure24-1.png",
        "caption": "Figure 24 . A ribbon covered with an azo-doped LCE is arranged around two pulleys. Upon irradiation with UV light on one side and visible light on the other, the fi lm starts to contract on one side and expands on the other, and it starts to rotate. Reprinted with permission. [ 144 ] Copyright 2008, Wiley-VCH.",
        "texts": [
            " The controlled use of this effect will be discussed below. In the following we will present examples from the literature that move these fascinating materials closer to real-world applications. Many of these systems use more densely crosslinked LC polymers, which show smaller deformations, but are mechanically more robust then the classical LCEs. Yamada et al. succeeded in constructing a motor based on an LCE material that transforms light energy into motion. [ 144 ] For this an LCE-covered ribbon was placed around two pulleys ( Figure 24 ). On the right side the belt is irradiated with UV light (leading to a trans\u2013cis isomerization) and on the left \u00a9 2010 WILEY-VCH Verlag G mbH & Co. KGaA, Weinheim Adv. Mater. 2010, 22, 3366\u20133387 side with visible light ( cis\u2013trans re-isomerization). Contraction on one side and expansion on the other leads to a rotation of the pulleys. An LCE-driven silicon gripper was introduced by S\u00e1nchez-Ferrer and et al. [ 145 ] The preparation of the actuating elastomer by photocrosslinking was integrated into a fabrication process for mMicro-electromechanical systems (MEMS)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure2.30-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure2.30-1.png",
        "caption": "Fig. 2.30 Calculating the Jacobian: We consider the motion of the end effector that results from minute movements in each joint.",
        "texts": [
            " In the previous section we introduced the Jacobian which gives you the relationship between small joint movements and spatial motion. Through the Jacobian we can also calculate the torque requirements of the joints in order to generate external forces through the hands and feet. As this is used extensively in robot control, papers on robotics that do not have a Jacobian somewhere in them are a rare thing indeed19. 19 We can find out how the Jacobian is used in robotics by reading Dr Yoshikawa\u2019s textbook [144]. 58 2 Kinematics Below we will go over the actual procedure of calculating the Jacobian. Figure 2.30 shows a chain with N links floating in space. We will assume that they are numbered in order from the base (link 1) to the end (link N). We will assume the end effector (robot hand or foot) is attached to the Nth link. Furthermore, we will assume that the forward kinematics have been calculated and the position and attitude of each link is already stored (jth link has pj , Rj). In this chain of links, let us assume we kept all the joints fixed except for the 2nd joint, which we turned by a small angle, \u03b4q2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000286_978-94-017-0359-8-FigureD.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000286_978-94-017-0359-8-FigureD.6-1.png",
        "caption": "Figure D.6. NASA's dream of a future flight vehicle.",
        "texts": [
            " Enhancing Group and Societal Outcomes Inside the cockpit compartment, the pilot sees everything on a 3-D display that shows local weather, accentuates obstacles, all near-by aircraft, and the safest flight path. The on-board clear air turbulence sensor uses lasers to detect unsteady air well ahead of the aircraft to assure a smooth ride. When approaching a major airport, the lingering vortices that were shed from the wingtips of larger aircraft and that can upset a smaller one, can be easily avoided. This is a long-term vision, but emerging technology can make it real. A key to achieving this vision is a fusion of nanoscale technology with biology and information technology (Figure D.6). An example is intelligent multifunctional material systems consisting of a number of layers, each used for a different purpose. The outer layer would be selected to be tough and durable to withstand the harsh space environment, with an embedded network of sensors, electrical carriers, and actuators to measure temperature, pressure, and radiation and to trigger a response whenever needed. The network would be intelligent. It would automatically reconfigure itself to bypass damaged components and compensate for any loss of capability"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-FigureC.10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-FigureC.10-1.png",
        "caption": "FIGURE C.10 Partial inductance for two rectangular conductors.",
        "texts": [],
        "surrounding_texts": [
            "Lp12 = \ud835\udf070 4\ud835\udf0b 1 12 4\u2211 k=1 4\u2211 \ud835\udcc1=1 4\u2211 m=1 (\u22121)k+\ud835\udcc1+m+1 [ \u2212 b\ud835\udcc1 cm a3 k 6 tan\u22121 ( b\ud835\udcc1 cm ak R ) PARTIAL INDUCTANCE FORMULAS FOR ORTHOGONAL GEOMETRIES 395 \u2212 b\ud835\udcc1 c3 m ak 6 tan\u22121 ( b\ud835\udcc1 ak cm R ) \u2212 b3 \ud835\udcc1 cm ak 6 tan\u22121 ( cm ak b\ud835\udcc1 R ) + ak ( b2 \ud835\udcc1c2 m 4 \u2212 b4 \ud835\udcc1 24 \u2212 c4 m 24 ) log \u239b\u239c\u239c\u239c\u239d ak + R\u221a b2 \ud835\udcc1 + c2 m \u239e\u239f\u239f\u239f\u23a0 + b\ud835\udcc1 ( c2 m a2 k 4 \u2212 c4 m 24 \u2212 a4 k 24 ) log \u239b\u239c\u239c\u239c\u239d b\ud835\udcc1 + R\u221a c2 m + a2 k \u239e\u239f\u239f\u239f\u23a0 + cm ( b2 \ud835\udcc1a2 k 4 \u2212 b4 \ud835\udcc1 24 \u2212 a4 k 24 ) log \u239b\u239c\u239c\u239c\u239d cm + R\u221a a2 k + b2 \ud835\udcc1 \u239e\u239f\u239f\u239f\u23a0 + 1 60 (b4 \ud835\udcc1 + c4 m + a4 k \u2212 3b2 \ud835\udcc1c2 m \u2212 3c2 ma2 k \u2212 3a2 kb2 \ud835\udcc1) R ] (C.40) with a1 = xs2 \u2212 xe1, a2 = xe2 \u2212 xe1 (C.41a) a3 = xe2 \u2212 xs1, a4 = xs2 \u2212 xs1 (C.41b) b1 = ys2 \u2212 ye1, b2 = ye2 \u2212 ye1 (C.41c) b3 = ye2 \u2212 ys1, b4 = ys2 \u2212 ys1 (C.41d) c1 = zs2 \u2212 ze1, c2 = ze2 \u2212 ze1 (C.41e) c3 = ze2 \u2212 zs1, c4 = zs2 \u2212 zs1 (C.41f) 1 = (ye1 \u2212 ys1)(ze1 \u2212 zs1), 2 = (ye2 \u2212 ys2)(ze2 \u2212 zs2) (C.41g) and R(ak, b\ud835\udcc1 , cm) = \u221a a2 k + b2 \ud835\udcc1 + c2 m. (C.41h) We added this formula for completeness. Unfortunately, even using double precision arithmetic leads only to limited accuracy for reasonably large parameters in this formulation. For this reason, using thin layer approximations in Section C.1.5 with numerical integration along the thickness leads to more stable results. C.1.9 1\u2215R3 Kernel Integral for Parallel Rectangular Sheets Partial elements are of importance for other formulation such as the ones in Chapter 11 so that other material properties can be taken into account. In some of these formulations, integrals with a higher order Green\u2019s function need to be solved. Specifically, the integral for a 1\u2215R3 kernel is important. Here, we consider the geometry to be two rectangles shown in Fig. C.11. 396 COMPUTATION OF PARTIAL INDUCTANCES Ip12 = \u222b xe1 xs1 \u222b ye1 ys1 \u222b xe2 xs2 \u222b ye2 ys2 (z1 \u2212 z2) R3 dy2 dx2 dy1 dx1, (C.42) where, R3 = [(x1 \u2212 x2)2 + (y1 \u2212 y2)2 + (z1 \u2212 z2)2]3\u22152. Then the final formulation of Ip12 for two orthogonal current sheets is Ip12 = 2\u2211 m=1 2\u2211 n=1 2\u2211 i=1 2\u2211 j=1 (\u22121)m+n+i+j(z1 \u2212 z2) \u23a7\u23aa\u23a8\u23aa\u23a9 \u2212 Rmnij + (xvm \u2212 xvn) ln[(xvm \u2212 xvn) + Rmnij] + (xvm \u2212 xvn)(yvi \u2212 yvj) tan\u22121 [ (xvm \u2212 xvn)(z1 \u2212 z2) (yvi \u2212 yvj)2 + (z1 \u2212 z2)2 + (yvi \u2212 yvj)Rmnij ] + (xvm \u2212 xvn)(yvi \u2212 yvj) tan\u22121 [ z1 \u2212 z2 xm \u2212 xn ] + (z1 \u2212 z2)2(yvi \u2212 yvj) \u22c5 ln \u23a1\u23a2\u23a2\u23a2\u23a3 \u221a ((yvi \u2212 yvj)2 + (z1 \u2212 z2)2 + Rmnij(yvi \u2212 yvj))2 + (xvm \u2212 xvn)2(z1 \u2212 z2)2\u221a (xvm \u2212 xvn)2 + (z1 \u2212 z2)2 \u23a4\u23a5\u23a5\u23a5\u23a6 \u23ab\u23aa\u23ac\u23aa\u23ad (C.43) Rmnij = \u221a (xvm \u2212 xvn)2 + (yvi \u2212 yvj)2 + (z1 \u2212 z2)2, where the subscripts are for m = 1 \u2236 vm = s1; and for m = 2 \u2236 vm = e1; (C.44) n = 1 \u2236 vn = s2; and for n = 2 \u2236 vn = e2; (C.45) PARTIAL INDUCTANCE FORMULAS FOR ORTHOGONAL GEOMETRIES 397 i = 1 \u2236 vi = s1; and for i = 2 \u2236 vn = e1; (C.46) j = 1 \u2236 vj = s2; and for j = 2 \u2236 vn = e2. (C.47) C.1.10 1\u2215R3 Kernel Integral for Orthogonal Rectangular Sheets A second important integral with a 1\u2215R2 Green\u2019s function is for two orthogonal rectangles shown in Fig. C.12. For this case, we want to give a solution for the following integral Ip12 = \u222b xe1 xs1 \u222b ye1 ys1 \u222b xe2 xs2 \u222b ze2 zs2 z1 \u2212 z2 R3 dx1 dy1 dx2 dz2, (C.48) where, R3 = [(x1 \u2212 x2)2 + (y1 \u2212 y2)2 + (z1 \u2212 z2)2]3\u22152. Then the final mutual high order coupling formulation Ip12 for two orthogonal current sheets is Ip12 = 2\u2211 m=1 2\u2211 n=1 2\u2211 i=1 2\u2211 j=1 (\u22121)m+n+i+j { \u2212 0.5(y2 \u2212 yvi)Rmnij + (y2 \u2212 yvi)(xvn \u2212 xvm) tanh\u22121 ( Rmnij xvn \u2212 xvm ) + 0.5[(xvn \u2212 xvm)2 \u2212 (zvj \u2212 z1)2] ln [(y2 \u2212 yvi) + Rmnij] + (xvn \u2212 xvm)(zvj \u2212 z1) tan\u22121 (zvj \u2212 z1) y2 \u2212 yvi \u2212 (xvn \u2212 xvm)(zvj \u2212 z1) tan\u22121 (zvj \u2212 z1)2 + (xvn \u2212 xvm)2 + Rmnij(xvn \u2212 xvm) (y2 \u2212 yvi)(zvj \u2212 z1) } (C.49) 398 COMPUTATION OF PARTIAL INDUCTANCES where the subscripts are for m = 1 \u2236 vm = s1; and for m = 2 \u2236 vm = e1; (C.50) n = 1 \u2236 vn = s2; and for n = 2 \u2236 vn = e2; (C.51) i = 1 \u2236 vi = s1; and for i = 2 \u2236 vi = e1; (C.52) j = 1 \u2236 vj = s2; and for j = 2 \u2236 vj = e2. (C.53) And Rmnij = \u221a (xvn \u2212 xvm)2 + (y2 \u2212 yvi)2 + (zvj \u2212 z1)2. C.2 PARTIAL INDUCTANCE FORMULAS FOR NONORTHOGONAL GEOMETRIES This section is dedicated to the computation of nonorthogonal partial inductance values. The general integrals for the computation of partial inductance for nonorthogonal such as nonorthogonal hexahedral elements are derived in Chapter 7, equation (7.18) as Lpaa\u2032 = \ud835\udf070\u222bc\u222bb\u222ba \u222bc\u2032\u222bb\u2032\u222ba\u2032 (a\u0302 \u22c5 a\u0302\u2032) ha\u2032 g(rg; r\u2032g) da\u2032 db\u2032 dc\u2032ha da db dc. (C.54) It is clear that analytical solutions are even more challenging for nonorthogonal geometries. Still, analytical integration needs to be used as much as possible, especially for the longer dimensions of the cells. Since many problems are in this class, we want to efficiently solve problems with relatively long cells such as a thin wire-type structure. To accomplish this, we pursue a mixed solution using analytical results wherever possible. Few formulas will provide a simple, complete answer but rather are important building blocks that lead to a combined analytical/numerical solution as is shown in Appendix E. C.2.1 Rotation for Different Nonorthogonal Conductor Orientations The formulas for the translation and rotation of the coordinate systems are given at the beginning of this chapter in (C.2)\u2013(C.5). Many models given in this section are constructed using non-orthogonal wire models. In these cases we need to rotate the orientation of the structures. The problem at hand has to be rotated into the orientation of the solution given in this text. To achieve this, we have to use the following steps: \u2022 The coordinate system is shifted such that one wire is placed at the right distance from the origin. Next, find the angle \ud835\udf13 of the second wire\u2019s projection on the xOy plane, where O is the origin. \u2022 Rotate the original coordinate system around the z\u0302 axis by the angle \u2212\ud835\udf13 such that the second wire is placed on the yOz plane of the new coordinate system. \u2022 Shift the coordinate system along the z\u0302 axis so that one end of the first wire is moved the right distance from the xOy plane. PARTIAL INDUCTANCE FORMULAS FOR NONORTHOGONAL GEOMETRIES 399 C.2.2 Lp for Arbitrary Oriented Wires in the Same Plane z= 0 We provide several solutions for filaments since wire filaments are useful basic elements for non-orthogonal Lp\u2019s. These filaments are used for non orthogonal cells in conjunction with numerical integrations in the cross section. The first situation consists of two wires in the z = 0 plane as shown in Fig. C.13. The integrals for two filaments are Lp12 = \ud835\udf070 4\ud835\udf0b\u222b\ud835\udcc11 \u222b\ud835\udcc12 t\u0302\ud835\udfcf \u22c5 t\u0302\ud835\udfd0 R1,2 d\ud835\udcc12 d\ud835\udcc11 (C.55) where \ud835\udcc11 and \ud835\udcc12 are the two wire segments shown in Fig. C.13 and the dot product between the tangential vectors is t\u0302\ud835\udfcf \u22c5 t\u0302\ud835\udfd0. This first result is derived from [1, 2]. The partial inductance between the filaments 1 and 2 is given by Lp12 = \ud835\udf070 4\ud835\udf0b CE [ 2 (T2 + \ud835\udcc12) tanh\u22121 ( \ud835\udcc11 R1 + R2 ) + 2 (T1 + \ud835\udcc11) tanh\u22121 ( \ud835\udcc12 R1 + R4 ) \u2212 2 T2 tanh\u22121 ( \ud835\udcc11 R3 + R4 ) \u22122 T1 tanh\u22121 ( \ud835\udcc12 R2 + R3 ) \u2212 W d SE ] . (C.56) where W = [ tan\u22121 ( CDQE + (T2 + \ud835\udcc12)(T1 + \ud835\udcc11) SEQ R1 SED ) \u2212 tan\u22121 ( CDQE + (T2 + \ud835\udcc12) T1 SEQ R2 SED ) 400 COMPUTATION OF PARTIAL INDUCTANCES + tan\u22121 ( CDQE + T1 T2 SEQ R3 SED ) \u2212tan\u22121 ( CDQE + (T1 + \ud835\udcc11) T2 SEQ R4 SED )] (C.57) The following definitions are used for the two filaments: a1 = xe2 \u2212 xe1 , a2 = xe2 \u2212 xs1 (C.58) a3 = xs2 \u2212 xs1 , a4 = xe1 \u2212 xs2 (C.59) a5 = xe1 \u2212 xs1 , a6 = xe2 \u2212 xs2 (C.60) b2 = (ye2 \u2212 ys2) (1 \u2212 10\u22128) , (C.61) \ud835\udcc11 = a5 , \ud835\udcc12 2 = a2 6 + b2 2 (C.62) R2 1 = a2 1 + y2 e2 , R2 2 = a2 2 + y2 e2 (C.63) R2 3 = a2 3 + y2 s2 , R2 4 = a2 4 + y2 s2 (C.64) S2 1 = R2 4 \u2212 R2 3 + R2 2 \u2212 R2 1 , S2 2 = R2 2 \u2212 R2 3 \u2212 \ud835\udcc12 2 (C.65) S2 3 = R2 4 \u2212 R2 3 \u2212 \ud835\udcc12 1 , S2 4 = 4 \ud835\udcc12 1 \ud835\udcc12 2 \u2212 (S2 1) 2 (C.66) The translations for the start of the wires are given by T1 = \ud835\udcc11 2 \ud835\udcc12 2 S2 3 + S2 1 S2 2 S2 4 , T2 = \ud835\udcc12 2 \ud835\udcc12 1 S2 2 + S2 1 S2 3 S2 4 . (C.67) were we use the notation for the angle \ud835\udf00: CE = cos (\ud835\udf00) = S2 1 2 \ud835\udcc11 \ud835\udcc12 , SEQ = sin2(\ud835\udf00) = 1 \u2212 C2 E (C.68a) SE = \u221a SEQ, CDQE = d2 CE, SED = d SE (C.68b) where d2 = R2 3 \u2212 T2 2 \u2212 T2 1 + 2 T1 T2 CE (like in equation C. 79). Again, this filament-to-filament partial mutual inductance is used in combination with numerical integration for different cross-sections, for example, [16]. This equation should be used if the wires are located in the same plane since this formula is simpler than the ones in the next sections. Please note that we did add a small number of the orientation of the wire such that parallel wires can be treated by this formula. PARTIAL INDUCTANCE FORMULAS FOR NONORTHOGONAL GEOMETRIES 401 C.2.3 Lp for Wire Filaments with an Arbitrary Direction In this section, the filaments are oriented in any mutual orientation as shown in Fig. C.14. This allows an arbitrary relative orientation of the filaments. Using the rotation and translation operations in (C.2) to (C.5), we can place the two wires in any location in the global rectangular coordinate system. Hence, this result will be ideally suited for the filament representation of non-orthogonal quadrilateral and hexahedral shapes. The integral for this case can be set up as Lp12 = \ud835\udf070 cos \ud835\udf00 4\ud835\udf0b \u222b xe1 xs1 \u222b \ud835\udefc=1 \ud835\udefc=0 d\ud835\udefc dx|r\ud835\udfcf \u2212 r2| (C.69) where the angle \u03b5 is defined as in [1] for the dot product and the vector is r\ud835\udfd0 = (1 \u2212 \u03b1) rs2 + \u03b1 re2 (C.70) where rs2 and re2 are the start and end point of filament 2, respectively, in the global coordinate system. It is important to minimize the number of divisions and multiplications. For this reason, we present the formulation in a more compute friendly form Lp12 = \ud835\udf070 4\ud835\udf0b CE [ 2 (T2 + \ud835\udcc12) tanh\u22121 ( \ud835\udcc11 R1 + R2 ) + 2 (T1 + \ud835\udcc11) tanh\u22121 ( \ud835\udcc12 R1 + R4 ) \u2212 2 T2 tanh\u22121 ( \ud835\udcc11 R3 + R4 ) \u22122 T1 tanh\u22121 ( \ud835\udcc12 R2 + R3 ) \u2212 W d SE ] (C.71) 402 COMPUTATION OF PARTIAL INDUCTANCES where W = [ tan\u22121 ( CDQE + (T2 + \ud835\udcc12)(T1 + \ud835\udcc11) SEQ R1 SED ) \u2212 tan\u22121 ( CDQE + (T2 + \ud835\udcc12) T1 SEQ R2 SED ) + tan\u22121 ( CDQE + T1 T2 SEQ R3 SED ) \u2212tan\u22121 ( CDQE + (T1 + \ud835\udcc11) T2 SEQ R4 SED )] (C.72) The following definitions are used for the arbitrary filaments: a1 = xe2 \u2212 xe1 , a2 = xe2 \u2212 xs1 a3 = xs2 \u2212 xs1 , a4 = xs2 \u2212 xe1 (C.73) a5 = xe1 \u2212 xs1 , a6 = xe2 \u2212 xs2 \ud835\udcc11 = a5 , \ud835\udcc12 2 = a2 6 + b2 2 + c2 2 (C.74) b2 = ye2 \u2212 ys2 , c2 = ze2 \u2212 zs2 R2 1 = a2 1 + y2 e2 + z2 e2 , R2 2 = a2 2 + y2 e2 + z2 e2 R2 3 = a2 3 + y2 s2 + z2 s2 , R2 4 = a2 4 + y2 s2 + z2 s2 (C.75) S2 1 = R2 4 \u2212 R2 3 + R2 2 \u2212 R2 1 , S2 2 = R2 2 \u2212 R2 3 \u2212 \ud835\udcc12 2 (C.76) S2 3 = R2 4 \u2212 R2 3 \u2212 \ud835\udcc12 1 , S2 4 = 4\ud835\udcc12 1\ud835\udcc1 2 2 \u2212 (S2 1) 2 (C.77) The translations for the start of the wires are given by T1 = \ud835\udcc11 2 \ud835\udcc12 2 S2 3 + S2 1 S2 2 S2 4 , T2 = \ud835\udcc12 2 \ud835\udcc12 1 S2 2 + S2 1 S2 3 S2 4 (C.78) Another definition from Grover we use is d2 = R2 3 \u2212 T2 2 \u2212 T2 1 + 2 T1 T2 CE (C.79) were we use the notation for the angle \ud835\udf00: CE = cos (\ud835\udf00) = S2 1 2 \ud835\udcc11 \ud835\udcc12 , SEQ = sin2(\ud835\udf00) = 1 \u2212 C2 E (C.80a) SE = \u221a SEQ, CDQE = d2 CE, SED = d SE (C.80b) PARTIAL INDUCTANCE FORMULAS FOR NONORTHOGONAL GEOMETRIES 403 Again, this filament-to-filament partial mutual inductance is used in combination with numerical integration for different cross-sections, for example, [16]. Please note that a small number could be added to the coordinates for parallel wires as we did in the formula for the in plane case in the previous section. C.2.4 Lp for Two Cells or Bars with Same Current Direction In this example, we assume that all the current filaments are in the same direction, even if conductor cross sections are not parallel to the y\u2013z coordinates as shown in Figure C.15. In this case, the unit vectors are the same or, x\u0302 = a\u0302. Hence, the wire-to-wire formula (C.8) or the sheet-to-wire inductance in Section C.1.3 can be utilized. Then, we can represent the partial inductance in the form Lp12 = \ud835\udf070 4\ud835\udf0b \u222b xe1 xs1 \u222b ye1 ys1 \u222b ze1 zs1 \u222bx2 \u222bb2 \u222bc2 1 R1,2 dx1 dy1 dz1 dx2 db2 dc2, (C.81) where R1,2 = R[r1(x1, y1, z2), r2(x2, b2, c2)]. In this specific example, conductor 1 is in orthogonal coordinates and conductor 2 is in local coordinates. Of course, if we apply the partial inductance for a sheet and a wire in (C.69), then the numerical integration for the sheet results in the numerical integration in the z-direction for conductor 1 and in the b\u2032 and c\u2032 direction for conductor 2. C.2.5 Lp for Arbitrary Hexahedral Partial Self-Inductance Progress in this area is also based on the Gauss law (3.33). Specifically, it was shown in Ref. [11] that the result is similar to quadrilateral surfaces (Section D.2.2). The resultant surface integral is obtained by starting with 1 R(r, r\u2032) = \u22121 2 \u2207\u2032 \u00d7 \u2207 R(r, r\u2032). (C.82) 404 COMPUTATION OF PARTIAL INDUCTANCES With this, the integration over the surface of the hexahedral shape replaces the integration over the volume as in Refs. [12] and [13] similar to (7.34) for surfaces Ip11 = \u22121 2 6\u2211 i=1 6\u2211 j=1 \u222b\u2032 i \u222bj R (u\u0302j \u22c5 u\u0302\u2032 i) dj d \u2032 i , (C.83) where in this case, u\u0302i is the normal to the surface i. We should note that the dot product between the current directions for the partial self-terms is given by currents in the same direction. The Gauss law was applied in the local coordinate domain in Ref. [14] and good results have been obtained using the Gauss law as well as other results using approximate shapes for some of the integrals yielded very good results in Ref. [13]. C.2.6 Lp for Arbitrary Hexahedral Partial Mutual Inductance The partial self- and mutual inductances for arbitrary hexahedral shapes are more challenging. We can use local coordinates to compute the partial mutual inductance for two arbitrarily placed hexahedral elements. An example for such elements is given in Refs [15, 16]. Figure C.16 shows two volume conductor cells. We assume that the local coordinate system for the first conductor is (a, b, c) and (a\u2032, b\u2032, c\u2032) for the second one. A vector in the global coordinates located on the first conductor is r = (x, y, z) for the first hexahedral element, and another one is r\u2032 = (x\u2032, y\u2032, z\u2032). The current directions are indicated by dashed line are pointing in the a\u0302 and a\u0302\u2032 directions as is shown in Fig. C.16 and Lp12 = \ud835\udf070\u222ba\u222bb\u222bc\u222ba\u2032\u222bb\u2032\u222bc\u2032 a\u0302 \u22c5 a\u0302\u2032 |||| \ud835\udf15r \ud835\udf15a |||| |||| \ud835\udf15r\u2032 \ud835\udf15a\u2032 |||| g(r, r\u2032) dv dv\u2032, (C.84) where g is the scalar Green\u2019s function g(r, r\u2032) = e\u2212j\ud835\udefd|r\u2212r\u2032| 4\ud835\udf0b|r \u2212 r\u2032| or, for a quasi-static case g(r, r\u2032) = 1 4\ud835\udf0b|r \u2212 r\u2032| . (C.85) From the local coordinate system, we can determine a set of filaments in the global coordinate using the transform equations in section (8.2.6) to find the end points of each wire filament. Section D.2.2 can be used together with the numerical Gaussian integration. In Appendix E, the numerical integration is discussed which can be applied to the cross-sections to yield the partial inductance of interest. To give more details, the local coordinate system presented in Chapter 8 is used to represent the hexahedral element shown in Fig. C.16. The purpose of the local coordinates is to identify the location of the points for the filaments in terms of the variables a, b, c where a \u2208 [\u22121,+1] and where a can represent a = a, b, c. The purpose is to uniquely map a point a, b, c into a point in the global coordinates r. The local coordinates for the filament are used to map them to the global x, y, z coordinates. Mapping a point in the above hexahedron PARTIAL INDUCTANCE FORMULAS FOR NONORTHOGONAL GEOMETRIES 405 from a local coordinate point a, b, c to a global coordinate point x, y, z is described by x = 7\u2211 k=0 Nk(a, b, c) xk, (C.86) which is applied for x = x, y, z. The coefficients in (C.86) are repeated here as N0 = 1\u22158(1 \u2212 a)(1 \u2212 b)(1 \u2212 c) (C.87a) N1 = 1\u22158(1 \u2212 a)(1 \u2212 b)(1 + c) (C.87b) N2 = 1\u22158(1 \u2212 a)(1 + b)(1 \u2212 c) (C.87c) N3 = 1\u22158(1 \u2212 a)(1 + b)(1 + c) (C.87d) N4 = 1\u22158(1 + a)(1 \u2212 b)(1 \u2212 c) (C.87e) N5 = 1\u22158(1 + a)(1 \u2212 b)(1 + c) (C.87f) N6 = 1\u22158(1 + a)(1 + b)(1 \u2212 c) (C.87g) N7 = 1\u22158(1 + a)(1 + b)(1 + c), (C.87h) where a \u2208 [\u22121,+1] and again a = a, b, c. The close relation to the binary variables has been given before. With this, we are in a position to also express the tangential vectors with respect to the local coordinates as \ud835\udf15r \ud835\udf15a = \ud835\udf15x \ud835\udf15a x\u0302 + \ud835\udf15y \ud835\udf15a y\u0302 + \ud835\udf15z \ud835\udf15a z\u0302, (C.88) 406 COMPUTATION OF PARTIAL INDUCTANCES where the derivatives are found from (C.86). Finally, the magnitude of the tangential vector ha = |\ud835\udf15rg\u2215\ud835\udf15a| where the position\u2013dependent unit vectors can be determined from a\u0302 = (\ud835\udf15rg\u2215\ud835\udf15a)\u2215ha where again a = a, b, c. With this transformation, we can simplify the sixfold integration into a two-area integration over the current directions a\u0302 and a\u0302\u2032, respectively Lpaa\u2032 = \ud835\udf070\u222ba\u222bb\u222bc\u222ba\u2032\u222bb\u2032\u222bc\u2032 \ud835\udeff(b \u2212 b0, b \u2032 \u2212 b\u20320, c \u2212 c0, c \u2032 \u2212 c\u20320) a\u0302 \u22c5 a\u0302\u2032 |||| \ud835\udf15r \ud835\udf15a |||| |||| \ud835\udf15r\u2032 \ud835\udf15a\u2032 |||| g(r,r\u2032) dv dv\u2032 = Lpff \u2032 (b0, b \u2032 0, c0, c \u2032 0), (C.89) where f and f \u2032 represent the two filaments. This is the partial mutual inductance between the two filaments pointing in a\u0302 and a\u0302\u2032 directions. Hence, for the volumes of the two conductors, the total partial mutual inductance between two hexahedrons is Lpaa\u2032 = \u222bb\u222bc\u222bb\u2032\u222bc\u2032 Lpff \u2032 (b, b\u2032, c, c\u2032) db db\u2032 dc dc\u2032, (C.90) where the integration over the cross sections is performed with Gaussian numerical integration. This section is added to point out the difference between the local and global coordinates. Compute time can be saved if the numerical integration is mostly used for the smaller dimensions of the cell sides. Local coordinates are efficient for the mapping of the dimensions to the normalized units used for numerical integration methods. REFERENCES 407 Local coordinates are only required for nonorthogonal conductor cell dimensions such as quadrilateral or hexahedral shapes. For the example in Fig. C.17, we can use mixed coordinates. Global coordinates are used for conductor cell 1 while local coordinates are used for conductor 2. The integrals are Lp12 = \ud835\udf070 4\ud835\udf0b \u222b xe1 xs1 \u222b ye1 ys1 \u222b ze1 zs1 \u222ba2 \u222bb2 \u222bc2 1 R1,2 dx1 dy1 dz1 da2 db2 dc2. (C.91) Besides the change in the coordinate systems, the filament representation is the same as for the case in previous section. Representation is the same as for the case in the previous section. More details on nonorthogonal systems is given in Chapter 7."
        ]
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure6.4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure6.4-1.png",
        "caption": "Fig. 6.4. Pin insertion task",
        "texts": [
            " High-level user instructions in such systems are automatically translated into sequences of more elementary operations that can be directly executed by the robot controller. However, transformations of the task-oriented commands into low-level manipulation commands can result in a disagreement between the intended action and a generated code. In such cases, interaction between the system and the user is necessary, and the importance of additional simulation and debugging tools is therefore significantly increased. For illustration, let us examine a simplified problem of inserting pins into hole~ in some assembly process, as shown in Fig. 6.4 - a pin insertion task. The pins are acquired from a feeder which ensures that all pins are picked up from the fixed position. Manipulator configurations during this course of insertion can vary due to tolerances of the assembly parts. As the insertion trajectory is not precisely known at the programming time, it should be determined using the robot sensors. One possible strategy for using the sensory information is by a spiral search around the hole: if the sensors do not confirm that the pin is seated in the hole, succeeding insertion tries are made at break points of a quasi-spiral path in a plane normal to the direction of insertion (see Fig. 6.4). Once the pin has fallen into the hole, the insertion should be performed as a motion with compliance to external forces. 198 6 Industrial Robot Programming Systems In the following pages, we shall see how this insertion task can be pro grammed in two different robot languages, WAVE and AML. The basic philosophy implemented in WAVE is that only minor deviations from the planned trajectory can be expected during the robot operation. With this assumption, it is possible to transform a source W AVE program into an executable object code that contains precalculated joint angles and dynamic effects as a function of time",
            "6 Robot Programming Examples 199 Contrary to its poor program control capabilities, WAVE provides some very high-level functions that are rarely found in commercial robot pro gramming languages. For example, a spiral search may be specified with the help of two commands: SEARCH, which allows selection of the search par ameters, and AOJ, which performs a jump to the next search step. Compliant motion is also supported: commands FREE and SPIN are provided for speci fying translatory and rotational degrees of freedom of the manipulator in relation to the reference frame. Now that we have some brief knowledge of the W AVE commands; let us return to the problem from Fig. 6.4. The whole insertion task can be divided into the following steps: (a) moving the hand to the pin picking position (free motion with the obvious requirement to have determined a safe trajectory preventing collision between the manipulator and surrounding objects); (b) closing the gripper's fingers around the pin and checking whether the pin is successfully acquired; (c) placing the pin above the expected position of the hole (free motion as in step (a)); (d) searching for the actual hole position (guarded motion is performed in every search step); (e) inserting (compliant motion)",
            " It is assumed here that the robot fingers are equipped with touch sensors; AML global identifiers for pinch force sensors of the left and the right fingers are SLP and SRP respectively, and the identifiers for the corresponding tip force sensors are SL T and SRT. Indices \"1\" in the return values of PINCH_FORCE and TIP _FORCE specify that the stopping conditions become satisfied if either the (left or right) sensor value exceeds the range 0 to F. Subroutines APPROACH, RETREAT, GRASP and RELEASE may be further used as building elements in programming more complex functions that implement particular steps of the insertion task in Fig. 6.4. Let us assume that the objects relevant to this task (e.g. pins and holes) are described by aggregates, and that every such aggregate has three components: expected position, diameter and height (or depth). If we decide to use distinct subroutines for obtaining the selected features of the objects, as shown in Fig. 6.8, then the functions that implement realize picking a pin, finding a hole, and inserting, may be programmed as in Fig. 6.9. AML subroutines like PICKUP and INSERT _PIN in Fig. 6.9 are actually high-level robot commands (clearly, Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.44-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.44-1.png",
        "caption": "FIGURE 8.44. A non-steerable wheel must have two DOF.",
        "texts": [],
        "surrounding_texts": [
            "The suspension is what links the wheels to the vehicle body and allows relative motion. This chapter covers the suspension mechanisms, and discusses the possible relative motions between the wheel and the vehicle body. The wheels, through the suspension linkage, must propel, steer, and stop the vehicle, and support the associated forces. 8.1 Solid Axle Suspension The simplest way to attach a pair of wheels to a vehicle is to mount them at opposite ends of a solid axle, such as the one that is shown in Figure 8.1. The solid axle must be attached to the body such that an up and down motion in the z-direction, as well as a roll rotation about the x-axis, is possible. So, no forward and lateral translation, and also no rotation about the axle and the z-axis, is allowed. There are many combinations of links and springs that can provide the kinematic and dynamic requirements. The simplest design is to clamp the axle to the middle of two leaf springs with their ends tied or shackled to the vehicle frame as shown schematically in Figure 8.1. A side view of a multi-leaf spring and solid axle is shown in Figure 8.2. A suspension with a solid connection between the left and right wheels is called dependent suspension. 456 8. Suspension Mechanisms The performance of a solid axle with leaf springs suspension can be improved by adding a linkage to guide the axle kinematically and provide dynamic support to carry the non z-direction forces. The solid axle with leaf spring combination came to vehicle industry from horse-drawn vehicles. Example 299 Hotchkiss drive. When a live solid axle is connected to the body with nothing but two leaf springs, it is called the Hotchkiss drive, which is the name of the car that used it first. The main problems of a Hotchkiss drive, which is shown in Figure 8.2, are locating the axle under lateral and longitudinal forces, and having a low mass ratio \u03b5 = ms/mu, where ms is the sprung mass and mu is the unsprung mass. Sprung mass refers to all masses that are supported by the spring, such as vehicle body. Unsprung mass refers to all masses that are attached to and not supported by the spring, such as wheel, axle, or brakes. Example 300 Leaf spring suspension and flexibility problem. The solid axle suspension systems with longitudinal leaf springs have many drawbacks. The main problem lies in the fact that springs themselves act as locating members. Springs are supposed to flex under load, but their flexibility is needed in only one direction. However, it is the nature of leaf springs to twist and bend laterally and hence, flex also in planes other than the tireplane. Leaf springs are not suited for taking up the driving and braking traction forces. These forces tend to push the springs into an S-shaped profile, as shown in Figure 8.3. The driving and braking flexibility of leaf springs, generates a negative caster and increases instability. Long springs provide better ride. However, long sprigs exaggerate their bending and twisting under different load conditions. Example 301 Leaf spring suspension and flexibility solution. To reduce the effect of a horizontal force and S-shaped profile appearance in a solid axle with leaf springs, the axle may be attached to the chassis by a longitudinal bar as Figure 8.4(a) shows. Such a bar is called an anti-tramp 8. Suspension Mechanisms 457 bar, and the suspension is the simplest cure for longitudinal problems of a Hotchkiss drive. A solid axle with an anti-tramp bar may be kinematically approximated by a four-bar linkage, as shown in Figure 8.4(b). Although an anti-tramp bar may control the shape of the leaf spring, it introduces a twisting angle problem when the axle is moving up and down, as shown in Figure 8.5. Twisting the axle and the wheel about the axle is called caster. The solid axle is frequently used to help keeping the wheels perpendicular to the road. Example 302 Leaf spring location problem. The front wheels need room to steer left and right. Therefore, leaf springs cannot be attached close to the wheel hubs, and must be placed closer to the middle of the axle. That gives a narrow spring-base, which means that a small side force can sway or tilt the body relative to the axle through a considerable roll angle due to weight transfer. This is uncomfortable for the vehicle passengers, and may also produce unwanted steering. The solid axle positively prevents the camber change by body roll. The wheels remain upright and hence, do not roll on a side. However, a solid axle shifts laterally from its static plane and its center does not remain on the vehicle\u2019s longitudinal axis under a lateral force. 458 8. Suspension Mechanisms A solid axle produces bump-camber when single-wheel bump occurs. If the right wheel goes over a bump, the axle is raised at its right end, and that tilts the left wheel hub, putting the left wheel at a camber angle for the duration of deflection. Example 303 Triangular linkage. A triangulated linkage, as shown in Figure 8.6, may be attached to a solid axle to provide lateral and twist resistance during acceleration and braking. Example 304 Panhard arm. High spring rate is a problem of leaf springs. Reducing their stiffness by narrowing them and using fewer leaves, reduces the lateral stiffness and increases the directional stability of the suspension significantly. A Panhard arm is a bar that attaches a solid axle suspension to the chassis laterally. Figure 8.7 illustrates a solid axle and a Panhard arm to guide the axle. Figure 8.8 shows a triangular linkage and a Panhard arm combination for guiding a solid axle. A double triangle mechanism, as shown in Figure 8.9, is an alternative design to guide the axle and support it laterally. Example 305 Straight line linkages. There are many mechanisms that can provide a straight line motion. The simplest mechanisms are four-bar linkages with a coupler point moving straight. Some of the most applied and famous linkages are shown in Figure 8.10. By having proper lengths, the Watt, Robert, Chebyshev, and Evance linkages can make the coupler point C move on a straight line vertically. Such a mechanism and straight motion may be used to guide a solid axle. Two Watt suspension mechanisms with a Panhard arm are shown in Figures 8.11 and 8.12. 460 8. Suspension Mechanisms 8. Suspension Mechanisms 461 Figures 8.13, 8.14, and 8.15 illustrate three combinations of Robert suspension linkages equipped with a Panhard arm. Example 306 Solid axle suspension and unsprung mass problem. A solid axle is counted as an unsprung member, and hence, the unsprung mass is increased where using solid axle suspension. A heavy unsprung mass ruins both, the ride and handling of a vehicle. Lightening the solid axle makes it weaker and increases the most dangerous problem in vehicles: axle breakage. The solid axle must be strong enough to make sure it will not break under any loading conditions at any age. As a rough estimate, 90% of the leaf spring mass may also be counted as unsprung mass, which makes the problem worse. The unsprung mass problem is worse in front, and it is the main reason that they are no longer used in street cars. However, front solid axles are still common on trucks and buses. These are heavy vehicles and solid axle 8. Suspension Mechanisms 463 464 8. Suspension Mechanisms suspension does not reduce the mass ratio \u03b5 = ms/mu very much. When a vehicle is rear-wheel-drive and a solid axle suspension is used in the back, the suspension is called live axle. A live axle is a casing that contains a differential, and two drive shafts. The drive shafts are connected to the wheel hubs. A live axle can be three to four times heavier than a dead I-beam axle. It is called live axle because of rotating gears and shafts inside the axle. Example 307 Solid axle and coil spring. To decrease the unsprung mass and increase vertical flexibility of solid axle suspensions, it is possible to equip them with coil springs. A sample of a solid axle suspension with coil spring is shown in Figure 8.16. The suspension mechanism is made of four longitudinal bars between the axle and chassis. The springs may have some lateral or longitudinal angle to introduce some lateral or longitudinal compliance. Example 308 De Dion axle. When a solid axle is a dead axle with no driving wheels, the connecting beam between the left and right wheels may have different shapes to do different jobs, usually to give the wheels independent flexibility. We may also modify the shape of a live axle to attach the differential to the chassis and reduce the unsprung mass. De Dion design is a modification of a beam axle that may be used as a dead axle or to attach the differential to the chassis and transfer the driving power to the drive wheels by employing universal joints and split shafts. Figure 8.17 illustrates a De Dion suspension. 8.2 Independent Suspension Independent suspensions is introduced to let a wheel to move up and down without affecting the opposite wheel. There are many forms and designs of independent suspensions. However, double A-arm and McPherson strut suspensions are the simplest and the most common designs. Figure 8.18 illustrates a sample of a double A-arm and Figure 8.19 shows a McPherson suspension. Kinematically, a double A-arm suspension mechanism is a four-bar linkage with the chassis as the ground link, and coupler as the wheel carrying link. A McPherson suspension is an inverted slider mechanism that has the chassis as the ground link and the coupler as the wheel carrying link. A double A-arm and a McPherson suspension mechanism on the left and right wheels are schematically shown in Figures 8.20 and 8.21 respectively. Double A-arm, is also called double wishbone, or short/long arm suspension. McPherson also may be written as MacPherson. Example 309 Double A-arm suspension and spring position. Consider a double A-arm suspension mechanism. The coil spring may be between the lower arm and the chassis, as shown in Figure 8.18. It is also possible to install the spring between the upper arm and the chassis, or between the upper and lower arms. In either case, the lower or the upper arm, which supports the spring, is made stronger and the other arm acts as a connecting arm. Example 310 Multi-link suspension mechanism. When the two side bars of an A-arm are attached to each other with a joint, as shown in Figure 8.22, then the double A-arm is called a multi-link mechanism. A multi-link mechanism is a six-bar mechanism that may have a better coupler motion than a double A-arm mechanism. However, multi- 466 8. Suspension Mechanisms 8. Suspension Mechanisms 467 link suspensions are more expensive, less reliable, and more complicated compare to a double A-arm four-bar linkage. There are vehicles with more than six-link suspension with possibly better kinematic performance. Example 311 Swing arm suspension. An independent suspension may be as simple as a triangle shown in Figure 8.23. The base of the triangle is jointed to the chassis and the wheel to the tip point. The base of the triangle is aligned with the longitudinal axis of the vehicle. Such a suspension mechanism is called a swing axle or swing arm. The variation in camber angle for a swing arm suspension is maximum, compared to the other suspension mechanisms. Example 312 Trailing arm suspension. Figure 8.24 illustrates a trailing arm suspension that is a longitudinal arm with a lateral axis of rotation. The camber angle of the wheel, supported by a trailing arm, will not change during the up and down motion. Trailing arm suspension has been successfully using in a variety of frontwheel-drive vehicles, to suspend their rear wheels. Example 313 Semi-trailing arm Semi-trailing arm suspension, as shown in Figure 8.25, is a compromise between the swing arm and trailing arm suspensions. The joint axis may have any angle, however an angle not too far from 45 deg is more applied. Such suspensions have acceptable camber angle change, while they can handle both, the lateral and longitudinal forces. Semi-trailing design has successfully applied to a series of rear-wheel-drive cars for several decades. Example 314 Antiroll bar and roll stiffness. Coil springs are used in vehicles because they are less stiff with better 470 8. Suspension Mechanisms ride comfort compared to leaf springs. Therefore, the roll stiffness of the vehicle with coil springs is usually less than in vehicles with leaf springs. To increase the roll stiffness of such suspensions, an antiroll bar must be used. Leaf springs with reduced layers, uni-leaf, trapezoidal, or nonuniform thickness may also need an antiroll bar to compensate for their reduced roll stiffness. The antiroll bar is also called a stabilizer. Figure 8.26 illustrates an anti-roll bare attached to a solid axle with coil springs. Example 315 Need for longitudinal compliance. A bump is an obstacle on the road that opposes the forward motion of a wheel. When a vehicle goes over a bump, the first action is a force that tends to push the wheel backward relative to the rest of the vehicle. So, the lifting force has a longitudinal component, which will be felt inside the vehicle unless the suspension system has horizontal compliance. There are situations in which the horizontal component of the force is even higher than the vertical component. Leaf springs can somewhat absorb this horizontal force by flattening out and stretching the distance from the forward spring anchor and the axle. Such a stretch is usually less than 1/2 in \u2248 1 cm. 8.3 Roll Center and Roll Axis The roll axis is the instantaneous line about which the body of a vehicle rolls. Roll axis is found by connecting the roll center of the front and rear suspensions of the vehicle. Assume we cut a vehicle laterally to disconnect the front and rear half of the vehicle. Then, the roll center of the front or rear suspension is the instantaneous center of rotation of the body with respect to the ground. 8. Suspension Mechanisms 471 472 8. Suspension Mechanisms Figure 8.27 illustrates a sample of the front suspensions of a car with a double A-arm mechanism. To find the roll center of the body with respect to the ground, we analyze the two-dimensional kinematically equivalent mechanism shown in Figure 8.28. The center of tireprint is the instant center of rotation of the wheel with respect to the ground, so the wheels are jointed links to the ground at their center of tireprints. The instant center I18 is the roll center of the body with respect to the ground. To find I18, we apply the Kennedy theorem and find the intersection of the line I12I28 and I13I38 as shown in Figure 8.29. The point I28 and I38 are the instant center of rotation for the wheels with respect to the body. The instant center of rotation of a wheel with respect to the body is called suspension roll center. So, to find the roll center of the front or rear half of a car, we should determine the suspension roll centers, and find the intersection of the lines connecting the suspension roll centers to the center of their associated tireprints. The Kennedy theorem states that the instant center of every three relatively moving objects are colinear. Example 316 McPherson suspension roll center. A McPherson suspension is an inverted slider crank mechanism. The instant centers of an example of inverted slider crank mechanism are shown in Figure 8.30. In this figure, the point I12 is the suspension roll center, which is the instant center of rotation for the wheel link number 2 with respect to the chassis link number 1. A car with a McPherson suspension system is shown in Figure 8.31. The kinematic equivalent mechanism is depicted in Figure 8.32. Suspension roll centers along with the body roll center are shown in Figure 8.33. To 8. Suspension Mechanisms 473 find the roll center of the front or rear half of a car, we determine each suspension roll center and then find the intersection of the lines connecting the suspension roll centers to the center of the associated tireprint. Example 317 Roll center of double A-arm suspension. The roll center of an independent suspension such as a double A-arm can be internal or external. The kinematic model of a double A-arm suspension for the front left wheel of a car is illustrated in Figure 8.34. The suspension roll center in Figure 8.34(a) is internal, and in Figure 8.34(b) is external. An internal suspension roll center is toward the vehicle body, while an external suspension roll center goes away from the vehicle body. A suspension roll center may be on, above, or below the road surface, as shown in Figure 8.35(a)-(c) for an external suspension roll center. When the suspension roll center is on the ground, above the ground, or below the ground, the vehicle roll center would be on the ground, below the ground, and above the ground, respectively. Example 318 F Camber variation of double A-arm suspension. When a wheel moves up and down with respect to the vehicle body, de- 474 8. Suspension Mechanisms 8. Suspension Mechanisms 475 pending on the suspension mechanism, the wheel may camber. Figure 8.36 illustrates the kinematic model for a double A-arm suspension mechanism. The mechanism is equivalent to a four-bar linkage with the ground link as the vehicle chasis. The wheel is always attached to a coupler point C of the mechanism. We set a local suspension coordinate frame (x, y) with the x-axis indicating the ground link MN . The x-axis makes a constant angle \u03b80 with the vertical direction. The suspension machanism has a length a for the upper A-arm, b for the coupler link, c for the lower A-arm, and d for the ground link. The configuration of the suspension is determined by the angles \u03b82, \u03b83, and \u03b84, all measured from the positive direction of the x-axis. When the suspension is at its equilibrium position, the links of the double A-arm suspension make initial angles \u03b820 \u03b830, and \u03b840 with the x-axis. The equilibrium position of a suspension is called the rest position. To determine the camber angle during the fluctuation of the wheel, we should determine the variation of the coupler angle \u03b83, as a function of vertical motion z of the coupler point C. Using \u03b82 as a parameter, we can find the coordinates (xC , yC) of the 476 8. Suspension Mechanisms coupler point C in the suspension coordinate frame (x, y) as xC = a cos \u03b82 + e cos (p\u2212 q + \u03b1) (8.1) yC = a sin \u03b82 + e sin (p\u2212 q + \u03b1) (8.2) where, q = tan\u22121 a sin \u03b82 d\u2212 a cos \u03b82 (8.3) p = tan\u22121 q 4b2f2 \u2212 (b2 + f2 \u2212 c2) 2 b2 + f2 \u2212 c2 (8.4) f = p a2 + d2 \u2212 2ad cos \u03b82. (8.5) The position vector of the coupler point is uC uC = xC \u0131\u0302+ yC j\u0302 (8.6) and the unit vector in the z-direction is u\u0302z = \u2212 cos \u03b80\u0131\u0302\u2212 sin \u03b80j\u0302. (8.7) Therefore, the displacement z in terms of xC and yC is: z = uC \u00b7 u\u0302z = \u2212xC cos \u03b80 \u2212 yC sin \u03b80 (8.8) 8. Suspension Mechanisms 477 The initial coordinates of the coupler point C and the initial value of z are: xC0 = a cos \u03b820 + e cos (p0 \u2212 q0 + \u03b1) (8.9) yC0 = a sin \u03b820 + e sin (p0 \u2212 q0 + \u03b1) (8.10) z0 = \u2212xC0 cos \u03b80 \u2212 yC0 sin \u03b80 (8.11) and hence, the vertical displacement of the wheel center can be calculated by h = z \u2212 z0 (8.12) The initial angle of the coupler link with the vertical direction is \u03b80 \u2212 \u03b830. Therefore, the camber angle of the wheel would be \u03b3 = (\u03b80 \u2212 \u03b83)\u2212 (\u03b80 \u2212 \u03b830) = \u03b830 \u2212 \u03b83 (8.13) The angle of the coupler link with the x-direction is equal to \u03b83 = 2 tan \u22121 \u00c3 \u2212E \u00b1 \u221a E2 \u2212 4DF 2D ! (8.14) where, D = J5 \u2212 J1 + (1 + J4) cos \u03b82 (8.15) E = \u22122 sin \u03b82 (8.16) F = J5 + J1 \u2212 (1\u2212 J4) cos \u03b82 (8.17) and J1 = d a (8.18) J2 = d c (8.19) J3 = a2 \u2212 b2 + c2 + d2 2ac (8.20) J4 = d b (8.21) J5 = c2 \u2212 d2 \u2212 a2 \u2212 b2 2ab . (8.22) Substituting (8.14) and (8.13), and then, eliminating \u03b82 between (8.13) and (8.8) provides the relationship between the vertical motion of the wheel, z, and the camber angle \u03b3. 478 8. Suspension Mechanisms Example 319 F Camber angle and wheel fluctuations. Consider the double A-arm suspension that is shown in Figure 8.36. The dimensions of the equivalent kinematic model are: a = 22.4 cm b = 22.1 cm c = 27.3 cm d = 17.4 cm \u03b80 = 24.3 deg (8.23) The coupler point C is at: e = 14.8 cm \u03b1 = 54.8 deg (8.24) If the angle \u03b82 at the rest position is at \u03b820 = 121.5 deg (8.25) then the initial angle of the other links are: \u03b830 = 18.36 deg \u03b840 = 107.32 deg (8.26) At the rest position, the coupler point is at: xC0 = \u221222.73 cm yC0 = 9.23 cm z0 = 16.92 cm (8.27) We may calculate h and \u03b3 by varying the parameter \u03b82. Figure 8.37 illustrates h as a function of the camber angle \u03b3. For this suspension mechanism, the wheel gains a positive camber when the wheel moves up, and gains a negative camber when the it moves down. The mechanism is shown in Figure 8.38, when the wheel is at the rest position and has a positive or a negative displacement. 8.4 F Car Tire Relative Angles There are four major wheel alignment parameters that affect vehicle dynamics: toe, camber, caster, and trust angle. 8. Suspension Mechanisms 479 8.4.1 F Toe When a pair of wheels is set so that their leading edges are pointed toward each other, the wheel pair is said to have toe-in. If the leading edges point away from each other, the pair is said to have toe-out. Toe-in and toe-out front wheel configurations of a car are illustrated in Figure 8.39. The amount of toe can be expressed in degrees of the angle to which the wheels are not parallel. However, it is more common to express the toein and toe-out as the difference between the track widths as measured at the leading and trailing edges of the tires. Toe settings affect three major performances: tire wear, straight-line stability, and corner entry handling. For minimum tire wear and power loss, the wheels on a given axle of a car should point directly ahead when the car is running in a straight line. Excessive toe-in causes accelerated wear at the outboard edges of the tires, while too much toe-out causes wear at the inboard edges. Toe-in increases the directional stability of the vehicle, and toe-out increases the steering response. Hence, a toe-in setting makes the steering function lazy, while a toe-out makes the vehicle unstable. With four wheel independent suspensions, the toe may also be set at the rear of the car. Toe settings at the rear have the same effect on wear, directional stability, and turn-in as they do on the front. However, we usually do not set up a rear-drive race car toed out in the rear, because of excessive instability. When driving torque is applied to the wheels, they pull themselves forward and try to create toe-in. Furthermore, when pushed down the road, a non-driven wheel or a braking wheel will tend to toe-out. Example 320 Toe-in and directional stability. Toe settings have an impact on directional stability. When the steering 480 8. Suspension Mechanisms 8. Suspension Mechanisms 481 wheel is centered, toe-in causes the wheels to tent to move along paths that intersect each other in front of the vehicle. However, the wheels are in balance and no turn results. Toe-in setup can increase the directional stability caused by little steering fluctuations and keep the car moving straight. Steering fluctuations may be a result of road disturbances. If a car is set up with toe-out, the front wheels are aligned so that slight disturbances cause the wheel pair to assume rolling directions that approach a turn. Therefore, toe-out encourages the initiation of a turn, while toe-in discourages it. Toe-out makes the steering quicker. So, it may be used in vehicles for a faster response. The toe setting on a particular car becomes a trade-off between the straight-line stability afforded by toe-in and the quick steering response by toe-out. Toe-out is not desirable for street cars, however, race car drivers are willing to drive a car with a little directional instability, for sharper turn-in to the corners. So street cars are generally set up with toe-in, while race cars are often set up with toe-out. Example 321 Toe-in and toe-out in the front and rear axles. Front toe-in: slower steering response, more straight-line stability, greater wear at the outboard edges of the tires. Front toe-zero: medium steering response, minimum power loss, minimum tire wear. Front toe-out: quicker steering response, less straight-line stability, greater wear at the inboard edges of the tires. Rear toe-in: straight-line stability, traction out of the corner, more steerability, higher top speed. 482 8. Suspension Mechanisms 8.4.2 F Caster Angle Caster is the angle to which the steering pivot axis is tilted forward or rearward from vertical, as viewed from the side. Assume the wheel is straight to have the body frame and the wheel frame coincident. If the steering axis is turned about the wheel yw-axis then the wheel has positive caster. If the steering axis is turned about the wheel \u2212yw-axis, then the wheel has negative caster. Positive and negative caster configurations on the front wheel of a car are shown in Figure 8.40. Negative caster aids in centering the steering wheel after a turn and makes the front tires straighten quicker. Most street cars are made with 4\u22126 deg negative caster. Negative caster tends to straighten the wheel when the vehicle is traveling forward, and thus is used to enhance straight-line stability. Example 322 Negative caster of shopping carts. The steering axis of a shopping cart wheel is set forward of where the wheel contacts the ground. As the cart is pushed forward, the steering axis pulls the wheel along, and because the wheel drags along the ground, it falls directly in line behind the steering axis. The force that causes the wheel to follow the steering axis is proportional to the distance between the steering axis and the wheel-to-ground contact point, if the caster is small. This distance is referred to as trail. The cars\u2019 steering axis intersects the ground at a point in front of the tireprint, and thus the same effect as seen in the shopping cart casters is achieved. While greater caster angles improves straight-line stability, they also cause an increase in steering effort. Example 323 Characteristics of caster in front axle. Zero castor provides: easy steering into the corner, low steering out of the corner, low straight-line stability. 8. Suspension Mechanisms 483 Negative caster provides: low steering into the corner, easy steering out of the corner, more straight-line stability, high tireprint area during turn, good turn-in response, good directional stability, good steering feel. When a castered wheel rotates about the steering axis, the wheel gains camber. This camber is generally favorable for cornering. 8.4.3 F Camber Camber is the angle of the wheel relative to vertical line to the road, as viewed from the front or the rear of the car. Figure 8.41 illustrates the wheel number 1 of a vehicle. If the wheel leans in toward the chassis, it is called negative camber and if it leans away from the car, it is called positive camber. The cornering force that a tire can develop is highly dependent on its angle relative to the road surface, and so wheel camber has a major effect on the road holding of a car. A tire develops its maximum lateral force at a small camber angle. This fact is due to the contribution of camber thrust, which is an additional lateral force generated by elastic deformation as the tread rubber pulls through the tire/road interface. To optimize a tire\u2019s performance in a turn, the suspension should provide a slight camber angle in the direction of rotation. As the body rolls in a turn, the suspension deflects vertically. The wheel is connected to the chassis by suspension mechanism, which must rotate to allow for the wheel deflection. Therefore, the wheel can be subject to large camber changes as the suspension moves up and down. So, the more the wheel must deflect from its static position, the more difficult it is to maintain an ideal camber angle. Thus, the relatively large wheel travel and soft roll stiffness needed to provide a smooth ride in passenger cars presents a difficult design challenge, while the small wheel travel and high roll stiffness inherent in racing cars reduces the problem. Example 324 Castor versus camber. Camber doesn\u2019t improve turn-in as the positive caster does. Camber is not generally good for tire wear. Camber in one wheel does not improve directional stability. Camber adversely affects braking and acceleration efforts. 8.4.4 F Trust Angle The trust angle \u03c5 is the angle between vehicle\u2019s centerline and perpendicular to the rear axle. It compares the direction that the rear axle is aimed with the centerline of the vehicle. A nonzero angle configuration is shown in Figure 8.42. Zero angle confirms that the rear axle is parallel to the front axle, and the wheelbase on both sides of the vehicle are the same. A reason for nonzero 484 8. Suspension Mechanisms 8. Suspension Mechanisms 485 trust angle would have unequal toe-in or toe-out on both sides of the axle. Example 325 Torque reaction. There are two kinds of torque reactions in rear-whel-drive: 1\u2212 the reaction of the axle housing to rotate in the opposite direction of the crown wheel rotation, and 2\u2212 the reaction of axle housing to spin about its own center, opposite to the direction of pinion\u2019s rotation. The first reaction leads to a lifting force in the differential causing a wind-up in springs. The second reaction leads to a lifting force on the right wheels. 8.5 Suspension Requirements and Coordinate Frames The suspension mechanism should allow a relative motion between the wheel and the vehicle body. The relative motions are needed to pass the road irregularities and steering. To function properly, a suspension mechanism should have some kinematic and dynamics requirements. 8.5.1 Kinematic Requirements To express the motions of a wheel, we attach a wheel coordinate system W (oxwywzw) to the center of the wheel. A wheel, as a rigid body, has six degrees-of-freedom with respect to the vehicle body: three translations and three rotations, as shown in Figure 8.43. 486 8. Suspension Mechanisms The axes xw, yw, and zw indicate the direction of forward, lateral, and vertical translations and rotations. In the position shown in the figure, the rotation about the xw-axis is the camber angle, about the yw-axis is the spin, and about the zw-axis is the steer angle. Consider a non-steerable wheel. Translation in zw-direction and spin about the yw-axis are the only two DOF allowed for such a wheel. So, we need to take four DOF. If the wheel is steerable, then translation in the zw-direction, spin about the yw-axis, and steer rotation about the zw-axis are the three DOF allowed. So, we must take three DOF of a steerable wheel. Kinematically, non-steerable and steerable wheels should be supported as shown in Figures 8.44 and 8.45 respectively. Providing the required freedom, as well as eliminating the taken DOF, are the kinematic requirements of a suspension mechanism. 8.5.2 Dynamic Requirements Wheels should be able to propel, steer, and stop the vehicle. So, the suspension system must transmit the driving traction and deceleration braking forces between the vehicle body and the ground. The suspension members must also resist lateral forces acting on the vehicle. Hence, the wheel suspension system must make the wheel rigid for the taken DOF. However, there must also be some compliance members to limit the untaken DOF. The most important compliant members are spring and dampers to provide returning and resistance forces in the z-direction. 8. Suspension Mechanisms 487 8.5.3 Wheel, wheel-body, and tire Coordinate Frames Three coordinate frames are employed to express the orientation of a tire and wheel with respect to the vehicle: the wheel frame W , wheel-body frame C, and tire frame T . A wheel coordinate frame W (xw, yw, zw) is attached to the center of a wheel. It follows every translation and rotation of the wheel except the spin. Hence, the xw and zw axes are always in the tire-plane, while the yw-axis is always along the spin axis. A wheel coordinate frame is shown in Figure 8.43. When the wheel is straight and the W frame is parallel to the vehicle coordinate frame, we attach a wheel-body coordinate frame C (xc, yc, zc) at the center of the wheel parallel to the vehicle coordinate axes. The wheelbody frame C is motionless with respect to the vehicle coordinate and does not follow any motion of the wheel. The tire coordinate frame T (xt, yt, zt) is set at the center of the tireprint. The zt-axis is always perpendicular to the ground. The xt-axis is along the intersection line of the tire-plane and the ground. The tire frame does not follows the spin and camber rotations of the tire however, it follows the steer angle rotation about the zc-axis. Figure 8.46 illustrates a tire and a wheel coordinate frames. Example 326 Visualization of the wheel, tire, and wheel-body frames. Figure 8.47 illustrates the relative configuration of a wheel-body frame C, a tire frame T , and a wheel frame W . If the steering axis is along the zc-axis then, the rotation of the wheel about the zc-axis is the steer angle \u03b4. Rotation about the xt-axis is the camber angle \u03b3. 488 8. Suspension Mechanisms Generally speaking, the steering axis may have any angle and may go through any point of the ground plane. Example 327 Wheel to tire coordinate frame transformation. If TdW indicates the T -expression of the position vector of the wheel frame origin relative to the tire frame origin, then having the coordinates of a point P in the wheel frame, we can find its coordinates in the tire frame using the following equation. T rP = TRW W rP + TdW (8.28) If W rP indicates the position vector of a point P in the wheel frame, W rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 (8.29) then the coordinates of the point P in the tire frame T rP are T rP = TRW W rP + Td = TRW W rP + TRW W T dW = \u23a1\u23a3 xP yP cos \u03b3 \u2212Rw sin \u03b3 \u2212 zP sin \u03b3 Rw cos \u03b3 + zP cos \u03b3 + yP sin \u03b3 \u23a4\u23a6 (8.30) 8. Suspension Mechanisms 489 where, WT dW is the W -expression of the position vector of the wheel frame in the tire frame, Rw is the radius of the tire, and TRW is the rotation matrix to go from the wheel frame W to the tire frame T . TRW = \u23a1\u23a3 1 0 0 0 cos \u03b3 \u2212 sin \u03b3 0 sin \u03b3 cos \u03b3 \u23a4\u23a6 (8.31) W T dW = \u23a1\u23a3 0 0 Rw \u23a4\u23a6 . (8.32) As an example, the center of the wheel W rP = W ro = 0 is the origin of the wheel frame W , that is at T ro = TdW = TRW W T dW = \u23a1\u23a3 0 \u2212Rw sin \u03b3 Rw cos \u03b3 \u23a4\u23a6 (8.33) in the tire coordinate frame T . 490 8. Suspension Mechanisms Example 328 F Tire to wheel coordinate frame transformation. If rP indicates the position vector of a point P in the tire coordinate frame, T rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 (8.34) then the position vector W rP of the point P in the wheel coordinate frame is W rP = WRT T rP \u2212 W T dW (8.35) = \u23a1\u23a3 xP yP cos \u03b3 + zP sin \u03b3 zP cos \u03b3 \u2212Rw \u2212 yP sin \u03b3 \u23a4\u23a6 because WRT = \u23a1\u23a3 1 0 0 0 cos \u03b3 sin \u03b3 0 \u2212 sin \u03b3 cos \u03b3 \u23a4\u23a6 (8.36) WdT = \u23a1\u23a3 0 0 Rw \u23a4\u23a6 (8.37) and we may multiply both sides of Equation (8.28) by TRT W to get TRT W T rP = W rP + TRT W TdW (8.38) = W rP + W T dW W rP = WRT T rP \u2212 W T dW . (8.39) As an example, the center of tireprint in the wheel frame is at W rP = \u23a1\u23a3 1 0 0 0 cos \u03b3 \u2212 sin \u03b3 0 sin \u03b3 cos \u03b3 \u23a4\u23a6T \u23a1\u23a3 0 0 0 \u23a4\u23a6\u2212 \u23a1\u23a3 0 0 Rw \u23a4\u23a6 = \u23a1\u23a3 0 0 \u2212Rw \u23a4\u23a6 . (8.40) Example 329 F Wheel to tire homogeneous transformation matrices. The transformation from the wheel to tire coordinate frame may also be expressed by a 4\u00d7 4 homogeneous transformation matrix TTW , T rP = TTW W rP = \u2219 TRW TdW 0 1 \u00b8 W rP (8.41) where TTW = \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 cos \u03b3 \u2212 sin \u03b3 \u2212Rw sin \u03b3 0 sin \u03b3 cos \u03b3 Rw cos \u03b3 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 . (8.42) 8. Suspension Mechanisms 491 The corresponding homogeneous transformation matrix WTT from the tire to wheel frame would be WTT = \u2219 WRT WdT 0 1 \u00b8 = \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 cos \u03b3 \u2212 sin \u03b3 0 0 sin \u03b3 cos \u03b3 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 . (8.43) It can be checked that WTT = TT\u22121W , using the inverse of a homogeneous transformation matrix rule. TT\u22121W = \u2219 TRW TdW 0 1 \u00b8\u22121 = \u2219 TRT W \u2212 TRT W TdW 0 1 \u00b8 = \u2219 WRT \u2212WRT TdW 0 1 \u00b8 (8.44) Example 330 F Tire to wheel-body frame transformation. The origin of the tire frame is at CdT in the wheel-body frame. CdT = \u23a1\u23a3 0 0 \u2212Rw \u23a4\u23a6 (8.45) The tire frame can steer about the zc-axis with respect to the wheel-body frame. The associated rotation matrix is CRT = \u23a1\u23a3 cos \u03b4 \u2212 sin \u03b4 0 sin \u03b4 cos \u03b4 0 0 0 1 \u23a4\u23a6 (8.46) Therefore, the transformation between the tire and wheel-body frames can be expressed by Cr = CRT T r+ CdT (8.47) or equivalently, by a homogeneous transformation matrix CTT . CTT = \u2219 CRT CdT 0 1 \u00b8 = \u23a1\u23a2\u23a2\u23a3 cos \u03b4 \u2212 sin \u03b4 0 0 sin \u03b4 cos \u03b4 0 0 0 0 1 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 (8.48) As an example, the wheel-body coordinates of the point P on the tread of a 492 8. Suspension Mechanisms negatively steered tire at the position shown in Figure 8.48, are: Cr = CTT T rP = \u23a1\u23a2\u23a2\u23a3 cos\u2212\u03b4 \u2212 sin\u2212\u03b4 0 0 sin\u2212\u03b4 cos\u2212\u03b4 0 0 0 0 1 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 \u23a1\u23a2\u23a2\u23a3 Rw 0 Rw 1 \u23a4\u23a5\u23a5\u23a6 = \u23a1\u23a2\u23a2\u23a3 Rw cos \u03b4 \u2212Rw sin \u03b4 0 1 \u23a4\u23a5\u23a5\u23a6 (8.49) The homogeneous transformation matrix for tire to wheel-body frame TTC is: TTC = CT\u22121T = \u2219 CRT CdT 0 1 \u00b8\u22121 = \u2219 CRT T \u2212CRT T CdT 0 1 \u00b8 = \u2219 CRT T \u2212 T CdT 0 1 \u00b8 = \u23a1\u23a2\u23a2\u23a3 cos \u03b4 sin \u03b4 0 0 \u2212 sin \u03b4 cos \u03b4 0 0 0 0 1 Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 (8.50) Example 331 F Cycloid. Assume that the wheel in Figure 8.48 is turning with angular velocity \u03c9 and has no slip on the ground. If the point P is at the center of the tireprint 8. Suspension Mechanisms 493 when t = 0, MrP = \u23a1\u23a3 0 0 \u2212Rw \u23a4\u23a6 (8.51) then we can find its position in the wheel frame at a time t by employing another coordinate frame M . The frame M is called the rim frame and is stuck to the wheel at its center. Because of spin, the M frame turns about the yw-axis, and therefore, the rotation matrix to go from the rim frame to the wheel frame is: WRM = \u23a1\u23a3 cos\u03c9t 0 sin\u03c9t 0 1 0 \u2212 sin\u03c9t 0 cos\u03c9t \u23a4\u23a6 (8.52) So the coordinates of P in the wheel frame are: W rP = WRM MrP = \u23a1\u23a3 \u2212Rw sin t\u03c9 0 \u2212Rw cos t\u03c9 \u23a4\u23a6 (8.53) The center of the wheel is moving with speed vx = Rw\u03c9 and it is at Gr =\u00a3 vxt 0 Rw \u00a4 in the global coordinate frame G on the ground. Hence, the coordinates of point P in the global frame G, would be GrP = W rP + \u23a1\u23a3 vxt 0 Rw \u23a4\u23a6 = \u23a1\u23a3 Rw (\u03c9t\u2212 sin t\u03c9) 0 Rw (1\u2212 cos t\u03c9) \u23a4\u23a6 . (8.54) The path of motion of point P in the (X,Z)-plane can be found by eliminating t between X and Z coordinates. However, it is easier to expressed the path by using \u03c9t as a parameter. Such a path is called cycloid. In general case, point P can be at any distance from the center of the rim frame. If the point is at a distance d 6= Rw, then its path of motion is called the trochoid. A trochoid is called a curtate cycloid when d < Rw and a prolate cycloid when d > Rw. Figure 8.49(a)-(c) illustrate a cycloid, curtate cycloid, and prolate cycloid respectively. Example 332 F Wheel to wheel-body frame transformation. The homogeneous transformation matrix CTW to go from the wheel frame 494 8. Suspension Mechanisms to the wheel-body frame can be found by combined transformation. CTW = CTT TTW (8.55) = \u23a1\u23a2\u23a2\u23a3 c\u03b4 \u2212s\u03b4 0 0 s\u03b4 c\u03b4 0 0 0 0 1 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 c\u03b3 \u2212s\u03b3 \u2212Rw sin \u03b3 0 s\u03b3 c\u03b3 Rw cos \u03b3 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 = \u23a1\u23a2\u23a2\u23a3 cos \u03b4 \u2212 cos \u03b3 sin \u03b4 sin \u03b3 sin \u03b4 Rw sin \u03b3 sin \u03b4 sin \u03b4 cos \u03b3 cos \u03b4 \u2212 cos \u03b4 sin \u03b3 \u2212Rw cos \u03b4 sin \u03b3 0 sin \u03b3 cos \u03b3 Rw cos \u03b3 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 If rP indicates the position vector of a point P in the wheel coordinate frame, W rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 (8.56) then the homogeneous position vector CrP of the point P in the wheel-body 8. Suspension Mechanisms 495 coordinate frame is: CrP = CTW W rP = \u23a1\u23a2\u23a2\u23a3 xP cos \u03b4 \u2212 yP cos \u03b3 sin \u03b4 + (Rw + zP ) sin \u03b3 sin \u03b4 xP sin \u03b4 + yP cos \u03b3 cos \u03b4 \u2212 (Rw + zP ) cos \u03b4 sin \u03b3 \u2212Rw + (Rw + zP ) cos \u03b3 + yP sin \u03b3 1 \u23a4\u23a5\u23a5\u23a6 (8.57) The position of the wheel center W r = 0, for a cambered and steered wheel is at Cr = CTW W r = \u23a1\u23a2\u23a2\u23a3 Rw sin \u03b3 sin \u03b4 \u2212Rw cos \u03b4 sin \u03b3 \u2212Rw(1\u2212 cos \u03b3) 1 \u23a4\u23a5\u23a5\u23a6 (8.58) The zc = Rw (cos \u03b3 \u2212 1) indicates how much the center of the wheel comes down when the wheel cambers. If the wheel is not steerable, then \u03b4 = 0 and the transformation matrix CTW reduces to CTW = \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 cos \u03b3 \u2212 sin \u03b3 \u2212Rw sin \u03b3 0 sin \u03b3 cos \u03b3 Rw (cos \u03b3 \u2212 1) 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 (8.59) that shows CrP = CTW W rP = \u23a1\u23a2\u23a2\u23a3 xP yP cos \u03b3 \u2212Rw sin \u03b3 \u2212 zP sin \u03b3 zP cos \u03b3 + yP sin \u03b3 +Rw (cos \u03b3 \u2212 1) 1 \u23a4\u23a5\u23a5\u23a6 (8.60) Example 333 F Tire to vehicle coordinate frame transformation. Figure 8.50 illustrates the first and fourth tires of a 4-wheel vehicle. There is a body coordinate frame B (x, y, z) attached to the mass center C of the vehicle. There are also two tire coordinate frames T1 (xt1 , yt1 , zt1) and T4 (xt4 , yt4 , zt4) attached to the tires 1 and 4 at the center of their tireprints. The origin of the tire coordinate frame T1 is at Bd1 BdT1 = \u23a1\u23a3 a1 \u2212b1 \u2212h \u23a4\u23a6 (8.61)"
        ]
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure8.8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure8.8-1.png",
        "caption": "Fig. 8.8. The experimental set up. 1 \u2014 manometer, 2 \u2014 AFPM machine, 3 \u2014 discharge duct, 4 \u2014 prime mover (drive machine), 5 \u2014 pressure tapping point, 6 \u2014 wind speed probe. Photo courtesy of the University of Stellenbosch, South Africa.",
        "texts": [
            " The characteristic curve of the whole machine may be obtained by adding flow rate at the same pressure, which is similar to two identical fans in parallel. Flow and Pressure Measurements Due to the nature and complexity of thermofluid analysis, the form of the system characteristics curve can at best be established by test. Depending on the machine topologies and size, the measurements may be taken either 264 8 Cooling and Heat Transfer at the air inlet or outlet. The AFPM machine under test is normally driven by another motor. Fig. 8.8 shows the experimental arrangements of the flow measurements at the machine outlet, in which a discharge duct is set up to provide good conditions for observing the flow. Along one side of the duct, several tapping points were made for measuring the static pressure with a pressure gauge (manometer). Near the outlet of the duct, provision was made for measuring velocity using a hot-wire anemometer probe. To vary the flow rate, the test duct was fitted at its outer end with an obstruction. The test was started with no obstruction at the end of the discharge duct"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000081_cr020719k-Figure11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000081_cr020719k-Figure11-1.png",
        "caption": "Figure 11. One-dimensional model diagram for supported laccase cathode.170",
        "texts": [
            " A model of such structures has been proposed that captures transport phenomena of both substrates and redox cosubstrate species within a composite biocatalytic electrode.170 The model is based on macrohomogeneous and thin-film theories for porous electrodes and accounts for Michaelis-Menton enzyme kinetics and one-dimensional diffusion of multiple species through a porous structure defined as a mesh of tubular fibers.85,171,172 In addition to the solid and aqueous phases, the model also allows for the presence of a gas phase (of uniformly contiguous morphology), as shown in Figure 11, allowing the treatment of high-rate gas-phase reactant transport into the electrode. When applied to a laccase-based oxygen-reducing electrode, the model predicted a maximum current density of 9.2 mA/cm2 at 0.6 V vs SHE for a 300-\u00b5mthick electrode relying on transport of oxygen by diffusion in the liquid phase. However, when either convective transport in the liquid phase or gas-phase diffusive transport was introduced in the electrode, current densities exceeding 50 mA/cm2 were predicted for air-saturated systems"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000801_j.finel.2014.04.003-Figure10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000801_j.finel.2014.04.003-Figure10-1.png",
        "caption": "Fig. 10. Substrate heating and quiet element wall model (1C, case sq1).",
        "texts": [
            " Typically, different values are used in front and behind the melt pool for the longitudinal ellipsoid axis c. However, here the same value is used here since the melt pool is nearly spherical in the LENS process. f is a scaling factor; vw is the heating spot travel speed; and t is time. The values used here are: P\u00bc425 W, \u03b7\u00bc .45, a\u00bc1.5 mm, b\u00bc .9 mm, c\u00bc1.5 mm, and f\u00bc1. Eq. (16) is applied on the entire model and beyond the edges of the ellipsoid (defined at the 5% value [46]) and thus no correction for cut-off is needed. FEA mesh: Fig. 9 shows a symmetric model of the substrate only and Fig. 10 shows a symmetric model of the substrate and wall. The substrate model (Fig. 9) is composed of 8480 hex8 elements and 10,086 nodes. The substrate and wall model contains 14,432 hex8 elements and 19,200 nodes. The mesh density is targeted towards obtaining temperature results for follow on mechanical analyses for computing residual stress and distortion. Finer meshes would be required for solidification studies. Time incrementation: During heating, the time increment is set to 0.0885827 s, which results into a 0",
            " Material properties are linearly interpolated between the values listed on the table, and kept constant beyond the minimum and maximum listed values. The density is 4.43 10 6 kg/mm3. The latent heat of fusion 365 kj/kg and is spread over a temperature range from 1600 1C to 1670 1C. In this test, heating of the substrate only with no metal deposition is modeled. A model of the substrate only (Fig. 9, case s1) is used as reference to evaluate the accuracy of models with active elements on the substrate and quiet or inactive elements on the wall (cases sq1 and si1). For the quiet element model (shown in Fig. 10), the conductivity and specific heat scaling factors at the quiet elements are set as sk\u00bc .000001 and sCp \u00bc :01. Temperature results along line AA as illustrated in Figs. 9 and 10 are extracted and plotted in Fig. 11. Convection and radiation are also applied on the interface between active and inactive (or quiet) elements. The maximum error for the quiet element and inactive methods is .0390% and 6.375e 5%, respectively (Table 4). It is noted that no elements are switched from quiet to active in this test. Fig. 10 shows the symmetric model of the substrate and wall used for this test. A time period of 1000 s is simulated. The metal deposition lasts 286.95 s and requires 3223 time increments. Further cool down for the remaining of the simulation requires 9 additional time increments. An average of 6 iterations per time increment is required to achieve convergence, which is defined as minimization of the norm of the global residual by 7 orders of magnitude. Tables 5 and 6 list the analysis cases performed for the wall build test and the respective wall CPU run times on a 16 3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-FigureC.16-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-FigureC.16-1.png",
        "caption": "FIGURE C.16 General nonorthogonal hexahedral elements.",
        "texts": [
            " [14] and good results have been obtained using the Gauss law as well as other results using approximate shapes for some of the integrals yielded very good results in Ref. [13]. C.2.6 Lp for Arbitrary Hexahedral Partial Mutual Inductance The partial self- and mutual inductances for arbitrary hexahedral shapes are more challenging. We can use local coordinates to compute the partial mutual inductance for two arbitrarily placed hexahedral elements. An example for such elements is given in Refs [15, 16]. Figure C.16 shows two volume conductor cells. We assume that the local coordinate system for the first conductor is (a, b, c) and (a\u2032, b\u2032, c\u2032) for the second one. A vector in the global coordinates located on the first conductor is r = (x, y, z) for the first hexahedral element, and another one is r\u2032 = (x\u2032, y\u2032, z\u2032). The current directions are indicated by dashed line are pointing in the a\u0302 and a\u0302\u2032 directions as is shown in Fig. C.16 and Lp12 = \ud835\udf070\u222ba\u222bb\u222bc\u222ba\u2032\u222bb\u2032\u222bc\u2032 a\u0302 \u22c5 a\u0302\u2032 |||| \ud835\udf15r \ud835\udf15a |||| |||| \ud835\udf15r\u2032 \ud835\udf15a\u2032 |||| g(r, r\u2032) dv dv\u2032, (C.84) where g is the scalar Green\u2019s function g(r, r\u2032) = e\u2212j\ud835\udefd|r\u2212r\u2032| 4\ud835\udf0b|r \u2212 r\u2032| or, for a quasi-static case g(r, r\u2032) = 1 4\ud835\udf0b|r \u2212 r\u2032| . (C.85) From the local coordinate system, we can determine a set of filaments in the global coordinate using the transform equations in section (8.2.6) to find the end points of each wire filament. Section D.2.2 can be used together with the numerical Gaussian integration. In Appendix E, the numerical integration is discussed which can be applied to the cross-sections to yield the partial inductance of interest. To give more details, the local coordinate system presented in Chapter 8 is used to represent the hexahedral element shown in Fig. C.16. The purpose of the local coordinates is to identify the location of the points for the filaments in terms of the variables a, b, c where a \u2208 [\u22121,+1] and where a can represent a = a, b, c. The purpose is to uniquely map a point a, b, c into a point in the global coordinates r. The local coordinates for the filament are used to map them to the global x, y, z coordinates. Mapping a point in the above hexahedron PARTIAL INDUCTANCE FORMULAS FOR NONORTHOGONAL GEOMETRIES 405 from a local coordinate point a, b, c to a global coordinate point x, y, z is described by x = 7\u2211 k=0 Nk(a, b, c) xk, (C"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000908_j.jmatprotec.2013.01.020-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000908_j.jmatprotec.2013.01.020-Figure2-1.png",
        "caption": "Fig. 2. CAD model of gyroid and diamond lattice structures.",
        "texts": [
            " The large overhang strut resulting from these designs caused a serious deformation or sagging to occur. Additionally, the design of lattice structure should permit the easy removal of the loose powder trapped inside of these structures after the build is completed. This is a necessary step before the part is heat treated or wire-cut by EDM. To address these issues, we have previously taken steps to improve lattice design (Hao et al., 2011) and evaluate the manufacturability and mechanical properties of lattice structures (Yan et al., 2012). These advanced lattice structures which are shown in Fig. 2 are called the Schoen Gyroid and Schwartz Diamond lattice types. These structures are implemented into and generated by the ScanIP and +CAD software from Simpleware Ltd. This software allows lattice structures to be introduced into a pre-defined geometry, specified by either image or CAD data, and subsequently saved as an STL file. The structures themselves can be generated to a specific controlled volume fractions and unit cell sizes and are defined by triple periodic implicit functions, such as the Schoen (1970) and Schwarz (1890)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000613_978-94-015-9261-1-Figure8.18-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000613_978-94-015-9261-1-Figure8.18-1.png",
        "caption": "Figure 8.18. A car in the ditch",
        "texts": [
            "17), where u = 2, while the other parameters are the same as in the previous case. Similarly the swinging algorithms for the case of a suspension point moving in the horizontal position can be derived. Other energy based algo rithms for a horizontally driven pendulum were suggested in [13, 260J. In this section we consider, following [160J a simple yet illustrative example of application of the speed gradient method to the control of Hamiltonian systems. Suppose the rear wheels of a rar are jammed in a ditch 01' in a. hole (see Figure 8.18). The control goal is to draw the car out of the hole by the controlling (external) force applied to the front of the car. The level of the controlling force should not exceed some pre-specified value. Let us investigate this problem in detail. For the sake of simplicity we neglect inertia of the wheels and assurne also that there are no damping forces, i.e., the system is conservative. Let C denote the center of mass of the car. It can be characterized by two coordinates: Yl and Y2. We suppose that the profile of the hole is such that the relation between the coordinates Yl and Y2 can be described by a smooth unimoda"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000551_s0893-6080(00)00077-0-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000551_s0893-6080(00)00077-0-Figure2-1.png",
        "caption": "Fig. 2. Swinging up an inverted pendulum by a LS-SVM controller with local stabilization in its upright position. Around the target point the controller is behaving as a LQR controller. (Top) inverted pendulum system; (Bottom) simulation result which visualizes the several pole positions in time.",
        "texts": [
            " The controller is generalizing well with respect to other initial conditions than the origin (for which it has been trained) and beyond the time horizon of N 20: The method (22) without the use of Mercer's condition gave similar results by taking the centers {ci} equal to {xk}. In this example we illustrate the LS-SVM control method of Section 5 on the problem of swinging up an inverted pendulum with local stabilization at the endpoint, which has been investigated in Suykens et al. (1994). A nonlinear state space model for the inverted pendulum (Fig. 2) system is given by _x F x 1 G x u 34 with F x x2 4 3 mlx2 4 sin x3 2 mg 2 sin 2x3 4 3 mt 2 m cos2 x3 x4 mtg sin x3 2 ml 2 x2 4 sin 2x3 l 4 3 mt 2 m cos2 x3 2666666666664 3777777777775 ; G x 0 4 3 1 4 3 mt 2 m cos2 x3 0 2 cos x3 l 4 3 mt 2 m cos2 x3 26666666664 37777777775 : The state variables x1, x2, x3, x4 are respectively position and velocity of the cart, angle of the pole with the vertical and rate of change of the angle. The input signal u is the force applied to the cart's center of mass"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure13.56-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure13.56-1.png",
        "caption": "Figure 13.56 (a) Sideways-supported mast ABC with (b) its support reactions.",
        "texts": [
            " The largest bending moment in an absolute sense is the support moment at B: MB = 96 kNm ( ). In addition, there are extreme field moments at E and G three metres from the supports, where the shear force is zero (see Figure 13.55c). The easiest way to find their magnitudes is from the hatched area of the V diagram: ME = MG = 1 2 \u00d7 3 \u00d7 36 = 54 kNm ( ). MG is also equal to the maximum bending moment in the simply supported beam SC with uniformly distributed full load: MG = 1 8 \u00d7 12 \u00d7 62 = 54 kNm ( ). Note that the M diagram has mirror symmetry about B. The mast ABC in Figure 13.56a is supported sideways by a number of bars. Dimensions and load are shown in the figure. Questions: a. Determine the support reactions at A and D. b. Determine the forces in the bars 1 to 3, with the correct signs for tension and compression. c. Isolate beam ABC, and draw all the forces acting on it. d. For beam ABC draw the M , V and N diagram, with the deformation 594 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM symbols. At A, B and C, also draw the tangents to the M diagram. e. Determine the extreme moments in ABC",
            " Solution (units kN and m): a. The vertical support reaction Dv (\u2193) at D follows from the moment equilibrium of the entire structure about A: \u2211 T|A = \u2212(12 \u00d7 5) \u00d7 6 + Dv \u00d7 6 = 0 \u21d2 Dv = 60 kN (\u2193). Bar (1) is a two-force member so that the line of action of the support reaction at D coincides with DE. The horizontal component Dh is therefore Dh = 4 6 \u00d7 Dv = 40 kN (\u2190). The support reactions at A follow from the force equilibrium of the structure as a whole: Ah = 20 kN (\u2190), Av = 60 kN (\u2191). The support reactions are shown in Figure 13.56b. b. The support reactions at D show that there is a tensile force in bar (1): N(1) = + \u221a 402 + 602 = +20 \u221a 13 kN (= +72.11 kN). The (normal) forces in bars (2) and (3) can now be determined from the force equilibrium of joint E. To do so we have to draw the force polygon for joint E (see Figure 13.57a). The force F (1) E = N(1) = 20 \u221a 13 kN, which bar (1) exerts on joint E, is known. We close the force polygon with the forces F (2) E and F (3) E , parallel to the two-force members (2) and (3). Figure 13"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000403_j.matdes.2013.05.070-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000403_j.matdes.2013.05.070-Figure4-1.png",
        "caption": "Fig. 4. Comparison of temperature distribution after the 1st track scan (a) built on a powder bed and (b) built on solid substrate.",
        "texts": [
            " This can be attributed to the fact that the rapidly cooling molten material has greater conductive properties than the untreated powder in front of the laser. The temperature distribution in the layer is very much affected by the energy density which is a function of laser power, spot size, scanning speed, hatch spacing and scanning strategy. Additionally, the temperature gradient in the layer is similarly influenced by conductivity of the material underneath the deposited layer being a loose powder bed, previously re-melted layer, support structure or a solid substrate. Fig. 4a and b shows temporal variation of temperature of single track deposited on powder bed and solid substrate respectively. For the powder bed, the melted layer thickness depends on the energy input from the laser, while for substrate a thin layer of 75 lm is deposited on solid material and then melted with the laser beam. A fast cooling rate can be observed when the track is built on a solid substrate. Because the solid bulk material has higher conductivity compared to powder bed allowing more heat sink through conduction in a shorter period of time and subsequent solidification to room temperature"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure11.14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure11.14-1.png",
        "caption": "FIGURE 11.14 PEEC equivalent circuit for magnetic surface equation.",
        "texts": [
            "77), which results in voltage sources in the equivalent circuit in Fig. 11.13. It is evident that this results in a coupled model for the electrical type. Similarly, we also have to take care of the MFIE based on the magnetic current and the field boundary condition in (11.74), or t\u0302 \u22c5 [Hinc 1 (r, t)] = t\u0302 \u22c5 [ \ud835\udf15F1(r, t) \ud835\udf15t \u2212 \ud835\udf15F2(r, t) \ud835\udf15t ] + t\u0302 \u22c5 [\u2207\ud835\udf19m,1(r, t) \u2212 \u2207\ud835\udf19m,2(r, t)] \u2212 t\u0302 \u22c5 [ 1 \ud835\udf071 \u2207 \u00d7 A1 \u2212 1 \ud835\udf072 \u2207 \u00d7 A2 ] , (11.78) where t\u0302 is again the tangential unit vector. We show the PEEC circuit model corresponding to (11.78) for the magnetic circuit in Fig. 11.14. It is evident that we can choose the same ground node at infinity for both the electric and magnetic interface PEEC model. Further, we did show that the MFIE based in (11.77) is solved by multiplying (11.78) by \ud835\udf07\u2215\ud835\udf16. This corresponds to the use of reciprocal medium in the formulation where \ud835\udf07 is replaced by \ud835\udf16 and vice versa. The surface formulation requires a few observations. First, it is evident from (11.77) that region 1 and region 2 equivalent circuits are connected in series. The inductive surface SURFACE MODELS FOR MAGNETIC AND DIELECTRIC MATERIAL SOLUTIONS IN PEEC 305 cells lead to the partial self-inductances for region 1 and the source v1 L1 is a reminder that the partial mutual inductances are coupled only to cells in region 1 and not region 2",
            "4 Single-loop inductance Write a Matlab program to compute the inductance of a square-shaped single loop borrowed from the problems in Chapter 5 of a size of 3 cm \u00d7 5 cm with a cross section 0.1 mm \u00d7 0.1 mm. A magnetic body with \ud835\udf07r = 160 is placed at a distance of 1 mm under the loop. The centered magnetic body also of size 3 cm \u00d7 5 cm and a thickness of 1 mm is placed under the loop. Compute the inductance of the loop with the magnetic sheet. 11.5 PEEC model for MFIE Explain each electrical component given in the PEEC circuit model shown in Fig. 11.14 for the magnetic field integral equation (MFIE). 308 PEEC MODELS FOR MAGNETIC MATERIAL 1. L. V. Bewley. Flux Linkages and Electromagnetic Induction. Dover Publications, New York, 1964. 2. J. A. Brandao Faria. Electromagnetic Foundations of Electrical Engineering. John Wiley and Sons, Ltd, Hoboken, NJ, 2nd edition, 2008. 3. J. R. Reitz and F. J. Milford. Foundation of Electromagnetic Theory. Pergamon, New York, 1960. 4. A. E. Ruehli and D. Ellis. Numerical calculation of magnetic fields in the vicinity of a magnetic body"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure10.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure10.3-1.png",
        "caption": "FIGURE 10.3. Top view of a moving vehicle to show the yaw angle \u03c8 between the x and X axes, the sideslip angle \u03b2 between the velocity vector v and the x-axis, and the crouse angle \u03b2 + \u03c8 between with the velocity vector v and the X-axis.",
        "texts": [
            " Analysis of the vehicle motion is equivalent to expressing the position and orientation of B(Cxyz) in G(OXY Z). Figure 10.2 shows how a moving vehicle is indicated by a body frame B in a global frame G. The angle between the x and X axes is the yaw angle \u03c8 and is called the heading angle. The velocity vector v of the vehicle makes an angle \u03b2 with the body x-axis which is called sideslip angle or attitude angle. The vehicle\u2019s velocity vector v makes an angle \u03b2 + \u03c8 with the global X-axis that is called the cruise angle. These angles are shown in the top view of a moving vehicle in Figure 10.3. There are many situations in which we need to number the wheels of a vehicle. We start numbering from the front left wheel as number 1, and then the front right wheel would be number 2. Numbering increases sequentially on the right wheels going to the back of the vehicle up to the rear right wheel. Then, we go to the left of the vehicle and continue numbering the wheels from the rear left toward the front. Each wheel is indicated by a position vector ri Bri = xii+ yij + zik (10.6) expressed in the body coordinate frame B. Numbering of a four wheel vehicle is shown in Figure 10.3. Example 375 Wheel numbers and their position vectors. Figure 10.4 depicts a six-wheel passenger car. The wheel numbers are indicated besides each wheel. The front left wheel is wheel number 1, and the front right wheel is number 2. Moving to the back on the right side, we count the wheels numbered 3 and 4. The back left wheel gets number 5, 586 10. Vehicle Planar Dynamics and then moving forward on the left side, the only unnumbered wheel is the wheel number 6. If the global position vector of the car\u2019s mass center is given by Gd = \u2219 XC YC \u00b8 (10"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002056_j.automatica.2016.07.038-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002056_j.automatica.2016.07.038-Figure1-1.png",
        "caption": "Fig. 1. Circuit diagram of Buck converter.",
        "texts": [
            " The main difference is that the mismatched perturbations of the nonlinear system in Cruz-Zavala and Moreno (2014) and Estrada and Fridman (2010a,b) should be suppressed, while the mismatched term of the sliding mode dynamics in this paper is in purpose introduced in order to reduce the uncertainties in the control channel. The Buck converters are one of the most important switched mode DC\u2013DC converters, which have been widely used in applications, such as mobile power supply equipment, photovoltaic system, DC supply system, etc. Generally, Buck converter consists of a DC voltage source (Vin), a controllable switch (Sw), a diode (D), an inductor (L), a capacitor (C) and a load resistor (R), properly connected as shown in Fig. 1. The mathematical model of DC\u2013DC Buck converter can usually be described as (Erickson & Maksimovic, 1999) diL dt = 1 L (uVin \u2212 v0), dv0 dt = 1 C  iL \u2212 v0 R  (17) where v0 is the output voltage and u is the controller. The task here is to design a controller u such that the output voltage v0 will well track a desired reference voltage vref . To design a SOSM controller for Buck converter control system (17), the first thing is to choose a sliding variable. We now define the sliding variable s (i"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure5.12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure5.12-1.png",
        "caption": "FIGURE 5.12 Long loop represented by four bars.",
        "texts": [
            " The proper solution of this problem involves two aspects. One aspect that relates to the PEEC model is considered in this section, whereas the second issue that relates to the evaluation of the partial inductances is considered in the following section. Not surprisingly, it is difficult to evaluate partial inductance with a high accuracy for extreme geometries. As an example geometry, it is sufficient to consider a typical loop shown in Fig. 5.4. Here, we consider a physically long version shown in Fig. 5.12. The circuit solution for the loop inductance is given in (5.16). We regroup the terms of the equation in pairs in the form L = (Lp11 \u2212 Lp13) + (Lp22 \u2212 Lp24) + (Lp33 \u2212 Lp31) + (Lp44 \u2212 Lp42). (5.23) It is apparent from Fig. 5.12 that we assume that the length \ud835\udcc1 of the loop is large compared to the spacing d. The length consists of the partial self inductances Lp11 and Lp33, while the orthogonal end branches have the partial self-inductances Lp22 and Lp44. Note that due to the orthogonality of the conductors, the only nonzero partial mutual inductances are Lp13 = Lp31 and Lp24 = Lp42. COMPUTATION OF PARTIAL INDUCTANCES 103 Now, we want to consider the case where the two bars with Lp11 and Lp33 are very long, or \ud835\udcc1 \u226b d. Approximations can be used for the end branch inductances since they are very short compared to the long branches of length \ud835\udcc1",
            " The exact partial self- and mutual inductances are computed with the formula (C.24). We use the equation (C.19) for both the partial self- and the partial mutual inductances with the same asymptotic behavior for Lpij as for Lpii. The zero thickness loop problem using formula (5.24) for two conductors of the same width also yields the same difference inductance Ld11 = Lp11 \u2212 Lp13, where I1 = \u2212I3. Results are presented for the long loop example with the three different cross sections corresponding to Fig. 5.12, where the short return conductors are ignored. The cross section of the round conductors is 252.3 \u03bcm. The zero thickness conductors have a width w = 1 mm. Finally, the thickness for the rectangular conductors is t = 50 \u03bcm and the length for all conductors is increased over a wide range from \ud835\udcc1 = 1 mm to \ud835\udcc1 = 3 m. Hence, the smallest to largest dimension is 1 : 3000. We have the zero and finite thickness conductors touching each other on the width side as would be the case for a VFI skin-effect model",
            " We should mention that the example of extreme aspect ratios for the dimensions does not only apply to conductor length, but cells or conductors with other extreme dimension ratios can also lead to similar problems. One of the advantages of the partial inductance concept is the ability to break a complicated three-dimensional problem into its constituent interactions resulting for general inductance calculations. In this section, we use a small PEEC model to address the accuracy issue that is encountered in dense problems. This is done using the two- conductor example from Fig. 5.12. In this approximate model, we represent each conductor by a partial inductance. If we ignore the vertical connections at the two ends, its inductance is approximated as L = 2(Lp11 \u2212 Lp12) = 2 Ld11. The inherent source of errors is due to the subtraction of the partial inductances for the closely located conductors. For close neighbor conductors, the first two digits in the partial inductances may be the same for Lp11 and Lp12 and only the third digit may contain information relevant for the result"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000065_s1005-0302(12)60016-4-Figure10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000065_s1005-0302(12)60016-4-Figure10-1.png",
        "caption": "Fig. 10 Schematic model for development of Cu2O precipitates in continuous melt zones created by x\u2013y electron beam scanning in EBM of Cu precursor powder (Fig. 1(c)). Beam scan p(x)\u2013p(y) correspond to x\u2013y in Fig. 9. (From Ramirez et al.[17])",
        "texts": [
            " 8 illustrates the (110) orientation texture in the vertical reference plane parallel to the build direction indicated in Fig. 7. Fig. 9 shows a simple sketch illustrating the regular melt pool geometry resulting from x-y beam scan which can characterize both EBM and SLM layer building. The melt pool dimension (Fig. 9), which may conceptually apply for the EBM Cu fabrication in Fig. 7, is roughly 2\u20133 \u03bcm, but the actual layer melt configuration does not appear to be as orthogonally regular or extensive as depicted in Fig. 9. Fig. 10 extends this orthogonal zone model in three dimensions showing the extension of melt pools creating columnar arrays of Cu2O precipitates and dislocation arrays as shown in Figs. 7 and 8. These precipitates (Cu2O) arise from oxygen absorbed during Cu atomization in purified Ar as a consequence of the affinity for oxygen of finely atomized Cu powder particles. Optical microscopy examination of the interior of atomized Cu precursor powder as in Fig. 1(b) has shown Cu2O precipitates at grain boundaries and these precipitates either reform (solutionize) or are reorganized within the melt pool structures creating columnar precipitate architectures shown in Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.53-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.53-1.png",
        "caption": "FIGURE 8.53. Top view of a steered wheel with a steer axis coincident with zw.",
        "texts": [
            " Substituting \u03b4 = 0 simplifies the rotation matrix CRW and the position vector CdW to I and 0 CRW = \u23a1\u23a3 1 0 0 0 1 0 0 0 1 \u23a4\u23a6 (8.89) CdW= \u23a1\u23a3 0 0 0 \u23a4\u23a6 (8.90) indicating that at zero steering, the wheel frameW and wheel-body frame C are coincident. Example 336 F Steer angle transformation for zero lean and caster. Consider a wheel with a steer axis coincident with zw. Such a wheel has no lean or caster angle. When the wheel is steered by the angle \u03b4, we can find the coordinates of a wheel point P in the wheel-body coordinate frame using transformation method. Figure 8.52 illustrates a 3D view, and Figure 8.53 a top view of such a wheel. 8. Suspension Mechanisms 501 502 8. Suspension Mechanisms Assume W rP = [xw, yw, zw] T is the position vector of a wheel point, then its position vector in the wheel-body coordinate frame C is CrP = CRW W rP = Rz,\u03b4 W rP = \u23a1\u23a3 cos \u03b4 \u2212 sin \u03b4 0 sin \u03b4 cos \u03b4 0 0 0 1 \u23a4\u23a6\u23a1\u23a3 xw yw zw \u23a4\u23a6 = \u23a1\u23a3 xw cos \u03b4 \u2212 yw sin \u03b4 yw cos \u03b4 + xw sin \u03b4 zw \u23a4\u23a6 . (8.91) We assumed that the wheel-body coordinate is installed at the center of the wheel and is parallel to the vehicle coordinate frame. Therefore, the transformation from the frame W to the frame C is a rotation \u03b4 about the wheel-body z-axis"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure3.20-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure3.20-1.png",
        "caption": "Fig. 3.20: The kinematic chain of a general 3R manipulator and its design parameters.",
        "texts": [
            " In this case, in fact, the algebraic formulation can be used to build a system of algebraic design equations where the axial and radial reaches are given by the prescribed workspace data and the design unknowns are the ring or hyper-ring coefficients. Once these coefficients are solved, the chain parameters can be calculated by using the definitions of the coefficients. In order to outline a design procedure, the case of a 3R manipulator is discussed in detail. Fundamentals of Mechanics of Robotic Manipulation 109 A general open-chain 3R manipulator with three revolute joints is sketched in Fig. 3.20, in which the design parameters are represented as the H\u2013D parameters a1, a2, a3, d2, d3, \u03b11, \u03b12, and \u03b83 is the Z3 joint variable. With the hypothesis \u03b11 \u2260 k \u03c0/2 (k = 0,1,...,4), the workspace boundary of a threerevolute open-chain manipulator can be expressed as a function of the radial reach r1 and axial reach z1 with respect to the manipulator base frame X1Y1Z1 as ( ) 2/1 1 111112 111 E FGDzC zAr +++\u2212= 1 1 11 1/2 11 1 C D CK QL z \u2212\u00b1\u2212= (3.1.63) with so-called \u2018structural coefficients\u2019 as 2 22 2 2 2 11 )d+z(+r+a=A 2 2 2 11 ra4-=B (3",
            "66) 111 GFL = ( )2 11 2 11 2 11 EB+FK-L=Q Thus, the workspace boundary of a general three-revolute open-chain manipulator can be evaluated by scanning the angle \u03b83 and plotting r1, z1. This gives numerically the workspace boundary shape with its hole and voids, and its volume. It should be noted that the workspace formulation of Eqs (3.1.63)\u2013(3.1.66) is an algebraic function of the \u03b83 joint variable and all design parameters. Assuming as additional design parameters the position and orientation vectors s and k of the manipulator base with respect to the fixed world frame XYZ, Fig. 3.20, the workspace design equations (3.1.63) can be modified to include the reference change. To accomplish this we take x as the position vector of a boundary point with respect to XYZ, and we use the expression 11 R sxx \u2212= (3.1.67) where r1, z1 are the components of x1 with respect to the manipulator base frame X1Y1Z1; R is the 3 \u00d7 3 rotation matrix describing the fixed frame with respect to the manipulator base frame; the position vector of robot base is s1 as measured in X1Y1Z1 and s = Rt s1 (R t is the transpose of R) as measured in XYZ",
            " Several workspace characteristics can be used to formulate the objective function but however workspace volume and manipulator length are usually preferred. These characteristics can also be conveniently combined to formulate a performance index for manipulators. The general optimization formulation of Eqs (3.1.84) to (3.1.86) can be better illustrated by referring to a specific case for a manipulator optimum design. Thus, the optimum design of a general three-revolute manipulator deals with the synthesis of the parameters a1, a2, a3, d2, d3, \u03b11, \u03b12, (d1 is not meaningful since it shifts up and down the workspace only), Fig. 3.20, when the workspace characteristics are considered to fulfil the design requirements which are expressed in Fig. 3.19. A specific optimum design can be formulated as an optimization problem in the form Chapter 3: Fundamentals of the Mechanics of Robots118 3 * L V -min (3.1.92) subject to minz(z)min \u2265 maxz(z)max \u2264 (3.1.93) minr(r)min \u2265 maxr(r)max \u2264 where V* indicates a measure of the workspace volume; L is the total dimension of the manipulator given as in Eq. (3.1.88); z and r are the axial and radial reaches of the boundary points of the manipulator workspace"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure10.28-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure10.28-1.png",
        "caption": "Figure 10.28 (a) Dimensions of a concrete column of the building used in Section 6.4. (b) The column is loaded on top by a force of 46.4 kN. The dead weight is to be considered as a uniformly distributed (line)load along the member axis of 1.92 kN/m.",
        "texts": [
            " The bending moments on both sectional planes are in equilibrium with the couple of 80 kNm. In straight members a (distributed) longitudinal load does not produce bending moments or shear forces. In these cases, there are only normal forces. The variation of the normal force is elaborated for two examples: 1. a column subject to its dead weight; 2. a simply supported member subject to a uniformly distributed axial load over three-quarters of its length. 410 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Example 1 For the example in Figure 10.28, we will use the concrete column from the building that we looked at in Section 6.4. Question: Determine the N diagram. Solution: The column is loaded on top by a force of 46.4 kN (see Figure 10.28b). The dead weight is a uniformly distributed (line) load along the member axis. With a specific weight of concrete of 24 kN/m3, and the cross-sectional dimensions given in Figure 10.28a, the dead weight is (0.4 m)(0.2m)(24kN/m3) = 1.2kN/m. The model for the column and load is shown in Figure 10.29a. In Figure 10.29b, a segment with length x has been isolated at the top of the column. In the section, the as yet unknown normal force N is shown according to its positive direction (that of a tensile force). For this segment, the equation for the force equilibrium in the x direction is \u2211 Fx = (46.4 kN) + (1.92 kN/m)(x m) + N = 0 from which it follows that (x expressed in m) N = (\u221246"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003413_ac0010882-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003413_ac0010882-Figure6-1.png",
        "caption": "Figure 6. Square-wave voltammetry, a, and cyclic voltammetry, b, recordings of 100 nM DA and 0.3 mM AA at an OPPD-CFME: medium, 0.05 M phosphate buffer pH 7.4. Scan rate (CV), 100 mV/ s. SWV settings: frequency, 10 Hz; potential step, 10 mV, pulse amplitude, 50 mV; accumulation condition, 45 s at -0.2 V.",
        "texts": [
            " Thus, AA cannot reduce the product of DA anodic oxidation back to DA, which would result in an increase in the DA current signal, dependent on the actual AA concentration. Such behavior was observed at electrodes coated with, for example, a negatively charged Nafion film.15,44 This property of the OPPD film further contributes to the AAinterference-free detection of DA at the OPPD-CFME. Analytical Performance of the OPPD-CFME for Simultaneous Sensing of DA and AA. When the SWV detection mode with the OPPD-coated microprobes was applied, a dramatic improvement in the resolution and heights of DA and AA voltammetric peaks was observed, as clearly displayed in Figure 6, which is of particular importance in the detection of nanomolar DA. The fast-pulsing potential ramp used in SWV, coupled with the advantageous diffusion-profiling properties of microelectrodes, apparently proves its strong ability to effectively eliminate the background and enhance the response currents. On the basis of this observation and backed by the findings of other authors,45 an optimized SWV detection mode was employed instead of CV (42) Chen, P.; McCreery, R. L. Anal. Chem. 1996, 68, 3958-3965"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002406_s11665-021-05919-6-Figure9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002406_s11665-021-05919-6-Figure9-1.png",
        "caption": "Fig. 9 Coble s geometrical models for (a) intermediate-stage and (b) final-stage sintering. Source: Ref 31",
        "texts": [
            " The steps of solid-state sintering consist of solid-state atomic diffusion, recrystallization, and grain growth, while six different mechanisms are involved with mass transfer, including surface diffusion, evaporation condensation, grainboundary diffusion, lattice diffusion, viscous flow, and plastic flow (Ref 31). The primary means of sintering is to achieve maximum density and strength based on the metallurgical bonds formed between neighboring particles. The bridge formed between metallurgically bonded particles is called a neck (Fig. 8). Pores begin to take shape in the intermediate and final sintering stages. Initially, pores form interconnected channels along three grain edges (Fig. 9). As the sintering process progresses, the pore channels disconnect, and isolated pores form upon the dihedral angles exceeding 60 and nonuniform shrinkage. Coble proposed the two geometric models illustrated in Fig. 9\u2014the channel pore model and the isolated pore model\u2014to describe pore morphology evolution (Ref 32, 33). The closed pore encompassed between the particles is dependent on the number of neighboring particles. In selective laser sintering, a laser beam irradiates each layer of powder that has been spread evenly on the previous layer and fuses the powder particles together to a high density (>90%) to form the component. The resulting temperature gradient results in faster grain coalescence at the surface than the subsurface layers"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure11.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure11.3-1.png",
        "caption": "FIGURE 11.3. The force system at the tireprint of tire number 1, and their resultant force system at C.",
        "texts": [
            "33) results in the force system. \u23a1\u23a3 Fx Fy 0 \u23a4\u23a6 = m \u23a1\u23a3 v\u0307x \u2212 \u03c9zvy v\u0307y + \u03c9zvx \u03c9xvy \u23a4\u23a6 (11.40) \u23a1\u23a3 Mx 0 Mz \u23a4\u23a6 = \u23a1\u23a3 \u03c9\u0307xI1 \u03c9x\u03c9zI1 \u2212 \u03c9x\u03c9zI3 \u03c9\u0307zI3 \u23a4\u23a6 (11.41) 11. F Vehicle Roll Dynamics 671 11.3 F Vehicle Force System To determine the force system on a rigid vehicle, we define the force system at the tireprint of a wheel. The lateral force at the tireprint depends on the sideslip angle. Then, we transform and apply the tire force system on the rollable model of the vehicle. 11.3.1 F Tire and Body Force Systems Figure 11.3 depicts wheel number 1 of a vehicle. The components of the applied force system on the rigid vehicle, because of the generated forces at the tireprint of the wheel number i, are Fxi = Fxwi cos \u03b4i \u2212 Fywi sin \u03b4i (11.42) Fyi = Fywi cos \u03b4i + Fxwi sin \u03b4i (11.43) Mxi = Mxwi + yiFzi \u2212 ziFyi (11.44) Myi = Mywi + ziFxi \u2212 xiFzi (11.45) Mzi = Mzwi + xiFyi \u2212 yiFxi (11.46) where (xi, yi, zi) are body coordinates of the wheel number i. It is possible to ignore the components of the tire moment at the tireprint, Mxwi , Mywi , Mzwi , and simplify the equations"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.8-1.png",
        "caption": "FIGURE 7.8. Trapezoidal steering triangle ABC.",
        "texts": [
            "6 illustrates a symmetric four-bar linkage, called a trapezoidal steering mechanism, that has been used for more than 100 years. The mechanism has two characteristic parameters: angle \u03b2 and offset arm length d. A steered position of the trapezoidal mechanism is shown in Figure 7.7 to illustrate the inner and outer steer angles \u03b4i and \u03b4o. The relationship between the inner and outer steer angles of a trapezoidal steering mechanism is sin (\u03b2 + \u03b4i) + sin (\u03b2 \u2212 \u03b4o) = w d + r\u00b3w d \u2212 2 sin\u03b2 \u00b42 \u2212 (cos (\u03b2 \u2212 \u03b4o)\u2212 cos (\u03b2 + \u03b4i)) 2. (7.23) To prove this equation, we examine Figure 7.8. In the triangle 4ABC we can write (w \u2212 2d sin\u03b2)2 = (w \u2212 d sin (\u03b2 + \u03b4i)\u2212 d sin (\u03b2 \u2212 \u03b4o)) 2 +(d cos (\u03b2 \u2212 \u03b4o)\u2212 d cos (\u03b2 + \u03b4i)) 2 (7.24) and derive Equation (7.23) with some manipulation. The functionality of a trapezoidal steering mechanism, compared to the associated Ackerman condition, is shown in Figure 7.9 for x = 2.4m \u2248 7. 87 ft and d = 0.4m \u2248 1.3 ft. The horizontal axis shows the inner steer angle and the vertical axis shows the outer steer angle. It depicts that for 386 7. Steering Dynamics 7"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure3.61-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure3.61-1.png",
        "caption": "Figure 3.61 The cube, which has been halved diagonally, is subject to a couple of 80 kNm in plane ABEF and a couple of 60 kNm in plane BCDE. The body is kept in equilibrium by a couple on the diagonal plane ACDF.",
        "texts": [
            "60 depicts the forces (in kN) as they are acting on the cube in reality. The forces F3, F5 and F6 act in directions opposite to those shown in Figure 3.58. By using alternative equilibrium equation (g), equation (b) for the force equilibrium in y direction was not used, and can be used as a check. With the forces expressed in kN this gives \u2211 Fy = F1 + 4 5F2 + F3 + 4 5F4 + F5 + 4 5F6 = 0 + 4 5 \u00d7 20 \u2212 4 + 4 5 \u00d7 40 \u2212 28 \u2212 4 5 \u00d7 20 = 0. The conditions for force equilibrium in y direction are met. 98 Example 3 The cube that has been halved diagonally in Figure 3.61 is subject to a couple of 80 kNm in plane ABEF and a couple of 60 kNm in plane BCDE. The directions are shown in the figure. The body is kept in equilibrium by a couple on the diagonal plane ACDF. Question: Determine the magnitude of that couple and resolve it into a component in plane ACDF and a component perpendicular to plane ACDF. Solution: There is moment equilibrium if the moment vectors form a closed polygon. The polygon in Figure 3.62a shows that a couple of 100 kNm is acting on plane ACDF"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure5.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure5.3-1.png",
        "caption": "FIGURE 5.3 Example loop k for flux and vector potential computations.",
        "texts": [
            " We give the definition of inductance for this simple case. LOOP INDUCTANCE COMPUTATIONS 91 Definition 1 (Inductance for N loops) The N2 inductances for a system of N loop are defined by Lkm = \ud835\udf13km Im = Vk s Im . (5.1) Where Im is the current flowing in the mth loop and VK is the voltage induced in the kth loop. As done in Ref. [3], we start the inductance computation by considering the flux\ud835\udf13 generated by loop m and picked up by a loop k similar to the ones in (5.1). The example geometry is the loop in Fig. 5.3. The unit vector n\u0302 is perpendicular to the loop area, and the magnetic flux through loop k is \ud835\udf13km = \u222bk Bm \u22c5 n\u0302 dk, (5.2) where k is the area of the kth loop. We would like to remind the reader again that we use  for a surface area or a flux area and c for a conductor current carrying cross section. Hence, we integrate the normal component of the magnetic field over the loop area. Using (\u222bc dc)\u2215c for the finite cross section of the conductors, we also average the integral over the conductor cross-sectional area depicted by ck in Fig. 5.3. We should note that the same result can also be obtained from an energy point of view (see Chapter 11, Refs [12] and [2]). The integration results in the flux in loop k \ud835\udf13km = 1 ck \u222bck \u222bk Bm \u22c5 n\u0302 dk dck, (5.3) where again n\u0302 is the unit vector normal to the loop surface. We continue the derivation by using (3.2a) for a homogeneous material with a relative permeability, which we assume to be \ud835\udf07r = 1 such that B = \ud835\udf070H. We start from Maxwell\u2019s equations (3.1b) for the quasistatic case, or \u2207 \u00d7 B = \ud835\udf070J (5",
            " This individually represents each segment of the loop over which we integrate the flux area \ud835\udf13km = 4\u2211 i=1 1 ck \u222bck \u222b\ud835\udcc1k,i Am \u22c5 t\u0302k,i d\ud835\udcc1k,i dck, (5.10) where, in general, the cross section of each conductor ck for the four sides can be different. A second important step in the derivation is to replace the vector potential Am, in (5.10) by (3.50), given in Chapter 3 as Am (r, t) = \ud835\udf070 4\ud835\udf0b\u222bV\u2032 Jve (r\u2032, t\u2032)|r \u2212 r\u2032| d \u2032 . (5.11) LOOP INDUCTANCE COMPUTATIONS 93 We want to apply (5.11) to a current filament in the cross section in the conductors in Fig. 5.3. In this example, both the observation point location r and the source point rm are located on loop conductors. We assume that the current in the loop conductors is uniform in the cross section and perform the averaging integration over the cross section to get the total current. This leads to the vector potential Am(r) = \ud835\udf070 4\ud835\udf0b Icm cm \u222bcm \u222b\ud835\udcc1m t\u0302m,j Rrm d\ud835\udcc1m dcm, (5.12) where Rrm = |r \u2212 rm| and Icm is the total current in the bar m. Note that the vector potential Am is in the same direction as the current Icm",
            "15) We observe that we compute the partial self-inductance if we apply this result to only one part where m = k. With this, we have a form of the equations that leads to the partial inductances in the following section. We use the loop inductance derived in the previous section as a starting point for a much more powerful circuit-oriented general approach for inductance computations using partial inductances. By considering Fig. 5.1, we see that we approximate each section of the loops by a rectangular bar. As a first example to the general case for inductance computations, we consider the loop in Fig. 5.3. We recognize that we can simplify the loop inductance equation in (5.15) into Lloop = 4\u2211 i=1 4\u2211 j=1 Lpij, (5.16) where Lpij = 1 k m \ud835\udf070 4\ud835\udf0b\u222bk \u222bm \u222b\ud835\udcc1k,i \u222b\ud835\udcc1m,j t\u0302k,i \u22c5 t\u0302m,j Rkm d\ud835\udcc1m d\ud835\udcc1k dm dk. (5.17) From this example, we find the following definition of a partial inductance Definition 2 (Partial inductance) A partial self-inductance Lpii is defined for a single piece of conductor as Vi = Lpii sIi, where Vi is the voltage along the conductor i and Ii is the current in the conductor i. Similarly, the partial mutual inductance is defined between two pieces of conductors i and j as Vi = Lpij s Ij (5",
            " The correct coupling signs are obtained if we also label all the partial inductances Lpkm in the matrix as positive. This rule automatically leads to the correct solution. After the solution is obtained, the actual sign of the elements indicates whether the current flows in the assigned direction. As usual, a positive current indicates that the current flows in the assigned direction, whereas a negative sign indicates current flows in the opposite direction. The flux that contributes to the inductance is confined inside the loop as shown in Fig. 5.3. We need to show how the flux is also confined using partial inductances. This issue is investigated in Refs [2, 3, 13]. In Fig. 5.6a, we consider two black, short parallel wires that we assume to be two parts of the same loop. We next want to answer the question how the flux is confined to the interior of the loop that is between the two wires. We can associate a flux area to the partial inductance perpendicular to the ends such that the perpendicular sides of the area do not contribute to the partial inductance loop"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure4.42-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure4.42-1.png",
        "caption": "Figure 4.42 (a) A block supported by three bars; (b) the support is kinematically indeterminate as no moment equilibrium about D is possible.",
        "texts": [
            " The last degree of freedom, the rotation about RC, can be removed with a third bar support, for example at C (see Figure 4.38d). The three bars now prevent all possible movement. However the body is pulled or pushed, it remains where it is. This is referred to as the body having an immovable or kinematically determinate support. Three bar supports (at least) are required for an immovable or kinematically 1 The fact that the centre of rotation RC is a fixed point is true only if the rotation is still small. When considering Figure 4.42, one should not be confused by the fact that the displacements have been drawn to a large scale with respect to the dimensions of the structure. 132 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM determinate support of a rigid body. The bars may not all intersect in one point, or all be parallel, as is shown in Figure 4.39. In the support in Figure 4.39a, all bars intersect at the rotation centre RC, allowing the body to rotate. This support is movable or kinematically indeterminate. The support in Figure 4",
            " A more general answer is found in linear algebra: there is a unique solution if the determinant of the coefficient matrix is not equal to zero. This is indeed the case in this example: Det = 1 2 \u221a 2 \u00b7 b = 0. The set of equations cannot be solved if the determinant of the coefficient matrix is zero. The figures in the coefficient matrix are determined by the manner in which the body is supported. The fact that the determinant is zero means that, from a physical perspective, the support is kinematically indeterminate. In order to illustrate this, the bar support (3) in Figure 4.42 has been placed at an angle. With this type of support, the three equilibrium conditions can be represented by \u23a1 \u23a2\u23a3 1 2 \u221a 2 0 0 1 2 \u221a 2 1 + 1 2 \u221a 2 0 0 0 \u23a4 \u23a5\u23a6 \u23a1 \u23a2\u23a3 F1 F2 F3 \u23a4 \u23a5\u23a6 = \u23a1 \u23a2\u23a3 Rx Ry R \u00b7 a \u23a4 \u23a5\u23a6 . The determinant of the coefficient matrix is now zero. In the last equilibrium equation, the moment equilibrium about D, the condition R \u00b7 a = 0 cannot be met. Neither can the support reactions for R \u00b7 a = 0 be determined. The method of support in Figure 4.42 allows a 4 Structures 135 rotation about D and is therefore kinematically indeterminate. This is in line with what we determined in the previous section for a support on three bars that pass through a single point. The support in Figure 4.43, on three parallel bars, is also kinematically indeterminate. No force equilibrium is possible in the direction normal to the bars and the block is able to move in that direction. With less than three support reactions, there are more equilibrium equations than unknowns"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure6.5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure6.5-1.png",
        "caption": "FIGURE 6.5. Limit position for a four-bar linkage.",
        "texts": [
            " A crank-crank mechanism is also called a drag-link. Example 224 Limit positions for a four-bar linkage. When the output link of a four-bar linkage stops while the input link can turn, we say the linkage is at a limit position. It happens when the angle between the input and coupler links is either 180 deg or 360 deg. Limit positions of a four-bar linkage, if there are any, must be determined by the designer to make sure the linkage is designed properly. A limit position for a four-bar linkage is shown in Figure 6.5. 320 6. Applied Mechanisms We show the limit angle of the output link by \u03b84L1 , \u03b84L2 , and the corresponding input angles by \u03b82L1 , \u03b82L2 . They can be calculated by the following equations: \u03b82L1 = cos\u22121 \" (a+ b) 2 + d2 \u2212 c2 2d (a+ b) # (6.78) \u03b84L1 = cos\u22121 \" (a+ b)2 \u2212 d2 \u2212 c2 2cd # (6.79) \u03b82L2 = cos\u22121 \" (b\u2212 a)2 + d2 \u2212 c2 2d (b\u2212 a) # (6.80) \u03b84L2 = cos\u22121 \" (b\u2212 a) 2 \u2212 d2 \u2212 c2 2cd # (6.81) The sweep angle of the output link would be \u03c6 = \u03b84L2 \u2212 \u03b84L1 . (6.82) Example 225 Dead positions for a four-bar linkage"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure15.10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure15.10-1.png",
        "caption": "Figure 15.10 During the change in displacement the forces do not change direction. Since N performs no work the virtual displacement \u03b4u has to take place along the tangent to the prescribed path.",
        "texts": [
            " (b) 15 Virtual Work 717 After the virtual displacement, it applies that f (x + \u03b4ux, y + \u03b4uy) = (x + \u03b4ux)2 + (y + \u03b4uy)2 \u2212 r2 = 0. (c) If we combine equations (c) and (b) we find the following relationship between \u03b4ux and \u03b4uy : 2x\u03b4ux + (\u03b4ux)2 + 2y\u03b4uy + (\u03b4uy)2 = 0. (d) Since this equation is determined by the geometry of the prescribed path, it is known as a geometric equation. During the variation of the displacement, the forces do not change direction. The same is true for N . Since N does not perform work, the virtual displacement has to occur along the tangent of the prescribed path, as shown in Figure 15.10. This means that the geometric relationship between \u03b4ux and \u03b4uy has to be linear. Ignoring the quadratic (higher order) terms in the geometric equation (d) can be physically interpreted as a demand that the virtual displacements have to be small. If we remove the quadratic terms in (d)1 we find 2x\u03b4ux + 2y\u03b4uy = 0 or \u03b4uy = \u2212x y \u03b4ux. (e) When the particle is in equilibrium, \u03b4A = 0. With (a) and (e) the virtual work equation becomes 1 This is referred to as the linearisation of the geometric equation (d)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure1.5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure1.5-1.png",
        "caption": "Fig. 1.5. Double-sided AFPM brushless machine with internal salient-pole stator and twin external rotor [185]: (a) construction; (b) stator; (c) rotor. 1 \u2014 PM, 2 \u2014 rotor backing steel disc, 3 \u2014 stator pole, 4 \u2014 stator coil.",
        "texts": [
            " The first choice has the advantage of using fewer PMs at the expense of poor winding utilisation while the second one is considered as a particularly advantageous machine topology [38]. The diverse topologies of AFPM brushless machines may be classified as follows: \u2022 single-sided AFPM machines \u2013 with slotted stator (Fig. 1.4a) \u2013 with slotless stator \u2013 with salient-pole stator \u2022 double-sided AFPM machines \u2013 with internal stator (Fig. 1.4b) \u00b7 with slotted stator \u00b7 with slotless stator \u00b7 with iron core stator \u00b7 with coreless stator (Fig. 1.4d) \u00b7 without both rotor and stator cores \u00b7 with salient pole stator (Fig. 1.5) \u2013 with internal rotor (Fig. 1.4c) \u00b7 with slotted stator \u00b7 with slotless stator \u00b7 with salient pole stator (Fig. 1.6) \u2022 multi-stage (multidisc) AFPM machines (Fig. 1.7) 1.5 Topologies and Geometries 7 The air gap of the slotted armature AFPM machine is relatively small. The mean magnetic flux density in the air gap decreases under each slot opening due to increase in the reluctance. The change in the mean magnetic flux density caused by slot openings corresponds to a fictitious increase in the air gap [121]",
            " Since there are no slots, Carter coefficient kC = 1. Compared to a conventional slotted winding, the slotless armature winding has advantages such as simple stator assembly, elimination of the cogging torque and reduction of rotor surface losses, magnetic saturation and acoustic noise. The disadvantages include the use of more PM material, lower winding inductances sometimes causing problems for inverter-fed motors and significant eddy current losses in slotless conductors [49]. In the double-sided, salient-pole AFPM brushless machine shown in Fig. 1.5, the stator coils with concentrated parameters are wound on axially laminated poles. To obtain a three-phase self-starting motor, the number of the stator poles should be different from the number of the rotor poles, e.g. 12 stator poles and 8 rotor poles [173, 174, 185]. Fig. 1.6 shows a double-sided AFPM machine with external salient pole stators and internal PM rotor. There are nine stator coils and eight rotor poles for a three-phase AFPM machine. Depending on the application and operating environment, slotless stators may have ferromagnetic cores or be completely coreless",
            " As with overlap windings non-overlap windings can be of single or double layer, concentrated or distributed, integral or fractional and air-cored or iron-cored. AFPM machines with non-overlap windings can be of single-sided or double-sided construction (Figs. 1.5, 1.6, 2.12). In double-layer, non-overlap slotted iron-cored windings, two coils are sharing a slot (two coil layers per slot), which means that all teeth are wound. Consequently these windings are sometimes called tooth windings. In air-cored, double-layer non-overlap windings, the stator coils lie side-by-side against each other (see e.g. Fig. 1.5b). Note that the term slot can also be visualized when applied to air-cored windings. In single-layer, non-overlap slotted iron-cored windings, only one coil layer is in a slot and therefore every second tooth is wound (alternate teeth wound); 2.2 Windings 39 air-cored, single-layer non-overlap windings have their coils displaced from each other as shown e.g. in Fig. 2.11. In double-layer, non-overlap windings, the number of stator coils is always equal to the number of stator slots, i.e. Qc = ncm1 = s1, where nc is the number coils per phase",
            " Electromagnetic analysis of AFPM machines with non-overlap stator windings are the same as those with overlap windings provided that the winding factors are calculated according to the following equations: the distribution factor for the fundamental space harmonic is calculated by [240] kd1 = sin(\u03c0/2m1) z sin[(\u03c0/(2m1z)] (2.14) 40 2 Principles of AFPM Machines For single and double layer slotted iron-cored non-overlap windings, the pitch factor for the fundamental space harmonic is calculated by using eqn (2.9) or else kp1 = sin( \u03c0 2 2p s1 ) = sin(\u03b8m/2) (2.15) where \u03b8m is the slot pitch angle (also corresponding to the coil span angle in the case of double layer windings shown in Fig. 1.5b or corresponding to the coil pitch angle in the case of single layer windings as shown in Fig. 2.11) given by \u03b8m = 2\u03c0p s1 (2.16) For air-cored, non-overlap windings, the pitch factor for the fundamental space harmonic is derived in [146]. For single layer non-overlap windings the pitch factor is given by kp1 = sin(\u03b8m/2) sin(\u03b8re/2) \u03b8re/2 (2.17) while for double layer, non-overlap windings, the pitch factor is given by kp1 = sin[(\u03b8m \u2212 \u03b8re)/2] sin(\u03b8re/2) \u03b8re/2 (2.18) where \u03b8re is the electrical angle corresponding to the coil-layer width. For example, for a double layer, air-cored non-overlap winding with m1 = 3, p = 12, s1 = Qc = 18 (thus nc = 6) and \u03b8re = 0.1 \u03b8m, the number of winding sections is F = 6, the number of coils distributed in a coil group is z = 1, the angular coil span is 1.333\u03c0, the distribution factor is kd1 = 1 and the pitch factor is kp1 = sin[(1.333 \u2212 0.1333)\u03c0/2]sin(0.1333\u03c0/2)/(0.1333\u03c0/2) = 0.944. Fig. 1.5b shows a double layer, three-phase, 12-coil non-overlap stator winding of a double-sided AFPM machine with 2p = 8 rotor poles. Figures 1.6 and 2.11 show a single layer, 9-coil (Qc = 9, s1 = 2Qc = 18) non-overlap winding, while Fig. 2.12 shows a double layer 9-coil (s1 = Qc = 9) non-overlap stator winding. The difference in the number of the stator coils and rotor poles provides the starting torque for motors and the reduction of torque pulsations. Since the dimensions of AFPM machines are functions of the radius, the electromagnetic torque is produced over a continuum of radii, not just at a constant radius as in cylindrical machines"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001467_j.mechmachtheory.2010.05.001-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001467_j.mechmachtheory.2010.05.001-Figure2-1.png",
        "caption": "Fig. 2. An RSS leg.",
        "texts": [
            " It can be represented as where unit v \u03bbi = j$Ti\u2218$I ij j$Ti\u2218$I ijmax \u00f08\u00de $Ti and $Ii represent the unit TWS and the unit input twist screw of the i-th leg, respectively. This index is the same as the where modified manipulability factor in [20]. It may be seen from Eq. (8) that \u03bbi is frame-free and its value ranges from 0 to 1. For different legs, their transmissionwrenches and input joints will be different. The input transmission indices of some typical legs used in parallel manipulators are studied below. Fig. 2 shows an RSS (R and S stand for revolute and spherical joints, respectively; the underlined joint symbol means the joint is actuated) leg that contains one actuating R joint and two passive S joints. Parameters a and b are the lengths of the input link and the coupler link, respectively. With respect to the coordinate system o\u2212xyz, the unit input twist screw $I can be expressed as $I = w12 ;0\u00f0 \u00de = 1; 0; 0; 0; 0; 0\u00f0 \u00de \u00f09\u00de The transmission wrench is a pure force along the coupler link; then the TWS can be represented as $T = f12 ;a \u00d7 f12\u00f0 \u00de \u00f010\u00de a represents a vector along the input link and its norm is a"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-FigureD.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-FigureD.6-1.png",
        "caption": "FIGURE D.6 Two parallel rectangular conducting sheets.",
        "texts": [
            "5 leads to a relatively simple formula. PARTIAL POTENTIAL COEFFICIENTS FOR ORTHOGONAL GEOMETRIES 415 The result is Pp11 = 1 6 \ud835\udf0b\ud835\udcc1\ud835\udf16 [ 3 log (u + \u221a u2 + 1) + u2 + 1 u + 3 u log ( 1 u + \u221a 1 u2 + 1 ) \u2212 ( u4\u22153 + 1 u2\u22153 )3\u22152 ] , (D.21) where u = \ud835\udcc1\u2215(ye1 \u2212 ys1) and \ud835\udcc1 = xe1 \u2212 xs1. Importantly, this result is computationally useful since it is simple. It also eliminates the singularity problem for the partial self-potential term. D.1.7 Pp12 for Two Parallel Rectangular Sheet Cells The geometry for two parallel conducting sheets is shown in Fig. D.6, which was used in Ref. [3]. Here, we transformed it into our standard coordinate system. The solution for this case is given by Pp12 = 1 4\ud835\udf0b\ud835\udf16 1 12 \u222b ye1 ys1 \u222b xe1 xs1 \u222b ye2 ys2 \u222b xe2 xs2 1 R1,2 dx2dy2dx1dy1, (D.22) 416 COMPUTATION OF PARTIAL COEFFICIENTS OF POTENTIAL where S1 = (xe1 \u2212 xs1)(ye1 \u2212 ys1) (D.23a) S2 = (xe2 \u2212 xs2)(ye2 \u2212 ys2) (D.23b) R1,2 = \u221a (x1 \u2212 x2)2 + (y1 \u2212 y2)2 + (z1 \u2212 z2)2. (D.23c) Then, the closed form of this potential coefficient is Pp12 = 1 4\ud835\udf0b\ud835\udf16 1 S1S2 4\u2211 k=1 4\u2211 m=1 (\u22121)m+k [ b2 m \u2212 Z2 2 ak log (ak + rkm + \ud835\udf16 ) + a2 k \u2212 Z2 2 bm log (bm + rkm + \ud835\udf16 ) \u2212 1 6 (b2 m \u2212 2Z2 + a2 k)rkm \u2212 bm Z ak tan\u22121 ( ak bm rkm Z )] , (D"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.51-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.51-1.png",
        "caption": "FIGURE 7.51. The optimal multi-link steering mechanism in a positive turning position.",
        "texts": [
            " The behavior of the multi-link steering mechanism for different values of x, is shown in Figure 7.48. The Ackerman condition is also plotted to 7. Steering Dynamics 433 434 7. Steering Dynamics compare with the optimal multi-link mechanism. The optimality of x = \u22120.824m may be more clear in Figure 7.49 that shows the difference \u2206 = \u03b42 \u2212 \u03b4Ac for different values of x. The optimal multi-link steering mechanism along with the length of its links is shown in Figure 7.50. The mechanism and the meaning of negative value for x are shown in Figure 7.51 where the mechanism is in a positive turning position. 7.7 F Trailer-Truck Kinematics Consider a car pulling a one-axle trailer, as shown in Figure 7.52. We may normalize the dimensions such that the length of the trailer is 1. The 7. Steering Dynamics 435 positions of the car at the hinge point and the trailer at the center of its axle are shown by vectors r and s. Assuming r is a given differentiable function of time t, we would like to examine the behavior of the trailer by calculating s, and predict jackknifing"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000047_5.4400-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000047_5.4400-Figure6-1.png",
        "caption": "Fig. 6. Path (dotted line) of state trajectory of system with control when perturbed slightly below the asymptote.",
        "texts": [
            " With the switch in the lower position, the feedback becomes positive and the system's free motion satisfies E:] = [\" 3 2 '1 p]. x2 Switching of course i s not random. It occurs with respect to a sliding or switching surface, generically denoted as U = 0. To illustrate this notion, consider the surface defined as U = ul(xl, x,) = slxl + x, = 0 with s1 > 1. If the feedback is switched according to -3, if ul(xl, x,)xl > 0 i 2, if ul(xl, xz)xl < 0 k(x1) = a behavior illustrated in the phase plane plot of Fig. 6 results. The \"unstable\" equilibrium point (0,O) i s now a saddle point with asymptotes x2 = 3x1 and x, = -xl, as shown in Fig. 5(b). Observe from the dotted line trajectory that if the state vec- tor i s perturbed below the surface, ul(xl, x,) = slxl + x, = 0, at time to, it circles to the point tl before intercepting the surface again. On the other hand, if the switching surface is u2(xl, x,) = slxl + x, = 0 with s1 < 1, then a perturbation off the surface i s always immediatelyforced back to the surface since the phase-plane velocity vectors always point towards the surface"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure3.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure3.1-1.png",
        "caption": "FIGURE 3.1. A vertically loaded stationary tire.",
        "texts": [
            " Assume m = 1800 kg a1 = 1100mm a2 = 1240mm a3 = 1500mm k1 = 12800N/m k2 = 14000N/m k3 = 14000N/m and find the axles load on a level road when the car is moving with no acceleration. 3 The tire is the main component interacting with the road. The performance of a vehicle is mainly influenced by the characteristics of its tires. Tires affect a vehicle\u2019s handling, traction, ride comfort, and fuel consumption. To understand its importance, it is enough to remember that a vehicle can maneuver only by longitudinal, vertical, and lateral force systems generated under the tires. Figure 3.1 illustrates a model of a vertically loaded stationary tire. To model the tire-road interactions, we determine the tireprint and describe the forces distributed on the tireprint. 3.1 Tire Coordinate Frame and Tire Force System To describe the tire-road interaction and force system, we attach a Cartesian coordinate frame at the center of the tireprint, as shown in Figure 3.2, assuming a flat and horizontal ground. The x-axis is along the intersection line of the tire-plane and the ground. Tire plane is the plane made by narrowing the tire to a flat disk"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001514_j.engfailanal.2015.06.004-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001514_j.engfailanal.2015.06.004-Figure8-1.png",
        "caption": "Fig. 8. Crack model at gear tooth root [36].",
        "texts": [
            " Moreover they also analyzed the TVMS of a cracked planetary gear pair based on the potential energy method. Regardless of the number of the teeth and the size of the root circle diameter, Ma et al. [35] presented an improved method to calculate TVMS of a cracked gear pair considering the effects of the gear tooth clamped on the root circle and the accurate transition curve between the tooth profile and root circle (see Fig. 7). Assuming that the crack propagates along both the tooth width (crack width) and tooth thickness (crack depth) for simulating early tooth root crack (see Fig. 8), Chen and Shao [36] developed an analytical model for TVMS calculation. Based on Chen\u2019s model [36], Mohammed et al. [37] discussed the validity of the RMS and kurtosis for diagnosing the crack levels under different crack propagation forms. Assuming the propagation path of the spatial crack as a quasi-parabolic form, Yu et al. [38] presented a TVMS calculation model for a gear pair with spatial tooth crack. The TVMS calculation models for the cracked spur gear pair have also been applied to the cracked planetary gear pair after the appropriate correction [39\u201341]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000824_j.jtbi.2005.04.004-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000824_j.jtbi.2005.04.004-Figure4-1.png",
        "caption": "Fig. 4. (a) The point-mass ricochetal brachiation model from (Bertram et between flight and an effective bounce at the branch-holding swing phase. T swing path at the transition. (b) A particle sliding on a frictionless corrugat surface can be simulated by an infinite number of massless legs which contac infinitesimal collision. Note, to match the path the legs have various lengths. downwards translation are phased so that ground contact point has zero veloc line. (d) If launched with the right spin, a sphere with an eccentric center-of",
        "texts": [
            ", 2002a, b; Kram and Taylor, 1990; Kuo, 2001, 2002) have shown a substantial metabolic cost for the \u2018\u2018step-to-step transition\u2019\u2019 in human walking. Like the models here, this step-tostep transition research was inspired by analysis of passive-dynamic-based biomimetic robots. The walking \u2018\u2018step-to-step transition\u2019\u2019 is one example of the \u2018\u2018downto-up\u2019\u2019 redirection we discuss here. The walking portion of the paper here formalizes and generalizes some portions of this work. In contrast to common notions about running overground, brachiation (arm swinging from branch to branch) of gibbons uses a mechanism of apparent \u2018\u2018bouncing\u2019\u2019 (Fig. 4a) that has no essential dependence on elasticity (Bertram et al., 1999). Ricochetal brachiation, analogous to upside-down running, has a flight phase between hand holds. If the velocity of the body at the end of free-flight is tangent to the circular path of the body at the beginning of the swing, then the redirection from downwards to upwards motions has no essential collisional loss. The ape effectively bounces, ricocheting along the curve determined by her extended swing arm. That is, the essence of the pendulum in ricochetal brachiation is not to exchange gravitational and potential energy, but to redirect the body motion from generally down to generally up",
            " The corrugated t in sequence. Each leg is orthogonal to the path and produces only an (c) A football-like body can bounce with no elasticity if its rotation and ity just before contact. The center-of-mass motion is shown as a dotted -mass can hop on a rigid frictionless substrate, even with no elasticity. ARTICLE IN PRESS A. Ruina et al. / Journal of Theoretical Biology 237 (2005) 170\u2013192 181 We initiated this research project by looking for a ground-support version of ricochetal brachiation (Fig. 4a). Fig. 4b illustrates such a system. A pointmass slides on and skips over a frictionless corrugated surface. For both models the free-flight and sliding motion paths are tangent at the time of contact and there is no collisional dissipation. Now we try to implement the idea using legs. As is intuitively acceptable and demonstrated by Eq. (20) an infinite number of infinitesimal glancing collisions, even if each is fully plastic, is equivalent to frictionless sliding. The right part of Fig. 4b shows that the frictionless surface could be simulated by an infinite number of massless legs each orthogonal to the path of the pointmass and each providing a compressional force along the leg. That is, a sequence of concave-down parabolic arcs approximates a polygon. The long hindmost leg lands first and the sequence of collisions proceeds through shorter and then longer legs to final lift-off mediated by the long foremost leg. Even without a corrugated frictionless surface or an infinite number of varying-length legs, an oval-shaped rigid body can effectively bounce off a rigid flat surface without elastic recoil (Fig. 4c). The occasional high bounce of an end-over-end tumbling rugby or American football seems to roughly illustrate this phenomenon. In flight the lower-most point of the football oscillates up and down relative to the center of mass. With the right timing such a retraction can exactly cancel the downwards center of mass motion and the lower surface will touch the ground with zero downwards velocity. If the horizontal motion is also matched then the new contact is made with zero relative velocity. With no velocity discontinuity at contact there is no collisional dissipation",
            " In contrast, the mass-spring model predicts the opposite, that mid-stance is the time of minimum forward velocity. The horse canter is also the opposite of the normal spring-mediated-running paradigm for the phasing of kinetic and potential energy. Rather, the horse\u2019s energy phasing is like that of human walking, as predicted by the sequential collision model here. The legs make contact from rear to front, not simultaneously as in a pronk. This is clear in Figs. 1 and 7. Although our point-mass model cannot distinguish front legs from rear legs, the analogy with the rolling egg (Fig. 4) suggests that extended-body considerations make landing on the rear most matic strobe of a horse shows mid-contact of the trailing rear limb t together in canter, light shading) and lead fore limb (last contact erkens et al., 1993) average ground reaction force for the contact of feet ground contacts (the two lightly shaded arrows) correspond to eight fluctuations h for a small horse galloping at 6.83m/s (redrawn and pasted) with corresponding approximate footfall timing (as per ) limb. Horizontal velocity initially decreases slightly, but increases the diagonal pair of left rear (LR) and right front (RF) limbs also the h reaches its minimum height"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001261_j.mprp.2016.12.062-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001261_j.mprp.2016.12.062-Figure5-1.png",
        "caption": "FIGURE 5 Schematic showing direct metal deposition (DMD) technology (courtesy: DM3D Technology).",
        "texts": [
            " The laser or electron beam scans the powder bed surface following the toolpath pre-calculated from the CAD data of the component being built. .doi.org/10.1016/j.mprp.2016.12.062 3 FIGURE 4 Schematic showing powder bed fusion technology (courtesy: Jim Sears). S P E C IA L F E A T U R E The above process is repeated for the next and subsequent layer until the build is complete. Directed energy deposition Directed energy deposition technologies use material injection in to the meltpool instead of scanning on a powder bed (AMS specification 4999A for Ti-6Al-4V). Figure 5 shows a schematic of the DMD technology (laser based metal deposition). The process steps for the directed energy deposition are; A substrate or existing part is placed on the work table. Similar to powder bed fusion, the machine chamber is closed and filled with inert gas (for laser processing) or evacuated (for electron beam processing) to reduce oxygen level in the chamber to the desired level (AMS 4999A specifies below 1200 ppm). The DMD process also offers local shielding and does not require inert gas chamber for less reactive metals than Titanium, such as, steels, Ni-alloys, Co-alloys, etc"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000457_j.sna.2017.12.028-Figure9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000457_j.sna.2017.12.028-Figure9-1.png",
        "caption": "Fig. 9. Graphene transistors on flexible substrate shown in 3-D (a), Optical (b), cross-sectional (c) and AFM (d) images [39].",
        "texts": [
            " The gate and metal contacts of arying compositions (Au (100 nm) / Ti (10 nm), Au (38 nm) / Ti 2 nm), Au, Al (40 nm) / Ni (50 nm) /Au (300 nm), Cr (1 nm) / Au 60%) \u2013 Pd (40%) alloy (20 nm)) [38\u201342] are made by employing one f the following processes: E-beam, photolithography, and inkjet rinting techniques followed by plasma etching in an atmosphere f O2 at specific conditions (300 W/5 mbar/5 min) at room temperture. The graphene is washed and transferred from CVD or other rocesses (inkjet printing) over the transistor\u2019s length over the gate ielectric (HfO2, nanoscale polyimide, Al2O3). The graphene forms he channel also, in the process covering the contact materials. The abricated graphene transistor of the varying gate length, channel idth, and channel length is finally removed by mechanical peelng off the flexible substrate or dissolving a flexible substrate by sing acetone. Fig. 9 shows the schematic diagram of the fabrication f graphene transistors in different views [72]. The combination f nanowire field-effect transistors (NW-FETs) and GFETs interace well with the electrogenic cells resulting in a highly sensitive esponse towards cell membranes [179]. Other uses of graphene in electrical sensors involve its uses n lithium-ion batteries, photodetectors, inverters and optoelecronics. The inclusion of graphene in batteries is done by forming lusters with lithium or mainly by forming hybrids with other comounds"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001079_j.automatica.2016.06.020-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001079_j.automatica.2016.06.020-Figure1-1.png",
        "caption": "Fig. 1. Earth-fixed frame and body-fixed frame (Fossen, 2011).",
        "texts": [
            " In this paper, simultaneously considering unknown timevarying disturbances and input saturation, we develop a robust nonlinear control law for the DP system. To the best of the authors\u2019 knowledge, it is the first time in the literature that unknown timevarying disturbances and input saturation are simultaneously dealt with in the DP control design. A disturbance observer is constructed to estimate unknown time-varying disturbances and an auxiliary dynamic system is employed to handle input saturation, on the basis of which the DP control law is designed by using the DSC technique. Two right-hand coordinate frames are defined as indicated in Fig. 1. The earth-fixed frame OX0Y0Z0 is an inertial coordinate frame. The origin O of the earth-fixed frame can be chosen as any point on the earth\u2019s surface. The axis OX0 is directed to the north, OY0 is directed to the east, and OZ0 points towards the center of the earth. The body-fixed frame AXYZ is amoving coordinate frame which is fixed to the ship. The origin A of the body-fixed frame is located at the gravity center of the ship. The axisAX is directed from aft to fore, AY is directed to starboard, and AZ is directed from top to bottom"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure7.5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure7.5-1.png",
        "caption": "Fig. 7.5. HI-T-HAND Expert-1",
        "texts": [
            " The information obtained is introduced in feedback and then the actuators correct the values of internal coordinates. Better performances can be obtained if, by means of a special construction, a passive compliance of the robot gripper (without the engagement of actuators) is achieved. The well-known solution for passive compliance is the Remote Centre Compliance (RCC) developed by researchers in the Charles Stark Draper laboratory which will be described later. An example of the successful application of force sensors on robots is the HI-T-HAND Expert-l system [3]. Figure 7.5 shows the cooperation of two robots, an auxiliary one which picks up the parts with a hole from the supply device and brings them to the working position, and the main robot which picks up the shafts and performs the insert operation. The arm and the gripper of the main robot are connected via flexible construction with four strain gauges 7.2 Environment Senso's 215 capable of detecting the displacement in XYZ directions. Both robots are connected to the single controller, which uses the sensor information for correcting the centre position and direction"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001803_icra.2015.7139759-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001803_icra.2015.7139759-Figure1-1.png",
        "caption": "Fig. 1: Schematic representation of the hexarotor described in this paper.",
        "texts": [],
        "surrounding_texts": [
            "A standard hexarotor possesses six propellers that are all rotating about six parallel axes. Even though this choice increases redundancy and payload, such configuration has an underactuated dynamics similar to a standard quadrotor. In fact, the six propellers create an input force that is always parallel to that axis, no matter the values of the six rotational speeds. In this case a change of the direction of the input force in world frame can only be obtained by reorienting the whole vehicle. As a consequence, the output trajectory can only be defined by a 4-dimensional output, namely the center of mass (CoM) 3D position plus the yaw angle, despite the presence of 6 control inputs. In fact in [17] it has been proven that such kind of systems are exactly linearizable with a dynamic feedback using as linearizing output, i.e. the CoM position and the yaw angle. Feedback linearizability also implies differential flatness of the system taking as flat output the linearizing one [18]. The remaining two configuration variables, i.e., the roll and pitch angles, cannot be chosen at will, since they are being determined by the desired trajectory of the CoM, the yaw angle, and their derivatives. On the converse, the goal of the hexarotor modeling approach presented here is to exploit at best the six available inputs, thus resulting in a system that is fully actuated, i.e., linear and angular accelerations can be set independently acting on the six inputs. In order to obtain full actuation, we remove the constraint for the propellers to rotate about six parallel axes, so that a force in any direction can be generated regardless of the vehicle orientation. Thanks to full actuation, this hexarotor can track 6-DoFs trajectories comprising both the CoM position and, independently, the vehicle orientation described, e.g., by roll, pitch, and yaw, or by a rotation matrix. Even though a reallocation and reorientation of the six propellers allows for more design flexibility it also increases the number of design parameters thus increasing the design complexity. In order to find a good compromise between full actuation and low number of model parameters, we decide to add the following constraint on the parameters: \u2022 the CoM and the six propeller centers are coplanar, like in a standard hexarotor; This design choice simplify the design complexity while still allowing a full spectrum of actuation capabilities, as it will be shown in the paper. The main symbols used in the paper are shown in Table I,"
        ]
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure6.23-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure6.23-1.png",
        "caption": "FIGURE 6.23. The McPherson strut suspension is an inverted slider mechanism.",
        "texts": [
            "208), for b\u0308 and \u03b14 b\u0308 = C7C12 \u2212 C9C10 C7C11 \u2212 C8C10 (6.210) \u03b14 = C9C11 \u2212 C8C12 C7C11 \u2212 C8C10 (6.211) where, C7 = sin \u03b84 C8 = b cos \u03b84 + e sin \u03b84 C9 = a\u03b12 sin \u03b82 + a\u03c922 cos \u03b82 \u2212 2b\u0307\u03c94 cos \u03b84 +b\u03c924 sin \u03b84 \u2212 e\u03c924 cos \u03b84 C10 = cos \u03b84 C11 = \u2212b sin \u03b84 + e cos \u03b84 C12 = a\u03b12 cos \u03b82 \u2212 a\u03c922 sin \u03b82 + 2b\u0307\u03c94 sin \u03b84 +b\u03c924 cos \u03b84 + e\u03c924 sin \u03b84. (6.212) 346 6. Applied Mechanisms Example 242 Application of inverted slider mechanism in vehicles. The McPherson strut suspension is a very popular mechanism for independent front suspension of street cars. Figure 6.23 illustrates a McPherson strut suspension and its equivalent kinematic model. We attach the wheel to a coupler point at C. The piston rod of the shock absorber serves as a kingpin axis at the top of the strut. At the bottom, the shock absorber pivots on a ball joint on a single lower arm. The McPherson strut, also called the Chapman strut, was invented by Earl McPherson in the 1940s. It was first introduced on the 1949 Ford Vedette, and also adopted in the 1951 Ford Consul, and then become one of the dominating suspension systems because it\u2019s compactness and has a low cost"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure5.46-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure5.46-1.png",
        "caption": "Figure 5.46 (a) A structure loaded by a vertical force of 40 kN on the tie rod, with (b) its support reactions. For self-contained structures the support reactions do not change if one shifts a loading force along its line of action; on the other hand the interaction forces do change.",
        "texts": [
            " In a force polygon, one can now determine the forces at D and E that ensure equilibrium with the support reaction at B (see Figure 5.44b). Check: The left-hand rafter ACE must also be in equilibrium. You can see immediately that there is force equilibrium in Figure 5.45. To check the moment equilibrium, write down the moment equation for all the forces about an arbitrary point. Figure 5.43 To see how rafter AEB and bar CD exert forces on one another, they have been isolated. 182 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Example 3 Figure 5.46a uses the same structure as in Example 2, except that this time, the vertical force F = 40 kN has been shifted along its line of action to a point of application on member CD. Questions: a. Determine the support reactions. b. Determine the forces acting on the isolated parts ACE, BDE, and DE. c. Perform a graphical check of the moment equilibrium for each of the parts. Solution: a. The support reactions are the same as those in example 2. They are shown in Figure 5.46b. Note that for a self-contained structure, the support reactions do not change if one shifts a force along its line of action. The forces within the structure do change, however, as is shown below. b. In Figure 5.47a, the various structural parts have been isolated, and all the interaction forces are shown. First look at the equilibrium of CD. From the moment equilibrium about C follows Dv = 10 kN. From the moment equilibrium about D follows Cv = 30 kN. The horizontal force equilibrium gives Ch = Dh"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure14.39-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure14.39-1.png",
        "caption": "Figure 14.39 The isolated anchor chain with a uniformly distributed load along the chain. The anchor chain has the shape of a catenary.",
        "texts": [
            " The anchor chain has a dead weight of 24 N/m. The upward water pressure on the chain is 3 N/m. Question: Determine the (horizontally measured) length for which the chain is free from the bottom, and the maximum force in the chain. Solution: The (uniformly) distributed load on the chain is equal to the dead weight minus the upward water pressure: q = (24 N/m) \u2212 (3 N/m) = 21 N/m. This load is a force per length measured along the chain. The anchor chain will therefore assume the shape of a catenary.1 In Figure 14.39, the anchor chain has been isolated. At A, the cable is tangent to the bottom, and only a horizontal force H acts. The horizontal force equilibrium gives H = 3.5 kN. For the further calculation, we use the formulas derived in Section 14.1.5. For the catenary in the coordinate system given in Figure 14.39, it applies 1 We ignore the fact that the upward water pressure is missing over the small part that the chain is above the water. 668 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM that z = \u2212H q ( cosh qx H \u2212 1 ) or cosh qx H = 1 \u2212 qz H so that x = H q cosh\u22121 ( 1 \u2212 qz H ) . At B, x = ; z = \u221230.9 m applies, which gives the following (be aware of the units!): = 3500 N 21 N/m cosh\u22121 ( 1 \u2212 (21 N/m)(\u221230.9 m) 3500 N ) = 100 m For the vertical force at B Bv = \u2212Vx= = H sinh q H = (3500 N) sinh ( 21 N/m)(100 m) 3500 N ) = 2228 N \u2248 2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure4.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure4.6-1.png",
        "caption": "Fig. 4.6. Stator slots of AFPM brushless machines: (a) rectangular semi-open slot; (b) rectangular open slot; (c) oval semi-open slot.",
        "texts": [
            "8) where N1 is the number of turns per phase, k1X is the skin-effect coefficient for leakage reactance, p is the number of pole pairs, q1 = s1/(2pm1) is the number of stator slots s1 per pole per phase (2.2), l1in is the length of the stator winding inner end connection, l1out is the length of the stator winding outer end connection, \u03bb1s is the coefficient of the slot leakage permeance (slotspecific permeance), \u03bb1ein is the coefficient of the leakage permeance of the inner end connection, \u03bb1eout is the coefficient of the leakage permeance of the outer end connection and \u03bb1d is the coefficient of the differential leakage. The coefficients of leakage permeances of the slots shown in Fig. 4.6 are: \u2022 for a rectangular semi-open slot (Fig. 4.6a): \u03bb1s = h11 3b11 + h12 b11 + 2h13 b11 + b14 + h14 b14 (4.9) \u2022 for a rectangular open slot (Fig. 4.6b): \u03bb1s \u2248 h11 3b11 + h12 + h13 + h14 b11 (4.10) \u2022 for an oval semi-open slot (Fig. 4.6c): \u03bb1s \u2248 0.1424 + h11 3b11 + h12 b12 + 0.5 arcsin[ \u221a 1\u2212 (b14/b12)2] + h14 b14 (4.11) The coefficients of leakage permeances for other shapes of slots than those shown in Fig. 4.6 are given, e.g. in [159]. The above specific-slot permeances 130 4 AFPM Machines With Iron Cores (4.9), (4.10) and (4.11) are for single-layer windings. To obtain the specific permeances of slots containing double-layer windings, it is necessary to multiply eqns (4.9), (4.10) and (4.11) by the factor 3\u03b2 + 1 4 (4.12) where \u03b2 is according to eqn (2.7). Such an approach is justified if 2/3 \u2264 \u03b2 \u2264 1.0. The specific permeance of the end connection (overhang) is estimated on the basis of experiments. For double-layer, low-voltage, small and mediumpower machines the specific permeances of the inner and outer end connections are: \u2022 inner end connection \u03bb1ein \u2248 0",
            "83, double-sided disc PM synchronous motor has the following dimensions of the magnetic circuit: PM outer diameter Dout = 0.28 m, PM inner diameter Din = 0.16 m, thickness of the rotor (PMs) 2hM = 8 mm, single-sided mechanical clearance g = 1.5 mm. A surface configuration 4.8 Finite Element Calculations 141 of uniformly distributed PMs has been used. The rotor does not have any soft ferromagnetic material. The rotor outer and inner diameters correspond to the outer and inner outline of PMs and the stator stack. The dimensions of rectangular semi-closed slots (Fig. 4.6b) are: h11 = 11 mm, h12 = 0.5 mm, h13 = 1 mm, h14 = 1 mm, b11 = 13 mm and b14 = 3 mm. The number of stator slots (one unit) is s1 = 24, the number of armature turns of a single stator per phase is N1 = 456, the diameter of the stator copper bare conductor is 0.511 mm (AWG 24), the number of stator parallel wires is aw = 2 and the air gap magnetic flux density is Bmg = 0.65 T. The rotational losses are \u2206Prot = 80 W and the core and PM losses are \u2206P1Fe +\u2206PPM = 0.05Pout. There are two winding layers in each stator slot"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000876_j.surfcoat.2004.06.029-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000876_j.surfcoat.2004.06.029-Figure1-1.png",
        "caption": "Fig. 1. The comparison of coaxial (left) and side (right) laser cladding set-up above a moving substrate.",
        "texts": [
            " In this process, the carrier gas is used to form a cladding powder stream blown under the laser beam while it scans the surface of the substrate generating a melt pool with a depth that corresponds to the thickness of a single clad generated in one step. A fully dense layer can be achieved when single tracks consecutively overlap side by side. The process requires minimum surface preparation and solves the problem of application on complex geometries. There are two ways of feeding powder: from a side or coaxially to the laser beam, see Fig. 1. The technique of side-cladding has been extensively described in previous works [1,3]. When the powder stream is injected off-axis from the laser beam, the change of substrate movement direction leads to completely different local cladding conditions. For instance, a so-called bagainst hillQ cladding condition takes place when the powder stream is applied from the side from which the substrate moves. In this case, the clad powder is trapped temporarily in a corner formed between the molten track and the flat substrate"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001777_j.optlastec.2016.04.009-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001777_j.optlastec.2016.04.009-Figure3-1.png",
        "caption": "Fig. 3. Temperature contour plots during SLM processing (v\u00bc100 mm/s and P\u00bc125 W). (a), (c) and (e): top surface of the molten pool of Point 1, Point 2 and Point 3, respectively; (b), (d) and (f): longitudinal view of the molten pool of Point 1, Point 2 and Point 3, respectively.",
        "texts": [
            " The processing parameters in the experiments are same as those in the simulation listed in the Table 1. The samples for metallographic examinations were cut, ground and polished according to standard procedures strictly. The cross-sectional microstructures of the SLM-produced TiC/Inconel 718 parts were characterized using a PMG3 optical microscopy (Olympus Corporation, Japan). The characteristic surface morphologies of the parts were characterized by scanning electron microscopy (SEM; Hitachi model S-4800, Japan) in secondary electron mode at 10 kV. Fig. 3 shows the transient temperature distribution on the top surface and longitudinal view of the molten pool as laser beam reaches Point 1, Point 2 and Point 3, respectively, with v of 100 mm/s and P of 125 W (Fig. 2(b)). At the center of the first track (Point 1, Fig. 2(b)), the isotherm curves on the top surface of the molten pool are similar to a series of ellipses and, the ellipses of fore part are more intensive than those at the back-end of it. Meantime, it is asymmetry of isotherms along the laser scanning direction. However, this phenomenon is not conspicuous enough as shown in temperature contour pots (Fig. 3). The dashed line circle shown in the temperature contour plots presents the melting temperature of Inconel 718 (1300 \u00b0C). The temperature in this dashed line circle is higher compared with the melting point of Inconel 718, which induces a small molten pool within this region. The predicted operative temperature of the molten pool reduces from 2113 \u00b0C in the center of the melt to 1182 \u00b0C at the edge of the molten pool as the laser beam reaches Point 1. Three dimensions of the molten pool are approximately 95.5 mm (width), 106.5 mm (length) and 61.3 mm (depth), respectively (Fig. 3(a) and (b)). Then the laser spot further moves to the Point 2, the working temperature of the molten pool decreases from 2291 \u00b0C in the center to 1290 \u00b0C at the edge of the molten pool. There, then, is a cooling time of 10-ms after the fabrication of n layer is completed by scanning track by track and, then, the SLM process of n\u00fe1 layer continues. The predicted working temperature of the molten pool ranges from 2348 \u00b0C in the center to 1345 \u00b0C at the edge of the pool as the laser beam reaches Point 3. Comparing three dimensions of molten pool in Point 2 with these three figures of the pool in Point 1, the width (109.3 mm), length (120.7 mm) and depth (67.8 mm) of the melt pool at Point 2 increases by 14.5%, 13.3% and 10.6%, respectively, to these at Point 1 (Fig. 3(c) and (d)). In addition, comparing these three figures of Point 3 with these of Point 1, the width (124.1 mm), length (132.2 mm) and depth (73.1 mm) of the molten pool at Point 3 are also slightly larger than these at Point 2, increasing by 13.5%, 9.5% and 7.8%, respectively (Fig. 3 (e) and (f)). The simulation temperature contour plots show that the isotherm ellipses of the fore-part of molten pool (unscanned region) arrange more intensive than these at back-end of the melt (scanned region). This is mainly attributed to the change of thermal conductivities of TiC/Inconel 718, which are listed partly in Tables 2 and 3, due to the transition from powder to solid. This kind of transformation from powder to block contributes to the heat transmission in the powder layer. Meantime, there is inconspicuous asymmetry of isotherms along the laser scanning direction since the special thermal-physical properties of materials, especially the thermal conductivities of TiC/Inconel 718"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000182_j.electacta.2008.03.005-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000182_j.electacta.2008.03.005-Figure3-1.png",
        "caption": "Fig. 3. Scheme displaying the construction and the transduction principle of a sandw from Ref. [53]).",
        "texts": [
            " This example shows that nanoparticles can be used not only as medium to retain biomolecules with a high stability, but also to provide versatile labels for the amplification of biosensing events. The secondary dissolution of the captured nanoparticles enables the amplified detection of the respective analyte by the release of many ions/molecules as a result of a single recognition event. This methodology was applied in a sensitive immunosensor for human IgG based on a sandwich-type assay using colloidal gold as electrochemical label [53]. As it is displayed in Fig. 3, the capture protein was immobilized on a carbon paste electrode through passive adsorption to bind quantitatively with the corresponding antigen and colloidal gold-labeled antibody. In order to detect the amount of colloidal gold captured on the electrode surface, it was oxidized electrochemically to produce AuCl4\u2212 ions which were strongly adsorbed on the electrode surface and determined by adsorptive voltammetry. a Nafion monolayer; (b) adsorption of thionine; (c) formation of gold nanoparticles A highly sensitive electrochemical impedance immunosensor was also developed using an amplification procedure with a colloidal gold-labeled antibody as the primary amplifying probe, and a multistep amplification sequence"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure3.54-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure3.54-1.png",
        "caption": "Figure 3.54 A junction of three coplanar tubes that are rigidly connected at equal angles of 120\u25e6. The tubes are loaded (by torsion) by the couples TA, TB and TC.",
        "texts": [
            " When determining the resultant of this system of parallel forces, only the force sum in the z direction and the moment sum about the x and y axis are relevant: \u2211 Fy = +25 \u2212 15 + 20 \u2212 45 + 15 = 0 kN,\u2211 Tx = 25 \u00d7 0 + 15 \u00d7 1 \u2212 20 \u00d7 3 \u2212 15 \u00d7 4 + 15 \u00d7 4 + 45 \u00d7 4 = +75 kN,\u2211 Tz = +25 \u00d7 6 \u2212 15 \u00d7 2 + 20 \u00d7 5 \u2212 45 \u00d7 3 + 15 \u00d7 1 = +100 kN. There is no resultant force, but there is a resultant couple T of which the moment vector is in the xy plane (see Figure 3.53b). Its magnitude is T = \u221a 752 + 1002 = 125 kNm. The resultant couple acts in a plane perpendicular to the moment vector. 92 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Example 3 Figure 3.54 shows a junction of three coplanar tubes that are rigidly connected at equal angles of 120\u25e6. The tubes are loaded (by torsion) by the couples TA, TB and TC. The resultant couple on the junction is zero. Question: How large are the couples TA and TB if TC = 75 Nm? Solution: In Figure 3.55a, the couples are represented by their moment vectors. The three vectors are in the xy plane, the plane in which the tubes are located. The resultant moment on the junction is zero if the three vectors form a closed polygon, analogous to the closed force polygon for force equilibrium"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001618_j.actamat.2016.11.018-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001618_j.actamat.2016.11.018-Figure5-1.png",
        "caption": "Fig. 5. (a)The meshed region indicates the remelted zone in the powder and substrate layer and the instantaneous melt pool is progressively dark colored. (b) Meltpool cross section as obtained from re-solidified region mass fraction fL \u00bc 0.5. The higher temperature region (dark color) indicates instantaneous melt pool. The height from the substrate indicates the height of the melt pool.",
        "texts": [
            " The model used two materials 1 Instead of defining the surface traction at the interface, one can define an appropriate boundary condition on the free liquid surface: m vui vz \u00bc ds dT$ vT vxi \u00bc 2$vTvxi , where i \u00bc 1, 2 corresponds to x and y-directions accordingly. (Argon and IN718) in the domain with no interphase mass transfer between them to compute the temperature gradient andmelt-pool dimension. The simulation starts from initial solid substrate and powder layer at the temperature of 300 K and with argon is at 500 K. The predicted longitudinal cross-section of the melt pool is shown in Fig. 5a. The moving heat source creates the instantaneous melt pool. The scan direction coincides with x-axis and the build direction with z-axis. The meshed region is overall penetration obtained due to laser scan and is obtained using transient history of temperature. The melt pool width obtained due to laser scan is shown in Fig. 5b the light colored region is the instantaneous melt pool as seen from the top view. The mushy zone size is numerically estimated assuming that there is all material in the temperature range between solidus and liquidus temperatures. The dark colored volume represents the instantaneous melt pool followed by the mushy zone at the trailing edge. Numerical results show that the volume of mushy zone reaches steady state value after approximately 8 10 4 s. The predicted volume of the steady state mushy zone is about 1"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002528_aisy.201900171-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002528_aisy.201900171-Figure2-1.png",
        "caption": "Fig. 2. Examples of soft robotic systems by (A) Hellebrekers et al. [17] and (B) Han et al.",
        "texts": [
            " Controlled dynamic experiments along with a recurrent neural network [18], [22] and controlled static experiments paired with A cc ep te d A rti cl e A cc ep te d A rti cl e A cc ep te d A rti cl e A cc ep te d A rti cl e A cc ep te d A rti cl e A cc ep te d A rti cl e A cc ep te d A rti cl e A cc ep te d A rti cl e A cc ep te d A rti cl e A cc ep te d A rti cl e A cc ep te d A rti cl e A cc ep te d A rti cl e A cc ep te d A rti cl e A cc ep te d A rti cl e A cc ep te d A rti cl e A cc ep te d A rti cl e A cc ep te d A rti cl e A cc ep te d A rti cl e A cc ep te d A rti cl e A cc ep te d A rti cl e A cc ep te d A rti cl e A cc ep te d A rti cl e This article is protected by copyright. All rights reserved feed-forward neural networks have been shown to estimate magnitude (in N) and location of touch (in mm) [17] and object orientation [23] (Figure 2 A,B). The second, more common, technique avoids direct controlled characterization and instead prioritizes higher-level labels, such as successful grasp recognition, slip detection, or object classification. We attribute this largely to the discrepancy in individual sensor characterization and practical use. Specifically, the inherent limitation that soft material sensor response will change with the properties of neighboring structures and mounting equipment. Additionally, the real-world values provided by the first method are not necessarily understood by the robotic system"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002646_j.compositesb.2020.108344-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002646_j.compositesb.2020.108344-Figure4-1.png",
        "caption": "Fig. 4. (a\u2013c) Experimental setup of SMP TCO metamaterial under axial compressive loading, (d) The boundary conditions of the finite element model.",
        "texts": [
            " The theory equation of the compression twist deformation on TCO metamaterial is given as follows. \u03c6= \u2211m i=1 (\u2212 1)p\u03c6i= \u2211m i=1 (\u2212 1)p \u22c5 \u23a7 \u23a8 \u23a9 2arcsin \u23a1 \u23a3 \u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305 4ri 2sin2(\u03b8i/2)+hi 2 \u2212 (hi \u2212 \u0394zi) 2 \u221a 2ri \u23a4 \u23a6\u2212 \u03b8i \u23ab \u23ac \u23ad (7) When the twisting direction of the i-th layer and the first layer is the same, p is 0, otherwise p is 1. The thickness t of the metamaterial paper in this work is 1 mm. Using the obtained compression twist deformation equation (7), the deformation mode of TCO metamaterial can be predicted and designed. In Fig. 4, a quasi-static uniaxial compression test is performed on the R. Tao et al. Composites Part B 201 (2020) 108344 SMP TCO unit structure after the 3D printing is completed, and the indenter speed is set to 10 mm/min. A thermostat with adjustable temperature is installed outside the compressor to control the temperature of the sample. An observation window is set on the outside of the thermostat, and a high-definition camera is installed to record the complete deformation and deployment process. The end of the sample is bonded to flat thrust ball bearings to release twist deformation",
            " To analyze the mechanical behavior of SMP TCO metamaterials, we used the commercial software ABAQUS (3DS Dassault Syste\u0300ms, France) for finite element simulation. At present, there are two main SMP deformation theories. One is based on the classical viscoelastic theory, and the other is based on the phase transition theory, in which includes a glass phase and a rubber phase [39,40]. In this work, we used the viscoelastic constitutive model proposed and finite element method by Tao et al. [41]. In Fig. 4d, to capture the deformation behavior of SMP metamaterial and apply the same loading conditions as the experiment. The SMP constitutive model, finite element method and material parameters are described in detail in the supporting information. SMP TCO metamaterials integrate reversibility and automatically mechanically tunable properties, which is a function not available in other metamaterials manufactured in the past. In order to characterize the mechanically tunable properties, a unit structure of SMP TCO metamaterial is compressed at 30 \u25e6C, 40 \u25e6C, 50 \u25e6C and 60 \u25e6C. The parameters of this unit structure t, h, r, \u03b8 and n are 1 mm, 40 mm, 30 mm, 30\u25e6, and 6, respectively. Considering the compression-twist coupling deformation effect, the experimental method in Fig. 4 can measure the twist angle of the TCO unit structure. Fig. 5a is the force-displacement curves of the unit structure during loading at different temperatures. The finite element analysis and experimental results have good consistency. The force-displacement curves exhibits three stages, that is, linear elastic deformation stage, platform stress stage, and densification stage. The platform stress stage of this unit structure is very long and stable. This feature can be used to absorb impact energy and vibration isolation"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003628_91.705510-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003628_91.705510-Figure6-1.png",
        "caption": "Fig. 6. Structure of an inverted pendulum system.",
        "texts": [
            " In this section, we shall demonstrate that the decoupled fuzzy sliding-mode controller is applicable to the cart-pole system [19], [22] and the ball-beam system [23] to support the theoretical development. The closed-loop system was simulated by means of Euler\u2019s method with constant step length on an i486 PC using MATLAB [24]. In order to demonstrate the external disturbance rejection capability of the proposed method, a uniformly distributed disturbance is applied on all cases. A. Inverted Pendulum Physical Interpretation: Assume the inverted pendulum system is initially at the origin with a pole angle (see Fig. 6). To balance the pole, the system should exert a force to the right of the cart and then the cart starts going to the left-hand direction. As time progresses, the pole will reach to the right side of the vertical line. Again, to balance the pole a force to the left of the cart should be generated to move the cart to the right-hand direction. To keep the pole in an upright position, it is necessary to move the cart back and forth along the horizontal direction. In terms of mathematical analysis, this can be achieved by making the pole trace a signal, which represents a decaying oscillation of pole angle, and will be determined later. The structure of an inverted pendulum is illustrated in Fig. 6 and its dynamics are described below: (28) where the angle of the pole with respect to the vertical axis; the angular velocity of the pole with respect to the vertical axis; the position of the cart; the velocity of the cart; the apply force to move the cart and the form of are shown in Appendix A. It is possible to define the following: (29) (30) (31) (32) Since the controlling of the pole angle is realized by changing the position of the cart, in the inverted pendulum, the main control target is to keep the pole balanced while the cart is moved back to the original position"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001401_tcyb.2015.2405616-Figure12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001401_tcyb.2015.2405616-Figure12-1.png",
        "caption": "Fig. 12. Trajectory of the system output y.",
        "texts": [
            " 9\u201311, it can be seen that the desired control performance can be guaranteed even if the system is affected by random disturbances. To further explain that the restrictive condition for quantized parameter in [15] can be removed, we consider the control performance for \u2200\u03b4 \u2208 [0.1, 0.8] and \u03bcmin = 0.5. The design parameters are chosen as k1 = k2 = 8, a1 = 0.8, a2 = 1, \u03bb1 = 20, \u03bb2 = 25, \u03b31 = 1, and \u03b32 = 2. The simulation is run with the initial conditions [x1(0), x2(0)]T = [1, 0.5]T , and [\u03b8\u03021(0), \u03b8\u03022(0)]T = [0.1, 0.2]T . The simulation results are displayed in Figs. 12\u201315. Fig. 12 displays the trajectory of the system output y. It can be seen that, for \u2200\u03b4 \u2208 [0.1, 0.8], the system output can track the reference signal yd to a bounded compact set. Figs. 13 and 14 show that the state variables x2 and the adaptive parameters \u03b8\u03021 and \u03b8\u03022 are bounded. Fig. 15 shows the control signal u and the quantized input q(u). The simulation results show that although the system is subjected to random disturbances and fi(x\u0304i) are not linear parametric, the system output still follows the given reference signal closely and the closed-loop signals remain bounded for \u2200\u03b4 \u2208 [0"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure2.23-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure2.23-1.png",
        "caption": "Figure 2.23 A rough sketch of the closed force polygon for the force equilibrium of joint A.",
        "texts": [
            " Here we have two equations with two unknowns, namely the magnitude of 1 It is incorrect to say that \u201cthe forces neutralise one another\u201d as the forces continue to exist. 38 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM the forces F1 and F2. We can write the equations as 0.0523 \u00d7 F1 \u2212 0.9397 \u00d7 F2 = 0, 0.9986 \u00d7 F1 \u2212 0.3420 \u00d7 F2 = 8800 N, so that: F1 = 8984 N, F2 = 500 N. Alternative solution: We also can calculate the forces on the basis of the closed force polygon for the equilibrium of junction A. A rough sketch of the force polygon, such as that in Figure 2.23, suffices. According to the sine rule this gives F1 sin 110\u25e6 = F2 sin 3\u25e6 = F3 sin(180\u25e6 \u2212 110\u25e6 \u2212 3\u25e6) = F3 sin 67\u25e6 . With F3 = 8800 N this means that F1 = F3 \u00b7 sin 110\u25e6 sin 67\u25e6 = 8983 N, F2 = F3 \u00b7 sin 3\u25e6 sin 67\u25e6 = 500 N. The fact that F1 is 1 N less than before is the result of rounding off the goniometric function values (to four decimal places) in the previous solution. 2 Statics of a Particle 39 Compounding coplanar forces (Sections 2.1.1 and 2.1.1) 2.1 Which combination of forces has the smallest resultant"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.36-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.36-1.png",
        "caption": "FIGURE 7.36. A symmetric four-wheel steering vehicle.",
        "texts": [
            "119) The position of turning center for a FWS vehicle is at xO = \u2212a2 yO = 1 2 wf + l tan \u03b4if (7.120) 418 7. Steering Dynamics and for a RWS vehicle is at xO = a1 yO = 1 2 wr + l tan \u03b4ir . (7.121) Example 283 F Curvature. Consider a road as a path of motion that is expressed mathematically by a function Y = f(X), in a global coordinate frame. The radius of curvature R\u03ba of such a road at point X is R\u03ba = \u00a1 1 + Y 02\u00a23/2 Y 00 (7.122) where Y 0 = dY dX (7.123) Y 00 = d2Y dX2 . (7.124) Example 284 F Symmetric four-wheel steering system. Figure 7.36 illustrates a symmetric 4WS vehicle that the front and rear wheels steer opposite to each other equally. The kinematic steering condition for a symmetric steering is simplified to cot \u03b4o \u2212 cot \u03b4i = wf l + wr l (7.125) 7. Steering Dynamics 419 and c1 and c2 are reduced to c1 = 1 2 l (7.126) c2 = \u22121 2 l. (7.127) Example 285 F c2/c1 ratio. Longitudinal distance of the turning center of a vehicle from the front axle is c1 and from the rear axle is c2. We show the ratio of these distances by cs and call it the 4WS factor"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003482_rob.4620120607-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003482_rob.4620120607-Figure1-1.png",
        "caption": "Figure 1. Two-wheeled planar mobile robot.",
        "texts": [
            " MATHEMATICAL PRELIMINARIES In this section we describe some mathematical tools that will be used in the sequel. We restrict our treatment to certain special cases of the general theory when appropriate. Our goal is to provide practical formulas; we include references to more complete discussions of the underlying mathematics. We begin with an example. Kelly and Murray: Geometric Phases and Robotic Locomotion 419 2.1. Example Consider the dynamics of a two-wheeled mobile robot that is able to drive in the direction in which it points and spin about its center, as shown in Figure 1. Let $l and $2 denote the angles of rotation of the two wheels (with respect to arbitrary initial states); the position of the robot is given by the xy location of its center and the heading angle 8. Balance is maintained by a small castor whose effect we shall otherwise ignore. Thus 9 = ($,, $2, x, y, 0) denotes the configuration of the system. The motion of the system is governed by Lagrange\u2019s equations, taking into account the constraints that describe the contact of the wheels with the ground"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure4.49-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure4.49-1.png",
        "caption": "Fig. 4.49: An example of studying human grasp: a) at the grasp with marks on the finger articulations; b) a kinematic model; c) a sequence for approaching and grasping a cylindrical object.",
        "texts": [
            " as due to the palm connection. The scheme also shows the proportions of the phalanges that are usually also used in robotic designs. Because of this architecture, a human hand has great versatility in grasping and manipulation. Basic movements of the fingers are shown in Fig. 4.48 as typical for many possibilities of grasp. Indeed, the study of the hand grasp is fundamental for developing suitable designs and also efficient grasping strategy. The kinematic study of a human grasp can be attached as shown in Fig. 4.49 in which a kinematic model is determined by using suitable marks on finger articulations, Fig. 4.49a), to obtain the sketch of Fig. 4.49b). Then, in Fig. 4.49c) the motion sequence for approaching and grasping an object is recorded in order to evaluate both the grasping operation and finger motion. The scheme of Fig. 4.49b) refers to a planar grasp that is obtained by the thumb and index finger, whose motion is properly described through the corresponding joint angles \u03b81, \u03b82, and \u03b83. Similar schemes can be determined for each of the many grasping configurations that a human hand can adapt to different objects. The kinematics of a finger motion can also be studied as referring to a scheme of the finger as a planar 3R manipulator. Fundamentals of the Mechanics of Robotic Manipulation 301 Chapter 4 Fundamentals of the Mechanics of Grasp302 Alternatively, planar mechanism can be used even to obtain 1 d"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure2.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure2.6-1.png",
        "caption": "FIGURE 2.6 Example PEEC model for three fundamental loops, which includes the capacitive part.",
        "texts": [
            " Finally, if a branch is not connected to a node, a zero results in both cases. Hence, we understand why the A matrix is related to AT . We also want to observe that an MNA stamp does, for each branch, automatically use the specific entries into A for the current equations and AT for the nodal connections to the circuit. So, it takes over the functions of both since it implicitly uses the KCL and KVL. We consider a circuit that is relevant for the PEEC model in most chapters, which is shown in Fig. 2.6. This also serves as an example for the application of the matrix KCL for the circuits of interest to us in this text. We can treat the currents for the capacitance as Ic, for the internal capacitive current sources Is, the inductances as I\ud835\udcc1 , and finally the input current sources as Ii. Note that an independent current source could be attached to each node in Fig. 2.6. However, possible independent current sources are not shown in this figure. We should note that inspecting the stamps given in Appendix B for resistors can either be included with the inductance currents as is the case in (2.13) or stamped similar to the capacitor admittances stamps using G. The KCL matrix that is split up into submatrices corresponding to the different components, which is AcIc + AsIs + A\ud835\udcc1I\ud835\udcc1 + AiIi = 0, (2.37) where Ac, As, A\ud835\udcc1 , and Ai are the KCL connectivity matrices corresponding to capacitive, controlled current sources, inductive, and input currents. So, in a sense, we want to stamp all elements of the same type simultaneously to make the process more efficient. Of course, this works best for structured circuits such as a PEEC model. It is evident how we subdivide the matrix current law according to the element types shown in Fig. 2.6. This results in all the elements of the same type that can be stamped into the matrix together. We still can add additional elements to the circuit matrix such as load resistors, etc., by stamping them at an element-by-element basis. The total capacitive current for each capacitive source current for each node is given by IT = Ic \u2212 Is. The details of the derivation have been assigned to Problem 2.3. Since all capacitor nodes are connected to ground, Ac is an identity matrix and As is minus an identity matrix",
            " Note that the input current pulse has an amplitude of 10 mA with a rise time and a fall time of 0.1 ns. The width of the pulse roof is 0.2 ns. Hence, it is a trapezoidal pulse shape. 2.3 Capacitance model Derive in detail that the MNA equation (2.45) that is based on the special capacitance current relations (2.39) and (2.38). Start from the relation for the potential (4.10), or \ud835\udebdn = Pp Q for a two-conductor capacitance problem and take the time derivative on both sides of the equation since i = dQ\u2215dt. Also, verify the capacitance equivalent circuit in Fig. 2.6 is correct. 2.4 Equivalent circuit synthesis Synthesize the equivalent circuit of the open-ended input impedance of a transmission line whose residues and poles are listed in Table 2.11. 2.5 State-space model identification For the pole\u2013residue set of the previous problem, find the state-space matrices according to Section 2.9.4. 1. J. Vlach and K. Singhal. Computer Methods for Circuit Simulation. Van Nostrand Reinhold Co., New York, 1983. 2. A. Ruehli Ed. Circuit Analysis, Simulation and Design, Part 1"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001928_j.medengphy.2012.04.007-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001928_j.medengphy.2012.04.007-Figure1-1.png",
        "caption": "Fig. 1. Basic phases in needle insertion: (a) no interaction; (b) bound",
        "texts": [
            " The large range of forces s mainly due to atypical needle diameters (30 m [49] to 11 mm 18]). The median is used because it is insensitive to these extrema. .2. Needle insertion phases During needle insertion into soft-tissue, the motion of the neele relative to the surrounding tissue needs to be considered, rather han the absolute motion of the needle. By looking at the position f a needle relative to a tissue boundary, it is possible to distinguish three basic phases of interaction, as depicted in Fig. 1. These phases may be repeated when the needle encounters internal structures or variations in tissue properties. Similar penetration phases are described in the German industry standard for hypodermic needle tips, DIN 13097. 2.2.1. Phase 1: Boundary displacement The first phase (Fig. 1b) starts when the needle comes into contact with the tissue boundary, and ends when the tissue boundary is breached. The actual breaching of the boundary is referred to as the puncture event. During the boundary displacement phase, the tissue boundary deflects under the influence of the load applied by the needle tip, but the needle tip does not penetrate the tissue (the boundary moves along with the needle). This phenomenon is known as \u201ctenting\u201d [51,54]. A typical force\u2013time curve is shown in Fig",
            " The work of Shergold and Fleck [71] and that of Azar and Hayward [37] suggests that the shape of this crack depends on the shape of the needle tip. A planar crack is initiated when using sharp bevel needles or conical needles. Diamond tips were found to create star shaped cracks, and needles with a very large bevel angle were found to create ring cracks, as illustrated by Fig. 3. This result was found for artificial gels as well as for porcine skin and porcine liver ex vivo, and for human skin in vivo. Once a crack has been initiated, phase 2 commences. 2.2.2. Phase 2: Tip insertion The second phase (Fig. 1c) starts when the tissue boundary is breached, and ends when the tissue-boundary slides from the tip onto the shaft. During this phase, as the needle is advanced, the crack in the tissue-boundary is enlarged. The cut made by the sharp edges of the tip is wedged open by the increasing cross-sectional area of the tip, as described by e.g. Mueller [24]. The crack growth process can be either gradual, stable crack growth (cutting), or sudden, unstable crack growth (rupture), depending on the local properties of the tissue, such as rupture 668 D",
            " The transition from tip to shaft may also give rise to an increase in axial force, due to the hole in the tissue boundary being wedged open. The magnitude of this effect depends on needle type, as will be discussed later. lar diamond tip, small bevel angle, very large bevel angle. D.J. van Gerwen et al. / Medical Engineer \u00a9 2006 IEEE. 2 s t b r a f i l t o i i i s 2 l a f t h c o rectional motion, is used by Abolhassani et al. [31] (turkey muscle), ct on the hub in all directions, but focus here is only on the axial force since this is onsidered the most important. .2.3. Phase 3: Tip and shaft insertion The third phase (Fig. 1d) starts just after the transition from tip to haft and ends when the needle is stopped or when a new (internal) issue boundary is encountered. During this phase, the contact area etween tip and tissue and the size of the hole at the boundary emain more or less constant. Only the contact area between shaft nd tissue increases as the needle is advanced. During this phase the needle is subject to cutting (or rupture) orces at the tip, and to a varying friction force that is due to the ncreasing contact area between shaft and tissue"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000355_20.195559-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000355_20.195559-Figure3-1.png",
        "caption": "Fig. 3. Features of \u201c2-d\u201d analytical models. (a) Model for calculating magnetic field produced by magnets. (b) Model for calculating magnetic field produced by windings. (c) Model for calculating relative permeance.",
        "texts": [
            " MAGNETIC FIELD DISTRIBUTION IN BRUSHLESS PERMANENT MAGNET DC MOTORS, PART 111 145 slot openings are neglected and the relative permeance function of the slotted airgap region. Similarly, the armature reaction field distribution is calculated from the product of the magnetic field produced by the stator windings when stator slot openings are neglected and the same relative permeance function, i.e., - Bopen-nrcuit(r7 Q, a m a ) = Bmognet(r, a, ama) (r , a) (124 - Barmature-reaction(r, a, a m a ) = Bwmdmg(r, a? a m a ) ( r , a) * (12b) The principle is illustrated in Fig. 3, whilst the calculation of the field components Bmagner(r, a, am& and Bwinding(r, a ama) have been described in parts I and I1 [3], [4], The calculation of the relative permeance function X (r, CY) now follows. 2) Calculation ofi; ( r , a) The permeance X (r, a) i s calculated from the flux density distribution B\u2019 (r, a) assuming unit difference in magnetic potential between the stator and roto? iron of the slotted machine, shown in Fig. 3(c). Since X ( r , a) is defined as the relative permeance with a unity maximum value it is obtained by dividing X ( r , a) by a reference permeance A , equal to the one-dimensional permeance of an equivalent slotless stator motor, i.e., Therefore, where X (r, a) equals the flux density distribution B\u2019 ( r , a ) in the magnetlairgap region. 3) Calculation of Bmagner ( , a, ama) and Bwinding (r , a m a ) Bmagnet (19 a, Qmo) and Bwrndrng ( r , a, Qma) are defined as the radial components of flux density due to the magnets and the stator windings, respectively, in the equivalent slotless motor, as shown in Figs"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000206_978-0-387-74315-8-Figure2-1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000206_978-0-387-74315-8-Figure2-1-1.png",
        "caption": "Figure 2-1: Rigid Body Velocity",
        "texts": [
            " A rigid body has six degrees of freedom -- three translational and three rotational -- so the location, displacement, or vclocity of a rigid body can be described complctely by six numbers: three describing the translational aspect and threc the rotational aspect. Conventionally, the velocity of a rigid body would be described by a linear and an angular velocity vector. The question is, how should they be assembled to form a spatial vector? The instantaneous velocity of a rigid body may be described by the linear velocity v p of some point Pin the rigid body and its angular velocity w (see Figure 2-1). w applies to the body as a whole, and is independent of the choice of P, but v p applies only to the point P. The instantaneous velocity v Q of any other point Q in the rigid body can be expressed in terms of w and vpas -+ vQ=vp+QPXw. 15 The instantaneous velocity of the rigid body is described completely by the pair of vectors ( w, v p) given the point P; and the body may be considered to be rotating with angular velocity w about an axis passing through Pwhilst simultaneously translating with linear velocity v p "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000995_revmodphys.79.643-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000995_revmodphys.79.643-Figure8-1.png",
        "caption": "FIG. 8. Coordinates for analyzing the shape of the d-cone after Cerda and Mahadevan 2005a . The unit vector u\u0302 s points from the vertex to a point on the unit circle at a distance s from the center of the buckled region. The Cartesian basis vectors e\u03021 , e\u03022 , e\u03023 are shown. The polar angle and azimuthal angle and the polar basis vectors e\u0302 , e\u0302 , e\u03023 are indicated. The inset shows the region near s viewed along the radial line u\u0302. The tangent unit vector t\u0302 is shown on the main picture and the inset. The normal unit vector n\u0302 is shown in the inset only. The inset defines the angle between t\u0302 and e\u0302 , as well as the n\u0302 unit vector needed to form a basis with u\u0302 and e\u0302 .",
        "texts": [
            " We may see this by drawing a circle of unit radius in the sheet concentric with the forcing point, following the discussion of Cerda and Mahadevan 2005a . The angular position of a point on this circle is denoted s. When the d-cone is formed, the director lines are radial lines. The only curvature occurs in the azimuthal direction. We denote this azimuthal curvature by c s . Determining this c s thus determines the shape of the surface. To describe the constraint imposed by the container, we now consider the position of this circle in space in the deformed sheet as shown in Fig. 8. We place our coordinate origin at the vertex, so that the point at s has a position u\u0302 s where u\u0302 is a unit vector. Evidently u\u0302 s maps out a curve in space. We define a vertical, polar coordinate axis perpendicular to the plane of the constraining ring. We define the polar angle of u\u0302 s as s and denote the azimuthal angle by s . The ring itself Rev. Mod. Phys., Vol. 79, No. 2, April\u2013June 2007 has a polar angle 0. Since the surface must lie within the ring, s must be less than or equal to 0 for all s"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001372_diacare.5.3.245-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001372_diacare.5.3.245-Figure2-1.png",
        "caption": "FIG. 2. Overall optical system.",
        "texts": [
            " In the absence of glucose, a certain fraction of the FITC-dextran will be bound to immobilized Con A on the wall of the fiber, governed by the equilibrium binding constant for Eq. (2). Exposure of the transducer to a solution containing glucose results in the diffusive penetration of glucose into the chamber. Competition of glucose for Con A binding sites will displace FITC-dextran from the fiber wall and increase the free FITC-dextran concentration. The change in free dextran concentration is detected by measuring the fluorescence of the bulk solution by means of the optical fiber and associated equipment, as shown in Figure 2. DIABETES CARE, VOL. 5 NO. 3, MAY-JUNE 1982 245 OPTICfiL FIBER HOLLOW DIflLYSIS TIBER \\ IMMOBILIZED CON H \\ EXCITflTION \\ EMISSION FITC-DEXTRRN \u2022\u00b0 9 GLUCOSE 3 mm .3 .37 mm Principal advantages of the affinity sensor approach are (1) the response is determined by the competitive equilibrium between the metabolite (glucose) and the signal producing ligand (FITC-dextran). Thus kinetic factors, such as membrane permeability and membrane fouling, do not affect the magnitude of the response. Membrane fouling is a serious problem in enzyme and catalytic sensors",
            " The efficiency of immobilization of Con A can be readily TABLE 4 Hollow-fiber diffusive permeability at 37\u00b0C from saline solution Solute Urea Glucose Sucrose Vitamin B12 Inulin Dextran From ref. 5. MW 60 180 342 1355 5200 16000 Membrane permeability Cordis 6.7 2.3 1.2 0.26 0.037 0.0097 Enka B2-AH 8.1 2.9 1.9 0.53 0.089 0.0042 x 104 (cm/s) Amicon PMD 8.2 4.0 3.2 1.5 0.28 0.012 monitored by exposing the fiber to a solution of FITC-dextran and then visualizing the extent of FITC-dextran binding with a fluorescent microscope. O ptical system. A schematic diagram of the optical system is shown in Figure 2. In this design an optical fiber is used for transmitting the excitation light to the hollow fiber as well as returning the fluorescent emission to the light detector. A single optical fiber has the advantage of an efficient optical configuration; the field of view of the detector is the same as the field of illumination, and a short path length is adequate for the collection of fluorescent light. The feasibility of using fiberoptics for fluorescence measurements has been demonstrated by Vurek and Bowman12 and Goldstein et al"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure3.4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure3.4-1.png",
        "caption": "Fig. 3.4: Manipulator architectures for industrial robots.",
        "texts": [
            " But the revolute joints also allow a kinematic chain and then a mechanical design with small size, since a manipulator does not need a large frame link and additionally its structure can be of small size in a work-cell. In addition, it is possible to also obtain operation of other kinematic pairs with revolute joints only, when they are assembled in a proper way and sequence. For example, three revolute joints can obtain a spherical joint and depending on the assembling sequence they may give different practical spherical joints. Fundamentals of Mechanics of Robotic Manipulation 75 The most common design architectures for industrial manipulators are shown in Fig. 3.4 by using a suitable combination of revolute and prismatic joints. In particular, referring to Fig. 3.4 they can be characterized as follows: - Cartesian manipulators, which refer to a kinematic construction that makes use of the axes of a Cartesian reference frame through prismatic joints that give motions along those axes; they are characterized by a position workspace that is a bulk parallelepiped volume; - Cylindrical manipulators, which refer to a kinematic construction that makes use of the axes of a cylindrical reference frame through one revolute joint and two prismatic Chapter 3: Fundamentals of the Mechanics of Robots76 joints that give motion of an extremity point on a cylindrical surface",
            " They are characterized by a position workspace that is a bulk spherical volume; - Vertical articulated manipulators or so-called anthropomorphic manipulators, which refer to a kinematic human-like construction that makes use of revolute joints only; they are characterized by a position workspace that is a bulk solid of revolution volume; - Horizontal articulated manipulators or so-called SCARA (Selective Compliance Arm for Robot Assembly) manipulators, which refer to a kinematic construction that makes use of the two parallel revolute joints and one prismatic joint; they are characterized by a position workspace that is a bulk cylindrical volume. In Fig. 3.4 joint 4 indicates wrist capability as a common joint for any manipulation. The architectures in Fig. 3.4 are the most common in industrial robots, but many others are used, even by combining revolute and prismatic joints in alternative sequences. Another way for manipulator identification consists of using a denomination for the sequence of joint types in the manipulator chain, when revolute joints are indicated as R and prismatic joints as P. Thus, an anthropomorphic manipulator can also be identified as a spatial RRR or 3R manipulator, and a spherical telescopic manipulator as RRP. It is worthy of note that suitable combinations and proper design architectures with R and P joints can be equivalent to other types of kinematic pairs"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002212_j.msea.2017.03.105-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002212_j.msea.2017.03.105-Figure4-1.png",
        "caption": "Fig. 4. (A) Euler EBSD maps representative location on sample, (B) Euler EBSD maps of 5 locations along build direction, all images are at the same magnification and same location in the corresponding axis, (C) Average grain size and number of grains (location correlates with the EBSD map locations in A and B). (For interpretation of the references to color in this figure legend, the reader is referred to the web version of this article.)",
        "texts": [
            " A Tescan Lyra FIB-FESEM equipped with an EDAX Hikari Super EBSD camera and an Octane Elite Silicon Drift Detector performed the EBSD, electron dispersive X-ray spectroscopy (EDX) and fractography of the as-deposited and fractured samples, respectively. The EBSD scans were performed at the same magnification along the build direction over an area of approximately 40\u00d740 \u00b5m, with a step size of 0.1 \u00b5m for the maps in the individual layers and 0.08 \u00b5m for the maps at the interfaces between layers. The scans were performed at 20 kV and beam current of 5.5 nA. Images were taken in the three different directions, as shown in Fig. 4A, to analyze the grain size and geometry to understand the microstructure evolution in the build direction, long transverse direction, and short transverse direction in the specimens. The EBSD data was post-processed using the grain dilation algorithm in the EDAX software with settings of a minimum grain size of 2 pixels and requiring that a grain must contain multiple rows of pixels. The grain diameter was determined measuring the area of each grain using the OIM software, and then calculating the average grain diameter assuming circular grain morphology",
            " The as-received material grain size of the solid filler material provided by the manufacturer was reported to have an average grain size of approximately 30 \u00b5m. Fig. 3A and B shows a comparison between the microstructure of the asreceived material and the as-built AFS material, respectively. Fig. 3B displays significant grain refinement for the AFS processed as-built material when compared to that as-received microstructure. Quantification of grain size for the as-built IN625 was quantified with EBSD and is discussed subsequently. EBSD performed along the build direction (Fig. 4) provided a thorough understanding of the initial state microstructure of the tensile specimens. In Fig. 4A, the orientation of the EBSD scans in relation to a representative tensile specimen are shown since EBSD scans were performed after receiving the machined specimens from Aeroprobe. The EBSD inverse pole figure (IPF) maps in Fig. 4B correlate to the locations depicted in Fig. 4A. The EBSD identified fine equiaxed grain (0.5 \u00b5m weighted average, with a 1.3 \u00b5m max) structures with even finer grain structures (0.26 \u00b5m) forming at the layer interfaces. Fig. 4C correlates the average grain diameter and the number of grains identified by EBSD for the locations in Fig. 4A and B. Fig. 5 provides a higher detailed analysis of three locations identified in Fig. 4C. This analysis was performed for all locations, but since the behavior was similar, only three locations are shown here for brevity. Fig. 5A, D, and G are the magnified Euler EBSD IPF map shown previously in Fig. 4B. Grain boundary maps in Fig. 5B, E, and H show a dominant fraction of high angle grain boundaries. The EBSD also observed that the fraction of high angle boundary increases from 0.82, 0.84 and 0.91 for locations 1, 2, and 3, respectively. Also, the measured grain size decreases toward the interface region (Fig. 5B, E, and H, respectively). Grain reference orientation deviation (GROD) maps (Fig. 5C, F, and I) show that the amount of intragranular misorientation decreases from the center of a layer towards a layer-layer-interface, GROD maps will show low levels of intragranular misorientation (~0"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure5.22-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure5.22-1.png",
        "caption": "FIGURE 5.22. Screw motion of a rigid body.",
        "texts": [
            " When the screw is not central and u\u0302 is not passing through the origin, a screw motion to move p to p00 is denoted by p00 = (p\u2212 s) cos\u03c6+ (1\u2212 cos\u03c6) (u\u0302 \u00b7 (p\u2212 s)) u\u0302 +(u\u0302\u00d7 (p\u2212 s)) sin\u03c6+ s+ hu\u0302 (5.463) or p00 = GRB (p\u2212 s) + s+ hu\u0302 = GRB p+ s\u2212 GRB s+ hu\u0302 (5.464) and therefore, p00 = s\u030c(h, \u03c6, u\u0302, s)p = [T ]p (5.465) where [T ] = \u2219 GRB Gs\u2212 GRB Gs+ hu\u0302 0 1 \u00b8 = \u2219 GRB Gd 0 1 \u00b8 . (5.466) The vector Gs, called location vector, is the global position of the body frame before screw motion. The vectors p00 and p are global positions of a point P after and before screw, as shown in Figure 5.22. The screw axis is indicated by the unit vector u\u0302. Now a body point P moves from its first position to its second position P 0 by a rotation about u\u0302. Then it moves to P 00 by a translation h parallel to u\u0302. The initial position of P is pointed by p and its final position is pointed by p00. A screw motion is a four variable function s\u030c(h, \u03c6, u\u0302, s). A screw has a line of action u\u0302 at Gs, a twist \u03c6, and a translation h. The instantaneous screw axis was first used by Mozzi (1730 \u2212 1813) in 1763 although Chasles (1793\u2212 1880) is credited with this discovery"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000099_tac.1980.1102367-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000099_tac.1980.1102367-Figure3-1.png",
        "caption": "Fig. 3. (a) Position and orientation vectors of the hahd. @) Euler angles of orientation.",
        "texts": [
            " This paper presents an alternative technique for position control of the manipulator in tem of the position and orientation of the hand The method extends the idea of resolved-rate control except that now the desired acceleraton is also considered. In addition, the method adopts the concept of \"inverse problem\" except that the computation of J,(q), V,,fC(q,q), and &(q) is avoided and thus the satisfaction of (3)-(6) for the purpose of convergence is immaterial. n. POSITION AND ORIJ3NTATION OF THE HAND The hand of a manipulator is defined in Appendix I. The position and orientation of the hand are briefly sketched in Fig. 3(a), and can be represented by a 4X 4 matrix [4], [q where, as indicated in Fig. 3(a), p is the position vector of the hand; n, s, and a are the unit normal, unit slide, and unit approach vectors of the hand, respectively. The last three orthonormal vectors, which specify the orientation of the hand as shown in Figs. 1 and 3(a), are defined in the right-hand sense of 8 = a X n, where X denotes the cross product. Thus, only two of the three unit vectors need to be specified. For convenience, the orientation is defined in terms of Euler angles of rotation [SI, [9] with respect to the base coordinates (x,,,ym rd). By a translation corresponding to p , the origins of (x,,,ymzo) and (n,s,a) coincide with each other as shown in Fig. 3@). Also, (x,,yo,zd) will coincide with (n,s,a) if: 1) first, x, is rotated a radians about 20 so that Xn aligns no where no is a vector H( r ) = where Z is the identity matrix. From (IO) and (1 I), one obtains cosacosy-sinacosgsiny sinasinB m. POSITION AND ORIFBTATION ERRORS OF THE HAND The position error is defined as the difference between the Wed and the actual position of the hand, The orientation error of the hand is defined in a similar manner. Thus, the control of the manipulator in terms of the position and orientation of the hand is achieved by means of reducing their errors to zero",
            " Each of these points contains all the TRANSACTIONS ON AuToMATfC CONTROL VOL. AC-25, NO. 3, JUNE 1980 - TOP 473 - 4 0 -30 - P O - 1 0 0 I O 2 0 30 4 0 5 0 G O 7 0 I I I I I I 1 - r, \\ CONVEYOR 15 crn/sec + 55 5 4 +a Fi& 6. An example for stimulation. TABLE I INPUT DATA FOR A SBIULATION PROBLEM 1- 225 1 8 0 information of the hand matrix H(f ) at that p i n t s f = [ P : , P , ~ ~ , Q ~ , B ~ , Y ~ ] ~ in which the first three components specify the position, and the last three the orientation of the hand according to (10) and (12), and Fig. 3@). The input data for si are specified in Table I. The manipulator is initially at rest at point SI. Let ti be the time required for the hand to travel from point si to si+', and T be the intercept time, Le., the amount of time before the object wil reach point s4. Thus, T- t l + tz+ t,. Since the conveyor is moving with a velocity of 15 cm/s, then the hand must have the same constant velocity along the line segment between s4 and 2. This is accomplished by setting t4=lls5-s411/15=4 s. Other time intervals are specified as t,=4"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001346_j.actamat.2020.06.026-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001346_j.actamat.2020.06.026-Figure3-1.png",
        "caption": "Fig. 3. Schematic illustration of sampling from the SLMed parts, with the coordinate system showing the orientations.",
        "texts": [
            " The loss in Ti concentraion was resulted from the evaporation of Ti during SLM due to the igh surface areas of the nanoparticles. But, the contents of other olutes showing in Table 1 indicate much lower evaporation rates uring SLM compared with the actual composition of the 2024 Al lloy powder. It is considered that the similar evaporation rates, Al, u and Mn in particular, retained the composition. For anisotropy xamination, samples were sectioned along both longitudinal (on Z or XZ plane) and transverse (on XY plane) directions from the LMed parts as schematically depicted in Fig. 3 . In addition to use the optimized processing parameters as bovementioned, extra samples were built using altering the laser ower (20 0, 30 0, 325 and 350 W) and the scanning speed 150 mm/s and 1200 mm/s) to evaluate the densification of the I-2024 alloy under a broad SLM processing window. .3. Heat treatment To further improve the properties, some as-built (AB) TI-2024 amples were heat treated to T6 condition in accordance with the onventional wrought 2024 alloy [7] . First, the AB samples were olution treated (ST) at 495 \u00b0C for 1 h in a muffle furnace, followed y water-quench to room temperature"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure10.9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure10.9-1.png",
        "caption": "FIGURE 10.9. Angular orientation of a moving tire along the velocity vector v at a sideslip angle \u03b1 and a steer angle \u03b4.",
        "texts": [
            "97) Transforming the force system of each tire to the body coordinate frame B, located at the body mass center C, generates the total force system applied on the vehicle BF = X i Fxi \u0131\u0302+ X i Fyi j\u0302 (10.98) BM = X i Mzi k\u0302 + X i Bri \u00d7 BFwi (10.99) where, Bri is the position vector of the wheel number i. Bri = xi\u0131\u0302+ yij\u0302+ zik\u0302 (10.100) Expanding Equations (10.98) and (10.99) provides the total planar force system. BFx = X i Fxw cos \u03b4i \u2212 X i Fyw sin \u03b4i (10.101) BFy = X i Fyw cos \u03b4i + X i Fxw sin \u03b4i (10.102) BMz = X i Mzi + X i xiFyi \u2212 X i yiFxi (10.103) 600 10. Vehicle Planar Dynamics 10.3.2 Tire Lateral Force Figure 10.9(a) illustrates a tire, moving along the velocity vector v at a sideslip angle \u03b1. The tire is steered by the steer angle \u03b4. If the angle between the velocity vector v and the vehicle x-axis is shown by \u03b2, then \u03b1 = \u03b2 \u2212 \u03b4. (10.104) The lateral force, generated by a tire, is dependent on sideslip angle \u03b1 that is proportional to the sideslip for small \u03b1. Fy = \u2212C\u03b1 \u03b1 = \u2212C\u03b1 (\u03b2 \u2212 \u03b4) (10.105) Proof. A tire coordinate frame Bw(xw, yw) is attached to the tire at the center of tireprint as shown in Figure 10.9(a). The orientation of the tire frame is measured with respect to another coordinate frame, parallel to the vehicle frame B(x, y). The angle between the x and xw axes is the tire steer angle \u03b4, measured about the z-axis. The tire is moving along the tire velocity vector v. The angle between the xw-axis and v is the sideslip angle 10. Vehicle Planar Dynamics 601 \u03b1, and the angle between the body x-axis is the global sideslip angle \u03b2. The angles \u03b1, \u03b2, and \u03b4 in Figure 10.9(a) are positive. The Figure shows that \u03b1 = \u03b2 \u2212 \u03b4. (10.106) Practically, when a steered tire is moving forward, the relationship between the angles \u03b1, \u03b2, and \u03b4 are such that the velocity vector sits between the x and xw axes. A practical situation is shown in Figure 10.9(b). A steer angle will turn the heading of the tire by a \u03b4 angle. However, because of tire flexibility, the velocity vector of the tire is lazier than the heading and turns by a \u03b2 angle, where \u03b2 < \u03b4. So, a positive steer angle generates a negative sideslip angle. Analysis of Figure 10.9(b) and using the definition for positive direction of the angles, shows that under a practical situation we have the same relation (10.104). According to (3.131), the existence of a sideslip angle is sufficient to generate a lateral force Fy, which is proportional to \u03b1 when the angle is small. Fy = \u2212C\u03b1 \u03b1 (10.107) 10.3.3 Two-wheel Model and Body Force Components Figure 10.10 illustrates the forces in the xy-plane acting at the tireprints of a front-wheel-steering four-wheel vehicle. When we ignore the roll motion of the vehicle, the xy-plane remains parallel to the road\u2019s XY -plane, and we may use a two-wheel model for the vehicle"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001116_s0022-0728(80)80002-7-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001116_s0022-0728(80)80002-7-Figure4-1.png",
        "caption": "Fig. 4. Curves log k 1 = f(log p) for z = 0.5 when t = 50 s.",
        "texts": [
            " The curves were ana lyzed as fol lows. First a hor i zon ta l line {ordinate \u00d71) is drawn in the figures, e.g. in Fig. 2a and b. Its in tersec t ion wi th a given curve yields a value z~ of z cor responding to the value of kl of the curve (Fig. 2b). A p lo t is made of these co r respond ing values o f kl and zl (Fig. 3); such a graph allows an in te rpo la ted value o f k~ for a given value of z (e.g. 0.5) to be obta ined. These values o f kl are p lo t t ed as a func t ion o f p unde r the fo rm log kl = f ( l o g p ) as in Fig. 4. If k i t is no t t oo small, we obta in straight lines whose slope is 2; thus, we have k l = c t \u00d7 p2 = c t \u00d7 12/e 2, i.e. e2k~ = c t \u00d7 l 2, which is cons t an t since l is cons tan t . It can be conc luded f rom the preceding discussion tha t k 1, hence k2, represents on ly an appa ren t cons tan t , whose value is p ropo r t i ona l to the real value k'l and k~ of kl and k2. In the case o f d i f fus ion in a solid, a re la t ionship be tween k'l and D is o f the fo rm D = ctk2k'~, where ?~ is the dis tance be tween the react ing sites [ 7] "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure6.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure6.2-1.png",
        "caption": "FIGURE 6.2. A four-bar linkage.",
        "texts": [
            " It is an angle for a revolute joint, and a distance for a prismatic joint. A set of connecting links to do a function is called amechanism. A linkage is made by attaching, and fixing, one link of a mechanism to the ground. The fixed link is called the ground link. There are two types of linkages, 310 6. Applied Mechanisms closed loop or parallel, and open loop or serial. In vehicle subsystems we usually use closed-loop linkages. Open-loop linkages are used in robotic systems where an actuator controls the joint variable at each joint. A four-bar linkage is shown in Figure 6.2. Link number 1 is the ground link MN . The ground link is the base and used as a reference link. We measure all the variables with respect to the ground link. Link number 2 \u2261MA is usually the input link which is controlled by the input angle \u03b82. Link number 4 \u2261 NB is usually the output link with angular position \u03b84, and link number 3 \u2261 AB is the coupler link with angular position \u03b83 that connects the input and output links together. The angular position of the output and coupler links, \u03b84 and \u03b83, are functions of the links\u2019 length and the value of the input variable \u03b82"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure3.40-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure3.40-1.png",
        "caption": "Fig. 3.40: A model for stiffness evaluation of a PUMA-like robot: a) the mechanical design; b) a stiffness model with lumped spring parameters.",
        "texts": [
            " The stiffness of a robotic multibody system is given by the stiffness of the components of the system that is described by means of a suitable model of elastic response of those components. A main source of compliance for a manipulator can be considered the links, joints, and actuators with their mechanical transmissions. A suitable model can refer to lumped stiffness parameters for each component that can be identified by means of suitable linear and torsion springs. An example of such a model is shown in Fig. 3.40 in which the main compliance source of a PUMA-like robot are identified and represented as linear springs for translation deformations and as torsion springs for angular deformations. A linear spring will take into account mainly axial deformation of a link body and the corresponding spring coefficient k is the lumped parameter that describes the elastic characteristic of the body. Transversal deformations can be considered by using a cantilever model to compute the compliant displacement of an extremity point that can be modeled through a suitable torsion spring. Thus, a torsion spring is identified to consider both angular deformations and transversal compliant displacements of a link body of a manipulator. A model for stiffness evaluation will be defined by analyzing each manipulator component to identify the spring coefficients and by determining the manipulator configuration with kinematic parameters affecting the compliant response of a manipulator. In the example of Fig. 3.40 a PUMA-like robot is analyzed for stiffness evaluation by Fundamentals of Mechanics of Robotic Manipulation 171 identifying the lumped springs in a kinematic diagram. Mass distribution can be considered by determining the center of mass for links and actuators. Thus, a mechanical model for stiffness evaluation will consist of: - a kinematic diagram; - an identification of lumped springs and their relative locations in the kinematic diagram; - a distribution of masses with indication of the positions of the concentrated masses in the link kinematic diagram"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001017_robot.1986.1087552-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001017_robot.1986.1087552-Figure8-1.png",
        "caption": "Figure 8. Cut t ing a loop a t j o i n t ( j )",
        "texts": [],
        "surrounding_texts": [
            "4.2. Open-loop Robots\nThe system is composed of n+l l inks , l ink (0) is the f ixed base , wh i l e l i nk (n ) is the te rmina l l ink . J o i n t (i) connects l ink ( i - I ) and l ink (i) . L e t :\nR. the f i x e d f rame wi th respec t to l ink (i)\nZ . t h e a x i s o f j o i n t (i)\nx. w i l l be def ined on the common perpendicular of\n1\n-1\n-1 Z . and 2. (Fig. 5 ) . -1 -1+ 1 The fol lowing parameters are required t o def ine the frame (i) w i t h r e s p e c t t o frame (i-1) : (Fig. 5)\nai angle between Z and 2. about zi-l d.. d is tance between 0 and Z i- 1 -i r . d i s t a n c e between 0 and &ii-l i 8. angle between a d X . about gi The v a r i a b l e o f j o i n t (i) denoted by q. is 8 i f (i) is ro ta t iona l and r i f (i) is prismatic. Hence -i- 1 -1 -1\ni i\nqi = 8,(1 - a . ) + ri ai where a , = B i f j o i n t (i) is ro t a t iona l and U.= 1 i f jo i .n t (i) is pr i smat ic .\nThe t ransformation matr ix i - l ~ . is equal t o :\ni-lT. = Rot(X,ai)Trans(X,d.)Rot(Z,8.)Trans(Z,ri)\ncosBi; -sin\u20ac$' I @'I di\nc o s a . s i n 8 . ~ c o s a . c o s e . l -sins. ,-r s i n a 1 1 1 1 1 1 i i - ( 7 )\ns ina . s in8 ' s i na . cose .1 cosa. ' r cosu 1 i l 1 1, 1 1 i i - - - - - + - - - - ----I----\nI -\n0 ' @ I @ ; 1 i\nI t i s to be no ted tha t :\n- R can be defined always such that R I R1 when\n51 = a which means t h a t : a1 = d l = q1 =@,where q. = 8 . 0. + r . ( 1 - a . ) - X can be taken along X i f a= 0, which means\nthat S = s. The proposed notation i s similar t o t h a t o f D-H. I n f a c t it is t o b e n o t e d t h a t ( 8 . , r .\n-\n1 1 1 1\n-n -n- 1\nn\n1 r ' a i + l r d i + l ) of the\nmodi f ied nota t ion a re respec t ive ly (Oi , r i , a . ,d i ) o f D-K . It has been no ted tha t t h i s de f in i t i on o f t he frame R . f i x e d w i t h l i n k i s u c h t h a t Z . is a long the ax i s 0% j o i n t (i) , l e a d s a l s o t o sim&fy t h e Zynamic model o f t he robo t [9 ] .\nThe geometric model of the robo t w i l l be obtained by the success ive mul t ip l i ca t ions o f t he t r ans fo rma t ion mat r ices : O T ~ = O T ~ ' T ~ ... n- 1 Tn ( 8 ) 4.3. Tree-Structure Robots\nThe system t o b e c o n s i d e r e d i n t h i s c a s e is composed of n+l l inks , n j o in t s and m end ef fec tors .\nThe l i n k s and the joints w i l l be numbered as fo l lows (Fig. 6 )\nthe base w i l l be considered as l i n k @ t h e numbers of l i n k s a n d j o i n t s a r e i n c r e a s i n g a t each branch when t ravers ing f rom the base t o an end effector l i n k (i) is a r t i c u l a t e d on j o i n t ( i ) , i .e. j o i n t (i) connects the l ink ( a ( i ) ) and l i nk ( i) , where a ( i ) is the number of the l i n k a n t e c e d e n t t o l i n k (i) when coming from the base frame (i) is de f ined f ixed wi th r e spec t t o l i n k (i) , and Z is the a x i s o f j o i n t (i) .\n-i I n t h e case of a l i nk wi th two j o i n t s t h e frame a t the e n d o f t h i s l i n k , s a y j , w i l l b e d e f i n e d w i t h r e s p e c t t o the frame a t the begin ing of the l i n k , say i = a ( j ) , e x a c t l y a s i n t h e case of open-loop robot descr ibed previous ly , i.e by the a id o f 4 parameters ( a . d O.,r ). X . is t h e common perpendicu-\nl a r on Z and, Z . I n the case o f l i nks w i t h more than two j o i n t s (Fig. 7) , w e def ine the d i f fe ren t f rames as fol lows\n- f i n d t h e common p e r p e n d i c u l a r s t o Z_. and each of the succeeding axis ,on the same l i n k ti), Z j where i = a ( j ) ( j = k. ,R,m ,... ). - l e t cne of these common perpendiculars be the Zi a x i s , it is p r e f e r e d t o t a k e X . as that corresponding t o t h e common perpendicular of the joint on which is a r t i cu la t ed the longes t b ranch , say k ,\nJ ' j ' 1. j T.\ni-j\n-1\n1177",
            "- t he o the r pe rpend icu la r s w i l l be denoted by X : X!', ... Thus some other auxi l ia r f rames R ! ( 0 ! X ' Y !\nZ , ) w i l l be de f ined f ixed wi th r e spec t t o l i nk (i) . -1 The mat r ix T w i l l be def ined by the use of 4 parameters ( %,%,rk,!k) as i n Eq. ( 7 ) . The other succeeding frames R . ( 3 = k,m, -. w i l l be def ined in genera l by the'following parameters (Fig. 7 ) :\n-I' f -1 J'...l t\ni k\ny . angle between X . and X! about Z , 7 -1 --I -1 E , d i s t a n c e between 0 , and 01 7 a . angle between Z , and Z . about X! 7 -1 -3 d , d i s t a n c e between O ! and 2 , 7 8, angle between X! and. X . about 2 , -3\n7 -1 -3 -3 r . distance between 0 and X!\n3 I n t h i s c a s e T . w i l l be def ined by : i\n-1\ni j\nT . = T. , T . 7 1 3\ni i ' I\nwhere\nIT.$ = Rot(Z,y,) Trans(Z,E , I 3 3\ncosy , - s h y . # (d 3 1 = s i n y . cosy , # # 3 7\n# # I E .\n# er 6 1\n- - - - - - - - - 3\nand T , is i d e n t i c a l t o T . given i.n (Eq. 7 ) . The\nparameters of T . w i l l have the subscr ipt ( j ) .\nThe matr ix T i s the un i ty ma t r ix when both E\nand y , are equa l t o ze ro .\nI t i s to be noted tha t the f rames Ri, R'.', D. a r e\nused on ly to de f ine the new parameters when more than two jo in ts a re connec ted on the same l ink , bu t w i l l n o t be used i n developpiny the mathematical models where we d i r e c t l y u s e i T\nThe descr ip t ion of an end ef fec tor i n t h e f i x e d frame R w i l l be obtained by the success ive mul t ip l i - ca t ion g f t he ma t r i ces l ead ing from t h e b a s e t o t h a t end e f f ec to r .\ni ' i- 1\ni ' 1\ni ' i ' 3\n1\ni\nj \"\n4.4. Closed-Loop Robots\nThe system i n t h i s c a s e i s composed of n+l l inks, m end e f f ec to r s , and k j o i n t s . Thus t h e number of closed-loop is e q u a l t o b=R-n.\nTo ge t the t ransformat ion mat r ices , w e def ine at f i r s t a t r ee - s t ruc tu re equ iva len t t o the sys t em by cutting each closed-loop through one of its j o i n t s . Many methods are a v a i l a b l e t o show where the cut t ing process may take p lace [ 5 p 10,111. For the purpose of t h i s pape r t he cu t t i ng p rocess can be assumed a r b i - t r a r i l y .\nOnce the equ iva len t t r ee - s t ruc tu re is def ined the l i n k s and j o i n t s w i l l be numbered a s d e s c r i b e d i n s e c t i o n 4 . 3 , wh i l e t he cu t j o in t s w i l l be numbered from ( n + l ) t o R .\nThe l i nks f ixed f r ames o f t he equ iva len t t r ee s t ruc - t u r e w i l l be def ined as i n s e c t i o n 4.3. Addit ional 2b frames w i l l be def ined f ixed on t h e l i n k s which are connected by t h e c u t j o i n t s . I f ( j ) i s the numbe r o f an opened jo in t , t he co r re spond ing add i t iona l frames defined on t h e l i n k s on bo th s ides o f t ha t j o i n t w i l l be denoted by R . and R , ( t h e i r 5 a x i s i s a l o n g t h e j o i n t ( j ) a x i s ) ( F i g . 8 ) . 3 I +b\nI f t h e a d d i t i o n a l frame R . i s f i r e d on the l i nk ( i ) , and R . i s f ixed on l ink ' (k) , then R . and R . W i l l\nbe def ined wi th respec t o R . and a s usual , and the parameters w i l l g e t t h e s u b s c r l p t ( J ) and ( j+b) respect ively. These parameters are constants , furthe rmore 0 , , \u20acI . can always be tak.en equal\nt o z e r o by taking X . = X . and Xj+b = X\nWe can define the transformation matrix T . from\none s ide of a c u t j o i n t t o t h e o t h e r s i d e , by t h e use of two parameters , one bee ing the jo in t va r i ab le q . . These parameters will be denoted by :\n3 p , = angle between X . and Ej+b about 2 .\n3 -1 -1 T , = distance between 0 , and 0 . along Z .\nThe t ransformation matr ix ' T , i s def ined by :\nI +b I I +b\n1 R k <\n3 3+bfr j ' r j+b\n-7 -1 -k' . 3\nI +b\n3 3 I +b -1\n3 +b\nc o s p . - s i n p . @ # 3 1 = s i n p . c o s p , 0 # # 7\n0 # I ?\n1\n- - _ _ j",
            "I f j o i n t (i) is r o t a t i o n a l p. = q . and i f (i) i s\np r i s m a t i c T~ = qj .\nThe r e l a t i o n s between t h e j o i n t s v a r i a b l e s o f a closed-loop w i l l be obtained by expressing that the product of all the t ransformation matr ices around the loop i s equal t o uni ty .\nThe transformation matrix between any two frames can be obta ined by the mul t ip l ica t ion of a l l the t ransformation matrices connecting these frames.\n3 1\n5. EXAMPLES\nI n t h i s s e c t i o n w e give the geometr ic parameters corresponding t o a serial robot (Stanford) and a c losed-loop robot (Hitachi HPR) . 5.1. The geometric parameters of the Stanford manipu la tor , F ig . 9\nT h i s r o b o t i s a six degree of freedom robot, it has an open-loop structure. Applying the technique devel o p e d i n s e c t i o n 4.1, we ge t the parameters Of the robot as fol lows :\ni 0 r a d 8\n1 0\nd 0 e 5 0 -90 5 0 e b d pr o 4\nO l d @ d 2\n1 3 1 P( 0 3 9 0 0 R L ~ 8 2 0 -90\n6 0 d 66 # +90 ~\n5.2. The geometric parameters of Hitachi HPR Robot\nThis robot conta ins a s ing le c losed loop ( f ig . 10) apply ing the new n o t a t i o n , the parameters Of the robot are obtained as fol lows :\"The loop has been opened a t j o i n t 8 between link 4 and l i n k 5. P s is qs and 1 8 is e q u a l t o 0.\"\nStanford Manipulator\n6. CONCLUSION\nThis paper presents a new n o t a t i o n which can be used to descr ibe the open-loop robots and the closed-loop robots wi th a minimum of parameters and without amb i g u i t i e s o r d i f f i c u l t i e s . The method is der ived from the popular D-H method.\nThe paper shows how to ge t t he t r ans fo rma t ion matrices of any robot ic mechanism by the use of the proposed method. A FORTRAN program has been developed t o der ive the symbolic t ransformation matrices automatically between any two frames of the system.\nREFERENCES\n[ I ] P. C o i f f e t , Robot Technology, vol.1 :\"Modeling and Control\\: Prentice-Hall, inc. , Englewood CLIFFS, N J , 1983. [ 2 ] J. Denavit, R.S. Hartenberg,('A Kinematic notat i on fo r l ower -pa i r mechanism based on matrices: Journal Of Applied Mechanism, vo1.22, pp. 215-221, 1955. [3] R. Pau1,\"Robot manipulators mathematics, programming and contro2; MIT Press 1981,, [4] P.N. Sheth, J.J. Uicker,'A generalized symbolic n o t a t i o n f o r mechanisms;' Journal of Engineer ing for Indus t ry , Transac t ions of the ASME, Vo1.93, 1971,\n[5] S. Megahed, M. Renaud, Dynamic modelling of robots manipulators containing closed kinematic chains': Advanced Software in Robotics , Liege, Belgium, May 4-6, 1983. [6] D.L. Pieper,\"The kinematics of manipulators under computer control'; Ph. D. S tanford Univers i ty , Stanford, October 1968. [7] B. Gorla, M. Renaud, Modeles des Robots Mani- (C p u l a t e u r s , a p p l i c a t i o n 5 l e u r commande, Cepadues ed i t ions , France , 1984. 181 B. :?th,\"Overview on advanced robot ics : manip u l a t i o n , ' 8 5 ICAR, Tokyo, Japan, 1985, pp. 559-570. [9] W. Khal i l , M. Gaut ier ,\"On the der ivat ion of the dynamic models of robots': '85 ICAR, Tokyo, Japan, 1985, pp. 243-250.\n[ l o ] J. Wittenburg,\"Dynamics of systems of r i g i d bodies: G.B. Teubner , S tu t tgar t , 1977. [ 1 1 ] D.A. Smith, M.A. Chace, A.C. Rubens ,\"The automatic genera t ion of a mathematical model for nachinery systems: Journal of Engineering for industry,\npp. 102-112. (1\nMay 1973, pp. 629-635. x x 6A 7\nFig.10 The HPR Robot\nY 1179"
        ]
    },
    {
        "image_filename": "designv10_0_0001803_icra.2015.7139759-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001803_icra.2015.7139759-Figure4-1.png",
        "caption": "Fig. 4: CAD model of the preliminary prototype of the hexarotor with tilted propellers. It is composed of: (1) Micro controller, (2) Brushless controller, (3) Lander, (4) Propeller motor, (5) Tilting set-up.",
        "texts": [
            " Now assuming that Rd(t) \u2208 C\u03043 and \u03c9d = [RT d R\u0307d]\u2228, where [\u00b7]\u2228 represents the inverse (vee) operator from so(3) to R3, the attitude tracking error eR \u2208 R3 is defined similarly to [20] as eR = 1 2 [RT d WRB \u2212WRT BRd]\u2228, (20) and the tracking error of the angular velocity e\u03c9 \u2208 R3 is given by e\u03c9 = \u03c9B \u2212WRT BRd\u03c9d. (21) In order to obtain an asymptotic convergence to 0 of the rotational error eR one can choose the following controller vR = \u03c9\u0307d \u2212KR1 e\u03c9 \u2212KR2 eR \u2212KR3 \u222b t t0 eR (22) where the diagonal positive definite gain matrixes KR1 , KR2 , KR3 define Hurwitz polynomials also in this case. In this section we present the design of a preliminary prototype obtained instantiating the general model introduced in Section II in a more particular case. A CAD of the prototype is shown in Fig. 4. First of all, to reduce the complexity and for the sake of symmetry, we have chosen \u03bbi = (i \u2212 1)\u03c03 and Lxi = 0.4 m \u2200 i = 1 . . . 6. With this choice, the origin OPi of each propeller frame is equally spaced with 60\u25e6 between each other from the center of the body frame OB to have a symmetric configuration in normal hovering position. Furthermore, as shown in Fig. 5 each propeller is mounted in an arc frame which is free to rotate in XPi and Y Pi , so that the tilt angle of \u03b1i and \u03b2i can be fixed as desired"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003457_j.bioelechem.2004.12.002-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003457_j.bioelechem.2004.12.002-Figure2-1.png",
        "caption": "Fig. 2. Cyclic voltammograms of (a) bare Au, (b) Au/dithiol, (c) Au/dithiol/ au, (d) Au/dithiol/au/cystamine, and (e) Au/dithiol/au/cystamine/GOx electrodes in 0.1 M KCl solution containing 5 mM Fe(CN)6 4 /3 . Scan rate: 50 mV/s.",
        "texts": [
            "8) containing of 20 AM IO4 oxidized GOx for 1 h at 4 8C. The assembly process is shown in Scheme 1. Au cystamine GOx SH SH S S S S IO4 --Oxidized GOx S S S S dithiol C= OO=C H H H HC C O O Scheme 1. Stepwise assembly of dithiol, gold nanoparticles Scheme 1 shows the stepwise fabrication process of biosensor, which was followed by cyclic voltammetry using a ferrocyanide as redox probe. Cyclic voltammetry of ferrocyanide is a valuable and convenient tool to monitor the barrier of the modified electrode. Fig. 2 SH SH SH SH S S S S NH2 NH2 NH2 NH2 N=CH N=CH N=CH N=CH NH 2 NH 2 NH2 NH 2 gold nanoparticles NH2 GOx GOx GOx GOx , cystamine, IO4 oxidized GOx on a gold electrode. shows CV response of modified electrodes in 0.1 M KCl aqueous solution containing 5 mM Fe(CN)6 4 /3 at a scan rate of 50 mV s 1. At bare gold electrode, a reversible electrochemical response for Fe(CN)6 4 /3 with peak potential separation (DEp, the potential difference between the oxidation peak potential and reduction peak potential, which is inversely proportional to the electron transfer rate [21]) of 93 mV is observed (Fig. 2a). After the electrode was dipped in the dithiol solution (marked as Au/dithiol), the CV curve changes drastically from an electrochemically reversible shape to a capacitive shape (Fig. 2b) due to the sluggish electron-transfer kinetics through the dithiol film [22]. However, when gold nanoparticles were selfassembled on the dithiol film (marked as Au/dithiol/au), a quasi-reversible CV with a DEp of 130 mV was obtained (Fig. 2c), which further restored to 110 mV (Fig. 2d) after cystamines were chemisorbed onto the gold nanoparticles (marked as Au/MPTS/au/cystamine). Obviously, the gold nanoparticles on the dithiol film have a marked influence on the interface property of the modified electrode and improve the electron transfer between Fe(CN)6 4 /3 and the electrode. It is worth mentioning that the gold nanoparticles are highly negative charged species as a result of the absorption of citrate in the fabrication process [23], and they repulse the negatively charged Fe(CN)6 4 /3 and interfere with the diffusion of ferricyanide toward the electrode surface. Self-assembly of cystamine monolayer on gold nanoparticles replaces most of negative charges and diminishes the repulse of the electrode interface to the redox probe, Fe(CN)6 4 /3 . As a result, the electron transfer of the modified electrode was further improved. The CV obtained from Au/dithiol/au/cystamine electrode (Fig. 2d) almost coincided with that obtained from bare gold electrode (Fig. 2a) and its peak current is proportional to the square root of scan rate (data not shown here), which imply that these nanoparticles not only provided the necessary conduction pathways in promoting the electron transfer between the analyte and the electrode surface, but also could act like nano-scaled electrode for the subsequent assembly of biomolecules. Cystamine monolayers have widely been used for the covalent attachment of different enzymes and, in particular, glucose oxidase [24]. It is well known that the reaction between amine group and aldehyde group proceeds easily in moderate condition with no contamination and significant deactivation of the enzyme. Based on this reaction, periodate oxidized GOx was covalently attached to the aminated gold nanoparticle surface (marked as Au/dithiol/au/cystamine/GOx). CV of ferrocyanide confirmed this process. Greatly diminished ferricyanide response and enlarged DEp (Fig. 2e) indicate that an insulating layer of GOx is introduced. The assembly process of the gold electrode was also monitored by EIS experiments. EIS is an effective method for probing the features of surface-modified electrodes [25\u2013 27]. Fig. 3 shows the impedance features, presented as Nyquist plots (ZW vs. ZV), of electrodes at different modification steps. Significant differences in the impedance spectra were observed during stepwise modification. The bare Au electrode exhibited an almost straight line (Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003702_s0043-1648(96)07467-4-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003702_s0043-1648(96)07467-4-Figure6-1.png",
        "caption": "Fig. 6. Points on the flanks represented in a coordinate system.",
        "texts": [
            " The data used in the simulations are summarized in Journal: WEA (Wear) Article: 7467 val. The interaction between the teeth are assumed to be quasi static. The Points on the flanks are aligned in such a way that they, when in mesh, can be considered to be on a straight line perpendicular to the line of action, according to Fig. 7. The first point is at the base diameter, the last point at the tip diameter and the points in between on an involute curve. The points can in a coordinate system according to Fig. 6 be defined as: d b \u00df s (cos b qb sin b ) pi i i i2 d b c s (sin b yb cos b ) pi i i i2 b s(see Fig. 6)i The flanks are then mirrored and rotated to fit the desired configuration. The simulatedwear of the pinion and the gear are presented in the Fig. 8(a) and (b). The wear depths are varying over the teeth flanks with the maximum wear at the root of the pinion and gear respectively. Throughout the wear process, the change between one and two tooth pairs is significantly noticeable in the wear curves, although the effect slowly vanishes away as the wear depth increases. The explanation of this is that the load transmitted by each tooth pair is influenced by the wear according to Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure1.17-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure1.17-1.png",
        "caption": "Fig. 1.17. Steering wheel mechanism",
        "texts": [
            "2 General Features of Robotic Mechanism~ and Their Classification 13 second type two rotations, and pairs of the third type only one relative rotational motion. As well as the pairs, in which relative motions of members are mutually independent, there are pairs with interconnected motion. The simplest example is the screw-nut kinematic pair, in which the linear and rotational motions are linearly dependent [4]. In kinematic schemes, symbolic presentation for various kinematic pairs is adopted (Fig. 1.18). Kinematic pairs form kinematic chains. One example of spatial mechanism of the steering wheel is presented in Fig. 1.17. The mechanism consists offour kinematic pairs, two each entering into two kinematic pairs only. One part of mechanism (in the figure, member 1) is fixed. This is usually designated the basis. The kinematic chain presented in Fig. 1.17 represents a simple closed chain. In a closed chain, each member enters in two kinematic pairs [4]. Another example of spatial mechanism is presented in Fig. 1.19. This is a mechanism of a five degrees-of-freedom manipulator. The chain consists of six kinematic pairs; only the last member, differing from the previous example, enters into one kinematic pair. This is an example of a simple kinematic chain. In the examples considered, each mechanism member enters into one kinematic pair only. However, mechanisms are known in which one member can belong to several pairs"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002584_tcst.2020.3001117-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002584_tcst.2020.3001117-Figure1-1.png",
        "caption": "Fig. 1. Quadrotor with body-fixed reference system and moment arm definitions.",
        "texts": [
            " This poses an important challenge in control design. In addition, accurate tracking of a reference trajectory with fast-changing acceleration requires considering its higher order time derivatives, i.e., jerk and snap. In contrast, control design for rotary-wing, vertical take-off and landing (VTOL) aircraft at low speeds typically neglects both aerodynamics and higher order derivatives. In this article, we propose a novel control design for accurate tracking of aggressive trajectories using a quadcopter aircraft, such as the one shown in Fig. 1. The proposed controller generates feedforward control inputs based on differential flatness of the quadcopter dynamics and uses incremental nonlinear dynamic inversion (INDI) to handle external disturbances, such as aerodynamic drag. Nonlinear dynamic inversion (NDI), also called feedback linearization, enables the use of a linear control law by transforming the nonlinear dynamics into a linear input\u2013output map [2]\u2013[4]. Although variants of NDI were quickly developed for flight control [5]\u2013[9], it is well known that exact dynamic inversion inherently suffers from lack of robustness [10]",
            " Analysis in Section IV illustrates the robustness of INDI and the effect of the feedforward control inputs through response modeling. Finally, we give experimental results from real-life flights in Section V. In this section, we describe the quadrotor dynamics model and its differential flatness property. Specifically, we show how the control system utilizes this property to track the reference trajectory jerk and snap through feedforward angular rate and angular acceleration inputs. We consider a six-degree-of-freedom (DOF) quadrotor, as shown in Fig. 1. The unit vectors shown in the figure are the basis of the body-fixed reference frame and form the rotation matrix R = [bx by bz] \u2208 SO(3), which gives the transformation from the body-fixed reference frame to the inertial reference frame. The basis of the north-east-down (NED) inertial reference frame consists of the columns of the identity matrix [ix iy iz]. The vehicle translational dynamics are given by x\u0307 = v (1) v\u0307 = giz + \u03c4bz + m\u22121fext (2) where x and v are the position and velocity in the inertial reference frame, respectively",
            " The third term of (4) accounts for the conservation of angular momentum. Each propeller axis is assumed to be aligned perfectly with the bz-axis so that all motor speeds are described by the four-element vector \u03c9 > 0. The total thrust T and control moment vector in body-reference frame \u03bc are given by[ \u03bc T ] = G1\u03c9 \u25e62 + G2\u03c9\u0307 (5) where \u25e6 indicates the Hadamard power G1 = \u23a1 \u23a2\u23a2\u23a3 lyk\u03c4 \u2212lyk\u03c4 \u2212lyk\u03c4 lyk\u03c4 lxk\u03c4 lx k\u03c4 \u2212lx k\u03c4 \u2212lxk\u03c4 \u2212k\u03bcz k\u03bcz \u2212k\u03bcz k\u03bcz \u2212k\u03c4 \u2212k\u03c4 \u2212k\u03c4 \u2212k\u03c4 \u23a4 \u23a5\u23a5\u23a6 (6) with lx and ly the moment arms shown in Fig. 1, k\u03c4 the propeller thrust coefficient, and k\u03bcz the propeller torque coefficient, and G2 = \u23a1 \u23a2\u23a2\u23a3 0 0 0 0 0 0 0 0 \u2212Jrz Jrz \u2212Jrz Jrz 0 0 0 0 \u23a4 \u23a5\u23a5\u23a6 (7) with Jrz the rotor and propeller moment of inertia. The second term in (5) represents the control torque directly due to motor torques. Due to their relatively small moment of inertia, the contribution of the motors to the total vehicle angular momentum may be neglected. The controller aims to accurately track the reference trajectory defined by the following function: \u03c3 ref(t) = [xref(t) T \u03c5ref(t)]T (8) which consists of four differentially flat outputs, i",
            " Furthermore, we examine the effect of the feedforward inputs based on the reference trajectory jerk and snap. We establish the independence of any model-based drag estimate by attaching a drag-inducing cardboard plate that more than triples the frontal area of the vehicle. Robustness against external disturbance forces is further displayed by pulling on a string attached to the quadcopter in hover. Finally, we compare the proposed nonlinear INDI angular acceleration control to its linearized counterpart. Experiments were performed in an indoor flight room using the quadcopter shown in Fig. 1. The quadrotor body is machined out of carbon fiber composite with balsa wood core. The propulsion system consists of T-Motor F35A ESCs and F40 Pro II Kv 2400 KV motors with Gemfan Hulkie 5055 propellers. Adjacent motors are mounted 18 cm apart. The quadcopter is powered by a single 4S LiPo battery. Its total flying mass is 609 g. Control computations are performed at 2000 Hz using an onboard STM32H7 400-MHz microcontroller running custom firmware. On this platform, the total computation time of a control update at 32-bit floating-point precision is 16 \u03bcs"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure6.32-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure6.32-1.png",
        "caption": "FIGURE 6.32. Instant center of rotation Q, for planar motion of a rigid body.",
        "texts": [
            " Then, the velocity of any point P of a rigid body is a superposition of the velocity Gd\u0307B of another arbitrary point o and the angular velocity G\u03c9B \u00d7 G BrP of the points P around o. The relative velocity vector GvP/o is perpendicular to the relative position vector G BrP . Employing the same concept we can say that the velocity of points P and o with respect to another point Q are perpendicular to G BrP/Q and G Bro/Q respectively. We may search for a point Q, as the instantaneous center of rotation, at which the velocity is zero. Points o, P , and Q are shown in Figure 6.32. 354 6. Applied Mechanisms Assuming a position vector Gro/Q for the instant center point Q, let\u2019s define Gro/Q = aQ Gd\u0307B + bQ G\u03c9B \u00d7 Gd\u0307B (6.225) then, following (6.223), the velocity of point Q can be expressed by GvQ = Gd\u0307B + G\u03c9B \u00d7 GrQ/o = Gd\u0307B \u2212 G\u03c9B \u00d7 Gro/Q (6.226) = Gd\u0307B \u2212 G\u03c9B \u00d7 \u00b3 aQ Gd\u0307B + bQ G\u03c9B \u00d7 Gd\u0307B \u00b4 = Gd\u0307B \u2212 aQ G\u03c9B \u00d7 Gd\u0307B \u2212 bQ G\u03c9B \u00d7 \u00b3 G\u03c9B \u00d7 Gd\u0307B \u00b4 = 0. (6.227) Now, using the following equations G\u03c9B = \u03c9K\u0302 (6.228) G\u03c9B \u00d7 \u00b3 G\u03c9B \u00d7 Gd\u0307B \u00b4 = \u00b3 G\u03c9B \u00b7 Gd\u0307B \u00b4 G\u03c9B \u2212 \u03c92 Gd\u0307B (6.229) G\u03c9B \u00b7 Gd\u0307B = 0 (6",
            "231) Because Gd\u0307B and G\u03c9B \u00d7 Gd\u0307B must be perpendicular, Equation (6.231) 6. Applied Mechanisms 355 provides 1 + bQ \u03c92 = 0 (6.232) aQ = 0 (6.233) and therefore, GrQ/o = 1 \u03c92 \u00b3 G\u03c9B \u00d7 Gd\u0307B \u00b4 . (6.234) Example 248 F Instantaneous center of acceleration. For planar motions of rigid bodies, it is possible to find a body point with zero acceleration. Such a point may be called the instantaneous center of acceleration. When a rigid body is in a planar motion, we can be express the acceleration of a body point P , such as shown in Figure 6.32, as GaP = Gd\u0308B + G\u03b1B \u00d7 \u00a1 GrP \u2212 GdB \u00a2 +G\u03c9B \u00d7 \u00a1 G\u03c9B \u00d7 \u00a1 GrP \u2212 GdB \u00a2\u00a2 = Gd\u0308B + G\u03b1B \u00d7 G BrP + G\u03c9B \u00d7 \u00a1 G\u03c9B \u00d7 G BrP \u00a2 . (6.235) The term G\u03c9B \u00d7 \u00a1 G\u03c9B \u00d7 G BrP \u00a2 is the centripetal acceleration, and the term G\u03b1B \u00d7 G BrP is the tangential acceleration and is perpendicular to G BrP . Because the motion is planar, the angular velocity vector is always in parallel to k\u0302 and K\u0302 unit vectors. G\u03c9B = \u03c9K\u0302 (6.236) G\u03b1B = \u03b1K\u0302 (6.237) Therefore, the velocity GvP and acceleration GaP can be simplified to GaP = Gd\u0308B + G\u03b1B \u00d7 G BrP \u2212 \u03c92 GBrP "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000571_elan.1140070104-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000571_elan.1140070104-Figure5-1.png",
        "caption": "Fig. 5. Reaction sequence for an oxidation reaction catalyzed by a hydrogen peroxide producing enzyme (glucose oxidase, GOD) coupled to electrocatalytic reduction of hydrogen peroxide catalyzed by a peroxidase (HRP).",
        "texts": [
            " Peroxidasemodified carbon paste electrodes have therefore been used for the determination of hydrogen peroxide and organic peroxides with the use of soluble or immobilized electron donors acting as mediators. Examples of such mediators are o-phenylenediamine [113, 1781, ferrocyanide [92, 1561, and ferrocene [112, 121, 1571. Peroxidase-based carbon pastes have also been used in mediatorless fashion and together with hydrogen peroxide producing oxidases, see Tables 1 and 3 and below. The reaction sequence for such a coupling is illustrated in Figure 5. Figure 6 is included to illustrate where on the potential scale the approximate Eo' values are found of the various groups of redox enzymes most commonly used in conjunction with carbon paste electrodes. The E o value of glucose/gluconolactone is also included for comparison as well as the potential range offering the most sensitive and interference-free detection. Recently an increasing number of articles have reported amperometric enzyme sensors based on mediatorless electrodes where a direct electric communication between the active site of the enzyme and the electrodes results in appreciable current densities in the presence of the enzyme substrate"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure13.27-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure13.27-1.png",
        "caption": "Figure 13.27 Beam on the ground. Due to the uniformly distributed load q, the ground pressure is distributed linearly.",
        "texts": [
            " Here we will determine the extreme value Mmin for field BC. In the shear force diagram the distance from A to the zero in BC is (see Figure 13.26c) (0.4 m) + 1 kN (1 kN) + (1.5 kN) \u00d7 (0.2 m) = 0.48 m. The magnitude of Mmin is now found most easily from the area of the V diagram: Mmin = 1 2 \u00d7 (0.48 m)(1 kN) = 0.24 kNm ( ). Figure 13.26 (a) The distributed loads on the railway sleeper replaced by (b) their resultants. (c) Shear force diagram and (d) bending moment diagram. 13 Calculating M, V and N Diagrams 571 Figure 13.27 shows a beam AB lying on the ground, of which the dead weight can be ignored. A uniformly distributed load q is acting on the righthand side of the beam over a length a. Due to this load, the earth pressure on the underside of the beam varies linearly, from 0 at A to 48 kN/m at B. Questions: a. From the equilibrium of the beam, determine length a and load q . b. Make a good sketch of the V diagram and the M diagram for the beam. c. At which cross-section is the shear force an extreme? Write down the extreme values for the V diagram. For these cross-sections, also include the tangents to the M diagram. d. At which cross-section is the bending moment an extreme? Determine this value, and include it with the M diagram. Solution: a. The resultant of the earth pressure and the resultant of the q load must have the same line of action (moment equilibrium of a body subject to two forces). The distance from B to the line of action of both resultants is (see Figure 13.27) 1 2a = 1 3 \u00d7 (1.80 m) \u21d2 a = 1.20 m. On the basis of the vertical force equilibrium, both resultants must be of equal magnitude: qa = 1 2 \u00d7 (1.80 m)(48 kN/m) = 43.2 kN so that q = 43.2 kN 1.20 m = 36 kN/m. 572 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM b. In Figure 13.28a, all the loads on beam AB are shown. In Figure 13.28b, the beam has been modelled as a line element, and the resulting load is shown. Three fields with a triangular load can be distinguished. The figure also shows the resultants of these triangular loads"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001046_j.ijheatmasstransfer.2014.09.014-Figure12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001046_j.ijheatmasstransfer.2014.09.014-Figure12-1.png",
        "caption": "Fig. 12. A close up view of the material density at the laser beam front at time t = 0.001 s with a layer of 50 lm powder.",
        "texts": [
            " Comparing HH with LH that has the same speed, HH needs a longer distance to reach the steady state as there is more energy input in it. Similarly, LL requires a longer distance than LH because LL is at a lower speed, which has more energy input per unit distance. Also from Fig. 10 it can be observed that steady state is reached after 570 lm for all three cases. This distance is used for the subsequent parameter sweep. Fig. 11 shows the material densities of LH at different times in the simulation. A close up view at t = 0.001 s is as shown in Fig. 12 where heat transfer has reached steady state and evaporation and melting are occurring under the laser beam. Furthermore, Fig. 12 shows that the powder turns into bulk solid as it cools after 0 W-1140 mm/s) with a layer of 50 lm powder. Fig. 13. The side view of the propa melting. The density of the material removed and evaporated (bright) is set to that of aluminium gas and remains as gas even as it cools. Fig. 13 shows the propagation of the melt pool\u2019s front with the laser beam movement in LH at steady state in its side view at six consecutive time points. The orange region represents the molten material, blue represents the material removed or evaporated while the red arrows represent the magnitudes and position of the laser beam"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000724_j.matdes.2015.08.086-Figure11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000724_j.matdes.2015.08.086-Figure11-1.png",
        "caption": "Fig. 11. Tensile testing. (a) Dimensional sizes and manufactured by SLM. (b) The specimens machined into desired shape. (c) The specimens after testing.",
        "texts": [
            " Particle B is however much larger than the powder layer thickness and cannot be totally melted by the laser beam, thereby creating inclusions on the SLM part, as shown in Fig. 10c. These inclusions not only reduce the relative density of SLM parts, but also connect badly to the parts, consequently turning into a source of fracture during tensile testing [35], as shown in Fig. 10d. Tensile test pieces were designed in accordance with the Chinese National Standard GB/T 228\u20132002 [36], as shown in Fig. 11a. They were then manufactured to the process parameters as shown in Table 3. The pieces were manufactured to a thickness of 8 mm, then each piece was cut and machined in order to obtain three specimens, regarded as a group, as shown in Fig. 11b. The tensile properties (including ultimate tensile strength (UTS), yield strength (\u03c30.2) and elongation (EL))weremeasured on aWDS-20H type tensile testingmachine at room temperature, the moving speed of the crosshead was set at 0.6 mm/min. The obtained results are shown in Fig. 12 and Table 4. Fig. 12 demonstrates that both groups of specimens aremainly characterized by ductile fracture. However Table 4 shows that the tension properties of specimens manufactured with contaminated powder are far inferior to those manufactured with fresh powder"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001866_msec2016-8819-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001866_msec2016-8819-Figure2-1.png",
        "caption": "Figure 2. Model geometric information and boundary conditions.",
        "texts": [
            ", powder to solid, was considered and controlled by user defined subroutine. The conductivity of powder material with 50% porosity is estimated according to the ratio of measured Ti64 powder to solid conductivity [19]. All the material properties have been summarized in Figure 1. Finite element model configuration A 3D FEM thermomechanical sequentially coupled model was developed to simulate the process temperature distribution, thermal stress and deformation in a 3-layer SLM deposition process. Figure 2 shows the geometry and boundary condition of the model, which has a solid substrate dimension of 8 \u00d7 8 \u00d7 1 mm (length, width, and height) as previously deposited material. The scanning domain is in a 6 \u00d7 6 mm square. However, actual SLM metal parts would greatly exceed the size of this numerical model and can be sliced into hundreds of thousands of layers, which is hard to perform simulation considering the computational cost. Thus, a smaller size model is considered in this study. The powder layers were In718 particles with each layer of 30 \u00b5m thickness"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure13.49-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure13.49-1.png",
        "caption": "Figure 13.49 Three-hinged frame with a tie rod.",
        "texts": [
            " The V diagram has the same slope in all fields, equal to the distributed load of 16 kN/m. The dashed M diagram due to the load resultants gives the tangents at A, D, E and G. The parabola in field DE passes through hinge S, since M = 0. Here, in the middle of field DE, the tangent is parallel to the chord k between the M values at D and E. Note that the M diagram at C and G has no horizontal tangents as the shear force is not zero. Check M diagram: Per field, p = 1 8q 2 applies for the rise p of the parabolic M diagram. 13 Calculating M, V and N Diagrams 589 Figure 13.49 shows a three-hinged frame ASB with tie rod AB. Girder CSD is carrying a uniformly distributed full load of 25 kN/m. The dimensions are shown in the figure. Questions: a. Determine the support reactions at A and B and the force in tie rod AB. b. Isolate frame ASB, and draw all the forces acting on it at A and B. c. For ASB, draw the M , V and N diagrams, with the deformation sym- bols. Include relevant values. At D, S and C, draw the tangents to the M diagram. d. Determine the maximum bending moment in field CSD"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure1.7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure1.7-1.png",
        "caption": "FIGURE 1.7. Illustration of a sample radial tire interior components and arrangement.",
        "texts": [
            " The vehicle will be bouncy and hard to steer because the tireprint is small and only the center portion of the tireprint is contacting 1. Tire and Rim Fundamentals 11 the road surface. In a properly-inflated tire, approximately 95% of the vehicle weight is supported by the air pressure in the tire and 5% is supported by the tire wall. 1.2 Tire Components A tire is an advanced engineering product made of rubber and a series of synthetic materials cooked together. Fiber, textile, and steel cords are some of the components that go into the tire\u2019s inner liner, body plies, bead bundle, belts, sidewalls, and tread. Figure 1.7 illustrates a sample of tire interior components and their arrangement. The main components of a tire are explained below. Bead or bead bundle is a loop of high strength steel cable coated with rubber. It gives the tire the strength it needs to stay seated on the wheel rim and to transfer the tire forces to the rim. Inner layers are made up of different fabrics, called plies. The most common ply fabric is polyester cord. The top layers are also called cap plies. Cap plies are polyesteric fabric that help hold everything in place",
            " Tire and Rim Fundamentals Belts or belt buffers are one or more rubber-coated layers of steel, polyester, nylon, Kevlar or other materials running circumferentially around the tire under the tread. They are designed to reinforce body plies to hold the tread flat on the road and make the best contact with the road. Belts reduce squirm to improve tread wear and resist damage from impacts and penetration. The carcass or body plies are the main part in supporting the tension forces generated by tire air pressure. The carcass is made of rubber-coated steel or other high strength cords tied to bead bundles. The cords in a radial tire, as shown in Figure 1.7, run perpendicular to the tread. The plies are coated with rubber to help them bond with the other components and to seal in the air. A tire\u2019s strength is often described by the number of carcass plies. Most car tires have two carcass plies. By comparison, large commercial jetliners often have tires with 30 or more carcass plies. The sidewall provides lateral stability for the tire, protects the body plies, and helps to keep the air from escaping from the tire. It may contain additional components to help increase the lateral stability",
            "3 Radial and Non-Radial Tires Tires are divided in two classes: radial and non-radial, depending on the angle between carcass metallic cords and the tire-plane. Each type of tire 1. Tire and Rim Fundamentals 15 construction has its own set of characteristics that are the key to its performance. The radial tire is constructed with reinforcing steel cable belts that are assembled in parallel and run side to side, from one bead to another bead at an angle of 90 deg to the circumferential centerline of the tire. This makes the tire more flexible radially, which reduces rolling resistance and improves cornering capability. Figure 1.7 shows the interior structure and the carcass arrangement of a radial tire. The non-radial tires are also called bias-ply and cross-ply tires. The plies are layered diagonal from one bead to the other bead at about a 30 deg angle, although any other angles may also be applied. One ply is set on a bias in one direction as succeeding plies are set alternately in opposing directions as they cross each other. The ends of the plies are wrapped around the bead wires, anchoring them to the rim of the wheel"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure5.9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure5.9-1.png",
        "caption": "FIGURE 5.9. A rotating rigid body B(Oxyz) with a fixed point O in a global frame G(OXY Z).",
        "texts": [
            " (5.164) Since matrix multiplication can be performed in any order, we find\u23a1\u23a3 I\u0302 J\u0302 K\u0302 \u23a4\u23a6 = \u23a1\u23a3 I\u0302 \u00b7 \u0131\u0302 I\u0302 \u00b7 j\u0302 I\u0302 \u00b7 k\u0302 J\u0302 \u00b7 \u0131\u0302 J\u0302 \u00b7 j\u0302 J\u0302 \u00b7 k\u0302 K\u0302 \u00b7 \u0131\u0302 K\u0302 \u00b7 j\u0302 K\u0302 \u00b7 k\u0302 \u23a4\u23a6\u23a1\u23a3 \u0131\u0302 j\u0302 k\u0302 \u23a4\u23a6 = GRB \u23a1\u23a3 \u0131\u0302 j\u0302 k\u0302 \u23a4\u23a6 (5.165) where, GRB = \u23a1\u23a3 I\u0302 J\u0302 K\u0302 \u23a4\u23a6 \u00a3 \u0131\u0302 j\u0302 k\u0302 \u00a4 . (5.166) Following the same method we can show that BRG = \u23a1\u23a3 \u0131\u0302 j\u0302 k\u0302 \u23a4\u23a6 \u00a3 I\u0302 J\u0302 K\u0302 \u00a4 . (5.167) 248 5. Applied Kinematics 5.7 Angular Velocity Consider a rotating rigid body B(Oxyz) with a fixed point O in a reference frame G(OXY Z) as shown in Figure 5.9. The motion of the body can be described by a time varying rotation transformation matrix between the global and body frames to map the instantaneous coordinates of any fixed point in body frame B into their coordinates in the global frame G Gr(t) = GRB(t) Br. (5.168) The velocity of a body point in the global frame is Gr\u0307(t) = Gv(t) (5.169) = GR\u0307B(t) Br (5.170) = G\u03c9\u0303B Gr(t) (5.171) = G\u03c9B \u00d7 Gr(t) (5.172) where G\u03c9B is the angular velocity vector of B with respect to G. It is equal to a rotation with angular rate \u03c6\u0307 about an instantaneous axis of rotation u\u0302. \u03c9 = \u23a1\u23a3 \u03c91 \u03c92 \u03c93 \u23a4\u23a6 = \u03c6\u0307 u\u0302 (5.173) The angular velocity vector is associated with a skew symmetric matrix 5. Applied Kinematics 249 G\u03c9\u0303B called the angular velocity matrix \u03c9\u0303 = \u23a1\u23a3 0 \u2212\u03c93 \u03c92 \u03c93 0 \u2212\u03c91 \u2212\u03c92 \u03c91 0 \u23a4\u23a6 (5.174) where G\u03c9\u0303B = GR\u0307B GRT B (5.175) = \u03c6\u0307 u\u0303. (5.176) Proof. Consider a rigid body with a fixed point O and an attached frame B(Oxyz) as shown in Figure 5.9. The body frame B is initially coincident with the global frame G. Therefore, the position vector of a body point P is Gr(t0) = Br. (5.177) The global time derivative of Gr is Gv = Gr\u0307 = Gd dt Gr(t) = Gd dt \u00a3 GRB(t) Br \u00a4 = Gd dt \u00a3 GRB(t) Gr(t0) \u00a4 = GR\u0307B(t) Br. (5.178) Eliminating Br between (5.168) and (5.178) determines the velocity of the point in global frame Gv = GR\u0307B(t) GRT B(t) Gr(t). (5.179) We denote the coefficient of Gr(t) by \u03c9\u0303 G\u03c9\u0303B = GR\u0307B GRT B (5.180) and write Equation (5.179) as Gv = G\u03c9\u0303B Gr(t) (5"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure5.56-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure5.56-1.png",
        "caption": "Figure 5.56 A beam (a) suspended from and (b) leaning upon a strengthening bar system.",
        "texts": [
            " The components of the hinge force S follow from the force equilibrium of the part to the left or to the right of S. 186 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM d. In Figure 5.55a, joint D has been isolated from SD and DB. Three interaction forces are acting in the rigid connections. The forces exerted by joint D on SD and DB can be found from the equilibrium of these parts. Equal and opposite forces act on joint D (see Figure 5.55b). Check: Joint D is in force equilibrium and in moment equilibrium. The strengthened beams in Figure 5.56 are in many ways comparable to trussed beams. An important difference is that in here the strengthening bar system is supported outside the beam. In Figure 5.56a the beam is suspended from the strengthening bar system, in Figure 5.56b the beam is leaning upon it. These structures are used in bridges. They are used also as auxiliary structures during building activities. The structures in Figure 5.56 are statically indeterminate to the first degree. In the following we will address only statically determinate examples. Example The structure in Figure 5.57 is loaded by a vertical force of 40 kN. Questions: a. Determine the support reactions at A and B. b. Determine the forces in bars (1) to (3) and (a) to (d). c. Draw the forces acting on the hinged joint S. Solution (units in kN and m): a. This compound structure has five support reactions: 5 Calculating Support Reactions and Interaction Forces 187 Figure 5"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003874_ar9002015-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003874_ar9002015-Figure5-1.png",
        "caption": "FIGURE 5. The FreeStyle test strip containing the 300 nL microcoulometric cell.",
        "texts": [
            " A change in mediator was made when the PQQ-GDH was changed to flavin adenine dinucleotide (FAD)-GDH. The structure of the current redox mediator, with a potential of -160 mV vs Ag/AgCl, is shown in Figure 4. This mediator also allows a fast 5 s testing time, while providing a higher average current in conjunction with FAD-GDH, than does the mediator of Figure 3. The volume-dependent coulometric assay required the reproducible mass manufacture of the 50 \u00b5m gap thin-layer electrochemical cell of Figure 5 with a coefficient of variation of 2-3% or less. The gap was maintained initially by using a spacer with cut windows, then by applying a surprisingly reproducible film of a pressure-sensitive adhesive. We credit our manufacturing engineer, Phil Plante, one of the first people Ephraim hired at TheraSense, with their mass-production. He was ably assisted by Joe Vivolo, who expanded and improved the production. The painless coulometric monitor, fondly named Colossus in internal TheraSense documents, was released as the FreeStyle blood glucose monitoring system"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure4.8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure4.8-1.png",
        "caption": "Fig. 4.8. Phasor diagram of an overexcited salient pole synchronous generator for RL load.",
        "texts": [
            " The shaft power Psh (input power Pin for generators, output power Pout for motors) is calculated by adding to the electromagnetic power 4.7 Performance Calculation 135 Pelm (generators) or subtracting from the electromagnetic power (motors) the rotational losses \u2206Prot according to eqn (2.70). The shaft torque is \u2022 for a motor (developed by the motor itself) Tsh = Pout 2\u03c0n (4.30) \u2022 for a generator (developed by the prime mover) Tsh = Pin 2\u03c0n (4.31) and the efficiency is \u03b7 = Pout Pin (4.32) The phasor diagram of a salient pole synchronous generator with RL load is shown in Fig. 4.8. The input voltage projections on the d and q axes according to Fig. 4.8 are V1 sin \u03b4 = IaqXsq \u2212 IadR1 V1 cos \u03b4 = Ef \u2212 IadXsd \u2212 IaqR1 (4.33) and V1 sin \u03b4 = IadRL \u2212 IaqXL V1 cos \u03b4 = IaqRL + IadXL (4.34) where ZL = RL + jXL is the load impedance per phase across the output terminals. The d- and q-axis currents are \u2022 from eqns (4.33) Iad = EfXsq \u2212 V1(Xsq cos \u03b4 +R1 sin \u03b4) XsdXsq +R2 1 (4.35) Iaq = V1(Xsd sin \u03b4 \u2212R1 cos \u03b4) + EfR1 XsdXsq +R2 1 (4.36) 136 4 AFPM Machines With Iron Cores \u2022 from eqns (4.34) Iad = V1(RL sin \u03b4 +XL cos \u03b4) R2 L +X2 L (4.37) Iaq = V1(RL cos \u03b4 \u2212XL sin \u03b4) R2 L +X2 L (4",
            "34), the d- and q-axis currents are independent of the load angle \u03b4, i.e. 4.7 Performance Calculation 137 Iad = Ef (Xsq +XL) (Xsd +XL)(Xsq +XL) + (R1 +RL)2 (4.40) Iaq = Ef (R1 +RL) (Xsd +XL)(Xsq +XL) + (R1 +RL)2 (4.41) The angle \u03a8 between the current Ia and q-axis and the angle \u03c6 between the current Ia and voltage V1 are respectively, \u03a8 = arccos ( Iaq Ia ) = arccos  Iaq\u221a I2 ad + I2 aq  (4.42) \u03c6 = arccos ( IaRL V1 ) = arccos ( RL ZL ) (4.43) where Ia = \u221a I2 ad + I2 aq (see also eqn (2.89)). The output electrical power on the basis of phasor diagram (Fig. 4.8) and eqn (4.33) is Pout = m1V1Ia cos\u03c6 = m1V1(Iaq cos \u03b4 + Iad sin \u03b4) = m1[EfIaq \u2212 IadIaq(Xsd \u2212Xsq)\u2212 I2 aR1] (4.44) Including only the stator winding losses, the internal electromagnetic power of the generator is Pelm = Pout +\u2206P1w = m1[EfIaq \u2212 IadIaq(Xsd \u2212Xsq)] (4.45) The square (trapezoidal) wave AFPM brushless machine has already been introduced in Section 2.10.2. Square-wave machines are characterized by their trapezoidal or quasi-square back EMF waveforms with the conduction angle from 100 to 150 electrical degrees, depending on the construction"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure13.14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure13.14-1.png",
        "caption": "FIGURE 13.14. Full car model for a vehicle vibrations.",
        "texts": [
            "271) Iy \u03b8\u0308 \u2212 a1cf \u00b3 x\u0307\u2212 x\u03071 + b1\u03d5\u0307\u2212 a1\u03b8\u0307 \u00b4 \u2212 a1cf \u00b3 x\u0307\u2212 x\u03072 \u2212 b2\u03d5\u0307\u2212 a1\u03b8\u0307 \u00b4 +a2cr \u00b3 x\u0307\u2212 x\u03073 \u2212 b1\u03d5\u0307+ a2\u03b8\u0307 \u00b4 + a2cr \u00b3 x\u0307\u2212 x\u03074 + b2\u03d5\u0307+ a2\u03b8\u0307 \u00b4 \u2212a1kf (x\u2212 x1 + b1\u03d5\u2212 a1\u03b8)\u2212 a1kf (x\u2212 x2 \u2212 b2\u03d5\u2212 a1\u03b8) +a2kr (x\u2212 x3 \u2212 b1\u03d5+ a2\u03b8) + a2kr (x\u2212 x4 + b2\u03d5+ a2\u03b8) = 0 (13.272) 866 13. Vehicle Vibrations mf x\u03081 \u2212 cf \u00b3 x\u0307\u2212 x\u03071 + b1\u03d5\u0307\u2212 a1\u03b8\u0307 \u00b4 \u2212 kf (x\u2212 x1 + b1\u03d5\u2212 a1\u03b8) \u2212kR 1 w \u00b5 \u03d5\u2212 x1 \u2212 x2 w \u00b6 + ktf (x1 \u2212 y1) = 0 (13.273) mf x\u03082 \u2212 cf \u00b3 x\u0307\u2212 x\u03072 \u2212 b2\u03d5\u0307\u2212 a1\u03b8\u0307 \u00b4 \u2212 kf (x\u2212 x2 \u2212 b2\u03d5\u2212 a1\u03b8) +kR 1 w \u00b5 \u03d5\u2212 x1 \u2212 x2 w \u00b6 + ktf (x2 \u2212 y2) = 0 (13.274) mrx\u03083 \u2212 cr \u00b3 x\u0307\u2212 x\u03073 \u2212 b1\u03d5\u0307+ a2\u03b8\u0307 \u00b4 \u2212kr (x\u2212 x3 \u2212 b1\u03d5+ a2\u03b8) + ktr (x3 \u2212 y3) = 0 (13.275) mrx\u03084 \u2212 cr \u00b3 x\u0307\u2212 x\u03074 + b2\u03d5\u0307+ a2\u03b8\u0307 \u00b4 \u2212kr (x\u2212 x4 + b2\u03d5+ a2\u03b8) + ktr (x4 \u2212 y4) = 0 (13.276) Proof. Figure 13.14 shows a better vibrating model of the system. The body of the vehicle is assumed to be a rigid slab. This slab has a mass m, which is the total body mass, a longitudinal mass moment of inertia Ix, and a lateral mass moment of inertia Iy. the moments of inertia are only the body mass moments of inertia not the vehicle\u2019s mass moments of inertia. The wheels have a mass m1, m2, m3, and m4 respectively. However, it is common to have m1 = m2 = mf (13.277) m3 = m4 = mr. (13.278) The front and rear tires stiffness are indicated by ktf and ktr respectively"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000252_j.automatica.2006.10.008-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000252_j.automatica.2006.10.008-Figure1-1.png",
        "caption": "Fig. 1. Convergence of various 2-sliding homogeneous controllers.",
        "texts": [
            " Indeed, if, for example, 1, 1 > 0, then a 1-sliding mode can easily be organized on the line 1\u0307 + 1| |1/2sign = 0. Sliding mode conditions being fulfilled at one point of the line, they automatically hold along the whole line due to the homogeneity properties. If for each i one of the coefficients is zero, the twisting controller (Levant, 1993) u = \u2212(r1 sign + r2 sign \u0307), is built. Its convergence condition is (r1 + r2)Km \u2212 C > (r1 \u2212 r2)KM + C, (r1 \u2212 r2)Km > C. A typical trajectory in the plane , \u0307 is shown in Fig. 1a. Controller (12) may be considered as a generalization of the twisting controller, when the switching takes place on parabolas 1\u0307 + 1| |1/2 sign = 0 instead of the coordinate axes. Thus, the resulting controller satisfies Theorem 1, and its discrete-sampling version u = \u2212 r1 sign( 1 i + 1 | i |1/2 sign i ) \u2212 r2 sign( 2 i + 2 | i |1/2 sign i ) (13) provides for the accuracy described in Theorem 4, i.e. \u223c 2, \u0307 \u223c . Similarly, the noisy measurements lead to the accuracy provided by Theorems 2 and 3",
            " Differentiating the function =\u0307+ | |1/2 sign along the trajectory, obtain \u0307 \u2208 [\u2212C, C] \u2212 [Km, KM] sign + 1 2 \u0307| |\u22121/2. Checking the condition \u0307 sign < const < 0 in a vicinity of each point on the curve =0, obtain using \u0307=\u2212 | |1/2 sign that the 1-sliding-mode existence condition holds at each point except of the origin, if Km \u2212 C > 2/2. The trajectories of the inclusion inevitably hit the curve =0 due to geometrical reasons. Indeed, each trajectory, starting with > 0, terminates sooner or later at the semi-axis = 0, \u0307 < 0, if u = \u2212 sign keeps its constant value \u2212 (Fig. 1b). Thus, on the way it inevitably hits the curve =0. The same is true for the trajectory starting with < 0. Since that moment the trajectory slides along the curve = 0 towards the origin and reaches it in finite time. Obviously, each trajectory starting from a disk centered at the origin comes to the origin in a finite time, the convergence time being uniformly bounded in the disk. Consider the region confined by the lines \u0307=\u00b1 and the trajectories of the differential equations \u0308 = \u2212C + Km with initial conditions = 2/ 2, \u0307 = \u2212 , and \u0308 = C \u2212 Km with initial conditions = \u2212 2/ 2, \u0307 = (Fig. 1b). No trajectory starting from the origin can leave . Since can be taken arbitrarily small, the trajectory cannot leave the origin. The same reasoning proves the Lyapunov stability of the origin. The 2-sliding stability analysis is greatly simplified by the fact that all the trajectories in the plane , \u0307 which pass through a given continuity point of u = ( , \u0307) are confined between the properly chosen trajectories of the homogeneous differential equations \u0308=\u00b1C +KM ( , \u0307) and \u0308=\u00b1C +Km ( , \u0307). These border trajectories cannot be crossed by other paths, if is locally Lipschitzian, and may be often chosen as boundaries of appropriate dilation-retractable regions",
            " Hence, the condition = \u0307=\u00b7 \u00b7 \u00b7= (r\u22121) =0 is never fulfilled in practice with r 2, and the control remains continuous function of time all the time. As a result, the chattering is significantly reduced. Following is the 2-sliding controller from such a family of arbitrary-order sliding controllers (Levant, 2005b): u = \u2212 \u0307 + | |1/2 sign |\u0307| + | |1/2 . (15) This control is continuous everywhere except of the origin. It vanishes on the parabola \u0307 + | |1/2 sign = 0. With sufficiently large there are such numbers 1, 2, 0 < 1 < < 2 that all the trajectories enter the region between the curves \u0307 + | |1/2 sign = 0 and cannot leave it (Fig. 1c). Note that no explicit conditions on the parameter choice are obtained in Levant (2005b). Proposition 3. Let , > 0, Km \u2212 C > 0 (16) and the inequality Km \u2212 C \u2212 2 Km + \u2212 1 2 2 > 0 (17) hold for some positive > (which is always true with sufficiently large ), then controller (15) provides for the establishment of the finite-time-stable 2-sliding mode \u2261 0. Conditions of the proposition can be solved for , but the resulting expressions are redundantly cumbersome. Proof. Denote =\u2212\u0307/| |1/2. Calculations show that u= ( \u2212 )/(| | + ) and due to the symmetry of the problem, it is enough to consider the case > 0, \u2212\u221e < < \u221e",
            " Thus, conditions (16), (17) provide for the establishment and keeping of the inequality 1 < < 2. Following is another example of a quasi-continuous 2-sliding controller: u = { min{ , max[\u2212 , \u2212 (\u0307/| |1/2 + sign )]}, = 0, \u2212 sign \u0307, = 0, (18) where , , > 0, Km \u2212 C > 2/2, > . The definition of the control u with = 0 is made by continuity and does not influence the system trajectories, since it only influences values on a zero-measure set. Enlarging one compels the trajectories to get closer to the parabola \u0307 + | |1/2 sign = 0 without increasing the control magnitude (Fig. 1d). Also here the discontinuity is concentrated at = \u0307 = 0. Thus, in the presence of measurement errors the motion takes place in some vicinity of the mode = \u0307 = 0 without entering it, and the control signal turns out to be continuous. Moreover, let (y) be any monotonously growing positive continuous function of a non-negative argument, then the following controller generalizes (18): u = min{ , max[\u2212 , \u2212 (|\u0307|/| |1/2) sign \u0307 + ( ) sign ( )]}, u = { u ( , \u0307), = 0, \u2212 sign \u0307, = 0. (19) Proposition 4. Let , , > 0, Km \u2212 C > ( )2/2, ( ) > and be sufficiently large, then controller (19) provides for the establishment of the finite-time-stable 2-sliding mode \u2261 0. The proof is very similar to that of Proposition 3. Enlarging one compels the trajectories to get closer to the parabola \u0307 + ( )| |1/2 sign = 0 without increasing the control magnitude (Fig. 1d). Like other listed controllers, this one also provides for the finite-time convergence to the 2-sliding mode and the accuracies corresponding to Theorems 2\u20134. Chattering attenuation: The standard problem of classical (first order) sliding-mode control is attenuation of the chattering effect (Fridman, 2003; Furuta & Pan, 2000; Slotine & Li, 1991). 2-sliding-mode control provides effective tools for the reduction or even practical elimination of the chattering without compromising the benefits of the standard sliding mode (Boiko & Fridman, 2005; Boiko, Fridman, Iriarte, Pisano, & Usai, 2006)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000613_978-94-015-9261-1-Figure4.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000613_978-94-015-9261-1-Figure4.1-1.png",
        "caption": "Figure 4.1. Geometrical objects in state and output spaces",
        "texts": [
            " in the course of the system design, reduced to those of sys tem stabilization in the state space. The most descriptive examples are presented by: i) output regulation problem, where a zero (or constant) output vector y is produced by a collection of the system trajectories x(l. xo) in the state space ]Rnwhich belang to an invariant hypersurface Z'\" (a so-called zero dynarnics subrnanifold, see Sections 2.5 and 5.2); 127 ii) curve-following problems for which the given curve S; of the output space IRm defined by the equation <py(y) = 0 is a projection of the hypersurface S* c IRn (Figure 4.1). In this chapter we introduce the main notions related to the properties of regular hypersurfaces of being an invariant set and non trivial attractor of a dynamical system. Then, a methodology of t.he systcm analysis and design tools for solving the problems of set stabilization are presented for nonlinear multivariable systems having several controlling inputs. The contents of the chapter is the basis of further analysis of multi input/multi-output systems (Chapter 5), for which similar problems im mediately appear when passing to analysis of the system behavior in the state space",
            "4, such a system can be stabilized by using the linear controllaw ue = -Ke(x*)e, (4.155) where K e(x*) is the matrix ofvarying feedback gains chosen by using an ad equate design tool and providing asymptotic stability of the nonstationary closed loop system (4.156) and Aec = Ae + BeKe. Thus, under the relevant conditions, the control law (4.149), (4.155) provides the required asymptotic stabilization of the transversal dynamics model (4.148). The resulting controllaw (4.147) solving the general problem of state tracking (Problem 4.3) takes the form (Figure 4.1:3) u = L(x,x*)v+Ueo(x*)-Ke(x*)e, (4.157) where L is a solution of (4.146), the control term UeOis defined by equation (4.151) and the error vector eis computed by using (4.14 :3). Example 4.5. Consider an n-channel plant. described by Xi = fi(X)+9i(X)Ui, (4.158) NONLINEAR CONTROL OF MULTIVARIABLE SYSTEM S 179 v Re fer ence m odel '1'* ( .) UeO( ' ) u C ontrol x t.- L system - Ue x , - X* e Fiqure 4.13. State tracking system where x = {xd E jRn is t he state , U = {ud is t he con trol, i = 1, .. "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure13.79-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure13.79-1.png",
        "caption": "Figure 13.79 (a) The forces on joint C and (b) the closed force polygon for the force equilibrium (scale: 1 square = 2.5 kN).",
        "texts": [
            " Determine the support reactions at A and D. d. Determine the M , V and N diagrams for the entire structure, with the deformation symbols. Solution (units kN and m): a. The two parts ABCS and DS provide e = 2 \u00d7 3 = 6 equilibrium equations (see Figure 13.78). The number of unknown support reactions at A and D is r = 3 + 3 = 6. The number of unknown joining forces at S is v = 2. The degree of static indeterminacy is: n = r + v \u2212 e = 6 + 2 \u2212 6 = 2. The structure is therefore statically indeterminate to the second degree. b. In Figure 13.79a, joint C has been isolated and all forces acting on it are shown. The bending moments acting on the joint are not shown! Figure 13.79b shows the closed force polygon for the force equilibrium of the joint (scale: 1 square = 2.5 kN). The force polygon gives NBC = +7.5 \u221a 2 kN, NCS = +5 kN. c. With NCS = +5 kN the vertical equilibrium of CSD gives Dv = 5 kN (\u2191). 13 Calculating M, V and N Diagrams 607 With V CS = 10 kN the horizontal equilibrium of ABC gives Ah = 10 kN (\u2190). From the force equilibrium of the structure as a whole follows Dh = 10 kN (\u2190), Av = 15 kN (\u2191). Finally, the fixed-end moment reactions at A and D follow from the moment equilibrium about S of DS and ABCS respectively"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002097_978-1-4939-3017-3-Figure1.9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002097_978-1-4939-3017-3-Figure1.9-1.png",
        "caption": "Figure 1.9.3. The rigid body with rotor.",
        "texts": [
            " The dynamics of the uncontrolled system is described, for example, in McMillan [1936] (see also Bloch, Krishnaprasad, Marsden, and Murray [1996], Jurdjevic [1993], Koon and Marsden [1997b], and Krishnaprasad, Yang and Dayawansa [1991]). We will use the coordinates x, y for the linear horizontal displacement and P \u2208 SO(3) for the angular displacement of the ball. Thus P gives the orientation of the ball with respect to inertial axes e1, e2, e3 fixed in the plane, where the ei are the standard basis vectors aligned with the x-, y-, and z-axes, respectively. See Figure 1.9.1. Let the ball have radius a and mass m and let \u03c9 \u2208 R 3 denote the angular velocity of the ball with respect to the inertial axes. In particular, the ball may spin freely about the z-axis, and the z-component of angular momentum is conserved. If J denotes the inertia tensor of the ball with respect to the body axes (i.e., fixed in the body), then J = PTJP denotes the inertia tensor of the ball with respect to the inertial axes (i.e., fixed in space) and J\u03c9 is the angular momentum of the ball with respect to the inertial axes",
            " A useful classical system for testing control-theoretic ideas is the inverted pendulum on a cart, the goal being to stabilize the pendulum about the vertical using a force acting on the cart. In this book we will use this system to illustrate stabilization using the energy methods as discussed in Bloch, Marsden, and Alvarez [1997] and Bloch, Leonard, and Marsden [1997] (see Chapter 9 for further references). Here we just write down the equations of motion. First, we compute the Lagrangian for the cart\u2013pendulum system. Let s denote the position of the cart on the s-axis and let \u03c6 denote the angle of the pendulum with the upright vertical, as in Figure 1.9.2. Here, the configuration space is Q = G\u00d7S = R\u00d7S1 with the first factor being the cart position s, and the second factor being the pendulum angle \u03c6. The velocity phase space TQ has coordinates (s, \u03c6, s\u0307, \u03c6\u0307). The velocity of the cart relative to the lab frame is s\u0307, while the velocity of the pendulum relative to the lab frame is the vector vpend = (s\u0307+ l cos\u03c6 \u03c6\u0307,\u2212l sin\u03c6 \u03c6\u0307). (1.9.20) The kinetic energy of the coupled cart-pendulum system is given by K ( s, \u03c6, s\u0307, \u03c6\u0307 ) = 1 2 (s\u0307, \u03c6\u0307) ( M +m ml cos\u03c6 ml cos\u03c6 ml2 )( s\u0307 \u03c6\u0307 ) ",
            " The equations of motion of the cart\u2013pendulum system with a control force u acting on the cart (and no direct forces acting on the pendulum) are, since s is a cyclic variable (i.e., L is independent of s), d dt \u2202L \u2202s\u0307 = u , d dt \u2202L \u2202\u03c6\u0307 \u2212 \u2202L \u2202\u03c6 = 0 , that is, d dt ps = d dt (\u03b3s\u0307+ \u03b2 cos\u03c6\u03b8\u0307) = u , d dt p\u03c6 + \u03b2 sin\u03c6s\u0307\u03c6\u0307+D sin\u03c6 = d dt (\u03b1\u03c6\u0307+ \u03b2 cos\u03c6s\u0307) + \u03b2 sin\u03c6s\u0307\u03c6\u0307+D sin\u03c6 = 0 . Rigid Body with a Rotor. Following the work of Krishnaprasad [1985], Bloch, Krishnaprasad, Marsden, and Alvarez [1992], and Bloch, Leonard, and Marsden[1997, 2000], we consider a rigid body with a rotor aligned along the third principal axis of the body as in Figure 1.9.3. This is a model for a satellite. The rotor spins under the influence of a torque u acting on the rotor. The configuration space is Q = SO(3) \u00d7 S1, with the first factor being the rigid body attitude and the second factor being the rotor angle. The Lagrangian is total kinetic energy of the system (rigid carrier plus rotor), with no potential energy. Again, this system will be used in Section 9.2 to illustrate the energy method in analyzing stabilization and stability. The Lagrangian for this system (see Bloch, Krishnaprasad, Marsden, and Alvarez [1992] and Bloch, Leonard, and Marsden [2001]) is L = 1 2 (\u03bb1\u03a9 2 1 + \u03bb2\u03a9 2 2 + I3\u03a9 2 3 + J3(\u03a93 + \u03b1\u0307)2) , (1"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure1.27-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure1.27-1.png",
        "caption": "Fig. 1.27. Mechanism of a six-legged loco motion machine",
        "texts": [
            "20 as an example, it is possible to isolate the chains of \"legs\", \"body\" and \"arms\"; here, the chain consisting of \"legs\" is independent since the chains of \"body\" and \"arms\" impose no kinematic constraints on it. For the mechanism presented in Fig. 1.26, if the motion of member 1 is known, the chain I is independent, with all the remaining ones being satellites. The sequence to be followed in performing kinematic and dynamic analyses has thus been determined; namely, the basic independent chains should be analysed first, and the satellite (guided) chains second. Figure 1.27 illustrates the kinematic chains of a six-legged walking machine, while in Fig. 1.28 the scheme of a four-legged locomotion machine is presented. [IJ Vukobratovic, M., \"Applied Dynamics of Manipulation Robots\", Springer-Verlag, 1988. [2J Vukobratovic, M., Potkonjak V., \"Applied Dynamic and CAD of Manipulation Robots\", Monograph, Springer-Verlag, 1985. [3J Robototechnika (in Russian), edited by E. P. Popov and E. I. Yureevitch, Mashinostroenie, Moscow, 1984. [4J Stepanenko, Y., \"Dynamics of Spatial Mechanisms\", (in Russian), Mathematical Institute, Beograd, 1974"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001674_j.mechmachtheory.2012.10.008-FigureA.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001674_j.mechmachtheory.2012.10.008-FigureA.2-1.png",
        "caption": "Fig. A.2. Contact Radii Rj for Involute\u2013Tip rounding contacts.",
        "texts": [
            " Next, a procedure for the calculation of the radius Rr n as a function of geometric parameters and the position of the gears is shown. In the first case of involute \u2013involute contact (see Fig. A.1): Rr j \u00bc CrPr j \u00bc ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi CrQr j 2 \u00fe QrPr j 2 r \u00f0A 1\u00de QrPrj \u00bc Qr0Pr0 \u00fe \u03c1r \u03c6T\u2212\u03c60 \u00fe \u03c8j\u2212\u03b8r j ; r \u00bc 1;2 \u00f0A 2\u00de Qr0Pr0 \u00bc \u03c1rtg\u03c60; r \u00bc 1;2 \u00f0A 3\u00de CrQr j \u00bc \u03c1r ; r \u00bc 1;2 \u00f0A 4\u00de \u03c6T \u00bc arccos \u03c11 \u00fe \u03c12 dT \u00f0A 5\u00de \u03c8j \u00bc \u2212 arctan x1\u2212x2 d0 \u00fe y1\u2212y2 \u00f0A 6\u00de In the second case of involute \u2013 tip rounding contact (see Fig. A.2.): Rr j \u00bc CrPr j \u00bc ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi CrQ r2 r j 2 \u00fe Qr2 r Pr j 2 r \u00f0A 7\u00de Q1 r2P1 j \u00bc Q10P10 \u00fe \u03c11 \u03c6T r2\u2212\u03c60 \u00fe \u03c8j\u2212\u03b81 j \u00f0A 8\u00de Q2 r2P2 j \u00bc C2Cr2 sen \u03c6T r2\u2212\u03bb2 \u00fe Cr2P2 \u00f0A 9\u00de C2Cr2 \u00bc RExt2\u2212Rtip ; C1Q1 r2 j \u00bc \u03c11 \u00f0A 10\u00de C2Q2 r2 j \u00bc dT cos \u03c6T r2 \u2212\u03c11 \u00f0A 11\u00de \u03c6T r2 \u00bc \u03bc1\u2212\u03be1 \u00f0A 12\u00de where \u03bc1 \u00bc arccos \u03c11 C1Cr2 ! \u00f0A 13\u00de \u03bb2 \u00bc tan \u03c6A\u00f0 \u00de\u2212 tan \u03c60\u00f0 \u00de \u00fe \u03c60\u2212\u03c6CR2 \u2212 \u03c8j\u2212\u03b82j \u00f0A 14\u00de \u03c6A \u00bc arc cos \u03c12 C2A 0 B@ 1 CA \u00f0A 15\u00de \u03c6CR2 \u00bc arc cos \u03c12 C2Cr2 "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure12.4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure12.4-1.png",
        "caption": "FIGURE 12.4 Electric field sensors for all three directions.",
        "texts": [
            " FAR-FIELD RADIATION MODELS BY USING SENSORS 313 We first want to remind the reader that we do need to include the delays, or the retardation times in the field calculations to the sensors. The delay time \ud835\udf0fr,r\u2032 is the delay between the locations of the observation point r and the source point r\u2032, respectively. As considered in (2.11) for (FW)PEEC models \ud835\udf0fr,r\u2032 = \u2223 r \u2212 r\u2032 \u2223 c , (12.4) where c is the speed of light. The eSensor approach is based on work presented in Ref. [6]. An electric field sensor used for this approach is shown in Fig. 12.4. Essentially, the potential difference between two end points in the far field region is computed to find the field. The evaluation of the field is based on the conventional orthogonal PEEC models in Chapter 6 and/or the nonorthogonal ones in Chapter 7. The sensor is located at a point x, y, z, where the field needs to be computed. We only need to have three orthogonal sensors aligned in x\u0302, y\u0302, and the z\u0302 directions for the general case. Further, a useful observation is that the sensor solution is independent of the rest of the PEEC model",
            " Another positive issue for the solution is that multiple sensors, for different locations, do not interact with each other. Another useful observation is that the structure of the coupling mechanisms from the PEEC circuit to the eSensor is similar to the coupling inside a PEEC model. Specifically, the partial element calculation facilities, which are part of the solver, can be used for the PEEC to sensors computations while the back coupling is set to zero. The electric field sensors shown in Fig. 12.4 use three bars, one for the field in each direction. Since the method is the same for all three bars, it is sufficient to describe the sensor solution for 1 bar only. We chose the derivation for the z-directed sensor for Ez. 314 INCIDENT AND RADIATED FIELD MODELS We should remember that the integral equation for PEEC is based on (3.25), which computes the E-field. We use this to specifically compute electric fields, for the z-directed sensor bar as (E \u22c5 z\u0302)\u0394z = (Einc \u22c5 z\u0302)\u0394z \u2212 \u2211 k Lpoz,k dik(t \u2212 \ud835\udf0foz,k) dt \u2212 \u2211 m \u222bz (\u2207\ud835\udebde \u22c5 z\u0302) dz, (12",
            "1) is not coupled since we do not have a galvanic connection. Therefore, we did not include it in the sensor equations (12.5). If necessary, this term can be included directly in the field calculation. The second right hand side term represents (12.1) the coupled voltage by the partial inductances. However, the very interesting situation is that we do not need a partial self-inductance coupling at the sensor to compute the inductively induced voltages. We must also compute the coupling to the z-directed part of the sensor in Fig. 12.4 for the inductive coupled voltage to a zero thickness sensor filament of a length \u0394z centered at the observation point O in Fig. 12.4. This represents one branch for the coupling of a partial mutual inductance. The contribution of the inductively coupled voltage from the cell k to the sensor is similar to (6.9) as Vez = Ez\u0394z = dI\u2032z(t \u2212 \ud835\udf0foz,k) dt \ud835\udf07 4 \ud835\udf0b \u2032 k \u222b e b \u222b\u2032 k \u222bzk\u2032 1|r \u2212 r\u2032|dz\u2032k d \u2032 k dz, (12.6) where I\u2032z is the current in the z-directed partial element of the PEEC model. We recognize that the integrals on the right-hand side represent the coupling using the contribution by a partial inductance coupled to a z-directed filament representing the sensor",
            " Note that these issues are the same as the updating of the right-hand-side elements for the time domain MNA equation solution (2.28) for systems with delay. FAR-FIELD RADIATION MODELS BY USING SENSORS 315 The computation of the delayed currents in the frequency domain is rather straightforward. The equivalent of (12.6) in the frequency domain can be approximated as Vez = Ez\u0394z = sIz\u2032 (s)e\u2212s \ud835\udf0f(oz,k) \ud835\udf07 4 \ud835\udf0b \u2032 k \u222b e b \u222b\u2032 k \u222bzk\u2032 1|r \u2212 r\u2032|dz\u2032k d\u2032 k dz, (12.8) where \ud835\udf0foz,k is the same delay as in the time domain. Recognizing that we can use a partial inductance that couples to the filament in Fig. 12.4, (12.8) can be simply evaluated as Vez = Ez\u0394z = sIz\u2032 (s)e\u2212s \ud835\udf0f(oz,k)Lpoz,k. (12.9) It is clear that, in general, the length of the sensor \u0394z is chosen to be small compared to the shortest wavelength \ud835\udf06min, like \ud835\udf06min\u221540. But on the other hand it shall be large enough not to lead to numerical errors. If finite dielectric bodies are present as considered in (10.62), the currents and the inductive couplings also need to be taken into account in exactly the same way. The equations for the contributions Vez for the dielectrics are the same as (12.6) for the time domain or (12.9) in the frequency domain. This is due to the fact that finite dielectric models also include partial inductances. We evaluate the last part in (12.5), \u2207\u03a6e(r, t). The electric field part of the sensed field is given by Ez \u0394z = \u2212\u222b e b \u2207\u03a6e \u22c5 z\u0302 dz, (12.10) where the potential evaluation points are shown as squares at the ends of the sensor in Fig. 12.4. The gradient is found by evaluating the potential \u03a6 at the two end point squares of the sensor. The gradient is numerically approximated as Ez(x, y, z, t) = \u03a6(+) \u2212 \u03a6(\u2212) \u0394z . (12.11) We note again that the sensors only involve mutual couplings. We have to establish a coupling mechanism from all the capacitive cells in the PEEC model to evaluate the potentials \u03a6(+) and \u03a6(\u2212) at the two end points. To accomplish this, we have to compute the coupled potentials based on Section 6.3.4 for the m capacitive cells in the problem at hand: \u03a6(+) = \u2211 m Ppks+ Ppkk \u03a6(t \u2212 \ud835\udf0fk,s+) (12"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure3.58-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure3.58-1.png",
        "caption": "FIGURE 3.58. Friction ellipse.",
        "texts": [
            " However, when the parameters A, B, C, D, C1, and C2 are determined for a tire, the equation expresses the tire behavior well enough. Figure 3.57 illustrates how the parameters can be determined from a test force-slip experimental result. Example 117 Friction ellipse. When the tire is under both longitudinal and sideslips, the tire is under combined slip. The shear force on the tireprint of a tire under a combined slip can approximately be found using a friction ellipse model.\u00b5 Fy FyM \u00b62 + \u00b5 Fx FxM \u00b62 = 1 (3.187) A friction ellipse is shown in Figure 3.58. Proof. The shear force Fshear, applied on the tire at tireprint, parallel to the ground surface, has two components: the longitudinal force Fx and the lateral force Fy. Fshear = Fx \u0131\u0302+ Fy j\u0302 (3.188) Fx = Cs sFz (3.189) Fy = \u2212C\u03b1 \u03b1 (3.190) 156 3. Tire Dynamics These forces cannot exceed their maximum values FyM and FxM . FyM = \u03bcy Fz FxM = \u03bcx Fz The tire shown in Figure 3.58 is moving along the velocity vector v at a sideslip angle \u03b1. The x-axis indicates the tire-plane. When there is no sideslip, the maximum longitudinal force is FxM = \u03bcx Fz = \u2212\u2192 OA. Now, if a sideslip angle \u03b1 is applied, a lateral force Fy = \u2212\u2212\u2192 OE is generated, and the longitudinal force reduces to Fx = \u2212\u2212\u2192 OB. The maximum lateral force would be FyM = \u03bcy Fz = \u2212\u2212\u2192 OD when there is no longitudinal slip. In presence of the longitudinal and lateral forces, we may assume that the tip point of the maximum shear force vector is on the following friction ellipse: \u00b5 Fy FyM \u00b62 + \u00b5 Fx FxM \u00b62 = 1 (3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure3.4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure3.4-1.png",
        "caption": "Fig. 3.4 Ground Reaction Force of 2D Model",
        "texts": [
            " On the other hand, when a human moves dynamically as shown in Fig. 3.3(b), the ground projection of CoM may exist outside the support polygon. However, the ZMP never exists outside the support polygon. In the following, we will explain the reason why the ZMP is always included in the support polygon. 1 ZMP in 2D In Fig. 3.1, although only the vertical component of the ground reaction force is shown, the horizontal component of it also exists due to the friction between the ground and the soles of the feet. In Fig. 3.4(a) and (b), we separately show the vertical component\u03c1(\u03be) and the horizontal component \u03c3(\u03be) of the ground reaction force per unit length of the sole. These forces simultaneously act on a humanoid robot. 72 3 ZMP and Dynamics Let us replace the forces distributed over the sole by a equivalent force and moment at a certain point in the sole. The force (fx and fz) and the moment (\u03c4(px)) at the point px in the sole can be expressed by fx = \u222b x2 x1 \u03c3(\u03be)d\u03be (3.1) fz = \u222b x2 x1 \u03c1(\u03be)d\u03be (3.2) \u03c4(px) = \u2212 \u222b x2 x1 (\u03be \u2212 px)\u03c1(\u03be)d\u03be"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure2.7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure2.7-1.png",
        "caption": "Fig. 2.7 A Rotating Cylinder. We define its angular velocity vector to be [0 0 1]T [rad/s].",
        "texts": [
            " If we calculate product of the transpose RT and R we get, RTR = \u23a1 \u23a3 eTx eTy eTz \u23a4 \u23a6 [ex ey ez] = \u23a1 \u23a3 eTx ex eTx ey eTx ez eTy ex eTy ey eTy ez eTz ex eTz ey eTz ez \u23a4 \u23a6 = \u23a1 \u23a3 1 0 0 0 1 0 0 0 1 \u23a4 \u23a6 = E (3\u00d73 identity matrix). From this we can safely say that RTR = E. If we multiply both sides of this equation by R\u22121 from the right, RT = R\u22121. (2.15) Therefore, when we transpose a rotation matrix we get its inverse. Matrices which have this characteristic are called orthogonal matrices. 2.2 Characteristics of Rotational Motion 29 Next we will define a way to notate rotational speed in three dimensional space. The simplest example is shown in Fig. 2.7. This figure shows a cylinder rotating around the z axis at 1 [rad/s]. In this case the rotational velocity of this object, \u03c9, can be described by \u03c9 = \u23a1 \u23a3 0 0 1 \u23a4 \u23a6 . (2.16) This is called an Angular Velocity Vector. Each element in this vector has the unit of [rad/s]. An Angular Velocity Vector(\u03c9) has the following characteristics. 1 \u03c9 is defined as a unit vector \u00d7 scalar value Let a be a unit vector on the rotating axis and q\u0307 be the rotational speed (a scalar value). Then the angular velocity vector of the object is their product \u03c9 = aq\u0307"
        ],
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    },
    {
        "image_filename": "designv10_0_0003443_0278364903022001005-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003443_0278364903022001005-Figure3-1.png",
        "caption": "Fig. 3. Tsai\u2019s mechanism.",
        "texts": [
            " The three constraint couples perpendicular to the base plane form a common constraint and \u03bb = 1. The other three constraint couples are coplanar and only two of them are linearly independent, thus forming a 2-system. Hence, one redundant constraint exists and \u03bd = 1. Accordingly, there are three linearly independent constraint couples, thereby constraining three rotational DoFs. The mechanism has three translational freedoms. Using eq (4), we have M = 5(11 \u2212 12 \u2212 1) + 12 + 1 = 3. (10) Similarly, for Tsai\u2019s mechanism as shown in Figure 3, the four-bar parallelogram in each limb with four revolute joints in each vertex corresponds to a 1-DoF generalized translational pair. Thus, the limb twist system in the general non-singular configuration is given by at OhioLink on October 23, 2014ijr.sagepub.comDownloaded from at OhioLink on October 23, 2014ijr.sagepub.comDownloaded from $i1 = (1 0 0 ; 0 0 0) $i2 = (1 0 0 ; 0 b2 c2) $g i3 = (0 0 0 ; l3 m3 n3) $i4 = (1 0 0 ; 0 b4 c4). (11) The limb constraint system is $r l1 = (0 0 0 ; 0 1 0) $r l2 = (0 0 0 ; 0 0 1)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002360_j.ijheatmasstransfer.2019.05.003-Figure15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002360_j.ijheatmasstransfer.2019.05.003-Figure15-1.png",
        "caption": "Fig. 15. Thermal field and melt region during the scanning of the third layer of the powder.",
        "texts": [
            " the melting front), as shown in the pressure contours in Fig. 14 in the x-z plane. This high-pressure front formed ahead of the laser beam drives the hot fluid (metal) backwards to the already molten zones (see Fig. 13(a)), thus widening the melt track. As shown in Fig. 14, the coexistence of a high pressure zone in front of the melt pool and a low-pressure one at its rear, forms a local pressure gradient which serves as a driver of hot metal flow back to the relatively colder melted regions. The temperature field for the third layer is also shown in Fig. 15 for different times. According to Fig. 15, the bulk material (the combination of two sintered layers) has an average temperature of around 950 K, which is due to the stored energy gained from the scanning of the previous layers. As a result, while the hot second layer and the cold powder layer come into contact with each other, the powder starts to get warmed up and since it has a lower mass and consequently lower heat capacity, compared to the bulk of the previous solidified layer, its temperature changes rapidly. Now that the third layer has higher initial temperature, higher temperatures are formed which consequently leads to the formation of deeper depression zones"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure12.47-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure12.47-1.png",
        "caption": "FIGURE 12.47. An oscillating platform hung from point A.",
        "texts": [
            " Assume that the spring has no tension or compression. When the system rests on the ground as shown in Figure 12.46(b), the spring is compressed by a static deflection \u03b4s = mg/k because of gravity. We may determine the natural frequency of the system by measuring \u03b4s \u03c9n = r g \u03b4s (12.465) because \u03b4s = mg k = g \u03c92n . (12.466) Example 464 Moments of inertia determination. Mass moments of inertia are important characteristics of a vehicle that affect its dynamic behavior. The main moments of inertia Ix, Iy, and Iz can be calculated by an experiment. Figure 12.47 illustrates an oscillating platform hung from point A. Assume the platform has a mass M and a moment of inertia I0 about the pivot point A. Ignoring the mass of cables, we can write the Euler equation about point A X My = I0 \u03b8\u0308 = \u2212Mgh1 sin \u03b8 (12.467) 802 12. Applied Vibrations and derive the equation of motion. I0 \u03b8\u0308 +Mgh1 sin \u03b8 = 0 (12.468) If the angle of oscillation \u03b8 is very small, then sin \u03b8 \u2248 \u03b8 and therefore, Equation (12.468) reduces to a linear equation \u03b8\u0308 + \u03c92n \u03b8 = 0 (12.469) \u03c9n = r Mgh1 I0 (12"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure6.25-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure6.25-1.png",
        "caption": "FIGURE 6.25. A 3-link mechanism with the ground as link number 1, and two moving links, numbers 2 and 3.",
        "texts": [
            " Applied Mechanisms 347 if their magnitudes are known, the instantaneous center of rotation I is at the intersection of AB with the line joining the extremities of the velocity vectors. Such a situation is shown in Figure 6.24(b). There is an instant center of rotation between every two links moving with respect to each other. The instant center is a point common to both bodies that has the same velocity in each body coordinate frame. The three instant centers , I12, I23, and I13 between three links numbered 1, 2, and 3 lie on a straight line. This statement is called the Kennedy theorem for three instant centers. Proof. Consider the two bodies shown in Figure 6.25. The ground is link number 1, links number 2 and 3 are pivoted to the ground at points M and N , and are rotating with angular velocities \u03c92 and \u03c93. The two links are contacted at point C. The revolute joint at M is the instant center I12 and the revolute joint at N is the instant center I13. The velocity of point C as a point of link 2 is vC2 , perpendicular to the radius MC. Similarly, the velocity of point C as a point of link 3 is vC2 , perpendicular to the radius NC. The instant center of rotation I13 must be a common point with the same velocity in both bodies"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003908_978-1-84882-730-1-Figure14.8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003908_978-1-84882-730-1-Figure14.8-1.png",
        "caption": "Figure 14.8 The VTOL aircraft",
        "texts": [
            "1 Control Objective A scaled mathematical model of a VTOL aircraft can be described as, see [153]: Px1 D x2; Px2 D u1 sin. /C \"u2 cos. /; Py1 D y2; Py2 D u1 cos. /C \"u2 sin. / g; P D !; P! D u2; (14.92) where x1, y1, and denote position of the aircraft center of mass and roll angle, x2, y2, and ! denote linear and roll angular velocities of the aircraft, respectively, u1 and u2 are the vertical control force and rotational moment, g >0 is the gravitational acceleration, and \" is the constant coupling between the roll moment and the lateral force, see Figure 14.8. The aircraft model (14.92) is underactuated if we want to use two available control inputs (u1 and u2) to control three outputs (x1, y1, and ). Furthermore, the 14.2 Vertical Take-off and Landing Aircraft 369 zero-dynamics of the aircraft model (14.92) is nonminimum phase for \"\u00a4 0 at the steady state when considering .x1;y1/ as the output and as an internal state. This phenomenon can be seen from (14.92) by setting x1 D y1 D x2 D y2 D 0. We assume that the reference trajectory to be tracked is generated by Px1r D x2r ; Px2r D u1r sin"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000554_978-3-540-85629-0-Figure8.12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000554_978-3-540-85629-0-Figure8.12-1.png",
        "caption": "Fig. 8.12. Definition of a Frenet frame on a parametric curve.",
        "texts": [],
        "surrounding_texts": [
            "8.2 Orientation of the Tool 353\naxes9 \u23a1 \u23a3wx\nwy wz\n\u23a4 \u23a6 = \u23a1 \u23a30 0.05 0.05 \u22120.77\n1 \u22121.10 0.32 \u22120.62 0 \u22120.99 \u22120.94 0.10\n\u23a4 \u23a6\nand by the angles [\u03b8t] = [0.90, 0.95, 3.10, 0.85].\nIn each segment the axis is kept constant, while the angle varies from 0 to \u03b8tk by means of a 5-th degree polynomial function of u (the same independent variable of the B-spline for the position trajectory) which allows to set initial (for u = u\u0304k, being u\u0304k the time instant in which the point qk is crossed) and final (for u = u\u0304k+1) velocities and accelerations equal to zero. The components of the trajectory for the position and the orientation are shown in Fig. 8.11. The RPY angles corresponding to the local frame at each point of the trajectory are used. Note in particular that the profile of an angle is discontinuous since the RPY angles are defined in the range [\u2212\u03c0, \u03c0].\n8.2.2 Case of position and orientation coupled\nIn many tasks, positioning and orientation problems are coupled at the Cartesian-coordinate level. In this case, a technique can be applied that allows to specify the orientation of the end effector on the basis of the orientation of the path at a given point. As a matter of fact, if the parametric form of the (regular) curve to be tracked, eq. (8.1), is expressed in terms of the curvilinear coordinate s (which measures the arc length10)\n\u0393 : p = p(s), s \u2208 [0, l] (8.6)\nit is possible to define a coordinate frame directly tied to the curve, the Frenet frame, represented by three unit vectors:\n\u2022 The tangent unit vector et, lying on the line tangent to the curve and is oriented according to the positive direction induced on the curve by s. \u2022 The normal unit vector en, lying on the line passing through the point p, and orthogonal to et; the orientation of en is such that in a neighborhood of p the curve is completely on the side of en with respect to the plane passing through et and normal to en. \u2022 The binormal unit vector eb, defined in a such way that the three vectors (et,en,eb) form a right handed frame.\n9 For each pair of frames (Rk, Rk+1) the axis and the rotation angle are computed by means of (8.4) and (8.5).\n10 Therefore the parameter l in (8.6) represents the total length of the curve.",
            "354 8 Multidimensional Trajectories and Geometric Path Planning\nThe value of the Frenet vectors at a generic point p can be deduced from the expression of the curve \u0393 by means of simple relations:\net = dp\nds , en = 1\u2223\u2223\u2223\u2223d2p ds2 \u2223\u2223\u2223\u2223 d2p ds2 , eb = et \u00d7 en.\nNote that if the curve is characterized by the arc-length parameterization s and not by a generic parameter u, the tangent vector et has unit length. In those applications in which the tool must have a fixed orientation with respect to the motion direction, e.g. in arc welding, the Frenet vectors implicitly define such an orientation. It is therefore sufficient to define the position trajectory function to obtain in each point the orientation of the tool.\nExample 8.5 A helicoidal trajectory is shown in Fig. 8.13, with the associated Frenet frames. The trajectory is described by the parametric form\np =\n\u23a1 \u23a3 r cos(u)\nr sin(u) d u\n\u23a4 \u23a6 (8.7)\nwith u \u2208 [0, 4\u03c0], which leads to the frame11\nRF = [et,en,eb] =\n\u23a1 \u23a3\u2212c sin(u) \u2212 cos(u) l sin(u)\nc cos(u) \u2212 sin(u) \u2212l cos(u) l 0 c\n\u23a4 \u23a6 (8.8)\nwhere c = r\u221a r2 + d2 and l = d\u221a r2 + d2 .\n11 Since the parameter u is not the curvilinear coordinate, the associated (time\nvarying) Frenet frames are computed as et = dp/du |dp/du| , en = det/du |det/du| , eb = et\u00d7en.",
            "8.2 Orientation of the Tool 355\nExample 8.6 Figure 8.14 shows a circular trajectory (for the sake of simplicity a planar case is considered) with the tool frame oriented in a fixed manner with respect to the Frenet frame. By considering the parametric description of the curve, expressed by (8.10) with o\u2032 = 0 and R = I3, the Frenet frame is\nRF (u) = R\n\u23a1 \u23a3\u2212 sin(u) \u2212 cos(u) 0\ncos(u) \u2212 sin(u) 0 0 0 1\n\u23a4 \u23a6 .\nThe desired tool frame differs from the Frenet frame by a rotation of \u03b1 = 30o about the zF = eb axis. Such a rotation can be represented by the constant matrix\nR\u03b1 =\n\u23a1 \u23a3 cos(\u03b1) \u2212 sin(\u03b1) 0\nsin(\u03b1) cos(\u03b1) 0 0 0 1\n\u23a4 \u23a6 = \u23a1 \u23a30.866 \u22120.500 0\n0.500 0.866 0 0 0 1\n\u23a4 \u23a6 .\nTherefore, the matrix representing the orientation in each point of the position trajectory is obtained simply by pre-multiplying the Frenet matrix by R\u03b1\nRT (u) = R\u03b1RF (u).\nThe tool frames computed for u = k\u03c0/4, k = 0, . . . , 8 are shown in Fig. 8.14."
        ]
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure5.16-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure5.16-1.png",
        "caption": "Fig. 5.16 Model of Humanoid Robot",
        "texts": [],
        "surrounding_texts": [
            "When you have knowledge of accurate physical parameters of the Humanoid Robot, you can generate full body motion for the robot by specifying the amount of change in translational and angular momentum and using this as a basis for calculating the joint angle speeds. This is called Split Momentum Control [108]. First you model the humanoid robot using a six degree of freedom body link that has four open links, which is shown in 5.16. By doing this you can use the following equations to calculate the goal speeds vref i , goal angular velocities \u03c9ref i body coordinates \u03a3B velocity vtrg B , angular velocity \u03c9trg B When given the the goal velocity of the end of the limbs vref i and goal angular velocity of the end of the limbs \u03c9ref i in limb end coordinates \u03a3i, defined within the world coordinate \u03a3W which is fixed to the floor, and the target velocity vtrg B and angular velocity \u03c9trg B in the body local coordinate \u03a3B, the goal joint angles of the ith limb, \u03b8\u0307 ref i can be calculated using the following. \u03b8\u0307 ref i = J\u22121 i {[ vref i \u03c9ref i ] \u2212 ( E \u2212r\u0302B\u2192i 0 E )[ vtrg B \u03c9trg B ]} (5.1) Here, J\u22121 i is the general inverse matrix of the Jacobian of the ith limb. E is a 3\u00d7 3 unit vector, rB\u2192i is the vector which starts from the origin of the body coordinates and ends at the origin of the operational coordinate \u03a3i. 5.4 Remote Operation of Humanoid Robots 173 The operator\u02c6is the skew symmetric matrix which is equivalent to the cross product. The vref i and \u03c9ref i in eq. 5.1 is defined by the remote operation to be done. For example, if the operator wants to make a translational movement with the wrist the operator uses the levers of the remote controller to openly specify the exact amount of movement to set to the translational velocity vector vref 1 . During this operation all the elements of the rotational velocity vector of the right hand, \u03c9ref 1 are set to zero together with the rotational and translational velocity vectors of all the other limbs, vref i , \u03c9ref i 6. vtrg B and \u03c9trg B can be calculated using split momentum control using the following equations. They rely on the humanoid robot\u2019s goal translational momentum Pref , goal angular momentum around it\u2019s center of gravity Lref and vref B , \u03c9ref B which are calculated by autonomously by the robot or specified by the remote operator. [ vtrg B \u03c9trg B ] = A\u2020S {[Pref Lref ] \u2212 4\u2211 i=1 ( M i H i ) J\u22121 i [ vref i \u03c9ref i ]} +(E6 \u2212A\u2020A) [ vref B \u03c9ref B ] (5.2) C A \u2261 S {( ME \u2212M r\u0302B\u2192c 0 I ) \u2212 4\u2211 i=1 ( M i Hi ) J\u22121 i ( E \u2212r\u0302B\u2192i 0 E )} M i and Hi are inertia matrices that indicates the effect induced by the total translational momentum and angular momentum of the robot to the ith link. M is the robot\u2019s total mass and I is the inertia tensor around the robot\u2019s center of gravity, rB\u2192C is a potential vector based in the robot\u2019s body coordinate system \u03a3B. E6 is a 6 \u00d7 6 unit matrix and \u2020 denotes a pseudo inverse matrix. S, which gives you to the moment element that you want to control, is a n\u00d7 6 selection matrix (0 \u2264 n \u2264 6) that is described as follows: S \u2261 [eS1 . . . eSn ]T Here, eSi is a 6\u00d7 1 column vector which has the elements that correspond to the moment elements you want to control set to 1 and the rest to 07. 6 Within the split momentum control framework it is possible to move the remaining limbs to satisfy the required motion. However, as this usually leads to unpredictable behaviour from the robot it is common to keep the remaining limbs still. 7 If you set this to control all moment elements A\u2020 will not have an area set to zero so vref B and \u03c9ref B cannot be controlled by equations 5.1 and 5.2. 174 5 Generation of Whole Body Motion Patterns In eq. 5.2, vRef I and \u03c9Ref I are specified by the remote operator. PRef , LRef , vRef B and \u03c9Ref B can be autonomously calculated by the robot\u2019s control system. For instance, PRef and LRef can be used to keep the robot balanced. If we define a coordinate system \u03a3F as being parallel to the World coordinate system \u03a3W with it\u2019s origin at the center of the supporting polygon of the robot, the relationship between the total translational momentum P and the translational velocity in the \u03a3F coordinate system r\u0307F\u2192c can be described as follows: P = M r\u0307F\u2192c (5.3) Therefore by calculating the the goal translational momentum Pref using the next equation, the center of gravity rF\u2192c can be moved to a given position rref F\u2192c. Pref = Mk(rref F\u2192c \u2212 rF\u2192c) (5.4) The k above is a preset gain. By adding equation 5.4 to the autonomous controller within the robot, the operator can specify motion while the robot keeps it\u2019s balance without the operator having to think about the difference between kinematics and dynamics8."
        ]
    },
    {
        "image_filename": "designv10_0_0000206_978-0-387-74315-8-Figure2-2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000206_978-0-387-74315-8-Figure2-2-1.png",
        "caption": "Figure 2-2: Definition of a line vector",
        "texts": [
            " A line vector has magnitude and direction, and is constrained to lie in a 16 definite line (i.e., a line in a particular position). Line vectors arise naturally in mechanics to represent quantities like angular velocity, where there is a definite axis of rotation, and forces acting on a rigid body, which have a definite line of action but may act at any point along that line. quantities like linear velocity and couple which have no definite line or axis associated with them. magnitude and direction given by the vector a and its line of action passes through the point A (see Figure 2-2). Since the direction of the line is the same as that of the line vector, we may say that the vector has magnitude and a line of action. Its magnitude is that of a, and the equation of the line of action is given by -+ ( r - OA ) X a = O , where r is the position of any point on the line. This equation IS more normally stated in the form 17 r X a= ao ' (2.2) ....... where ao = OA X a. a and a o are known as the PI~cker vectors of the line and determine it completely. a is called the resultant vector and a O the moment vector"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure4.14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure4.14-1.png",
        "caption": "Figure 4.14 (a) The bar support has (b) two degrees of freedom (a rotation and a displacement perpendicular to the bar) and gives (c) one support reaction (a force in the direction of the bar).",
        "texts": [
            " The free (freely adjustable) motions are called the degrees of freedom at the support; the free (freely adjustable) force is the support reaction. A bar support therefore has two degrees of freedom and generates one support reaction. If a motion is prescribed, the associated force is unknown, and vice versa. This is true not only for bar supports but also for all other supports discussed below. The total number of degrees of freedom and support reactions is therefore always three for a support (in a plane). In Figure 4.14 the degrees of freedom and support reactions are shown. Sometimes, motions are depicted by means of open arrows, while forces are depicted by closed arrows. Figure 4.15 is a schematic representation of a roller support. For a roller support at A, the body can move parallel to the so-called roller track, and can also rotate freely about A. Only motion of A perpendicular to the roller track is prevented; this is the direction in which the interaction force is exerted. For the roller support in Figure 4"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002401_j.ymssp.2017.12.008-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002401_j.ymssp.2017.12.008-Figure1-1.png",
        "caption": "Fig. 1. A two-stage compound planetary gearbox set: (a) first stage and (b) second stage.",
        "texts": [
            " The organization of this paper is as follows: Section 2 presents the studied planetary gearbox; Section 3 describes the fundamentals of AMMFmethod; Section 4 gives the detailed steps of MHPE method and validates the superiority of MHPE using simulation signals; Section 5 introduces the LS for the feature ranking. Section 6 illustrates the procedures of the proposed feature extraction method based on AMMF and MHPE. Section 7 discusses experimental test results obtained using the proposed method. Finally, conclusions are drawn in Section 8. In this paper, we conduct our research on a two-stage compound planetary gearbox as shown in Fig. 1. To further show its special characteristics, Fig. 2 illustrates the engineering drawing and the transmission mechanism of the two-stage planetary gearbox. We can find that the gear set consists of two gear stages, which are connected via a common carrier G. The ring gears are made into one part denoted as C. The first gear stage is a basic planetary gear set with three planets denoted by B. Each planet is in mesh with the sun gear A and the left part of the ring gear C. The second gear stage is a meshedplanet planetary gear set and has three planet-planet mesh pairs E-F"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000531_tpas.1965.4766135-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000531_tpas.1965.4766135-Figure1-1.png",
        "caption": "Fig. 1. A 2-pole 2-phase symmetrical machine. The flux-linkage equa-tion can be written",
        "texts": [
            " Krause is with the University of Wisconsin, Madison, Wisc. press the mutual coupling between it and an equivalenlt C.iH.aThomasWisc wihteAlsCamr'MnfcuigCmay rotor coil as a vsinusoidal function of the angular displace- 10)38 KRAUSE AND THOMAS: SIMULATION OF SYMMETRICAL MACHINES 1039 ment between their magnetic axes. If the induction ma- chine has either a squirrel-cage rotor or a coil-wound rotor br-axis bs-axis ~~~with the same number of phases as the stator, the rotor can be considered as havinig equivalent coils (shown in Fig. 1). The modifications, which are necessary to include ar-axis a double-cage rotor or a rotor wound with a different number of phases than the stator, are straightforwvard and will br Gr ~~~~~not be considered in this development. bs ROTOR- - as-axis The stator windings are identical, i.e., both windings ROTATION have an identical niumber of effective turns ATs, identical resistance r5,, identical leakage inductance L,,, and identical or ~~~~~~~self-inductance Ls. Similarly, equivalent rotor windings are identical which have the same effective turnis N7, resiss ~~~~~~~tance r,, leakage inductance Ll,, and self-inductance Lr. [Xas] IXbsI JXarJ br-axis as' [ s 0 L87 cos Or -Lsr Sin Or Fiasi L,co L,7si Lsr Sin Or cos1Or b bi or-axis L -Lrsn Or Lsr CoOS 1r0L LibrJ r ROTATION ~~er(5 STATOR ROTOR as9:..~ where L,r iS the amplitude of the mutual inductance be- cr ~~~~~tween stator and rotor windinigs and 0, is the angular disbr s ~~~~~~~~placement between the stator and rotor axes (Fig. 1). b s or ~~~~~~~~~~Inmost cases, the stator and rotor of a 3-phase induction machine are connected as 3-wire systems. In the case of s ~~~~~~~asquirrel-cage machine, the rotor windings can be considered equivalent to a 3-wire system. A symmetrical 3- cs- axisr-OX'iS ~~~~phase machine of this type is shown in Fig. 2. Four-wire 1bs i connections will not be considered. The line-to-neutral stator voltages are + N5 Nr r br Vas =- PXas + iasrs (6) N Ns r~ r-PS+'bl 1040 IEEE TRANSACTIONS ON POWER APPARATUS AND SYSTEMS NOVEMBER Since the stator and the rotor are 3-wire systems, the flux- in which linkage equations are given in (12)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003605_027836499601500604-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003605_027836499601500604-Figure6-1.png",
        "caption": "Fig. 6. Excavator with calibrator.",
        "texts": [
            " Circle Point Analysis The main screw axis measurement method has been circle point analysis (CPA), where each joint is moved individually in a circle (Mooring et al. 1991). The normal of the plane in which the circle lies and the center of the circle define the joint screw. In the past, position measurements of a point on the end effector have been used to define each circle (Stone 1987). In biomechanics, CPA is often performed to find the instantaneous center of rotation of a joint (Woltring et al. 1985). In Khoshzaban et al. (1992), a teleoperated log-loader with four unsensed joints (joints 1--~ in Figure 6) was at University of Waikato Library on July 13, 2014ijr.sagepub.comDownloaded from 587 calibrated using an additional 7-DOF linkage termed a calibrator. Joints 5-8 of the calibrator are sensed, where joint 5 is attached to the grapple; the ball-joint attachment to the base is a 3-DOF unsensed joint. The log-loader arm joints are moved one at a time for calibration. The authors were successful in formulating a recursive closedloop procedure to effect the calibration, working from distal to proximal log-loader joints"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure2.22-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure2.22-1.png",
        "caption": "FIGURE 2.22. A cresting car over a circular hill.",
        "texts": [
            "302) we can find the critical minimum speed vc to start flying. There are two critical speeds vc1 and vc2 for losing the contact of the front and rear wheels respectively. vc1 = s 2gRH \u00b5 h l sin\u03c6+ 1 2 cos\u03c6 \u00b6 (2.303) vc2 = s \u22122gRH \u00b5 h l sin\u03c6\u2212 1 2 cos\u03c6 \u00b6 (2.304) For any car, the critical speeds vc1 and vc2 are functions of the hill\u2019s radius of curvature RH and the angular position on the hill, indicated by \u03c6. The angle \u03c6 cannot be out of the tilting angles given by Equation (2.141). \u2212a1 h \u2264 tan\u03c6 \u2264 a2 h (2.305) Figure 2.22 illustrates a cresting car over a circular hill, and Figure 2.23 depicts the critical speeds vc1 and vc2 at a different angle \u03c6 for \u22121.371 rad \u2264 \u03c6 \u2264 1.371 rad. The specifications of the car and the hill are: l = 2272mm h = 230mm a1 = a2 a = 0m/ s2 RH = 100m. At the maximum uphill slope \u03c6 = 1.371 rad \u2248 78.5 deg, the front wheels can leave the ground at zero speed while the rear wheels are on the ground. When the car moves over the hill and reaches the maximum downhill slope \u03c6 = \u22121.371 rad \u2248 \u221278.5 deg the rear wheels can leave the ground at zero speed while the front wheels are on the ground"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-FigureC.11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-FigureC.11-1.png",
        "caption": "FIGURE C.11 Zero thickness conductors example.",
        "texts": [
            "5 with numerical integration along the thickness leads to more stable results. C.1.9 1\u2215R3 Kernel Integral for Parallel Rectangular Sheets Partial elements are of importance for other formulation such as the ones in Chapter 11 so that other material properties can be taken into account. In some of these formulations, integrals with a higher order Green\u2019s function need to be solved. Specifically, the integral for a 1\u2215R3 kernel is important. Here, we consider the geometry to be two rectangles shown in Fig. C.11. 396 COMPUTATION OF PARTIAL INDUCTANCES Ip12 = \u222b xe1 xs1 \u222b ye1 ys1 \u222b xe2 xs2 \u222b ye2 ys2 (z1 \u2212 z2) R3 dy2 dx2 dy1 dx1, (C.42) where, R3 = [(x1 \u2212 x2)2 + (y1 \u2212 y2)2 + (z1 \u2212 z2)2]3\u22152. Then the final formulation of Ip12 for two orthogonal current sheets is Ip12 = 2\u2211 m=1 2\u2211 n=1 2\u2211 i=1 2\u2211 j=1 (\u22121)m+n+i+j(z1 \u2212 z2) \u23a7\u23aa\u23a8\u23aa\u23a9 \u2212 Rmnij + (xvm \u2212 xvn) ln[(xvm \u2212 xvn) + Rmnij] + (xvm \u2212 xvn)(yvi \u2212 yvj) tan\u22121 [ (xvm \u2212 xvn)(z1 \u2212 z2) (yvi \u2212 yvj)2 + (z1 \u2212 z2)2 + (yvi \u2212 yvj)Rmnij ] + (xvm \u2212 xvn)(yvi \u2212 yvj) tan\u22121 [ z1 \u2212 z2 xm \u2212 xn ] + (z1 \u2212 z2)2(yvi \u2212 yvj) \u22c5 ln \u23a1\u23a2\u23a2\u23a2\u23a3 \u221a ((yvi \u2212 yvj)2 + (z1 \u2212 z2)2 + Rmnij(yvi \u2212 yvj))2 + (xvm \u2212 xvn)2(z1 \u2212 z2)2\u221a (xvm \u2212 xvn)2 + (z1 \u2212 z2)2 \u23a4\u23a5\u23a5\u23a5\u23a6 \u23ab\u23aa\u23ac\u23aa\u23ad (C"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure13.10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure13.10-1.png",
        "caption": "Figure 13.10 (a) Simply supported beam with total length and overhang of length a. (b) If the maximum bending moment at E is equal to the bending moment at B, the hatched areas in the V diagram are also equal.",
        "texts": [
            "3 23.3 + 26.2 \u00d7 (16.5 m) = 7.77 m. The bending moment at D can be found from the moment equilibrium of the isolated segment AD or, as shown below, from the area of the V diagram for AD: Mmax = 1 2 (7.77 m)(23.3 kN) = 90.5 kNm. Another extreme bending moment is the support moment1 at B. Note that this moment can be found from the area of the V diagram for segment BC (see Figure 13.9d): Mmin = 1 2 (4 m)(12 kN) = 24 kNm. c. Let the total length of the pile be and the length of the overhang be a (see Figure 13.10a). Figure 13.10b shows a sketch of the V diagram. The shear force to the right of B is equal to qa. The slope of the V diagram is the same everywhere. The extreme bending moments occur at E and B. The bending moment at B is equal to the hatched area of the V diagram between B and C: MB = 1 2qa2. The bending moment at E is equal to the hatched area of the V diagram between A and E. The bending moment in the pile is least when the extreme bending moments at E and B are equal: ME = MB = 1 2qa2. 1 A support moment is the bending moment in the beam at a support"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure11.7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure11.7-1.png",
        "caption": "FIGURE 11.7 Interface between the two regions with different permeability.",
        "texts": [
            " In this case, the solution for \u03a6m becomes \u03a6m(r) = 1 4\ud835\udf0b\u222b \u2032 qvm(r\u2032)|r \u2212 r\u2032| d \u2032 + 1 4\ud835\udf0b\u222b\u2032 qsm(r\u2032)|r \u2212 r\u2032| d \u2032. (11.38) For the case of interest where we assume that the magnetization is uniform inside  , the first term in (11.38) is zero being qvm = 0 and only the contribution of surface magnetic charge qsm holds. We get the magnetic field H(r) = \u2212\u2207\u03a6m(r), from (11.38) H(r) = 1 4\ud835\udf0b\u222b\u2032 qsm(r\u2032) (r \u2212 r\u2032)|r \u2212 r\u2032|3 d \u2032 + Hinc(r), (11.39) where Hinc(r) is a potential externally applied magnetic field. We show the interface between two materials with a different permeability in Fig. 11.7. The continuity of the magnetic field across the interface yields \ud835\udf072 Hn2 \u2212 \ud835\udf071 Hn1 = 0. (11.40) The unit vectors n\u03022 and n\u03021 are normal to the interface between the two materials with the permeability \ud835\udf072 and \ud835\udf071. The surface S is assumed to be sufficiently smooth. We subdivide 294 PEEC MODELS FOR MAGNETIC MATERIAL the surface as shown in Fig. 11.7 into a small disk of radius \ud835\udf0cwith the center at the projection of r onto the surface S. The radius of the disk \ud835\udf0c\u2192 0 such that \ud835\udeff = o(\ud835\udf0c). Here, \ud835\udeff is the normal distance between the point r and the center of the disk. In the limit as \ud835\udeff \u2192 0, the total magnetic field on either side of the interface S is given by Hn(r) = \u00b1Hi(r) + 1 4\ud835\udf0b\u222bPV qsm(r\u2032) (r \u2212 r\u2032)|r \u2212 r\u2032|3 d \u2032 + Hinc, (11.41) where Hinc is the usual incident field. The contribution of the magnetic surface charge close to the surface is dominated by the local normal field in both directions given by Hi(r) = qsm\u22152"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure6.20-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure6.20-1.png",
        "caption": "Figure 6.20 (a) Part of a roof, modelled as a line element, is loaded by (b) its dead weight, (c) snow, and (d) wind. The resultant and support reactions due to (e) the dead weight, (f) snow load, and (g) wind load.",
        "texts": [
            "19d, it follows that R1 = 1 3 \u00d7 3 \u00d7 4 = 6 kN, R2 = 6 \u00d7 8 = 48 kN,\u2211 T |B = 0 \u21d2 Av = \u2212 5 6R1 + 3 6R2 + 1 6R1 = \u22125 + 24 + 1 = 20 kN,\u2211 T |A = 0 \u21d2 Bv = \u2212 1 6R1 + 3 6R2 + 5 6R1 = \u22121 + 24 + 5 = 28 kN. Irrespective of how the load diagram is split, the support reactions are always the same. Figure 6.19 A simply supported beam with the trapezoidal load split up in four different ways into triangles and/or rectangles to determine the support reactions. 6 Loads 225 Example 3 Part of a roof modelled as the line element in Figure 6.20a is supported by a hinge at A, and a roller with vertical roller track at B. The member is loaded by three uniformly distributed (line) loads, of which the load diagrams are shown in Figures 6.20b to 6.20d: \u2022 the dead weight qdw (vertical force per length measured along the member axis); \u2022 a snow load qsn (vertical force per horizontally measured length); \u2022 a wind load qw normal to the member axis (force per length measured along the member axis). Unlike dead weight and wind load, the snow load is given as a load per length projected on the horizontal ground plane. The load diagram for snow in Figure 6.20c is drawn differently therefore. Question: Determine the support reactions at A and B for all three loads. Solution: When calculating the support reactions, we can replace the distributed loads by their resultants. The dead weight and the wind load act over a length of 15a, while the snow load acts over a length of 12a, so that Rdw = 15aqdw, Rsn = 12aqsn, Rw = 15aqw. The resultants and their lines of action are shown in Figures 6.20e to 6.20g. The same figures also show the associated support reactions",
            " Therefore, when looking at the equilibrium of the hinged beam as a whole, we can use the resultant of the entire distributed load (see Figure 6.21b). For the directions assumed for the support reactions, this gives \u2211 F (ASD) x = \u2212Ah = 0 \u21d2 Ah = 0,\u2211 F (ASD) z = \u2212Av \u2212 Bv \u2212 Cv + (36 kN) = 0, (a) \u2211 T ((ASD) y |A = +Bv \u00d7 (4 m) + Cv \u00d7 (8 m) \u2212 (36 kN)(5 m) = 0. (b) The two equations (a) and (b) are not sufficient to determine all vertical support reactions. The additional equation required is found from the moFigure 6.20 (a) Part of a roof, modelled as a line element, is loaded by (b) its dead weight, (c) snow, and (d) wind. The resultant and support reactions due to (e) the dead weight, (f) snow load, and (g) wind load. The reader is asked to verify the correctness of the support reactions. 6 Loads 227 ment equilibrium of parts SA or SD about hinge S. In this case it is not possible to work with the resultant in Figure 6.21b; this resultant has to be replaced by the resultants of the distributed loads on the individual parts (see Figure 6"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure12.34-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure12.34-1.png",
        "caption": "FIGURE 12.34. Free body diagram of an eccentric excitated single-DOF system.",
        "texts": [
            "210) The absolute displacement responses of the system is x = A4 sin\u03c9t+B4 cos\u03c9t (12.211) = X sin (\u03c9t\u2212 \u03d5e) (12.212) which has an amplitude X, and phases \u03d5e X e\u03b5 = r2q (1\u2212 r2)2 + (2\u03ber)2 (12.213) \u03d5e = tan \u22121 2\u03ber 1\u2212 r2 (12.214) Phase \u03d5e indicates the angular lag of the response x with respect to the excitationmee\u03c9 2 sin\u03c9t. The frequency responses forX and \u03d5e as a function of r and \u03be are plotted in Figures 12.32 and 12.33. 770 12. Applied Vibrations 12. Applied Vibrations 771 Proof. Employing the free body diagram of the system, as shown in Figure 12.34, and applying Newton\u2019s method in the x-direction generate the equation of motion mx\u0308 = \u2212c x\u0307\u2212 kx+mee\u03c9 2 sin\u03c9t. (12.215) Equation (12.208) can be transformed to (12.209) by dividing over m and using the following definitions for natural frequency, damping ratio, and frequency ratio. \u03c9n = r k m (12.216) \u03be = c 2 \u221a km (12.217) r = \u03c9 \u03c9n (12.218) The parameter \u03b5 = me m is called the mass ratio and indicates the ratio between the eccentric mass me and the total mass m. The steady-state solution of Equations (12"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001467_j.mechmachtheory.2010.05.001-Figure11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001467_j.mechmachtheory.2010.05.001-Figure11-1.png",
        "caption": "Fig. 11. A 6-UPS parallel manipulator.",
        "texts": [
            ", \u03b7i=0) and the LTI vanishes, there exists a motion represented by $Oi, along which no power can be transmitted to the output. In other words, the output member of themanipulator cannotmove along the screw $Oi by driving its inputs. In this case, themanipulator is said to be singular, and this kind of singularity is referred to as output singularity. If both the input and output singularities happen at the same time, the manipulator is considered as being at its double singularity. A 6-DOF 6-UPS parallel manipulator shown in Fig. 11 is taken as an example. With respect to the global coordinate system R: O\u2212xyz attached to the base, the position of the center O\u2032 of the moving platform can be described as (xO\u2032, yO\u2032, zO\u2032), and the orientation of the moving platform can be denoted by the Tilt-and-Torsion (T&T) angles (\u03d5, \u03b8, \u03c3). Parameters r1 and r2 represent the radiuses of the base and the moving platform, respectively. The geometric relation between the passive joints can be expressed as jSiSi + 1j = jUiUi + 1j = r3 i = 1;3;5\u00f0 \u00de \u00f039\u00de According to the results in Section 3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001351_j.ijfatigue.2019.03.025-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001351_j.ijfatigue.2019.03.025-Figure4-1.png",
        "caption": "Fig. 4. Radial measurement locations on fatigue bar gage section.",
        "texts": [
            " The surface features from each method are then related to the fatigue life to create a relationship that can be used by process developers to improve fatigue performance. Measurements of the surfaces of all geometries were performed using a Keyence VR-3200 wide area 3D measurement macroscope, which uses angled SL scanning to capture the surface heights. For the fatigue bars, eight measurements were evenly spaced around the circumference of the gage section of each bar in order to capture the entire surface as seen in Fig. 4. Stitched line measurements with a length of 7mm at each radial location were used to capture the surface along the majority of the length of the gage section while avoiding the filleted regions on each end. Each measurement had a width of 0.5mm to avoid extreme curvature. The secant curvature correction was applied in the Keyence software to correct for the cylindrical shape of the fatigue bars. The raw height data was then exported to allow for calculation of the surface roughness metrics for all radial locations"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003841_08905458708905124-FigureI-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003841_08905458708905124-FigureI-1.png",
        "caption": "Fig. I Two degree-of-freedom relative coordinates.",
        "texts": [],
        "surrounding_texts": [
            "BAE AND HAUG\nI. INTRODUCTION\nAnalysis of multi-body systems requires formulation of mixed systems of differential equations of motion and nonlinear algebraic equations of constraint. Two basically different kinds of generalized coordinates can be used in deriving equations of motion. One is joint or relative coordinates between bodies and the other is Cartesian or absolute coordinates of each body. Even though Cartesian coordinates are quite general and treat open and closed loop systems in the same way, a maximal set of coordinates and associated kinematic constraint equations must be introduced. Relative coordinates are excellent for open loop systems, but require some extension to treat closed loops. As a way to combine these two approaches, a velocity transformation method was presented by Jerkovsky (11 and extended in several subsequent papers [2, 31, using linear relationships between Cartesian and relative coordinate velocities.\nA variational form of the equations of motion was used by Wittenburg [4] to reduce equations of motion to explicit form in terms of relative coordinates. A related variational-vector calculus approach was used in Ref. 5, where a two body subsystem was used to illustrate systematic derivation of variational (virtual work) equations of motion with Cartesian and relative coordinates. The basic approach in the present paper is to use a systematic reduction of equations of motion from one coordinate set to another, as in Ref. 5, but to exploit properties of the variational form of the equations of motion and treat more general classes of systems.\nIn robotics, Hollerbach and Luh [6, 71 developed an efficient recursive algorithm for inverse dynamics of open-loop manipulators. Walker and Orin [8] used this algorithm for inverse dynamics, as a basis for their procedures to calculate direct dynamics problems. Featherstone [9] recently presented a recursive algorithm to calculate acceleration of robot arms with revolute and translational joints, using screw notation.\nKinematic couplings between pairs of contiguous bodies, with one rotational and one translational degree of freedom, are treated in this paper. This formulation includes revolute, translational, and cylindrical joints. Open loop multibody systems; i.e., systems with tree graph structure are treated in this paper. A recursive algorithm for reducing the variational equations is developed, using a variational-vector calculus approach.",
            "RECURSIVE FORMULATION FOR MECHANICAL DYNAMICS\nII. RELATIVE COORDINATE KINEMATICS\nA pair of coupled bodies whose kinematics are defined in terms of Cartesian coordinates of one of the bodies and rotational and translational relative coordinates is presented in Ref. 5. Bodies a and c in Ref. 5 are denoted bodies i and j here, as shown in Fig. 1, as a more systematic notation for multi-body systems. Relative translational and rotational coordinates between bodies i and j are a,, and @;,. Vectors that locate joint attachment points in bodies i and j are denoted sjj and s,,, respectively. Unit vectors along translational and rotational axes of the joint, in the 6-y,y-z,!' joint definition frame, are denoted d,; and h,;, respectively. Similarly, orthogonal matrices Cij, Cji, and A,; are transformations from the \" frames to the centroidal ' frames on bodies i and j and from the \" frame on body j to the \" frame on body i, respectively.\nVectors in the global reference frame are used in this paper to express ve-",
            "362 BAE AND HAUG\nlocity, virtual displacement, and acceleration relations, in order to avoid appearance of many transformation matrices. In particular, from Fig. 1 ,\nwhere the translational unit vector djj and body fixed vector s,; in the global frame are obtained from their body-fixed constant values by the transformations\nwhere s> is a fixed vector in the 0; frame. Note from Eq. 4 that the orientation matrix for body j is\nThe angular velocity of body j is\nwhere\nh. = AiC..h!' 'I 'I (7)\nand h,; = [ O , 0, 1IT, as shown in Fig. 1. The time derivative of Eq. 1, using Eq. 2, is\nSince vectors s;, and dij are fixed in body i , their time derivatives are 1.51\nwhere the matrix-vector product notation used is defined by\nSimilarly,"
        ]
    },
    {
        "image_filename": "designv10_0_0003413_ac0010882-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003413_ac0010882-Figure3-1.png",
        "caption": "Figure 3. Cyclic voltammograms of 1 mM potassium hexacyanoferrate(III) in a 0.1 M KCl solution at (a) a PPD-, (b) an OPPD-modified CFME, (c) an overoxidized bare CFME, and (d, inset) an untreated bare CFME. Scan rate, 100 mV/s.",
        "texts": [
            " At low pHs (below 6), low OH- concentration apparently does not allow overoxidation to start, and at pHs that are too high (beyond 7.5), the degree of overoxidation is too high and, thus, appears to suppress the response signals to both of the analytes at the resulting OPPD film. Characterization of the OPPD Film with Respect to DA and AA Sensing. To demonstrate the difference in electrochemical behavior between the PPD and the new OPPD coating, CV responses to 1 mM potassium hexacyanoferrate(III) at four different electrode surfaces were recorded (Figure 3): (i) at a PPD-coated CFME, (ii) at an OPPD-modified CFME, (iii) at a bare CFME that was overoxidized using the same procedure as for preparing the OPPD film, and (iv) at an untreated bare CFME. The PPD-CFME with no signal at all (curve a) and the untreated bare CFME with a well-defined voltammogram (d, inset) clearly displayed their expected behavior, which is in agreement with the observations of other authors that the PPD, prepared under similar conditions, is a strongly adhering nonconducting polymer.35,39 Obviously, the examined redox species could not penetrate through the insulating PPD film to reach the carbon fiber surface. In contrast, the OPPD film shows a distinct current response (Figure 3, curve b). At the OPPD film, low reversibility behavior with a cathodic-to-anodic peak separation of \u223c120 mV is observed. The large difference in current responses that was obtained at the OPPD-CFME and the untreated bare CFME (curves b and d with a cathodic current ratio of \u223c1:20) is primarily due to a negative charge at the oxygen-rich OPPD surface repelling the ferricyanide anion. For comparison, a similarly suppressed but ill-defined response at an overoxidized bare CFME is also shown (curve c)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003794_robot.1990.126059-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003794_robot.1990.126059-Figure3-1.png",
        "caption": "Fig. 3: CEBOT model fo r deflection calculation.",
        "texts": [
            "3 Position Accuracy of Manipulator End Effector 5.3.1 Position Inaccuracies Caused by Gravity, Deflection and Torsion CEBOT consists of many connected cells. We consider t he e n t i r e manipulator a s a r ig id structure, take in account the manipulator weight and payload and the compliance of the coupling mechanism. With d i f f e ren t i a l d e f l e c t i o n and torsion a t the connecting mechanism we calculate t h e p o s i t i o n i n g e r r o r v e c t o r E d f o r t h e manipulator model shown in Fig. 3. The differential transformation vector for cel l i and the force/moment vector Fi are defined as: center of cell k of cel l k t o cel l i (T-matrix). where d i j : d i f f e ren t i a l t r a n s l a t i o n in j - aij : differential rotation about j-axis. f i j : force in j-direction. : moment about j-axis. We mkh rewrite equation (10) and (11) into direction. A i = Kil Fi (12) W e define the matrix A; which combines the (13) three matrices Bi. Ci and Di (see Fig. 4) as: A; = B i * C i * D i where Bi: transformation matrix from center point of c e l l (i-1) t o connecting s u r f a c e , both have t h e same orientation, Ci: differential transformation matrix from connecting surface of cell (i-1) t o next cel l i"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001056_s11465-015-0341-2-Figure11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001056_s11465-015-0341-2-Figure11-1.png",
        "caption": "Fig. 11 Parts fabricated by SLM in HUST-RMC",
        "texts": [
            "1 Manufacture of holistic structures for difficult-to-process materials Titanium alloys and nickel-based superalloys are widely applied in aerospace because of their high strength and ability to work at elevated temperature. However, these materials are difficult to be processed using casting and forging. SLM can avoid harsh forming conditions, such as extreme pressure and temperature. So far, SLM can fabricate titanium alloys (Ti-6Al-4V [70], Ti55 [79]) and nickel-based superalloys (Ni718 [80], Ni625 [81]). Figure 11 shows the parts fabricated in HUST-RMC. NASAMarshall Space Flight Center has manufactured the rocket nozzle parts of superalloys with holistic structure using SLM, as shown in Fig. 12. The fabrication only took approximately 40 hours instead of several months using traditional methods, thereby significantly saving time and cost. Moreover, ignition test was performed, and combustion temperature reached 3315 \u00b0C [82]. 4.2 Manufacture of part with complex cooling channels Similar to other additive manufacturing processes, SLM has the obvious advantage for fabricating complex structures compared with traditional processes"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure10.38-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure10.38-1.png",
        "caption": "Figure 10.38 Bending symbols.",
        "texts": [
            " In manual calculations, we will therefore use deformation symbols: the bending symbol for bending moments, and the shear symbol for shear forces. The bending symbol and shear symbol symbolise the deformation of the member axis due to a bending moment and a shear force respectively. These deformation symbols can be used to set the direction of the section forces unequivocally, regardless of a coordinate system. We always use the plus and minus sign for normal forces. The bending symbol and shear symbol will be explained in more detail below. 418 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Figure 10.38 (a) A small member segment subject to bending moments will lengthen at the tension side and shorten at the compression side. The member segment will bend. (b) The small arc that represents the deformation is used as deformation symbol for the bending moment and is known as the bending symbol. Figure 10.39 (a) In a member segment subject to shear forces, one sectional plane will try to shift with respect to the other. (b) This effect can be visualised by introducing an (imaginary) slide joint in the segment, so that both sectional planes can move with respect to one another. (c) The step formed by the moved member axes is used as the deformation symbol for shear forces and is known as the shear symbol. \u2022 Bending symbol (deformation symbol for bending moments) Figure 10.38a shows a small member segment subject to bending moments. The member segment will lengthen at the side being pulled, and shorten at the side being compressed. The member segment will bend. Since it is possible to determine the bending moment from the bent shape of the member axis, we use the small arc as deformation symbol for the bending moment (see Figure 10.38b). \u2022 Shear symbol (deformation symbol for shear forces) Figure 10.39a again shows a small member segment, but now with shear forces. When subject to shear forces, one sectional plane will try to shift with respect to the other. This effect can be visualised by applying an (imaginary) slide joint within the segment, so that both parts can move with respect to one another (see Figure 10.39b). The step change formed by the moved member axes is used as the deformation symbol for shear forces (see Figure 10"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure14.32-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure14.32-1.png",
        "caption": "Figure 14.32 Cables BC and CD isolated from tower C and foundation block D.",
        "texts": [
            " Solution: Since the load is a force per horizontally measured length, the cable shapes in the middle span and the end spans are parabolas. 14 Cables, Lines of Force and Structural Shapes 663 a. For middle span BC pBC k = (254 m) \u2212 (70 m) = 184 m. This gives H BC = 1 8q( BC)2 pBC k = 1 8 (250 kN/m)(1624 m)2 184 m = 448 MN. b. The vertical component of the tensile force in cable BC is a maximum at B and C, as the cable is steepest here: V BC max = 1 2q BC = 1 2 (250 kN/m)(1624 m) = 203 MN. The cable force is also a maximum here: NBC max = \u221a (H BC)2 + (V BC max) 2 = \u221a (448 MN)2 + (203 MN)2 = 492 MN. c. In Figure 14.32, cables BC and CD have been isolated at C from the tower. If there is no bending in the tower, the resulting horizontal force on the tower must be zero. This means that the horizontal component of the cable force in an end span is equal to that in the middle span: H CD = H BC = 448 MN. In Figure 14.33, cable CD has been isolated. The resultant R of the uniformly distributed load is R = (250 kN/m)(536 m) = 134 MN. Figure 14.34 The forces on tower C and foundation block D. The moment equilibrium of cable CD gives \u2211 T |D = (448 MN)(200 m) \u2212 V CD C (536 m) + (134 MN)(268 m) = 0 so that V CD C = (448 MN)(200 m) + (134 MN)(268 m) 536 m = 234 MN"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure7.23-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure7.23-1.png",
        "caption": "Fig. 7.23. Distance measurement with a laser",
        "texts": [
            " The application oflasers with appropriate optical detectors on robots is mainly directed to distance measurement which can be easily extended to 3D scanning of the working space. Lasers that are present on the market are not suitable for use on robots because of the sensitive and complex mechanics needed for directing the laser beam and the high price of the device. Nevertheless, the possibilities offered by lasers indicate a wide range of applications in future robotics. A simple technique for distance measurement using a laser is presented in Fig. 7.23. Laser source Semits a narrow beam perpendicularly to the axis of rotation of the mirror M on its surface. The mirror rotates with angular speed w, and the reflected beam, according to the laws of reflection, rotates with 2w. The lens L, 228 7 Sensors in Robotics focusing the beam on the detector D, rotates with 2w, and is synchronized with the mirror so that the angle rx between the line B and the reflected beam is equal to the angle between the optical axis of the lens L and line B in every time instant. The supposition is that the scanned object has the appropriate coef ficient of reflection such that the place of the beam incidence can be detected. According to Fig. 7.23, the distance R is given by relation: R=B/2tanrx The angle rx is calculated by measuring the elapsed time t between the path source- mirror-detector and the path source-mirror- object-detector. The speed of rotation of the reflected beam 2w implies: rx=2wt A detailed analysis shows what relations among w, Rand B result in some predefined accuracy, if the minimal measurable time delay is known. It follows from the second relation that the stability of the speed of rotation has negligible influence if it is stable at small time intervals, comparable to t"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000637_978-3-642-83006-8-Figure2.15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000637_978-3-642-83006-8-Figure2.15-1.png",
        "caption": "Fig. 2.15. The four-linked mechanism",
        "texts": [
            " in the case when J the driving torques Pi are known. ..i When an inverse problem is considered (with the known accelerations q and unknown torques Pi)' then, the system of 26 equations (2.5.23) con tains 32 unknown quantities (torques P and vector 0). In that case, equations (2.5.25) cannot be uniquely solved. One can prescribe 6 arbitrary torques of the first chain joints and, then, determine the other moments and & from (2.5.23). This can be easily explained. Let us consider the simplest four-linked mechanism (Fig. 2.15). only one value of the torque should be prescribed. The analogous case 86 also arises with the more complex kinematic chains, where it is neces* * sary to prescribe (m-n ) torques; m and n being the numbers of d.o.f. of the equivalent and closed chain, respectively. Since in the dynamics analysis we actually consider an open chain which is equivalent to the closed chain, it is possible to use basically the same procedure as for open chains, but with certain modifications. Thus, the algorithm should be supplied with an iterative procedure for cal culation of the position (see the flow-chart in Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003443_0278364903022001005-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003443_0278364903022001005-Figure6-1.png",
        "caption": "Fig. 6. 2R3T 5-DoF 5 \u2212 xRuPxyUN yR PM.",
        "texts": [
            " To meet this condition, the axes of revolute pairs most adjacent to the base in all limbs must be set parallel and so are the axes of the revolute pairs most adjacent to the moving platform. Because of this, the mechanism of this type is only limb-symmetrical. The characteristics of constraint and structure of such 5- DoF PMs are shown in Table A2 in the Appendix. Accordingly, we can obtain the limb kinematic chain by linear combination of the five twists in eq (17). The enumeration of such PMs is shown in Table 2. Consider a xRuPxRyRyR limb. Note that xRyR can form a universal joint xyUN. Figure 6 shows such a 5-DoF 5\u2212xRuPxyUN yR PM. The universal joint plane in the ith limb is denoted byU34, at OhioLink on October 23, 2014ijr.sagepub.comDownloaded from at OhioLink on October 23, 2014ijr.sagepub.comDownloaded from which is parallel to the base plane in the initial configuration. The geometrical arrangement of the five limbs guarantees that the five universal joint planes are always parallel. In this initial configuration, such a single xRuPxyUN yR limb exerts a constraint couple as that in eq (18) on the moving platform and restricts the rotation about the normal of U34"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure8.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure8.2-1.png",
        "caption": "Fig. 8.2. Exploded view of an AFPM machine: 1 \u2014 rotor disc, 2 \u2014 stator winding, 3 \u2014 PM, and 4 \u2014 epoxy core.",
        "texts": [
            "10) where Dout is the outer diameter of the rotor disc, the average Nusselt number is given by Nu = 0.133 Re2/3 D Pr1/3 (8.11) and Reynolds number at the disc periphery is ReD = \u2126D2 out/\u03bd (8.12) Note that a uniform temperature distribution in the cylinder is normally assumed when eqn (8.10) is used. Since h\u0304p is proportional to the angular speed \u2126, it may be concluded that the rotor periphery plays an increasingly important role in the heat dissipation as \u2126 increases. Rotor-stator system As seen in Fig. 8.2, an AFPM machine consists of a number of rotating and stationary discs. The heat transfer relations between a rotating and a stationary disc are of paramount importance in the thermal calculations. Due 256 8 Cooling and Heat Transfer to centrifugal effects, there is a forced flow between the two discs, which increases the local heat transfer rate compared with that of a free disc. The relative increase will depend on the gap ratio, G = g/R, where g is the clearance between the rotor and the stator and R is the radius of the disc, the mass flow rate and the rotational speed of the system [209]",
            " From a cooling perspective, AFPM machines may be classified into two categories as follows: \u2022 machines with self-ventilation, in which cooling air is generated by a rotating disc, PM channels or other fan-alike devices incorporated with the rotating part of the machine, and 8.3 Cooling of AFPM Machines 257 \u2022 machines with external ventilation, in which the cooling medium is circulated with the aid of external devices, for an example, a fan or a pump. The majority of AFPM machines are air-cooled. Compared with conventional electrical machines, a particularly advantageous feature of disc-type AFPM machines from a cooling perspective is that they possess inherent selfventilation capability. Fig. 8.2 shows the layout and active components of a typical AFPM machine. A close examination of the machine structure reveals that an air stream will be drawn through the air inlet holes into the machine and then forced outwards into the radial channel as the rotor discs rotate. The PMs in fact act as impeller blades. The fluid behaviour of the AFPM machine is much like that of a centrifugal fan or compressor. The Ideal Radial Channel According to the theory of an ideal impeller, a number of assumptions have to be made to establish the one-dimensional model of the ideal radial channel [78, 232]: (a) there are no tangential components in the flow through the channel; (b) the velocity variation across the width or depth of the channel is zero; (c) the inlet flow is radial, which means that air enters the impeller without pre-whirl; 258 8 Cooling and Heat Transfer (d) the pressure across the blades can be replaced by tangential forces acting on the fluid; (e) the flow is treated as incompressible and frictionless",
            "24)\u2013\u2206pfr shown in Fig. 8.7, is significantly higher than the experimental one. The shaded area in Fig. 8.7 represents the shock and leakage losses. It can be seen that at low flow rates the shock and leakage losses are greater but tend to zero at the maximum flow rate. This has been discussed and experimentally validated in [260]. The derived characteristics describes the pressure relations of a single rotating PM disc facing a stator. For a double-sided AFPM machine with two identical coaxial rotating discs (Fig. 8.2) operating on the same stator, the characteristic curve presented in Fig. 8.7 represents only half of the AFPM machine. The characteristic curve of the whole machine may be obtained by adding flow rate at the same pressure, which is similar to two identical fans in parallel. Flow and Pressure Measurements Due to the nature and complexity of thermofluid analysis, the form of the system characteristics curve can at best be established by test. Depending on the machine topologies and size, the measurements may be taken either 264 8 Cooling and Heat Transfer at the air inlet or outlet",
            " After many subsequent cycles, the machine\u2019s temperature variation becomes uniform and tends to be within a certain limited range (see Fig. 8.14c). 274 8 Cooling and Heat Transfer Under the same load and cooling condition, the maximum stablised temperature of a machine in intermittent operation duty is smaller than that in continuous duty. Hence, similar to short-time operation duty, a machine operated in intermittent duty has overload capacity. Numerical Example 8.1 A self-cooled 8-pole, 16-kW AFPM generator with an ironless stator as shown in Fig. 8.2 will be considered. The outer diameter isDout = 0.4 m and the inner diameter is Din = 0.23 m. The magnet width\u2013to\u2013pole pitch ratio \u03b1i = 0.8 and thickness of a rotor disc d = 0.014 m. The measured flow characteristic curves are shown in Fig. 8.15. At rated speed 1260 rpm, the total losses are 1569 W, of which (i) rotational losses \u2206Prot = 106 W; (ii) eddy current losses in the stator \u2206Pe = 23 W; (iii) stator winding losses (rated) \u2206P1w = 1440 W. Find: (a) Convective heat transfer coefficients of the disc system (b) Steady-state temperatures at different parts of the machine"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003483_s0890-6955(02)00163-3-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003483_s0890-6955(02)00163-3-Figure3-1.png",
        "caption": "Fig. 3. Multiple powder feeding system. (a) Schematic of multiple powder feeding system. (b) Newly designed powder feeder and controller. (c) Part built by multiple powder feeding systems.",
        "texts": [
            " A system control software supports TCP/IP, and serial port communication protocols are running on the central PC. Sending the commands through Ethernet or RS485 serial connections, the desired delivery rate of each kind of powder is controlled by the central PC in real time. Up to four powder feeders can be controlled simultaneously by the central PC. Functionally graded material can be fabricated by synchronizing the motion, laser operating parameters, and powder delivery rates according to the design of the material. Fig. 3 shows the schematic and real setup of multiple powder feeding systems. A coaxial infrared image sensing setup with respect to the metal powder delivery nozzle has been developed. As shown in Fig. 4 (a), the laser head consists of a partial reflective mirror and lens set. The Nd: YAG laser beam inducted by the optical fiber is reflected from the partial reflective mirror, and it is focused on the substrate by the set of lenses arranged in the laser head. The optical part of the laser head forms another optical path for the observation of the laser processing"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002318_301-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002318_301-Figure2-1.png",
        "caption": "Fig. 2. View of the sample.",
        "texts": [
            " The velocity dependence of rubber friction makes it very difficult to measure the static coefficient of friction, for it is almost impossible to determine the exact value of the tangential stress under which the sample begins to move. We decided therefore to measure the dynamic coefficient of friction. The experiments were carried out with equipment very similar to that of Roth, Driscoll and Holt but in our case the track was supported on vertical cantilever springs by the deflection of which the frictional force was measured. The track was covered with plate glass which was carefully cleaned with acetone by a standard technique after each run. T h e samples had an area of about 3.4 cm2 and had the form shown in fig. 2 , which is a perspective view of the sample as seen from underneath. p = c'al W, . . . . * , (2) Taking into account the results of fig. 1 we p = c w-113, . . . . , . ( 3) The surface has been divided into two runners in order to produce as broad a supporting area for the weights constituting the normal load as was possible i n the circumstances. When the sample moves it is strained in simple shear by the frictional force and if no precautions are taken the rear edge of the sample will trail b&lnd and be under a lower pressure than the rest of the surface, the pressure on which increases correspondingly. T o obviate this source of error the ends of the samples were chamfered as shown in fig. 2. The heights of the runners and of the supporting plate were each in. I n order to obtain a standard rugosity of the rubber surface the samples were moulded against a ground glass plate, as was done by Roth, Driscoll and Holt. 2 x-2 660 A. Schallamach The velocity at which the measurements were made was in all caseS 2.16 x cmjsec. It had to have a comparatively low value if vibrations and stick-slip phenomena were to be rigidly excluded. It is invariably found that during the course of an experiment the frictional force increases as the sample travels along the track"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000354_9.119645-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000354_9.119645-Figure1-1.png",
        "caption": "Fig. 1. The ball and beam system.",
        "texts": [],
        "surrounding_texts": [
            "I. INTRODUCTION The conditions for feedback linearization of nonlinear systems are restrictive and it is of practical interest to investigate situations where these conditions fail but do so only slightly. Continuing the work of Krener [ 11, who gave conditions for approximate full state linearization of nonlinear multiinput systems, we discuss approximate input-output linearization of single-input single-output systems which fail to have relative degree. In contrast to extended linearization [2] and pseudolinearization [3], our technique does not approximate the system by a linear system or family of linear systems but rather by a single nonlinear system that is input-output linearizable.\nOur approach to the tracking problem differs from the recent work of Isidori and Byrnes [4] who provide (fragile) conditions under which one can exactly track the output of a finite-dimensional exosystem. In contrast, our goal is to provide approximate tracking of a large class of output signals that is valid under a wide class of nonlinear perturbations in the system model.\nIn Section 11, we begin with an example drawn from undergraduate control laboratories, the ball and beam experiment, and show how the exact input-output linearization approach may fail. In Section U, we use the same example to motivate our approximate input-output linearization technique and provide a comparison with the standard Jacobian linearization. In Section IV, we define robust relative degree and present a method of approximate input-output linearization for SISO nonlinear systems.\nManuscript received December 14, 1989; revised July 23, 1990 and September 14, 1990. Paper recommended by Past Associate Editor, M. A. Shayman. This work was supported in part by NASA under Grant NAG2243, the U.S. Army under Grant DAAL-88-K0572, the U.S. Air Force Office of Scientific Research under Grant AFOSR 90-0011, the National Science Foundation under Grant ECS 88-18166, The Schlumberger Foundation, and the Berkeley Engineering Fund.\nJ. Hauser is with the Department of Electrical Engineering-Systems, University of Southern California, Los Angeles, CA 90089-2563.\nS. Sastry is with the Department of Electrical Engineering and Computer Science, University of California, Berkeley, CA 94720.\nP. Kokotovii is with the Center for Control Engineering and Computation, University of California, Santa Barbara, CA 93 106-9560,\nIEEE Log Number 9104386.\nLet the moment of inertia of the beam be J , the mass and moment of inertia of the ball be M and Jb, respectively, the radius of the ball be R , and the acceleration of gravity be G. Choosing the beam angle 0 and the ball position r as generalized position coordinates for the system, the Lagrangian equations of motion are given by\n7 = ( M r 2 + J + J b ) B i + 2 M r j e + M G r c o s e (2.1)\nwhere 7 is the torque applied to the beam. Defining B := M / ( Jb / R 2 + M) and changing the coordinates in the input space using the invertible nonlinear transformation\n7 = 2 M r i . 8 + M G r c o s O + ( M r 2 + J + J b ) u (2.2)\nto define a new input U the system can be written in state-space form as\nx2\nB( x, xi - G sin x3) + [!]U\n, - g ( x ) / ( x )\nY = XI (2.3) v h ( x ) .\nw h e r e x = ( x , , x 2 , x 3 . x 4 ) T := ( r , i , e , e ) \u2018 and y = h ( x ) :=r . The system output y ( t ) is to track a desired trajectory yd(t) , i.e., y ( t ) --* yd(t) as t + 00. Following the usual procedure [5], we differentiate the output until the input appears\nY = X I ,\nY = x2,\ny = Bx, xt - BG sin x 3 ,\ny\u20183\u2019 = B x ~ x ~ - BGx, COS x3 + ~Bx,x ,u . (2.4) 3 -\nb ( x ) a( x )\nIf the coefficient of U, a( x) , were nonzero in the region of interest, we could use a control law of the form U = (- b( x ) + u)/a( x) to yield a linear input-output system described by y(3) = U. Unfortunately, the control coefficient a ( x ) is zero whenever the beam angular velocity x4 = 8 or ball position x I = r are zero. Therefore, the relative degree [6] of the ball and beam system is not well defined! Thus exact input-output linearization approach is not applicable to this problem.\nNext, to try our hand at fully linearizing the state of this system, we apply the necessary and sufficient conditions given by Jakubczyk and Respondek [7] and by Hunt, Su, and Meyer [8]. After some calculations, we discover that the ball and beam system violates the\n0018-9286/92$03.00 0 1992 IEEE",
            "IEEE TRANSACTIONS ON AUTOMATIC CONTROL, VOL. 37, NO. 3, MARCH 1992 393\nL small nonlinear terms\nFig. 2. Approximate input-output linearization: A chain of integrators perturbed by small nonlinear terms.\nintegrability condition which requires the involutivity of\ns p a n { g , a d f g , - - . , a d ; - 2 g } (2 .5)\nsince [g, ad:g] = (2Bx1 -2Bx, 0 O)=does not lie in ( 2 . 5 ) . Here ad$g denotes the iterated Lie bracket [ f , * . . [f, g ] . . . 1. Thus, it is not possible to fully linearize the ball and beam system.\nIII. APPROXIMATE INPUT-OUTPUT LINEARIZATION: THE BALL AND BEAM SYSTEM\nIn this section, we show that, by appropriate choice of vector fields close to the system vector fields, we can design a feedback control law to achieve approximate output tracking. The control law will, in fact, be the exact output tracking control law for an approximate system. We construct two input-output linearizable approximations for the ball and beam system. In each case, we construct a nonlinear change of coordinates E = + ( x ) , and a state feedback U ( X , U) = a ( x ) + p(x)u , to make the system look like a chain of integrators perturbed by higher order terms $ ( x , U) as depicted in Fig. 2. Each successive component of the coordinate change is constructed by differentiating the previous one along the system trajectories and discarding some higher order terms. (The first component is chosen to be the system output.) The performance of these designs is compared to the performance of a linear controller based on the standard Jacobian approximation.\nApproximate tracking is achieved for each design by choosing U\nto equate U = b( x) + a( x)u with\nmaking the error system into an exponentially stable linear system perturbed by small nonlinear terms. Here, the ai are chosen so that s4 + a3s3 + a2s2 + a,s + a. is a Hurwitz polynomial.\nFor each approximation, we present simulation results depicting the output error y d ( t ) - 4 1 ( x ( t ) ) , the angle of the beam 0 ( t ) = x 3 ( t ) , and the neglected nonlinearity $(x, U ) , and the input torque T(t ) , for three desired trajectories given by y d ( t ) = A cos mt/5, with A = 1 , 2 , and 3 and initial conditions for ( r , 0) of (1,0.0564), (2,0.1129), and (3,0.1698), respectively. The system parameters\nused for simulation are M = 0.05 kg, R = 0.01 m, J = 0.02 kg m2, Jb = 2 x l o p 6 kg m2, and G = 9.81 m/s2 (thus B = 0.7143). All closed-loop poles are placed at -2 . Note that the desired trajectories require much greater deviations than possible with most experimental setups.\nApproximation 1 (Modification off) Since the system fails to have a well defined relative degree because of the centrifugal acceleration term Bx, x i = Br02, we design our first approximate system by simply neglecting the term Bx,xi . Let = q5,(x) = h ( x ) . Then, choosing 4i(.) at each step, we have\nE', = x2 - \u20ac 2 = dZ(X)\n= -BGs inx , + Bx,xi - \u20ac 3 = @,(I) 3 d X )\nE'3 = -BGx4 COS x3\n54 = d 4 ( d\ni4 = BGxj sin x3 + ( -BG cos x 3 ) U - '\nb ( x ) a( x )\n\u20ac 1 = E2\n\u20ac 2 = E 3 + J / z ( x )\nor E'3 = '$4\nE'4 = b ( x ) + ~ ( x ) u\n=: u ( x , U).\nAs expected, by neglecting the centrifugal term (which is higher order), we obtain an approximate system with a well-defined relative degree. Note that the choice of what to neglect (i.e., $ z ( x ) ) leads to a specification of the coordinate transformation @ ( x ) . In this case, the approximate system has been obtained by a simple modification of the f vector field (i.e., by neglecting & ( a ) ) .\nThe simulation results in Fig. 3 show that the closed-loop system provides good tracking. The tracking error increases in a nonlinear fashion as the amplitude ( A ) of the desired trajectory is increased. This is expected since the neglected term $ , ( x ) is a nonlinear function of the state. A good a priori estimate of the mismatch of the approximate system for a desired trajectov can be calculated using $ ( W 1 ( y d , j d , J d , ~ 2 ) ) ) where @ - I : ( +, x is the inverse of the coordinate transformation. This in turn may be a useful way to define a class of trajectories that the system can track with small error.\nApproximation 2 (Modi3cation of g)\nFor this approximation, we only discard terms that we must in order to obtain an approximate system with a well-defined relative degree, that is, we modify the g vector field. Again, let = 4 , ( x )",
            "394\n= h(x) . We have\n- 0 0\nG(G cos2 x3 - 2x ,x4 cos x3 - 2 x l x j sin x 3 )\n- 2 x , x,' sin x3\nG cos2 x3 - 2 x2 x4 cos x3 - 2 x , x: sin x3\n=\n- -\nIEEE TRANSACTIONS ON AUTOMATIC CONTROL, VOL. 37, NO. 3, MARCH 1992\n(3 .4 )\nil = F2\ns'z = t 3\n5 3 = 4,(x) or s'3 = t 4 + $3(x9 U) i3 = -BGx4 COS x 3 + BX,X: + ~ B x , x ~ u\ni4 = B 2 x l x ; + B ( l - B ) x j sin x3 + ( - B G COS x3 + ~ B x ~ x , ) U\n$, = b ( x ) + a ( x ) u , - \u20ac4 = 44(x) +.,(x, U ) =: \" ( X , U ) .\nY\n(3.3)\nb ( x )\nIn order to guarantee a well-defined relative degree, we were forced to discard $3(x , U ) = 2Bx,x4u since x 1 x 4 is zero at x = 0 but not identically zero in a neighborhood of x = 0 .\nThe advantage of writing the system in mixed x and 4 coordinates as in (3.3) is that it is easy to identify the terms in the g vector field that should be dropped (in this case, $3). However, when this modification is expressed completely in x (or t ) coordinates it is, in fact, very complicated. For example, in x coordinates, dropping G3 corresponds to subtracting\nfrom the original g vector field. The system with g modified in this manner is input-output linearizable and is an approximation to the original system since A g u is higher order in x 2 , x 3 , x 4 , and U (for every x l , in fact). The simulation results in Fig. 4 show that the tracking error is somewhat smaller than that obtained by Approximation 1 .\nApproximation 3 (Jacobian Approximation)\nTo provide a basis for comparison, we calculate a linear control law based on the standard Jacobian approximation of the original system (2.1) at x = 0, 7 = 0. As with the previous approximations, define t , = I#I1(x) = h ( x ) = x 1 and write (2.1) as\na( x )\nThe Jacobian approximation is, of course, obtained by replacing the f vector field by its linear approximation and the g vector field by its constant approximation, that is, by neglecting $2 and G4. Fig. 5 shows the simulation results from the Jacobian approximation, indicating that the control system is no longer stable for A = 3.\nTable I provides a direct comparison of the maximum error magnitude 1 e I = I yd - 1#11 1 between 30 and 40 s for the three approximations. Note that Approximation 2 provides somewhat better tracking for this class of inputs than Approximation 1 . It is of practical interest that Approximations 1 and 2 continue to approximate the ball and beam system for trajectories with A more than two times as large as the value at which the Jacobian approximation resulted in an unstable system. Initial conditions for the simulation with A = 6 are ( r , 0) = (6,0.345).\nIV. ANALYSIS OF APPROXIMATE LINEAFUZATTON\nIn this section, we will consider single-input single-output systems of the form\nX = f ( x ) + g ( x ) u\nY = h ( x ) (4.1) where XEW\", U , Y E W, f and g are smooth vector fields on W\" (i.e., ~ ( x ) E T,W\" = R\", X E R \" ) , and h:R\" -+ R is a smooth function (smooth is understood to mean as differentiable as needed). For simplicity, we assume that x = 0 is an equilibrium point of the undriven system, i.e., f ( 0 ) = 0, and that g(0) # 0.\nIf the control objective is tracking, the input-output linearization proceeds as follows: differentiate the output repeatedly until the input appears for the first time on the right-hand side\nY = L f h ( x ) ,\nj ; = L$h( x ) ,"
        ]
    },
    {
        "image_filename": "designv10_0_0003838_7.953252-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003838_7.953252-Figure4-1.png",
        "caption": "Fig. 4. Angle \u00b1 corresponding to displacement of a phase in relation to another.",
        "texts": [
            " Therefore, the torque will be in the direction of the nearest aligned position. The parameters used to simulate the 6/4 SR motor were previously obtained by a finite element analysis (FEM) in [1]. Fig. 3 shows its linear inductance profile L(\u00b5) with each phase inductance displaced by an angle \u00b5s given by \u00b5s = 2\u00bc \u00b5 1 Nr \u00a1 1 Ns \u00b6 (6) where Nr and Ns are the number of rotor and stator poles, respectively. When the motor has equal rotor and stator pole arcs, \u00afr = \u00afs, one has the following angle relations \u00b5x = \u00b5 \u00bc Nr \u00a1\u00afr \u00b6 (7) \u00b5y = \u00bc Nr (8) which are indicated in Fig. 3. Fig. 4 shows the angle \u00b1 corresponding to the displacement of a phase in relation to another, and given by \u00b1 = 2\u00a6 \u00b5 1 Nr \u00a1 1 Ns \u00b6 : (9) The 6/4 SRM has the following parameters: Lmin = 8 mH, Lmax = 60 mH, and \u00afr = \u00afs = 30 \u00b1. Thus, from (7) and (8), one gets \u00b5x = 15 \u00b1 and \u00b5y = 45 \u00b1. The electric equation of each phase is given by d\u00aai(\u00b5,Ii) dt +RIi = V with i= f1,2,3g: (10) While excluding saturation and mutual inductance effects, the flux in each phase is given by the linear equation \u00aai(\u00b5,Ii) = L(\u00b5)Ii: (11) The total energy associated with the three phases (n= 3) is given by Wtotal = 1 2 3X i=1 L(\u00b5+(n\u00a1 i\u00a1 1)\u00b5s)I2i (12) and the motor total torque by \u00a1 = 1 2 3X i=1 dL(\u00b5+(n\u00a1 i\u00a1 1)\u00b5s) d\u00b5 I2i : (13) The mechanical equations are J d"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002041_tfuzz.2014.2360954-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002041_tfuzz.2014.2360954-Figure1-1.png",
        "caption": "Fig. 1: Non-symmetric nonlinear dead-zone model",
        "texts": [],
        "surrounding_texts": [
            "1063-6706 (c) 2013 IEEE. Personal use is permitted, but republication/redistribution requires IEEE permission. See http://www.ieee.org/publications_standards/publications/rights/index.html for more information.\n> IEEE TRANSACTIONS ON FUZZY SYSTEMS<\n2\napproaches were designed in [31-34] for linear systems with input saturation and input disturbances. Two adaptive fuzzy output feedback control schemes were obtained in [35, 36] for nonlinear SISO strict-feedback systems with dead-zones. The problem of adaptive fuzzy control for SISO systems in non-strict feedback form was addressed in [37]. The considered systems contain unknown dead-zone and unknown constant time delays and can be stabilized by introducing a variable separation approach and employing the unique structural property of fuzzy logic systems. Based on the mean-value theorem, an adaptive fuzzy control was studied in [38] for uncertain pure-feedback nonlinear stochastic SISO systems with input saturation. To solve the problem of immeasurable states, an adaptive fuzzy control using output-feedback was studied in [39] for uncertain nonlinear pure-feedback systems with unknown dead zone inputs, in which a state filter observer is designed for estimating the unmeasured states. More recently, an adaptive NN control was developed in [40] for nonlinear SISO systems with a nonlinear dead-zone and multiple time-delays. In addition, the proposed control approach is also applied to a turntable servo system with permanent-magnet synchronous motor. An adaptive neural network control was given in [41] for nonlinear stochastic nonstrict-feedback SISO systems with backlash-like hysteresis. Note that the mentioned control approaches focus on SISO nonlinear systems and they cannot solve the control problem of MIMO nonlinear systems with non-smooth input nonlinearities.\nIt is well known that a MIMO system is more general than a SISO one. Furthermore, from the viewpoint of engineering practice, most of the plants have the characteristic of uncertainty, multivariable and nonaffine nonlinearity. Therefore, studying the control design problem for uncertain MIMO nonlinear system is a very important topic. For a class of uncertain MIMO nonlinear systems with non-symmetric input constraints, an adaptive control technique was given in [42] by introducing an auxiliary design system analyzing the effect of input constraints. Based on the NN approximation, a robust adaptive control problem was explored in [43] for MIMO nonlinear systems with unknown coefficient matrices and non-symmetric input nonlinearities by combining variable structure control and backstepping design. To control MIMO strict-feedback nonlinear systems with non-symmetric dead-zone inputs, the adaptive fuzzy tracking controllers with an observer were designed in [44] where the observer is used to estimate the unmeasured states. Subsequently, the control problem of nonlinear large-scale strict-feedback systems with unknown dead zones was dealt with by designing adaptive fuzzy decentralized controllers based on the backstepping [45]. However, the main restrictions in [42-45] are that the considered systems belong to nonlinear systems in strict feedback form and it is difficult to employ these approaches for control MIMO systems in pure-feedback form. A pure-feedback system represents a more general class of triangular systems than a strict-feedback system. In the controller design, because there is no affine appearance of the variables to be used as\ncontrollers in pure-feedback system, it is quite difficult to find the explicit controllers for stabilizing the pure-feedback systems.\nIn this paper, the controller design and the stability problem are considered for a class of uncertain MIMO nonlinear systems in pure feedback form. The considered MIMO systems contain unknown nonlinear functions and nonlinear dead-zone inputs. In the controller design, the mean value theorem is utilized to transform the systems into affine structure, but there still are nonaffine functions in this structure. The unknown functions are approximated by using the fuzzy logic systems. To compensate for the effects of nonlinear dead-zone inputs, the auxiliary control terms are designed. Based on the backstepping design technique and adaptive control theory, a new adaptive fuzzy control scheme is synthesized. The stability of the closed-loop system and the convergence of the tracking errors are proved by using the Lyapunov stability theorem. A simulation example is utilized to verify the effectiveness of the approach.\nCompared with the previous works, the main advantages of this paper are summarized as follows: 1) The nonlinear MIMO systems considered in this paper are the nested pure- feedback lower triangular systems, not the nonlinear systems studied in [35, 36, 38-40, 42-45], which are in strict-feedback form. Therefore, the control design problem of this paper can not be solved by employing the previous works directly; 2) The proposed approach can compensate for nonlinear dead-zones inputs and most of the existing results are required to satisfy linear dead-zone inputs which are a special case of nonlinear dead-zone inputs; 3) The proposed adaptive control method can not only guarantee the stability of the control system, but also compensate for the effects of nonlinear dead-zone on the control performance effectively.\nII. SYSTEM DESCRIPTION AND PRELIMINARIES\nConsider nonlinear MIMO systems, which are described by\nthe following form\n( )\n( )\n, , 1 1 1 1 ,1 , 1\n, , 1 1 1 1\n,1\n, , , , , , , ,\n1, , 1\n, , , , , ,\n, 1, ,\ni i i\ni i\ni k i k i i i i k\ni i\ni m i m i i i i\ni i\nx x u x u x x\nk m\nx x u x u x u\ny x i n\n\u03d5\n\u03d5\n\u2212 \u2212 +\n\u2212 \u2212\n =   = \u2212  =  = = & L L L & L L\n(1)\nwhere ,1 ,, , i\ni\nT m\ni i i mx x x R = \u2208 L and iy R\u2208 are the state\nvector and the output of the thi subsystem, respectively;\n( ), . , 1, , , 1, , ii k i ik m i n\u03d5 = =L L are unknown nonlinear smooth functions; ( )i iu v= \u03a4 is a nonlinear dead-zone to be expressed\nas follows [43]\n( ) ( )\n( )\n,\n0,\n,\nir i i ir\ni i il i ir\nil i i il\nf v v b\nu v b v b\nf v v b\n\u2265 \n= \u03a4 = \u2212 < <  \u2264 \u2212\n(2)\nwhere ( )i iu v= \u03a4 is the output of the i th dead-zone, iv is\nthe input of the i th dead-zone, 0irb > and 0ilb > are the",
            "1063-6706 (c) 2013 IEEE. Personal use is permitted, but republication/redistribution requires IEEE permission. See http://www.ieee.org/publications_standards/publications/rights/index.html for more information.\n> IEEE TRANSACTIONS ON FUZZY SYSTEMS<\n3\nunknown parameters of the i th dead-zone, ir ilb b\u2260 , ( )ir if v and ( )il if v are unknown nonlinear functions. The\nFor non-symmetric nonlinear dead-zone in(2), we make the\nfollowing assumption.\nAssumption 1 [43]: Assume that ( )ri if v and ( )li if v are\nsmooth, and there exist positive constants 0rik , 1rik , 0lik and 1lik such that ( ) [ )0 10 , ,ir ir i ir i irk f v k v b\u2032< \u2264 \u2264 \u2208 +\u221e and 00 ilk< \u2264\n( ) 1,il i ilf v k\u2032 \u2264 ( ],i ilv b\u2208 \u2212\u221e \u2212 where ( ) ( ) / | iir i ir z vf v df z dz =\u2032 =\nand ( ) ( ) / | iil i il z vf v df z dz =\u2032 = .\nRemark 1: Many practical systems can be expressed by or transformed into the pure-feedback systems in(1), such as Duffing oscillator, aircraft systems and biochemical processes etc. Different from nonlinear systems in strict-feedback form, a nonlinear pure-feedback system is difficult to find a clear variable to be viewed as virtual or actual controller due to the presence of nonaffine function.\nRemark 2: In(2), ( )ir if v and ( )il if v are nonlinear\nfunctions of dead-zone input iv . Most of the existing methods are obtained for nonlinear systems with linear dead-zones. Specifically, when nonlinear dead-zone inputs are embedded in nonlinear MIMO pure-feedback systems, it has still not practicable strategy. An effective way is to separate nonlinear dead-zone inputs by using the differential mean value theorem.\nDefine\n( ) ( )\n( ) ( )\n, 1 1 1 1 ,1 , 1\n, 1 1 1 1 ,1 , 1\n, 1\n, 1 1 1 1\n, 1 1 1 1\n, , , , , , ,\n, , , , , , , ,\n, , , , , ,\n, , , , , ,\ni i\ni i\ni\ni\ni\ni k i i i i k\ni k i i i i k\ni k\ni m i i i i\ni m i i i i\ni\ng x u x u x x\nx u x u x x\nx\ng x u x u x u\nx u x u x u\nu\n\u03d5\n\u03d5\n\u2212 \u2212 +\n\u2212 \u2212 +\n+\n\u2212 \u2212\n\u2212 \u2212\n   \u2202  =  \u2202    \u2202 = \u2202 L L L L L L\n(3)\nBy using the mean value theorem and(3), we can obtain\n( ) ( ) ( )\n( ) ( )\n* , 1 , 1\n, 1 1 1 1 ,1 , 1\n, 1 1 1 1 ,1 ,\n, 1 1 1 1 ,1 , 1\n, 1\n, 1 1 1 1\n, 1 1 1 1\n,\n, , , , , , ,\n, , , , , , , ,0\n, , , , , , , |\n, , , , , ,\n, , , , , ,0\ni i\ni i\ni i i k i ki i\ni\ni\ni\ni\ni k i i i i k\ni k i i i i k\ni k i i i i k x x\ni k\ni m i i i i\ni m i i i\ni m\nx u x u x x\nx u x u x x\ng x u x u x x\nx\nx u x u x u\nx u x u x\ng\n\u03d5\n\u03d5\n\u03d5\n\u03d5\n+ +\n\u2212 \u2212 +\n\u2212 \u2212\n\u2212 \u2212 + =\n+\n\u2212 \u2212\n\u2212 \u2212\n=\n+\n\u00d7\n=\n+\nL L\nL L\nL L\nL\nL\n( ) *1 1 1 1, , , , , , | i i i i i i iu u x u x u x u u\u2212 \u2212 =              \u00d7   L\n(4)\nwhere * , 1ii kx + is some points between 0 and , 1ii kx + , and * iu is\nsome points between 0 and iu .\nLet\n( ) ( )\n( ) ( )\n( )\n* , 1 , 1\n, 1 1 1 1 ,1 ,\n, 1 1 1 1 ,1 ,\n*\n, 1 1 1 1 ,1 , , 1\n, 1 1 1 1 ,1 , 1\n, 1 1 1 1\n, , , , , , ,\n, , , , , , , ,0 ,\n, , , , , , , ,\n, , , , , , , |\n, , , , ,\ni i\ni i\ni i i\ni i i k i ki i\ni\ni k i i i i k\ni k i i i i k\ni k i i i i k i k\ni k i i i i k x x\ni m i i i\nx u x u x x\nx u x u x x\nG x u x u x x x\ng x u x u x x\nx u x u x\n\u03d5\n\u03d5\n+ +\n\u2212 \u2212\n\u2212 \u2212\n\u2212 \u2212 +\n\u2212 \u2212 + =\n\u2212 \u2212\n\u03a6\n=\n=\n\u03a6\n=\nL L\nL L\nL L\nL L\nL\n( )\n( ) ( ) *\n, 1 1 1 1\n*\n, 1 1 1 1\n, 1 1 1 1\n, , , , , ,0 ,\n, , , , , ,\n, , , , , , |\ni\ni\ni i i\ni m i i i\ni m i i i i\ni m i i i i u u\nx u x u x\nG x u x u x u\ng x u x u x u\n\u2212 \u2212\n\u2212 \u2212\n\u2212 \u2212 =\n               =\nL\nL\nL\n(5)\nFor convenience, , ii k\u03a6 and , ii kG will be used to denote\n( ), . ii k\u03a6 and ( ), . ii kG , respectively, where 1, ,i ik m= L , 1, ,i n= L .\nUsing (4) and (5), (1) can be rewritten as\n, , , , 1\n, , ,\n,1\n, 1, , 1\n, 1, ,\ni i i i\ni i i\ni k i k i k i k i i\ni m i m i m i\ni i\nx G x k m\nx G u\ny x i n\n+ = \u03a6 + = \u2212  = \u03a6 +  = = & L & L\n(6)\nAssumption 2 [39]: There exist the constants , ii k\u03be \u2265\n, 0, 1, , , 1, , ii k i ik m i n\u03be > = =L L such that , ii k\u03be \u2264 , ,i ii k i kG \u03be\u2264 .\nThe above assumption implies that , ii kG are strictly either positive or negative. Without losing generality, we shall assume , , ,i i ii k i k i kG\u03be \u03be\u2264 \u2264 .\nRemark 3: It can be seen from (5) that , ii kG is still a nonaffine\nfunction in character because , ii kG is a function with respect to\n* , 1ii kx + or * iu . Due to this character, it will bring about a very\ncomplicated design. The main reason is that , ii kG can not be\ncontained in the approximated functions. When , ii kG is approximated, it is not easy to determine the input variables of the approximators because *\n, 1ii kx + or * iu are uncertain.",
            "1063-6706 (c) 2013 IEEE. Personal use is permitted, but republication/redistribution requires IEEE permission. See http://www.ieee.org/publications_standards/publications/rights/index.html for more information.\n> IEEE TRANSACTIONS ON FUZZY SYSTEMS<\n4\nAccording to the differential mean value theorem, we have\n( ) ( )( ) ( ) ( ) ( ), ,il i il il i i il il i irf v f v v b v b\u03b6 \u03b6\u2032= + \u2208 \u2212\u221e \u2212 (7)\n( ) ( )( )( ) ( ) ( ), ,ir i ir ir i i ir ir i ilf v f v v b v b\u03b6 \u03b6\u2032= \u2212 \u2208 +\u221e (8)\nDefine ( ) ( ) ( ), T\ni ir ilt p t p t\u03a1 =    and\n( ) ( )( )( ) ( )( )( ), T\ni ir ir i il il it f v t f v t\u03b6 \u03b6 \u2032 \u2032\u039d =   where\n( ) ( ) ( ) 1, 0, i il ir\ni il\nv t b p t\nv t b\n> \u2212 = \n\u2264 \u2212 and ( )\n( ) ( ) 1, 0, i ir il\ni ir\nv t b p t\nv t b\n< = \n\u2265 . Based on\nAssumption 1, the dead-zone (2) can be rewritten as follows\n( ) ( ) ( ) ( )T i i i i i i iu v t t v v\u03b4= \u03a4 = \u039d \u03a1 + (9)\nwhere ( )\n( )( ) ( )( ) ( )( )\n( )( )\n,\n= ,\n,\nir ir i ir i ir\ni i il il i ir ir i il i ir\nil il i il i il\nf v b v b\nv f v f v b v b\nf v b v b\n\u03b6\n\u03b4 \u03b6 \u03b6\n\u03b6\n \u2032\u2212 \u2265   \u2032 \u2032\u2212 + \u2212 < <    \u2032\u2212 \u2264 \u2212\nand ( )i i iv\u03b4 \u03b4\u2264 with unknown positive constant\n( ) { }1 1 max ,i ir il ir ilk k b b\u03b4 = + \u2212 .\nThe control objective of the paper is to design an adaptive fuzzy approach such that all the signals in the closed-loop system are bounded and the system outputs can track the\nbounded reference signals ,i ry as closely as possible.\nBecause unknown functions are included in the considered systems, it will result in a difficult task for coping with unknown functions. Fortunately, the fuzzy logic systems are effective way for handling unknown functions. It has been proven that the fuzzy systems can approximate an arbitrarily continuous function to a given accuracy [10]. We introduce the following Lemma.\nLemma 1 [10]: For any given continuous function ( )f x on\na compact set n U R\u2208 and an arbitrary 0\u03b5 > , there exist the\nfuzzy systems such that ( ) ( )sup x U f x y x \u03b5 \u2208 \u2212 < .\nUsing the fuzzy approximation, the continuous function\n( )f x can be expressed as\n( ) ( ) ( )* *T f x S x x\u03b8 \u03b5= + (10)\nwhere * MR\u03b8 \u2208 and ( )* x\u03b5 are the optimal fuzzy parameter\nvector and the fuzzy approximation error, respectively.\nAssumption 3 [10]: On compact set \u2126 , there exist the constants 0\u03b8 > and 0\u03b5 > such that *\u03b8 \u03b8\u2264 and\n( )* x\u03b5 \u03b5\u2264 .\nIII. ADAPTIVE FUZZY DESIGN AND STABILITY\nBased on the fuzzy approximation, it can be assumed that\nunknown functions ( ), ,i ii k i k\u03c7\u03d2 are expressed as\n( ) ( ) ( )* *\n, , , , , , ,i i i i i i i\nT\ni k i k i k i k i k i k i kS\u03c7 \u03b8 \u03c7 \u03b5 \u03c7\u03d2 = + (11)\nwhere * , ii k\u03b8 , ( ), ,i ii k i kS \u03c7 , ( )* , ,i ii k i k\u03b5 \u03c7 and , ,i ii k i k\u03c7 \u2208 \u2126 are the\noptimal fuzzy parameter vector, the fuzzy basis function vector, the minimal approximation error, and the input vector of the fuzzy approximation systems, respectively; and ( ), ,i ii k i k\u03c7\u03d2\nand , ii k\u03c7 will be defined later on. Define ( )2 1 * , , , , i i ii k i k i k\u03c1 \u03be \u03b8\u2212=\n1, , 1i ik m= \u2212L and ( )2 1 * , , ,i i ii m i m irl i mk\u03c1 \u03be \u03b8\u2212= . It can be seen\nthat , ii k\u03c1 is an unknown constant and ( ), \u02c6 0 0 ii k\u03c1 > denotes the estimation of , ii k\u03c1 . Let , , , \u02c6 i i ii k i k i k\u03c1 \u03c1 \u03c1= \u2212% .\nFor conciseness of presentation, the detailed design\nprocedure of the thi subsystem is proposed here. To this end, assume that the controller , 1, , 1ju j i= \u2212L is designed for the\nthj subsystem where ju can be viewed as a function of\n, 1 1, ,, , , , , jj m j r j rx x y y\u03c7  =  L L with\n( ) , , , ,, , , j\nT m\nj r j r j r j ry y y y =    & L . Then, based on the equation (6)\nand using the backstepping design, an adaptive fuzzy controller\niv for the thi subsystem can be designed in the following.\nStep 1:i. Define the tracking error ,1 ,1 ,i i i rz x y= \u2212 and ,1iz&\ncan represent from (6) as\n,1 ,1 ,1 ,2 ,i i i i i rz G x y= \u03a6 + \u2212 && (12)\nDefine ( ),1 ,1 ,1 ,i i i i ry\u03c7\u03d2 = \u03a6 \u2212 & and\n,1 1 1 1, 1, ,1 , ,1, , , , , , , T i i r i r i i r ix x y y x y\u03c7 \u2212 \u2212 = \u2208\u2126 &L L is the input of\nthe fuzzy systems.\nDesign the virtual controller as\n( ) 2\n,1 ,1 ,1 ,1 ,1 ,1 ,1\n,1\n1 \u02c6\n2 i i i i i i i\ni\nc z z S Z\u03c0 \u03c1 \u03c4 = \u2212 \u2212 (13)\nwhere ,1ic and ,1i\u03c4 are positive design parameters, and ,1ic\nwill be specified later on.\nIt is obvious that ( ) 2 2 2\n,1 ,1 ,1 ,1 ,1 ,1 ,1 ,1\n,1\n1 \u02c6\n2 i i i i i i i i\ni\nz c z z S\u03c0 \u03c1 \u03c7 \u03c4 = \u2212 \u2212\n0\u2264 . Then, using Assumption 2, it has\n( ) 2 2 2\n,1 ,1 ,1 ,1 ,1 ,1 ,1 ,1 ,1 ,1\n,1\n1 \u02c6\n2 i i i i i i i i i i\ni\nz G c z z S\u03c0 \u03be \u03c1 \u03c7 \u03c4   \u2264 \u2212 \u2212  \n  (14)\nBy introducing ,2 ,2 ,1i i iz x \u03c0= \u2212 and using (11) with 1ik = ,\nwe have\n( ) ( )* *\n,1 ,1 ,1 ,1 ,1 ,1 ,1 ,2 ,1 ,1\nT\ni i i i i i i i i iz S G z G\u03b8 \u03c7 \u03b5 \u03c7 \u03c0= + + +& (15)\nConsider the Lyapunov function ,12 2\n,1 ,1 ,1\n1\n2 2\ni i i iV z \u03be \u03c1= + % and\ndifferentiating it yields\n,1 ,1 ,1 ,1 ,1 ,1 \u02c6 i i i i i iV z z \u03be \u03c1 \u03c1= \u2212 && %& (16)\nUsing (14) and(15), (16) becomes\n( )2 *\n,1 ,1 ,1 ,1 ,1 ,1 ,2 ,1 ,1 ,1 ,1\nT\ni i i i i i i i i i iV c z z G z z S\u03be \u03b8 \u03c7\u2264 \u2212 + +&\n( ) ( ) 2 2 *\n,1 ,1 ,1 ,1 ,1 ,1 ,1 ,1 ,1 ,1 ,1\n,1\n1 \u02c6 \u02c6\n2 i i i i i i i i i i i\ni\nz S z\u03be \u03c1 \u03c7 \u03be \u03c1 \u03c1 \u03b5 \u03c7 \u03c4 \u2212 \u2212 +&% (17)"
        ]
    },
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        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure1.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure1.3-1.png",
        "caption": "Fig. 1.3 Example of inverse kinematics of a biped robot; (a) initial configuration, (b) that at which the right foot is raised by 0.3 [m] and rotated by 20 [deg] about y-axis",
        "texts": [],
        "surrounding_texts": [
            "1 Introduction 3\nbiped legs with a sufficient stability and mount a computer and battery on the body. Honda published P3, 160 cm height and 130 kg weight, in 1997, and ASIMO, 120 cm weight and 43 kg weight, in 2000. The pictures of P2, P3 and ASIMO are shown in Fig.1.2.\nBefore P2 was published, the majority in robotics community were pessimistic about the development of a biped humanoid robot that can walk stably. This is the reason why Honda P2 astonished the community. Then what are the major differences between the conventional humanoid robots and P2? Let us examine the hardware at first.\nMost humanoid robots developed at universities were made by graduate students or by a small manufacturer. Then the mechanical links of the robots had to be made by bending or cutting and the whole structure was not rigid enough. The reduction mechanisms were implemented by heavy gears with large backlash. By contrast with the old robots, Honda humanoid robots use casted mechanical links with high rigidity and light weight using most advanced mechanical CAD. It was obvious that the casted links should have such properties, but the links were too expensive to be developed by university projects. Honda humanoid robots use harmonic drives which have no backlash. The conventional harmonic drives could transform too little torque to be applied to biped walking, so Honda developed harmonic drives with high torque capacity. After Honda P2 was revealed, most advanced humanoid robots have comparable configurations with those of Honda humanoid robots.\nLet us consider the sensors for the robots. Biped walking may not be stable due to disturbance even when the desired walking pattern is planned to make the walking stable. Then the walking should be stabilized by a feedback control that demands appropriate sensors. The humanoid robots developed in the early stage did not have necessary sensors, but Honda humanoid robots have accelerometers and gyroscopes, to find the orientation of bodies and sixaxes force/torque sensors to find the contact force/torque between the feet and the floor.",
            "4 1 Introduction\nThe goal of this textbook is to give the theoretical background for the development of the software to control the well-designed hardware described above.\nChapter 2 overviews the kinematics of humanoid robots. A representation of the motions of the robots is presented after the representation of rotations in three dimensional space, angular velocity vector and the relationship between the derivatives of the rotation matrices and the angular velocity vector are described. It is presented how to find the position and orientation of a link such as a hand or a foot of the robot from given joint angles. The method is called forward kinematics. Then it is explained how to find the corresponding joint angles from given position and orientation of a specific link. The computation is the inverse of the forward kinematics, and is called inverse kinematics. An example of inverse kinematics problems is illustrated in 1.3. When the configuration of the robot shown in 1.3(a) is given, the problem is\nhow to find the corresponding joint angles at which the right foot is raised by 0.3 [m] and rotated by 20 [deg] about Y axis shown in 1.3(b).\nGenerally speaking, the postion and orientation of a link and the joint angles are represented by nonlinear equations, since most joints of robots are rotational ones. The inverse kinematics problem can be solve by finding solutions of the nonlinear equations analytically, but it is unlikely to solve the nonlinear equations with many variables and a high Bezout number even if the rotation is parametarized algebraically. However, the relationship between the derivatives of the postion and rotation of a link and those of the joint angles can be represented by linear equations, and the inverse kinematics problem can be solved by finding solutions of the linear equations and integrating the solutions. The coefficient matrix of the linear equations is called Jacobian, which is an important concept in many fields including robotics.",
            "1 Introduction 5\nChaper 3 explains the concept of ZMP (Zero-Moment Point) that plays an important role in the motion control of humanoid robots. When the robot is falling down, the sole of the supporting foot should not contact with the ground any more. The ZMP (Zero Moment Point) proposed by Vukobratovic\u0301 et al. is a criterion to judge if the contact between the sole and the ground can be kept without solving the corresponding equations of motions. The contact is kept if the ZMP is an internal point on the sole. When the robot does not move, the contact is kept when the projection of the center of the mass of the robot onto the ground is an internal point of the sole. See Fig.1.4. The\nZMP can be considered to be a dynamic extension of the projection. The ZMP can be used to plan motion patterns that can make the robot walk while keeping the contact between the sole of the supporting foot and the ground. The walking patterns of the majority of humanoid robots have been generated based on the ZMP after Honda P2 appeared.\nThe robot may not fall down even when the sole of the supporting foot left the ground. It can keep walking or standing by controlling the swing leg and changing the touch down positions. The ZMP criterion is a sufficient condition to prevent the robot from falling down, and not a necessary one. It can neither judge rigorously if the contact is kept when the robot walks on a rough terrain or a stair. When the robot walks while catching a handrail, the contact may be more stable but the ZMP criterion can not tell how much the contact should be made stable. Several trials have been made to extend the criterion, but the generic and rigorous criterion has not been established so far. Chapter 3 presents the concept of the ZMP, the relationship between the contact force and the ZMP, the sensing of the ZMP and an algorithm to compute the ZMP based on the forward dynamics of humanoid robots, which is overviewed as well.\nChapter 4 describes how the walking patterns of biped robots can be generated and the walking can be controlled. Generally, the walking patterns are planned to make the robots walk with no disturbance at first, and a feedback"
        ]
    },
    {
        "image_filename": "designv10_0_0000554_978-3-540-85629-0-FigureB.9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000554_978-3-540-85629-0-FigureB.9-1.png",
        "caption": "Fig. B.9. Local modification of a Nurbs trajectory by acting on a weight.",
        "texts": [
            " wj = w\u0304 = 0, \u2200j, then7 Np j (u) = Bp j (u); therefore, B-splines are a special case of Nurbs curves. All the properties stated for the B-spline hold for Nurbs: 1. Endpoints interpolation: n(umin) = p0 and n(umax) = pm. 2. The curve is invariant under affine transformations, i.e. translations, ro- tations, scalings, shears, that can be applied to n(u) by applying them to the control points. 3. Local modifications: the change of a control point pj or of a weight wj modifies n(u) only in the interval [uj , uj+p+1], see Fig. B.9. 4. The control polygon represents a piecewise linear approximation of the curve. In general, the lower the degree of the Nurbs, the closer the curve follows the control polygon. An efficient way to represent (and evaluate) Nurbs curves is based on homogeneous coordinates. In the three-dimensional case, for a given set of control points pj = [px,j , py,j , pz,j ]T and weights wj , it is possible to construct the weighted control points pw j = [wjpx,j , wjpy,j , wjpz,j , wj ]T \u2208 IR4, which define the nonrational B-spline 7 In this case, (B"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000554_978-3-540-85629-0-Figure7.34-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000554_978-3-540-85629-0-Figure7.34-1.png",
        "caption": "Fig. 7.34. Scheme of the mechanical system considered for the design of a polydyne function.",
        "texts": [
            "1 Polydyne and splinedyne functions The first attempts of shaping an input motion profile by taking into account the dynamic model of the system were made in the field of mechanical cam design [58, 59]. In 1953, the term polydyne was introduced as a contraction of polynomial and dynamic, [60]. It describes a design technique for mechanical cams, originally proposed in 1948 [61], that takes into account the dynamics of the cam/follower system in order to define a cam profile able to \u201ccompensate\u201d for the dynamic vibrations of the follower train, at least at one particular cam speed. Given, for example, the simple mechanical chain shown in Fig. 7.34, it is possible to write the differential equation 306 7 Dynamic Analysis of Trajectories mx\u0308(t) + d1x\u0307(t) + (k1 + k2)x(t) = k1y(t). (7.10) As in Sec. 7.3, in the sequel it is assumed that an ideal actuation system is present, and therefore y(t) = q(t). The basic idea of this approach is to use (7.10) for the definition of the function q(t)(= y(t)) in order to obtain the desired profile x(t) or, in equivalent terms, to \u201cinvert\u201d the dynamic model (7.10) in order to define the proper trajectory q(t) for the actuator, given the desired motion profile x(t) of the mass m (this type of approach will be considered also in the following Sec",
            "17) (for the sake of simplicity t0 = 0 is assumed) one obtains q(t) = 1 k [ n\u2211 i=0 (i + 1) (i + 2) h T i+2 ai+2 ti + 2\u03b4\u03c9n n\u2211 i=0 (i + 1) h T i+1 ai+1 ti+ +\u03c92 n ( q0 + n\u2211 i=0 ai h T i ti )] = \u03c92 n k q0 + h k n\u2211 i=0 ( (i + 1) (i + 2) ai+2 T i+2 + 2\u03b4\u03c9n (i + 1) ai+1 T i+1 + \u03c92 n ai T i ) ti = \u03c92 n k q0 + n\u2211 i=0 bi ti (7.19) where the coefficients bi are bi = h k ( \u03c92 n ai T i + 2\u03b4\u03c9n(i + 1) ai+1 T i+1 + (i + 2)(i + 1) ai+2 T i+2 ) (7.20) with an+1 = an+2 = 0. With (7.19) and (7.20) it is possible to compute directly the polydyne function q(t) from the coefficients ai of the normalized polynomial xN(\u03c4), the desired duration T and displacement h, and obviously the relevant parameters of the dynamic model. Example 7.2 Let us consider the mechanism represented in Fig. 7.34 with m = 5 kg, k1 = 5000 N/m, k2 = 1000 N/m, d1 = 25 Ns/m. 312 7 Dynamic Analysis of Trajectories The desired displacement is a periodic motion from x0 = 0 to x1 = 50 (h = 50) and back in a period of time T = Tr + Td = 1, with Tr = Td = 0.5 for the rise (Tr) and for the return (Td) part. If the Peisekah polynomial is chosen, the polydyne function q(t) is computed by using (7.15) and its derivatives in (7.17) and then in (7.11) or, equivalently, from eq. (7.19) and (7.20). From the parameters of the system, it results \u03c9n = \u221a k1 + k2 m = 34",
            "20) one obtains qrise(t) = h k [ a11 T 11 r \u03c92 nt11 + ( a10 T 10 r \u03c92 n + 22 a11 T 11 r \u03b4\u03c9n ) t10 + ( a9 T 9 r \u03c92 n + 20 a10 T 10 r \u03b4\u03c9n + 110 a11 T 11 r ) t9 + ( a8 T 8 r \u03c92 n + 18 a9 T 9 r \u03b4\u03c9n + 90 a10 T 10 r ) t8 + ( a7 T 7 r \u03c92 n + 16 a8 T 8 r \u03b4\u03c9n + 72 a9 T 9 r ) t7 + ( a6 T 6 r \u03c92 n + 14 a7 T 7 r \u03b4\u03c9n + 56 a8 T 8 r ) t6 + ( a5 T 5 r \u03c92 n + 12 a6 T 6 r \u03b4\u03c9n + 42 a7 T 7 r ) t5 + ( 10 a5 T 5 r \u03b4\u03c9n + 30 a6 T 6 r ) t4 + 20 a5 T 5 r t3 ] and, after the substitutions, qrise(t) = 51609600 t11 \u2212 139560960 t10 + 162247680 t9 \u2212106122240 t8 + 42822144 t7 \u2212 10937472 t6 +1737792 t5 \u2212 168000 t4 + 10752 t3 for t \u2208 [0, Tr]. Because of symmetry (Tr = Td), the computation of qreturn(t) in the interval [Tr, T ], can be simply obtained as qreturn(t) = qrise(Tr) \u2212 qrise(t \u2212 Tr), t \u2208 [Tr, T ]. The desired motion x(t) for the mechanical system of Fig. 7.34, obtained from (7.17) as x(t) = h xN(\u03c4), with \u03c4 = t Tr , t \u2208 [0, Tr] for the rise (and similarly for the return) period is reported in Fig. 7.37. The overall function q(t) for the rise and return periods and its derivatives are shown in Fig. 7.38. 7.4 Frequency Modifications of Trajectories 313 In Fig. 7.39, a comparison is reported between the motion of the mass when q(t) is computed as described above, taking into consideration the dynamics of the mechanism, and when the input profile q\u2032(t) is the (smooth) profile defined by the polynomial function of degree 11 in (7",
            " The continuity of y(t) requires that the target trajectory profile q(t) belongs to the class of C\u03c1 bounded functions. If the mechanical system has no zeros, the computation of the feedforward action is particularly simple, being the integral terms in (7.26) not present. As a matter of fact, in this case the signal y(t) which guarantees a perfect tracking is a linear combination of q(t) and of its first \u03c1 derivatives, as already discussed in Sec. 7.4.1 with the polydyne approach. Example 7.7 Consider the linear and time-invariant mechanical system of Fig. 7.34, whose transfer function is G(s) = k1 ms2 + d1s + (k1 + k2) . In this case, the feedforward filter is F (s) = G\u22121(s) = ms2 + d1s + (k1 + k2) k1 and, if the desired motion q(t) is known along with its derivative up to the second order, the equation (7.26) provides the expression of the desired input signal: y(t) = m k1 q\u0308(t) + d1 k1 q\u0307(t) + k1 + k2 k1 q(t). (7.27) Note that the \u201cs\u201d operator in the Laplace domain is equivalent to the derivative action in the time domain. From (7.27), it results that a continuous input function y(t) can be obtained if the trajectory q(t) is continuous at least up to its second derivative (note that the relative degree of the system is \u03c1 = 2)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001394_tro.2010.2043757-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001394_tro.2010.2043757-Figure3-1.png",
        "caption": "Fig. 3. Forces acting on a supporting body (e.g., the right foot). Establishing the balance of moments on each contact body allows us to determine the position of contact CoP\u2019s.",
        "texts": [
            " For a number of ns links in contact, we associate ns contact CoP\u2019s. Each contact CoP is defined as the 2-D point on the contact surface where tangential moments are equal to zero. Therefore, 2ns coordinates describe all contact pressure points. In Fig. 2, we illustrate a multicontact scenario involving three supporting contacts on the robot\u2019s feet and left hand, and an operational task designed to interact with the robot\u2019s right hand. We focus on the analysis of the forces and moments taking place on a particular contact body (see Fig. 3). Based on the studies of Vukobratovic\u0301 and Borovac [33], we abstract the influence of the robot\u2019s body above the kth supporting extremity by the inertial and gravity force fsk , and the inertial and gravity moment msk acting on a sensor point Sk . For simplicity, we assume that the sensor is located at the mechanical joint of the contact extremity. Here, Pk represents the contact CoP of the kth contact extremity, frk is the vector of reaction forces acting on Pk , and mrk is the vector of reaction moments acting on Pk "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003750_(sici)1099-1239(199809)8:11<995::aid-rnc373>3.0.co;2-w-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003750_(sici)1099-1239(199809)8:11<995::aid-rnc373>3.0.co;2-w-Figure5-1.png",
        "caption": "Figure 5. Simplified helicopter",
        "texts": [
            " The tracking outputs are the (x, y) co-ordinates of the centre of mass. Analogous to References 7 and 18, the flat outputs are x & \"x! J m x r sin h, y & \"y# J m y r cos h (8) Note that these outputs are not fixed in body co-ordinates. The variable h can be expressed in terms of the flat outputs as tan h\" !m x x~ & m y y~ & #m ' g (9) From h and the flat outputs we can find the other states and the inputs. Example 3 (simplified helicopter model ) Consider the simplified helicopter depicted in Figure 5. At some level of abstraction we can look at the helicopter as a rigid body actuated by the thrust of the main rotor and the tail rotor. ( 1998 John Wiley & Sons, Ltd. Int. J. Robust Nonlinear Control 8, 995\u20141020 (1998) The tail rotor exerts a thrust along the body y-axis and a torque along the body z-axis. The tail rotor force is small compared to the thrust of the main rotor and we neglect it. The main thrust is roughly aligned with the body z-axis. We can measure the X\u00bdZ Euler angles (/, h,t)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003881_mssp.2001.1414-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003881_mssp.2001.1414-Figure6-1.png",
        "caption": "Figure 6. Geometry of contact between gears.",
        "texts": [
            " The results presented here are based on FEA modelling and the tooth sti!ness changes appear to be much more subtle than those used by Choy [4]. The direction, value and position of the frictional force between the gear teeth need to be determined in order to correctly introduce the frictional force into equations (4), (5), (14) and (17). The friction force between sliding teeth is in a direction perpendicular to the normal contacting force, which for involute gears is always along the line of contact ae as shown in Fig. 6. The relative velocity of the teeth in the direction of the line of contact ae is always zero. This result can be obtained by using velocity analysis as shown below. As the meshing position traverses the line of contact, the line of action of the sliding friction force between teeth in mesh also changes. The particular direction of the sliding friction force can be determined by analysing the relative velocity of the teeth at each position along the line of contact. When the contacting point is the pitch point p as shown in Fig. 6, the relative velocity between the contacting teeth will be zero. This can be shown from the geometry where qh/of\"qp/op\"u p /u g (19) in which u p represents the angular velocity of the pinion and u g the angular velocity of the gear. The relative velocity of the contacting point of the pinion with respect to the contacting point of the gear is v p \"of]u p !qh]u g \"0. (20) The sliding friction force at the contacting point p of the pinion will thus be zero. When the contacting point is on the left-hand side of the pitch point p as shown in Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000901_ja00216a040-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000901_ja00216a040-Figure2-1.png",
        "caption": "Figure 2. Cyclic voltammograms obtained with glucose oxidase modified with 13 f 1 ferrocylacetamide functions at different glucose concentrations; 3-mm-diameter glassy carbon electrodes; scan rate, 2 mV/s. Glucose (mM): a, 0; b, 0.8; c, 2; d, 4; e, 8.",
        "texts": [
            ", the currents observed a t sufficiently oxidizing potentials and a t high substrate concentrations, increase, in general, with the number of relays per enzyme molecule (Figures 2-6 and Table I, column The number of relays per enzyme molecule, the number of enzyme daltons per relay, the observed redox potentials of the enzyme-bound relays, the potential difference between the flavoenzymes FAD/FADH2 centers and those of the bound relays, and the period to 10% loss in substrate-dependent current 100 mV positive of the bound relay\u2019s reversible redox potential are 2). (22) Scheller, F.; Strand, G.; Neumann, B.; Kuhn, M.; Ostrowski, W. Bioelectrochem. Bioenerg. 1979, 6, 117. summarized in Table 11. The glucose concentration dependence of the current of a relay-modified glucose oxidase solution a t 0.43 V (vs SHE), derived from the data in Figure 2, is shown in Figure 7. The enzyme used in this measurement was modified by forming amides between 13 f 1 of its free amines and ferrocenylacetic acid. The limiting currents obtained for the same enzyme batch, modified through the same experimental procedures, were found M O D I F i C C T l O h C X 1 9 ' . : m i s e t - . J . Am. Chem. SOC., Vol. 110, No. 8, 1988 2619 COMMEhlTS ( Y W 3 i ) R u N s N = h - \" R O T F l N ~ 4 3 f C 4 I 2 R E L b l S DER ENZYME M O L E C J L E YONE DEC ~ I I 4 1 I 0 4 NA- i E EUZYME 3 2 : 0 3 I T R E C T W E N "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003908_978-1-84882-730-1-Figure3.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003908_978-1-84882-730-1-Figure3.2-1.png",
        "caption": "Figure 3.2 Motion variables for an ocean vessel",
        "texts": [
            " The following briefly presents the ocean vessel equations of motion based on the results in [11]. The resulting nonlinear model presented in this section is mainly intended for designing control systems in the next chapters. For a detailed and comprehensive derivation of the model, the reader is referred to [11,12,25]. The physical and control 3.3 Modeling of Ocean Vessels 43 properties of the model are also presented for control design and stability analysis. In this section, we use the notation, see Table 3.1 and Figure 3.2, that complies with the Society of Naval Architects and Marine Engineers (SNAME) [26]. For an ocean vessel moving in six degrees of freedom, six independent coordinates are required to determine its position and orientation. The first three coordinates .x;y;z/ and their first time derivatives correspond to the position and translational motion along the x-, y- and z-axes, while the last three coordinates . ; ; / and their first time derivatives describe orientation and rotational motion. According to SNAME, the six different motion components are defined as surge, sway, heave, roll, pitch, and yaw. To determine the equations of motion, two reference frames are considered: the inertial or fixed to earth frame OEXEYEZE that may be taken to coincide with the vessel fixed coordinates in some initial condition and the body-fixed frame ObXbYbZb see Figure 3.2. Since the motion of the Earth hardly affects ocean vessels (different from air vehicles), the earth-fixed frame OEXEYEZE can be considered to be inertial. For ocean vessels in general, the 44 3 Modeling of Ocean Vessels most commonly adopted position for the body-fixed frame is such that it gives hull symmetry about theObXbZb-plane and approximate symmetry about theObYbZbplane. In this sense, the body axesObXb ,ObYb , andObZb coincide with the principal axes of inertia and are usually defined as follows: ObXb is the longitudinal axis (directed from aft to fore); ObYb is the transverse axis (directed to starboard); and ObZb is normal axis (directed from top to bottom"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003438_s0921-5093(01)01179-0-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003438_s0921-5093(01)01179-0-Figure1-1.png",
        "caption": "Fig. 1. Beam and substrates.",
        "texts": [
            " A three dimensional FEM was developed to examine the deposition pattern effect on distortions induced during the SDM process. This model is similar to models developed to investigate thermal stresses in welding and a general treatment of the subject is found in the book, \u2018\u2018Analysis of Welded Structures\u2019\u2019, by Masubuchi [12]. In addition, a survey of advances in modeling thermal stresses in welding was done by Chandra [13] in 1985. An uncoupled thermal and mechanical analysis was performed. As shown in Fig. 1a 15.2\u00d72.5\u00d70.6 cm3 1117 steel beam substrate and a 15.2\u00d715.2\u00d70.6 cm3 1117 steel plate substrate were considered. To reduce the computation time for both geometries the deposition pattern was approximated with a pattern that has two symmetry planes. With this approximation only 1/4 of the beam or the plate needed to be analyzed. A uniform and relatively coarse mesh of 60\u00d710\u00d710 cm3 elements for the beam and 45\u00d745\u00d75 cm3 elements for the plate model was used. This mesh density was chosen to keep the computation time under 100 h using an Ultra 5 Sun Sparc Station",
            " Since the aluminum block was four times thicker than the substrate, the distortions of the aluminum were small compared to the distortions of the steel substrates. The aluminum block was therefore modeled as a rigid surface. The bolts were modeled by constraining the material near the bolts from moving off of the rigid surface. After the substrate cooled to room temperature, the boundary conditions were re- defined to remove the restraint due to the bolts and the deflection was measured as shown in Fig. 1. Room temperature material properties for annealed 1117 steel were used. The value for the modulus of elasticity was 200 GPa, Poisson\u2019s ratio was 0.3, the yield strength was 230 MPa, the ultimate tensile strength was 430 MPa, the elongation was 23 percent, and the expansion coefficient was 12 m mK\u22121 [14]. All material properties were assumed to be temperature independent except for the yield strength, which varies linearly from its room temperature value to 10 percent of its room temperature value at the melting temperature"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000783_1.1898523-Figure9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000783_1.1898523-Figure9-1.png",
        "caption": "Fig. 9. A single two-dimensional strip of weight W per unit length in the direction perpendicular to the page floating at an interface, with the hatched area indicating the area of the displaced fluid, the weight of which is equal to the buoyancy force on the strip. The asymmetric equilibrium position here is possible because of the off-center position for the strip\u2019s center of mass.",
        "texts": [
            " 20 For spherical particles of radius less than 0.3 mm or B 0.01), R 0.1 throughout the motion, which is sufficiently small that our approach is self-consistent. VII. AMPHIPHILIC STRIPS WITHOUT CHEMISTRY As a final illustration of the principles that we have applied to the problem of the attraction of interfacial objects, we consider briefly the equilibrium of a single two- dimensional strip, say of plastic of weight W per unit length into the page and width 2b) floating horizontally at an interface, as shown in Fig. 9. This problem was studied extensively in Ref. 11, where the interactions between two such strips also was investigated. Here we content ourselves with studying the simpler problem of a single strip but with a slight twist: we take the center of mass line of the strip to be displaced from the strip\u2019s center line by a distance (1 )b so that (1 ) is a measure of the offset of the center of mass of the strip . As we shall see, an offset center of mass is enough to break the symmetry of the problem, and thus allow the strip to float at an angle to the horizontal",
            "1 and half-width b/Lc 2. 823 823Am. J. Phys., Vol. 73, No. 9, September 2005 D. Vella and L. Mahadevan This article is copyrighted as indicated in the article. Reuse of AAPT content is subject to the terms at: http://scitation.aip.org/termsconditions. Downloaded to IP: 129.49.23.145 On: Thu, 18 Dec 2014 14:23:52 W sin 1 sin 2 g 2b cos z1 z2 2 22 for vertical equilibrium. The second term on the right-hand side of Eq. 22 is the weight of liquid displaced by the strip, which is shown as the hatched area in Fig. 9. In addition, there is the condition of torque balance, which we did not encounter for the equilibrium of the floating sphere it is automatically satisfied for objects of circular cross-section . Here this condition is crucial because it gives a third equation by which to determine the three unknowns. By taking moments about the center of mass, we have: 0 b 2 sin 2 sin 1 g b (2 )b z s sds , 23 where s is an arclength coordinate measured along the strip from the center of mass. In principle, these equations can be solved even for interfacial deformations that are not small following the strategy outlined in Ref"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000730_s0022112006002631-Figure26-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000730_s0022112006002631-Figure26-1.png",
        "caption": "Figure 26. Effect of bottom-heaviness on the trajectories of two squirmers with \u03b2 =5 and Gbh =0, 5 or 50. The initial condition is \u03b80 = 0 (two-dimensional orientation) and \u03b4y = 5. Gravity acts in the \u2212x-direction.",
        "texts": [
            " All cases show a similar tendency, and the interaction is increased by decreasing \u03b4y . Lastly the effect of bottom-heaviness was investigated. The relevant parameter Gbh is defined in (4.7). If one assumes that the micro-organisms swim in water at 20 body lengths per second with their centre of mass a half-radius from the geometric centre, Gbh is about 5 for micro-organisms with radius 10 \u00b5m and about 50 for micro-organisms with radius 100 \u00b5m. The trajectories of two bottom-heavy squirmers under the initial conditions \u03b80 = 0 and \u03b4y =5 are shown in figure 26. Gravity acts in the \u2212x-direction. The interaction in the near field inclines the trajectories to the right in all cases; however, the final direction is almost upwards if Gbh = 0. The two squirmers may come closer if Gbh is increased. The interaction of two squirmers has been calculated analytically for the limits of small and large separation and has also been calculated numerically using a boundaryelement method. In the far-field analysis, the translational\u2013rotational velocities and the stresslet due to the interaction were derived, and the results correspond well with the numerical results"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure12.25-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure12.25-1.png",
        "caption": "Figure 12.25 (a) A structure loaded by a vertical force on the left frame leg. (b) The isolated member AEB. (c) Bending moment diagram.",
        "texts": [
            " Check 1: We can read from the M diagram that the bending moment at the position of the point load is 20 kNm, with tension at the underside of the beam. This can be checked using the moment equilibrium of the isolated part BDE. Check 2: Note that the magnitude of the shear forces and the deformation symbols agree with the slopes of the M diagram. This relationship between the M and V diagram represents a simple and fast way of checking their correctness. Example 6 All the forces on the isolated members ACE and BDE for the structure in Figure 12.25a were calculated in Section 5.5, Example 2. To draw the M diagram we have to calculate only the bending moment at the three points C, D and G. To do so, it is sufficient to know the support reactions and the normal force in member CD (see Figure 12.25b): 504 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM MC = (25 kN)(3 m) = 75 kNm, MG = (25 kN)(4.5 m) \u2212 (22.5 kN)(2 m) = 67.5 kNm, MD = (15 kN)(3 m) = 45 kNm. The M diagram is shown in Figure 12.25c. Since there is only a normal force in the two-force member CD, this member has been omitted to simplify the figure. The shear forces can be determined directly from the slopes of the M diagram. For example: V (AC) = M(AC) (AC) = 75 kNm 5 m = 15 kN, V (CG) = M(CG) (CG) = (75 kNm) \u2212 (67.5 kNm) 2.5 m = 3 kN, V (GE) = M(GE) (GE) = 67.5 kNm 2.5 m = 27 kN. The associated deformation symbols follow from the \u201csteps\u201d in the M diagram (they are not shown here). The complete V diagram is shown in Figure 12.25d. Here too the two-force member CD has been omitted. The N diagram is shown in Figure 12.25e. Determining the N diagram is relatively laborious as we have to resolve all the forces on the members into components normal to the member axis (step changes in the V diagram) and components parallel to the member axis (step changes in the N diagram). For example, for the horizontal force of 22.5 kN at C, the component normal to the axis of member ACE is 12 Bending Moment, Shear Force and Normal Force Diagrams 505 4 5 \u00d7 (22.5 kN) = 18 kN and the component parallel to the member axis is 3 5 \u00d7 (22"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure8.28-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure8.28-1.png",
        "caption": "Fig. 8.28. Gripper solution for various diameter round objects",
        "texts": [
            " These added jaw parts are adapted to the form and material of the work pieces. Fig. 8.26. Scissor-type grippers Fig. 8.27. Parallel jaw-motion grippers Frequently gripping over two surfaces, or even along two lines, is not sufficient to ensure transferring and handling of the workpiece, as in the case of hard material which tends to slip out of the jaws. For these special solutions like the one from the Japanese firm Fanuc are applied for secure gripping of round work pieces of different diameters (see Fig. 8.28). By parallelogram linking of each of the three jaw elements, their movement which runs parallel to the gripped surface under a constant angle is ensured, so that twisting of the workpiece during gripping is avoided which can be important in some cases. There are often unusual solutions for grippers handling fragile, breakable, weak or workpieces with a specially treated (e.g. polished) surface. Thus the gripper illustrated in Fig. 8.29 is executed in the form of two or more (usually three) \"fingers\" made of a special rubber compound which by using solution of an internal plane and external ribbed surface have the property of bending when subdued to internal air prt1ssure and so gripping the workpiece \"softly\""
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure6.14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure6.14-1.png",
        "caption": "FIGURE 6.14. The final design of the windshield mechanism at the initial and final positions.",
        "texts": [
            "13 illustrates the finalized four-bar linkage of the windshield wiper at the initial position. Example 229 F Designing a dyad to attach a motor. The main four-bar linkage of a windshield wiper is a rocker-rocker mechanism because both the input and the output links must oscillate between two specific limits. To run the wipers and lock them at the limits, a two-link dyad can be designed. First we set the point of installing a rotary motor according to the physical conditions. Let point P , as shown in Figure 6.14, be the point at which we install the electric motor to run the mechanism. The next step would be to select a point on the input link to attach the second link of the dyad. Although joint B is usually the best choice, we select a point on the extension of the input link, indicated by D. There must be a dyad between joints D and P with lengths p and q. When the mechanism 330 6. Applied Mechanisms is at the initial position, joint D is at the longest distance form the motor P , and when it is at the final position, joint D is at the shortest distance form the motor P ",
            " l = longest distance between P and D s = shortest distance between P and D p, q = dyad lengths between P and D When P and D are at the maximum distance, the two-link dyad must be along each other, and when P and D are at the minimum distance, the two-link dyad must be on top of each other. Therefore, l = q + p (6.117) s = q \u2212 p (6.118) where p is the shortest link, and q is the longest link of the dyad. Solving Equations (6.117) and (6.118) for p and q provides p = l \u2212 s 2 (6.119) q = l + s 2 . (6.120) In this example we measure l = 453.8mm s = 312.1mm (6.121) and calculate for p and q p = 70.8mm q = 382.9mm. (6.122) The final design of the windshield mechanism and the running motor is shown in Figure 6.14 at the initial and final positions. The shorter link of the running dyad, p, must be attached to the motor at P , and the larger link, q, connects joint D to the shorter link at C. The motor will turn the shorter link, PC continuously at an angular speed \u03c9, while the longer link, CD, will run the mechanism and protect the wiper links to go beyond the initial and final angles. Example 230 Application of four-bar linkage in a vehicle. The double A arm suspension is a very popular mechanism for independent suspension of street cars"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.21-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.21-1.png",
        "caption": "FIGURE 7.21. Two possible angle \u03b8 for a set of (Rt, b1, b2).",
        "texts": [
            " Steering Dynamics tation is related to the vehicle\u2019s geometry by Rt = s\u00b5 l cot \u03b4i + 1 2 w \u00b62 + b21 \u2212 b22 (7.58) Rt = s\u00b5 l cot \u03b4o \u2212 1 2 w \u00b62 + b21 \u2212 b22 (7.59) Rt = q R2 \u2212 a22 + b21 \u2212 b22. (7.60) Using the equation Rt sin \u03b8 = b1 + b2 cos \u03b8 (7.61) and employing trigonometry, we may calculate the angle \u03b8 between the trailer and the vehicle as (7.55). The minus sign, in case b1 \u2212 b2 6= 0, is the usual case in forward motion, and the plus sign is a solution associated with a backward motion. Both possible configuration \u03b8 for a set of (Rt, b1, b2) are shown in Figure 7.21. The \u03b82 is called a jackknifing configuration. 7. Steering Dynamics 401 Example 270 F Two possible trailer-vehicle angles. Consider a four-wheel vehicle that is pulling a one-axle trailer with the following dimensions: l = 103.1 in \u2248 2.619m w = 61.6 in \u2248 1.565m b1 = 24 in \u2248 0.61m b2 = 90 in \u2248 2.286m \u03b4i = 12deg \u2248 0.209 rad (7.62) The kinematic steering characteristics of the vehicle would be \u03b4o = cot\u22121 \u00b3w l + cot \u03b4i \u00b4 = 0.186 rad \u2248 10.661 deg (7.63) Rt = s\u00b5 l cot \u03b4i + 1 2 w \u00b62 + b21 \u2212 b22 = 509.57 in \u2248 12"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure3.9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure3.9-1.png",
        "caption": "FIGURE 3.9. Illustration of laterally and longitudinally tire deformation.",
        "texts": [
            " They are measured by the slope of the experimental stiffness curves 102 3. Tire Dynamics in the (Fx,4x) and (Fy,4y) planes. kx = lim 4x\u21920 \u2202f \u2202 (4x) (3.10) ky = lim 4y\u21920 \u2202f \u2202 (4y) (3.11) When the longitudinal and lateral forces increase, parts of the tireprint creep and slide on the ground until the whole tireprint starts sliding. At this point, the applied force saturates and reaches its maximum supportable value. Generally, a tire is most stiff in the longitudinal direction and least stiff in the lateral direction. kx > kz > ky (3.12) Figure 3.9 illustrates tire deformation under a lateral and a longitudinal force. Example 77 F Nonlinear tire stiffness. In a better modeling, the vertical tire force Fz is a function of the normal tire deflection 4z, and deflection velocity 4z\u0307. Fz = Fz (4z,4z\u0307) (3.13) = Fzs + Fzd (3.14) In a first approximation we may assume Fz is a combination of a static and a dynamic part. The static part is a nonlinear function of the vertical tire deflection and the dynamic part is proportional to the vertical speed of the tire"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003657_02783649922066475-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003657_02783649922066475-Figure2-1.png",
        "caption": "Fig. 2. The Dante II mechanism and sensors.",
        "texts": [
            " Section 2 begins with an overall description of the Dante II system, followed by a summary of the approach that guided our design and development process. A complete technical description of the Dante II robot system, including associated remote and support equipment, is then presented. Design rationale are given throughout the technical description. In Section 3, technical results from the extensive testing program and mission are described. Finally, Section 4 enumerates technical and programmatic lessons that the project can teach or demonstrate. Dante II, shown in Figure 2, is a framewalker; its eight pantographic legs are arranged in two groups of four, on inner and at East Tennessee State University on June 6, 2015ijr.sagepub.comDownloaded from outer frames. Each leg can individually adjust its position vertically to avoid obstacles and adapt to rough terrain. Body translation is actuated by a single drive train that moves the frames with respect to each other. The frames can turn relative to one another, to change heading. The maximum turn per step is 7.5\u25e6, so it is best to avoid obstacles in advance to minimize repeated turning"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.11-1.png",
        "caption": "FIGURE 8.11. A Watt suspension mechanisms with a Panhard arm.",
        "texts": [],
        "surrounding_texts": [
            "The suspension is what links the wheels to the vehicle body and allows relative motion. This chapter covers the suspension mechanisms, and discusses the possible relative motions between the wheel and the vehicle body. The wheels, through the suspension linkage, must propel, steer, and stop the vehicle, and support the associated forces. 8.1 Solid Axle Suspension The simplest way to attach a pair of wheels to a vehicle is to mount them at opposite ends of a solid axle, such as the one that is shown in Figure 8.1. The solid axle must be attached to the body such that an up and down motion in the z-direction, as well as a roll rotation about the x-axis, is possible. So, no forward and lateral translation, and also no rotation about the axle and the z-axis, is allowed. There are many combinations of links and springs that can provide the kinematic and dynamic requirements. The simplest design is to clamp the axle to the middle of two leaf springs with their ends tied or shackled to the vehicle frame as shown schematically in Figure 8.1. A side view of a multi-leaf spring and solid axle is shown in Figure 8.2. A suspension with a solid connection between the left and right wheels is called dependent suspension. 456 8. Suspension Mechanisms The performance of a solid axle with leaf springs suspension can be improved by adding a linkage to guide the axle kinematically and provide dynamic support to carry the non z-direction forces. The solid axle with leaf spring combination came to vehicle industry from horse-drawn vehicles. Example 299 Hotchkiss drive. When a live solid axle is connected to the body with nothing but two leaf springs, it is called the Hotchkiss drive, which is the name of the car that used it first. The main problems of a Hotchkiss drive, which is shown in Figure 8.2, are locating the axle under lateral and longitudinal forces, and having a low mass ratio \u03b5 = ms/mu, where ms is the sprung mass and mu is the unsprung mass. Sprung mass refers to all masses that are supported by the spring, such as vehicle body. Unsprung mass refers to all masses that are attached to and not supported by the spring, such as wheel, axle, or brakes. Example 300 Leaf spring suspension and flexibility problem. The solid axle suspension systems with longitudinal leaf springs have many drawbacks. The main problem lies in the fact that springs themselves act as locating members. Springs are supposed to flex under load, but their flexibility is needed in only one direction. However, it is the nature of leaf springs to twist and bend laterally and hence, flex also in planes other than the tireplane. Leaf springs are not suited for taking up the driving and braking traction forces. These forces tend to push the springs into an S-shaped profile, as shown in Figure 8.3. The driving and braking flexibility of leaf springs, generates a negative caster and increases instability. Long springs provide better ride. However, long sprigs exaggerate their bending and twisting under different load conditions. Example 301 Leaf spring suspension and flexibility solution. To reduce the effect of a horizontal force and S-shaped profile appearance in a solid axle with leaf springs, the axle may be attached to the chassis by a longitudinal bar as Figure 8.4(a) shows. Such a bar is called an anti-tramp 8. Suspension Mechanisms 457 bar, and the suspension is the simplest cure for longitudinal problems of a Hotchkiss drive. A solid axle with an anti-tramp bar may be kinematically approximated by a four-bar linkage, as shown in Figure 8.4(b). Although an anti-tramp bar may control the shape of the leaf spring, it introduces a twisting angle problem when the axle is moving up and down, as shown in Figure 8.5. Twisting the axle and the wheel about the axle is called caster. The solid axle is frequently used to help keeping the wheels perpendicular to the road. Example 302 Leaf spring location problem. The front wheels need room to steer left and right. Therefore, leaf springs cannot be attached close to the wheel hubs, and must be placed closer to the middle of the axle. That gives a narrow spring-base, which means that a small side force can sway or tilt the body relative to the axle through a considerable roll angle due to weight transfer. This is uncomfortable for the vehicle passengers, and may also produce unwanted steering. The solid axle positively prevents the camber change by body roll. The wheels remain upright and hence, do not roll on a side. However, a solid axle shifts laterally from its static plane and its center does not remain on the vehicle\u2019s longitudinal axis under a lateral force. 458 8. Suspension Mechanisms A solid axle produces bump-camber when single-wheel bump occurs. If the right wheel goes over a bump, the axle is raised at its right end, and that tilts the left wheel hub, putting the left wheel at a camber angle for the duration of deflection. Example 303 Triangular linkage. A triangulated linkage, as shown in Figure 8.6, may be attached to a solid axle to provide lateral and twist resistance during acceleration and braking. Example 304 Panhard arm. High spring rate is a problem of leaf springs. Reducing their stiffness by narrowing them and using fewer leaves, reduces the lateral stiffness and increases the directional stability of the suspension significantly. A Panhard arm is a bar that attaches a solid axle suspension to the chassis laterally. Figure 8.7 illustrates a solid axle and a Panhard arm to guide the axle. Figure 8.8 shows a triangular linkage and a Panhard arm combination for guiding a solid axle. A double triangle mechanism, as shown in Figure 8.9, is an alternative design to guide the axle and support it laterally. Example 305 Straight line linkages. There are many mechanisms that can provide a straight line motion. The simplest mechanisms are four-bar linkages with a coupler point moving straight. Some of the most applied and famous linkages are shown in Figure 8.10. By having proper lengths, the Watt, Robert, Chebyshev, and Evance linkages can make the coupler point C move on a straight line vertically. Such a mechanism and straight motion may be used to guide a solid axle. Two Watt suspension mechanisms with a Panhard arm are shown in Figures 8.11 and 8.12. 460 8. Suspension Mechanisms 8. Suspension Mechanisms 461 Figures 8.13, 8.14, and 8.15 illustrate three combinations of Robert suspension linkages equipped with a Panhard arm. Example 306 Solid axle suspension and unsprung mass problem. A solid axle is counted as an unsprung member, and hence, the unsprung mass is increased where using solid axle suspension. A heavy unsprung mass ruins both, the ride and handling of a vehicle. Lightening the solid axle makes it weaker and increases the most dangerous problem in vehicles: axle breakage. The solid axle must be strong enough to make sure it will not break under any loading conditions at any age. As a rough estimate, 90% of the leaf spring mass may also be counted as unsprung mass, which makes the problem worse. The unsprung mass problem is worse in front, and it is the main reason that they are no longer used in street cars. However, front solid axles are still common on trucks and buses. These are heavy vehicles and solid axle 8. Suspension Mechanisms 463 464 8. Suspension Mechanisms suspension does not reduce the mass ratio \u03b5 = ms/mu very much. When a vehicle is rear-wheel-drive and a solid axle suspension is used in the back, the suspension is called live axle. A live axle is a casing that contains a differential, and two drive shafts. The drive shafts are connected to the wheel hubs. A live axle can be three to four times heavier than a dead I-beam axle. It is called live axle because of rotating gears and shafts inside the axle. Example 307 Solid axle and coil spring. To decrease the unsprung mass and increase vertical flexibility of solid axle suspensions, it is possible to equip them with coil springs. A sample of a solid axle suspension with coil spring is shown in Figure 8.16. The suspension mechanism is made of four longitudinal bars between the axle and chassis. The springs may have some lateral or longitudinal angle to introduce some lateral or longitudinal compliance. Example 308 De Dion axle. When a solid axle is a dead axle with no driving wheels, the connecting beam between the left and right wheels may have different shapes to do different jobs, usually to give the wheels independent flexibility. We may also modify the shape of a live axle to attach the differential to the chassis and reduce the unsprung mass. De Dion design is a modification of a beam axle that may be used as a dead axle or to attach the differential to the chassis and transfer the driving power to the drive wheels by employing universal joints and split shafts. Figure 8.17 illustrates a De Dion suspension. 8.2 Independent Suspension Independent suspensions is introduced to let a wheel to move up and down without affecting the opposite wheel. There are many forms and designs of independent suspensions. However, double A-arm and McPherson strut suspensions are the simplest and the most common designs. Figure 8.18 illustrates a sample of a double A-arm and Figure 8.19 shows a McPherson suspension. Kinematically, a double A-arm suspension mechanism is a four-bar linkage with the chassis as the ground link, and coupler as the wheel carrying link. A McPherson suspension is an inverted slider mechanism that has the chassis as the ground link and the coupler as the wheel carrying link. A double A-arm and a McPherson suspension mechanism on the left and right wheels are schematically shown in Figures 8.20 and 8.21 respectively. Double A-arm, is also called double wishbone, or short/long arm suspension. McPherson also may be written as MacPherson. Example 309 Double A-arm suspension and spring position. Consider a double A-arm suspension mechanism. The coil spring may be between the lower arm and the chassis, as shown in Figure 8.18. It is also possible to install the spring between the upper arm and the chassis, or between the upper and lower arms. In either case, the lower or the upper arm, which supports the spring, is made stronger and the other arm acts as a connecting arm. Example 310 Multi-link suspension mechanism. When the two side bars of an A-arm are attached to each other with a joint, as shown in Figure 8.22, then the double A-arm is called a multi-link mechanism. A multi-link mechanism is a six-bar mechanism that may have a better coupler motion than a double A-arm mechanism. However, multi- 466 8. Suspension Mechanisms 8. Suspension Mechanisms 467 link suspensions are more expensive, less reliable, and more complicated compare to a double A-arm four-bar linkage. There are vehicles with more than six-link suspension with possibly better kinematic performance. Example 311 Swing arm suspension. An independent suspension may be as simple as a triangle shown in Figure 8.23. The base of the triangle is jointed to the chassis and the wheel to the tip point. The base of the triangle is aligned with the longitudinal axis of the vehicle. Such a suspension mechanism is called a swing axle or swing arm. The variation in camber angle for a swing arm suspension is maximum, compared to the other suspension mechanisms. Example 312 Trailing arm suspension. Figure 8.24 illustrates a trailing arm suspension that is a longitudinal arm with a lateral axis of rotation. The camber angle of the wheel, supported by a trailing arm, will not change during the up and down motion. Trailing arm suspension has been successfully using in a variety of frontwheel-drive vehicles, to suspend their rear wheels. Example 313 Semi-trailing arm Semi-trailing arm suspension, as shown in Figure 8.25, is a compromise between the swing arm and trailing arm suspensions. The joint axis may have any angle, however an angle not too far from 45 deg is more applied. Such suspensions have acceptable camber angle change, while they can handle both, the lateral and longitudinal forces. Semi-trailing design has successfully applied to a series of rear-wheel-drive cars for several decades. Example 314 Antiroll bar and roll stiffness. Coil springs are used in vehicles because they are less stiff with better 470 8. Suspension Mechanisms ride comfort compared to leaf springs. Therefore, the roll stiffness of the vehicle with coil springs is usually less than in vehicles with leaf springs. To increase the roll stiffness of such suspensions, an antiroll bar must be used. Leaf springs with reduced layers, uni-leaf, trapezoidal, or nonuniform thickness may also need an antiroll bar to compensate for their reduced roll stiffness. The antiroll bar is also called a stabilizer. Figure 8.26 illustrates an anti-roll bare attached to a solid axle with coil springs. Example 315 Need for longitudinal compliance. A bump is an obstacle on the road that opposes the forward motion of a wheel. When a vehicle goes over a bump, the first action is a force that tends to push the wheel backward relative to the rest of the vehicle. So, the lifting force has a longitudinal component, which will be felt inside the vehicle unless the suspension system has horizontal compliance. There are situations in which the horizontal component of the force is even higher than the vertical component. Leaf springs can somewhat absorb this horizontal force by flattening out and stretching the distance from the forward spring anchor and the axle. Such a stretch is usually less than 1/2 in \u2248 1 cm. 8.3 Roll Center and Roll Axis The roll axis is the instantaneous line about which the body of a vehicle rolls. Roll axis is found by connecting the roll center of the front and rear suspensions of the vehicle. Assume we cut a vehicle laterally to disconnect the front and rear half of the vehicle. Then, the roll center of the front or rear suspension is the instantaneous center of rotation of the body with respect to the ground. 8. Suspension Mechanisms 471 472 8. Suspension Mechanisms Figure 8.27 illustrates a sample of the front suspensions of a car with a double A-arm mechanism. To find the roll center of the body with respect to the ground, we analyze the two-dimensional kinematically equivalent mechanism shown in Figure 8.28. The center of tireprint is the instant center of rotation of the wheel with respect to the ground, so the wheels are jointed links to the ground at their center of tireprints. The instant center I18 is the roll center of the body with respect to the ground. To find I18, we apply the Kennedy theorem and find the intersection of the line I12I28 and I13I38 as shown in Figure 8.29. The point I28 and I38 are the instant center of rotation for the wheels with respect to the body. The instant center of rotation of a wheel with respect to the body is called suspension roll center. So, to find the roll center of the front or rear half of a car, we should determine the suspension roll centers, and find the intersection of the lines connecting the suspension roll centers to the center of their associated tireprints. The Kennedy theorem states that the instant center of every three relatively moving objects are colinear. Example 316 McPherson suspension roll center. A McPherson suspension is an inverted slider crank mechanism. The instant centers of an example of inverted slider crank mechanism are shown in Figure 8.30. In this figure, the point I12 is the suspension roll center, which is the instant center of rotation for the wheel link number 2 with respect to the chassis link number 1. A car with a McPherson suspension system is shown in Figure 8.31. The kinematic equivalent mechanism is depicted in Figure 8.32. Suspension roll centers along with the body roll center are shown in Figure 8.33. To 8. Suspension Mechanisms 473 find the roll center of the front or rear half of a car, we determine each suspension roll center and then find the intersection of the lines connecting the suspension roll centers to the center of the associated tireprint. Example 317 Roll center of double A-arm suspension. The roll center of an independent suspension such as a double A-arm can be internal or external. The kinematic model of a double A-arm suspension for the front left wheel of a car is illustrated in Figure 8.34. The suspension roll center in Figure 8.34(a) is internal, and in Figure 8.34(b) is external. An internal suspension roll center is toward the vehicle body, while an external suspension roll center goes away from the vehicle body. A suspension roll center may be on, above, or below the road surface, as shown in Figure 8.35(a)-(c) for an external suspension roll center. When the suspension roll center is on the ground, above the ground, or below the ground, the vehicle roll center would be on the ground, below the ground, and above the ground, respectively. Example 318 F Camber variation of double A-arm suspension. When a wheel moves up and down with respect to the vehicle body, de- 474 8. Suspension Mechanisms 8. Suspension Mechanisms 475 pending on the suspension mechanism, the wheel may camber. Figure 8.36 illustrates the kinematic model for a double A-arm suspension mechanism. The mechanism is equivalent to a four-bar linkage with the ground link as the vehicle chasis. The wheel is always attached to a coupler point C of the mechanism. We set a local suspension coordinate frame (x, y) with the x-axis indicating the ground link MN . The x-axis makes a constant angle \u03b80 with the vertical direction. The suspension machanism has a length a for the upper A-arm, b for the coupler link, c for the lower A-arm, and d for the ground link. The configuration of the suspension is determined by the angles \u03b82, \u03b83, and \u03b84, all measured from the positive direction of the x-axis. When the suspension is at its equilibrium position, the links of the double A-arm suspension make initial angles \u03b820 \u03b830, and \u03b840 with the x-axis. The equilibrium position of a suspension is called the rest position. To determine the camber angle during the fluctuation of the wheel, we should determine the variation of the coupler angle \u03b83, as a function of vertical motion z of the coupler point C. Using \u03b82 as a parameter, we can find the coordinates (xC , yC) of the 476 8. Suspension Mechanisms coupler point C in the suspension coordinate frame (x, y) as xC = a cos \u03b82 + e cos (p\u2212 q + \u03b1) (8.1) yC = a sin \u03b82 + e sin (p\u2212 q + \u03b1) (8.2) where, q = tan\u22121 a sin \u03b82 d\u2212 a cos \u03b82 (8.3) p = tan\u22121 q 4b2f2 \u2212 (b2 + f2 \u2212 c2) 2 b2 + f2 \u2212 c2 (8.4) f = p a2 + d2 \u2212 2ad cos \u03b82. (8.5) The position vector of the coupler point is uC uC = xC \u0131\u0302+ yC j\u0302 (8.6) and the unit vector in the z-direction is u\u0302z = \u2212 cos \u03b80\u0131\u0302\u2212 sin \u03b80j\u0302. (8.7) Therefore, the displacement z in terms of xC and yC is: z = uC \u00b7 u\u0302z = \u2212xC cos \u03b80 \u2212 yC sin \u03b80 (8.8) 8. Suspension Mechanisms 477 The initial coordinates of the coupler point C and the initial value of z are: xC0 = a cos \u03b820 + e cos (p0 \u2212 q0 + \u03b1) (8.9) yC0 = a sin \u03b820 + e sin (p0 \u2212 q0 + \u03b1) (8.10) z0 = \u2212xC0 cos \u03b80 \u2212 yC0 sin \u03b80 (8.11) and hence, the vertical displacement of the wheel center can be calculated by h = z \u2212 z0 (8.12) The initial angle of the coupler link with the vertical direction is \u03b80 \u2212 \u03b830. Therefore, the camber angle of the wheel would be \u03b3 = (\u03b80 \u2212 \u03b83)\u2212 (\u03b80 \u2212 \u03b830) = \u03b830 \u2212 \u03b83 (8.13) The angle of the coupler link with the x-direction is equal to \u03b83 = 2 tan \u22121 \u00c3 \u2212E \u00b1 \u221a E2 \u2212 4DF 2D ! (8.14) where, D = J5 \u2212 J1 + (1 + J4) cos \u03b82 (8.15) E = \u22122 sin \u03b82 (8.16) F = J5 + J1 \u2212 (1\u2212 J4) cos \u03b82 (8.17) and J1 = d a (8.18) J2 = d c (8.19) J3 = a2 \u2212 b2 + c2 + d2 2ac (8.20) J4 = d b (8.21) J5 = c2 \u2212 d2 \u2212 a2 \u2212 b2 2ab . (8.22) Substituting (8.14) and (8.13), and then, eliminating \u03b82 between (8.13) and (8.8) provides the relationship between the vertical motion of the wheel, z, and the camber angle \u03b3. 478 8. Suspension Mechanisms Example 319 F Camber angle and wheel fluctuations. Consider the double A-arm suspension that is shown in Figure 8.36. The dimensions of the equivalent kinematic model are: a = 22.4 cm b = 22.1 cm c = 27.3 cm d = 17.4 cm \u03b80 = 24.3 deg (8.23) The coupler point C is at: e = 14.8 cm \u03b1 = 54.8 deg (8.24) If the angle \u03b82 at the rest position is at \u03b820 = 121.5 deg (8.25) then the initial angle of the other links are: \u03b830 = 18.36 deg \u03b840 = 107.32 deg (8.26) At the rest position, the coupler point is at: xC0 = \u221222.73 cm yC0 = 9.23 cm z0 = 16.92 cm (8.27) We may calculate h and \u03b3 by varying the parameter \u03b82. Figure 8.37 illustrates h as a function of the camber angle \u03b3. For this suspension mechanism, the wheel gains a positive camber when the wheel moves up, and gains a negative camber when the it moves down. The mechanism is shown in Figure 8.38, when the wheel is at the rest position and has a positive or a negative displacement. 8.4 F Car Tire Relative Angles There are four major wheel alignment parameters that affect vehicle dynamics: toe, camber, caster, and trust angle. 8. Suspension Mechanisms 479 8.4.1 F Toe When a pair of wheels is set so that their leading edges are pointed toward each other, the wheel pair is said to have toe-in. If the leading edges point away from each other, the pair is said to have toe-out. Toe-in and toe-out front wheel configurations of a car are illustrated in Figure 8.39. The amount of toe can be expressed in degrees of the angle to which the wheels are not parallel. However, it is more common to express the toein and toe-out as the difference between the track widths as measured at the leading and trailing edges of the tires. Toe settings affect three major performances: tire wear, straight-line stability, and corner entry handling. For minimum tire wear and power loss, the wheels on a given axle of a car should point directly ahead when the car is running in a straight line. Excessive toe-in causes accelerated wear at the outboard edges of the tires, while too much toe-out causes wear at the inboard edges. Toe-in increases the directional stability of the vehicle, and toe-out increases the steering response. Hence, a toe-in setting makes the steering function lazy, while a toe-out makes the vehicle unstable. With four wheel independent suspensions, the toe may also be set at the rear of the car. Toe settings at the rear have the same effect on wear, directional stability, and turn-in as they do on the front. However, we usually do not set up a rear-drive race car toed out in the rear, because of excessive instability. When driving torque is applied to the wheels, they pull themselves forward and try to create toe-in. Furthermore, when pushed down the road, a non-driven wheel or a braking wheel will tend to toe-out. Example 320 Toe-in and directional stability. Toe settings have an impact on directional stability. When the steering 480 8. Suspension Mechanisms 8. Suspension Mechanisms 481 wheel is centered, toe-in causes the wheels to tent to move along paths that intersect each other in front of the vehicle. However, the wheels are in balance and no turn results. Toe-in setup can increase the directional stability caused by little steering fluctuations and keep the car moving straight. Steering fluctuations may be a result of road disturbances. If a car is set up with toe-out, the front wheels are aligned so that slight disturbances cause the wheel pair to assume rolling directions that approach a turn. Therefore, toe-out encourages the initiation of a turn, while toe-in discourages it. Toe-out makes the steering quicker. So, it may be used in vehicles for a faster response. The toe setting on a particular car becomes a trade-off between the straight-line stability afforded by toe-in and the quick steering response by toe-out. Toe-out is not desirable for street cars, however, race car drivers are willing to drive a car with a little directional instability, for sharper turn-in to the corners. So street cars are generally set up with toe-in, while race cars are often set up with toe-out. Example 321 Toe-in and toe-out in the front and rear axles. Front toe-in: slower steering response, more straight-line stability, greater wear at the outboard edges of the tires. Front toe-zero: medium steering response, minimum power loss, minimum tire wear. Front toe-out: quicker steering response, less straight-line stability, greater wear at the inboard edges of the tires. Rear toe-in: straight-line stability, traction out of the corner, more steerability, higher top speed. 482 8. Suspension Mechanisms 8.4.2 F Caster Angle Caster is the angle to which the steering pivot axis is tilted forward or rearward from vertical, as viewed from the side. Assume the wheel is straight to have the body frame and the wheel frame coincident. If the steering axis is turned about the wheel yw-axis then the wheel has positive caster. If the steering axis is turned about the wheel \u2212yw-axis, then the wheel has negative caster. Positive and negative caster configurations on the front wheel of a car are shown in Figure 8.40. Negative caster aids in centering the steering wheel after a turn and makes the front tires straighten quicker. Most street cars are made with 4\u22126 deg negative caster. Negative caster tends to straighten the wheel when the vehicle is traveling forward, and thus is used to enhance straight-line stability. Example 322 Negative caster of shopping carts. The steering axis of a shopping cart wheel is set forward of where the wheel contacts the ground. As the cart is pushed forward, the steering axis pulls the wheel along, and because the wheel drags along the ground, it falls directly in line behind the steering axis. The force that causes the wheel to follow the steering axis is proportional to the distance between the steering axis and the wheel-to-ground contact point, if the caster is small. This distance is referred to as trail. The cars\u2019 steering axis intersects the ground at a point in front of the tireprint, and thus the same effect as seen in the shopping cart casters is achieved. While greater caster angles improves straight-line stability, they also cause an increase in steering effort. Example 323 Characteristics of caster in front axle. Zero castor provides: easy steering into the corner, low steering out of the corner, low straight-line stability. 8. Suspension Mechanisms 483 Negative caster provides: low steering into the corner, easy steering out of the corner, more straight-line stability, high tireprint area during turn, good turn-in response, good directional stability, good steering feel. When a castered wheel rotates about the steering axis, the wheel gains camber. This camber is generally favorable for cornering. 8.4.3 F Camber Camber is the angle of the wheel relative to vertical line to the road, as viewed from the front or the rear of the car. Figure 8.41 illustrates the wheel number 1 of a vehicle. If the wheel leans in toward the chassis, it is called negative camber and if it leans away from the car, it is called positive camber. The cornering force that a tire can develop is highly dependent on its angle relative to the road surface, and so wheel camber has a major effect on the road holding of a car. A tire develops its maximum lateral force at a small camber angle. This fact is due to the contribution of camber thrust, which is an additional lateral force generated by elastic deformation as the tread rubber pulls through the tire/road interface. To optimize a tire\u2019s performance in a turn, the suspension should provide a slight camber angle in the direction of rotation. As the body rolls in a turn, the suspension deflects vertically. The wheel is connected to the chassis by suspension mechanism, which must rotate to allow for the wheel deflection. Therefore, the wheel can be subject to large camber changes as the suspension moves up and down. So, the more the wheel must deflect from its static position, the more difficult it is to maintain an ideal camber angle. Thus, the relatively large wheel travel and soft roll stiffness needed to provide a smooth ride in passenger cars presents a difficult design challenge, while the small wheel travel and high roll stiffness inherent in racing cars reduces the problem. Example 324 Castor versus camber. Camber doesn\u2019t improve turn-in as the positive caster does. Camber is not generally good for tire wear. Camber in one wheel does not improve directional stability. Camber adversely affects braking and acceleration efforts. 8.4.4 F Trust Angle The trust angle \u03c5 is the angle between vehicle\u2019s centerline and perpendicular to the rear axle. It compares the direction that the rear axle is aimed with the centerline of the vehicle. A nonzero angle configuration is shown in Figure 8.42. Zero angle confirms that the rear axle is parallel to the front axle, and the wheelbase on both sides of the vehicle are the same. A reason for nonzero 484 8. Suspension Mechanisms 8. Suspension Mechanisms 485 trust angle would have unequal toe-in or toe-out on both sides of the axle. Example 325 Torque reaction. There are two kinds of torque reactions in rear-whel-drive: 1\u2212 the reaction of the axle housing to rotate in the opposite direction of the crown wheel rotation, and 2\u2212 the reaction of axle housing to spin about its own center, opposite to the direction of pinion\u2019s rotation. The first reaction leads to a lifting force in the differential causing a wind-up in springs. The second reaction leads to a lifting force on the right wheels. 8.5 Suspension Requirements and Coordinate Frames The suspension mechanism should allow a relative motion between the wheel and the vehicle body. The relative motions are needed to pass the road irregularities and steering. To function properly, a suspension mechanism should have some kinematic and dynamics requirements. 8.5.1 Kinematic Requirements To express the motions of a wheel, we attach a wheel coordinate system W (oxwywzw) to the center of the wheel. A wheel, as a rigid body, has six degrees-of-freedom with respect to the vehicle body: three translations and three rotations, as shown in Figure 8.43. 486 8. Suspension Mechanisms The axes xw, yw, and zw indicate the direction of forward, lateral, and vertical translations and rotations. In the position shown in the figure, the rotation about the xw-axis is the camber angle, about the yw-axis is the spin, and about the zw-axis is the steer angle. Consider a non-steerable wheel. Translation in zw-direction and spin about the yw-axis are the only two DOF allowed for such a wheel. So, we need to take four DOF. If the wheel is steerable, then translation in the zw-direction, spin about the yw-axis, and steer rotation about the zw-axis are the three DOF allowed. So, we must take three DOF of a steerable wheel. Kinematically, non-steerable and steerable wheels should be supported as shown in Figures 8.44 and 8.45 respectively. Providing the required freedom, as well as eliminating the taken DOF, are the kinematic requirements of a suspension mechanism. 8.5.2 Dynamic Requirements Wheels should be able to propel, steer, and stop the vehicle. So, the suspension system must transmit the driving traction and deceleration braking forces between the vehicle body and the ground. The suspension members must also resist lateral forces acting on the vehicle. Hence, the wheel suspension system must make the wheel rigid for the taken DOF. However, there must also be some compliance members to limit the untaken DOF. The most important compliant members are spring and dampers to provide returning and resistance forces in the z-direction. 8. Suspension Mechanisms 487 8.5.3 Wheel, wheel-body, and tire Coordinate Frames Three coordinate frames are employed to express the orientation of a tire and wheel with respect to the vehicle: the wheel frame W , wheel-body frame C, and tire frame T . A wheel coordinate frame W (xw, yw, zw) is attached to the center of a wheel. It follows every translation and rotation of the wheel except the spin. Hence, the xw and zw axes are always in the tire-plane, while the yw-axis is always along the spin axis. A wheel coordinate frame is shown in Figure 8.43. When the wheel is straight and the W frame is parallel to the vehicle coordinate frame, we attach a wheel-body coordinate frame C (xc, yc, zc) at the center of the wheel parallel to the vehicle coordinate axes. The wheelbody frame C is motionless with respect to the vehicle coordinate and does not follow any motion of the wheel. The tire coordinate frame T (xt, yt, zt) is set at the center of the tireprint. The zt-axis is always perpendicular to the ground. The xt-axis is along the intersection line of the tire-plane and the ground. The tire frame does not follows the spin and camber rotations of the tire however, it follows the steer angle rotation about the zc-axis. Figure 8.46 illustrates a tire and a wheel coordinate frames. Example 326 Visualization of the wheel, tire, and wheel-body frames. Figure 8.47 illustrates the relative configuration of a wheel-body frame C, a tire frame T , and a wheel frame W . If the steering axis is along the zc-axis then, the rotation of the wheel about the zc-axis is the steer angle \u03b4. Rotation about the xt-axis is the camber angle \u03b3. 488 8. Suspension Mechanisms Generally speaking, the steering axis may have any angle and may go through any point of the ground plane. Example 327 Wheel to tire coordinate frame transformation. If TdW indicates the T -expression of the position vector of the wheel frame origin relative to the tire frame origin, then having the coordinates of a point P in the wheel frame, we can find its coordinates in the tire frame using the following equation. T rP = TRW W rP + TdW (8.28) If W rP indicates the position vector of a point P in the wheel frame, W rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 (8.29) then the coordinates of the point P in the tire frame T rP are T rP = TRW W rP + Td = TRW W rP + TRW W T dW = \u23a1\u23a3 xP yP cos \u03b3 \u2212Rw sin \u03b3 \u2212 zP sin \u03b3 Rw cos \u03b3 + zP cos \u03b3 + yP sin \u03b3 \u23a4\u23a6 (8.30) 8. Suspension Mechanisms 489 where, WT dW is the W -expression of the position vector of the wheel frame in the tire frame, Rw is the radius of the tire, and TRW is the rotation matrix to go from the wheel frame W to the tire frame T . TRW = \u23a1\u23a3 1 0 0 0 cos \u03b3 \u2212 sin \u03b3 0 sin \u03b3 cos \u03b3 \u23a4\u23a6 (8.31) W T dW = \u23a1\u23a3 0 0 Rw \u23a4\u23a6 . (8.32) As an example, the center of the wheel W rP = W ro = 0 is the origin of the wheel frame W , that is at T ro = TdW = TRW W T dW = \u23a1\u23a3 0 \u2212Rw sin \u03b3 Rw cos \u03b3 \u23a4\u23a6 (8.33) in the tire coordinate frame T . 490 8. Suspension Mechanisms Example 328 F Tire to wheel coordinate frame transformation. If rP indicates the position vector of a point P in the tire coordinate frame, T rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 (8.34) then the position vector W rP of the point P in the wheel coordinate frame is W rP = WRT T rP \u2212 W T dW (8.35) = \u23a1\u23a3 xP yP cos \u03b3 + zP sin \u03b3 zP cos \u03b3 \u2212Rw \u2212 yP sin \u03b3 \u23a4\u23a6 because WRT = \u23a1\u23a3 1 0 0 0 cos \u03b3 sin \u03b3 0 \u2212 sin \u03b3 cos \u03b3 \u23a4\u23a6 (8.36) WdT = \u23a1\u23a3 0 0 Rw \u23a4\u23a6 (8.37) and we may multiply both sides of Equation (8.28) by TRT W to get TRT W T rP = W rP + TRT W TdW (8.38) = W rP + W T dW W rP = WRT T rP \u2212 W T dW . (8.39) As an example, the center of tireprint in the wheel frame is at W rP = \u23a1\u23a3 1 0 0 0 cos \u03b3 \u2212 sin \u03b3 0 sin \u03b3 cos \u03b3 \u23a4\u23a6T \u23a1\u23a3 0 0 0 \u23a4\u23a6\u2212 \u23a1\u23a3 0 0 Rw \u23a4\u23a6 = \u23a1\u23a3 0 0 \u2212Rw \u23a4\u23a6 . (8.40) Example 329 F Wheel to tire homogeneous transformation matrices. The transformation from the wheel to tire coordinate frame may also be expressed by a 4\u00d7 4 homogeneous transformation matrix TTW , T rP = TTW W rP = \u2219 TRW TdW 0 1 \u00b8 W rP (8.41) where TTW = \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 cos \u03b3 \u2212 sin \u03b3 \u2212Rw sin \u03b3 0 sin \u03b3 cos \u03b3 Rw cos \u03b3 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 . (8.42) 8. Suspension Mechanisms 491 The corresponding homogeneous transformation matrix WTT from the tire to wheel frame would be WTT = \u2219 WRT WdT 0 1 \u00b8 = \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 cos \u03b3 \u2212 sin \u03b3 0 0 sin \u03b3 cos \u03b3 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 . (8.43) It can be checked that WTT = TT\u22121W , using the inverse of a homogeneous transformation matrix rule. TT\u22121W = \u2219 TRW TdW 0 1 \u00b8\u22121 = \u2219 TRT W \u2212 TRT W TdW 0 1 \u00b8 = \u2219 WRT \u2212WRT TdW 0 1 \u00b8 (8.44) Example 330 F Tire to wheel-body frame transformation. The origin of the tire frame is at CdT in the wheel-body frame. CdT = \u23a1\u23a3 0 0 \u2212Rw \u23a4\u23a6 (8.45) The tire frame can steer about the zc-axis with respect to the wheel-body frame. The associated rotation matrix is CRT = \u23a1\u23a3 cos \u03b4 \u2212 sin \u03b4 0 sin \u03b4 cos \u03b4 0 0 0 1 \u23a4\u23a6 (8.46) Therefore, the transformation between the tire and wheel-body frames can be expressed by Cr = CRT T r+ CdT (8.47) or equivalently, by a homogeneous transformation matrix CTT . CTT = \u2219 CRT CdT 0 1 \u00b8 = \u23a1\u23a2\u23a2\u23a3 cos \u03b4 \u2212 sin \u03b4 0 0 sin \u03b4 cos \u03b4 0 0 0 0 1 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 (8.48) As an example, the wheel-body coordinates of the point P on the tread of a 492 8. Suspension Mechanisms negatively steered tire at the position shown in Figure 8.48, are: Cr = CTT T rP = \u23a1\u23a2\u23a2\u23a3 cos\u2212\u03b4 \u2212 sin\u2212\u03b4 0 0 sin\u2212\u03b4 cos\u2212\u03b4 0 0 0 0 1 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 \u23a1\u23a2\u23a2\u23a3 Rw 0 Rw 1 \u23a4\u23a5\u23a5\u23a6 = \u23a1\u23a2\u23a2\u23a3 Rw cos \u03b4 \u2212Rw sin \u03b4 0 1 \u23a4\u23a5\u23a5\u23a6 (8.49) The homogeneous transformation matrix for tire to wheel-body frame TTC is: TTC = CT\u22121T = \u2219 CRT CdT 0 1 \u00b8\u22121 = \u2219 CRT T \u2212CRT T CdT 0 1 \u00b8 = \u2219 CRT T \u2212 T CdT 0 1 \u00b8 = \u23a1\u23a2\u23a2\u23a3 cos \u03b4 sin \u03b4 0 0 \u2212 sin \u03b4 cos \u03b4 0 0 0 0 1 Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 (8.50) Example 331 F Cycloid. Assume that the wheel in Figure 8.48 is turning with angular velocity \u03c9 and has no slip on the ground. If the point P is at the center of the tireprint 8. Suspension Mechanisms 493 when t = 0, MrP = \u23a1\u23a3 0 0 \u2212Rw \u23a4\u23a6 (8.51) then we can find its position in the wheel frame at a time t by employing another coordinate frame M . The frame M is called the rim frame and is stuck to the wheel at its center. Because of spin, the M frame turns about the yw-axis, and therefore, the rotation matrix to go from the rim frame to the wheel frame is: WRM = \u23a1\u23a3 cos\u03c9t 0 sin\u03c9t 0 1 0 \u2212 sin\u03c9t 0 cos\u03c9t \u23a4\u23a6 (8.52) So the coordinates of P in the wheel frame are: W rP = WRM MrP = \u23a1\u23a3 \u2212Rw sin t\u03c9 0 \u2212Rw cos t\u03c9 \u23a4\u23a6 (8.53) The center of the wheel is moving with speed vx = Rw\u03c9 and it is at Gr =\u00a3 vxt 0 Rw \u00a4 in the global coordinate frame G on the ground. Hence, the coordinates of point P in the global frame G, would be GrP = W rP + \u23a1\u23a3 vxt 0 Rw \u23a4\u23a6 = \u23a1\u23a3 Rw (\u03c9t\u2212 sin t\u03c9) 0 Rw (1\u2212 cos t\u03c9) \u23a4\u23a6 . (8.54) The path of motion of point P in the (X,Z)-plane can be found by eliminating t between X and Z coordinates. However, it is easier to expressed the path by using \u03c9t as a parameter. Such a path is called cycloid. In general case, point P can be at any distance from the center of the rim frame. If the point is at a distance d 6= Rw, then its path of motion is called the trochoid. A trochoid is called a curtate cycloid when d < Rw and a prolate cycloid when d > Rw. Figure 8.49(a)-(c) illustrate a cycloid, curtate cycloid, and prolate cycloid respectively. Example 332 F Wheel to wheel-body frame transformation. The homogeneous transformation matrix CTW to go from the wheel frame 494 8. Suspension Mechanisms to the wheel-body frame can be found by combined transformation. CTW = CTT TTW (8.55) = \u23a1\u23a2\u23a2\u23a3 c\u03b4 \u2212s\u03b4 0 0 s\u03b4 c\u03b4 0 0 0 0 1 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 c\u03b3 \u2212s\u03b3 \u2212Rw sin \u03b3 0 s\u03b3 c\u03b3 Rw cos \u03b3 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 = \u23a1\u23a2\u23a2\u23a3 cos \u03b4 \u2212 cos \u03b3 sin \u03b4 sin \u03b3 sin \u03b4 Rw sin \u03b3 sin \u03b4 sin \u03b4 cos \u03b3 cos \u03b4 \u2212 cos \u03b4 sin \u03b3 \u2212Rw cos \u03b4 sin \u03b3 0 sin \u03b3 cos \u03b3 Rw cos \u03b3 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 If rP indicates the position vector of a point P in the wheel coordinate frame, W rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 (8.56) then the homogeneous position vector CrP of the point P in the wheel-body 8. Suspension Mechanisms 495 coordinate frame is: CrP = CTW W rP = \u23a1\u23a2\u23a2\u23a3 xP cos \u03b4 \u2212 yP cos \u03b3 sin \u03b4 + (Rw + zP ) sin \u03b3 sin \u03b4 xP sin \u03b4 + yP cos \u03b3 cos \u03b4 \u2212 (Rw + zP ) cos \u03b4 sin \u03b3 \u2212Rw + (Rw + zP ) cos \u03b3 + yP sin \u03b3 1 \u23a4\u23a5\u23a5\u23a6 (8.57) The position of the wheel center W r = 0, for a cambered and steered wheel is at Cr = CTW W r = \u23a1\u23a2\u23a2\u23a3 Rw sin \u03b3 sin \u03b4 \u2212Rw cos \u03b4 sin \u03b3 \u2212Rw(1\u2212 cos \u03b3) 1 \u23a4\u23a5\u23a5\u23a6 (8.58) The zc = Rw (cos \u03b3 \u2212 1) indicates how much the center of the wheel comes down when the wheel cambers. If the wheel is not steerable, then \u03b4 = 0 and the transformation matrix CTW reduces to CTW = \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 cos \u03b3 \u2212 sin \u03b3 \u2212Rw sin \u03b3 0 sin \u03b3 cos \u03b3 Rw (cos \u03b3 \u2212 1) 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 (8.59) that shows CrP = CTW W rP = \u23a1\u23a2\u23a2\u23a3 xP yP cos \u03b3 \u2212Rw sin \u03b3 \u2212 zP sin \u03b3 zP cos \u03b3 + yP sin \u03b3 +Rw (cos \u03b3 \u2212 1) 1 \u23a4\u23a5\u23a5\u23a6 (8.60) Example 333 F Tire to vehicle coordinate frame transformation. Figure 8.50 illustrates the first and fourth tires of a 4-wheel vehicle. There is a body coordinate frame B (x, y, z) attached to the mass center C of the vehicle. There are also two tire coordinate frames T1 (xt1 , yt1 , zt1) and T4 (xt4 , yt4 , zt4) attached to the tires 1 and 4 at the center of their tireprints. The origin of the tire coordinate frame T1 is at Bd1 BdT1 = \u23a1\u23a3 a1 \u2212b1 \u2212h \u23a4\u23a6 (8.61)"
        ]
    },
    {
        "image_filename": "designv10_0_0000983_j.matdes.2013.09.006-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000983_j.matdes.2013.09.006-Figure7-1.png",
        "caption": "Fig. 7. Planar scanning experimen",
        "texts": [
            " In view of the clad height and width obtained in the best condition through the orthogonal experiment, H = 0.46 and W = 2 were brought into the optimization algorithm, so the optimal scanning space and the overlapping percentage can be figured out, namely S = 1.38 and gc = 31%. In order to prove the feasibility of planned experiment method, the critical optimal processing parameters gained by orthogonal experimental design and the optimum scanning space derived from ideal overlapping model were selected to perform the planar scanning experiment. The schematic diagram is depicted in Fig. 7(a), and the experiment specimen is displayed in Fig. 7(b). As can be seen from Fig. 7(b), relatively, the surface of cladding layer is smooth, the space between cladding passes is uniform, and the appearance of specimen is free of obvious crack and porosity. As a result, the validity and reliability of orthogonal experimental design and ideal overlapping model could be authenticated. Based on the experiment mentioned above, the cross section of SS 316 clad created with the optimal parameters was analyzed for microstructure. Fig. 8 displays the morphology of bonding zone between clad and substrate"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003391_5.663540-Figure24-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003391_5.663540-Figure24-1.png",
        "caption": "Fig. 24. Base module (upper left) and example topology built in the COBRA system [81]. The mapping includes a tower of three base modules surrounded by three bus modules (top), a base module (center), a RAM module (bottom), and an I/O module (right).",
        "texts": [
            " The users of the system are free to combine whatever set of modules they desire, yielding a system capacity ranging from only two medium FPGA\u2019s (a single X210MOD-00) to 32 large FPGA\u2019s (16 X213MOD-82s) and a significant RAM capacity. Similar systems include DEEP [132] and Paradigm RP [147]. The Paradigm RP has locations to plug in up to eight boards. These boards can include FPGA\u2019s, memories, DSP\u2019s, and standard processors. In the G800 and Paradigm RP systems, cards with new functions or different types of chips can easily be accommodated. Even more flexible systems are possible. One example is the COBRA system [81]. As shown in Fig. 24, the system is made up of several types of modules. The standard base module has four FPGA\u2019s, each attached to an external connection at one of the board\u2019s edges. Boards can be attached together to build a larger system, expanding out in a 2-D mesh. Other module types can easily be included, such as modules bearing only RAM, or a host interface, or a standard processor. These modules attach together the same way as the base module and will have one to four connectors. One somewhat different type of module is a bus module"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001466_j.mechmachtheory.2014.02.001-Figure11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001466_j.mechmachtheory.2014.02.001-Figure11-1.png",
        "caption": "Fig. 11. Cracked tooth model when root circle is bigger than base circle.",
        "texts": [
            " The bending and shear stiffness of the cracked tooth are expressed as follows: 1 kb \u00bc Z \u03b1r \u03b13 12 sin\u03b1 N cos\u03b10 N\u22122:5 \u2212 cos\u03b1 \u00fe cos\u03b13\u2212 cos\u03b1r\u2212 sin\u03b13 sin\u03c5 \u2212 q2 Rr cos\u03c5 cos\u03b11 h i2 EL sin\u03b1\u2212 q2 Rr sin\u03c5 3 d\u03b1 \u00fe 4 1\u2212 N\u22122:5\u00f0 \u00de cos\u03b11 cos\u03b13 N cos\u03b10 h i3\u22124 1\u2212 cos\u03b11 cos\u03b12\u00f0 \u00de3 EL cos\u03b11 sin\u03b12\u2212 q2 Rb sin\u03c5 3 \u00fe Z \u03b12 \u2212\u03b1c 12 1\u00fe cos\u03b11 \u03b12\u2212\u03b1\u00f0 \u00de sin\u03b1\u2212 cos\u03b1\u00bd f g2 \u03b12\u2212\u03b1\u00f0 \u00de cos\u03b1 EL sin\u03b1 \u00fe \u03b12\u2212\u03b1\u00f0 \u00de cos\u03b1\u2212 q2 Rb sin\u03c5 h i3 d\u03b1 \u00f05:15\u00de 1 ks \u00bc Z \u03b1r \u03b13 2:4 1\u00fe \u03bd\u00f0 \u00de cos2\u03b11 sin\u03b1 EL sin\u03b1\u2212 q2 Rr sin\u03c5 3 da\u00fe 2:4 1\u00fe \u03bd\u00f0 \u00de cos2\u03b11 cos\u03b12\u2212 N\u22122:5 N cos\u03b10 cos\u03b13 EL sin\u03b12\u2212 q2 Rb sin\u03c5 \u00fe Z \u03b12 \u2212\u03b1c 2:4 1\u00fe \u03bd\u00f0 \u00de \u03b12\u2212\u03b1\u00f0 \u00de cos\u03b1 cos2\u03b11 EL sin\u03b1 \u00fe \u03b12\u2212\u03b1\u00f0 \u00de cos\u03b1\u2212 q2 Rb sin\u03c5 d\u03b1: \u00f05:16\u00de Fig. 11 shows the tooth crack model when the root circle is bigger than the base circle. The crack is modeled as a straight line starting from the intersection (point M) of the involute curve and the root circle. The point M is within the critical area of the gear tooth. Similar to Case 1, four conditions are considered in the derivation of the bending and shear stiffness. In each condition, the expressions of the tooth section area and the area moment of inertia are the same as derived for the condition when the root circle is smaller than the base circle"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001545_j.rcim.2015.01.003-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001545_j.rcim.2015.01.003-Figure4-1.png",
        "caption": "Fig. 4. The medial axis transformation (MAT) of an example shape. The dash lines represent the medial axis [28].",
        "texts": [
            " After this introductory section, Section 2 presents the Medial Axis Transform (MAT) and its computing algorithms. Section 3 introduces the methods for generating paths from MAT, followed by the implementation and the discussions of the proposed methodology in Section 4. The paper ends with the conclusions in Section 5. 2. Medial Axis Transformation (MAT) 2.1. Definition The Medial Axis Transformation (MAT) is a technique first proposed by Blum [29] to describe shapes with the medial axis which is defined as loci of centres of locally maximal balls inside an object. In two dimensions, as shown in Fig. 4, the MAT would be the loci of centres of locally maximal disks inside the region. The points on the medial axis are called medial axis points as represented by the dash line. The medial axis points are further classified into end points, normal points, and branch points depending on the number of points where the disk touches the boundary [28]. If the disk touches the boundary at either two connected segments, two points, or one segment and one point, the medial axis point is a normal point. For those disks touching the boundary at more than two points or connected segments, such medial axis points are branch points"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure2.26-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure2.26-1.png",
        "caption": "Fig. 2.26: The manipulation of a kinescope in the work-cell of Figs 2.22 and 2.23.",
        "texts": [
            " Moreover, the manipulation analysis has been carried out to verify for possible small changes in the work-cell which can improve the manipulation without settling a new workcell or requiring a lengthy stop in production in order to run the suggested modifications. The manipulation under examination can be considered as a \"pick up and place\" application since it consists of picking up a kinescope from a given location in the loading station and placing it at a given location in the unloading station, as shown in Fig. 2.26. Chapter 2: Analysis of manipulations 69 The term \u2018location\u2019 refers both to position and orientation of an object. Figure 2.26 shows the performed manipulation of a kinescope emphasizing the rotation, whose direction has been given to obtain a gyroscope torque to help in the bending motion of the robot. From Figs 2.23 and 2.26 the degrees of freedom that are required for the manipulative task can be easily evaluated to be four, since the motion of a kinescope consists of the three spatial translations and one rotation only. Particularly, a translation along a vertical plane can even be considered pleonastic when by rearranging quotes hB and hF in Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure9.65-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure9.65-1.png",
        "caption": "Figure 9.65 The truss crane with the support reactions as they are acting in reality.",
        "texts": [
            " Member 5 presses against the joint and is a compression member, member 6 pulls on the joint and is a tension member: N5 = \u2212F \u221a 5, N6 = +3F \u221a 5. In Figures 9.61 to 9.64, the other member forces are calculated using the same method. Table 9.4 provides a summary of all the member forces. 358 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM 9 Trusses 359 360 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM In Figures 9.63 and 9.64 the support reactions in G and H have also been calculated: Gv = 10F, Hh = 0, Hv = 6F. In Figure 9.65, the support reactions are shown with the directions in which they are acting. To check the calculation, we can look at the equilibrium of the truss as a whole: \u2211 Fx = Hh = 0,\u2211 Fy = Gv \u2212 Hv \u2212 4F = 10F \u2212 6F \u2212 4F = 0,\u2211 Tz|H = Gv \u00d7 2a \u2212 4F \u00d7 5a = 10F \u00d7 2a \u2212 4F \u00d7 5a = 0. The truss as a whole meets the equilibrium conditions. Example 2 The truss in Figure 9.66 is loaded at joint E by a vertical force of 120 kN. Question: Calculate the member forces, with the correct sign for tension and compression. Solution: In this truss, we cannot find a joint with only two unknown forces"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000730_s0022112006002631-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000730_s0022112006002631-Figure2-1.png",
        "caption": "Figure 2. The arrangement of a bottom-heavy squirmer. Gravity acts in the g-direction, and the squirmer has orientation vector e, radius a and centre-of-mass distance h from its geometrical centre.",
        "texts": [
            " Rotational velocities The rotational velocity of the first squirmer in the same arrangement as in the previous section can be given by Fa\u0301xen\u2019s second law (Russel et al. 1992): T 1 = 8\u03c0\u00b5a3 ( \u2212\u21261 + 1 2 \u2207 \u2227 V ( \u2212r, \u03b1a, e(2), B(2) )) , (2.5) where T 1 is the torque exerted on the first squirmer. If the squirmer is bottom heavy, there will be a torque acting on it and this must be equal and opposite to the hydrodynamic torque, so that the net torque on the squirmer is zero. If the distance of the centre of gravity to the centre of the squirmer is h, in the opposite direction to its swimming direction in undisturbed fluid (see figure 2), then there is an additional torque equal to 4 3 \u03c0a3\u03c1he \u2227 g, (2.6) where \u03c1 is the density and g is the gravitational acceleration. This torque must be matched by the hydrodynamic torque, T . Thus \u21261 = \u2212 \u03c1 6\u00b5 he1 \u2227 g + \u2207 \u2227 V ( \u2212 r, \u03b1a, e(2), B(2) ) . (2.7) The correction to the rotational velocity due to the effect of the movement of the first sphere on the second sphere and the consequent effect back on the first sphere is of order O (\u2223\u2223\u2223\u2223B (1) 2 a \u2223\u2223\u2223\u2223a6 r6 r \u2227 e ) , (2.8) which is much smaller",
            " The boundary-element method using single- or double-layer potentials only, the so-called generalized boundary-integral method, is explained in detail, with a derivation of the integral equations and the force and torque exerted on a particle, in the established text Pozrikidis (1992). It is supposed that squirmer m is subjected to known external forces Fm and torques Tm. The equilibrium conditions for squirmer m are Fm = \u222b Am q(x) dAm, (4.4) Tm = \u222b Am x \u2227 q(x) dAm. (4.5) Squirmers are assumed to be neutrally buoyant, so Fm =0. The centre of buoyancy of the squirmer may not coincide with its geometric centre (see figure 2), and in that case the torque Tm is given by (2.6). The stresslet of squirmer m can be expressed using the single-layer potential as Sm = \u222b Am [ 1 2 (qx + xq) \u2212 1 3 x \u00b7 qI ] dA. (4.6) A detailed derivation of the stresslet is given in Appendix B. These governing equations are non-dimensionalized using the radius a, the swimming velocity of a solitary squirmer, 2B1/3, and the fluid viscosity \u00b5. Let the ratio of second-mode squirming to first-mode squirming be \u03b2 , i.e. \u03b2 = B2/B1. In the case of a bottom-heavy squirmer, its behaviour is determined by a dimensionless parameter Gbh given by Gbh = 2\u03c0\u03c1gah \u00b5B1 "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001514_j.engfailanal.2015.06.004-Figure25-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001514_j.engfailanal.2015.06.004-Figure25-1.png",
        "caption": "Fig. 25. (a) 6 DOF model (b) 8 DOF model in Ref. [67].",
        "texts": [
            " Based on a similar model presented in Ref. [36] (see Fig. 24b), Mohammed et al. [37] investigated the influence of tooth root crack propagation on dynamic response of a gear system. Ma et al. [55] established a 6 DOF model of a perforated gear system, which is similar to the model in Ref. [36] except for ignoring the effect of tooth friction. Considering the torsional and lateral vibrations, Wu et al. [28] adopted a 6 DOF model of a gear system to analyze the effects of crack on the system vibration response [67] (see Fig. 25a). To study the influence of tooth friction, Bartelmus [67] presented an 8 DOF model (see Fig. 25b) where two lateral degrees of freedom (DOFs) are added into the 6 DOF model. The same models can also be found in Refs. [68,69]. Considering the effects of local tooth damage, changing load and speed, Chaari et al. [70] presented a two stage gearbox model with 12 DOFs (see Fig. 26a). Mohammed et al. [71] developed a 12 DOF gear dynamic model including the gyroscopic and friction effects (see Fig. 26b). In order to investigate dynamic characteristics of the system when the crack in the pinion grows, Zhou et al"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001498_tie.2010.2045322-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001498_tie.2010.2045322-Figure7-1.png",
        "caption": "Fig. 7. Nonadjacent double-phase (phase-b and phase-e) open-circuit fault.",
        "texts": [
            " It can be mentioned that the solution reported in [20] can produce 72% torque with 0.85% ripple. The solution in (24) produces higher torque with zero torque ripple. In a five-phase machine, two different types of double-phase fault may occur. In one case, the double-phase open-circuit fault may take place in two nonadjacent phases, and in the other case, the fault may occur in two adjacent phases. In this section, the fault-tolerant operations of a five-phase PM machine under both double-phase open-circuit faults are presented. Consider Fig. 7, in which a double-phase open-circuit fault has occurred in phases b and e. Therefore, the fault axis will be aligned with the phase-a axis. It can be noted that phases c and d are symmetrically located with respect to the fault axis, whereas phase a is located on the fault axis. The fundamental component of the fault-tolerant currents can be constrained to be equal to obtain higher output torque from the motor. Using the condition of mirror symmetry, the fault-tolerant currents in the remaining three healthy phases can be defined as follows: ia = I1"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000818_j.apsusc.2010.02.030-Figure10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000818_j.apsusc.2010.02.030-Figure10-1.png",
        "caption": "Fig. 10. Schematic drawing of the balling induced pores.",
        "texts": [
            " When a molten track possesses high viscosity or high wetting angle, it is very easy to lead to a large amount of agglomerates or balls, so as to minimize its surface energy under the capillary force. In this situation, a very rough surface with many metallic balls is liable to be formed, and combined with many pores between the metallic balls, as is shown in Fig. 9. Since SLM is a typical RM technology based on repeatedly rolling powder on a previous melted layer in every slice forming process, an obtained fluctuate surface, in turn, is unfavorable for fresh loose powders to fill in pore channels of a previous coarse layer, as schematically illustrated in Fig. 10. Moreover, even through a deep pore channel could be filled by some powders during the rolling powder process, the laser energy may not be high enough to penetrate a thicker thickness and melt the powder in deep pore channel. Therefore, a porous and coarse layer can lead to next porous layer, resulting in chains effect. Combined with the both high capillary and chains effect, a more and more porous and rough layer tends to be generated. More than that, when a significantly coarse layer is formed, the SLM experiment may not proceed under the jamming force between ripple agglomerates and the roller"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001207_acs.chemrev.7b00168-Figure37-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001207_acs.chemrev.7b00168-Figure37-1.png",
        "caption": "Figure 37. (a) Schematic representation of the preparation method of the aligned CNT/wax/polyimide strips and the dependence of their light-induced bending on the orientation of the CNTs. Photographs documenting the motion of strips with (b) longitudinal and (c) transverse CNT orientations upon exposure to visible light (100 mW/cm2) for the times indicated. Reproduced from ref 250. Copyright 2016 American Chemical Society.",
        "texts": [
            "250 Instead of cellulose fibrils and their ensuing inhomogeneous swelling, carbon nanotubes (CNTs) were employed, resulting in thermal expansion differences of the material. The CNTs were aligned and embedded in a paraffin wax matrix, which was deposited on a polyimide film. The resulting bilayer film was cut in strips at different angles to create actuators in which the alignment direction of the CNTs in the reinforced layer adopted different orientations with respect to the long axis of the strips (Figure 37). CNTs are known in the field of actuating materials for their ability to translate infrared irradiation into locally released heat, which can be used to optically stimulate a temperatureresponsive material.251 Previous studies on CNT-containing polymer films have shown the photothermally and electrothermally induced actuation of the material, albeit these were not correlated to an example of a natural actuator.252\u2212255 When irradiated with visible light of an intensity comparable to that of the sun (ca"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure5.27-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure5.27-1.png",
        "caption": "FIGURE 5.27 Round wire example for a retarded partial inductance evaluation.",
        "texts": [
            " We evaluate as an example the retarded partial inductance evaluation for relatively thin round wires using the derivations from the previous section, which uses the approximation of the exponential term. The partial inductance (5.17) is repeated in a simplified form with retardation in (5.64) which assumes that the current flows in the axial direction LpR 12 = 1 1 2 \ud835\udf070 4\ud835\udf0b\u222b1 \u222b2 e\u2212j\ud835\udefdR R d2 d1, (5.64) where we evaluate only one term in the inductance part in (5.61). We assume that the length of the wire (xe \u2212 xs) in Fig. 5.27 is at least two times the diameter d. We can either use the approximate formula (C.15) or the tube formula (C.17) for the quasi-static self term. The tube formula (C.17) is exact if (xe \u2212 xs) is much larger than the tube diameter d. From the previous section, we get the impedance model for a retarded partial self-inductance as ZR Lp,11 = j\ud835\udf14Lp11 + \ud835\udf14 2 v \ud835\udf070 4\ud835\udf0b (xe \u2212 xs)2, (5.65) where Lp11 is the quasistatic partial inductance computed with the above formula and the resistive part is given by (5.63)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure7.30-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure7.30-1.png",
        "caption": "Figure 7.30 A horizontal force of 1476 kN is acting on the door per metre in the circumferential direction. In other words, the horizontal water pressure on the door consists of a uniformly distributed load qw = 1476 kN/m in radial direction.",
        "texts": [
            " The horizontal water pressure on the door is independent of the shape of the door.1 The resultant of the horizontal water pressure on the 1-metre wide vertical strip is 1 2 \u00d7 (22 m)(225.5 kN/m) \u2212 1 2 \u00d7 (14 m)(143.5 kN/m) = 1476 kN. We have shown, therefore, that per metre in the circumferential direction, the door is subject to a force of 1476 kN. In other words, the horizontal water pressure on the door consists of a uniformly distributed load 1 See the previous example. 264 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM qw = 1476 kN/m in radial direction (see Figure 7.30). With an arc length of 209.5 m for AB and a radius of r = 250 m, the aperture angle \u03b1 is \u03b1 = arc length AB 2\u03c0r \u00b7 360\u25e6 = 209.5 m 2\u03c0 \u00d7 (250 m) \u00d7 360\u25e6 = 48\u25e6. The resultant R of the horizontal water pressure on arc AB is equal to the resultant of the horizontal water pressure on chord AB (see Section 7.2, Example 2). This gives: R = qwa = qw2r sin(\u03b1/2) = (1476 kN/m) \u00d7 2 \u00d7 (250 m) \u00d7 sin 24\u25e6 = 300 MN. Example 4 Figure 7.31 represents an element of a floating two-track metro tunnel, ready to be transported to its sinking site"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure14.21-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure14.21-1.png",
        "caption": "Figure 14.21 Cable with a uniformly distributed load q (force per horizontally measured length).",
        "texts": [
            " Finding the solution is far more complicated in that case. Hereafter, we assume that the horizontal component H is known. In Section 14.1.4, using the cable equation as basis, we determine the cable shape under a uniformly distributed load (force per horizontally measured length). The associated cable shape is a parabola. In Section 14.1.5, we calculate the cable shape due to its dead weight (force per length measured along the cable). The shape of the cable under its dead weight is a catenary. In Figure 14.21, cable AB, with span , carries a uniformly distributed load qz = q . The difference in elevation of the supports A and B is h. From the cable equation we find H d2z dx2 = \u2212q. After integrating once, we find H dz dx = \u2212qx + C1, 14 Cables, Lines of Force and Structural Shapes 651 while after integrating once more we find Hz = \u2212 1 2qx2 + C1x + C2. The integration constants C1 and C2 follow from the boundary conditions at supports A and B: x = 0; z = 0, x = ; z = h. Working out the boundary conditions gives C1 = H h + 1 2q , C2 = 0. The cable shape is a parabola: z = 1 2qx( \u2212 x) H + h x. This can be denoted as z = zk + h x in which zk is the distance from the chord to the cable (see Figure 14.21): zk = 1 2qx( \u2212 x) H = M H . 652 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM M = 1 2qx( \u2212 x) is the bending moment in a simply supported beam with a uniformly distributed load (see Section 10.2.2, Example 1). The cable has the same parabolic shape under the chord as the M diagram; the scale factor is H . Assume pk is the sag of the parabola under the chord, that is the distance between the parabola and the chord at the middle of the span (see Figure 14.22): pk = zk ( x = 1 2 ) = 1 8q 2 H . (17a) If the sag pk of the parabola under the chord is given, the horizontal component of the cable force follows from H = 1 8q 2 pk "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure3.9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure3.9-1.png",
        "caption": "Fig. 3.9: A kinematic scheme for a four-bar linkage transmission",
        "texts": [
            " The transmission ratio can assume a value that is bigger or less than 1, being constant or variable as function of \u03c8 and \u03b8, depending on the mechanical design of the transmission. In general, the transmissions that are used in robots can be grouped and classified referring to tr values as: - gear systems with tr < 1 or tr > 1; - harmonic drives with tr < 1; - belt transmissions with tr < 1 or tr > 1; - linkages and mechanisms with tr variable as function of \u03c8 and \u03b8. The computation of tr for each system can be performed by using common schemes for mechanism analysis. In the following, a general algorithm is outlined for the case of Fig. 3.9 in which a fourbar linkage is the transmission system between an actuator and manipulator joint. The kinematics of a four-bar linkage can be formulated as closed-form expressions by solving the position closure equation in the form 0lcoslcoslcosl 1432 =\u2212\u03b8+\u03b3+\u03c8 0sinlsinlsinl 432 =\u03b8+\u03b3+\u03c8 (3.1.24) Squaring and summing them, after algebraic manipulations it yields 321 KcosKsinK =\u03c8+\u03c8 (3.1.25) with \u03b8= sinK1 , \u03b8+= cos l l K 2 1 2 , 42 2 4 2 3 2 2 2 1 4 1 3 ll2 llll cos l l K +\u2212+ +\u03b8= Fundamentals of Mechanics of Robotic Manipulation 87 The solution of Eq"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001894_s11071-015-1958-8-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001894_s11071-015-1958-8-Figure1-1.png",
        "caption": "Fig. 1 Nonlinear input function",
        "texts": [
            " T , D, L , and Myy represent thrust, drag, lift-force and pitching moment, respectively, and have the following expressions: T = T\u03a6(\u03b1)\u03a6 + T0(\u03b1) \u2248 [\u03b21\u03a6 + \u03b22]\u03b13 + [\u03b23\u03a6 + \u03b24]\u03b12 + [\u03b25\u03a6 + \u03b26]\u03b1 + [\u03b27\u03a6 + \u03b28] D \u2248 q\u0304 S(C\u03b12 D \u03b1 2 + C\u03b1 D\u03b1 + C0 D) L = L0 + L\u03b1\u03b1 \u2248 q\u0304 SC0 L + q\u0304 SC\u03b1 L\u03b1 Myy = MT + M0(\u03b1)+ M\u03b4e\u03b4e \u2248 zT T + q\u0304 Sc\u0304(C\u03b12 M \u03b1 2 + C\u03b1 M\u03b1 + C0 M ) +q\u0304 Sc\u0304C\u03b4e M\u03b4e N1 = C\u03b12 N1 \u03b12 + C\u03b1 N1 \u03b1 + C0 N1 N2 = C\u03b12 N2 \u03b12 + C\u03b1 N2 \u03b1 + C0 N2 + C\u03b4e N2 \u03b4e q\u0304 = 1 2 \u03c1V 2, \u03c1 = \u03c10 exp [ \u2212h \u2212 h0 hs ] The detail of the parameter values can be found in [33]. While most papers [7,37,39] do not consider the actuator dynamics of elevator deflection, the input nonlinearity is described as in Fig. 1 with the following description: \u03b4e(u) = \u23a7 \u23aa\u23a8 \u23aa\u23a9 mr (u \u2212 br ) if u \u2265 br 0 if \u2212 bl < u < br mr (u + bl) if u \u2264 \u2212bl (7) Then we have \u03b4e(u) = a(t)u + b(t) (8) where a(t) = { ml if u \u2264 0 mr if u > 0 (9) and b(t) = \u23a7 \u23aa\u23a8 \u23aa\u23a9 \u2212mr br if u \u2265 br \u2212a(t)u if \u2212 bl < u < br mlbl if u \u2264 \u2212bl (10) Assumption 1 br and bl are unknown positive constants. There exist unknown positive constants m\u2212 r , m+ r , m\u2212 l , m+ l such that 0 < m\u2212 r < mr < m+ r , 0 < m\u2212 l < ml < m+ l . Remark 1 From Assumption 1, it is easy to deduce that a(t) and b(t) are bounded"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000801_j.finel.2014.04.003-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000801_j.finel.2014.04.003-Figure4-1.png",
        "caption": "Fig. 4. 1D Test problem geometry and boundary conditions.",
        "texts": [
            " Thus, it is common practice to neglect the surface convection and radiation on this interface. This simplification may be appropriate in weld modeling since the filler metal and therefore the size of the interface is negligible compared to the base metal. However, in additive manufacturing, the size of the deposited material could be significant compared to the substrate and thus neglecting surface convection and radiation on the interface between active and inactive elements can be a source of error as demonstrated in the examples section. The 1D bar problem of Fig. 4 is used to illustrate and quantify potential errors of the implementation of the quiet and inactive element methods. The length of the bar is set to 10 mm and the cross section to 1 mm2. The thermal conductivity k is set to .05 W/mm/1C, the specific heat Cp to 1000 kJ/kg/1C, and the density \u03c1 to 10 6 kg/mm3. The half left of the bar is set to be active at all time, while the right half is set to be inactive or quiet for times 0\u2013.1 s, and then switched to active. A heat source Q of 1 W/mm3 is applied to the left half of the bar for times 0\u2013"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.23-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.23-1.png",
        "caption": "FIGURE 8.23. A swing arm suspension.",
        "texts": [
            " A multi-link mechanism is a six-bar mechanism that may have a better coupler motion than a double A-arm mechanism. However, multi- 466 8. Suspension Mechanisms 8. Suspension Mechanisms 467 link suspensions are more expensive, less reliable, and more complicated compare to a double A-arm four-bar linkage. There are vehicles with more than six-link suspension with possibly better kinematic performance. Example 311 Swing arm suspension. An independent suspension may be as simple as a triangle shown in Figure 8.23. The base of the triangle is jointed to the chassis and the wheel to the tip point. The base of the triangle is aligned with the longitudinal axis of the vehicle. Such a suspension mechanism is called a swing axle or swing arm. The variation in camber angle for a swing arm suspension is maximum, compared to the other suspension mechanisms. Example 312 Trailing arm suspension. Figure 8.24 illustrates a trailing arm suspension that is a longitudinal arm with a lateral axis of rotation. The camber angle of the wheel, supported by a trailing arm, will not change during the up and down motion"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-FigureC.17-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-FigureC.17-1.png",
        "caption": "FIGURE C.17 Combined rectangular and quadrilateral cell example.",
        "texts": [
            "90) where the integration over the cross sections is performed with Gaussian numerical integration. This section is added to point out the difference between the local and global coordinates. Compute time can be saved if the numerical integration is mostly used for the smaller dimensions of the cell sides. Local coordinates are efficient for the mapping of the dimensions to the normalized units used for numerical integration methods. REFERENCES 407 Local coordinates are only required for nonorthogonal conductor cell dimensions such as quadrilateral or hexahedral shapes. For the example in Fig. C.17, we can use mixed coordinates. Global coordinates are used for conductor cell 1 while local coordinates are used for conductor 2. The integrals are Lp12 = \ud835\udf070 4\ud835\udf0b \u222b xe1 xs1 \u222b ye1 ys1 \u222b ze1 zs1 \u222ba2 \u222bb2 \u222bc2 1 R1,2 dx1 dy1 dz1 da2 db2 dc2. (C.91) Besides the change in the coordinate systems, the filament representation is the same as for the case in previous section. Representation is the same as for the case in the previous section. More details on nonorthogonal systems is given in Chapter 7. 1. F. W"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure14.20-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure14.20-1.png",
        "caption": "Figure 14.20 The cable force N is directed along the tangent to the cable: tan \u03b1 = dz/dx = V/H .",
        "texts": [
            " Divide each of these equations by x and proceed to the limit x \u2192 0; we generate three differential equations: dH dx = 0, (12) dV dx = \u2212qz, (13) H dz dx = V. (14) 14 Cables, Lines of Force and Structural Shapes 649 It follows from equation (12) that H = constant. The horizontal component H of cable force N is constant, or in other words, independent of x. This is in line with what we derived earlier in Section 14.1.1 for a cable subject to a system of vertical forces. From equation (14) we find that the cable force N is directed along the tangent to the cable, as in Figure 14.20: tan \u03b1 = dz dx = V H . The tensile force N in the cable is therefore N = \u221a H 2 + V 2 = H \u221a 1 + ( dz dx )2 = H \u221a 1 + (tan \u03b1)2. (15) Tensile force N is largest where the slope dz/dx of the cable is largest. If we differentiate (14), in which H is constant, we find H d2z dx2 = dV dx . By substituting (13) in the equation above, we arrive at the so-called cable equation: H d2z dx2 = \u2212qz. (16) This differential equation, derived from the equilibrium of a cable element, 650 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM forms a relationship between the horizontal component H of the cable force, the cable shape z = z(x), and the distributed load qz = qz(x)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002054_tie.2016.2593683-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002054_tie.2016.2593683-Figure5-1.png",
        "caption": "Fig. 5 FE model of stator. (a) stator core. (b) stator assembly.",
        "texts": [
            " It is not practical to model these two parts according to their actual structure. Thus, an equivalent stator model is required for the accurate prediction of vibration and noise. In order to validate the equivalent model of stator core and coils, modal tests of stator core and stator assembly are respectively implemented, as shown in Fig. 4. In the test a vibration exciter is used to yield excitation force with a wide frequency band and frequency response function is measured to estimate modal parameters. Fig. 5 shows the corresponding FE structural model which is built in ANSYS. In this motor, a coil winds around a single tooth and the end part beyond the stator core is short. Therefore, only the coil part within the slot is modeled. The material anisotropy should be considered according to the actual stator structure. Because the electromagnetic force distributes almost uniformly along the axial direction, the circumferential stator modes contribute the most to the vibration. Furthermore, according to the modal shapes of circumferential modes, the Young\u2019s modulus in x and y direction and the shear modulus in xy plane have the greatest influence on the corresponding modal frequencies"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000206_978-0-387-74315-8-Figure9-2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000206_978-0-387-74315-8-Figure9-2-1.png",
        "caption": "Figure 9-2: Piston-actuated joint",
        "texts": [],
        "surrounding_texts": [
            "v /1 X R k. Unlike the joint motion space, S, which represents both a sub space and a basis, R represents only a sub-space, so it is sometimes possible , , to give Ra simpler form than S. For example, the reaction-force space for a spherical joint is R- 100 O 1 O O O 1 O O O O O O O O O regardless of the choice of joint velocity variables. Calculations involving the loop Jacobians can be implemented using the LX N matrix of coefficients Eko = e 0- e 0, which define the circuit of , III pa each kinematic loop.18 The elements of this matrix are constants having the values +1, -1 or O, depending on whether tree joint i is on the path leading to link S k' the path leading to link Pk' or is not a part of the circuit of loop k. In terms of these coefficients, loop Jacobian j k is expressed as Jk = [Ek1S 1 \u2022\u2022\u2022 EkNSN ], , s ' and the product R k J k as \u00b7s, \u00b7s\u00b7 's' R k J k = [ Ekl R k SI\u00b7 .. EkN R k S N ] \u2022 Each individual sub-matrix R: Si need only be calculated if Ea is non-zero. 18This matrix is equivalent ta the matrix U in [53J. 165 If the computations are to be performed in link coordinates, then for each sub-matrix Ii; Si we have a choice as to whcther to compute it in i-coordinates or in the outer frame of Ioop joint k. The former involves transforming Rk around the links making up the Ioop, and the Iatter involves transforming the relevant Si to joint k's outer frame. Which is faster depends on the dimensions of the quantities involved and the size of the 100Pi but whichever way is chosen a large amount of transformation must be done, and the total amount of computation is likely to be more than for the corresponding calculations performed in absolute coordinates. Once Equation (9.17) has been set up, the next step is to solve for the unknown accelerations and constraint forces. Although it is possible for the matrix in Equation (9.17) to be singular, it is always possible to determine the accelerations. The accelerations comply with the motion constraints imposed by the loops, and can be integrated to get the position and velocity of the mechanism at the next time instant. Due to numerical errors in the calculation and integration of the accelerations, care must be taken to avoid the possibility that errors wilI accumulate in the position and velocity of the open-loop mechanism which are not consistent with the loop constraints. One way to prevent such errors from growing is to stabilise the constraints [5], [48], [76]. This process is effectively like putting weak springs and dampers in parallei with the loop closure constraints, so that small errors in position and velocity induce forces which tend to correct the errors. Another way is to identify n-r independent joint variables and calculate the dependent ones explicitly from the independent ones so that the constraints are satisfied exactly. In this case only the independent variables are integrated. 166 9.4.1. Redundant Constraints \u00b7s . It may happen that the matr\u00cex J k R k does not have full rank, or that when assembled into Equation (9.17) some of the columns are linearly dependent on those of other loops. In the first case it means that some of the loop-closure constraints are linearly dependent on constraints in the tree mechanism part of the loop; and in the second case it indicates that, although topologically independent, the loops are not algebraically independent. 'Whichever case pertains, the result is that the coefficient matrix in Equation (9.17) is singular and one or more of the loop-closure forces can not be determined. lf the indeterminate forces are removed rrom the equations, then the remainder can be solved for the rest of the unknowns. This state of affairs can be either transitory (Le., occurring only at certain positions) or permanent. In the latter case it is usually possible to identify and remove redundant constraints rrom the mechanism beforehand. For example, consider the piston-actuated rev.olute joint shown in Figure 167 9-2. If we cut the loop at the joint shown then the open-loop mechanism has three degrees of motion freedom -- two revolute and one prismatic -- but the loop joint imposes five motion constraints on this mechanism. In fact, three of these constraints are linearly dependent on constraints in the rest of the mcchanism -- they serve to prevent non-planar motion when the mechanism is already constrained to have planar motion -- so only two constraints have any effect, and the loop has one degree of motion freedom. The constraint forces for the redundant constraints are indeterminate, and the redundant constraints should be removed from the equations. This can be done by treating the loop joint as a sphere-in-cylinder joint, which imposes 'the correct number of constraints. A beneficial side-effect is that r is smaller, so less calculation is needed to fmd the accelerations."
        ]
    },
    {
        "image_filename": "designv10_0_0000554_978-3-540-85629-0-Figure3.40-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000554_978-3-540-85629-0-Figure3.40-1.png",
        "caption": "Fig. 3.40. Geometrical construction of the modified cycloidal trajectory with the Wildt method.",
        "texts": [
            " The amplitude of the sinusoidal profile, in the cycloidal motion, must be added to the constant velocity trajectory in the direction perpendicular to the axis t. A possible modification of the cycloidal trajectory is obtained by adding the sinusoidal profile in a direction perpendicular to the constant velocity profile, as shown in Fig. 3.39(b). This is the so-called \u201cAlt modification\u201d, after Herman Alt, a German kinematician who first proposed it, [7]. This modification allows to obtain a smoother profile, but implies an higher value of the maximum acceleration. Another modification, which aims at reducing the maximum acceleration value, is shown in Fig. 3.40(a), known as the \u201cWildt modification\u201d (after Paul Wildt, [7]). Point D is located at a distance equal to 0.57T 2 from T 2 , and then connected to M. The segment DM defines the direction along which the sinusoidal profile is added to the constant velocity trajectory. In this manner, the maximum acceleration is 5.88 h T 2 , similar to the modified trapezoidal trajectory, whereas the standard cycloidal trajectory has a maximum acceleration of 6.28 h T 2 . Therefore, a reduction of 6.8% of the maximum acceleration value is achieved. In Fig. 3.41 the three acceleration profiles for the cycloidal trajectory and the two modified profiles are shown. Further details can be found in [6] and [7]. The general expression of the modified cycloidal trajectory, where the base sinusoidal motion is projected along a direction specified by a generic angle \u03b3 (see Fig. 3.40(b)), is q(tm) = h [ tm T \u2212 1 2\u03c0 sin 2\u03c0tm T ] (3.52) where tm is defined by t = tm \u2212 \u03ba T 2\u03c0 sin 2\u03c0tm T (3.53) 3.9 Modified Cycloidal Trajectory 129 with \u03ba = tan \u03b4 tan \u03b3 and tan \u03b4 = h T . The angle \u03b3 determines the direction of the basis of the sinusoidal trajectory (\u03b3 is sometimes called the \u201cdistortion\u201d angle) . For example, for a pure cycloidal trajectory \u03b3 = \u03c0/2 (as shown in Fig. 3.39(a)). The velocity and the acceleration of the trajectory can be computed by differentiating q(t) with respect to the time6 t: q\u0307(t) = h T 1 \u2212 cos 2\u03c0t T 1 \u2212 \u03ba cos 2\u03c0t T q\u0308(t) = h T 2 2\u03c0(1 \u2212 \u03ba) sin 2\u03c0t T[ 1 \u2212 \u03ba cos 2\u03c0t T ]3 "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000693_cs200124a-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000693_cs200124a-Figure6-1.png",
        "caption": "Figure 6. Schematic diagram of an alumina membrane and silica alumina composite membrane. Reproduced with permission from ref 65. Copyright 2011 Royal Society of Chemistry.",
        "texts": [
            " A flow-through glucose sensor was fabricated from the composite fibers and shown to be more stable and active than the using just the CAF fibers or mesoporous silica containing glucose oxidase. 963 dx.doi.org/10.1021/cs200124a |ACS Catal. 2011, 1, 956\u2013968 Itoh et al prepared an artificial biomembrane by immobilizing formaldehyde dehydrogenase (FDH) in the mesopores of a silica film coated in the channels of a commercial anodic alumina membrane.65 The 200 nm pores of the alumina were coated by sucking a TEOS/F127 surfactant solution through the membrane as shown in Figure 6. After calcination at 500 C the pores were reduced to 100 nm by the mesoporous silica film. FDH was immobilized in the 13 nm pores of the silica and then the reduction of NAD+ to NADH across the membrane was tested. The activity of the enzyme was maintained even after 10 cycles. Itoh and co-workers have also filled the alumina membranes with mesoporous silica where the1-D channels \u223c8 nm in diameter run in the same direction as for the alumina support.66 In this case, a catalase enzyme was immobilized and tested for the decomposition of hydrogen peroxide"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000233_978-1-4614-0460-6-Figure5.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000233_978-1-4614-0460-6-Figure5.3-1.png",
        "caption": "Fig. 5.3 Reciprocity demonstration for cantilever beam cross FRF measurements",
        "texts": [
            " Also, a12 \u00bc X1 F2= is a cross FRF that relates X1 and F2. The FRFs X2 F2= and X2 F1= are depicted in Fig. 5.2. All together, our two degree of freedom system has four direct and cross FRFs (22). A three degree of freedom system has nine (32). 170 5 Two Degree of Freedom Forced Vibration From Eqs. 5.9 and 5.10, we see that a12 \u00bc a21 because A\u00bd 1 is symmetric. This attribute is referred to as reciprocity and can be observed by comparing cross FRFs measured on actual systems. Let\u2019s consider the cantilever, or fixed-free, beam shown in Fig. 5.3 for two cases. First, a harmonic force is applied at coordinate 1 (the free end) and the response is measured at coordinate 2 somewhere along the beam, let\u2019s say the midpoint. Second, the same force is applied at coordinate 2, but the response is measured at coordinate 1. In the first case, if we measured the response and converted both the force and response into the frequency domain (using the Fourier transform), we would obtain the cross FRF X2 F1= . In the second case, we would determine the cross FRF X1 F2= from our measurements"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003965_978-94-009-1718-7_40-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003965_978-94-009-1718-7_40-Figure2-1.png",
        "caption": "Figure 2. Joint angles and link lengths of limb i",
        "texts": [
            " Since the limbs are connected to the moving platform and the fixed base by universal joints, no bending moments will transmit to the limbs. That is the force acting on each limb is directed along the longitudinal axis of the limb and the only moment acting on each limb is a twisting moment. Hence, these limbs can be made of hollow cylindrical rods to produce a light weight, high stiffness, and high speed manipulator. To facilitate the analysis, two coordinate systems ~ and E are attached to the fixed base and moving platform, respectively, as shown in Fig. 1. Figure 2 shows the geometry of limb i, where WI denotes the based attached axis of the lower universal joint, W5 denotes the moving platform attached axis of the upper universal joint, W2 and W4 denote the axes of the universal joints that are respectively attached to lower and upper members of the limb, UI is a unit vector pointing from 0 to Ai, U2 is a unit vector normal to the plane defined by WI and W2, U4 is a unit vector pointing from Ai to Bi, U5 is a unit vector normal to the plane defined by W4 and W5, and U6 a unit vector pointing from Q to Bi",
            " The angles 8li is measured from UI to U2 about the WI axis, 82i is measured from U2 to U4 about the W2 axis, 84i is measured from U4 to U5 about the W4 axis, and 85i is measured from U5 to U6 about the W6 axis. Let ~ = (aix, aiy, aiz)T and hi = (bix , biy , biz)T be the position vectors of points Ai and Bi in coordinate systems ~ and E, respectively. Also let p = (Px,py,Pz)T be the position vector of the origin Q of E in ~, and R be the 3 x 3 orthogonal matrix describing the orientation of E in ~. Then, for the ith limb of the manipulator, we may write (p + Rbi -~) . (p + Rbi -~) = d~, (3) where di is the length of the ith limb. Using the joint angles defined in Fig. 2, it can be shown that the rotation matrix becomes an identity matrix, R = I, if the following two conditions are satisfied: (4) and (5) Hence, if the universal joints of the limbs are arranged in such a way that Eqs. (4) and (5) are satisfied, the manipulator will possess a three-DOF translational motion within its workspace. This unique characteristic is very useful in many applications such as a non-conventional X - Y - Z positioning device. With matrix R = I, Eq. (3) reduces to (6) 4. Inverse Kinematics For the inverse kinematics problem, the position vector p of E with respect to E is given and the limb lengths di , i = 1,2, and 3, are to be found"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000844_bfb0015081-Figure2.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000844_bfb0015081-Figure2.1-1.png",
        "caption": "Fig. 2.1. Two link robot",
        "texts": [],
        "surrounding_texts": [
            "Consider the Lagrangian formulation of the dynamics of an n-degree-offreedom mechanical system\nD(q)~ + C(q, (t)O + g(q) = B(q)7- (2.1)\nwhere q E R n is the vector of generalized coordinates, 7- E R m is the input generalized force (m < n), and B(q) E R nx'~ has full rank for all q.\nFor a suitable part i t ion of the vector q of generalized coordinates as qT = (qT, qT), where ql E R ~-m and q2 E R m we may write the system (2.1) as\ndl lql + d12q2 q- hl(ql,Ol,q2,(t2) \"b \u00a2l(ql ,q2) = 0 (2.2)\ndl2ql +d22q2 -'bh2(ql,ql,q2,(~2)+\u00a22(qx,q2) -~ b(qx,q2)7- (2.3)\nwhere hi include Coriolis and centrifugal terms, and \u00a2i contains the terms derived from the potential energy, such as gravitational and elastic generalized forces. The m \u00d7 m matr ix b(ql, q2) is assumed to be invertible.\nExample 2.1. T w o - l i n k r o b o t . Consider the two-link robot shown in Fig. (2.1):\nwhere\ndllql q- di2iiz + hi + \u00a21 = 7-1 (2.4)\nd12ql -~- d22~2 + h2 + \u00a22 = 7-2 (2.5)\ndn = mlg~l + m2(g~ + g2~2 + 2e1~c2 COS(q2)) -k- I1 + I2\nd22 = m292~2 + I 2\nd12 = m2(g~2 + gle~2cos(q2)) +/2 hi = -m2glgc2sin(q2)q~ - 2m~e162sin(q2)q2q1\nh2 = m2glg~2sin(q2)~ 2\n\u00a21 m (mlgcl -k-m2gl)gcos(ql) -}-m2~.c29Cos(ql +q2)\n\u00a22 = m2g~2gcos(qi +q2)\nIf ~-i ---- 0 this system represents the Acrobot [13, 5], while if 72 = 0 the system represents the Pendubot [39]. In addition, with \u00a21 = 0 --- \u00a22 and 7-2 = 0 one has the underactuated manipulator system considered in by several authors, such as [28, 9, 2].\nExample 2.2. C a r t - p o l e s y s t e m . The cart-pole system is one of the classic examples and yet it still holds some interesting challenges from the standpoint of global nonlinear control. Referring to Fig. (2.2) the dynamics are given by:\n(my + me)2 + mpg cosO0 - moo 2 sin0 = F\nrnpgcos02 + mpO- mpggsinO = 0",
            "For simplicity we normalize all constants to unity. To put the system in standard form, we set ql = 0, q2 = x, T = F, and write the equations as\nq l + c o s q l ~ 2 - s i n q l = 0 (2.6)\ncosq l~ l+2~2-01~s inq l = ~- (2.7)\nThe nature of the fixed points of (2.2), (2.3) is closely tied to the controllability of the system. Let ~- = e = constant. Then, since the terms hi are quadratic in the velocities ~, the equilibrium solutions satisfy\n\u00a2l (q l ,q2) = 0 (2.8)\n\u00a22(ql,q2) = b(ql,q2)~ (2.9)\nand may either be isolated fixed points for each fixed e, as in the case of the Acrobot, and Pendubot or they may be higher dimension as happens (for\n= 0) in systems without potential terms. For example, in the absence of gravity, the Pendubot dynamics satisfies\n\u00a2~(ql, q2) = 0 for i = 1, 2 (2.10)\nfor all (ql, q2) E Q, where Q denotes the two dimensional configuration space. In the first case, systems with potential terms are linearly controllable around (almost all) fixed points, i.e. the Taylor series linearization is a controllable linear system. Systems without potential terms are generally not linearly controllable. Their local controllability properties are therefore more subtle to determine.\nWe may interpret Eq. (2.2) as a (dynamic) constraint on the accelerations of the generalized coordinates. It is then interesting to ask whether these constraints are holonomic, i.e. integrable. For many of the most interesting cases, including underactuated robot manipulators [28], the Acrobot [37] and Pendubot [39], the PVTOL system [43], the TORA system, and underwater robots, these constraints turn out to be completely nonintegrable as shown in [30]. An important consequence is that the system (2.2), (2.3) is (strongly) accessible, since nonintegrabiIity of the second order constraint equations means that the dimension of the reachable set is not reduced.\nAccessibility does not imply stabilizability of an equilibrium configuration using time-invariant continuous state feedback (either static or dynamic). In fact, for systems without potential terms it is known [30] that such stabilizability is not possible. The proof of this follows from an application of Broekett 's Theorem [6]. The situation here is, therefore, quite similar to the case of control of nonholonomic mobile robots. Of course, systems with potential terms are exponentially stabilizable by linear time-invariant feedback."
        ]
    },
    {
        "image_filename": "designv10_0_0000233_978-1-4614-0460-6-Figure9.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000233_978-1-4614-0460-6-Figure9.2-1.png",
        "caption": "Fig. 9.2 Example showing rigid coupling of components I and II to form assembly III. The force F1 is applied to the assembly at coordinate X1 in order to determine H11 and H21",
        "texts": [
            " The result at the selected frequency is a i b a2\u00feb2 , which has a real part a a2\u00feb2 and an imaginary part b a2\u00feb2 . This computation is repeated at every frequency. Multiplying two FRFs means that we multiply the two complex numbers at each frequency, a\u00fe i b\u00f0 \u00de c\u00fe i d\u00f0 \u00de \u00bc ac bd\u00f0 \u00de \u00fe i bc\u00fe ad\u00f0 \u00de. 9.2 Two-Component Rigid Coupling 323 Similarly, we can predict the assembly response at another coordinate, not coincident with the coupling point, by defining the component receptance at the desired location. Consider Fig. 9.2, where the direct assembly response at X1 is again desired, but this location is now at another point on component I. We again assume x1 and X1 are collocated before and after coupling. The new coupling coordinates at the rigid coupling point are x2a and x2b. The component direct and cross receptances corresponding to Fig. 9.2 are h11 \u00bc x1 f1 , h2a2a \u00bc x2a f2a , h12a \u00bc x1 f2a , and h2a1 \u00bc x2a f1 for I and h2b2b \u00bc x2b f2b for II. The compatibility condition for the rigid coupling is x2b x2a \u00bc 0 and we can therefore write x2a \u00bc x2b \u00bc X2. Also, x1 \u00bc X1. The equilibrium conditions are f2a \u00fe f2b \u00bc 0 (because there is no external force at the coupling point in this case) and f1 \u00bc F1. To determine H11 \u00bc X1 F1 , we will first write the component displacements. For I, we now have two forces acting on the body, so the frequency-domain displacements are: x1 \u00bc h11f1 \u00fe h12af2a and x2a \u00bc h2a1f1 \u00fe h2a2af2a: (9",
            " k h1a1af1a h1b1bF1b \u00fe h1b1bf1a\u00f0 \u00de \u00bc f1a kh1a1af1a kh1b1bF1b \u00fe kh1b1bf1a \u00bc f1a h1a1a \u00fe h1b1b \u00fe 1 k f1a \u00bc h1b1bF1b f1a \u00bc h1a1a \u00fe h1b1b \u00fe 1 k 1 h1b1bF1b \u00f09:21\u00de Again applying the equilibrium condition, f1b \u00bc F1b f1a, we obtain f1b \u00bc 1 h1a1a \u00fe h1b1b \u00fe 1 k 1 h1b1b F1b. Substitution then gives the assembly direct and cross receptances due to F1b. H1b1b \u00bc X1b F1b \u00bc x1b F1b \u00bc h1b1bf1b F1b \u00bc h1b1b 1 h1a1a \u00fe h1b1b \u00fe 1 k 1 h1b1b ! F1b F1b H1b1b \u00bc h1b1b h1b1b h1a1a \u00fe h1b1b \u00fe 1 k 1 h1b1b \u00f09:22\u00de H1a1b \u00bc X1a F1b \u00bc x1a F1b \u00bc h1a1af1a F1b \u00bc h1a1a h1a1a \u00fe h1b1b \u00fe 1 k 1 h1b1bF1b F1b H1a1b \u00bc h1a1a h1a1a \u00fe h1b1b \u00fe 1 k 1 h1b1b \u00f09:23\u00de Similar to the rigid connection example depicted in Fig. 9.2, we can again add another coordinate, not located at the coupling location, and apply the external force at that point. See Fig. 9.7. The component displacements are again x1 \u00bc h11f1 \u00fe h12af2a and x2a \u00bc h2a1f1 \u00fe h2a2af2a for substructure I and x2b \u00bc h2b2bf2b for substructure II. The equilibrium conditions are f2a \u00fe f2b \u00bc 0 and f1 \u00bc F1. The compatibility condition is: k x2b x2a\u00f0 \u00de \u00bc f2b: (9.24) 9.3 Two-Component Flexible Coupling 329 As before, the component and assembly coordinates are coincident, so we have that x1 \u00bc X1, x2a \u00bc X2a, and x2b \u00bc X2b"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002585_j.tifs.2021.01.076-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002585_j.tifs.2021.01.076-Figure2-1.png",
        "caption": "Fig. 2. The type of difffused deposition modelling printer according to the configuration of displacement platforms (Sun et al., 2018).",
        "texts": [
            " The types of printers used in 3D food printing are mainly diffused deposition modelling (FDM), stereolithography, direct ink writing, inkjet, digital light processing, and select laser melting (Lipton, Cutler, Nigl, Cohen, & Lipson, 2015; Pallottino et al., 2016; Sun, Peng, Yan, Fuh, & Hong, 2015). Currently, FDM has become the most commonly used technology for 4D food printing due to high universality, low price and simple operation. According to the configuration of displacement platforms, FDM is divided into Cartesian, Delta, Polar and Scara (Fig. 2), among which Cartesian configuration is the most widely used (Sun, Zhou, Yan, Huang, & Lin, 2018). X. Teng et al. Trends in Food Science & Technology 110 (2021) 349\u2013363 Smart materials and design are the key factors in 4D food printing. Therefore, 4D printed food should be designed to achieve controllable results through strict control of mechanism, predicted behavior and required parameters according to material characteristics. This requires printers to use a variety of inks, achieve multi-material printing and design more complex and detailed structures (Choi, Kwon, Jo, Ju Lee, & Moon, 2015)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure6.34-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure6.34-1.png",
        "caption": "FIGURE 6.34. A double A arm suspension mechanism and its equivalent four-bar linkage kinematic model.",
        "texts": [
            " Applied Mechanisms For computer calculation ease, it is better to find \u03b2 from the trigonometric equation tan2 \u03b2 = sec2 \u03b2 \u2212 1 (6.259) after we substitute sec\u03b2 from Equation (6.257). \u03b2 = tan\u22121 q 4b2f2 \u2212 (b2 + f2 \u2212 c2)2 b2 + f2 \u2212 c2 (6.260) The angle \u03b3 can be found from a tan equation based on the vertical distance of point A from the ground link MN . \u03b3 = tan\u22121 a sin \u03b82 d\u2212 a cos \u03b82 (6.261) Therefore, the coordinates xC and yC can be calculated as two parametric functions of \u03b82 for a given set of a, b, c, d, e, and \u03b1. Example 249 A poorely designed double A arm suspension mechanism. Figure 6.34 illustrates a double A arm suspension mechanism and its equivalent four-bar linkage kinematic model. Points M and N are fixed joints on the body, and points A and B are moving joints attached to the wheel supporting coupler link. Point C is on the spindle and supposed to be the wheel center. When the wheel moves up and down, the wheel center moves on a the couple point curve shown in the figure. The wheel\u2019s center of proper suspension mechanism is supposed to move vertically, however, the wheel center of the suspension moves on a high curvature path and generates an undesired camber"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000811_j.jsv.2012.05.039-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000811_j.jsv.2012.05.039-Figure7-1.png",
        "caption": "Fig. 7. Planetary gearbox test rig.",
        "texts": [
            " These findings illustrate the capability of the proposed method to identify the characteristic frequencies of planetary gearbox vibration signals in industrial applications. In this section, we use two datasets to validate the performance of spectral analysis in diagnosing planetary gearbox faults according to our theoretical derivations. One is the experimental signals collected from a lab planetary gearbox with manually created pitting on a planet gear. The other one is the signals from the same gearbox with naturally developed gear wear. Fig. 7 shows the planetary gearbox test rig. We did experiments on the 2nd stage planetary gearbox, for both the manually created damage and the naturally developed fault. Table 4 lists the gear parameters of the 2nd stage planetary gearbox. For the manual pitting experiments, we introduced pitting of different levels (i.e. baseline, slight, moderate, and severe pitting) to the planet gears inside of the 2nd planetary gearbox. Fig. 8 shows the picture of planet gears with different pitting levels. In each experiment, a planet gear of different pitting level was used, whereas all the other gears were normal"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure8.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure8.6-1.png",
        "caption": "Fig. 8.6. System losses of an AFPM machine.",
        "texts": [
            " These losses can be accounted for by introducing a pressure loss term \u2206pl in eqn (8.24) as follows [232] \u2206p = \u03c1\u21262(ksR 2 out \u2212R2 in) + \u03c1 2 ( 1 A2 1 \u2212 1 A2 2 )Q2 \u2212\u2206pl (8.25) System losses As the air passes through the AFPM machine, the system pressure loss due to friction must be taken into account. The sum of these losses is given by: \u2206pfr = \u03c1Q2 2 n\u2211 i=1 ki A2 i (8.26) where ki and Ai are the loss coefficient and the cross section area of the flow path i respectively. There are a number of sections through which the air flows in the AFPM machine (see Fig. 8.6). They are: (1) entry into the rotor air inlet holes; (2) passage through rotation short tube; (3) bending of air from pipe to annulus (90o); (4) round bend air passage (90o); (5) contraction of air in the tapering passage; (6) bending of air through the 90o elbow; (7) entry to the permanent magnet channels; (8) expansion of air through the tapering enlargement; (9) abrupt expansion of the air on the exit of the channel; (10) expansion as the air leaves the opening of the parallel rotor discs. The loss coefficients associated with each section in eqn (8",
            "27\u03b3/D}\u22121}16 Y = {37530/Re}16 (8.27) where Re = \u03c1DhQ \u00b5A and where \u03b3 is the equivalent sand grain roughness [61]. Characteristics It is now possible to relate the theoretical prediction obtained from the ideal flow model to the actual characteristic by accounting for the various losses discussed above. Assuming that the AFPM machine (shown in Fig. 8.1) operates at a constant speed of 1200 rpm, the ideal developed pressure characteristic for a radial channel is a function described by eqn (8.22) as shown in Fig. 8.6. After introducing the slip factor, the resultant curve is shown as a dotted line as 8.3 Cooling of AFPM Machines 263 eqn (8.24). It was not possible to obtain a suitable correlation in the literature [248] for the pressure loss due to shock and leakage as was the case for the slip. The calculated characteristic curve without considering shock and leakage losses, i.e. eqn (8.24)\u2013\u2206pfr shown in Fig. 8.7, is significantly higher than the experimental one. The shaded area in Fig. 8.7 represents the shock and leakage losses"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003981_rob.4620080405-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003981_rob.4620080405-Figure1-1.png",
        "caption": "Figure 1. PUMA 560 Robot.",
        "texts": [
            " (16) and ( 2 3 ) , we obtain: w P PW2 = O r x ( & ) = w1 v 2 1 + w 2 v 2 2 The linear relation [eq. ( 2 3 ) ] takes the explicit form: w2 = w1p p = -v21 v22-' (27) P is not unique but must be chosen for V22 to be regular. Starting from the last row of V2 we extract the first regular (c-6) x (c-b) matrix, which gives the subscript of the columns W:,k to be deleted, and defines the matrix P. The numerical values of the base parameters are given by eq. ( 2 0 ) : APPLl CAT1 0 N A Serial-Link Manipulator A six-joint serial link manipulator, similar to the PUMA 560, is chosen (Fig. 1). The geometric parameters are given in Table I. There are 66 standard inertial parameters. The matrices d(q,i,q), do(q,q), and h(q,i) can be calculated directly, numerically. We have preferred to make use of the symbolic expressions, automatically computed using the software package SYMOR0.24 There are 11 functions hi which are constant, and which correspond to 11 zero columns of d or do. These columns and the corresponding parameters are eliminated because they do not affect the dynamic model"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure2.13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure2.13-1.png",
        "caption": "FIGURE 2.13 Example for impedance circuit.",
        "texts": [
            " Again, these models can be used in both the time and the frequency domain. Very similar to the approach in Section 2.9.1, where we started from some data in tables, we also can use different techniques such as VF to come up with a rational model. At this point, the approach differs from the equivalent circuit model in the last section by directly stamping the contribution into the MNA matrix. Hence, we show how the stamps are constructed. We start with a simple impedance model. The small circuit that we would like to include in an MNA matrix is shown in Fig. 2.13. To be clear, this small circuit could also be included INCLUDING FREQUENCY DOMAIN MODELS IN CIRCUIT SOLUTION 35 with the conventional MNA solution approach. However, we want to take the rational function in the frequency domain form Z(s) = s2 + s R1 L1 + 1 L1 C2 sC1 [ s2 + s R1 L1 + 1 L1 (C1+C2 C1C2 ) ] (2.65) into account. Importantly, the circuit representation in Fig. 2.13 is not necessary and it only serves as an example. The residue/pole representation for a general circuit can be obtained from VF in the above rational form Z(s) = d + Mr\u2211 m=1 rm s \u2212 pm + Mc\u2211 m=1 [ rm,r + j rm,i s \u2212 pm,r \u2212 j pm,i + rm,r \u2212 j rm,i s \u2212 pm,r + j pm,i ] , (2.66) where Mr are the real poles and Mc are the complex pole pairs. The constant d can be interpreted as a series resistance, while we can write the equivalent differential equation for the real poles and residues, or dxm\u2215dt \u2212 pmxm = rmi(t) where the current i(t) is the same for all the terms in series"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001593_j.mechmachtheory.2015.11.017-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001593_j.mechmachtheory.2015.11.017-Figure4-1.png",
        "caption": "Fig. 4. Schematic of the meshing tooth pairs.",
        "texts": [
            " In order to solve this problem, assuming that the mesh stiffness under double-tooth engagement is proportional to the stiffness under single-tooth engagement with a constant ratio, the ratio is determined by FE method, which is called correction coefficient in this paper. The fillet-foundation stiffness formula can be extended to double- or triple-tooth engagement condition by multiplying the corresponding correction coefficients. The detailed calculation flow chart is shown in Fig. 3. The schematic of the meshing tooth pairs is displayed in Fig. 4. In the figure, the solid lines and dashed lines denote the actual gear profiles and theoretical gear profiles, respectively. The difference between actual profiles and theoretical profiles is due to gear tooth errors, such as the manufacturing error and TPM. In the figure, AB and CD denote the theoretical double-tooth contact zone, and BC denotes the theoretical single-tooth contact zone. Sa and Sr are the separation distances of pairs 3 and 1 along the line of action, respectively. One can find more details about the expressions of Sa and Sr in [25]",
            " Yuzhu Guo from the University of Sheffield for the proofreading of the final version of the paper and offering suggestions on the improvement of presentation. Considering the effect of extended tooth contact, the mesh stiffness calculation in handover region can be divided into three cases [25,26,37], as illustrated in Fig. A.1. Case 1. : Occurrence of ETC in the single-tooth contact zone without overlap. The extended tooth contact only occurs in the single-tooth contact zone without overlap (see Fig. A.1a). Theoretically, tooth pair 1 is just leaving contact when the tooth pair 2 meshes at the point B (see Fig. 4). However, due to the flexibility of gear, tooth pair 1 is still in mesh under Sr \u2264 Er gear. Tooth pair 1 leaves contact when Sr N Er gear. When the tooth pair 2 enters into meshing from the single-tooth engagement to double-tooth engagement, the value of Sa decreases gradually. Tooth pair 3 enters contact ahead of theoretical start of contact when Sa \u2264 Er gear. In Fig. A.1a, points B and C denote the positions of theoretical start of contact and theoretical end of contact, and points B\u2032 and C\u2032 denote the positions of actual start of contact and actual end of contact"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001470_j.actamat.2009.08.027-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001470_j.actamat.2009.08.027-Figure1-1.png",
        "caption": "Fig. 1. Schematic overview of a typical SLM setup.",
        "texts": [
            " All rights reserved. Keywords: Laser treatment; Thermodynamics; Finite volume modeling; Phase transformations; Evaporation Selective Laser Melting (SLM) is a rapid manufacturing technique in which three-dimensional parts of complex shape are produced in a layer-by-layer fashion, typically in small series. Examples of applications of this technology are biomedical implants [1,2] and casting molds with sophisticated internal cooling channels [3,4]. A typical configuration of a SLM machine is shown in Fig. 1. Most machines are equipped with a Nd-yttrium aluminum garnet (YAG) fiber-laser or a CO2-laser. The laser beam is deflected by galvano mirrors, which control the movement of the laser source over the surface of the powder bed. In each layer the laser beam follows a certain scanning path. Upon absorption of the laser radiation, the powder particles heat up and after melting and solidifying, a solid structure is formed. In this way, SLM allows functional parts with near full density to be produced directly, in contrast to selective laser sintering (SLS) where post-processing is 1359-6454/$36"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000613_978-94-015-9261-1-Figure4.4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000613_978-94-015-9261-1-Figure4.4-1.png",
        "caption": "Figure 4.4. Neighborhoods of submanifold S'",
        "texts": [
            "5 for all x E S'\" th e mappings (4.38), (4.39) are local diffeomorphisms and , in the neighborhood of each point x E S* , t here exist smoot h inverse mappings . This property is easily generalized (see [267, 27]) for points of a eert ain neighborhood of S* and th e correspond ing neighborhoods [(P \"' '') = {( s\" , e) E!Rn : s\" ES\"}::::> P*\", 1\\ E [ of th e planes P \"' '' , frequently being th e Cartesian products [ (P*\") = S \" X R( E) , where R(E) is a spherical neighborhood of t he point e = 0: and E > 0 is th e radius (Figure 4.4). Proposition 4 .6. Suppose that Assumption 4.4 holds. Then ihere exis ts an atlas of cotnpaiible charts {S*\", w\" } and the neighborhood [t (S*) = U/,Jt (S\"''') such that f07' each r: it holds i) [or alt x E [ t (S \"' '') the matrix I~\" I is inoertible; ii) the mapping (4.38), (4.39) is dijfeomorphism [rom [t(S*\" ) onto the neighborhood [ t (P\"''') = S \" X R (Et ) with smooth inverse x = r\"( s\" , e) . (4.40) 142 CHAPTER 4 Denoting r:( s\") = r\"( s\", 0) we obtain formula (4.29) describing S* as an embedded submanifold , Note that boundedness of the hypersurface neighborhood E(S*\") and sets S \" , R is usually caused by the necessity to prevent probable singulari ties of the including space",
            " Then there exists a neighborhood [(S*) such that for all x E [(S*7r)n[(S*ID) ::J- 0, 7r, wE IT, and the relevant values of s\", sr;:], e it holds that i) (4.44) 144 ii) CHAPTER 4 The proposition establishes , in particular , that the differential ob jects defined in E(S*) are smooth in x and gives a reasonfor omitting the upper indexes in their notations. Now we consider metric properties of the space in the vicinity of S*. For, we introduce the the metric matrix (4.45) characterizing the distortion of the Euclidean metric when passing to the coordinates (s\" , e) (see Figure 4.4). Indeed, in a small enough neighborhood of the point x* E S*\" , one can write the linear approximation of equation (4.40) x ~ x* + R; (s\" - s*\") + Re e and find 11 s\" - s\"''' 11Ix - x*I ~ e Q'. (4.46) where s\"' = 1/J\" (x\"' ). By using equation (4.46) and the definition (4 .37) , we find the distance between the surface S* and an arbitrary point x E E(S*) , chosen close enough, to be (4.4 7) where the symmetrie positive semi-definite matrix Qe(s\") = (R~Re)ors(s\") is the sell of the metrie matrix Q"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure3.15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure3.15-1.png",
        "caption": "FIGURE 3.15. Direction of tangential stresses on the tireprint of a stationary vertically loaded tire.",
        "texts": [
            "2 Static Tire, Tangential Stresses Because of geometry changes to a circular tire in contact with the ground, a three-dimensional stress distribution will appear in the tireprint even for a stationary tire. The tangential stress \u03c4 on the tireprint can be decomposed in x and y directions. The tangential stress is also called shear stress or friction stress. The tangential stress on a tire is inward in x direction and outward in y direction. Hence, the tire tries to stretch the ground in the x-axis and compact the ground on the y-axis. Figure 3.15 depicts the shear stresses on a vertically loaded stationary tire. The force distribution on the tireprint is not constant and is influenced by tire structure, load, inflation pressure, and environmental conditions: The tangential stress \u03c4x in the x-direction may be modeled by the following equation. \u03c4x(x, y) = \u2212\u03c4xM \u00b5 x2n+1 a2n+1 \u00b6 sin2 \u00b3x a \u03c0 \u00b4 cos \u00b3 y 2b \u03c0 \u00b4 n \u2208 N (3.32) 3. Tire Dynamics 109 \u03c4x is negative for x > 0 and is positive for x < 0, showing an inward longitudinal stress. Figure 3.16 illustrates the absolute value of a \u03c4x distribution for n = 1"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003403_978-1-4615-5633-6-Figure6.12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003403_978-1-4615-5633-6-Figure6.12-1.png",
        "caption": "Figure 6.12. AC Network in Q Sequence",
        "texts": [
            " The worst case is when the operation of the HVDC link is in a radial mode. Based on these considerations, the simplest system that exhibits all the characteristics of the torsional interactions is shown in Fig.6.11. Here, AC fil ters at the converter buses are represented as capacitors. The AC line is shown to be series compensated. This enables comparison of the TI to HVDC system with that due to the fixed series compensation. The generator and ac network modelling have been discussed in Chapter 2 and current sources injected at the converter bus (see Fig.6.12). INTERACTIONS WITH HVDC CONVERTER CONTROL 149 I leba I I AC NETWORK J I I I Here, iga is the current injected by the generator, while ira and iia are the currents injected at the rectifier and inverter buses respectively. The D-Q components of the current at the converter bus can be derived from the phasor diagram shown in Fig.6.13. Ir and Ii are currents injected into the AC network at the rectifier and the inverter bus respectively. - Ir is the load current at the rectifier bus and lags the voltage v,"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.39-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.39-1.png",
        "caption": "FIGURE 7.39. A comparison among the different steering lengths.",
        "texts": [
            "38 shows that the turning radius of a negative 4WS vehicle can be determined from the same equation (7.133). Example 289 F FWS and 4WS comparison. The turning center of a FWS car is always on the extension of the rear axel, and its steering length ls is always equal to 1. However, the turning center of a 4WS car can be: 7. Steering Dynamics 421 1\u2212 ahead of the front axle, if ls < \u22121 2\u2212 for a FWS car, if \u22121 < ls < 1 or 3\u2212 behind the rear axle, if 1 < ls A comparison among the different steering lengths is illustrated in Figure 7.39. A FWS car is shown in Figure 7.39(a), while the 4WS systems with ls < \u22121, \u22121 < ls < 1, and 1 < ls are shown in Figures 7.39(b)-(d) respectively. Example 290 F Passive and active four-wheel steering. The negative 4WS is not recommended at high speeds because of high yaw rates, and the positive steering is not recommended at low speeds because of increasing radius of turning. Therefore, to maximize the advantages of a 4WS system, we need a smart system to allow the wheels to change the mode of steering depending on the speed of the vehicle and adjust the steer angles for different purposes"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001362_epe.2007.4417204-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001362_epe.2007.4417204-Figure4-1.png",
        "caption": "Fig. 4. Stator iron core",
        "texts": [
            " We define the following parameters : q : number of phases Nc : number of cells per phase \u03b8s : angular width of the cell to the stator \u03b8r : angular width of the cell to the rotor Ns : number of teeth to the stator Nr : number of teeth to the rotor ws : angular width of a stator tooth wr : angular width of a rotor tooth The relations making it possible to define a polyphase structure are as follows : \u03c0=\u03b8 2N rr and \u03c0=\u03b8 2N ss c s Nq 2\u03c0 =\u03b8 ; \u239f\u239f \u23a0 \u239e \u239c\u239c \u239d \u239b \u00b1\u03c0=\u03b8 q2 k12 elecs and elecr 2\u03c0=\u03b8 with k, natural entirety Then, \u239f\u239f \u23a0 \u239e \u239c\u239c \u239d \u239b \u00b1 \u03b8 =\u03b8 q2 k1 s r We also have, for reasons of symmetries : 4 w s s \u03b8 = The angular width of the rotor teeth is defined by : rrrw \u03b8\u03b2= with ] [1;0r\u2208\u03b2 In order to balance the radial efforts and to minimize the harmonic components of flows, the numbers of teeth to the stator (Ns) and the rotor (Nr) must even beings. For a three-phase machine (q = 3) and with Nc = 4, we obtain: Ns = 12 and Nr = 10. For a diphasic machine (q = 2) and with Nc = 4, we obtain: Ns = 8 and Nr = 6. We built a three-phase machine. On figures 4 and 5 we can see the stator. On figure 4, there is only the carcass out of aluminium with ferromagnetic sheets. On figure 5, we can see windings of the three phases and excitation circuit. Also, all the active parts are arranged on the static part (stator) which is beneficial to evacuating the copper and iron losses. In Fig. 6, we show that the no-load voltage is almost sinusoidal and that it is possible to modulate their amplitude (dexc is the current density of the wound excitation in A/mm2). This amplitude modulation is useful under driving operation and also under generating operation associated a bridge of diode"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001514_j.engfailanal.2015.06.004-Figure14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001514_j.engfailanal.2015.06.004-Figure14-1.png",
        "caption": "Fig. 14. The crack growth model in the pinion [46].",
        "texts": [
            " [44] proposed a potential energy method to calculate the TVMS of a cracked gear pair. Considering the combined effects of tooth crack and the plastic inclination, Shao and Chen [45] developed a tooth plastic inclination model for a spur gear pair by considering the cracked tooth as a variable cross-section cantilever beam (see Fig. 13). Considering the effects of the gear tooth root crack which propagates into the gear-body, Zhou et al. [46] presented a revised mathematical model based on the assumption of a linear propagation path (see Fig. 14). For a spur gear pair, Saxena et al. [47] established an analytical model by taking the influences of the shaft misalignment and the friction of tooth surface into account. Omar et al. [48] presented a TVMS expression for a gear pair with tooth root crack, which can be written as follows: km \u00bc XN 1 i\u00bc1 ki m \u00fe kcracked m ; \u00f01\u00de where ki m \u00bc f k xt 2p i 1 N \u00fe n 2p x i 1 N \u00fe n 6 t 6 2p x i 1 N \u00fe n \u00fe hf x 0 elsewhere ( ; \u00f02\u00de kcracked m is evaluated using Eq. (1) considering the corresponding crack size function fk in Ref"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure2.13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure2.13-1.png",
        "caption": "Fig. 2.13. Vectors relevant for evaluating the Jacobian matrix",
        "texts": [
            " Let us now consider the evaluation of the sub-matrix J c. Let us denote by vIi) the part of the total linear velocity Ii = [x Y i JT which is due to the joint rate qi. 2.4 Inverse Kinematic Problem 41 Then we have (2.66) If the joint i is a revolute one, the component v(i) is given by (2.67) where Zi _ 1 is the ort of the joint axis, ri _ 1, n is the distance vector between the centre of joint i(origin of coordinate system (i-i)) and the manipulator tip (origin of system n), both expressed in the base reference frame (Fig. 2.13). If joint i is a sliding one, the component vii) of the total linear velocity is given by (2.68) In order to obtain the linear velocity as a function of elements of the homo geneous matrices, we will first express the vectors in Eqs. (2.67) and (2.68) with respect to system (i-i) and then premultiply them by \u00b0Ai _ 1 rotation matrix. Since i -1 Zi _ 1 = [0 0 1 JT is valid and the distance vectors i -lri -1, n are in fact the fourth columns of the homogeneous transformations i -1 Tn' i = 1, ... , n i-lr"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure3.9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure3.9-1.png",
        "caption": "Fig. 3.9. Stabilization of a PM: (a) PM alone, (b) PM with pole shoes, (c) PM inside an external magnetic circuit, (d) PM with a complete external armature system.",
        "texts": [
            " The energy of a PM in the external space only exists if the reluctance of the external magnetic circuit is higher than zero. If a previously magnetized PM is placed inside a closed ideal ferromagnetic circuit, i.e. toroid, this PM does not show any magnetic properties in the external space, in spite of the fact that there is the magnetic flux \u03a6r = BrSM = BrwM lM (3.16) corresponding to the remanent flux density Br inside the PM. A PM previously magnetized and placed alone in an open space, as in Fig. 3.9a, generates a magnetic field. To sustain a magnetic flux in the external open space, an MMF developed by the magnet is necessary. The state of the PM is characterized by the point K on the demagnetization curve (Fig. 3.10). The location of the point K is at the intersection of the demagnetization curve with a straight line representing the permeance of the external magnetic circuit (open space): Gext = \u03a6K FK , tan\u03b1ext = \u03a6K/\u03a6r FK/Fc = Gext Fc \u03a6r (3.17) The permeance Gext corresponds to the flux \u03a6\u2014MMF coordinate system and is referred to as MMF at the ends of the PM",
            " In the \u03a6\u2014MMF coordinate 100 3 Materials and Fabrication system the remanent flux \u03a6r is according to eqn (3.16) and the MMF corresponding to the coercivity Hc is Fc = HchM (3.18) The magnetic energy per unit produced by the PM in the external space is wK = BKHK/2. This energy is proportional to the rectangle limited by the coordinate system and lines perpendicular to the \u03a6 and F coordinates projected from the point K. It is obvious that the maximum magnetic energy is for BK = Bmax and for HK = Hmax. If the poles are furnished with pole shoes (Fig. 3.9b) the permeance of the external space increases. The point which characterizes a new state of the PM in Fig. 3.10 moves along the recoil line from the point K to the point A. The recoil line KGM is the same as the internal permeance of the PM, i.e. GM = \u00b5rec wM lM hM = \u00b5rec SM hM (3.19) The point A is the intersection of the recoil line KGM and the straight line OGA representing the leakage permeance of the PM with pole shoes, i.e. GA = \u03a6A FA , tan\u03b1A = GA Fc \u03a6r (3.20) The energy produced by the PM in the external space decreases as compared with the previous case, i.e. wA = BAHA/2. 3.2 Rotor Magnetic Circuits 101 The next stage is to place the PM in an external ferromagnetic circuit as shown in Fig. 3.9c. The resultant permeance of this system is GP = \u03a6P FP , tan\u03b1P = GP Fc \u03a6r (3.21) which meets the condition GP > GA > Gext. For an external magnetic circuit without any electric circuit carrying the armature current, the magnetic state of the PM is characterized by the point P (Fig. 3.10), i.e. the intersection of the recoil line KGM and the permeance line OGP . When the external magnetic circuit is furnished with an armature winding and when this winding is fed with a current which produces an MMF magnetizing the PM (Fig. 3.9d), the magnetic flux in the PM increases to the value \u03a6N . The d-axis MMF F \u2032ad of the external (armature) field acting directly on the PM corresponds to \u03a6N . The magnetic state of the PM is described by the point N located on the recoil line on the right-hand side of the origin of the coordinate system. To obtain this point it is necessary to lay off the distance OF \u2032ad and to draw a line GP from the point F \u2032ad inclined by the angle \u03b1P to the F -axis. The intersection of the recoil line and the permeance line GP gives the point N "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000613_978-94-015-9261-1-Figure8.8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000613_978-94-015-9261-1-Figure8.8-1.png",
        "caption": "Figure 8.8. End link and curv e S;",
        "texts": [
            "52) These relations characterize a so-called direci (01' forward) kin ematics (DK) of a robot mechanism. Here we consider a 6-link manipulator with rotational joints. Being a 6-channel 12th-order MIMO control plant (Figure 8.7), it is described by CONTROL OF MECHANICAL SYSTEMS 441 equations (8.45), (8.46), (8.51), (8.52), where qj and Wj are the state vari ables, Uj, j = 1,2 , . . . ,6, are the inputs (controls), Yi and O'i, i = 1,2,3, are the plant outputs . We analyze the motion of the robot end-point in the external Cartesian space 1R3 , where the desired trajectory (a spatial curve S;, Figure 8.8) is given in the form (8 .14) and the path length (longitudinal variable) is defined by equation s = 'IjJ (y). (8.53) The smooth functions !.pi and 'IjJ are assumed to be chosen such that (see Section 8.1) , on the curve S;, the nominal Jacobian matrix Y\"(y) = 1'(0''') E SO(3), where 0''' = col(0']', oi, (3) is the vector of Euler angles . A desired angular orientation of the end-effector is given with respect to the curve-fixed frame 1'(a\") in the form (8 .18) . Then we can pose the basic problem of robot trajectory control (see Figure 8.8) through holonomic relationships (the coordination conditions) between the robot variables. The equations !.p(Y) = 0 (8.54) introduce relationships between the end-effector Cartesian coordinates Yi, while the equation 1'(0') = 1'(60') 1'(0'*), (8.55) where 6a is th e vector of the desired relative rotation and 1'(60') E SO(3) , corresponds to angle relationships . The task of maintaining the holonomic restrictions (8.54) and (8.55) induces the necessity of coordinating the robot control actions Uj"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001996_j.rcim.2015.12.004-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001996_j.rcim.2015.12.004-Figure1-1.png",
        "caption": "Fig. 1. Automated process planning for robotic WAAM system.",
        "texts": [
            " Wire and arc additive manufacturing (WAAM) using either the gas metal arc welding process (GMAW) or the gas tungsten arc welding process (GTAW) attracts great interest due to its high deposition rate, environmental friendliness, and cost-competitiveness [4\u20138]. In particular, WAAM becomes a promising alternative to conventional subtractive methods for fabricating large aerospace alloy components that feature high buy-to-fly ratio [9,10]. Generally, process planning for a WAAM system involves CAD modelling, 3D slicing, 2D path planning, weld bead modelling, weld setting, robot code generation, and post-process machining, as shown in Fig. 1. 3D CAD models are firstly sliced into a set of 2D layers. Then the path planning module generates deposition paths for each of the sliced layers. After the paths are generated, the desired bead geometries along the path are determined accordingly. Bead modelling, on one hand, controls the path planning variables. On the other hand, it determines the optimum weld settings corresponding to the desired bead geometry. Subsequently, the deposition path together with the selected welding parameters is transferred into an integrated robot code file through the robot code generation module"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-FigureC.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-FigureC.6-1.png",
        "caption": "FIGURE C.6 Two parallel zero thickness conductors example.",
        "texts": [
            "5 leads to a relatively simple formula. The result is Lp11 = \ud835\udf070\ud835\udcc1 6 \ud835\udf0b [ 3 log [u + (u2 + 1)1\u22152 ] + u2 + 1 u + 3 u log [ 1 u + ( 1 u2 + 1 )1\u22152 ] \u2212 [ u4\u22153 + (1 u )2\u22153 ]3\u22152 ] , (C.24) where \ud835\udcc1 = xe1 \u2212 xs1 and u = \ud835\udcc1\u2215(ye1 \u2212 ys1). We note that this relatively simple formula is very useful for semianalytical solutions in layered models. Importantly, it eliminates the singular behavior problem for the partial self-inductance. C.1.5 Lp12 for Two Parallel Zero Thickness Current Sheets Two zero thickness parallel current sheets are shown in Fig. C.6. Fortunately, several analytical formulas are available, for example, Ref. [2]. The integral to be solved is Lp12 = 1 (ye1 \u2212 ys1)(ye2 \u2212 ys2) \ud835\udf07 4\ud835\udf0b \u222b ye1 ys1 \u222b xe1 xs1 \u222b ye2 ys2 \u222b xe2 xs2 1 R1,2 dx2 dy2 dx1 dy1. (C.25) 390 COMPUTATION OF PARTIAL INDUCTANCES The closed solution of integral (C.25) is Lp12 = \ud835\udf070 4\ud835\udf0b 1 (ye1 \u2212 ys1) (ye2 \u2212 ys2) 4\u2211 k=1 4\u2211 m=1 (\u22121)m+k [ b2 m \u2212 Z2 2 ak log (ak + rkm + \ud835\udf16) + a2 k \u2212 Z2 2 bm log (bm + rkm + \ud835\udf16) \u2212 1 6 (b2 m \u2212 2 Z2 + a2 k) rkm \u2212 bm Z aktan\u22121 ( ak bm rkm Z )] , (C.26) where a1 = xs2 \u2212 xe1, a2 = xe2 \u2212 xe1 (C"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure3.11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure3.11-1.png",
        "caption": "Fig. 3.11 Foot of HRP-2 [65].",
        "texts": [
            "24) and (3.25) are the basis for measuring the position of the ZMP2. 2 When a foot does not contact the ground, the ZMP position cannot be determined since the denominators of (3.24) and (3.25) become zero. Therefore, when measuring the ZMP, we have to introduce a threshold and set px = py = 0 when the denominator is less than the threshold. 3.2 Measurement of ZMP 79 First, focusing on the contact between one foot and the ground, we measure the ZMP. 1 Measurement Using 6 Axis Force/Torque Sensor Figure 3.11 shows the foot of the humanoid robot HRP-2 [65]. The ground reaction force applied to the sole is transmitted to the sensor mount through rubber bushes and dampers. A 6 axis force/torque sensor is attached at the sensor mount, and the force is transmitted to the ankle of the robot through this sensor. The rubber bushes and the dampers are positioned to prevent large impulse forces from being transmitted to the robot. Since the displacement of them is small, we do not consider the displacement when calculating the ZMP"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure6.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure6.3-1.png",
        "caption": "FIGURE 6.3. Expressing a four-bar linkage with a vector loop.",
        "texts": [
            " The angles \u03b84 and \u03b83 can be calculated by the following functions \u03b84 = 2 tan\u22121 \u00c3 \u2212B \u00b1 \u221a B2 \u2212 4AC 2A ! (6.1) \u03b83 = 2 tan\u22121 \u00c3 \u2212E \u00b1 \u221a E2 \u2212 4DF 2D ! (6.2) where, A = J3 \u2212 J1 + (1\u2212 J2) cos \u03b82 (6.3) B = \u22122 sin \u03b82 (6.4) C = J1 + J3 \u2212 (1 + J2) cos \u03b82 (6.5) D = J5 \u2212 J1 + (1 + J4) cos \u03b82 (6.6) E = \u22122 sin \u03b82 (6.7) F = J5 + J1 \u2212 (1\u2212 J4) cos \u03b82 (6.8) 6. Applied Mechanisms 311 and J1 = d a (6.9) J2 = d c (6.10) J3 = a2 \u2212 b2 + c2 + d2 2ac (6.11) J4 = d b (6.12) J5 = c2 \u2212 d2 \u2212 a2 \u2212 b2 2ab . (6.13) Proof. We may show a closed loop, four-bar linkage by a vector loop as shown in Figure 6.3. The direction of each vector is arbitrary. However, the angle of each vector should be measured with respect to the positive direction of the x-axis. The vector expression of each link is shown in Table 6.1. 312 6. Applied Mechanisms The vector loop in global coordinate frame G is Gr4 + Gr1 + Gr2 \u2212 Gr3 = 0 (6.14) where, Gr1 = \u2212d \u0131\u0302 (6.15) Gr2 = a (cos \u03b82 \u0131\u0302+ sin \u03b82 j\u0302) (6.16) Gr3 = b (cos \u03b83 \u0131\u0302+ sin \u03b83 j\u0302) (6.17) Gr4 = c (cos \u03b84 \u0131\u0302+ sin \u03b84 j\u0302) (6.18) and the left superscript G reminds that the vectors are expressed in the global coordinate frame attached to the ground link",
            "51) where \u03c92 = \u03b8\u03072 \u03c93 = \u03b8\u03073 \u03c94 = \u03b8\u03074. (6.52) 316 6. Applied Mechanisms Assuming \u03b82 and \u03c92 are given values, and \u03b83, \u03b84 are known from Equations (6.1) and (6.2), we may solve Equations (6.50) and (6.51), for \u03c93 and \u03c94. \u03c94 = a sin (\u03b82 \u2212 \u03b83) c sin (\u03b84 \u2212 \u03b83) \u03c92 (6.53) \u03c93 = a sin (\u03b82 \u2212 \u03b84) b sin (\u03b84 \u2212 \u03b83) \u03c92 (6.54) Example 220 Velocity of moving joints for a four-bar linkage. Having the coordinates \u03b82, \u03b83, \u03b84 and velocities \u03c92, \u03c93, \u03c94 enables us to calculate the absolute and relative velocities of points A and B shown in Figure 6.3. The absolute velocity is referred to the ground link, and relative velocity refers to a moving point. The absolute velocity of points A and B are GvA = G\u03c92 \u00d7 Gr2 = \u23a1\u23a3 0 0 \u03c92 \u23a4\u23a6\u00d7 \u23a1\u23a3 a cos \u03b82 a sin \u03b82 0 \u23a4\u23a6 = \u23a1\u23a3 \u2212a\u03c92 sin \u03b82a\u03c92 cos \u03b82 0 \u23a4\u23a6 (6.55) GvB = G\u03c94 \u00d7 Gr4 = \u23a1\u23a3 0 0 \u03c94 \u23a4\u23a6\u00d7 \u23a1\u23a3 c cos \u03b84 c sin \u03b84 0 \u23a4\u23a6 = \u23a1\u23a3 \u2212c\u03c94 sin \u03b84c\u03c94 cos \u03b84 0 \u23a4\u23a6 (6.56) and the velocity of point B with respect to point A is GvB/A = GvB \u2212 GvA = \u23a1\u23a3 \u2212c\u03c94 sin \u03b84c\u03c94 cos \u03b84 0 \u23a4\u23a6\u2212 \u23a1\u23a3 \u2212a\u03c92 sin \u03b82a\u03c92 cos \u03b82 0 \u23a4\u23a6 = \u23a1\u23a3 a\u03c92 sin \u03b82 \u2212 c\u03c94 sin \u03b84 c\u03c94 cos \u03b84 \u2212 a\u03c92 cos \u03b82 0 \u23a4\u23a6 ",
            "62) \u03b13 = C3C5 \u2212 C2C6 C1C5 \u2212 C2C4 (6.63) where C1 = c sin \u03b84 C2 = b sin \u03b83 C3 = a\u03b12 sin \u03b82 + a\u03c922 cos \u03b82 + b\u03c923 cos \u03b83 \u2212 c\u03c924 cos \u03b84 C4 = c cos \u03b84 C5 = b cos \u03b83 C6 = a\u03b12 cos \u03b82 \u2212 a\u03c922 sin \u03b82 \u2212 b\u03c923 sin \u03b83 + c\u03c924 sin \u03b84. (6.64) 318 6. Applied Mechanisms Example 222 Acceleration of moving joints for a four-bar linkage. Having the angular kinematics of a four-bar linkage \u03b82, \u03b83, \u03b84, \u03c92, \u03c93, \u03c94, \u03b12, \u03b13, and \u03b14 is necessary and enough to calculate the absolute and relative accelerations of points A and B shown in Figure 6.3. The absolute acceleration is referred to as the ground link, and the relative acceleration refers to a moving point. The absolute acceleration of points A and B are GaA = G\u03b12 \u00d7 Gr2 + G\u03c92 \u00d7 \u00a1 G\u03c92 \u00d7 Gr2 \u00a2 = \u23a1\u23a3 \u2212a\u03b12 sin \u03b82 \u2212 a\u03c922 cos \u03b82 a\u03b12 cos \u03b82 \u2212 a\u03c922 sin \u03b82 0 \u23a4\u23a6 (6.65) GaB = G\u03b14 \u00d7 Gr4 + G\u03c94 \u00d7 \u00a1 G\u03c94 \u00d7 Gr4 \u00a2 = \u23a1\u23a3 \u2212c\u03b14 sin \u03b84 \u2212 c\u03c924 cos \u03b84 c\u03b14 cos \u03b84 \u2212 c\u03c924 sin \u03b84 0 \u23a4\u23a6 (6.66) where Gr2 = \u23a1\u23a3 a cos \u03b82 a sin \u03b82 0 \u23a4\u23a6 (6.67) Gr4 = \u23a1\u23a3 c cos \u03b84 c sin \u03b84 0 \u23a4\u23a6 (6.68) G\u03c92 = \u23a1\u23a3 0 0 \u03c92 \u23a4\u23a6 (6.69) G\u03c94 = \u23a1\u23a3 0 0 \u03c94 \u23a4\u23a6 (6"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003989_tcst.2008.924576-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003989_tcst.2008.924576-Figure1-1.png",
        "caption": "Fig. 1. Coordinate reference frames.",
        "texts": [
            " The remainder of this brief is organized as follows. Section II defines the different reference frames used and reviews the mathematical models of rigid-body dynamics and kinematics. The tracking controller design is performed in Section III. Simulation results of the ESEO with the derived controller are presented in Section IV, and conclusions and possibilities of future work comprises Section V. 1063-6536/$25.00 \u00a9 2008 IEEE The different reference frames used throughout the brief are depicted in Fig. 1. The earth-centered inertial (ECI) reference frame, denoted , has its origin located in the center of the Earth. The -axis is directed along the Earth\u2019s rotation axis towards the celestial north pole, and the -axis is directed towards the vernal equinox. The -axis completes a right-handed orthogonal frame. The orbit reference frame, denoted , has its origin located in the mass center of the satellite. The -axis is pointing towards the center of the Earth, and the -axis is parallel to the orbit momentum vector"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.40-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.40-1.png",
        "caption": "FIGURE 8.40. A positive and negative caster configuration on front wheel of a car.",
        "texts": [
            " Suspension Mechanisms 8.4.2 F Caster Angle Caster is the angle to which the steering pivot axis is tilted forward or rearward from vertical, as viewed from the side. Assume the wheel is straight to have the body frame and the wheel frame coincident. If the steering axis is turned about the wheel yw-axis then the wheel has positive caster. If the steering axis is turned about the wheel \u2212yw-axis, then the wheel has negative caster. Positive and negative caster configurations on the front wheel of a car are shown in Figure 8.40. Negative caster aids in centering the steering wheel after a turn and makes the front tires straighten quicker. Most street cars are made with 4\u22126 deg negative caster. Negative caster tends to straighten the wheel when the vehicle is traveling forward, and thus is used to enhance straight-line stability. Example 322 Negative caster of shopping carts. The steering axis of a shopping cart wheel is set forward of where the wheel contacts the ground. As the cart is pushed forward, the steering axis pulls the wheel along, and because the wheel drags along the ground, it falls directly in line behind the steering axis"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002151_j.engfailanal.2014.11.015-Figure14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002151_j.engfailanal.2014.11.015-Figure14-1.png",
        "caption": "Fig. 14. Experimental test rig.",
        "texts": [
            " fs and fp denotes the rotation frequency of the sun gear and the planet gear, respectively, while fscrack represents the characteristic frequency of the cracked sun gear, which can be calculated as follows [8]: f scrack \u00bc f m N=Zsun \u00f04:1\u00de where N represents the number of planet gears and Zsun denotes the teeth number of sun gear. Experimental signals are acquired from a gearbox test rig to verify the simulated resultant vibration signals and the fault symptoms revealed in this study. The configuration of the test rig is shown in Fig. 14 and the parameters of gears are listed in Table 2. All the experimental gears are standard spur gears without tooth profile modification. In addition, we assume that all gears are perfect without manufacturing errors. An acceleration sensor is installed in the vertical direction of the casing of the second stage planetary gearbox to acquire the vibration signals. The simulated planetary gear set has the same configuration and gear parameters as the second stage planetary gear set. The vibration data was acquired when the rotation frequency of the motor was 1200 r/min"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002652_j.jmapro.2019.12.009-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002652_j.jmapro.2019.12.009-Figure6-1.png",
        "caption": "Fig. 6. Sizes of the dog-bone specimens for tensile test.",
        "texts": [
            " The H13 steel laser cladding specimen was cut by wire electro-discharge machining cutting and then polished. Microstructural characteristic was observed using a LINT OLS4100 3D measurement laser microscope on the polished section etched with alcohol solution containing 4 % of nitric acid. Tensile tests were carried out at room temperature using an electronic universal testing machine with a maximum load of 100 kN, with a tensile rate ranging from 0.0005 mm/min to 1000 mm/min. In this experiment, a strain rate of 3 mm/min was used to fracture the specimens. Fig. 6 shows the sizes of the dog-bone specimen for tensile tests. It can be seen that there is a laser cladding repaired zone in the middle of the dog-bone specimen. Three specimens repaired using H13 steel powder were tested and their tensile testing results were compared with those of the nondestructive specimens and specimens repaired using 45 steel powder. For wear test, the H13 steel laser cladding specimen and substrate specimen were machined to obtain a flat surface of about 5 mm wide, then thoroughly cleaned and weighed"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001256_j.ijmachtools.2021.103729-Figure34-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001256_j.ijmachtools.2021.103729-Figure34-1.png",
        "caption": "Fig. 34. Illustration of the hot cracking mechanism in the Laser Powder Bed Fusion process [178], demonstrating crack formation and growth within a single melt pool. This shows that the difference in interdendritic liquid pressure between the dendrite tip and the root causes an insufficient feeding of molten material at the dendrite root, promoting void generation and therefore highly affecting the hot cracking behaviour of the part.",
        "texts": [
            " The detrimental effect on fatigue performance previously observed by Zhang et al. correlates well with the misorientation effect [177]. In a further study by Chauvet et al., it was found that grain boundaries with a high misorientation angle were prone to crack propagation in AB and HT EBM samples (Fig. 33) [146]. Han et al. obtained similar results in this regard [178]. Research also concluded that the difference in interdendritic liquid pressure between the dendrite tip and root, as illustrated in Fig. 34, would cause an insufficient feeding of molten material at the dendrite root, promoting void generation and therefore highly affecting the part hot cracking behaviour. Tomus et al. compared the grain morphology of LPBF Hastelloy X specimens processed with various post-processing techniques [179]. The HT consisted in a single solution step (1175 \u25e6C/2h), while HIP was performed using the same time and temperature, with an applied stress of 150 MPa. Fig. 35a,c,e,g display a series of EBSD images illustrating the grain morphology in the XZ plane for AB, HT, HIP and HIP + HT specimens, respectively"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003863_0022-0248(78)90293-2-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003863_0022-0248(78)90293-2-Figure2-1.png",
        "caption": "Fig. 2. Deformation of GaAIAs epitaxial layers (L) on (001) and (113) oriented GaAs substrates (S).",
        "texts": [
            " The deforma- between the lattice constants of AlAs and GaAs (z~a/ tion observed is symmetrical. However, when the a) = 14 X 10\u2014a). The difference between the results growth plane is (113) a shear deformation is observed of these measurements and the values of the which is clearly visible in the different values of ~p aluminium content obtained by either photolumiand ~d/d for the equally inclined planes (333) and nescence or electron microprobe measurements was (115). The results are schematically shown in fig. 2. never greater than 1% aluminium (5x = 0.01). Clearly, The crystal lattice is dilated more in the [001] direc- AlAs must have almost the same elastic constants as tion than in the [1111 direction due to the aniso- GaAs, because the correction factors were calculated tropic elasticity. This anisotropy manifests itself also for pure GaAs. The elastic strain model was conin the degree of deformation of the crystal lattice of firmed by calculating the values of ~opand L~.d/dfor layers with the same aluminium content grown on the each lattice plane from the measured (&i/a)i using (001) as compared to the (iii) surface"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000462_j.actamat.2015.05.017-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000462_j.actamat.2015.05.017-Figure1-1.png",
        "caption": "Fig. 1. A schematic illustration showing the orientation of the tensile bars and the axis system used.",
        "texts": [
            " A detailed transmission electron microscopy (TEM) investigation of the eutectic microstructure of the Al\u201312Si alloy which exhibits excellent mechanical properties (elongation 25% with a tensile strength 190 MPa) after solution heat treatment has also been conducted. The relationship between the SLM processing, eutectic microstructure and mechanical properties is discussed. A possible underlying mechanism for the formation of this novel eutectic microstructure is also proposed. This study provides new insights into the modification of Al\u2013Si alloys through SLM and solution heat treatment to achieve the desired mechanical properties without the need for the addition of other elements. Al\u201312Si specimens (4 4 4 mm3) and tensile bars (shown in Fig. 1) were produced on a ReaLizer SLM-100 machine (ReaLizer GmbH, Germany) which is equipped with a fibre laser, which has a laser wavelength of 1.06 lm and maximum power of 200 W at the part bed. Al\u201312Si (in wt.%) powder (d50 38 lm, TLS Technik, Germany) was used. An inert, high purity argon gas atmosphere was used during processing to minimise oxidation. The laser scan speed and laser power were 500 mm/s and 200 W, respectively, which has been shown to produce the best properties [33]. The powder layer thickness was fixed at 50 lm and the scan spacing at 150 lm, while the substrate was heated to 200 C",
            " The laser light was coupled into the microscope using a single-mode fibre and brought on to the sample using a dichroic mirror and a 100 microscope objective (NA 0.9). The spatial resolution is about 100 nm with a spectral resolution of 0.02 cm 1. Tensile tests were carried out on machined specimens (gauge length 4 6 15 mm3), using an Instron 5982 machine at a constant strain rate of 1 mm/min. Strain was measured with a 10 mm gauge length extensometer. Samples were aligned perpendicular to the build direction as shown in Fig. 1. Since the laser\u2013material interaction time is usually on the order of 100 ls and the interaction volume is less than 50 lm, it is not practical to measure the temperature distribution within the Al\u2013 12Si melt pool during SLM. Therefore, a finite element modelling (FEM) method using a multi-physics-based computational model based on the heat transfer theory in the heat transient mode of COMSOL\u2122 was used to estimate the temperature distribution. For Al\u201312Si alloys, the laser absorption is very low 5% at the 1"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001514_j.engfailanal.2015.06.004-Figure28-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001514_j.engfailanal.2015.06.004-Figure28-1.png",
        "caption": "Fig. 28. (a) 26 DOF model [51], (b) 21 DOF model of planetary gear set [45].",
        "texts": [
            " In order to investigate dynamic characteristics of the system when the crack in the pinion grows, Zhou et al. [46] adopted a 16 DOF mathematic model, which is developed by Howard et al. [50] for a one-stage gear system. In the model, the gear engagement is described by a TVMS function k(t) and a viscous damping c(t) (see Fig. 27a). Using a lumped parameter technique, Endo et al. [59] established a 16 DOF LMM (Fig. 27b) of a gear test rig to reproduce fault signals. Jia and Howard [51] presented a 26 DOF mathematic model including two pairs of meshing gears (see Fig. 28a) in which the vertical direction (xi) is parallel to the line of action in order to facilitate modeling. Liang et al. [34], Chen and Shao [39,40], Chen et al. [41] as well as Shao and Chen [45] presented a mathematic model of a one stage planetary-gear set with 21 DOFs (see Fig. 28b) to study the influences of the tooth root crack on the system vibration. Wan et al. [32] developed a FE model considering the effect of the lateral and torsional vibrations to calculate the vibration responses of a cracked gear rotor system (see Fig. 29). Aiming at the gear tooth crack fault in a one-stage reduction gearbox, Ma et al. [35,54] proposed a FE model of this system considering the TVMS of the cracked gear pair (see Fig. 30). Using the model, they investigated the effects of tooth crack parameters including crack depth, width, initial position as well as crack propagation direction on TVMS of the cracked gear pair and crack-induced vibration features"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure3.5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure3.5-1.png",
        "caption": "Fig. 3.5 Ground Reaction Forces and their Equivalent Force and Moment",
        "texts": [],
        "surrounding_texts": [
            "1 ZMP in 2D In Fig. 3.1, although only the vertical component of the ground reaction force is shown, the horizontal component of it also exists due to the friction between the ground and the soles of the feet. In Fig. 3.4(a) and (b), we separately show the vertical component\u03c1(\u03be) and the horizontal component \u03c3(\u03be) of the ground reaction force per unit length of the sole. These forces simultaneously act on a humanoid robot. 72 3 ZMP and Dynamics Let us replace the forces distributed over the sole by a equivalent force and moment at a certain point in the sole. The force (fx and fz) and the moment (\u03c4(px)) at the point px in the sole can be expressed by fx = \u222b x2 x1 \u03c3(\u03be)d\u03be (3.1) fz = \u222b x2 x1 \u03c1(\u03be)d\u03be (3.2) \u03c4(px) = \u2212 \u222b x2 x1 (\u03be \u2212 px)\u03c1(\u03be)d\u03be. (3.3) Focusing on (3.3) with respect to the moment, let us consider the point px where moment becomes zero. Considering \u03c4(px) = 0 for (3.3), px can be obtained as follows: px = \u222b x2 x1 \u03be\u03c1(\u03be)d\u03be\u222b x2 x1 \u03c1(\u03be)d\u03be . (3.4) Here, \u03c1(\u03be) is equivalent to the pressure since it is the vertical component of forces per unit length. Thus, px defined in (3.4) is the center of pressure and is the ZMP defined in the previous section. For 2D cases, since the ZMP 3.1 ZMP and Ground Reaction Forces 73 is a point where the moment of the ground reaction force becomes zero, it has become the origin of the name. 2 Region of ZMP in 2D Generally speaking, since the vertical component of the ground reaction force does not become negative unless magnets or suction cups are attached at the sole, \u03c1(\u03be) \u2265 0. Substituting this relationship into (3.4), we obtain x1 \u2264 px \u2264 x2. (3.5) Equation (3.5) means that the ZMP is included in the segment of contact between the sole and the ground and does not exist outside it. Figure 3.6 shows the relationship between the pressure distribution and the position of the ZMP. As shown in (a), when the reaction force is distributed over the sole almost equally, the ZMP exists at the center of the sole. On the other hand, as shown in (b), when the distribution is inclined to the front part of the sole, the ZMP exists at the front part of the sole. Furthermore, as shown in (c), when a point at the toe supports all the reaction forces, the ZMP also exists at the toe. In this case, since the surface contact between the sole and ground is not guaranteed any longer, the foot begins to rotate about the toe by only a slight external disturbance applied to the robot. To reduce the danger of falling down when a humanoid robot moves, it is desirable to have the ZMP located inside of the support polygon while maintaining a certain margin from the end."
        ]
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure3.13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure3.13-1.png",
        "caption": "Figure 3.13 The moment of the resultant couple is found by adding the moments of the couples to be compounded (algebraically).",
        "texts": [
            " The letter T is given the index z, which indicates the normal of the plane in which the couple acts. In Figure 3.12b the moment of the same couple is in another coordinate system: Ty = +Fa = +12 kNm. In Figure 3.12c, the couple is represented by a curved arrow. In this visual notation the arrow indicates the direction of rotation of the moment and includes a value. The same conventions apply as for the visual notation of 60 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM a force (see Section 1.3.6). Couples can be compounded in many different ways. In Figure 3.13, the couples operating in the xy plane, T1, T2 and T3, have been compounded by replacing them by equal couples of which the forces have common lines of action. If T1 = 12 kNm, T2 = 6 kNm and T3 = 10 kNm, and the distance a between the lines of action is 4 m, then, in the coordinate system shown, Tz;1 = +T1 = +12 kNm = F1a \u21d2 F1 = Tz;1/a = +3 kN, Tz;2 = +T2 = +6 kNm = F2a \u21d2 F2 = Tz;2/a = +1.5 kN, Tz;3 = \u2212T3 = \u221210 kNm = F3a \u21d2 F3 = Tz;3/a = \u22122.5 kN. Note: The force F3 has the value 2.5 kN and acts opposite to the direction shown in Figure 3.13. The moment of the resultant couple is Tz = Ra = (F1 + F2 + F3) \u00b7 a = 3\u2211 i=1 Tz;i = 8 kNm. If all the couples are exerted in the same plane, the moment of the resultant couple is found by compounding the couple moments simply by adding them together. The example shows that the couples form an equilibrium system if the sum of their moments is zero (because R = 0). The moment of a force about a point A is defined as the product of magnitude F of the force and the perpendicular distance a from point A to the line of action of the force"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure15.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure15.1-1.png",
        "caption": "Figure 15.1 Work is defined as the inner product of force F and displacement d u: dA = F \u00b7 u = | F | \u00b7 |d u| \u00b7 cos \u03b1 = F \u00b7 du cos \u03b1 = F cos \u03b1 \u00b7 du.",
        "texts": [
            " Support reactions and section forces can be derived directly from the equilibrium equations, but also with the principle of virtual work. We provide a number of examples in Section 15.5. The virtual work equation is especially useful for determining the influence lines for support reactions and section forces in statically determinate bar type structures; this is covered in Chapter 16. 710 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM In this section we look at the concepts work and strain energy. If the point of application of force F in Figure 15.1a undergoes an infinitesimal displacement d u along path s, this is referred to as the force performing an (infinitesimal) amount of work dA, defined as the inner product of the vectors F and d u: dA = F \u00b7 d u = Fx dux + Fy duy + Fz duz. Work is a scalar quantity. The inner product of two vectors can also be calculated as the product of the magnitude (modulus) of both vectors and the cosine of the enclosed angle: dA = | F | \u00b7 |d u| \u00b7 cos \u03b1 = F \u00b7 du \u00b7 cos \u03b1. This can be seen as the product of the force and the component of the displacement in the direction of the force, F and du cos \u03b1 respectively (see Figure 15.1b). It can also be seen as the product of the displacement and the component of the force in the direction of the displacement, du and F cos \u03b1 respectively (see Figure 15.1c). Note that the force F does not perform any work if it is normal to the displacement du (in that case \u03b1 = \u00b1\u03c0/2 and cos \u03b1 = 0). If the point of application of the force moves a finite distance along path s, 15 Virtual Work 711 the total amount of work is equal to the sum of the contributions of all the infinitesimal displacements. Mathematically this corresponds to integrating over the path length s: A = \u222b s F \u00b7 d u. The magnitude and direction of the force F may depend on the location on the route"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.47-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.47-1.png",
        "caption": "FIGURE 8.47. Illustration of tire, wheel, and body coordinate frames.",
        "texts": [
            " The wheelbody frame C is motionless with respect to the vehicle coordinate and does not follow any motion of the wheel. The tire coordinate frame T (xt, yt, zt) is set at the center of the tireprint. The zt-axis is always perpendicular to the ground. The xt-axis is along the intersection line of the tire-plane and the ground. The tire frame does not follows the spin and camber rotations of the tire however, it follows the steer angle rotation about the zc-axis. Figure 8.46 illustrates a tire and a wheel coordinate frames. Example 326 Visualization of the wheel, tire, and wheel-body frames. Figure 8.47 illustrates the relative configuration of a wheel-body frame C, a tire frame T , and a wheel frame W . If the steering axis is along the zc-axis then, the rotation of the wheel about the zc-axis is the steer angle \u03b4. Rotation about the xt-axis is the camber angle \u03b3. 488 8. Suspension Mechanisms Generally speaking, the steering axis may have any angle and may go through any point of the ground plane. Example 327 Wheel to tire coordinate frame transformation. If TdW indicates the T -expression of the position vector of the wheel frame origin relative to the tire frame origin, then having the coordinates of a point P in the wheel frame, we can find its coordinates in the tire frame using the following equation"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure7.9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure7.9-1.png",
        "caption": "FIGURE 7.9 Corner of cell for continuity equation.",
        "texts": [
            " The continuity equation for the rectangular PEEC models is considered in Section 6.3.1. The continuity equation must also be satisfied for the nonorthogonal case at the cell level for the currents and charges. Its differential form is given by (3.3) or \u2207 \u22c5 J + \ud835\udf15q \ud835\udf15t = 0 where again J is the current density and q is the surface charge density. The continuity equation needs to be applied at the location of each node corresponding to Figs. 7.3\u20137.5. Unlike in Fig. 6.3, for the model we only show a quarter of the area in Fig. 7.9 for which the continuity equation is applied. Here, we show only one quarter of the centre cell unlike in Fig. 6.3, which shows the cell for rectangular coordinates. Since only one quarter of the elements surrounding the node is shown, we assume that the surface element in Fig. 7.9 may be connected to other similar surfaces along the a\u2013c and the b\u2013c surfaces. Hence, the volume for which the continuity equation is applied consists of the corners that are involved in the geometry surrounding the node(s). It is sufficient to consider only the corner elements by ignoring the internal surfaces shown in Fig. 7.9 for simplicity. Integrating the continuity equation over the corner yields \u222b \u2207 \u22c5 Jd = \u2212 \ud835\udf15 \ud835\udf15t\u222b q d = \u222b J \u22c5 n\u0302 d , (7.25) EVALUATION OF PARTIAL ELEMENTS FOR NONORTHOGONAL PEEC CIRCUITS 169 where the divergence theorem (3.33) [20] is used in the last step and the vector n\u0302 is normal to the surface  . The volume integral part pertains to the charge on the top and bottom a \u2212 b surface corners connected to nodes 1 and 0, which are charged as indicated with Qs in Fig. 7.9. If the nodes 0 and 1 are shorted, then the charge density consists of two \ud835\udeff-functions at the surfaces of the conductors with the surface charge q(a, b, c) and the contributions at c = \u00b11 are Q\u00b1 = \u222b 0 a=\u22121 \u222b 0 b=\u22121 q\u00b1(a, b) |||| \ud835\udf15r \ud835\udf15a \u00d7 \ud835\udf15r \ud835\udf15b |||| da db (7.26) and where the surface charge can be found from (7.19) as q\u00b1(a, b) = Q\u00b1 /||| \ud835\udf15r \ud835\udf15a \u00d7 \ud835\udf15r \ud835\udf15b |||. Substituting this into (7.26) yields the charges Q\u00b1 on the corner surfaces. Similarly, the currents associated with the corner nodes 0,1 are flowing through the cross-sectional areas indicated by Ia in the a direction and Ib in the b direction"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure10.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure10.2-1.png",
        "caption": "FIGURE 10.2 Rectangular block of dielectric material.",
        "texts": [
            " In the following sections, equivalent circuits are presented for the Debye model, the Lorentz model, and other general combined models. We first consider the one-pole Debye model (10.13) for lossy dielectrics. The equivalent circuit is used from the dielectric PEEC model in Section 10.4.5. This one-pole model CIRCUIT ORIENTED MODELS FOR DISPERSIVE DIELECTRICS 255 is the most simple for dielectrics with loss resulting in the least number of unknowns or compute time. However, a single-pole model can only accommodate a limited frequency range. Figure 10.2 shows the geometry for the dielectric bar cell for which we apply the model. We assume that the crosshatched areas are the contact surfaces. The admittance of the block of dielectric material in Fig. 10.2 for a Debye model is Yc(s) = s\u2215d\ud835\udf000 [ \ud835\udf00\u221e + (\ud835\udf00S \u2212 \ud835\udf00\u221e) 1 + s \ud835\udf0f ] , (10.19) where the model parameters to be specified are \ud835\udf00\u221e, \ud835\udf00S and \ud835\udf0f. It is easy to identify the equivalent circuit parameters. The capacitive admittance can be rewritten as Yc(s) = sCD,\u221e + YRC(s), (10.20) where the capacitance at f \u2192 \u221e, CD,\u221e is given by CD,\u221e = \ud835\udf000 \ud835\udf00\u221e  d . (10.21) The RC series parameters are CD = \ud835\udf000 (\ud835\udf00S \u2212 \ud835\udf00\u221e)  d (10.22) and RD = \ud835\udf0f d  \ud835\udf000 (\ud835\udf00S \u2212 \ud835\udf00\u221e) . (10.23) The corresponding equivalent circuit is shown in Fig. 10.3. Again, this simple model can only represent a limited variation of the permittivity with frequency",
            " 256 PEEC MODELS FOR DIELECTRICS The Lorentz media equations add somewhat more flexibility over the Debye model, which is necessary for some materials since it involves possibly complex pole behavior. Applying the same reasoning as in the Debye model case, the equivalent circuit for the model is given in Fig. 10.4. The admittance for the model branch is given by Yc(s) = s\u2215d\ud835\udf000 [ \ud835\udf00\u221e + (\ud835\udf00S \u2212 \ud835\udf00\u221e) \ud835\udf142 0 s2 + 2 s \ud835\udeff + \ud835\udf142 0 ] , (10.24) where the shape of the dielectric block is the same as for the Debye model in Fig. 10.2. The first term in (10.24) is again the admittance of the excess capacitance at infinite frequency and the second one is treated as an admittance YRLC(s) for the series connection in the equivalent circuit in Fig. 10.4 such that we have Yc(s) = sCL,\u221e + YRLC(s), (10.25) where CL,\u221e = \ud835\udf000\ud835\udf00\u221e\u2215d. It is a small task in circuit analysis to find the values of the circuit elements. Some simple manipulations allow us to synthesize YRLC(s) as a series RLC equivalent circuit with the following values of circuit parameters: \u23a7\u23aa\u23a8\u23aa\u23a9 LL = 1\u2215(\ud835\udf000\u2215d(\ud835\udf00S \u2212 \ud835\udf00\u221e)\ud835\udf142 0) RL = 2\ud835\udeff\u2215(\ud835\udf000\u2215d(\ud835\udf00S \u2212 \ud835\udf00\u221e)\ud835\udf142 0) CL = \ud835\udf000\u2215d(\ud835\udf00S \u2212 \ud835\udf00\u221e) (10"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure3.42-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure3.42-1.png",
        "caption": "Figure 3.42 Both bodies are in force equilibrium; (a) there is moment equilibrium; (b) there is no moment equilibrium: the forces together form a couple (anti-clockwise) with magnitude 3Fa.",
        "texts": [
            "41a) or are all parallel (Figure 3.41b). If for all the forces on a body the lines of action intersect at a single point, the moment equilibrium of the body is assured (see Figure 3.41a). The force equilibrium needs further investigation. If for all the forces on a body the lines of action are parallel, the force equilibrium is assured in the direction perpendicular to the lines of action (see Figure 3.41b). The force equilibrium in other directions, and the moment equilibrium needs further investigation. Example In Figure 3.42, both bodies are in force equilibrium. If one investigates the moment equilibrium by determining the sum of the moments of the forces, for example about A, it turns out that in case (a) the system is in moment equilibrium, while it is not in moment equilibrium in case (b). In case (a) the forces form an equilibrium system. In case (b), the forces form a couple acting anti-clockwise with magnitude 3Fa. Nothing can be said about the sign associated with the direction of the couple until a coordinate system is chosen"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure6.17-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure6.17-1.png",
        "caption": "Fig. 6.17 Force and moment acting on the j-th link",
        "texts": [
            " Next we consider the force and the moment act on each link. For preparation, let us rewrite the equation of rigid body motion we obtained in section 6.3 by using the spatial velocity vector \u03be [ f \u03c4 ] = IS \u03be\u0307 + \u03be \u00d7 IS\u03be (6.33) where the operator \u00d7 represents the following calculation. 3 Thanks to the spatial velocity representation we get this simple equation. When we use a conventional speed of a reference point, we need a more complicated propagation rule of acceleration[99]. 198 6 Dynamic Simulation \u03be\u00d7 = [ vo \u03c9 ] \u00d7 \u2261 [ \u03c9\u0302 0 v\u0302o \u03c9\u0302 ] . (6.34) Figure 6.17 shows the force and the moment acting on link j. f j , \u03c4 j are the force and the moment from body side (mother) to link i. The environmental 6.4 Dynamics of Link System 199 force and moment directory acting on link i are fE j , \u03c4 E j . Effects from the mother link, environment and reaction effect from the child link cause the motion of the link i. The equation of motion becomes [ f j \u03c4 j ] + [ fE j \u03c4E j ] \u2212 [ f j+1 \u03c4 j+1 ] = IS j \u03be\u0307j + \u03bej \u00d7 IS j \u03bej . (6.35) By rewriting this, we obtain a recurrence equation to give the propagation of the force and moment [ f j \u03c4 j ] = IS j \u03be\u0307j + \u03bej \u00d7 IS j \u03bej \u2212 [ fE j \u03c4E j ] + [ f j+1 \u03c4 j+1 ] "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001046_j.ijheatmasstransfer.2014.09.014-Figure11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001046_j.ijheatmasstransfer.2014.09.014-Figure11-1.png",
        "caption": "Fig. 11. Material densities at different times of LH (15",
        "texts": [
            " However, the distance required to reach steady state differs with the cases. Comparing HH with LH that has the same speed, HH needs a longer distance to reach the steady state as there is more energy input in it. Similarly, LL requires a longer distance than LH because LL is at a lower speed, which has more energy input per unit distance. Also from Fig. 10 it can be observed that steady state is reached after 570 lm for all three cases. This distance is used for the subsequent parameter sweep. Fig. 11 shows the material densities of LH at different times in the simulation. A close up view at t = 0.001 s is as shown in Fig. 12 where heat transfer has reached steady state and evaporation and melting are occurring under the laser beam. Furthermore, Fig. 12 shows that the powder turns into bulk solid as it cools after 0 W-1140 mm/s) with a layer of 50 lm powder. Fig. 13. The side view of the propa melting. The density of the material removed and evaporated (bright) is set to that of aluminium gas and remains as gas even as it cools"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure4.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure4.3-1.png",
        "caption": "Figure 4.3 A (rigid) curved line element is called an arch.",
        "texts": [
            " Figure 4.2 represents the mechanical diagram of a structure constructed from line elements. Line elements with a straight axis as known by a wide range of names, such as bar, beam, joist, girder, column, post and member. The nomenclature sometimes relates to the position of the line element in the structure: horizontal (beam, joist, girder) or vertical (column, post, stay). Hereafter, we will refer to a line element in general as a member. An (inflexible) curved line element is known as an arch, see Figure 4.3. A line element without a particular shape is a cable: cables adapt to the loading. A surface element (Figure 4.1c) is a two-dimensional structural element: one dimension (the thickness) is small with respect to the other two dimensions (the length and width). The behaviour of this element, which in reality is three-dimensional, can be described sufficiently accurately by means of a two-dimensional model by making simplified assumptions with respect to the thickness. In the two-dimensional model, all the properties of the element are assigned to a plane"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure2.15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure2.15-1.png",
        "caption": "Figure 2.15 The (tensile) force F that the rope AC exerts on the foundation block in C.",
        "texts": [
            " The first example relates to resolving a force into its components (Section 2.2.1). The second example relates to compounding forces (Section 2.2.2). In order to be able to resolve a force into its x, y and z component, we first have to calculate the direction cosines. This is illustrated by means of an example. 32 Example In Figure 2.14, a mast is being kept upright by three ropes. Rope AC is subject to a tensile force of 35 kN. Question: Find the x, y and z component of force F that the rope exerts on the foundation block in C (see Figure 2.15). Note that here we use the formal vector notation. Solution: The force F that is working on the foundation block has the same direction as the vector CA (directed from C to A). This vector, which indicates the direction of F is hereby referred to as d ,1 see Figure 2.15. F and d have the same direction cosines, so that cos \u03b1x = Fx F = dx d , cos \u03b1y = Fy F = dy d , cos \u03b1z = Fz F = dz d . Figure 2.15 shows that dx = \u22128 m, dy = +24 m, dz = +12 m. The x component of d is negative as it is pointing in the negative x 1 Remember the d of direction. ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM 2 Statics of a Particle 33 direction. The magnitude (length) d of d is d = \u221a d2 x + d2 y + d2 z = \u221a (\u22128 m)2 + (24 m)2 + (12 m)2 = 28 m. Using this information, we can calculate the components of F : Fx = F dx d = (35 kN) \u00d7 \u22128 m 28 m = \u221210 kN, Fy = F dy d = (35 kN) \u00d7 24 m 28 m = +30 kN, Fz = F dz d = (35 kN) \u00d7 12 m 28 m = +15 kN"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001805_j.matdes.2019.108091-Figure9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001805_j.matdes.2019.108091-Figure9-1.png",
        "caption": "Figure 9. Comparisons of defect diameter distributions for four equal control volumes from different build heights using exponential probability plots, (a) Ti-6AL-4V M290 vertical annealed machined surface, (b) Ti-6AL-4V M290 vertical HIPed machined surface, (c) Ti-6AL4V AM250 vertical annealed machined surface, (d) 17-4 PH M290 vertical heat treated as-built surface, (e) Ti-6AL-4V AM250 vertical annealed as-built surface.",
        "texts": [
            " To inspect the variability of the defect characteristics distributions along the build direction as can be seen in Figure 4(a), the scanned Jo ur na l P re -p ro of volume was divided into four equal sub-volumes with 2.5 mm height and a 1 mm gap was considered in between every two sub-volumes. The exponential probability plots for the defect diameter, volume, projected area, and sphericity distributions were depicted to visualize the variability of these defect characteristics distributions along the height of each specimen. The exponential probability plots of defect diameter distributions for the four control volumes along the build direction for different conditions are shown in Figure 9. As discussed earlier, only the exponential probability plots for defect diameter are presented here as a representation of defects size although the K-S test results for all these characteristics are later presented in tables. F is the cumulative distribution function value which Jo ur na l P re -p ro of is used to determine the probability that a random observation taken from a population will be less than or equal to a certain value. These plots show slight differences in defect diameter distributions between four regions of each specimen which are more pronounced for larger defects"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003841_08905458708905124-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003841_08905458708905124-Figure2-1.png",
        "caption": "Fig. 2 Three body mechanism.",
        "texts": [
            ", The matrix form of the acceleration relationship between bodies j and i is obtained by taking the time derivative of Eq. 14, where For a revolute joint, a;,, bij, diij, and 6a, are zero. As a result, the relative rotational axis z , in body i and z;' in body j coincide. For a translational BAE AND HAUG joint, B,, = 0, in which case the transformation matrix A,; is the identity. A cylindrical joint requires both relative coordinates and the full generality of the kinematics derived in this section. In order to illustrate use of the equations presented, the three body mechanism of Fig. 2 is used. Joint reference frames are parallel to centroidal reference frames, so matrices Cij are identities. Joint definition points are unit distances from the centroidal reference frames for each body, to simplify vectors sij and sji. The position and orientation of body 2 can be expressed in terms of the position and orientation of body 1 and relative coordinate 812 as sin 41 cos +I cos\" -sin\"][-:] sin 42 cos + I 4~ = +I + e12 (18) Similarly, the position and orientation of body 3 can be obtained as [;] = [;I + [ y h - s i n & ] [ l ',aa] sin +, cos +Z RECURSIVE FORMULATION FOR MECHANICAL DYNAMICS 365 Equations 18 and 19 can be differentiated to obtain velocity relations between bodies as and Equations 20 and 21 can be expressed in matrix form as and where Yi = [x i , yi , &IT and matrices Bijk are defined as in Eqs",
            " The reduced variational equations for the next level of inboard junction nodes can be calculated by starting from first level junction nodes and, if any, tree end chains that are connected to each junction node. This procedure is repeated for every chain in working back to the base body. The reduced variational equation for the base body is thus obtained as in Eq. 44. The Lagrange multiplier theorem can be used to obtain the augmented base body equations of motion, similar to Eq. 48. VI. A N ELEMENTARY EXAMPLE In order to illustrate the derivation in Sections 1V and V, the mechanism in Fig. 2 is used . '~ot ion is planar, to simplify derivation of formulas. The base body is designated as body 1, which has no externally applied constraints. A revolute joint connects bodies 1 and 2. Body 3 translates with respect to body 2. Therefore, this system has 5 degrees-of-freedom, including 3 base body Cartesian degrees-of-freedom. The variational equation of motion for this system is where Mi = diag(mi, mi, J,!) and 6Zi must satisfy kinematic admissibility conditions for constraint equations corresponding to joints 1 and 2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002612_acsami.0c13863-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002612_acsami.0c13863-Figure1-1.png",
        "caption": "Figure 1. Magnetic multimaterial DIW(M3DIW) system and working mechanism. (a) Schematic of the M3DIW fabrication and material composition. (b) Material distribution and magnetization directions of a one-dimensional M-SMP/MSM stripe with four segments. (c) Four different deformation modes achieved by cooperatively controlling the temperature and magnetic field.",
        "texts": [
            " The functionality of the aforementioned multimaterials relies on not only the carefully designed material systems but also the fabrication methods to effectively implement the complex material integration, which has been greatly promoted by the recent development of 3D printing techniques such as stereolithography,1,34,35 fused deposition modeling,36\u221238 direct ink writing (DIW),39\u221242 and integrated technique.43 In this work, we present a magnetic multimaterial DIW (M3DIW) technique for the complex structural integration of photocurable M-SMP and MSM to explore their enhanced multimodal shape transformation with tunable properties and shiftable mechanical behaviors. Figure 1a schematically shows the M3DIW fabrication system and the main composition of the inks. Two types of magnetic composite inks, M-SMP and MSM, composed of uncured photocurable polymeric resins, photoinitiators, magnetized neodymium\u2212iron-boron (NdFeB) microparticles, and fumed silica nanoparticles as a rheological modifier, are loaded in the ultraviolet (UV) block syringes. The photocurable resins are prepared by two different recipes of monomers and cross-linker (see Materials and Methods for more details) for synthesizing materials with distinct temperature-dependent material properties",
            " During the printing, the programmed magnetization is achieved by reorienting the particles\u2019 polarities to the longitudinal direction of the nozzles, where a printing magnetic field around 180 mT is induced by a ring-shaped magnet. To protect the magnetization of the already-printed structure from the influence of the printing magnetic field, we added a steel magnetic shield at the nozzle tip to mitigate the intensity of the magnetic field. By controlling the switching between the two syringes as well as the printing directions, MSM and MSMP can be structurally integrated into the design. The direction of the printing magnetic field and the magnetization of the printed structure are shown in Figure 1a. An LED chip emitting 385 nm wavelength UV light is utilized to cure the resins after printing. After curing, MSM is soft at room temperature, whereas M-SMP has a Tg around 60\u221270 \u00b0C. The multimodal shape configuration of the M-SMP/MSM composites relies on the cooperative stimulation by both the magnetic field and temperature change. Here, we use a simple one-dimensional M-SMP/MSM stripe to illustrate the operation mechanism. The magnetization distribution and the four deformation modes are depicted in Figure 1b,c, respectively. At room temperature (lower than the Tg of MSMP), only the MSM parts can deform when applying an actuation magnetic field, whereas the M-SMP parts stay undeformed because of the high stiffness, forming the deformation Mode 1 with a \u201cU\u201d shape. Upon the removal of the magnetic field, the deformed shape returns to the undeformed configuration, demonstrating fast-transforming capability. When the operating temperature is above Tg, both materials can be actuated by the magnetic field, forming the Mode 2 with a \u201cW\u201d shape",
            " Next, by maintaining the magnetic field and cooling the structure to below Tg, the M-SMP parts lock into the deformed shape, whereas the MSM parts return to their undeformed configuration once the magnetic field is removed, leading to the shape in the Mode 3 with a peak at the center. Finally, applying a reversed magnetic field to the https://dx.doi.org/10.1021/acsami.0c13863 ACS Appl. Mater. Interfaces XXXX, XXX, XXX\u2212XXX B structure in the Mode 3 results in the wavy deformation in the Mode 4. Note that another set of four vertically symmetric deformation modes can be easily obtained by reversing all the directions of the external magnetic field in Figure 1c. By conducting finite-element (FE) simulations,44 the responses of the magnetic multimaterial structures can be precisely predicted, guiding the designs for desired multimodal actuation. \u25a0 RESULTS AND DISCUSSION Ink Characterization. The success of multimaterial DIW printing of the M-SMP/MSM composites depends on the printability of the ink and its curing efficiency and quality. In M3DIW, the inks are extruded from nozzles with predetermined diameters and then cured by UV light, where two major ink properties influence the printing and curing process, i"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.17-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.17-1.png",
        "caption": "FIGURE 8.17. Illustration of a De Dion suspension.",
        "texts": [
            " When a solid axle is a dead axle with no driving wheels, the connecting beam between the left and right wheels may have different shapes to do different jobs, usually to give the wheels independent flexibility. We may also modify the shape of a live axle to attach the differential to the chassis and reduce the unsprung mass. De Dion design is a modification of a beam axle that may be used as a dead axle or to attach the differential to the chassis and transfer the driving power to the drive wheels by employing universal joints and split shafts. Figure 8.17 illustrates a De Dion suspension. 8.2 Independent Suspension Independent suspensions is introduced to let a wheel to move up and down without affecting the opposite wheel. There are many forms and designs of independent suspensions. However, double A-arm and McPherson strut suspensions are the simplest and the most common designs. Figure 8.18 illustrates a sample of a double A-arm and Figure 8.19 shows a McPherson suspension. Kinematically, a double A-arm suspension mechanism is a four-bar linkage with the chassis as the ground link, and coupler as the wheel carrying link"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure6.7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure6.7-1.png",
        "caption": "Figure 6.7 In regulations, the mobile load is seen as a uniformly distributed surface load together with a limited number of concentrated loads.",
        "texts": [
            " For flat roofs, the possibility that the water cannot drain away has to be considered. This incurs the risk of water accumulation: the deflection of the roof due to the water allows for the storage of an increasing amount of water (see Figure 6.6). If the roof is not sufficiently rigid, this can result in its collapse at times of continuing rainfall. Vertical live loads on bridges due to traffic are referred to as moving loads. In regulations, this load is a uniformly distributed surface load together with a limited number of concentrated loads (see Figure 6.7). 6 Loads 213 The uniformly distributed load is a representation of the actual load that can occur over large lengths. This load becomes more important for longer spans. The system of point loads, with the underlying part of the uniformly distributed load, represents the load caused by a few very heavy trucks or locomotives. This load is important for bridge elements of limited length. It may occur that certain structural elements are loaded more unfavourably if the load is omitted over a certain length"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure14.9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure14.9-1.png",
        "caption": "Figure 14.9 A cable loaded by two vertical forces.",
        "texts": [
            " VOLUME 1: EQUILIBRIUM The forces at A and B on the isolated cable in Figure 14.8b can be seen as the support reactions of a cable without compression member, supported at two fixed points. The vertical support reactions Av and Bv are equal to those of a beam with the same span and load only if the supports are at equal elevations. With a difference h between both support elevations, the two horizontal support reactions H form a couple that leads to a change in the vertical support reactions of Hh/ . This is applied in the following two examples. Example 1 The cable in Figure 14.9, supported at the fixed points A and B, is loaded in C and D by vertical forces of 75 and 30 kN respectively. Only the location of point D is given for the cable shape: it is 4 metres lower than support A. Questions: a. Determine the cable shape. b. Determine the horizontal support reactions at A and B. c. Determine the vertical support reactions at A and B. d. Determine the maximum and minimum cable forces. Solution: a. Figure 14.10a shows a simply supported beam AB with the same (horizontal) span as the cable and the same vertical load. The support reactions are also shown. Figure 14.10b shows the associated M diagram. Under chord AB, the cable has the same shape as the M diagram; according to (11): Hzk = M. The scale factor H is the horizontal component of the tensile force in the cable. At D, the distance from the chord AB to the cable is (see Figure 14.9) 14 Cables, Lines of Force and Structural Shapes 643 zk;D = 2 3 (12 m) + (4 m) = 12 m. This gives H = MD zk;D = 900 kNm 12 m = 75 kN. At C, the distance between the chord and the cable is zk;C = MC H = 1200 kNm 75 kN = 16 m. The cable shape is now determined (see Figure 14.11). b. The horizontal support reactions at A and B are equal to the horizontal component H of the cable force determined above: Ah = Bh = H = 75 kN. The horizontal support reactions are shown in Figure 14.11. c. The vertical support reactions Av and Bv in Figure 14"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure6.38-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure6.38-1.png",
        "caption": "FIGURE 6.38. A centric and symmetric slider-crank mechanism.",
        "texts": [
            " The coordinates (xC , yC) for point C are xC = a cos \u03b82 + c cos\u03b2 (6.266) yC = a sin \u03b82 + c sin\u03b2. (6.267) 6. Applied Mechanisms 361 To calculate the angle \u03b3, we examine 4AEB and find sin \u03b3 = AE AB = a sin \u03b82 \u2212 e b (6.268) that finishes Equation (6.264). Therefore, the coordinates xC and yC can be calculated as two parametric functions of \u03b82 for a given set of a, b, c, e, and \u03b1. Example 250 A centric and symmetric slider-crank mechanism. Point C (xC , yC) is the coupler point of a centric and symmetric slidercrank mechanism shown in Figure 6.38. It is centric because e = 0, and is symmetric because a = b, and therefore, \u03b82 = \u03b84. Point C is on the coupler link AB and is at a distance kb from A, where 0 < k < 1. The coordinates of point C are xC = a cos \u03b82 + ka cos \u03b82 = a (1 + k) cos \u03b82 (6.269) yC = a sin \u03b82 \u2212 ka sin \u03b82 = a (1\u2212 k) sin \u03b82 (6.270) and therefore, cos \u03b82 = xC a (1 + k) (6.271) sin \u03b82 = yC a (1\u2212 k) . (6.272) Using cos2 \u03b82 + sin2 \u03b82 = 1, we can show that the coupler point C will 362 6. Applied Mechanisms move on an ellipse. x2C a2 (1 + k) 2 + y2C a2 (1\u2212 k) 2 = 1 (6"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001611_978-3-319-26054-9-Figure22-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001611_978-3-319-26054-9-Figure22-1.png",
        "caption": "Fig. 22 Initial exported tree structure",
        "texts": [
            " a Spherical tank used on the virtual environment. b A real sphere, with the same dimensions of the virtual one performing calculations somewhat similar to a forward kinematics, and keeping track of the robot\u2019s odometry, but it does not count slips on the wheels and this errors are cumulative\u2014this is known as Dead Reckoning\u2014, and lastly the LIDAR odometry, that has been covered in detail in the previous section. Wheel odometry is calculated based on the wheel\u2019s encoders. A very simple encoder is shown on Fig. 22. This disc-like structure is attached to the motor and for each full turn on the motor, the two beams are tampered sixteen times, making the encoder generate sixteen pulses. The encoder attached to the real motors has a resolution Localization and Navigation of a Climbing Robot Inside a LPG Spherical Tank \u2026 177 way higher, having 13.500 pulses for each wheel turn, given the gear ratio from the motor to the wheel itself and publish its informations on topic /to_odom. Given the diameter of each wheel and the number of pulses per turn, it is possible to calculate how much the robot have moved in a given time",
            " Once this Goal point is found, a curvature is computed to reach it as depicted in the Fig. 21. In order to localize the robot in the environment, Agvs can use either amcl along with a map created previously or magnetic landmarks installed along the path. While amcl provides higher flexibility and easy and fast mapping, magnetic landmarks implies a tedious installation and the mapping of every mark. However, magnetic landmarks provide much higher accuracy and independence of environment or map modifications (Fig. 22). Coordination amongst robots sharing the same environment required a higher level component to control all the movements in order to manage the fleet planning and scheduling, and efficient resource sharing while avoiding system interlocks. The planning problem is to some point similar to the planning and scheduling problem of an operating system. A set of n AGVS are commanded to perform a Robotnik\u2014Professional Service Robotics Applications with ROS 277 number of transport missions inside a building",
            " The last chain which makes up the rover is the rear Simulation of Closed Kinematic Chains in Realistic \u2026 587 Fig. 20 Transverse bogie an open kinematic chain Fig. 21 Base link and the longitudinal bogie assembly transverse bogie. This chain is open so it can be exported directly. Each longitudinal bogie consists of five links, excluding the wheels, as shown in Fig. 21. The base link will have five direct children links in total, four from the front bogies and one from the rear transverse bogie. Figure22 shows the complete tree structure with all the links, including the wheels, named (the numbers refer to Fig. 21 and have been added for clarity, they are not part of the description). Once all of the links and the joints have been created and named, the URDFmodel is ready for export. As the export process happens the plugin will create origins and axis for the joints. During the export, the plugin will create several folders, among them one called robots. Inside of the robots folder there is a URDF file that can be converted to a SDF file"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure15.15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure15.15-1.png",
        "caption": "Figure 15.15 A concentrated couple can be replaced by a statically equivalent pair of forces. The virtual work performed by the couple is equal to the product of couple and rotation.",
        "texts": [
            " The expression for virtual work \u03b4A is now in the same form as the expression for G; the quantities \u03bb1; \u03bb2; \u03bb3 have been replaced by the virtual displacements \u03b4ux0; \u03b4uy0; \u03b4\u03d5z0. The demand \u03b4A = \u03b4ux0 \u2211 i Fxi + \u03b4uy0 \u2211 i Fyi + \u03b4\u03d5z0 \u2211 i (xiFyi \u2212 yiFxi) = 0 for all arbitrary combinations of \u03b4ux0; \u03b4uy0; \u03b4\u03d5z0 (not all equal to zero) is a necessary and sufficient condition for equilibrium and is known as the principle of virtual work. For simplicity, concentrated couples were not addressed. A concentrated couple can be replaced by a statically equivalent pair of forces. Assuming 724 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM the pair of forces in Figure 15.15 we find \u03b4A = \u2212F \u00b7 b\u03b4\u03d5z0 + F \u00b7 (a + b)\u03b4\u03d5z0 = Fa \u00b7 \u03b4\u03d5z0 = Tz \u00b7 \u03b4\u03d5z0. The work performed by a couple is equal to the product of couple and rotation. The work is positive when the couple and rotation are in the same direction. In deriving the virtual work equation, we found that the geometric relationships had to be linear in the virtual displacements. Mathematically, this means that the first-order variation has to be assumed for these virtual displacements. We previously deduced (see Figure 15.13) that the following applies for the displacement of an arbitrary point i, due to a translation ux0; uy0 and a rotation \u03d5z0: uxi = ux0 \u2212 ri { cos \u03b1i \u2212 cos(\u03b1i + \u03d5z0) } , uyi = uy0 + ri { sin(\u03b1i + \u03d5z0) \u2212 sin \u03b1i } "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001103_j.jmst.2021.06.011-Figure10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001103_j.jmst.2021.06.011-Figure10-1.png",
        "caption": "Fig. 10. Processing parameters and alloy compositions as fundamental elements that determine additively manufactured microstructures of metals.",
        "texts": [
            " Owing to the rapid cooling rate in additive manufacturing, a suersaturated solid solution is obtained at the atomic scale, which rovides additional strengthening for the metallic materials. In adition, when the solid\u2013liquid interface reverses from fluxing forard, it leaves additional solute at the rear of the melt pool, form- ng a unique type of segregation structure. . Perspective In a broad sense, the process parameters and alloy composiions are two fundamental elements for determining the additively anufactured microstructures of metals, as shown in Fig. 10 . Durng additive manufacturing process, the input energy has a diect impact on the fusion of the metallic powder and thus deter- ines the final solidification microstructures through grain nucletion and growth in the molten bead; the overlap of melt pools nduces cyclic heat treatment, finally leading to construction of ost-solidification microstructures, including dislocation cells and anoscale precipitates; in addition, there also exist some factors uch as implicit time and residual stress owing to the intermittent eposition fashion of additive manufacturing, they also have signifcant effects on the microstructure evolution process, thus require ore attention in the future"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000554_978-3-540-85629-0-FigureC.4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000554_978-3-540-85629-0-FigureC.4-1.png",
        "caption": "Fig. C.4. RPY angles.",
        "texts": [
            "12) In both cases there are infinite solutions, since only the sum (difference) of \u03d5 and \u03b8 can be determined. In fact, being \u03b8 = 0, \u03c0, the rotations \u03d5,\u03c8 are about parallel axes and therefore it is not possible to distinguish between them. C.4 Roll-Pitch-Yaw Angles 493 C.4 Roll-Pitch-Yaw Angles The Roll-Pitch-Yaw (RPY) representation indicates three consecutive rotations about the axes of the base reference frame F0: of an angle \u03d5 (Yaw) about x, of an angle \u03b8 (Pitch) about y and of an angle \u03c8 (Roll) about z, see Fig. C.4. The rotation matrix RRPY corresponding to these three rotations is defined as RRPY (\u03d5, \u03b8, \u03c8) = \u23a1 \u23a3 c\u03c6c\u03b8 \u2212 s\u03c6c\u03c8 + c\u03c6s\u03b8s\u03c8 s\u03c6s\u03c8 + c\u03c6s\u03b8c\u03c8 s\u03c6c\u03b8 c\u03c6c\u03c8 + s\u03c6s\u03b8s\u03c8 \u2212 c\u03c6s\u03c8 + s\u03c6s\u03b8c\u03c8 \u2212s\u03b8 c\u03b8s\u03c8 c\u03b8c\u03c8 \u23a4 \u23a6 .(C.13) For the inverse problem, i.e given a generic rotation matrix R as in (C.5) define the three angles \u03d5, \u03b8, \u03c8, there are two possibilities: 1. r2 11 + r2 21 = 0 \u2192 cos \u03b8 = 0: one obtains\u23a7\u23aa\u23a8 \u23aa\u23a9 \u03d5 = atan2(r21, r11) \u03b8 = atan2(\u2212r31, \u221a r2 32 + r2 33) \u03c8 = atan2(r32, r33) with \u03b8 \u2208 [\u2212\u03c0/2, \u03c0/2], or\u23a7\u23aa\u23a8 \u23aa\u23a9 \u03d5 = atan2(\u2212r21,\u2212r11) \u03b8 = atan2(\u2212r31,\u2212 \u221a r2 32 + r2 33) \u03c8 = atan2(\u2212r32,\u2212r33) if \u03b8 \u2208 [\u03c0/2, 3\u03c0/2]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002770_tii.2021.3089340-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002770_tii.2021.3089340-Figure6-1.png",
        "caption": "Fig. 6. The bearing accelerated life test-rig.",
        "texts": [
            " Personal use is permitted, but republication/redistribution requires IEEE permission. See http://www.ieee.org/publications_standards/publications/rights/index.html for more information. Cores Conv 1 4 4 / 2 64 68 68 64  IN, ReLu Conv 2 4 4 / 2 128 34 34 128  IN, ReLu Conv 3 4 4 / 1 256 34 34 256  IN, ReLu Conv 4 4 4 / 1 512 34 34 128  IN, ReLu Conv 5 4 4 / 1 1 34 34 1  - IV. EXPERIMENTAL VERIFICATION The life-cycle bearing vibration samples used in this study are from the XJTU-SY accelerated life test datasets [40]. The bearing accelerated life test-rig is illustrated in Fig. 6. The test-rig is used to conduct the accelerated degradation test of rolling bearing. To acquire the run-to-failure data of the testing bearings, two unidirectional acceleration sensors (PCB 352c33) were used to measure the radial and axial vibration signals of the bearing respectively. The sampling frequency was 25.6 kHz, the sampling interval was 1 min, and the sampling length of one sample was 1.28 s. The type of the testing bearing is LDK UER204 and whose main parameters are shown in Table V"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure4.30-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure4.30-1.png",
        "caption": "Fig. 4.30: The kinematic chain and its design parameters for a simple widely-used gripping mechanism.",
        "texts": [
            " Once the numerical computations are convergent to a feasible solution, the velocity and the accuracy of the solution can be enhanced by a designer by updating the convergence parameters \u03b5f and \u03b5g, which refer to the objective function f and the constraint functions g, respectively. Of course, the optimum solution is affected by the initial guess. But the algebraic formulation of the objective function f and constraint g can be useful to obtain optimum design also when the initial gripping mechanism is far away from the prescribed limits or it even violates some of the prescribed requirements. In Fig. 4.30 a widely-used finger gripping mechanism is shown with its design parameters. It is of the slider-crank type and a design problem can be formulated as concerned with the dimensions of the links and stroke range, but in order to obtain suitable behavior within a certain range of grasping capability. Therefore, the optimization problem of Eqs (4.7.7) and (4.7.8) by using the G.I. can be conveniently applied with the numerical procedure of Fig. 4.29. The specific kinematic chain of the gripping mechanism allows the opportunity to exploit the expression (4"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure2.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure2.2-1.png",
        "caption": "Fig. 2.2. Configuration of double-sided AFPM brushless machine with internal disc rotor: 1 \u2014 rotor, 2 \u2014 PM, 3 \u2014 stator core, 4 \u2014 stator winding.",
        "texts": [
            " The single-sided motor according to Fig. 2.1a has a standard frame and shaft. It can be used in industrial, traction and servo electromechanical drives. The motor for hoist applications shown in Fig. 2.1b is integrated with a sheave (drum for ropes) and brake (not shown). It is used in gearless elevators [113]. In the double-sided machine with internal PM disc rotor , the armature winding is located on two stator cores. The disc with PMs rotates between two stators. An eight-pole configuration is shown in Fig. 2.2. PMs are embedded or glued in a nonmagnetic rotor skeleton. The nonmagnetic air gap is large, i.e. the total air gap is equal to two mechanical clearances plus the thickness of a PM with its relative magnetic permeability close to unity. A double-sided machine with parallel connected stators can operate even if one stator winding is broken. On the other hand, a series connection is preferred because it can provide equal but opposing axial attractive forces. 30 2 Principles of AFPM Machines A practical three-phase, 200 Hz, 3000 rpm, double-sided AFPM brushless motor with built-in brake is shown in Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001467_j.mechmachtheory.2010.05.001-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001467_j.mechmachtheory.2010.05.001-Figure5-1.png",
        "caption": "Fig. 5. A spherical RRR leg.",
        "texts": [
            " 4), where the prismatic joint between two passive joints is actuated, is usually employed in parallel manipulators. In such a leg, the position of the prismatic actuator is different from that in the PSS leg. However, the input transmission index of the SPS or UPS leg can still be obtained through the above result. Since the translational direction of the prismatic actuator is along the coupler link, the input transmission index of the SPS or UPS leg is constant and its value is equal to 1. As shown in Fig. 5, a spherical RRR leg contains three revolute joints, the axes (OR1, OR2, and OR3) of which intersect at a common point O. The revolute joint attached to the base is actuated. Parameter \u03b112 represents the angle between the axes OR1 and OR2, and \u03b123 stands for the angle between the axes OR2 and OR3. In such a spherical RRR leg, the transmission wrench is a pure torque, the axis of which is perpendicular to the plane OR2R3 and passes through the pointO. Based on the results in [20], the input transmission index of the spherical RRR leg can be obtained as \u03bbs = jsin\u2220R1 OR2 R3j \u00f016\u00de \u2220R1\u2212OR2\u2212R3 represents the angle between the planes OR1R2 and OR2R3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.2-1.png",
        "caption": "FIGURE 8.2. A side view of a multi-leaf spring and solid axle suspension.",
        "texts": [
            " The solid axle must be attached to the body such that an up and down motion in the z-direction, as well as a roll rotation about the x-axis, is possible. So, no forward and lateral translation, and also no rotation about the axle and the z-axis, is allowed. There are many combinations of links and springs that can provide the kinematic and dynamic requirements. The simplest design is to clamp the axle to the middle of two leaf springs with their ends tied or shackled to the vehicle frame as shown schematically in Figure 8.1. A side view of a multi-leaf spring and solid axle is shown in Figure 8.2. A suspension with a solid connection between the left and right wheels is called dependent suspension. 456 8. Suspension Mechanisms The performance of a solid axle with leaf springs suspension can be improved by adding a linkage to guide the axle kinematically and provide dynamic support to carry the non z-direction forces. The solid axle with leaf spring combination came to vehicle industry from horse-drawn vehicles. Example 299 Hotchkiss drive. When a live solid axle is connected to the body with nothing but two leaf springs, it is called the Hotchkiss drive, which is the name of the car that used it first. The main problems of a Hotchkiss drive, which is shown in Figure 8.2, are locating the axle under lateral and longitudinal forces, and having a low mass ratio \u03b5 = ms/mu, where ms is the sprung mass and mu is the unsprung mass. Sprung mass refers to all masses that are supported by the spring, such as vehicle body. Unsprung mass refers to all masses that are attached to and not supported by the spring, such as wheel, axle, or brakes. Example 300 Leaf spring suspension and flexibility problem. The solid axle suspension systems with longitudinal leaf springs have many drawbacks"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003881_mssp.2001.1414-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003881_mssp.2001.1414-Figure1-1.png",
        "caption": "Figure 1. Diagram of the 16-degree-of-freedom gear dynamic model. Note that the vertical direction is aligned with the pressure line of the gear mesh.",
        "texts": [
            "7 mm root crack in a single tooth has also been modelled, and it has been observed that the crack reduces the tooth sti!ness, particularly just prior to the next tooth coming into contact [6]. The model developed here is based on a single stage reduction gearbox, where a 1 : 1 ratio, with 23 teeth on each gear is used for simplicity. The major parameters that are included in the model are shown in Table 1. There are a total of 16 degrees of freedom in the model and a schematic diagram of the model is shown in Fig. 1. The major assumptions which the dynamic model is based upon are the following. (i) Resonances of the gearcase are neglected. (ii) Shaft mass and inertia are lumped at the bearings or the gears. (iii) Shaft transverse resonances are neglected. (iv) Shaft torsional sti!ness is ignored (#exible coupling torsional sti!ness is very low). (v) Gear teeth pro\"les are perfect involute curves, with no geometrical, pitch or runout errors. (vi) Static transmission error e!ects are very much smaller than the dynamic transmis- sion error e"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure3.9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure3.9-1.png",
        "caption": "Fig. 3.9 Force/Moment at the ZMP Expressed by Forces at Discretized Points",
        "texts": [
            " When a robot moves, \u03c4tz = 0 is not generally satisfied. Thus, the ZMP is not a point where all components of the moment becomes zero for 3D cases. The ZMP is defined as the point where the horizontal component of the moment of the ground reaction forces becomes zero for 3D cases. 76 3 ZMP and Dynamics 2 Region of ZMP in 3D Let us define the region of the ZMP for 3D cases. For simplicity, we consider the ground reaction forces f i = [fix fiy fiz ] T acting at the discretized points pi \u2208 S (i = 1, \u00b7 \u00b7 \u00b7 , N) as shown in Fig. 3.9. This approximation becomes more exact as the number of discretized points increases. Next, distributed N force vectors are replaced by a force and a moment vectors acting at the point p as f = N\u2211 i=1 f i (3.17) \u03c4 (p) = N\u2211 i=1 (pi \u2212 p)\u00d7 f i. (3.18) The position of the ZMP can be obtained by setting the first and the second elements of (3.18) be zero. This yields 3.2 Measurement of ZMP 77 p = \u2211N i=1 pifiz\u2211N i=1 fiz . (3.19) For ordinary humanoid robots without magnets or suction cups at the sole, the vertical component of the ground reaction forces becomes zero for all discretized points, i"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002011_1.4008346-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002011_1.4008346-Figure5-1.png",
        "caption": "Fig. 5",
        "texts": [
            "yields: 3 PtxR (81) Qy' = - f - d r w a r - d r v r ~ t2 sin 7 dtdq (82) The moment about an axis through the center of the ball, perpendicular to the line defining the contact angle and lying in the plane of x', z' is: dQR = [VR2 - x2 - VR2 - a2 + Vr For the whole ellipse: a2] cos 7 dF (83) v - d r w d y - ^ ) ' ] t2 cos 7 dtdq (84) The value of 7 is found from Fig. 4 as: po>, sin 6 - Vx tan 7 = tan 7 = pw, cos 6 + Vu aw, kt ? + V\u201e (85) (86) 6 / M A R C H 1 9 5 9 Transactions of the ASME Downloaded From: https://fluidsengineering.asmedigitalcollection.asme.org on 06/24/2019 Terms of Use: http://www.asme.org/about-asme/terms-of-use Equilibrium Conditions Fig. 5 shows the moments acting on a ball. Fig. 6 shows the forces acting on a ball. From Fig. 5: -Qn0 sin ft + Qeo COS ft + M,' + Qri sin ft - Qsi cos ft = 0 (87) -QRo COS ft - Qso sin ft + QRi cos ft + Qsi sin ft = 0 (88) J+o In Fig. 7 ft is the initial contact angle of the bearing. The race centers are originally at (0) and (3) while the ball is at (1). When equilibrium is attained under high-speed conditions the inner race curvature centers has moved axially from (3) to (4), an amount SH. The ball has moved from (1) to (2). These movements result from the elastic deformations 8\u201e and occurring at the outer and inner race contacts"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001988_j.eng.2017.05.023-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001988_j.eng.2017.05.023-Figure2-1.png",
        "caption": "Fig. 2. (a) Thermal boundary conditions: \u2460 laser heat input on the top surface; \u2461 heat losses due to convection and radiation; \u2462 insulated walls; \u2463 constant temperature at the bottom. (b) Bulk 3D geometry considered in the FE calculation.",
        "texts": [
            " Therefore, if the melt pool length is a major concern, the length of the calculation where hrad is the heat transfer coefficient for radiation, \u03c3 is the StefanBoltzmann constant, \u03b5 is the emissivity, and T\u221e is the ambient temperature. The boundary condition at the bottom of the substrate is maintained at room temperature, based on the experimental observation that the measured temperature at the substrate shows insignificant change during the process [6]. The side walls are assumed to be insulated. Fig. 2(a) summarizes the thermal boundary conditions used for numerical modeling in the present study, while Fig. 2(b) illustrates the bulk 3D geometry considered in the FE calculation. Rosenthal [8] originally developed the analytical method to predict the temperature history in fusion welding. Because of similarities between fusion welding and SLM, the Rosenthal equation is extended to predict the thermal characteristic in laser melting. The Rosenthal equation owes its merit to its simplicity and wide applicability. Moreover, this single formula is capable of predicting temperature history as a function of time, temperature gradient, cooling rate, and solidification rate"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003774_j.1469-185x.1974.tb01299.x-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003774_j.1469-185x.1974.tb01299.x-Figure6-1.png",
        "caption": "Fig. 6. Features of Paramecium. ( a ) Cycle of beat; the cilia in positions 4-7 are bent to the side away from the observer during the recovery stroke. (6) Metachronal wave and direction of propulsion (Machemer, 1972a).",
        "texts": [
            " During its beat each cilium performs a three-dimensional movement \u2018polarized\u2019 in both time and space - there is a more rapid and near-planar phase (effective stroke) and a slower phase in which the cilium moves back towards one side while a bend is propagated to the tip (recovery stroke); the tip of the cilium is seen from above to follow an anti-clockwise path (Machemer, 1972b, 1973; Tamm, 1972). The orientation of the effective stroke is directed backwards and obliquely (normally at about 20\u2019) towards the organism\u2019s left, while the metachronal waves travel forwards and obliquely towards the organism\u2019s left; the effective stroke is towards the right of, and perpendicular to, the direction of propagation of the metachronal wave, so that the metachronism is truly dexioplectic, although the organism as a whole exhibits an antiplectoid metachronal pattern at first sight (Fig. 6). Normally about ten of these oblique metachronal waves may be seen on the body at any instant; at normal temperatures the frequency of ciliary beating is 30-35 Hz and the metachronal wavelength is about 10 pm (Machemer, 1 9 7 2 ~ ) . The organism swims forward in a helical path, and the body also rotates, so that the oral groove may remain at the inside of the helix; swimming speeds in the range of 60o-z500,um/s are common (Gray, 1928; Machemer, 1972b). As in Opalina, changes in the orientation of ciliary beating (which changes in relation to ionic conditions and the cell\u2019s membrane potential) are reflected in changes in the direction of propagation of metachronal waves (Eckert, 1972)",
            " Conversely, during the effective stroke in symplectic metachronism, the cilia are closer together, so w(s, t ) is greater than I. During the recovery stroke, we need only interchange the two cases. If we are considering other types of metachronism, we would only consider the relative ciliary velocity component in the metachronal wave direction. The velocity field for the movement of a particular cilium can be obtained numerically. Calculations were made on the movements of cilia taken from the three organisms Oputina (Fig. 4), Paramecium (Fig. 6) and Pleurobrachia (Fig. 5 ) . In the case of Paramecium we have taken a three-dimensional beat pattern from Machemer (1972b). The time-averaged velocity fields are shown in Fig. 16. In solving the integral equations, non-dimensional variables y, K and T were defined as follows: where C, and CT are the normal and tangential resistance coefficients respectively (see (2)), cr is the radian frequency of the cilium, c is the metachronal wave speed, 6\" the angle of inclination of the direction of propagation of the metachronal waves to the effective stroke (i"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure10.37-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure10.37-1.png",
        "caption": "FIGURE 10.37. A two-wheel vehicle model with the vehicle body coordinate frame and the center of rotation O.",
        "texts": [
            " Furthermore, some other transient inputs may also be used to analyze the dynamic behavior of a vehicle. Single sine steering, linearly increasing steering, and half sine lane change steering are the most common transient steering inputs. Example 409 F Position of the rotation center. The position of the center of rotation O in the vehicle body coordinate is at x = \u2212R sin\u03b2 (10.482) y = R cos\u03b2 (10.483) 652 10. Vehicle Planar Dynamics 10. Vehicle Planar Dynamics 653 because \u03b2 is positive when it is about a positive direction of the z-axis. Figure 10.37 illustrates a two-wheel vehicle model, the vehicle body coordinate frame, and the center of rotation O. At the steady-state condition the radius of rotation can be found from the curvature response S\u03ba, and the angle \u03b2 can be found from the sideslip response S\u03b2. R = 1 \u03b4S\u03ba = vx (DrC\u03b2 \u2212 CrD\u03b2 +mvxD\u03b2) (C\u03b4D\u03b2 \u2212 C\u03b2D\u03b4) \u03b4 (10.484) \u03b2 = \u03b4S\u03b2 = D\u03b4 (Cr \u2212mvx)\u2212DrC\u03b4 DrC\u03b2 \u2212 CrD\u03b2 +mvxD\u03b2 \u03b4 (10.485) Therefore, the position of the center point O is at x = \u2212vx (DrC\u03b2 \u2212 CrD\u03b2 +mvxD\u03b2) (C\u03b4D\u03b2 \u2212 C\u03b2D\u03b4) \u03b4 \u00d7 sin D\u03b4 (Cr \u2212mvx)\u2212DrC\u03b4 DrC\u03b2 \u2212 CrD\u03b2 +mvxD\u03b2 \u03b4 (10"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003928_tpel.2006.872371-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003928_tpel.2006.872371-Figure5-1.png",
        "caption": "Fig. 5. Finite element field solution with phase overlap under short flux path excitation.",
        "texts": [
            " Notably, a part of the back iron, located between two excited stator phases, exhibits a very low flux density. This is due to the fact that fluxes generated by each phase oppose each other within this region. In the remaining 75% of the back iron, on the other hand, fluxes generated by two stator phases are added. This explains the existence of a relatively high flux density in this part of the back iron. This pattern of magnetization has been referred to as \u201clong flux path excitation (LFPE).\u201d Fig. 5 shows the distribution of flux density which changes once in every electrical cycle in the SRM for conventional winding in the stator coils under unipolar excitation. This pattern of magnetization is referred to as \u201cshort flux path excitation (SFPE).\u201d This means that in an 8/6 SRM, within every electrical cycle, there exist three LFPE and one SFPE. Since SRM drives are usually operated under heavy saturation, one can observe certain differences between the performances of the machine under each mode of operation"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003728_1.3099008-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003728_1.3099008-Figure4-1.png",
        "caption": "Fig 4. Reconstructed sledge chariot of Queen Shub-ad of Ur, 2500 BC (from Cole, 1956).",
        "texts": [
            " Corddriven drilling mechanisms (Fig 1) are depicted in several Egyptian bas-reliefs from around 1450 BC (Davison, 1957/58). Remains were also found in Tutankhamen's tomb, dated 1350 BC (Harrison, 1956). It is likely to be a much older technology. The urge to reduce friction accompanied advances in transportation technology. The sledge was in use around 7000 BC by hunter-fisher societies in northern Europe near the arctic circle. Presumably, this sledge was used in snow and pulled by dogs. Sledges (eg, Fig 4) were used on dry land by people of the Tripolye culture around 4000 BC in what is now southern Russia (Piggott, 1992). The Egyptians used lubricated sledges to transport colossi (Fig 5). A mural drawing from a grotto at El Bersheh (1880 BC) shows 172 men pulling such a sledge, with Downloaded From: http://appliedmechanicsreviews.asmedigitalcollection.asme.org/ on 10/19/2017 Terms of Use: http://www.asme.org/about-asme/terms-of-use Appl Mech Rev vol 51, no 5, May 1998 one man pouring lubricant in front of the runners"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003443_0278364903022001005-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003443_0278364903022001005-Figure5-1.png",
        "caption": "Fig. 5. 3R2Txy 5-DoF n \u2212 xPzR(iRjRkR)N PM.",
        "texts": [
            " By setting the first revolute pair axis of the 3R or 2R spherical subchain perpendicular to its anterior revolute pair axis and assuming the intersection of the two axes, we can obtain a universal joint. Similarly, by setting the first revolute pair axis of the 3R or 2R spherical subchain parallel to its anterior prismatic pair, we can obtain a cylindrical pair. A 3\u2212xPzRzR(iRjR)N PM is shown in Figure 4. Obviously, the mechanism is mechanism-symmetrical. Following the steps in Section 2.4, we can obtain the appropriate input selection. Any two prismatic pairs and one revolute pairs in each 2R subchain can be chosen as an actuated pair. Figure 5 shows another 5-DoF 3 \u2212 xPzR(iRjRkR)N PM. The two mechanisms are not instantaneous. With 2R Spherical Subchain With 3R Spherical Subchain p \u2212 zRzRzR(iRjR)N p \u2212 zRzRxP(iRjR)N p \u2212 zRzR(iRjRkR)N p \u2212 zRxPzR(iRjR)N p \u2212 xPyP(iRjRkR)N p \u2212 xPzRzR(iRjR)N p \u2212 zRxP(iRjRkR)N p \u2212 xPyPzR(iRjR)N p \u2212 xPzR(iRjRkR)N p \u2212 zRxPyP(iRjR)N p \u2212 xPzRyP(iRjR)N The 5-DoF PM with three translational DoFs and two rotational DoFs is denoted by 2R3T. We assume that the two rotational axes are in the XY plane in the initial configuration"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure10.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure10.6-1.png",
        "caption": "FIGURE 10.6. A rigid vehicle in a planar motion.",
        "texts": [
            " Vehicle Planar Dynamics 589 10.2 Rigid Vehicle Newton-Euler Dynamics A rigid vehicle is assumed to act similar to a flat box moving on a horizontal surface. A rigid vehicle has a planar motion with three degrees of freedom that are: translation in the x and y directions, and a rotation about the z-axis. The Newton-Euler equations of motion for a rigid vehicle in the body coordinate frame B, attached to the vehicle at its mass center C are: Fx = mv\u0307x \u2212m\u03c9z vy (10.26) Fy = mv\u0307y +m\u03c9z vx (10.27) Mz = \u03c9\u0307z Iz. (10.28) Proof. Figure 10.6 illustrates a rigid vehicle in a planar motion. A global coordinate frame G is attached to the ground and a local coordinate frame B is attached to the vehicle at the mass center C. The Z and z axes are parallel, and the orientation of the frame B is indicated by the heading angle \u03c8 between the x and X axes. The global position vector of the mass center is denoted by Gd. 590 10. Vehicle Planar Dynamics The velocity vector of the vehicle, expressed in the body frame, is BvC = \u23a1\u23a3 vx vy 0 \u23a4\u23a6 (10.29) where vx is the forward component and vy is the lateral component of v"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000554_978-3-540-85629-0-FigureC.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000554_978-3-540-85629-0-FigureC.3-1.png",
        "caption": "Fig. C.3. Euler angles.",
        "texts": [
            " Note that for \u03b8 = 0 the rotation axis w is arbitrary. C.3 Euler Angles The Euler angles, as well as the Roll-Pitch-Yaw angles, are minimum representations of rotations, in the sense that only three parameters (i.e. three angles \u03d5, \u03b8, \u03c8) are used to express an arbitrary rotation in the 3D space. Commonly, the ZYZ representation is chosen for the Euler angles, that is three consecutive rotations about the z0 (\u03d5), y1 (\u03b8), z2 (\u03c8) axes of the current reference frame (the frame obtained by applying the rotations), see Fig. C.3. The rotation matrix REuler corresponding to these three rotations is defined as REuler(\u03d5, \u03b8, \u03c8) = \u23a1 \u23a3 c\u03c6c\u03b8c\u03c8 \u2212 s\u03c6s\u03c8 \u2212 c\u03c6c\u03b8s\u03c8 \u2212 s\u03c6c\u03c8 c\u03c6s\u03b8 s\u03c6c\u03b8c\u03c8 + c\u03c6s\u03c8 \u2212 s\u03c6c\u03b8s\u03c8 + c\u03c6c\u03c8 s\u03c6s\u03b8 \u2212s\u03b8c\u03c8 s\u03b8s\u03c8 c\u03b8 \u23a4 \u23a6 . (C.8) 492 C Representation of the Orientation For the inverse problem, i.e given a generic rotation matrix R as in eq. (C.5) define the three angles \u03d5, \u03b8, \u03c8, there are two possibilities: 1. r2 13 + r2 23 = 0, then sin \u03b8 = 0. There are two sets of solutions depending on the value assigned to the sign of \u03b8. If 0 < \u03b8 < \u03c0, i"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003443_0278364903022001005-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003443_0278364903022001005-Figure7-1.png",
        "caption": "Fig. 7. 3TIRz 4-DoF 4 \u2212 zxUN uPxzUM PM.",
        "texts": [
            " (24) The limb constraint system reciprocal to eq (24) is $r i1 = (0 0 0 ; 0 1 0), (25) which is a constraint couple in the yi-axis. Thus, all the limb constraint couples are parallel to the XY plane. $i2, $i3 and $i4 can be transformed into revolute pairs by linear combination with $i1 or $i5. Thus, the revolute pairs fall into two groups. The axes of one group are parallel to the base plane. The axes of the other group are perpendicular to the base plane. To prevent the mechanism from being instantaneous, the axes of revolute pairs fixed to the base or the moving platform must be perpendicular to the base, as shown in Figure 7. Note that $r i1 is a couple which is actually perpendicular to plane formed by $i1 and $i5, namely, the xizi plane. Since all the zi-axes are parallel, we only need to make the xi-axes of local systems not parallel to one another. Hence, the revolute axes parallel to the base plane in each limb must not be parallel to one another. The characteristics of constraint and structure of such 3TIRz 4-DoF PMs are shown in Table A3 in the Appendix. An enumeration of such 4-DoF PMs is shown in Table 3, where g/p denotes the number of kinematic pairs in a limb and 2 \u2264 p \u2264 4. Consider a zRxRuPxRzR limb. zRxR can form a universal joint zxUN while xRzR can form a universal joint xzUM. Figure 7 shows a 4 \u2212 zxUN uPxzUM PM. Each zxUN uPxzUM branch exerts a constraint couple on the moving platform. This couple is perpendicular to the first universal joint plane. All four limbs exert four constraint couples on the moving platform in total. Because the four constraint couples are coplanar, only two of them are linearly independent and two redundant constraints exist, i.e., \u03bd = 2. The two independent constraint couples restrict two rotational DoF of the moving platform about the axes parallel to the base plane"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002551_s10853-019-03750-y-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002551_s10853-019-03750-y-Figure5-1.png",
        "caption": "Figure 5 Schematic representation of oxidation and reduction reaction on carbon nanohybrid-modified electrodes.",
        "texts": [
            " Their chemical robustness sets them apart from many other nanomaterials whose stability can be compromised by factors such as temperature, high ionic strength, pH and solvent stability. Subsequently, there has been a sustained interest in the use of carbon nanomaterials for sensing applications. In this section, we highlight some of the key milestones to date and survey the recent advances in the use of CNT and graphene, carbon quantum dots and their nanocomposites for nonenzymatic H2O2 sensing (Table 3). The oxidation and reduction mechanisms of H2O2 on carbon nanohybrid/nanocomposite-modified electrodes are given in Fig. 5. Carbon nanotubes The unique properties of CNT have been exploited to develop a wide range of biosensors. Functionalized CNT and modified electrodes containing CNT have been used to detect several molecules, including glucose, acetaminophen, dopamine, ascorbic acid and various organic vapors. CNT/Nafion-modified electrodes and CNT/Teflonmodified electrodes were employed for the electrocatalysis of H2O2 by Joseph Wang\u2019s group [225, 226]. For both electrodes, the oxidation and reduction peak current significantly occurred at approximately "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure10.15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure10.15-1.png",
        "caption": "Figure 10.15 The positive directions of the section forces that are transferred via normal stresses. N is the normal force, My is the bending moment in the xy plane and Mz is the bending moment in the xz plane.",
        "texts": [
            " 10 Section Forces 397 Note that indices y and z in My and Mz also occur under the integral symbol. This makes the formulas easy to memorise. In addition, y and z reoccur in the indication of the planes in which the bending moments act: My in the xy plane and Mz in the xz plane. The normal force N is positive as a tensile force and negative as a compressive force. The bending moments My and Mz are positive when a tensile stress (\u03c3xx >0) on a small elemental area A for y >0 makes a positive contribution to My or for z > 0 makes a positive contribution to Mz. Figure 10.15 shows the positive directions of N , My and Mz. These are the section forces that are transferred via normal stresses in the member cross-section. The resultant of the shear stresses on a small area A around a point P is a small shear force V , with components Vy and Vz: Vy = \u03c3xy A, Vz = \u03c3xz A. When assuming these small forces act in the member axis (by shifting them to the origin of the yz coordinate system), we have to add a small moment Mt in the cross-sectional plane (see Figure 10.16): Mt = y \u00b7 Vz \u2212 z \u00b7 Vy = (y\u03c3xz \u2212 z\u03c3xy) A"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003908_978-1-84882-730-1-Figure11.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003908_978-1-84882-730-1-Figure11.2-1.png",
        "caption": "Figure 11.2 Feasible initial conditions",
        "texts": [
            "4 Discussion of the Initial Condition 283 vuut 2e .t0/Cm11 Qu2.t0/Cm33 Qr2.t0/C 3X iD1 Q Ti .t0/ 1 i Q i .t0/C =\u0131 < 1 2 : (11.43) On the other hand, the term 3P iD1 Q Ti .t0/ 1 i Q i .t0/C \u0131 can be made arbitrarily small. Noticing that xe.t0/ D x.t0/ xd .s.t0// and ye.t0/ D y.t0/ yd .s.t0//, it can be seen that the initial value, s.t0/, can be adjusted such that (11.43) holds if: 1. the ship heads toward \u201calmost\u201d of the half-plane containing the initial part of the path to be followed, 2. the path satisfies Assumption 11.1, see Figure 11.2, 3. the initial velocities u.t0/; v.t0/ and r.t0/ are not too large, and 4. the design constants are chosen such that k1 is small and k2 is large. The angle \u01310 (see Figure 11.2) should be increased if the initial velocities u.t0/; v.t0/, and r.t0/ are large. Otherwise the ship might cross the edge line of the plane in question, which might result in e D \u02d90:5 . Similarly the last inequality of (11.32) is rewritten as q .x.t0/ xd .s.t0///2C .y.t0/ yd .s.t0///2 p1 X 1 .t0/ 1 k1 \u0131e: (11.44) Again s.t0/ can be adjusted such that (11.44) holds. It is noted that if the abovementioned conditions are not satisfied, one might generate an additional path segment such that it makes the above conditions hold and \u201csmoothly\u201d connects to the path to be followed"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure2.26-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure2.26-1.png",
        "caption": "FIGURE 2.26. A dipping vehicle at a point where the hill has a radius of curvature Rh.",
        "texts": [
            "24 shows the effect of hill radius of curvature RH on the critical speed vc for a car with h/l = 0.10123mm/mm and Figure 2.25 shows the effect of a car\u2019s high factor h/l on the critical speed vc for a circular hill with RH = 100m. 2.8.2 F Vehicles on a Dip Moving on the concave curve of a hill is called dipping. The normal force under the wheels of a dipping vehicle is more than the force on a flat inclined road with the same slope, because of the developed centrifugal 2. Forward Vehicle Dynamics 83 84 2. Forward Vehicle Dynamics force mv2/RH in the z-direction. Figure 2.26 illustrates a dipping vehicle at a point where the hill has a radius of curvature RH . The traction and normal forces under the tires of the vehicle are approximately equal to Fx1 + Fx2 \u2248 1 2 m (a+ g sin\u03c6) (2.306) Fz1 \u2248 1 2 mg \u2219\u00b5 a2 l cos\u03c6+ h l sin\u03c6 \u00b6\u00b8 \u22121 2 ma h l + 1 2 m v2 RH a2 l (2.307) Fz2 \u2248 1 2 mg \u2219\u00b5 a1 l cos\u03c6\u2212 h l sin\u03c6 \u00b6\u00b8 + 1 2 ma h l + 1 2 m v2 RH a1 l (2.308) l = a1 + a2. (2.309) Proof. To develop the equations for the traction and normal forces under the tires of a dipping car, we follow the same procedure as a cresting car"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure3.9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure3.9-1.png",
        "caption": "Fig. 3.9. a Mechanical scheme of anthropomorphic manipulator with six degrees of freedom, b manipulator task",
        "texts": [
            " Hence, we restrict our consideration here to observing just three degrees-of-freedom, the other three degrees-of-freedom (gripper) we consider as immobile (q4, Q5' Q6=const.). We consider the problem of synthesizing the nominal trajectory XO(t) and programmed control. It should be noticed, that in cases where the actuators do not change their subsystem order, the vector of the i-th subsystem consists of two coordinates, Xi = (Qi 4Y. In this case, there is a unique corre spondence between the angles of the three degrees-of-freedom of the anthropo morphic manipulator and the tip coordinates of minimal configuration (point D in Fig. 3.9) in Cartesian coordinates. In this way, it is obvious that the point D can be moved in advance along a desired prescribed trajectory with a unique change of angles. The relationship between the coordinates of the point D(x, y, z) and the angular coordinates can be presented in the form (3.40) In order to calculate the manipulator angle trajectories, when point D in the space is given, let us consider the small increments ofthe manipulator tip motion along a prescribed trajectory L1 S = (L1 x, L1y, L1Z)T",
            "43), the necessary angular velocity 4\u00b0(tl ) is calculated, etc. Thus, one obtains the nominal trajectories for all coordinates of the state vector XO(t), corresponding to the minimal configuration (three angles and three angular velocities are calculated). When the nominal trajectories XO(t) are calculated on the basis of the robot dynamics model, Eq. (3.23), the nominal driving torques pO(t) can be calculated. The motion of the manipulator UMS-1 for a specific task which we are considering is presented in Fig.3.9(b). The manipulator tip should be moved from the point A, defined by XO(O) = (q 1 (0), 41(0), ... , 43(OW =(0.1,0, -0.8,0, 0.1,0), to the point B, defined by XO('o)=( -0.4,0, -0.9,0, 1.9, O)T in the time interval 'Os = '0 = 1.8 s. The manipulator tip should move along a straight line between the points A and B. The tip acceleration should be constant and change its sign once during the movement. Namely, the manipulator tip acceleration should be { amax atip = -a max if 8\u00b0(t)~0.5(80('O)+80(0\u00bb if 8\u00b0(t) >0",
            "2 Mathematical Model of Manipulation Robot Dynamics 71 object, along a prescribed trajectory in space, it is necessary to achieve a particular orientation of the working object during (or at the end of) transfer, i.e. to ensure that some axis of the working object has a defined orientation with respect to the axes of the absolute coordinate system. In order to realize this task three degrees of freedom of the minimal configuration of the manipulator are not sufficient. Hence, for solving this task, we consider the manipulator with six degrees of freedom (Fig. 3.9). Let us consider now a problem ofthe nominal trajectory synthesis XO(t), i.e. a desired motion of the object with a prescribed orientation to the coordinate system axes. Determining the trajectories of the six angles and six angular velocities of the manipulator, when the trajectory and the orientation of the working object is given in space, is not simple. However, if the system is decoupled into two subsystems, so that one subsystem consists of the first three degrees of freedom (manipulator minimal configuration), and the other consists of three degrees of freedom of the gripper, then the problem of nominal trajectory synthesis qO(t) and qO(t) becomes simpler",
            " In this way, the nominal angles qO(t) and nominal velocities qO(t) can be calculated for both functional subsystems (positioning subsystem and orienting subsystem). From the prescribed velocity profile of the manipulator tip (Eq. (3.45)), the nominal, generalized accelerations q\u00b7O(t) for all six degrees of freedom can be calculated. Furthermore, on the basis of the model of the robot mechanism dynamics (Eq. (3.23)) the nominal driving torques pO(t) in robot joints can be calculated. Figures 3.12 and 3.13 present the results of the nominal trajectories synthesis and driving torques for one concrete task for the manipulation system UMS-1 (Fig. 3.9). The working object should move from position A to position B (Fig. 3. 12e). The object position at point A is defined by the manipulator coordinates (or gripper joint) with SA (0) = (0.425, 0.167, 0.571) T, while the gripper centre of gravity coordinates are S~A(0)=(0.455, 0.203, 0.530)T. The position of the object at the point B is given by the manipulator tip coordinates S~(r)=(0.367, 0.365, 0.355)T Em], while the coordinates of the gripper centre of gravity are S~B('t')=(0.391, 0.410, O.4lO)T Em]",
            " Thus, the curve Tmin(R), i.e. minimal execution time depend ing on R is obtained. Let us first consider the limitations (IHII) only. The procedure of reducing R is repeated until limitation m is violated, i.e. until the stresses in segments exceed the permitted values. If further reduction of R is required, T has to be increased. Consequently, minimum time (maximum velocity) appears in both limitations (I) and (II). Figure 3.20 illustrates these results (for the anthropomorphic manipulator in Fig. 3.9), i.e. the curve Tmin(R) and limitations (I) and (II). The minimum appears at the point M 1 designated by a circle. Let us now introduce the limitation of elasticity (III). We impose the condition that the manipulator tip linear deviation due to segment elasticity be 84 3 Dynamics and Dynamic Analysis of Manipulation Robots less than 0.001 m. In this case we consider only the quasi-static deflexion due to nominal dynamics. By introducing limitation (III) the permitted domain is narrowed and the minimum point moves into position M 2 , designated by a square in Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure12.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure12.1-1.png",
        "caption": "FIGURE 12.1. A mass m, spring k, and damper c.",
        "texts": [
            " Because a periodic fluctuations of kinetic energy appears as a periodic motion of a massive body, we call this energy transformation mechanical vibrations. The mechanical element that stores kinetic energy is calledmass, and the mechanical element that stores potential energy, is called spring. If the total value of mechanical energy E = K + V decreases during a vibration, there is a mechanical element that dissipates energy. The dissipative element is called . A mass, spring, and damper are illustrated as symbols in Figure 12.1. The amount of stored kinetic energy in a mass m is proportional to the square of its velocity, v2. The velocity v \u2261 x\u0307 may be a function of position 730 12. Applied Vibrations and time. K = 1 2 mv2 (12.1) The required force fm to move a mass m is proportional to its acceleration a \u2261 x\u0308. fm = ma (12.2) A spring is characterized by its stiffness k. A force fk to generate a deflection in spring is proportional to relative displacement of its ends. The stiffness k may be a function of position and time"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure6.18-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure6.18-1.png",
        "caption": "FIGURE 6.18. Two possible configurations of a slider-crank mechanism having the same input angle \u03b82.",
        "texts": [
            "139) To solve the possible configurations, we start by calculating the coefficients of the quadratic equation (6.133) G = \u22122a cos \u03b82 = \u22121.4142 (6.140) H = a2 + e2 \u2212 b2 \u2212 2ae sin \u03b82 = \u22123.4571. (6.141) Employing Equation (6.136) provides s = \u2212G\u00b1 \u221a G2 \u2212 4H 2 = \u00bd 2.696 \u22121.282 . (6.142) 6. Applied Mechanisms 335 The corresponding coupler angle \u03b83 can be calculated form either Equation (6.137) or (6.138). \u03b83 = sin\u22121 \u00b5 e\u2212 a sin \u03b82 \u2212b \u00b6 = cos\u22121 \u00b5 s\u2212 a cos \u03b82 \u2212b \u00b6 \u2248 \u00bd 3.037 rad \u2248 174 deg 0.103 rad \u2248 5.9 deg (6.143) Figure 6.18 depicts the two possible configurations of the mechanism for \u03b82 = 45deg. Example 232 Velocity analysis of a slider-crank mechanism. The velocity analysis of a slider-crank mechanism is possible by taking a time derivative of Equations (6.128) and (6.129), d dt (a sin \u03b82 \u2212 b sin \u03b83 \u2212 e) = a\u03c92 cos \u03b82 \u2212 b\u03c93 cos \u03b83 = 0 (6.144) d dt (a cos \u03b82 \u2212 b cos \u03b83 \u2212 s) = \u2212a\u03c92 sin \u03b82 + b \u03c93 sin \u03b83 \u2212 s\u0307 = 0 (6.145) where \u03c92 = \u03b8\u03072 \u03c93 = \u03b8\u03073. (6.146) Assuming \u03b82 and \u03c92 are given values, and s, \u03b83 are known from Equations (6"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002208_j.ymssp.2019.05.014-Figure9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002208_j.ymssp.2019.05.014-Figure9-1.png",
        "caption": "Fig. 9. Schematic of different foundation types: (a) different foundation types; (b) foundation parameters of gear pair 3; (c) foundation parameters of gear pair 4.",
        "texts": [
            " This part will focus on verifying the proposed method through calculating the mesh stiffness of spur gears with complex foundation structures and real crack propagation paths. The influence of different foundation types and crack propagation paths on the mesh stiffness is analyzed. Firstly, the mesh stiffness obtained from the proposed method is compared with the results of the potential energy method [40] and finite element method. Then, the mesh stiffness under different foundation types (see Fig. 9) and foundation parameters is calculated and the effects of foundation parameters (including the web thickness and the rim thickness) on the mesh stiffness are analyzed. Finally, considering the complex foundation and real crack propagation paths, the mesh stiffness of spur gears is calculated and the effects of different crack types (including the tooth fracture and rim facture) on the mesh stiffness are analyzed. Furthermore, the mesh stiffness obtained from the proposed method is verified by the three-dimensional finite element method. Different foundation types of spur gears are shown in Fig. 9 and the parameters of different gear pairs are listed in Table 1. In this part, the mesh stiffness obtained from the proposed method is compared with that obtained from the potential energy method [40] and the finite element method. Using the parameters of the gear pair 1 (see Table 1), the mesh stiffness of the three methods is shown in Fig. 10 and the calculation errors of the proposed and potential energy methods at the moments A and B are listed in Table 2 (the moments A and B are the middle moment of the double-tooth contact zone and the single-tooth contact zone, respectively)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002397_j.matdes.2016.09.043-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002397_j.matdes.2016.09.043-Figure1-1.png",
        "caption": "Fig. 1. The flat dog bone parts fabricated by SLM",
        "texts": [
            "02 max In this study, flat dog bone shaped samples are designed and fabricated by SLM processing by PXM Phenix/3D Systems (Toledo, Ohio). This equipment has the maximum laser power of 300 W, and employs a Ytterbium fiber laser with AC C EP U SC R IP T wavelength of 1070 nm. The diameter of the laser beam is approximately 80 \u00b5m and its profile is Gaussian (TEM00). This technique uses controllable amounts of argon gas at the oxygen level of below 500 ppm to avoid oxidation of these parts. The SLM fabricated 316L stainless steel parts are shown in Fig. 1. The surface quality of these parts shows that they are not the best-case in this fabrication, because they are not manufactured with the optimum laser parameters to get the higher density and better surface quality. The gauge length of the parts is 25 mm, gauge width is 6 mm, thickness is 2 mm, and overall length is 70 mm. In order to study the effect of laser parameters on the mechanical properties of the 316L stainless steel parts and to relate them to microstructural modeling, selected laser powers (129, 144, 159, and 189 W) and scanning velocities (1400, 1540, and 1680 mm/s) are used, while the hatch distance of 50 \u00b5m and layer thickness of 30\u00b5m are kept constant"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure2.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure2.6-1.png",
        "caption": "Fig. 2.6. Double-sided machine with one internal slotted stator and buried PMs. 1 \u2014 stator core with slots, 2 \u2014 PM, 3 \u2014 mild steel core (pole), 4 \u2014 nonmagnetic rotor disc.",
        "texts": [
            " There are a number of applications for medium and large power axial flux motors with external PM rotors, especially in electrical vehicles [101, 2.1 Magnetic Circuits 33 276]. Disc-type motors with external rotors have a particular advantage in low speed high torque applications, such as buses and shuttles, due to their large radius for torque production. For small electric cars the electric motor mounted directly into the wheel is recommended [101]. The ring-type stator core can also be made with slots (Fig. 2.6). For this type of motor, slots are progressively notched into the steel tape as it is passed from one mandrel to another and the polyphase winding is inserted [276]. In the case of the slotted stator the air gap is small (g \u2264 1 mm) and the air gap magnetic flux density can increase to 0.85 T [101]. The magnet volume is less than 50% that of the previous design, shown in Figs 2.4 and 2.5. AFPM machines with coreless stators have the stator winding wound on a non-magnetic and non-conductive supporting structure or mould",
            " Single-sided AFPM machines with stator ferromagnetic cores have a single PM rotor disc opposite to a single stator unit consisting of a polyphase winding and ferromagnetic core (Fig. 2.1). The stator ferromagnetic cores can be slotted or slotless. The stator winding is always made of flat wound coils (Fig. 2.8). The PMs can be mounted on the surface of the rotor or embedded (buried) in the rotor disc. In the case of a slotless stator the magnets are almost always surface mounted, while in the case of a slotted stator with a small air gap between the rotor and stator core, the magnets can be either surface mounted on the disc (Fig. 2.1) or buried in the rotor disc (Fig. 2.6). Large axial magnetic forces on bearings are the main drawback of single-sided AFPM machines with ferromagnetic stator cores. In double-sided AFPM machines with ideal mechanical and magnetic symmetry, the axial magnetic forces are balanced. Double-sided AFPM machines with stator ferromagnetic cores have either a single PM rotor disc with ironcored stators on both sides of the discs (Figs 1.4c and 2.3) or outer PM rotor discs with iron-cored stator fixed in the middle (Figs 2.4, 2.5 and 2.6). As with single-sided AFPM machines the stator ferromagnetic cores can be slotted or slotless, and the rotor magnets can be surface mounted, embedded or buried 124 4 AFPM Machines With Iron Cores (Fig. 2.6). Again, in the case of a slotless stator with a large air gap between the rotor and stator core the magnets are almost always surface mounted. The stator windings of double-sided AFPM machines can be flat wound (slotted or slotless) as shown in Fig. 2.8 or toroidally wound (normally slotless) as shown in Fig. 2.9. An example of a commercial double-sided AFPM servo motor with ferromagnetic core is shown in Figs. 4.1 and 4.2. External stators have slotted ring-shaped cores made of nonoriented electrotechnical steel ribbon"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure13.46-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure13.46-1.png",
        "caption": "Figure 13.46 Three-hinged shored frame.",
        "texts": [
            " VOLUME 1: EQUILIBRIUM To be able to draw the M , V and N diagrams for compound and associated structures, it is first necessary to determine the support reactions and the joining forces between the compound sections. Subsequently, the M , V and N diagrams can be determined and drawn for the constituent parts, in the same way as for the self-contained structures in Section 13.1. By then adding together the M , V and N diagrams of the constituent parts we can determine the requested M , V and N diagrams for the entire structure. The three-hinged shored frame ASB in Figure 13.46 is loaded over CDSEG by a uniformly distributed load of 16 kN/m. The dimensions are shown in the figure. Questions: a. Determine the support reactions at A and B. b. Determine the forces in shores DD\u2032 and EE\u2032, with the correct signs for tension and compression. c. Isolate CDSEG, and draw all the forces acting on it. d. For CDSEG, draw the N , V and M diagrams with the deformation symbols and the tangents at C, D, S, E and G to the M diagram. Include relevant values. Solution (units kN and m): a"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.7-1.png",
        "caption": "FIGURE 7.7. Steered configuration of a trapezoidal steering mechanism.",
        "texts": [
            " Steering Dynamics 385 In this example the width of the car wv and the track w are assumed to be equal. The width of vehicles are always greater than their track. wv > w (7.22) Example 260 Trapezoidal steering mechanism. Figure 7.6 illustrates a symmetric four-bar linkage, called a trapezoidal steering mechanism, that has been used for more than 100 years. The mechanism has two characteristic parameters: angle \u03b2 and offset arm length d. A steered position of the trapezoidal mechanism is shown in Figure 7.7 to illustrate the inner and outer steer angles \u03b4i and \u03b4o. The relationship between the inner and outer steer angles of a trapezoidal steering mechanism is sin (\u03b2 + \u03b4i) + sin (\u03b2 \u2212 \u03b4o) = w d + r\u00b3w d \u2212 2 sin\u03b2 \u00b42 \u2212 (cos (\u03b2 \u2212 \u03b4o)\u2212 cos (\u03b2 + \u03b4i)) 2. (7.23) To prove this equation, we examine Figure 7.8. In the triangle 4ABC we can write (w \u2212 2d sin\u03b2)2 = (w \u2212 d sin (\u03b2 + \u03b4i)\u2212 d sin (\u03b2 \u2212 \u03b4o)) 2 +(d cos (\u03b2 \u2212 \u03b4o)\u2212 d cos (\u03b2 + \u03b4i)) 2 (7.24) and derive Equation (7.23) with some manipulation. The functionality of a trapezoidal steering mechanism, compared to the associated Ackerman condition, is shown in Figure 7"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000538_acs.chemrev.8b00172-Figure41-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000538_acs.chemrev.8b00172-Figure41-1.png",
        "caption": "Figure 41. (1) Design of the customized 3-D holder and detail of the magnetic switch. On the left side, the holder design integrating a silicon sensor. A, the base of the main body of the holder; B, the cover of the main body of the holder; C and D furnish the magnetic switch. On the right side, details of the magnetic switch mechanism. (2) Amperometric immunosensor configuration for detection of myeloperoxidase (MPO), a marker for identification of patients at risk of cardiac events. Reproduced from ref 426. Copyright 2013 American Chemical Society.",
        "texts": [
            " The proposed iMEX features an overall low device cost (<$50), affordability, scalability, and comprehensive exosome analyses which is highly promising for deepening insights into tumor biology and accelerating effective cancer management in clinical applications. Barallat et al. developed a novel magnetic switch integrated with electrodes and electrical connectors for amperometric detection of myeloperoxidase (MPO), a marker for diagnosis and stratification of acute coronary syndrome, in human plasma (Figure 41).426 The vertical displacement of a permanent magnet under the transducer assists with operation of the magnetic field (on vs off) which assists in confining the magnetic nanoparticle\u2212analyte complex over the WE and facilitating electrode washing and regeneration. Finally, Torrente-Rodrig\u0301uez and colleagues have developed an amperometric magnetic biosensor for the simultaneous determination of two biomarkers associated with salivary oral cancer, protein IL-8 and its mRNA (IL-8 mRNA) in undiluted human saliva samples,427 which is a good example for typical application of magnetic beads (MBs) in electrochemical DOI: 10"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001692_1.3659004-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001692_1.3659004-Figure8-1.png",
        "caption": "Fig. 8 B o u n d a r y - l a y e r p ro f i l es o b t a i n e d a t p o s i t i o n s e v e r y 3 0 d e g a r o u n d t h e c y l i n d e r f o r a = 1 . T h e d o t t e d l i n e r e p r e s e n t s t h e a p p r o x i m a t e l i m i t fo r t h e t h r e e b o u n d a r y l a y e r s . T h e v e r t i c a l s c a l e o n t h e pro f i l es is t e n t i m e s t h a t f o r t h e c y l i n d e r . Sca les a r e v/toro v e r s u s y / r o .",
        "texts": [
            " The previously described boundary-layer origin is the most logical position at which to attempt to start a boundary-layer solution for a geometry where the wall is in a nonuniform motion with respect to the external flow. As the rotational speed is further increased (beyond a value corresponding to a ~ 0.5) the boundary-layer origin moves further back on the upper surface with a consequent increased length of flow attachment producing more Magnus lift. It is believed that both boundary layers are in the fully developed turbulent state at a velocity ratio of 1.0. An idea of the flow pattern may be obtained by referring to Fig. 8. Although the relative velocity between the surface and free stream is much 4 6 6 / S E P T E M B E R 1 9 6 1 Transactions of the AS ME Downloaded From: http://fluidsengineering.asmedigitalcollection.asme.org/ on 01/30/2016 Terms of Use: http://www.asme.org/about-asme/terms-of-use smaller in the upper boundary layer than in the lower boundary layer, the velocity gradients appear to be about the same at equal distances from the boundary-layer origin. The main indication that both the layers are fully developed turbulent at and beyond this point lies in the fact that the drag in this region of a is a minimum corresponding to the fully developed turbulent boundary layer over the nonrotating cylinder at R ~ 3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-FigureD.4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-FigureD.4-1.png",
        "caption": "FIGURE D.4 Partial potential coefficient for filament and sheet.",
        "texts": [
            " The solution of the integrals is based on the assumption that the length \ud835\udcc1 of the wire is longer than the diameter d = 2a. The result for the tube conductor is Pp11 = 1 4 \ud835\udf160\ud835\udcc1 [( k2 480 + k4 1280 + 1 3600 ) \ud835\udf0b 3 + ( 1 18 \u2212 k2 24 ) \ud835\udf0b + ( 2 \u2212 2 log(\ud835\udcc1) + 6 log(2) + 2 log(a) \u2212 4 log(k\ud835\udf0b) ) 1 \ud835\udf0b + 8a \ud835\udcc1\ud835\udf0b2 ] , (D.17) where k = d\u2215\ud835\udcc1 and \ud835\udcc1 = xe \u2212 xs. Hence, its evaluation is not difficult. D.1.5 Pp12 for a Sheet and a Filament Another important building block for the partial potential coefficient conductor is a filament and a uniformly charged conducting sheet or cell as shown in Fig. D.4. Of course, the wire 414 COMPUTATION OF PARTIAL COEFFICIENTS OF POTENTIAL may only be part of another surface. The integration for the partial potential coefficients is given by Pp12 = 1 1\ud835\udcc12 1 4\ud835\udf0b\ud835\udf16 \u222b ye1 ys1 \u222b xe1 xs1 \u222b xe2 xs2 1 R1,2 dx2 dx1 dy1. (D.18) The analytical solution for the above formula is Pp12 = 1 4\ud835\udf0b\ud835\udf16 1 1 (xe2 \u2212 xs2) 4\u2211 k=1 2\u2211 m=1 (\u22121)m+k [ a2 k \u2212 Z2 2 log(bm + rkm + \ud835\udf16 ) + ak bm log(ak + rkm + \ud835\udf16 ) \u2212 ak Z tan\u22121 ( ak bm Z rkm ) \u2212 bm 2 rkm ] (D.19) with a1 = xs2 \u2212 xe1, a2 = xe2 \u2212 xe1 (D"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure7.7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure7.7-1.png",
        "caption": "FIGURE 7.7 PEEC model for quad with four nodes in Fig. 7.5.",
        "texts": [
            "21) and by canceling the appropriate terms, we get for the normalized coefficients of potential Ppaa\u2032 = 1 \ud835\udf160\u222ba\u222bb \u222ba\u2032\u222bb\u2032 g(r, r\u2032) da\u2032 db\u2032 da db, (7.22) where the global coordinates r and r\u2032 are evaluated on the cell related to the cell of interest. We also want to give an example for the evaluation of a Pp. In Fig. 7.5, we chose the capacitive quarter cells, one is the crosshatched attached to corner 1 and to the quarter cell SPECIFICATION OF NONORTHOGONAL PARTIAL ELEMENTS 167 attached at corner 2. Pp(1, 2) = 1 \ud835\udf160 \u222b 0 a=\u22121 \u222b 1 b=0 \u222b 1 a\u2032=0 \u222b 0 b\u2032=\u22121 g(r, r\u2032) da\u2032 db\u2032 da db. (7.23) The PEEC circuit for a quadrilateral element is shown in Fig. 7.7. It consists of basic KVL loops. Specifically, a KVL loop involves two nodes with a partial inductance and a resistance in series that is closed by the capacitances to infinity. As is evident from Fig. 7.5, the part of the model represented by Fig. 7.7 includes the inductive half cells and the capacitive quarter cells. Of course, we can simplify neighboring quads with some partial elements that are connected to the same nodes in parallel. Specifically, the two inductive half cells connected to \ud835\udefc and \ud835\udefd in Fig. 7.6 could be combined. Again, the PEEC circuit topology for the orthogonal and the nonorthogonal cases are the same, with the exception of the circuit element values and the additional partial mutuals. Importantly, the same modified nodal analysis (MNA) circuit solver can be used for all cases",
            " Show that points (x, y, z) on the quadrilateral element can be converted into a point in the local a, b by the formulation in Section 7.1 to a point on the surfaces of the quadrilateral shape. We note that the quadrilateral surface is bounded by the cornerpoints points 0 to 3. Show that the points (a, b) in the \u22121, +1 range correspond to global coordinate points (x, y, z) by using equation (7.3). 7.2 Continuity equation Explain the function of the continuity equation in relation to the equivalent circuit in Fig. 7.7. Show that equation (7.27) is based on the continuity equation for the nonorthogonal geometry shown. 7.3 Resistance for nonorthogonal geometries Prove equation (7.14) for the resistance of the nonorthogonal geometries. Also, we don\u2019t consider possible resistive couplings due to the nonorthogonal overlap of resistive cells. Find a formula for the resistive coupling between overlapping cells, which is an extension of (7.14). An example for resistive coupling in Fig. 7.3 for two overlapping inductance half cells which are connected between nodes 0 and 2 and the other one is connected between nodes 0 and 1"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure9.33-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure9.33-1.png",
        "caption": "Fig. 9.33. DuraHeartR centrifugal rotary pump with axial flux PM brushless motor and magnetic levitation bearings: 1 - impeller, 2 - PM, 3 - stator winding, 4 - stator core, 5 - magnetic suspension, 6 - ceramic pivot, 7 - inlet, 8 - outlet.",
        "texts": [
            " Electromagnetic 1G pumps were driven by electromagnets, linear oscillating motors or linear short-stroke actuators. The pump, integrated with a linear actuator, is heavy, large and noisy. DuraHeartR 3G LVAD developed by Terumo Corporation, Tokyo, Japan, combines centrifugal pump with magnetic levitation technologies (Figs 9.33 and 9.34) [102]. Magnetic levitation allows the impeller to be suspended within the blood chamber by electromagnets and position sensors. The three-phase, 8-pole, axial flux PM brushless motor with slotless stator resembles a floppy disc drive spindle motor (Fig. 9.33). NdFeB PMs are integrated with the impeller. The output power of the motor is 4.5 W, speed 2000 rpm and torque 0.0215 Nm [226]. 9.11 Ventricular Assist Devices 315 An axial flux slotless motor integrated with a centrifugal blood pump is shown in Figs. 9.35 and 9.36. The so-called VentrAssistTM manufactured now by Australian company Ventracor is a new cardiac LVAD, which has only one moving part - a hydrodynamically suspended impeller integrated with a PM rotor. The hydrodynamic forces act on tapered edges of the four blades"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure13.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure13.6-1.png",
        "caption": "FIGURE 13.6. A chain pendulum.",
        "texts": [
            "75) The kinetic and potential energies make the following Lagrangean: L = K \u2212 V = 1 2 m1l 2 1 \u03b8\u0307 2 1 + 1 2 m2 \u00b3 l21 \u03b8\u0307 2 1 + l22 \u03b8\u0307 2 2 + 2l1l2\u03b8\u03071\u03b8\u03072 cos (\u03b81 \u2212 \u03b82) \u00b4 +m1gl1 cos \u03b81 +m2g (l1 cos \u03b81 + l2 cos \u03b82) (13.76) Employing Lagrange method (13.2) we find the following equations of motion: d dt \u00b5 \u2202L \u2202\u03b8\u03071 \u00b6 \u2212 \u2202L \u2202\u03b81 = (m1 +m2) l 2 1 \u03b8\u03081 +m2l1l2\u03b8\u03082 cos (\u03b81 \u2212 \u03b82) \u2212m2l1l2\u03b8\u0307 2 2 sin (\u03b81 \u2212 \u03b82) + (m1 +m2) l1g sin \u03b81 = 0 (13.77) 836 13. Vehicle Vibrations d dt \u00b5 \u2202L \u2202\u03b8\u03072 \u00b6 \u2212 \u2202L \u2202\u03b82 = m2l 2 2 \u03b8\u03082 +m2l1l2\u03b8\u03081 cos (\u03b81 \u2212 \u03b82) +m2l1l2\u03b8\u0307 2 1 sin (\u03b81 \u2212 \u03b82) +m2l2g sin \u03b82 = 0 (13.78) Example 478 F Chain pendulum. Consider an n-chain-pendulum as shown in Figure 13.6. Each pendulum has a massless length li with a concentrated point mass mi, and a generalized angular coordinate \u03b8i measured from the vertical direction. The xi and yi components of the mass mi are xi = iX j=1 lj sin \u03b8j (13.79) yi = \u2212 iX j=1 lj cos \u03b8j . (13.80) We find their time derivatives x\u0307i = iX j=1 lj \u03b8\u0307j cos \u03b8j (13.81) y\u0307i = iX j=1 lj \u03b8\u0307j sin \u03b8j (13.82) 13. Vehicle Vibrations 837 and the square of x\u0307i and y\u0307i x\u03072i = \u239b\u239d iX j=1 lj \u03b8\u0307j cos \u03b8j \u239e\u23a0\u00c3 iX k=1 lk\u03b8\u0307k cos \u03b8k ! = nX j=1 nX k=1 ljlk\u03b8\u0307j \u03b8\u0307k cos \u03b8j cos \u03b8k (13"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure8.51-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure8.51-1.png",
        "caption": "Fig. 8.51. Robotized painting booth in car factory",
        "texts": [
            "49, which illustrates guiding the real robot tip through the cycle, but with heavier robots and for easier operation a pilot robot is applied with the same dimensions as the original robot and of (Fig. 8.50) a much lighter structure with potentiometers in the joints, the signals of which are recorded and memorized as described. Fig. 8.48. \"TRALLFA\" spray-painting robot (NILS-UNDERHAUG - Norway) Fig. 8.49. Direct \"teaching\" of robot Fig. 8.50. Pilot-teaching robot 8.3 Examples of Industrial Robot Applications 259 How complex the installation of a robotized painting booth is, is presented in the simplified sketch in Fig. 8.51. The main components of the installation are: 1, workpiece-car body; 2 and 3, spray-painting robots; 4, supply duct of conditioned air to the cabin; 5, lights; 6, moisture separators; 7, water collector; 8, floor; 9, waste paint collector; 10, servicing opening; 11, service gangway gallery; 12, exhaust tubes; 13, fan; 14, inlet air heater; 15, flow damper; 16, inlet port; 17, conditioning cabin. Workpieces 1 (car bodies) move through the booth with a constant speed and robots 2, 3 perform painting during motion"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000448_s11740-009-0192-y-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000448_s11740-009-0192-y-Figure5-1.png",
        "caption": "Fig. 5 Temperatures and stresses (von Mises) during the solidification of the 1st layer",
        "texts": [
            " This loadcase is comparable to a scanning vector with a length of 10.0 mm, which is solidified by a scanning speed of 500 mm/s. Afterwards, the whole structure is cooled down for 4.0 s with the result of a nearly homogeneous temperature field. Through the approach of merging 20 real layers of 50 lm to one single layer of 1.0 mm in the simulation, this procedure of heating and cooling has to be calculated only once, so that the calculation time is considerably reduced. As an alternative, the model allows an even more detailed layer size of 50 lm as well. Figure 5 displays the transient temperature field during the solidification of the cantilevers first layer. For this time step, the actual condition of the first layer is liquid and thus strainless. Thermal strains within the upper surface of the structural panel lead to a bending of the vertical externals of 0.32 mm into the negative z-direction combined with equivalent stresses of 253 N/mm2. After the entire cooling to ambient temperature of 100 C after 4.0 s, the structure shows equivalent stresses of 280 N/mm2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure13.41-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure13.41-1.png",
        "caption": "Figure 13.41 (a) The snow load of 5 kN/m on the lean-to, resolved into components (b) normal to the roof plane and (c) parallel to the roof plane.",
        "texts": [
            " Snow load Over a length a measured horizontally, the resultant of the snow load is aqsn (see Figure 13.40). The components of this force normal to and parallel to the axis are respectively aqsn cos \u03b1 and aqsn sin \u03b1. They act on a member segment with length a/cos \u03b1. For the components of the distributed load normal to and parallel to the beam axis we now find qsn;tr = aqsn cos \u03b1 a/ cos\u03b1 = qsn cos2 \u03b1, = (5 kN/m) cos2 21.8\u25e6 = 4.310 kN/m, qsn;pa = aqsn sin \u03b1 a/ cos\u03b1 = qsn sin \u03b1 cos \u03b1, = (5 kN/m) sin 21.8\u25e6 cos 21.8\u25e6 = 1.724 kN/m. The distributed load qsn = 5 kN/m due to the snow (Figure 13.41a) has components of 4.310 kN/m normal to the beam axis (Figure 13.41b) and 1.724 kN/m parallel to the beam axis (Figure 13.41c). The bending moment in ACD is caused by the load of 4.310 kN/m normal to the beam axis. The M diagram is equal to that in Figure 13.33d, but now 4.310 times as large, so that the extreme values of the bending moment are: Msn;max = (2.558 kNm) \u00d7 4.310 = 11.02 kNm ( ), Msn;min = (2.320 kNm) \u00d7 4.310 = 10 kNm ( ). 582 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM The N diagram in Figure 13.42a due to the load of 4.310 kN/m normal to the beam axis is equal to the N diagram in Figure 13.33b, but with values that are 4"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000138_6.2007-6461-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000138_6.2007-6461-Figure3-1.png",
        "caption": "Figure 3. STARMAC II vehicle and its components.",
        "texts": [
            " The second and third requirements drove the selection of the sensor suite to be included on board the vehicles, which in turn drove the payload requirements. The fourth requirement drove the need for a broadband communication device and the last requirement resulted in the inclusion of significant computational power. Ultimately, the key tradeoff was between keeping the vehicle small enough to be flown without too many special precautions for ensuring safety, and making it large enough to be able to support the necessary payload to achieve the tasks required for the many applications envisaged for it. The vehicle frame, visible in Figure 3 was designed to be as light as possible, while maintaining sufficient stiffness to ensure accurate state measurement and control actuation. Constructed of carbon fiber tubes, honeycomb platforms, plastic fasteners and Delran motor mounts, the entire frame weighs approximately 150g, or less than 15% of the total mass of the vehicle in its lightest configuration. In order to minimize vibrational effects on inertial measurement and to ensure consistent thrust actuation, crossbraces were added which significantly improved vertical stiffness of the motor mount location, as well as torsional rigidity of the core"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure9.57-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure9.57-1.png",
        "caption": "Figure 9.57 The order in which the forces in a force polygon are plotted does not influence the result (vector addition is associative and commutative).",
        "texts": [
            " To do so, we first have to draw the forces found as they act on joint A, see Figure 9.56c. Only then we can see that FA;1 is a compressive force, and FA;2 is a tensile force, so that N(1) = \u2212FA;1 = \u22128F \u221a 2, N(2) = +FA;2 = +4F \u221a 5. Note that the forces in the force polygon have not been denoted as N . The force polygon provides information only on the magnitude of the member forces, and not on the sign for tension or compression. The order in which one writes down the forces in a force polygon does not influence the result (vector addition is associative and commutative). Figure 9.57 shows two equivalent force polygons. The first force polygon is created by ranking the various forces acting on joint A in an order that is associated with an anti-clockwise rotation about joint A: 4F \u21d2FA;2 \u21d2FA;1. 9 Trusses 355 The second force polygon arises from ranking the forces in a clockwise order, so that 4F \u21d2 FA;1 \u21d2 FA;2. In Figure 9.58, the order (a) to (h) shows how, per joint, we can consecutively calculate two member forces (and then the support reactions in G and H). The members for which the forces are known are shown in bold"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure2.13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure2.13-1.png",
        "caption": "Fig. 2.13. Location of the armature current Ia in d\u2014q coordinate system.",
        "texts": [
            " V1 = Ef + IaR1 + jIadXsd + jIaqXsq = Ef + Iad(R1 + jXsd) + Iaq(R1 + jXsq) (2.82) where 54 2 Principles of AFPM Machines Ia = Iad + Iaq (2.83) and Iad = Ia sin\u03a8 Iaq = Ia cos\u03a8 (2.84) The angle \u03a8 is between the q-axis and armature current Ia. When the current arrows are in the opposite direction, the phasors Ia, Iad, and Iaq are reversed by 180o. The same applies to the voltage drops. The location of the armature current Ia with respect to the d- and q-axis for generator and motor mode is shown in Fig. 2.13. Phasor diagrams for synchronous generators are constructed using the generator arrow system. The same system can be used for motors, however, the consumer arrow system is more convenient. An overexcited generator (Fig. 2.14a) delivers both active and reactive power to the load or utility grid. An underexcited motor (Fig. 2.14b) draws both active and reactive power from the line. For example, the load current Ia (Fig. 2.14b ) lags the voltage phasor V1 by the angle \u03c6. An overexcited motor, consequently, draws a leading current from the circuit and delivers reactive power to it"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure7.17-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure7.17-1.png",
        "caption": "Fig. 7.17. Element of a tactile sensor",
        "texts": [
            " current is applied to the primary coils, and the output is measured after the multiplexer. In this experiment only two levels of the output voltage were used, so the binary picture was produced. The tactile picture was scanned at the rate of 1 kHz. Circuit board The magnetoelastic material that was used showed promising results when the temperature sensitivity (a common problem with resistive elastomers), dynamic range, sensitivity, linearity and small hysteresis were considered. An interesting tactile sensor (Figure 7.17) with an electro-optical force transduction was developed in the Lord Company, USA [8]. The tactile matrix has 8 x 8 taxels. The forming of the tactile picture has two phases: first, the force acting on a taxel is converted into the deformation (the displacement of the elastic element), and then the displacement is measured by the optical transducer and converted into the electrical signal. The other phase produces some difficulties because the mechanical and electrical properties of the emitter detector pairs is not uniform over the sensing array, so it has to be compensated through the use of appropriate hardware and software"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000811_j.jsv.2012.05.039-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000811_j.jsv.2012.05.039-Figure1-1.png",
        "caption": "Fig. 1. Three transfer paths.",
        "texts": [
            " We have to consider the effect of vibration transfer paths, because the meshing location of sun\u2013planet or planet\u2013ring gear pairs are time-varying, with respect to the fixed sensor. In most cases, the ring gear of a planetary gearbox is stationary, and sensors are mounted on the gearbox casing which is connected to or fastened to the ring gear directly. We consider only this case in this paper. The damage induced vibration signal at the meshing location has three possible paths to go from its origin to the senor through solid mechanical components and their contacts, as shown in Fig. 1. Suppose the damage occurs on a planet gear. Through the first path, the damage induced vibration signal propagates from its origin to the ring gear firstly, then from the ring gear to the gearbox casing, and finally to the sensor. Whereas through the second path, the vibration signal follows a longer path, from its origin to the sun gear firstly, then from the sun gear to the shaft connected to the sun gear, next from the sun gear shaft to the supporting bearing, further from the bearing to the gearbox casing, and finally reaches the sensor"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001478_j.conengprac.2011.04.005-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001478_j.conengprac.2011.04.005-Figure3-1.png",
        "caption": "Fig. 3. Free body diagram of a quadrotor helicopter.",
        "texts": [
            " First, the vehicle\u2019s full nonlinear dynamics are presented. Then, the vehicle\u2019s input forces and moments are computed for non-zero free-stream velocities based on techniques from helicopter analysis. These inputs are used in the development of vehicle controllers in Section 5. The derivation of the nonlinear dynamics is performed in North\u2013East\u2013Down (NED) inertial and body-fixed coordinates. Let {eN,eE,eD} denote unit vectors along the respective inertial axes, and {xB, yB, zB} denote unit vectors along the respective body axes, as shown in Fig. 3. Euler angles to rotate from NED axes to body-fixed axes are the 3\u20132\u20131 sequence fc,y,fg, referred to as yaw, pitch, and roll, respectively. The current velocity direction unit vector is ev, in inertial coordinates. The direction of the projection of ev onto the xB\u2013yB plane defines the direction of elon in the body-fixed longitudinal, lateral, vertical frame, {elon, elat, ever}, as shown in Fig. 8. Due to blade flapping, the rotor plane does not necessarily align with the xB, yB plane, so for the jth rotor let fxRj ,yRj ,zRj g denote unit vectors aligned with the plane of the rotor and oriented with respect to the {elon, elat, ever} frame",
            ", and let xB be defined as the angular velocity of the aircraft in the body frame. The rotors, numbered 1\u20134, are mounted outboard on the xB-, yB-, xB and yB-axes, respectively, with position vectors rj with respect to the c.g. The thrust Tj produced by the jth rotor acts perpendicularly to the rotor plane along the zRj -axis, as defined in Fig. 4. The vehicle body drag force is Dbpv2 1, vehicle mass is m, acceleration due to gravity is g, and the inertia matrix is IBAR3 3. A free body diagram is depicted in Fig. 3, with is a depiction of the rotor forces and moments in Fig. 4. The total force, F, can be summed as F\u00bc Dbev\u00femgeD\u00fe X4 j \u00bc 1 \u00f0 TjRRj ,IzRj \u00de \u00f01\u00de where RRj ,I is the rotation matrix from the plane of rotor j to inertial coordinates. Similarly, the total moment, M, is M\u00bc X4 j \u00bc 1 \u00f0Mj\u00feMbf ,j\u00ferj \u00f0 TjRRj ,BzRj \u00de\u00de \u00f02\u00de where RRj ,B is the rotation matrix from the plane of rotor j to body coordinates. Note that the drag force was neglected in computing the moment. This force was found to cause a negligible disturbance on the total moment over the flight regime of interest, relative to blade flapping torques"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure12.21-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure12.21-1.png",
        "caption": "Figure 12.21 Joint E and member ED isolated.",
        "texts": [
            " As there are no distributed loads, the bending moment varies linearly along all members, and the shear force and normal force are constant in all members (rule 1). The M diagram is shown in Figure 12.19b. Since the variation of M along all members is linear, it is sufficient to determine the bending moments at the member ends to get the M diagram. The bending moments M (CA) C , M (CB) C and M (CE) C at joint C have already been calculated1 (see Figure 12.20). M (ED) E follows from the moment equilibrium of the isolated part ED (see Figure 12.21): M (CE) E = M (ED) E with tension on the upper side of ED. This value in the M diagram is therefore plotted at the upper side. The moment equilibrium of joint E in Figure 12.21 gives M (CE) E = M (ED) E . The bending moment \u201cgoes round the corner\u201d. This is emphasised in the M diagram in Figure 12.19b by means of a dotted arc at joint E. 1 The upper index refers to the member in which the bending moment acts and the lower index refers to the location. 12 Bending Moment, Shear Force and Normal Force Diagrams 501 The magnitude of the shear force and the associated deformation symbol follow from the slope of the M diagram.1 The V diagram is shown in Figure 12.19c. The N diagram is shown in Figure 12"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001735_tmech.2014.2329945-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001735_tmech.2014.2329945-Figure2-1.png",
        "caption": "Fig. 2. Quadrotor configuration with a free body diagram.",
        "texts": [
            " An accurate mathematical dynamic model of quadrotor, which includes model uncertainties, is essential for designing a nonlinear controller. In the following section, we will describe the dynamic model of quadrotor and introduce experimental platform of quadrotor. The dynamic model of quadrotor is derived using the EulerLagrangian approach. The full dynamics of attitude and position of the quadrotor is basically those of a rotating rigid body with six degrees of freedom. The structure of quadrotor including the free body diagram and coordinates system is shown in Fig. 2. Let q = (x, y, z, \u03c6, \u03b8, \u03c8) \u2208 R6 be the generalized coordinates where \u03be = (x, y, z) \u2208 R3 denotes the absolute position of the mass of the quadrotor relative to a fixed inertial frame. Euler angles, which are the orientation of the quadrotor, are expressed by \u03b7 = (\u03c6, \u03b8, \u03c8) \u2208 R3 , where \u03c6 is the roll angle around the x axis, \u03b8 is the pitch angle around the y axis, and \u03c8 is the yaw angle around the z axis. It is assumed that the Euler angles are bounded as follows: roll : \u2212\u03c0 2 \u2264 \u03c6 \u2264 \u03c0 2 pitch : \u2212\u03c0 2 \u2264 \u03b8 \u2264 \u03c0 2 yaw : \u2212\u03c0 \u2264 \u03c8 < \u03c0"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002159_0278364919894385-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002159_0278364919894385-Figure2-1.png",
        "caption": "Fig. 2. A typical walking gait that is used for filter comparisons. The Cassie bipedal robot is simulated using Simscape Multibody\u2122.",
        "texts": [
            " The discrete, simulated measurements were corrupted by additive white Gaussian noise, which are specified in Table 1 along with the initial state covariance values. The same values were used in both simulation and experimental convergence evaluations of the filters. The IMU bias estimation was turned off for these simulations. The simulation was performed with MATLAB and Simulink (Simscape Multibody\u2122) where the simulation environment models ground contact forces with a linear force law (having a stiffness and damping term) and a Coulomb friction model. A typical walking gait is shown in Figure 2. To compare the convergence properties of the two filters, 100 simulations of each filter were performed using identical measurements, noise statistics, initial covariance, and various random initial orientations and velocities. The initial Euler angle estimates were sampled uniformly from \u221230\u25e6 to 30\u25e6. The initial velocity estimates were sampled uniformly from \u22121.0 m/s to 1.0 m/s. The pitch and roll estimates as well as the (body frame) velocity estimates for both filters are shown in Figure 3. Although both filters converge for this set of initial conditions, the proposed RIEKF converges considerably faster than the standard quaternion-based EKF"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001059_chem.200900421-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001059_chem.200900421-Figure1-1.png",
        "caption": "Figure 1. The electrochemistry of carbon nanotubes resembles that of graphite. A) Schematic representation of a crystal of highly ordered pyrolytic graphite where the layers of graphite have an interlayer spacing of 3.35 . This corresponds to a single MWCNT in which edge-plane-like sites are shown at the end of the tube and along the tube axis. These similarities result in identical electrochemical responses for CNT-modified electrodes when compared with the edge-plane pyrolytic graphite electrode. B) Controllable introduction of defects to MWCNT walls for enhanced heterogeneous electron transfer. Reprinted with permission from references [32, 14], respectively.",
        "texts": [
            " Usually, however, no reliable explanation is given. Here, we wish to describe our understanding of the current \u201cstate of the art\u201d of the electrochemistry of carbon nanotubes. We also wish to describe and discuss important applications of CNTs to sensing, biosensing, and energy-storage devices. Electrochemistry on edges and walls of carbon nanotubes : The first indication of the factors explaining the electrochemistry of carbon nanotubes is their structure. CNTs are concentric cylinders of seamlessly closed rolled-up graphene sheets (see Figure 1). Depending on the number of such concentric rolled sheets cylinders of which they consist, they can be divided into two general categories: single-walled (SWCNT\u2014one rolled graphene sheet) or multiple-walled (MWCNT\u2014two or more rolled graphene sheets). MWCNTs with a small number of walls are generally called doublewalled (DWCNT\u2014two concentric tubes) or few-walled (FWCNT\u2014typically a mixture of 2-, 3-, or 4-walled CNTs). The interlayer distance in MWCNTs approximately 3.3 , which is close to the distance between the graphene layers in graphite. SWCNTs have a typical diameter of 1.2\u20135 nm, while that of MWCNTs are typically 10\u201350 nm. The most important conclusion that can be made from the description of the structure of carbon nanotubes is that they are basically rolled-up graphene sheets (see Figure 1A). While it is theoretically possible to justify that CNTs with very small diameters exhibit electronic properties that are different from those of graphite, there is no fundamental reason that multiwalled carbon nanotubes with a diameter larger than 10 nm should behave differently electrochemi[a] Dr. M. Pumera International Center for Materials Nanoarchitectonics and Biomaterials Center National Institute for Materials Science 1-1 Namiki; Tsukuba, Ibaraki 305-0044 (Japan) Fax: (+81) 29-860-4714 E-mail : PUMERA",
            " This was first demonstrated by Compton and co-workers,[9] who compared the electrochemical response of MWCNTs and graphite microparticle-modified basal-plane pyrolytic graphite (BPPG) electrodes to the redox properties of NADH, epinephrine, and norepinephrine. They found that the response of MWCNTs and graphite-modified BPPG is essentially the same. In their subsequent work, the same group investigated the reasons for this behavior. Comparing the responses of electrodes fabricated from highly oriented pyrolytic graphite in two orientations, such as BPPG and edge-plane pyrolytic graphite (EPPG), and those modified with MWCNTs, they found that the edge-plane sites and the tube ends are the electrochemically active sites (see Figure 1A).[10, 11] The pristine walls of carbon nanotubes behave as a basal plane of graphite with the heterogeneous electron-transfer constant limit to zero,[12] while the ends of the carbon nanotubes and the defects on their walls resemble the behavior and fast electron transfer of EPPG (Figure 1A).[11,13] It is possible to tune the electrochemical activity of MWCNTs by inducing such defects in the walls of the carbon nanotubes (Figure 1B), as we have recently shown.[14,15] The remaining question is does this situation also apply to SWCNTs? SWCNTs have a profoundly different electronic structure than 2D graphene sheets. The overlap between the p orbitals on the neighboring carbon atoms involved in p bonding is reduced, because of their small diameter and resulting greater curvature, suggesting increased chemical reactivity.[16] Therefore, in the case of SWCNTs, there is a possibility that not only their ends and the defects on their sides are electroactive, but the pristine walls can also be electroactive"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-FigureC.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-FigureC.2-1.png",
        "caption": "FIGURE C.2 Partial mutual inductance for two filament wires.",
        "texts": [
            " For these applications, at least 4\u20135 digits of accuracy is required. We should be aware that some of the approximate formulas in Ref. [1] have limited accuracy. For several analytical formulas in this appendices we include very small numbers in the order of 10\u221220 to 10\u221240 to avoid possible artificial singularities for some combinations of input parameters. This is a conventional numerical technique to avoid singularity problems. C.1.1 Lp12 for Two Parallel Filaments A simple important geometry is the wire-to-wire or filament-to-filament problem shown in Fig. C.2. Of course, the result is singular if the wires coincide. Filaments are very important building blocks, which we use for the combined analytical and numerical integration. The current direction in the wires is shown to be along the length. The partial inductance formula for Lp12 for two current filaments shown in Fig. C.2 reduces to Lp12 = \ud835\udf070 4\ud835\udf0b \u222b xe1 xs1 \u222b xe2 xs2 1 R1,2 dx2 dx1, (C.6) 386 COMPUTATION OF PARTIAL INDUCTANCES where R1,2 = \u221a (x1 \u2212 x2)2 + (y1 \u2212 y2)2 + (z1 \u2212 z2)2. (C.7) The analytical form of the partial inductance is given for this case by a simple integration step as Lp12 = \ud835\udf070 4\ud835\udf0b 4\u2211 k=1 (\u22121)k+1 [ ak log ( ak + \u221a a2 k + r2 12 ) \u2212 \u221a a2 k + r2 12 ] (C.8) with r12 = \u221a (y2 \u2212 y1)2 + (z2 \u2212 z1)2 (C.9) a1 = xs2 \u2212 xe1 (C.10) a2 = xe2 \u2212 xe1 (C.11) a3 = xe2 \u2212 xs1 (C.12) a4 = xs2 \u2212 xs1. (C.13) Note that e represents the end of the object and s is used for the start"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.3-1.png",
        "caption": "FIGURE 7.3. Equivalent bicycle model for a front-wheel-steering vehicle.",
        "texts": [
            " So, the turning center O is on the left, and the inner wheels are the left wheels that are closer to the center of rotation. The inner and outer steer angles \u03b4i and \u03b4o may be calculated 7. Steering Dynamics 381 from the triangles 4OAD and 4OBC as follows: tan \u03b4i = l R1 \u2212 w 2 (7.4) tan \u03b4o = l R1 + w 2 (7.5) Eliminating R1 R1 = 1 2 w + l tan \u03b4i = \u22121 2 w + l tan \u03b4o (7.6) provides the Ackerman condition (7.1), which is a direct relationship between \u03b4i and \u03b4o. cot \u03b4o \u2212 cot \u03b4i = w l (7.7) To find the vehicle\u2019s turning radius R, we define an equivalent bicycle model, as shown in Figure 7.3. The radius of rotation R is perpendicular to the vehicle\u2019s velocity vector v at the mass center C. Using the geometry shown in the bicycle model, we have R2 = a22 +R21 (7.8) cot \u03b4 = R1 l = 1 2 (cot \u03b4i + cot \u03b4o) (7.9) and therefore, R = q a22 + l2 cot2 \u03b4. (7.10) The Ackerman condition is needed when the speed of the vehicle is too small, and slip angles are zero. There is no lateral force and no centrifugal force to balance each other. The Ackerman steering condition is also called the kinematic steering condition, because it is a static condition at zero velocity"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure10.4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure10.4-1.png",
        "caption": "Figure 10.4 It is usual to choose the section across a member as a plane normal to the member axis. The centre of force is the intersection of the line of action of the stress resultant R with the plane of the cross-section.",
        "texts": [
            " Mathematically, we describe this phenomenon in the section by means of the concept stress (see Section 6.5).1 The distributions in magnitude and direction of the stresses in the section are as yet unknown. The equilibrium, however, shows that the stress resultant R, regardless of the shape of the section, must be 50 kN, and that the line of action of R must coincide with the line of action of the support reaction at A (see Figure 10.3). We usually do not give the section across a member an arbitrary shape, but rather choose one that is straight and normal to the member axis, as shown in Figure 10.4. This type of section is called a normal section or simply cross-section. Hereafter, when we refer to a section, we always mean a normal section. The intersection of the line of action of the stress resultant R and the crosssectional plane is known as the centre of force. The intersection of the member axis with the cross-sectional plane is the normal force centre, or normal centre, of the section. The normal (force) centre is indicated by the two-character symbol NC (see Figure 10.4). Consistent with the model used for a line element, it is usual to represent 1 In Section 10.1.2, the definition of the concept stress, as introduced in Section 6.5, is adapted to describe the interaction between the particles of matter. The member axis is by definition chosen through the normal centre NC of the cross-section. The location of the normal centre is covered in Volume 2 Stresses, Deformations, Displacements. In so-called homogeneous cross-sections (the whole cross-section consists of the same material) the normal centre coincides with the centroid of the cross-section"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000376_j.jallcom.2014.06.172-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000376_j.jallcom.2014.06.172-Figure1-1.png",
        "caption": "Fig. 1. Schematic cross-section of the Concept Laser M2 powder-bed system.",
        "texts": [
            " Recent work has highlighted the influence of this scan strategy and is presented here in terms of grain structure; crystallographic orientation and distribution of cracks within the as fabricated material. This study forms part of a larger project aimed at establishing a processing route to produce fully-dense and netshaped aerospace components from the nickel superalloy, CM247LC, via a Selective Laser Melting (SLM) powder-bed route. SLM powder-bed fabrication is an ALM technology where components can be netshape formed directly from metal powder. A schematic representation of a typical powder-bed system is shown in Fig. 1. Preparation involves virtually slicing a threedimensional shape, in the form of a CAD file, into a sequence of two-dimensional slices. For each slice fabricated the following sequence of events occurs: \u2013 The powder platform is raised, presenting a small amount of powder proud of the bed. \u2013 The flexible rubber recoater blade is moved across the bed spreading a thin layer of powder across the previously built layers (or build-plate in the case of the first layer). \u2013 The computer controlled laser scans the surface of the bed to selectively melt the current two-dimensional slice of the CAD file"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure2.7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure2.7-1.png",
        "caption": "Fig. 2.7. Double stator, triple rotor AFPM brushless motor with water cooling system [59]. 1 \u2014 PM, 2 \u2014 stator core, 3 \u2014 stator winding.",
        "texts": [
            " 34 2 Principles of AFPM Machines There is a limit on the increase of motor torque that can be achieved by enlarging the motor diameter. Factors limiting the single disc design are: (a) axial force taken by bearings; (b) integrity of mechanical joint between the disc and shaft; (c) disc stiffness. A more reasonable solution for large torques are double or triple disc motors. There are several configurations of multidisc motors [5, 6, 7, 59, 81, 82]. Large multidisc motors rated at least 300-kW have a water cooling system (Fig. 2.7) with radiators around the winding end connections [59]. To minimize the winding losses due to skin effect, variable cross section conductors may be used so that the cross section of conductors is bigger in the slot area than in the end connection region. Using a variable cross section means a gain of 40% in the rated power [59]. However, variable cross section of conductors means an increased cost of manufacturing. Owing to high mechanical stresses a titanium alloy is recommended for disc rotors",
            "29) where \u03d1hot is the average temperature of the elements that surround the heat pipe in the stator, \u03d1cold is the average temperature of the air cooling the finned surface, hhot is the convective heat transfer coefficient on the inside wall of the heat pipe in the stator, Ahot is the exposed area of the heat pipe in the stator, hcold is the convection heat transfer coefficient on the inside wall of the heat pipe in the finned area, Acold is the exposed area of the heat pipe at the finned surface, \u03b7fin is the efficiency of the finned surface, hfin is the convection heat transfer coefficient on the surface of the fins and Afin is the total exposed area of the finned surface. Direct Water Cooling Depending on the conditions at the site of operation, it is often necessary to use forced water circulation to cool the stator windings directly, especially for large power AFPM machines. An external water pump is required for forcing water circulation. A longitudinal section of a water cooled doubledisc motor has been shown in Fig. 2.7. For AFPM machines with an internal iron core stator, an ideal location of placing cooling channel is around the outer periphery of the stator disc, where the effective heat transfer area is the largest. For coreless winding AFPM machines, the winding coils may have a rhomboidal shape so that the space between the two active sides of each coil may be utilised for placing a cooling water duct [45]. The heat removed through cooling pipes can be calculated by using eqn (8.29). However, the heat transfer coefficients, hhot and hcold, are calculated using the following relationships: (i) for laminar flow, i"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001355_j.addma.2016.10.010-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001355_j.addma.2016.10.010-Figure1-1.png",
        "caption": "Fig. 1. (a) SEM image of pre-alloyed atomised Hastelloy-X powder showing spherical morphology and (b) particle size distribution.",
        "texts": [
            " The current study seeks to fill these gaps and to proide further contributions particularly in terms of understanding he surface roughness formation of inclined surfaces, and the diferences between powder and bulk solid material in terms of heat ransfer and their contribution to melt pool shape formation. . Experimental methods The SLM equipment used in this work is an EOS M280 machine, hich applies a Yb-fibre laser with a spot size of 100 m and a maxmum power output of 400 W. The building chamber was pre-filled ith argon gas to protect the parts from the effects of oxidation. Pre-alloyed gas atomised Hastelloy-X powder (from Carpenter owder Products, USA) with a composition of 48.64Ni-21.1Cr7.93Cr-8.8Mo-2.0W-0.36Si-0.06C was used. Fig. 1 shows a canning electron microscopy (SEM) image of the powder particles ith a volume median size of 28.3 m. Initial single laser scans were first produced to gain an under- Please cite this article in press as: Y. Tian, et al., Influences of process selective laser melting, Addit Manuf (2016), http://dx.doi.org/10.1016 tanding of laser-material interaction and establish usable melting egions (scan power and speed). Once the optimised scan power nd speed had been identified, vertical multi-layer samples were hen built with varied layer thickness and hatch distance (dis- PRESS uring xxx (2016) xxx\u2013xxx tance between adjacent scans)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure3.22-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure3.22-1.png",
        "caption": "FIGURE 3.22. Side view of a normal stress \u03c3z distribution and its resultant force Fz on a rolling tire.",
        "texts": [],
        "surrounding_texts": [
            "The tire is the main component interacting with the road. The performance of a vehicle is mainly influenced by the characteristics of its tires. Tires affect a vehicle\u2019s handling, traction, ride comfort, and fuel consumption. To understand its importance, it is enough to remember that a vehicle can maneuver only by longitudinal, vertical, and lateral force systems generated under the tires. Figure 3.1 illustrates a model of a vertically loaded stationary tire. To model the tire-road interactions, we determine the tireprint and describe the forces distributed on the tireprint. 3.1 Tire Coordinate Frame and Tire Force System To describe the tire-road interaction and force system, we attach a Cartesian coordinate frame at the center of the tireprint, as shown in Figure 3.2, assuming a flat and horizontal ground. The x-axis is along the intersection line of the tire-plane and the ground. Tire plane is the plane made by narrowing the tire to a flat disk. The z-axis is perpendicular to the ground, opposite to the gravitational acceleration g, and the y-axis makes the coordinate system a right-hand triad. To show the tire orientation, we use two angles: camber angle \u03b3 and sideslip angle \u03b1. The camber angle is the angle between the tire-plane and 96 3. Tire Dynamics the vertical plane measured about the x-axis. The camber angle can be recognized better in a front view as shown in Figure 3.3. The sideslip angle \u03b1, or simply sideslip, is the angle between the velocity vector v and the x-axis measured about the z-axis. The sideslip can be recognized better in a top view, as shown in Figure 3.4. The force system that a tire receives from the ground is assumed to be located at the center of the tireprint and can be decomposed along x, y, and z axes. Therefore, the interaction of a tire with the road generates a 3D force system including three forces and three moments, as shown in Figure 3.2. 1. Longitudinal force Fx. It is a force acting along the x-axis. The resultant longitudinal force Fx > 0 if the car is accelerating, and Fx < 0 if the car is braking. Longitudinal force is also called forward force. 2. Normal force Fz. It is a vertical force, normal to the ground plane. The resultant normal force Fz > 0 if it is upward. Normal force is also called vertical force or wheel load. 3. Lateral force Fy. It is a force, tangent to the ground and orthogonal to both Fx and Fz. The resultant lateral force Fy > 0 if it is in the y-direction. 4. Roll moment Mx. It is a longitudinal moment about the x-axis. The resultant roll moment Mx > 0 if it tends to turn the tire about the x-axis. The roll moment is also called the bank moment, tilting torque, or overturning moment. 3. Tire Dynamics 97 98 3. Tire Dynamics 5. Pitch moment My. It is a lateral moment about the y-axis. The resultant pitch momentMy > 0 if it tends to turn the tire about the y-axis and move forward. The pitch moment is also called rolling resistance torque. 6. Yaw moment Mz. It is an upward moment about the z-axis. The resultant yaw moment Mz > 0 if it tends to turn the tire about the z-axis. The yaw moment is also called the aligning moment, self aligning moment, or bore torque. The moment applied to the tire from the vehicle about the tire axis is called wheel torque T . Example 75 Origin of tire coordinate frame. For a cambered tire, it is not always possible to find or define a center point for the tireprint to be used as the origin of the tire coordinate frame. It is more practical to set the origin of the tire coordinate frame at the center of the intersection line between the tire-plane and the ground. So, the origin of the tire coordinate frame is at the center of the tireprint when the tire is standing upright and stationary on a flat road. Example 76 SAE tire coordinate system. The tire coordinate system adopted by the Society of Automotive Engineers (SAE) is shown in Figure 3.5. The origin of the coordinate system is at the center of the tireprint when the tire is standing stationary. The x-axis is at the intersection of the tire-plane and the ground plane. The z-axis is downward and perpendicular to the tireprint. The y-axis is on the ground plane and goes to the right to make the coordinate frame a righthand frame. The sideslip angle \u03b1 is considered positive if the tire is slipping to the right, and the camber angle \u03b3 is positive when the tire leans to the right. The SAE coordinate system is as good as the coordinate system in Figure 3.2 and may be used alternatively. However, having the z-axis directed downward is sometimes inefficient and confusing. Furthermore, in SAE convention, the camber angle for the left and right tires of a vehicle have opposite signs. So, the camber angle of the left tire is positive when the tire leans to the right and the camber angle of the right tire is positive when the tire leans to the left. 3.2 Tire Stiffness As an applied approximation, the vertical tire force Fz can be calculated as a linear function of the normal tire deflection 4z measured at the tire center. Fz = kz4z (3.1) 3. Tire Dynamics 99 The coefficient kz is called tire stiffness in the z-direction. Similarly, the reaction of a tire to a lateral and a longitudinal force can be approximated by Fx = kx4x (3.2) Fy = ky4y (3.3) where the coefficient kx and ky are called tire stiffness in the x and y directions. Proof. The deformation behavior of tires to the applied forces in any three directions x, y, and z are the first important tire characteristics in tire dynamics. Calculating the tire stiffness is generally based on experiment and therefore, they are dependent on the tire\u2019s mechanical properties, as well as environmental characteristics. Consider a vertically loaded tire on a stiff and flat ground as shown in Figure 3.6. The tire will deflect under the load and generate a pressurized contact area to balance the vertical load. Figure 3.7 depicts a sample of experimental stiffness curve in the (Fz,4z) plane. The curve can be expressed by a mathematical function Fz = f (4z) (3.4) however, we may use a linear approximation for the range of the usual application. Fz = \u2202f \u2202 (4z) 4z (3.5) 100 3. Tire Dynamics F1 < F2 < F3 3. Tire Dynamics 101 The coefficient \u2202f \u2202(4z) is the slope of the experimental stiffness curve at zero and is shown by a stiffness coefficient kz kz = tan \u03b8 = lim 4z\u21920 \u2202f \u2202 (4z) . (3.6) Therefore, the normal tire deflection 4z remains proportional to the vertical tire force Fz. Fz = kz4z (3.7) The tire can apply only pressure forces to the road, so normal force is restricted to Fz > 0. The stiffness curve can be influenced by many parameters. The most effective one is the tire inflation pressure. Lateral and longitudinal force/deflection behavior is also determined experimentally by applying a force in the appropriate direction. The lateral and longitudinal forces are limited by the sliding force when the tire is vertically loaded. Figure 3.8 depicts a sample of longitudinal and lateral stiffness curves compared to a vertical stiffness curve. The practical part of a tire\u2019s longitudinal and lateral stiffness curves is the linear part and may be estimated by linear equations. Fx = kx4x (3.8) Fy = ky4y (3.9) The coefficients kx and ky are called the tire stiffness in the x and y directions. They are measured by the slope of the experimental stiffness curves 102 3. Tire Dynamics in the (Fx,4x) and (Fy,4y) planes. kx = lim 4x\u21920 \u2202f \u2202 (4x) (3.10) ky = lim 4y\u21920 \u2202f \u2202 (4y) (3.11) When the longitudinal and lateral forces increase, parts of the tireprint creep and slide on the ground until the whole tireprint starts sliding. At this point, the applied force saturates and reaches its maximum supportable value. Generally, a tire is most stiff in the longitudinal direction and least stiff in the lateral direction. kx > kz > ky (3.12) Figure 3.9 illustrates tire deformation under a lateral and a longitudinal force. Example 77 F Nonlinear tire stiffness. In a better modeling, the vertical tire force Fz is a function of the normal tire deflection 4z, and deflection velocity 4z\u0307. Fz = Fz (4z,4z\u0307) (3.13) = Fzs + Fzd (3.14) In a first approximation we may assume Fz is a combination of a static and a dynamic part. The static part is a nonlinear function of the vertical tire deflection and the dynamic part is proportional to the vertical speed of the tire. Fzs = k14z + k2 (4z)2 (3.15) Fzd = k3z\u0307 (3.16) 3. Tire Dynamics 103 The constants k1 and k2 are calculated from the first and second slopes of the experimental stiffness curve in the (Fz,4z) plane, and k3 is the first slope of the curve in the (Fz, z\u0307) plane, which indicates the tire damping. k1 = \u2202 Fz \u22024z \u00af\u0304\u0304\u0304 4z=0 (3.17) k2 = 1 2 \u22022 Fz \u2202 (4z) 2 \u00af\u0304\u0304\u0304 \u00af 4z=0 (3.18) k3 = \u2202 Fz \u2202 z\u0307 \u00af\u0304\u0304\u0304 z\u0307=0 (3.19) The value of k1 = 200000N/m is a good approximation for a 205/50R15 passenger car tire, and k1 = 1200000N/m is a good approximation for a X31580R22.5 truck tire. Tires with a larger number of plies have higher damping, because the plies\u2019 internal friction generates the damping. Tire damping decreases by increasing speeds. Example 78 F Hysteresis effect. Because tires are made from rubber, which is a viscoelastic material, the loading and unloading stiffness curves are not exactly the same. They are similar to those in Figure 3.10, which make a loop with the unloading curve below the loading. The area within the loop is the amount of dissipated energy during loading and unloading. As a tire rotates under the weight of a vehicle, it experiences repeated cycles of deformation and recovery, and it dissipates energy loss as heat. Such a behavior is a common property of hysteretic material and is called hysteresis. So, hysteresis is a characteristic of a deformable material such as rubber, that the energy of deformation is greater than the energy of recovery. The amount of dissipated energy depends on the mechanical characteristics of the tire. Hysteretic energy loss in rubber decreases as temperature increases. The hysteresis effect causes a loaded rubber not to rebound fully after load removal. Consider a high hysteresis race car tire turning over road irregularities. The deformed tire recovers slowly, and therefore, it cannot push the tireprint tail on the road as hard as the tireprint head. The difference in head and tail pressures causes a resistance force, which is called rolling resistance. Race cars have high hysteresis tires to increase friction and limit traction. Street cars have low hysteresis tires to reduce the rolling resistance and low operating temperature. Hysteresis level of tires inversely affect the stopping distance. A high hysteresis tire makes the stopping shorter, however, it wears rapidly and has a shorter life time. 104 3. Tire Dynamics 3.3 Tireprint Forces The force per unit area applied on a tire in a tireprint can be decomposed into a component normal to the ground and a tangential component on the ground. The normal component is the contact pressure \u03c3z, while the tangential component can be further decomposed in the x and y directions to make the longitudinal and lateral shear stresses \u03c4x and \u03c4y. For a stationary tire under normal load, the tireprint is symmetrical. Due to equilibrium conditions, the overall integral of the normal stress over the tireprint area AP must be equal to the normal load Fz, and the integral of shear stresses must be equal to zero. Z AP \u03c3z(x, y) dA = Fz (3.20)Z AP \u03c4x(x, y) dA = 0 (3.21)Z AP \u03c4y(x, y) dA = 0 (3.22) 3.3.1 Static Tire, Normal Stress Figure 3.11 illustrates a stationary tire under a normal load Fz along with the generated normal stress \u03c3z applied on the ground. The applied loads on the tire are illustrated in the side view shown in Figure 3.12. For a stationary tire, the shape of normal stress \u03c3z(x, y) over the tireprint area depends on tire and load conditions, however its distribution over the tireprint is generally in the shape shown in Figure 3.13. 3. Tire Dynamics 105 106 3. Tire Dynamics The normal stress \u03c3z(x, y) may be approximated by the function \u03c3z(x, y) = \u03c3zM \u00b5 1\u2212 x6 a6 \u2212 y6 b6 \u00b6 (3.23) where a and b indicate the dimensions of the tireprint, as shown in Figure 3.14. The tireprints may approximately be modeled by a mathematical function x2n a2n + y2n b2n = 1 n \u2208 N. (3.24) For radial tires, n = 3 or n = 2 may be used, x6 a6 + y6 b6 = 1 (3.25) while for non-radial tires n = 1 is a better approximation. x2 a2 = y2 b2 = 1. (3.26) Example 79 Normal stress in tireprint. A car weighs 800 kg. If the tireprint of each radial tire is AP = 4\u00d7a\u00d7b = 4 \u00d7 5 cm \u00d7 12 cm, then the normal stress distribution under each tire, \u03c3z, 3. Tire Dynamics 107 must satisfy the equilibrium equation. Fz = 1 4 800\u00d7 9.81 = Z AP \u03c3z(x, y) dA = Z 0.05 \u22120.05 Z 0.12 \u22120.12 \u03c3zM \u00b5 1\u2212 x6 0.056 \u2212 y6 0.126 \u00b6 dy dx = 1.7143\u00d7 10\u22122\u03c3zM (3.27) Therefore, the maximum normal stress is \u03c3zM = Fz 1.7143\u00d7 10\u22122 = 1.1445\u00d7 10 5 Pa (3.28) and the stress distribution over the tireprint is \u03c3z(x, y) = 1.1445\u00d7 105 \u00b5 1\u2212 x6 0.056 \u2212 y6 0.126 \u00b6 Pa. (3.29) Example 80 Normal stress in tireprint for n = 2. The maximum normal stress \u03c3zM for an 800 kg car having an AP = 4\u00d7 a\u00d7 b = 4\u00d7 5 cm\u00d7 12 cm, can be found for n = 2 as Fz = 1 4 800\u00d7 9.81 = Z AP \u03c3z(x, y) dA = Z 0.05 \u22120.05 Z 0.12 \u22120.12 \u03c3zM \u00b5 1\u2212 x4 0.054 \u2212 y4 0.124 \u00b6 dy dx = 1.44\u00d7 10\u22122\u03c3zM (3.30) 108 3. Tire Dynamics \u03c3zM = Fz 1.44\u00d7 10\u22122 = 1.3625\u00d7 10 5 Pa. (3.31) Comparing \u03c3zM = 1.3625 \u00d7 105 Pa for n = 2 to \u03c3zM = 1.1445 \u00d7 105 Pa for n = 3 shows that maximum stress for n = 2 is \u00a1 1\u2212 1.1445 1.3625 \u00a2 \u00d7100 = 16% more than n = 3. 3.3.2 Static Tire, Tangential Stresses Because of geometry changes to a circular tire in contact with the ground, a three-dimensional stress distribution will appear in the tireprint even for a stationary tire. The tangential stress \u03c4 on the tireprint can be decomposed in x and y directions. The tangential stress is also called shear stress or friction stress. The tangential stress on a tire is inward in x direction and outward in y direction. Hence, the tire tries to stretch the ground in the x-axis and compact the ground on the y-axis. Figure 3.15 depicts the shear stresses on a vertically loaded stationary tire. The force distribution on the tireprint is not constant and is influenced by tire structure, load, inflation pressure, and environmental conditions: The tangential stress \u03c4x in the x-direction may be modeled by the following equation. \u03c4x(x, y) = \u2212\u03c4xM \u00b5 x2n+1 a2n+1 \u00b6 sin2 \u00b3x a \u03c0 \u00b4 cos \u00b3 y 2b \u03c0 \u00b4 n \u2208 N (3.32) 3. Tire Dynamics 109 \u03c4x is negative for x > 0 and is positive for x < 0, showing an inward longitudinal stress. Figure 3.16 illustrates the absolute value of a \u03c4x distribution for n = 1. The y-direction tangential stress \u03c4y may be modeled by the equation \u03c4y(x, y) = \u2212\u03c4yM \u00b5 x2n a2n \u2212 1 \u00b6 sin \u00b3y b \u03c0 \u00b4 n \u2208 N (3.33) where \u03c4y is positive for y > 0 and negative for y < 0, showing an outward lateral stress. Figure 3.17 illustrates the absolute value of a \u03c4y distribution for n = 1. 3.4 Effective Radius Consider a vertically loaded wheel that is turning on a flat surface as shown in Figure 3.18. The effective radius of the wheel Rw, which is also called a rolling radius, is defined by Rw = vx \u03c9w (3.34) where, vx is the forward velocity, and \u03c9w is the angular velocity of the wheel. The effective radius Rw is approximately equal to Rw \u2248 Rg \u2212 Rg \u2212Rh 3 (3.35) and is a number between the unloaded or geometric radius Rg and the loaded height Rh. Rh < Rw < Rg (3.36) 110 3. Tire Dynamics Proof. An effective radius Rw = vx/\u03c9w is defined by measuring a wheel\u2019s angular velocity \u03c9w and forward velocity vx. As the tire turns forward, each part of the circumference is flattened as it passes through the contact area. A practical estimate of the effective radius can be made by substituting the arc with the straight length of tireprint. The tire vertical deflection is Rg \u2212Rh = Rg (1\u2212 cos\u03d5) (3.37) and therefore Rh = Rg cos\u03d5 (3.38) a = Rg sin\u03d5. (3.39) If the motion of the tire is compared to the rolling of a rigid disk with radius Rw, then the tire must move a distance a = Rw\u03d5 for an angular rotation \u03d5. a = Rg sin\u03d5 = Rw\u03d5 (3.40) Hence, Rw = Rg sin\u03d5 \u03d5 . (3.41) Expanding sin\u03d5 \u03d5 in a Taylor series show that Rw = Rg \u00b5 1\u2212 1 6 \u03d52 +O \u00a1 \u03d54 \u00a2\u00b6 . (3.42) 3. Tire Dynamics 111 Using Equation (3.37) we may approximate cos\u03d5 \u2248 1\u2212 1 2 \u03d52 (3.43) \u03d52 \u2248 2 (1\u2212 cos\u03d5) \u2248 2 \u00b5 1\u2212 Rh Rg \u00b6 (3.44) and therefore, Rw \u2248 Rg \u00b5 1\u2212 1 3 \u00b5 1\u2212 Rh Rg \u00b6\u00b6 = 2 3 Rg + 1 3 Rh. (3.45) Because Rh is a function of tire load Fz, Rh = Rh (Fz) = Rg \u2212 Fz kz (3.46) the effective radius Rw is also a function of the tire load. The angle \u03d5 is called tireprint angle or tire contact angle. The vertical stiffness of radial tires is less than non-radial tires under the same conditions. So, the loaded height of radial tires, Rh, is less than the non-radials\u2019. However, the effective radius of radial tires Rw, is closer to their unloaded radius Rg. As a good estimate, for a non-radial tire, Rw \u2248 0.96Rg, and Rh \u2248 0.94Rg, while for a radial tire, Rw \u2248 0.98Rg, and Rh \u2248 0.92Rg. 112 3. Tire Dynamics Generally speaking, the effective radius Rw depends on the type of tire, stiffness, load conditions, inflation pressure, and wheel\u2019s forward velocity. Example 81 Compression and expansion of tires in the tireprint zone. Because of longitudinal deformation, the peripheral velocity of any point of the tread varies periodically. When it gets close to the starting point of the tireprint, it slows down and a circumferential compression results. The tire treads are compressed in the first half of the tireprint and gradually expanded in the second half. The treads in the tireprint contact zone almost stick to the ground, and therefore their circumferential velocity is close to the forward velocity of the tire center vx. The treads regain their initial circumferential velocity Rg\u03c9w after expanding and leaving the contact zone. Example 82 Tire rotation. The geometric radius of a tire P235/75R15 is Rg = 366.9mm, because hT = 235\u00d7 75% = 176.25mm \u2248 6.94 in (3.47) and therefore, Rg = 2hT + 15 2 = 2\u00d7 6.94 + 15 2 = 14.44 in \u2248 366.9mm. (3.48) Consider a vehicle with such a tire is traveling at a high speed such as v = 50m/ s = 180 km/h \u2248 111.8mi/h. The tire is radial, and therefore the effective tire radius Rw is approximately equal to Rw \u2248 0.98Rg \u2248 359.6mm. (3.49) After moving a distance d = 100 km, this tire must have been turned n1 = 44259 times because n1 = d \u03c0D = 100\u00d7 103 2\u03c0 \u00d7 359.6\u00d7 10\u22123 = 44259. (3.50) Now assume the vehicle travels the same distance d = 100 km at a low inflation pressure, such that the effective radius of the tire remained close to at the loaded radius Rw \u2248 Rh \u2248 0.92Rg = 330.8mm. (3.51) 3. Tire Dynamics 113 This tire must turn n2 = 48112 times to travel d = 100km, because, n2 = d \u03c0D = 100\u00d7 103 2\u03c0 \u00d7 330.8\u00d7 10\u22123 = 48112. (3.52) Example 83 F Radial motion of tire\u2019s peripheral points in the tireprint. The radial displacement of a tire\u2019s peripheral points during road contact may be modeled by a function d = d (x, \u03b8) . (3.53) We assume that a peripheral point of the tire moves along only the radial direction during contact with the ground, as shown in Figure 3.19. Let\u2019s show a radius at an angle \u03b8, by r = r (x, \u03b8). Knowing that cos \u03b8 = Rh r (3.54) cos\u03c6 = Rh Rg (3.55) we can find r = Rg cos\u03c6 cos \u03b8 . (3.56) 114 3. Tire Dynamics Thus, the displacement function is d = Rg \u2212 r (x, \u03b8) = Rg \u00b5 1\u2212 Rh Rg cos \u03b8 \u00b6 \u2212 \u03c6 < \u03b8 < \u03c6. (3.57) Example 84 Tread travel. Let\u2019s follow a piece of tire tread in its travel around the spin axis when the vehicle moves forward at a constant speed. Although the wheel is turning at constant angular velocity \u03c9w, the tread does not travel at constant speed. At the top of the tire, the radius is equal to the unloaded radius Rg and the speed of the tread is Rg\u03c9w relative to the wheel center. As the tire turns, the tread approaches the leading edge of tireprint, and slows down. The tread is compacted radially, and gets squeezed in the heading part of the tireprint area. Then, it is stretched out and unpacked in the tail part of the tireprint as it moves to the tail edge. In the middle of the tireprint, the tread speed is Rh\u03c9w relative to the wheel center. The variable radius of a tire during the motion through the tireprint is r = Rg cos\u03c6 cos \u03b8 \u2212 \u03c6 < \u03b8 < \u03c6 (3.58) where \u03c6 is the half of the contact angle, and \u03b8 is the angular rotation of the tire, as shown in Figure 3.19. The angular velocity of the tire is \u03c9w = \u03b8\u0307 and is assumed to be constant. Then, the radial velocity r\u0307 and acceleration r\u0308 of the tread with respect to the wheel center are r\u0307 = Rg\u03c9w cos\u03c6 sin \u03b8 cos2 \u03b8 (3.59) r\u0308 = 1 2 Rg\u03c9 2 w cos\u03c6 cos3 \u03b8 (3\u2212 cos 2\u03b8) . (3.60) Figure 3.20 depicts r, r\u0307, and r\u0308 for a sample car with the following data: Rg = 0.5m (3.61) \u03c6 = 15deg (3.62) \u03c9w = 60 rad/ s (3.63) 3.5 Rolling Resistance A turning tire on the ground generates a longitudinal force called rolling resistance. The force is opposite to the direction of motion and is proportional to the normal force on the tireprint. Fr = \u2212Fr \u0131\u0302 (3.64) Fr = \u03bcr Fz (3.65) 3. Tire Dynamics 115 The parameter \u03bcr is called the rolling friction coefficient. \u03bcr is not constant and mainly depends on tire speed, inflation pressure, sideslip and camber angles. It also depends on mechanical properties, speed, wear, temperature, load, size, driving and braking forces, and road condition. Proof.When a tire is turning on the road, that portion of the tire\u2019s circumference that passes over the pavement undergoes a deflection. Part of the energy that is spent in deformation will not be restored in the following relaxation. Hence, a change in the distribution of the contact pressure makes normal stress \u03c3z in the heading part of the tireprint be higher than the tailing part. The dissipated energy and stress distortion cause the rolling resistance. Figures 3.21 and 3.22 illustrate a model of normal stress distribution across the tireprint and their resultant force Fz for a turning tire. Because of higher normal stress in the front part of the tireprint, the resultant normal force moves forward. Forward shift of the normal force makes a resistance moment in the \u2212y direction, opposing the forward rotation. Mr = \u2212Mr j\u0302 (3.66) Mr = Fz\u2206x (3.67) The rolling resistance momentMr can be substituted by a rolling resistance 116 3. Tire Dynamics 3. Tire Dynamics 117 force Fr parallel to the x-axis. Fr = \u2212Fr \u0131\u0302 (3.68) Fr = 1 Rh Mr = \u2206x Rh Fz (3.69) Practically the rolling resistance force can be defined using a rolling friction coefficient \u03bcr. Fr = \u03bcr Fz (3.70) Example 85 A model for normal stress of a turning tire. We may assume that the normal stress of a turning tire is expressed by \u03c3z = \u03c3zm \u00b5 1\u2212 x2n a2n \u2212 y2n b2n + x 4a \u00b6 (3.71) where n = 3 or n = 2 for radial tires and n = 1 for non-radial tires. We may determine the stress mean value \u03c3zm by knowing the total load on the tire. As an example, using n = 3 for an 800 kg car with a tireprint AP = 4\u00d7 a\u00d7 b = 4\u00d7 5 cm\u00d7 12 cm, we have Fz = 1 4 800\u00d7 9.81 = Z AP \u03c3z(x, y) dA = Z 0.05 \u22120.05 Z 0.12 \u22120.12 \u03c3zm \u00b5 1\u2212 x6 0.056 \u2212 y6 0.126 + x 4\u00d7 0.05 \u00b6 dy dx = 1.7143\u00d7 10\u22122\u03c3zm (3.72) and therefore, \u03c3zm = Fz 1.7143\u00d7 10\u22122 = 1.1445\u00d7 10 5 Pa (3.73) Example 86 Deformation and rolling resistance. The distortion of stress distribution is proportional to the tire-road deformation that is the reason for shifting the resultant force forward. Hence, the rolling resistance increases with increasing deformation. A high pressure tire on concrete has lower rolling resistance than a low pressure tire on soil. To model the mechanism of dissipation energy for a turning tire, we assume there are many small dampers and springs in the tire structure. Pairs of parallel dampers and springs are installed radially and circumstantially. Figures 3.23 and 3.24 illustrate the damping and spring structure of a tire. 118 3. Tire Dynamics 3. Tire Dynamics 119 3.5.1 F Effect of Speed on the Rolling Friction Coefficient The rolling friction coefficient \u03bcr increases with a second degree of speed. It is possible to express \u03bcr = \u03bcr(vx) by the function \u03bcr = \u03bc0 + \u03bc1 v 2 x. (3.74) Proof. The rolling friction coefficient increases by increasing speed experimentally. We can use a polynomial function \u03bcr = nX i=0 \u03bci v i x (3.75) to fit the experimental data. Practically, two or three terms of the polynomial would be enough. The function \u03bcr = \u03bc0 + \u03bc1 v 2 x (3.76) is simple and good enough for representing experimental data and analytic calculation. The values of \u03bc0 = 0.015 (3.77) \u03bc1 = 7\u00d7 10\u22126 s2/m2 (3.78) are reasonable values for most passenger car tires. However, \u03bc0 and \u03bc1 should be determined experimentally for any individual tire. Figure 3.25 depicts a comparison between Equation (3.74) and experimental data for a radial tire. Generally speaking, the rolling friction coefficient of radial tires show to be less than non-radials. Figure 3.26 illustrates a sample comparison. Equation (3.74) is applied when the speed is below the tire\u2019s critical speed. Critical speed is the speed at which standing circumferential waves appear and the rolling friction increases rapidly. The wavelength of the standing waves are close to the length of the tireprint. Above the critical speed, overheating happens and tire fails very soon. Figure 3.27 illustrates the circumferential waves in a rolling tire at its critical speed. Example 87 Rolling resistance force and vehicle velocity. For computer simulation purposes, a fourth degree equation is presented to evaluate the rolling resistance force Fr Fr = C0 + C1 vx + C2 v 4 x. (3.79) The coefficients Ci are dependent on the tire characteristics, however, the following values can be used for a typical raided passenger car tire: C0 = 9.91\u00d7 10\u22123 C1 = 1.95\u00d7 10\u22125 (3.80) C2 = 1.76\u00d7 10\u22129 120 3. Tire Dynamics Example 88 Road pavement and rolling resistance. The effect of the pavement and road condition is introduced by assigning a value for \u03bc0 in equation \u03bcr = \u03bc0 + \u03bc1 v 2 x. Table 3.1 is a good reference. 122 3. Tire Dynamics Example 89 Tire information tips. A new front tire with a worn rear tire can cause instability. Tires stored in direct sunlight for long periods of time will harden and age more quickly than those kept in a dark area. Prolonged contact with oil or gasoline causes contamination of the rubber compound, making the tire life short. Example 90 F Wave occurrence justification. The normal stress will move forward when the tire is turning on a road. By increasing the speed, the normal stress will shift more and concentrate in the first half of the tireprint, causing low stress in the second half of the tireprint. High stress in the first half along with no stress in the second half is similar to hammering the tire repeatedly. Example 91 F Race car tires. Racecars have very smooth tires, known as slicks. Smooth tires reduce the rolling friction and maximize straight line speed. The slick racing tires are also pumped up to high pressure. High pressure reduces the tireprint area. Hence, the normal stress shift reduces and the rolling resistance decreases. Example 92 F Effect of tire structure, size, wear, and temperature on the rolling friction coefficient. The tire material and the arrangement of tire plies affect the rolling friction coefficient and the critical speed. Radial tires have around 20% lower \u03bcr, and 20% higher critical speed. Tire radius Rg and aspect ratio hT /wT are the two size parameters that affect the rolling resistance coefficient. A tire with larger Rg and smaller hT /wT has lower rolling resistance and higher critical speed. Generally speaking, the rolling friction coefficient decreases with wear in both radial and non-radial tires, and increases by increasing temperature. 3.5.2 F Effect of Inflation Pressure and Load on the Rolling Friction Coefficient The rolling friction coefficient \u03bcr decreases by increasing the inflation pressure p. The effect of increasing pressure is equivalent to decreasing normal load Fz. The following empirical equation has been suggested to show the effects of both pressure p and load Fz on the rolling friction coefficient. \u03bcr = K 1000 \u00b5 5.1 + 5.5\u00d7 105 + 90Fz p + 1100 + 0.0388Fz p v2x \u00b6 (3.81) The parameter K is equal to 0.8 for radial tires, and is equal to 1.0 for nonradial tires. The value of Fz, p, and vx must be in [ N], [ Pa], and [m/ s] respectively. 3. Tire Dynamics 123 Example 93 Motorcycle rolling friction coefficient. The following equations are suggested for calculating rolling friction coefficient \u03bcr applicable to motorcycles. They can be only used as a rough lower estimate for passenger cars. The equations consider the inflation pressure and forward velocity of the motorcycle. \u03bcr = \u23a7\u23aa\u23aa\u23a8\u23aa\u23aa\u23a9 0.0085 + 1800 p + 2.0606 p v2x vx \u2264 46m/ s (\u2248 165 km/h) 1800 p + 3.7714 p v2x vx > 46m/ s (\u2248 165 km/h) (3.82) The speed vx must be expressed in m/ s and the pressure p must be in Pa. Figure 3.28 illustrates this equation for vx \u2264 46m/ s (\u2248 165 km/h). Increasing the inflation pressure p decreases the rolling friction coefficient \u03bcr. Example 94 Dissipated power because of rolling friction. Rolling friction reduces the vehicle\u2019s power. The dissipated power because of rolling friction is equal to the rolling friction force Fr times the forward velocity vx. Using Equation (3.81), the rolling resistance power is P = Fr vx = \u2212\u03bcr vx Fz = \u2212K vx 1000 \u00b5 5.1 + 5.5\u00d7 105 + 90Fz p + 1100 + 0.0388Fz p v2x \u00b6 Fz. (3.83) 124 3. Tire Dynamics The resultant power P is in [W] when the normal force Fz is expressed in [ N], velocity vx in [m/ s], and pressure p in [ Pa]. The rolling resistance dissipated power for motorcycles can be found based on Equation (3.82). P = \u23a7\u23aa\u23aa\u23a8\u23aa\u23aa\u23aa\u23a9 \u00b5 0.0085 + 1800 p + 2.0606 p v2x \u00b6 vxFz vx \u2264 46m/ s (\u2248 165 km/h)\u00b5 1800 p + 3.7714 p v2x \u00b6 vxFz vx > 46m/ s (\u2248 165 km/h) (3.84) Example 95 Rolling resistance dissipated power. If a vehicle is moving at 100 km/h \u2248 27.78m/ s \u2248 62mi/h and each radial tire of the vehicle is pressurized up to 220 kPa \u2248 32 psi and loaded by 220 kg, then the dissipated power, because of rolling resistance, is P = 4\u00d7 K vx 1000 \u00b5 5.1 + 5.5\u00d7 105 + 90Fz p + 1100 + 0.0388Fz p v2x \u00b6 Fz = 2424.1W \u2248 2.4 kW. (3.85) To compare the given equations, assume the vehicle has motorcycle tires with power loss given by Equation (3.84). P = \u00b5 0.0085 + 1800 p + 2.0606 p v2x \u00b6 vxFz = 5734.1W \u2248 5.7 kW. (3.86) It shows that if the vehicle uses motorcycle tires, it dissipates more power. Example 96 Effects of improper inflation pressure. High inflation pressure increases stiffness, which reduces ride comfort and generates vibration. Tireprint and traction are reduced when tires are over inflated. Over-inflation causes the tire to transmit shock loads to the suspension, and reduces the tire\u2019s ability to support the required load for cornerability, braking, and acceleration. Under-inflation results in cracking and tire component separation. It also increases sidewall flexing and rolling resistance that causes heat and mechanical failure. A tire\u2019s load capacity is largely determined by its inflation pressure. Therefore, under-inflation results in an overloaded tire that operates at high deflection with a low fuel economy, and low handling. Figure 3.29 illustrates the effect of over and under inflation on tire-road contact compared to a proper inflated tire. Proper inflation pressure is necessary for optimum tire performance, safety, and fuel economy. Correct inflation is especially significant to the endurance and performance of radial tires because it may not be possible to find a 5 psi \u2248 35 kPa under-inflation in a radial tire just by looking. 3. Tire Dynamics 125 However, under-inflation of 5 psi \u2248 35 kPa can reduce up to 25% of the tire performance and life. A tire may lose 1 to 2 psi (\u2248 7 to 14 kPa) every month. The inflation pressure can also change by 1 psi\u2248 7 kPa for every 10 \u25e6F \u2248 5 \u25e6C of temperature change. As an example, if a tire is inflated to 35 psi \u2248 240 kPa on an 80 \u25e6F \u2248 26 \u25e6C summer day, it could have an inflation pressure of 23 psi \u2248 160 kPa on a 20 \u25e6F \u2248 \u22126 \u25e6C day in winter. This represents a normal loss of 6 psi \u2248 40 kPa over the six months and an additional loss of 6 psi \u2248 40 kPa due to the 60 \u25e6F \u2248 30 \u25e6C change. At 23 psi \u2248 160 kPa, this tire is functioning under-inflated. Example 97 Small / large and soft / hard tires. If the driving tires are small, the vehicle becomes twitchy with low traction and low top speed. However, when the driving tires are big, then the vehicle has slow steering response and high tire distortion in turns, decreasing the stability. Softer front tires show more steerability, less stability, and more wear while hard front tires show the opposite. Soft rear tires have more rear traction, but they make the vehicle less steerable, more bouncy, and less stable. Hard rear tires have less rear traction, but they make the vehicle more steerable, less bouncy, and more stable. 3.5.3 F Effect of Sideslip Angle on Rolling Resistance When a tire is turning on the road with a sideslip angle \u03b1, a significant increase in rolling resistance occurs. The rolling resistance force Fr would 126 3. Tire Dynamics then be Fr = Fx cos\u03b1+ Fy sin\u03b1 (3.87) \u2248 Fx \u2212 C\u03b1\u03b1 2 (3.88) where, Fx is the longitudinal force opposing the motion, and Fy is the lateral force. Proof. Figure 3.30 illustrates the top view of a turning tire on the ground under a sideslip angle \u03b1. The rolling resistance force is defined as the force opposite to the velocity vector of the tire, which has angle \u03b1 with the xaxis. Assume a longitudinal force Fx in \u2212x-direction is applied on the tire. Sideslip \u03b1 increases Fx and generates a lateral force Fy. The sum of the components of the longitudinal force Fx and the lateral force Fy makes the rolling resistance force Fr. Fr = Fx cos\u03b1+ Fy sin\u03b1 (3.89) For small values of the sideslip \u03b1, the lateral force is proportional to \u2212\u03b1 and therefore, Fr \u2248 Fx \u2212 C\u03b1\u03b1 2. (3.90) 3. Tire Dynamics 127 3.5.4 F Effect of Camber Angle on Rolling Resistance When a tire travels with a camber angle \u03b3, the component of rolling moment Mr on rolling resistance Fr will be reduced, however, a component of aligning moment Mz on rolling resistance will appear. Fr = \u2212Fr \u0131\u0302 (3.91) Fr = 1 Rh Mr cos \u03b3 + 1 Rh Mz sin \u03b3 (3.92) Proof. Rolling moment Mr appears when the normal force Fz shifts forward. However, only the component Mr cos \u03b3 is perpendicular to the tireplane and prevents the tire\u2019s spin. Furthermore, when a moment in zdirection is applied on the tire, only the component Mz sin \u03b3 will prevent the tire\u2019s spin. Therefore, the camber angle \u03b3 will affect the rolling resistance according to Fr = \u2212Fr \u0131\u0302 Fr = 1 h Mr cos \u03b3 1 h Mz sin \u03b3 (3.93) where Mr may be substituted by Equation (3.66) to show the effect of normal force Fz. Fr = \u2206x h Fz cos \u03b3 1 h Mz sin \u03b3 (3.94) 3.6 Longitudinal Force The longitudinal slip ratio of a tire is s = Rg\u03c9w vx \u2212 1 (3.95) where, Rg is the tire\u2019s geometric and unloaded radius, \u03c9w is the tire\u2019s angular velocity, and vx is the tire\u2019s forward velocity. Slip ratio is positive for driving and is negative for braking. To accelerate or brake a vehicle, longitudinal forces must develop between the tire and the ground. When a moment is applied to the spin axis of the tire, slip ratio occurs and a longitudinal force Fx is generated at the tireprint. The force Fx is proportional to the normal force, Fx = Fx \u0131\u0302 (3.96) Fx = \u03bcx (s) Fz (3.97) where the coefficient \u03bcx (s) is called the longitudinal friction coefficient and is a function of slip ratio s as shown in Figure 3.31. The friction coefficient 128 3. Tire Dynamics reaches a driving peak value \u03bcdp at s \u2248 0.1, before dropping to an almost steady-state value \u03bcds. The friction coefficient \u03bcx (s) may be assumed proportional to s when s is very small \u03bcx (s) = Cs s s << 1 (3.98) where Cs is called the longitudinal slip coefficient. The tire will spin when s & 0.1 and the friction coefficient remains almost constant. The same phenomena happens in braking at the values \u03bcbp and \u03bcbs. Proof. Slip ratio, or simply slip, is defined as the difference between the actual speed of the tire vx and the equivalent tire speeds Rw\u03c9w. Figure 3.32 illustrates a turning tire on the ground. The ideal distance that the tire would freely travel with no slip is denoted by dF , while the actual distance the tire travels is denoted by dA. Thus, for a slipping tire, dA > dF , and for a spinning tire, dA < dF . The difference dF \u2212 dA is the tire slip and therefore, the slip ratio of the tire is s = dF \u2212 dA dA . (3.99) To have the instant value of s, we must measure the travel distances in an infinitesimal time length, and therefore, s \u2261 d\u0307F \u2212 d\u0307A d\u0307A . (3.100) 3. Tire Dynamics 129 If the angular velocity of the tire is \u03c9w then, d\u0307F = Rg\u03c9w and d\u0307A = Rw\u03c9w where, Rg is the geometric tire radius and Rw is the effective radius. Therefore, the slip ratio s can be defined based on the actual speed vx = Rw\u03c9w, and the free speed Rg\u03c9w s = Rg\u03c9w \u2212Rw\u03c9w Rw\u03c9w = Rg\u03c9w vx \u2212 1 (3.101) A tire can exert longitudinal force only if a longitudinal slip is present. Longitudinal slip is also called circumferential or tangential slip. During acceleration, the actual velocity vx is less than the free velocity Rg\u03c9w, and therefore, s > 0. However, during braking, the actual velocity vx is higher than the free velocity Rg\u03c9w and therefore, s < 0. The frictional force Fx between a tire and the road surface is a function of normal load Fz, vehicle speed vx, and wheel angular speed \u03c9w. In addition to these variables there are a number of parameters that affect Fx, such as tire pressure, tread design, wear, and road surface. It has been determined empirically that a contact friction of the form Fx = \u03bcx(\u03c9w, vx)Fz can model experimental measurements obtained with constant vx, \u03c9w. Example 98 Slip ratio based on equivalent angular velocity \u03c9eq. It is possible to define an effective angular velocity \u03c9eq as an equivalent angular velocity for a tire with radius Rg to proceed with the actual speed vx = Rg\u03c9eq. Using \u03c9eq we have vx = Rg\u03c9eq = Rw\u03c9w (3.102) 130 3. Tire Dynamics and therefore, s = Rg\u03c9w \u2212Rg\u03c9eq Rg\u03c9eq = \u03c9w \u03c9eq \u2212 1. (3.103) Example 99 Slip ratio is \u22121 < s < 0 in braking. When we brake, a braking moment is applied to the wheel axis. The tread of the tire will be stretched circumstantially in the tireprint zone. Hence, the tire is moving faster than a free tire Rw\u03c9w > Rg\u03c9w (3.104) and therefore, s < 0. The equivalent radius for a braking tire is more than the free radius Rw > Rg. (3.105) Equivalently, we may express the condition using the equivalent angular velocity \u03c9e and deduce that a braking tire turns slower than a free tire Rg\u03c9eq > Rg\u03c9w. (3.106) The brake moment can be high enough to lock the tire. In this case \u03c9w = 0 and therefore, s = \u22121. It shows that the longitudinal slip would be between \u22121 < s < 0 when braking. \u22121 < s < 0 for a < 0 (3.107) Example 100 Slip ratio is 0 < s <\u221e in driving. When we drive, a driving moment is applied to the tire axis. The tread of the tire will be compressed circumstantially in the tireprint zone. Hence, the tire is moving slower than a free tire Rw\u03c9w < Rg\u03c9w (3.108) and therefore s > 0. The equivalent radius for a driving tire is less than the free radius Rw < Rg. (3.109) Equivalently, we may express the condition using the equivalent angular velocity \u03c9e and deduce that a driving tire turns faster than a free tire Rg\u03c9eq < Rg\u03c9w. (3.110) The driving moment can be high enough to overcome the friction and turn the tire on pavement while the car is not moving. In this case vx = 0 3. Tire Dynamics 131 and therefore, s = \u221e. It shows that the longitudinal slip would be between 0 < s <\u221e when accelerating. 0 < s <\u221e for a > 0 (3.111) The tire speed Rw\u03c9w equals vehicle speed vx only if acceleration is zero. In this case, the normal force acting on the tire and the size of the tireprint are constant in time. No element of the tireprint is slipping on the road. Example 101 Power and maximum velocity. Consider a moving car with power P = 100 kW \u2248 134 hp can attain 279 km/h \u2248 77.5m/ s \u2248 173.3mi/h. The total driving force must be Fx = P vx = 100\u00d7 103 77.5 = 1290.3N. (3.112) If we assume that the car is rear-wheel-drive and the rear wheels are driving at the maximum traction under the load 1600N, then the longitudinal friction coefficient \u03bcx is \u03bcx = Fx Fz = 1290.3 1600 \u2248 0.806. (3.113) Example 102 Slip of hard tire on hard road. A tire with no slip cannot create any tangential force. Assume a toy car equipped with steel tires is moving on a glass table. Such a car cannot accelerate or steer easily. If the car can steer at very low speeds, it is because there is sufficient microscopic slip to generate forces to steer or drive. The glass table and the small contact area of the small metallic tires deform and stretch each other, although such a deformation is very small. If there is any friction between the tire and the surface, there must be slip to maneuver. Example 103 Samples for longitudinal friction coefficients \u03bcdp and \u03bcds. Table 3.2 shows the average values of longitudinal friction coefficients \u03bcdp and \u03bcds for a passenger car tire 215/65R15. It is practical to assume \u03bcdp = \u03bcbp, and \u03bcds = \u03bcbs. 132 3. Tire Dynamics Example 104 Friction mechanisms. Rubber tires generate friction in three mechanisms: 1\u2212 adhesion, 2\u2212 deformation, and 3\u2212 wear. Fx = Fad + Fde + Fwe. (3.114) Adhesion friction is equivalent to sticking. The rubber resists sliding on the road because adhesion causes it stick to the road surface. Adhesion occurs as a result of molecular binding between the rubber and surfaces. Because the real contact area is much less than the observed contact area, high local pressure make molecular binding, as shown in Figure 3.33. Bound occurs at the points of contact and welds the surfaces together. The adhesion friction is equal to the required force to break these molecular bounds and separate the surfaces. The adhesion is also called cold welding and is attributed to pressure rather than heat. Higher load increases the contact area, makes more bounds, and increases the friction force. So the adhesion friction confirms the friction equation Fx = \u03bcx (s) Fz. (3.115) The main contribution to tire traction force on a dry road is the adhesion friction. The adhesion friction decreases considerably on a road covered by water, ice, dust, or lubricant. Water on a wet road prevents direct contact between the tire and road and reduces the formation of adhesion friction. The main contribution to tire friction when it slides on a road surface is the viscoelastic energy dissipation in the tireprint area. This dissipative energy is velocity and is time-history dependent. Deformation friction is the result of deforming rubber and filling the microscopic irregularities on the road surface. The surface of the road has many peaks and valleys called asperities. Movement of a tire on a rough 3. Tire Dynamics 133 surface results in the deformation of the rubber by peaks and high points on the surface. A load on the tire causes the peaks of irregularities to penetrate the tire and the tire drapes over the peaks. The deformation friction force, needed to move the irregularities in the rubber, comes from the local high pressure across the irregularities. Higher load increases the penetration of the irregularities in the tire and therefore increases the friction force. So the deformation friction confirms the friction equation (3.115). The main contribution to the tire traction force on a wet road is the deformation friction. The adhesion friction decreases considerably on a road covered by water, ice, dust, or lubricant. Deformation friction exists in relative movement between any contacted surfaces. No matter how much care is taken to form a smooth surface, the surfaces are irregular with microscopic peaks and valleys. Opposite peaks interact with each other and cause damage to both surfaces. Wear friction is the result of excessive local stress over the tensile strength of the rubber. High local stresses deform the structure of the tire surface past the elastic point. The polymer bonds break, and the tire surface tears in microscopic scale. This tearing makes the wear friction mechanism. Wear results in separation of material. Higher load eases the tire wear and therefore increases the wear friction force. So the wear friction confirms the friction equation (3.115). Example 105 Empirical slip models. Based on experimental data and curve fitting methods, some mathematical equations are presented to simulate the longitudinal tire force as a function of longitudinal slip s. Most of these models are too complicated to be useful in vehicle dynamics. However, a few of them are simple and accurate enough to be applied. The Pacejka model, which was presented in 1991, has the form Fx (s) = c1 sin \u00a1 c2 tan \u22121 \u00a1c3s\u2212 c4 \u00a1 c3s\u2212 tan\u22121 (c3s) \u00a2\u00a2\u00a2 (3.116) where c1, c2, and c3 are three constants based on the tire experimental data. The 1987 Burckhardt model is a simpler equation that needs three numbers. Fx (s) = c1 \u00a1 1\u2212 e\u2212c2s \u00a2 \u2212 c3s (3.117) There is another Burckhardt model that includes the velocity dependency. Fx (s) = \u00a1 c1 \u00a1 1\u2212 e\u2212c2s \u00a2 \u2212 c3s \u00a2 e\u2212c4v (3.118) This model needs four numbers to be measured from experiment. By expanding and approximating the 1987 Burckhardt model, the simpler model by Kiencke and Daviss was suggested in 1994. This model is Fx (s) = ks s 1 + c1s+ c2s2 (3.119) 134 3. Tire Dynamics where ks is the slope of Fx (s) versus s at s = 0 ks = lim s\u21920 4Fs 4s (3.120) and c1, c2 are two experimental numbers. Another simple model is the 2002 De-Wit model Fx (s) = c1 \u221a s\u2212 c2s (3.121) that is based on two numbers c1, c2. In either case, we need at least one experimental curve such as shown in Figure 3.31 to find the constant numbers ci. The constants ci are the numbers that best fit the associated equation with the experimental curve. The 1997 Burckhardt model (3.118) needs at least two similar tests at two different speeds. Example 106 F Alternative slip ratio. An alternative method for defining the slip ratio is s = \u23a7\u23aa\u23aa\u23a8\u23aa\u23a9 1\u2212 vx Rg\u03c9w Rg\u03c9w > vx driving Rg\u03c9w vx \u2212 1 Rg\u03c9w < vx braking (3.122) where vx is the speed of the wheel center, \u03c9w is the angular velocity of the wheel, and Rg is the tire radius. In another alternative definition, the following equation is used for longitudinal slip: s = 1\u2212 \u00b5 Rg\u03c9w vx \u00b6n where n = \u00bd +1 Rg\u03c9w \u2264 vx \u22121 Rg\u03c9w > vx (3.123) s \u2208 [0, 1] In this definition s is always between zero and one. When s = 1, then the tire is either locked while the car is sliding, or the tire is spinning while the car is not moving. Example 107 F Tire on soft sand. Figure 3.34 illustrates a tire turning on sand. The sand will be packed when the tire passes. The applied stresses from the sand on the tire are developed during the angle \u03b81 < \u03b8 < \u03b82 measured counterclockwise from vertical direction. It is possible to define a relationship between the normal stress \u03c3 and tangential stress \u03c4 under the tire \u03c4 = (c+ \u03c3 tan \u03b8) \u00b3 1\u2212 e r k [\u03b81\u2212\u03b8+(1\u2212s)(sin \u03b8)\u2212sin \u03b81] \u00b4 (3.124) 3. Tire Dynamics 135 where s is the slip ratio defined in Equation (3.122), and \u03c4M = c+ \u03c3 tan \u03b8 (3.125) is the maximum shear stress in the sand applied on the tire. In this equation, c is the cohesion stress of the sand, and k is a constant. Example 108 F Lateral slip ratio. Analytical expressions can be established for the force contributions in x and y directions using adhesive and sliding concept by defining longitudinal and lateral slip ratios sx and sy sx = Rg\u03c9w vx \u2212 1 (3.126) sy = Rg\u03c9w vy (3.127) where vx is the longitudinal speed of the wheel and vy is the lateral speed of the wheel. The unloaded geometric radius of the tire is denoted by Rg and \u03c9w is the rotation velocity of the wheel. At very low slips, the resulting tire forces are proportional to the slip Fx = Csx sx (3.128) Fy = Csy sy (3.129) where Csx is the longitudinal slip coefficient and Csy is the lateral slip coefficient. 3.7 Lateral Force When a turning tire is under a vertical force Fz and a lateral force Fy, its path of motion makes an angle \u03b1 with respect to the tire-plane. The angle 136 3. Tire Dynamics is called sideslip angle and is proportional to the lateral force Fy = Fy j\u0302 (3.130) Fy = \u2212C\u03b1 \u03b1 (3.131) where C\u03b1 is called the cornering stiffness of the tire. C\u03b1 = lim \u03b1\u21920 \u2202 (\u2212Fy) \u2202\u03b1 = \u00af\u0304\u0304\u0304 lim \u03b1\u21920 \u2202Fy \u2202\u03b1 \u00af\u0304\u0304\u0304 (3.132) The lateral force Fy is at a distance ax\u03b1 behind the centerline of the tireprint and makes a moment Mz called aligning moment. Mz = Mz k\u0302 (3.133) Mz = Fy ax\u03b1 (3.134) For small \u03b1, the aligning moment Mz tends to turn the tire about the z-axis and make the x-axis align with the velocity vector v. The aligning moment always tends to reduce \u03b1. Proof. When a wheel is under a constant load Fz and then a lateral force is applied on the rim, the tire will deflect laterally as shown in Figure 3.35. The tire acts as a linear spring under small lateral forces Fy = ky\u2206y (3.135) with a lateral stiffness ky. The wheel will start sliding laterally when the lateral force reaches a maximum value FyM . At this point, the lateral force approximately remains constant and is proportional to the vertical load FyM = \u03bcy Fz (3.136) 3. Tire Dynamics 137 where, \u03bcy is the tire friction coefficient in the y-direction. A bottom view of the tireprint of a laterally deflected tire is shown in Figure 3.36. If the laterally deflected tire is turning forward on the road, the tireprint will also flex longitudinally. A bottom view of the tireprint for such a laterally deflected and turning tire is shown in Figure 3.37. Although the tire-plane remains perpendicular to the road, the path of the wheel makes an angle \u03b1 with tire-plane. As the wheel turns forward, undeflected treads enter the tireprint region and deflect laterally as well as longitudinally. When a tread moves toward the end of the tireprint, its lateral deflection increases until it approaches the tailing edge of the tireprint. The normal load decreases at the tail of the tireprint, so the friction force is lessened and the tread can slide back to its original position when leaving the tireprint region. The point where the laterally deflected tread slides back is called sliding line. A turning tire under lateral force and the associated sideslip angle \u03b1 are shown in Figure 3.38. Lateral distortion of the tire treads is a result of a tangential stress distribution \u03c4y over the tireprint. Assuming that the tangential stress \u03c4y is proportional to the distortion, the resultant lateral force Fy Fy = Z AP \u03c4y dAp (3.137) is at a distance ax\u03b1 behind the center line. ax\u03b1 = 1 Fy Z AP x \u03c4y dAp (3.138) The distance ax\u03b1 is called the pneumatic trail, and the resultant moment 138 3. Tire Dynamics Mz is called the aligning moment. Mz = Mz k\u0302 (3.139) Mz = Fy ax\u03b1 (3.140) The aligning moment tends to turn the tire about the z-axis and make it align with the direction of tire velocity vector v. A stress distribution \u03c4y, the resultant lateral force Fy, and the pneumatic trail ax\u03b1 are illustrated in Figure 3.38. There is also a lateral shift in the tire vertical force Fz because of slip angle \u03b1, which generates a slip moment Mx about the forward x-axis. Mx = \u2212Mx \u0131\u0302 (3.141) Mx = Fz ay\u03b1 (3.142) The slip angle \u03b1 always increases by increasing the lateral force Fy. However, the sliding line moves toward the tail at first and then moves forward by increasing the lateral force Fy. Slip angle \u03b1 and lateral force Fy work as action and reaction. A lateral force generates a slip angle, and a slip angle generates a lateral force. Hence, we can steer the tires of a car to make a slip angle and produce a lateral force to turn the car. Steering causes a slip angle in the tires and creates a lateral force. The slip angle \u03b1 > 0 if the tire should be turned about the z-axis to be aligned with the velocity vector v. A positive slip angle \u03b1 generates a negative lateral force Fy. Hence, steering to the right about the \u2212z-axis makes a positive slip angle and produces a negative lateral force to move the tire to the right. A sample of measured lateral force Fy as a function of slip angle \u03b1 for a constant vertical load is plotted in Figure 3.39. The lateral force Fy is linear for small slip angles, however the rate of increasing Fy decreases 3. Tire Dynamics 139 for higher \u03b1. The lateral force remains constant or drops slightly when \u03b1 reaches a critical value at which the tire slides on the road. Therefore, we may assume the lateral force Fy is proportional to the slip angle \u03b1 for low values of \u03b1. Fy = \u2212C\u03b1 \u03b1 (3.143) C\u03b1 = lim \u03b1\u21920 \u2202 (\u2212Fy) \u2202\u03b1 (3.144) The cornering stiffness C\u03b1 of radial tires are higher than C\u03b1 for non-radial tires. This is because radial tires need a smaller slip angle \u03b1 to produce the same amount of lateral force Fy. Examples of aligning moments for radial and non-radial tires are illustrated in Figure 3.40. The pneumatic trail ax\u03b1 increases for small slip angles up to a maximum value, and decreases to zero and even negative values for high slip angles. Therefore, the behavior of aligning momentMz is similar to what is shown in Figure 3.40. The lateral force Fy = \u2212C\u03b1 \u03b1 can be decomposed to Fy cos\u03b1, parallel to the path of motion v, and Fy sin\u03b1, perpendicular to v as shown in Figure 3.41. The component Fy cos\u03b1, normal to the path of motion, is called cornering force, and the component Fy sin\u03b1, along the path of motion, is called drag force. Lateral force Fy is also called side force or grip. We may combine the lateral forces of all a vehicle\u2019s tires and have them acting at the car\u2019s mass center C. 140 3. Tire Dynamics 3. Tire Dynamics 141 Example 109 Effect of tire load on lateral force curve. When the wheel load Fz increases, the tire treads can stick to the road better. Hence, the lateral force increases at a constant slip angle \u03b1, and the slippage occurs at the higher slip angles. Figure 3.42 illustrates the lateral force behavior of a sample tire for different normal loads. Increasing the load not only increases the maximum attainable lateral force, it also pushes the maximum of the lateral force to higher slip angles. Sometimes the effect of load on lateral force is presented in a dimensionless variable to make it more practical. Figure 3.43 depicts a sample. Example 110 Gough diagram. The slip angle \u03b1 is the main affective parameter on the lateral force Fy and aligning moment Mz = Fyax\u03b1. However, Fz and Mz depend on many other parameters such as speed v, pressure p, temperature, humidity, and road conditions. A better method to show Fz and Mz is to plot them versus each other for a set of parameters. Such a graph is called a Gough diagram. Figure 3.44 depicts a sample Gough diagram for a radial passenger car tire. Every tire has its own Gough diagram, although we may use an average diagram for radial or non-radial tires. Example 111 Effect of velocity. The curve of lateral force as a function of the slip angle Fy (\u03b1) decreases as velocity increases. Hence, we need to increase the sideslip angle at higher velocities to generate the same lateral force. Sideslip angle increases by 142 3. Tire Dynamics 3. Tire Dynamics 143 increasing the steer angle. Figure 3.45 illustrates the effect of velocity on Fy for a radial passenger tire. Because of this behavior, a fixed steer angle, the curvature of a one-wheel-car trajectory, increases by increasing the driving speed. Example 112 F A model for lateral force. When the sideslip angle is not small, the linear approximation (3.131) cannot model the tire behavior. Based on a parabolic normal stress distribution on the tireprint, the following third-degree function was presented in the 1950s to calculate the lateral force at high sideslips Fy = \u2212C\u03b1 \u03b1 \u00c3 1\u2212 1 3 \u00af\u0304\u0304\u0304 C\u03b1 \u03b1 FyM \u00af\u0304\u0304\u0304 + 1 27 \u00b5 C\u03b1 \u03b1 FyM \u00b62! (3.145) where FyM is the maximum lateral force that the tire can support. FyM is set by the tire load and the lateral friction coefficient \u03bcy. Let\u2019s show the sideslip angle at which the lateral force Fy reaches its maximum value FyM by \u03b1M . Equation (3.145) shows that \u03b1M = 3FyM C\u03b1 (3.146) and therefore, Fy = \u2212C\u03b1 \u03b1 \u00c3 1\u2212 \u03b1 \u03b1M + 1 3 \u00b5 \u03b1 \u03b1M \u00b62! (3.147) Fy FyM = 3\u03b1 \u03b1M \u00c3 1\u2212 \u03b1 \u03b1M + 1 3 \u00b5 \u03b1 \u03b1M \u00b62! . (3.148) 144 3. Tire Dynamics Figure 3.46 shows the cubic curve model for lateral force as a function of the sideslip angle. The Equation is applicable only for 0 \u2264 \u03b1 \u2264 \u03b1M . Example 113 F A model for lateral stress. Consider a tire turning on a dry road at a low sideslip angle \u03b1. Assume the developed lateral stress on tireprint can be expressed by the following equation: \u03c4y(x, y) = c\u03c4yM \u00b3 1\u2212 x a \u00b4\u00b5 1\u2212 x3 a3 \u00b6 cos2 \u00b3 y 2b \u03c0 \u00b4 (3.149) The coefficient c is proportional to the tire load Fz sideslip \u03b1, and longitudinal slip s. If the tireprint AP = 4 \u00d7 a \u00d7 b = 4 \u00d7 5 cm \u00d7 12 cm, then the lateral force under the tire, Fy, for c = 1 is Fy = Z AP \u03c4y(x, y) dA = Z 0.05 \u22120.05 Z 0.12 \u22120.12 \u03c4yM \u00b3 1\u2212 x 0.05 \u00b4\u00b5 1\u2212 x3 0.053 \u00b6 cos2 \u00b3 y\u03c0 0.24 \u00b4 dy dx = 0.0144\u03c4yM . (3.150) If we calculate the lateral force Fy = 1000N by measuring the lateral acceleration, then the maximum lateral stress is \u03c4yM = Fz 0.014 4 = 69444Pa (3.151) and the lateral stress distribution over the tireprint is \u03c4y(x, y) = 69444 \u00b3 1\u2212 x 0.05 \u00b4\u00b5 1\u2212 x3 0.053 \u00b6 cos2 \u00b3 y\u03c0 0.24 \u00b4 Pa. (3.152) 3. Tire Dynamics 145 3.8 Camber Force Camber angle \u03b3 is the tilting angle of tire about the longitudinal x-axis. Camber angle generates a lateral force Fy called camber trust or camber force. Figure 3.47 illustrates a front view of a cambered tire and the generated camber force Fy. Camber angle is assumed positive \u03b3 > 0, when it is in the positive direction of the x-axis, measured from the z-axis to the tire. A positive camber angle generates a camber force along the \u2212y-axis. The camber force is proportional to \u03b3 at low camber angles, and depends directly on the wheel load Fz. Therefore, Fy = Fy j\u0302 (3.153) Fy = \u2212C\u03b3 \u03b3 (3.154) where C\u03b3 is called the camber stiffness of tire. C\u03b3 = lim \u03b3\u21920 \u2202 (\u2212Fy) \u2202\u03b3 (3.155) In presence of both, camber \u03b3 and sideslip \u03b1, the overall lateral force Fy on a tire is a superposition of the corner force and camber trust. Fy = \u2212C\u03b3 \u03b3 \u2212 C\u03b1 \u03b1 (3.156) Proof. When a wheel is under a constant load and then a camber angle is applied on the rim, the tire will deflect laterally such that it is longer in 146 3. Tire Dynamics the cambered side and shorter in the other side. Figure 3.48 compares the tireprint of a straight and a cambered tire, turning slowly on a flat road. As the wheel turns forward, undeflected treads enter the tireprint region and deflect laterally as well as longitudinally. However, because of the shape of the tireprint, the treads entering the tireprint closer to the cambered side, have more time to be stretched laterally. Because the developed lateral stress is proportional to the lateral stretch, the nonuniform tread stretching generates an asymmetric stress distribution and more lateral stress will be developed on the cambered side. The result of the nonuniform lateral stress distribution over the tireprint of a cambered tire produces the camber trust Fy in the cambered direction. Fy = Fy j\u0302 (3.157) Fy = Z AP \u03c4y dA (3.158) The camber trust is proportional to the camber angle for small angles. Fy = \u2212C\u03b3 \u03b3 (3.159) The camber trust Fy shifts a distance ax\u03b3 forward when the cambered tire turns on the road. The resultant moment Mz = Mz k\u0302 (3.160) Mz = Fy ax\u03b3 (3.161) 3. Tire Dynamics 147 is called camber torque, and the distance ax\u03b3 is called camber trail. Camber trail is usually very small and hence, the camber torque can be ignored in linear analysis of vehicle dynamics. Because the tireprint of a cambered tire deforms to be longer in the cambered side, the resultant vertical force Fz Fz = Z AP \u03c3z dA (3.162) that supports the wheel load, shifts laterally to a distance ay\u03b3 from the center of the tireprint. ay\u03b3 = 1 Fz Z AP y \u03c3z dAp (3.163) The distance ay\u03b3 is called the camber arm, and the resultant moment Mx 148 3. Tire Dynamics is called the camber moment. Mx = Mx k\u0302 (3.164) Mx = \u2212Fz ay\u03b3 (3.165) The camber moment tends to turn the tire about the x-axis and make the tire-plane align with the z-axis. The camber arm ay\u03b3 is proportional to the camber angle \u03b3 for small angles. ay\u03b3 = Cy\u03b3 \u03b3 (3.166) Figure 3.49 shows the camber force Fy for different camber angle \u03b3 at a constant tire load Fz = 4500N. Radial tires generate lower camber force due to their higher flexibility. It is better to illustrate the effect of Fz graphically to visualize the camber force. Figure 3.50 depicts the variation of camber force Fy as a function of normal load Fz at different camber angles for a sample radial tire. If we apply a slip angle \u03b1 to a turning cambered tire, the tireprint will distort similar to the shape in Figure 3.51 and the path of treads become more complicated. The resultant lateral force would be at a distance ax\u03b3 and ay\u03b3 from the center of the tireprint. Both distances ax\u03b3 and ay\u03b3 are functions of angles \u03b1 and \u03b3. Camber force due to \u03b3, along with the corner force due to \u03b1, give the total lateral force applied on a tire. Therefore, the lateral force can be calculated as Fy = \u2212C\u03b1 \u03b1\u2212 C\u03b3 \u03b3 (3.167) 3. Tire Dynamics 149 150 3. Tire Dynamics that is acceptable for \u03b3 . 10 deg and \u03b1 . 5 deg. Presence of both camber angle \u03b3 and slip angle \u03b1 makes the situation interesting because the total lateral force can be positive or negative. Figure 3.52 illustrates an example of lateral force as a function of \u03b3 and \u03b1 at a constant load Fz = 4000N. Similar to lateral force, the aligning moment Mz can be approximated as a combination of the slip and camber angle effects Mz = CM\u03b1 \u03b1+ CM\u03b3 \u03b3. (3.168) For a radial tire, CM\u03b1 \u2248 0.013Nm/deg and CM\u03b3 \u2248 0.0003Nm/deg, while for a non-radial tire, CM\u03b1 \u2248 0.01Nm/deg and CM\u03b3 \u2248 0.001Nm/deg. Example 114 Banked road. Consider a vehicle moving on a road with a transversal slope \u03b2, while its tires remain vertical. There is a downhill component of weight, F1 = mg sin\u03b2, that pulls the vehicle down. There is also an uphill camber force due to camber \u03b3 \u2248 \u03b2 of tires with respect to the road F2 = C\u03b3 \u03b3. The resultant lateral force Fy = C\u03b3 \u03b3 \u2212 mg sin\u03b2 depends on camber stiffness C\u03b3 and determines if the vehicle goes uphill or downhill. Since the camber stiffness C\u03b3 is higher for non-radial tires, it is more possible for a radial tire to go downhill and a non-radial uphill. The effects of cambering are particularly important for motorcycles that produce a large part of the cornering force by cambering. For cars and trucks, the cambering angles are much smaller and in many applications their effect can be negligible. However, some suspensions are designed to make the wheels camber when the axle load varies. 3. Tire Dynamics 151 Example 115 Camber importance and tireprint model. Cambering of a tire creates a lateral force, even though there is no sideslip. The effects of cambering are particularly important for motorcycles that produce a large part of the lateral force by camber. The following equations are presented to model the lateral deviation of a cambered tireprint from the straight tireprint, and expressing the lateral stress \u03c4y due to camber y = \u2212 sin \u03b3 \u00b3q R2g \u2212 x2 \u2212 q R2g \u2212 a2 \u00b4 (3.169) \u03c4y = \u2212\u03b3k \u00a1 a2 \u2212 x2 \u00a2 (3.170) where k is chosen such that the average camber defection is correct in the tireprint Z a \u2212a \u03c4y dx = Z a \u2212a y dx. (3.171) Therefore, k = 3 sin \u03b3 4a3\u03b3 \u00b5 \u2212a q R2g \u2212 a2 +R2g sin \u22121 a Rg \u00b6 (3.172) \u2248 3 4 Rg q R2g \u2212 a2 a2 (3.173) and \u03c4y = \u2212 3 4 \u03b3 Rg q R2g \u2212 a2 a2 \u00a1 a2 \u2212 x2 \u00a2 . (3.174) 3.9 Tire Force Tires may be considered as a force generator with two major outputs: forward force Fx, lateral force Fy, and three minor outputs: aligning moment Mz, roll moment Mx, and pitch moment My. The input of the force generator is the tire load Fz, sideslip \u03b1, longitudinal slip s, and the camber angle \u03b3. Fx = Fx (Fz, \u03b1, s, \u03b3) (3.175) Fy = Fy (Fz, \u03b1, s, \u03b3) (3.176) Mx = Mx (Fz, \u03b1, s, \u03b3) (3.177) My = My (Fz, \u03b1, s, \u03b3) (3.178) Mz = Mz (Fz, \u03b1, s, \u03b3) (3.179) Ignoring the rolling resistance and aerodynamic force, and when the tire is under a load Fz plus only one more of the inputs \u03b1, s, or \u03b3, the major 152 3. Tire Dynamics output forces can be approximated by a set of linear equations Fx = \u03bcx (s) Fz (3.180) \u03bcx (s) = Cs s Fy = \u2212C\u03b1 \u03b1 (3.181) Fy = \u2212C\u03b3 \u03b3 (3.182) where, Cs is the longitudinal slip coefficient, C\u03b1 is the lateral stiffness, and C\u03b3 is the camber stiffness. When the tire has a combination of tire inputs, the tire forces are called tire combined force. The most important tire combined force is the shear force because of longitudinal and sideslips. However, as long as the angles and slips are within the linear range of tire behavior, a superposition can be utilized to estimate the output forces. Driving and braking forces change the lateral force Fy generated at any sideslip angle \u03b1. This is because the longitudinal force pulls the tireprint in the direction of the driving or braking force and hence, the length of lateral displacement of the tireprint will also change. Figure 3.53 illustrates how a sideslip \u03b1 affects the longitudinal force ratio Fx/Fz as a function of slip ratio s. Figure 3.54 illustrates the effect of sideslip \u03b1 on the lateral force ratio Fy/Fz as a function of slip ratio s. Figure 3.55 and 3.56 illustrate the same force ratios as Figures 3.53 and 3.54 when the slip ratio s is a parameter. Proof. Consider a turning tire under a sideslip angle \u03b1. The tire develops a lateral force Fy = \u2212C\u03b1 \u03b1. Applying a driving or braking force on this tire will reduce the lateral force while developing a longitudinal force Fx = \u03bcx (s) Fz. Experimental data shows that the reduction in lateral force in presence of a slip ratio s is similar to Figure 3.54. Now assume the sideslip \u03b1 is reduced to zero. Reduction \u03b1 will increase the longitudinal force while decreasing the lateral force. Increasing the longitudinal force is experimentally similar to Figure 3.55. A turning tire under a slip ratio s develops a longitudinal force Fx = \u03bcx (s) Fz. Applying a sideslip angle \u03b1 will reduce the longitudinal force while developing a lateral force. Experimental data shows that the reduction in longitudinal force in presence of a sideslip \u03b1 is similar to Figure 3.53. Now assume the slip ratio s and hence, the driving or breaking force is reduced to zero. Reduction s will increase the lateral force while decreasing the longitudinal force. Increasing the lateral force is similar to Figure 3.54. Example 116 Pacejka model. An approximate equation is presented to describe force Equations (3.175) 3. Tire Dynamics 153 154 3. Tire Dynamics 3. Tire Dynamics 155 or (3.176). This equation is called the Pacejka model. F = A sin \u00a9 B tan\u22121 \u00a3 Cx\u2212D \u00a1 Cx\u2212 tan\u22121 (Cx) \u00a2\u00a4\u00aa (3.183) A = \u03bcFz (3.184) C = C\u03b1 AB (3.185) B,D = shape factors (3.186) The Pacejka model is substantially empirical. However, when the parameters A, B, C, D, C1, and C2 are determined for a tire, the equation expresses the tire behavior well enough. Figure 3.57 illustrates how the parameters can be determined from a test force-slip experimental result. Example 117 Friction ellipse. When the tire is under both longitudinal and sideslips, the tire is under combined slip. The shear force on the tireprint of a tire under a combined slip can approximately be found using a friction ellipse model.\u00b5 Fy FyM \u00b62 + \u00b5 Fx FxM \u00b62 = 1 (3.187) A friction ellipse is shown in Figure 3.58. Proof. The shear force Fshear, applied on the tire at tireprint, parallel to the ground surface, has two components: the longitudinal force Fx and the lateral force Fy. Fshear = Fx \u0131\u0302+ Fy j\u0302 (3.188) Fx = Cs sFz (3.189) Fy = \u2212C\u03b1 \u03b1 (3.190) 156 3. Tire Dynamics These forces cannot exceed their maximum values FyM and FxM . FyM = \u03bcy Fz FxM = \u03bcx Fz The tire shown in Figure 3.58 is moving along the velocity vector v at a sideslip angle \u03b1. The x-axis indicates the tire-plane. When there is no sideslip, the maximum longitudinal force is FxM = \u03bcx Fz = \u2212\u2192 OA. Now, if a sideslip angle \u03b1 is applied, a lateral force Fy = \u2212\u2212\u2192 OE is generated, and the longitudinal force reduces to Fx = \u2212\u2212\u2192 OB. The maximum lateral force would be FyM = \u03bcy Fz = \u2212\u2212\u2192 OD when there is no longitudinal slip. In presence of the longitudinal and lateral forces, we may assume that the tip point of the maximum shear force vector is on the following friction ellipse: \u00b5 Fy FyM \u00b62 + \u00b5 Fx FxM \u00b62 = 1 (3.191) When \u03bcx = \u03bcy = \u03bc, the friction ellipse would be a circle and Fshear = \u03bcFz. (3.192) 3. Tire Dynamics 157 Example 118 Wide tires. A wide tire has a shorter tireprint than a narrow tire. Assuming the same vehicle and same tire pressure, the area of tireprint would be equal in both tires. The shorter tireprint at the same sideslip has more of its length stuck to the road than longer tireprint. So, a wider tireprint generates more lateral force than a narrower tireprint for the same tire load and sideslip. Generally speaking, tire performance and maximum force capability decrease with increasing speed in both wide and narrow tires. Example 119 sin tire forces model. A few decades ago, a series of applied sine functions were developed based on experimental data to model tire forces. The sine functions, which are explained below, may be used to model tire forces, especially for computer purposes, effectively. The lateral force of a tire is Fy = A sin \u00a9 B tan\u22121 (C\u03a6) \u00aa (3.193) \u03a6 = (1\u2212E) (\u03b1+ \u03b4)\u03bcFz (3.194) C = C\u03b1 AB (3.195) C\u03b1 = C1 sin \u00b5 2 tan\u22121 Fz C2 \u00b6 (3.196) A,B = Shape factors (3.197) C1 = Maximum cornering stiffness (3.198) C2 = Tire load at maximum cornerin stiffness (3.199) 3.10 Summary We attach a coordinate frame (oxyz) to the tire at the center of the tireprint, called the tire frame. The x-axis is along the intersection line of the tire-plane and the ground. The z-axis is perpendicular to the ground, and the y-axis makes the coordinate system right-hand. We show the tire orientation using two angles: camber angle \u03b3 and sideslip angle \u03b1. The camber angle is the angle between the tire-plane and the vertical plane measured about the x-axis, and the sideslip angle \u03b1 is the angle between the velocity vector v and the x-axis measured about the z-axis. A vertically loaded wheel turning on a flat surface has an effective radius Rw, called rolling radius Rw = vx \u03c9w (3.200) where vx is the forward velocity, and \u03c9w is the angular velocity of the 158 3. Tire Dynamics wheel. The effective radius Rw is approximately equal to Rw \u2248 Rg \u2212 Rg \u2212Rh 3 (3.201) and is a number between the unloaded or geometric radius Rg and the loaded height Rh. Rh < Rw < Rg (3.202) A turning tire on the ground generates a longitudinal force called rolling resistance. The force is opposite to the direction of motion and is proportional to the normal force on the tireprint. Fr = \u03bcr Fz (3.203) The parameter \u03bcr is called the rolling friction coefficient and is a function of tire mechanical properties, speed, wear, temperature, load, size, driving and braking forces, and road condition. The tire force in the x-direction is a combination of the longitudinal force Fx and the roll resistance Fr. The longitudinal force is Fx = \u03bcx (s) Fz (3.204) where s is the longitudinal slip ratio of the tire s = Rg\u03c9w vx \u2212 1 (3.205) \u03bcx (s) = Cs s s << 1 (3.206) The wheel force in the tire y-direction, Fy, is a combination of the lateral force and the tire roll resistance Fr. The lateral force is Fy = \u2212C\u03b3 \u03b3 \u2212 C\u03b1 \u03b1 (3.207) where \u2212C\u03b3\u03b3 is called the camber trust and C\u03b1\u03b1 is called the sideslip force. 3. Tire Dynamics 159 3.11 Key Symbols a \u2261 x\u0308 acceleration a, b semiaxes of AP ax\u03b1 ax\u03b3 camber trail ay\u03b3 camber arm AP tireprint area c1, c2, c3, c4 coefficients of the function Fx = Fx (s) C0, C1, C2 coefficients of the polynomial function Fr = Fr (vx) Cs longitudinal slip coefficient Csx , Csy longitudinal and lateral slip coefficients C\u03b1 sideslip coefficient C\u03b3 camber stiffness d distance of tire travel dF no slip tire travel dA actual tire travel D tire diameter E Young modulus f function fk spring force Fr Fr rolling resistance force Fx longitudinal force, forward force Fy lateral force FyM pneumatic trail Fz normal force, vertical force, wheel load g g gravitational acceleration k stiffness k1, k2, k3 nonlinear tire stiffness coefficients keq equivalent stiffness ks slope of Fx (s) versus s at s = 0 kx tire stiffness in the x-direction ky tire stiffness in the y-direction kz tire stiffness in the z-direction K radial and non-radial tires parameter in \u03bcr = \u03bcr (p, vx) m mass Mr Mr rolling resistance moment Mx, Mx roll moment, bank moment, tilting torque, My pitch moment, rolling resistance torque Mz yaw moment, aligning moment, self aligning moment, bore torque n exponent for shape and stress distribution of AP n1 number of tire rotations p tire inflation pressure P rolling resistance power 160 3. Tire Dynamics r radial position of tire periphery r = \u03c9/\u03c9n frequency ratio Rg geometric radius Rh loaded height Rw rolling radius s longitudinal slip sy lateral slip T wheel torque v \u2261 x\u0307, v velocity x, y, z, x displacement x, y, z coordinate axes 4x tire deflection in the x-direction, rolling resistance arm 4y tire deflection in the y-direction 4z tire deflection in the z-direction z\u0307 tire deflection rate in the z-direction \u03b1 sideslip angle \u03b1M maximum sideslip angle \u03b2 transversal slope \u03b3 camber angle \u03b4 deflection 4x tire deflection in the x-direction, rolling resistance arm 4y tire deflection in the y-direction 4z tire deflection in the z-direction \u03b8 tire angular rotation \u03bc0, \u03bc1 nonlinear rolling friction coefficient \u03bcr rolling friction coefficient \u03bcx (s) longitudinal friction coefficient \u03bcdp friction coefficient driving peak value \u03bcds friction coefficient steady-state value \u03c3zM maximum normal stress \u03c3z(x, y) normal stress over the tireprint \u03c3zm normal stress mean value \u03c4x(x, y), \u03c4y(x, y) shear stresses over the tireprint \u03c4xM , \u03c4yM maximum shear stresses \u03d5 contact angle, angular length of AP \u03c9eq equivalent tire angular velocity \u03c9w angular velocity of a wheel \u03c9w actual tire angular velocity 3. Tire Dynamics 161"
        ]
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure12.5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure12.5-1.png",
        "caption": "Figure 12.5 Simply supported beam with V and M diagrams due to a uniformly distributed full load.",
        "texts": [
            " \u2022 Rule 9 If we replace the load per field by its resultant and we draw the bending moment diagram due to these resultants, this bending moment diagram is tangent to the actual bending moment at the field boundaries. From the deformation symbol in the M diagram we observe tension at the Figure 12.6 Simply supported beam with the V and M diagrams due to the resultant of a uniformly distributed full load. We will illustrate this with three examples. Example 1 The first example relates to a simply supported beam AB with length . In Figure 12.5, the beam is over its entire length subjected to a uniformly distributed load q . In Figure 12.6, the beam is subjected to a concentrated force R = q at midspan C, to be seen as the resultant of the previously mentioned uniformly distributed load q . Question: Verify the rules 8 and 9. Solution: In the case of Figure 12.5, the shear force varies linearly and the bending moment varies parabolically. At midspan C, it applies that V = dM dx = 0. Here the M diagram has a horizontal tangent. This means that the bending moment M at C is a maximum (rule 7): Mmax = 1 8q 2. In the case in Figure 12.6, the shear force at C is not zero (as is often wrongly said), but rather changes sign. Here too the bending moment is a maximum (rule 7): Mmax = 1 4R = 1 4q 2. This maximum bending moment, twice as large as in the case of the uniformly distributed load, is a boundary extreme at the joining of fields AC and BC. 12 Bending Moment, Shear Force and Normal Force Diagrams 475 If the distributed load q in Figure 12.5 is replaced by its resultant R in Figure 12.6, the V and M diagrams change. The values at A and B do not change. Since the shear forces at A in both cases are equal, the slopes of the M diagram at A are also equal in both cases. The same holds for B. It is now clear that the tangents at A and B for the distributed load can be found by drawing the M diagram due to the resultant R. Both tangents intersect, in accordance with rule 8, at the middle of AB, on the line of action R. For the value p indicated in the M diagram, p = 1 8q 2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.37-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.37-1.png",
        "caption": "FIGURE 7.37. Bicycle model for a positive 4WS vehicle.",
        "texts": [
            " The case 1 < ls happens when the turning center is behind the car, and the case ls < \u22121 happens when the turning center is ahead of the car. Example 287 F FWS and Ackerman condition. When a car is FWS vehicle, then the Ackerman condition (7.1) can be written as the following equation. cot \u03b4fr \u2212 cot \u03b4fl = w l (7.130) Writing the Ackerman condition as this equation frees us from checking the inner and outer wheels. Example 288 F Turning radius. To find the vehicle\u2019s turning radius R, we may define equivalent bicycle models as shown in Figure 7.37 and 7.38 for positive and negative 4WS 420 7. Steering Dynamics vehicles. The radius of rotation R is perpendicular to the vehicle\u2019s velocity vector v at the mass center C. Let\u2019s examine the positive 4WS situation in Figure 7.37. Using the geometry shown in the bicycle model, we have R2 = (a2 + c2) 2 +R21 (7.131) cot \u03b4f = R1 c1 = 1 2 (cot \u03b4if + cot \u03b4of ) (7.132) and therefore, R = q (a2 + c2) 2 + c21 cot 2 \u03b4f . (7.133) Examining Figure 7.38 shows that the turning radius of a negative 4WS vehicle can be determined from the same equation (7.133). Example 289 F FWS and 4WS comparison. The turning center of a FWS car is always on the extension of the rear axel, and its steering length ls is always equal to 1. However, the turning center of a 4WS car can be: 7"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002797_j.jsv.2021.116029-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002797_j.jsv.2021.116029-Figure4-1.png",
        "caption": "Fig. 4. Contact relationship for ball-inner raceway contact.",
        "texts": [
            " Masjedi [23] proposed a method for estimating the shear friction force between the raceways and ball, which is represented by f s = ( Q L a 100 ) f c + lim p h ( 1 \u2212 exp (\u2212\u03b7a v g u s \u03c4lim h c )) \u03c0ab (14) The shear friction torque is represented as [30] m = \u2212 \u222b b \u2212b \u222b a \u2212a \u03b7a v g h c ( u y x \u2212 u x y ) dy dx (15) where, Q denotes the contact force between the raceways and ball; f c descibes the friction coefficient of asperity; L a denotes the asperity load ratio; lim is the ultimate shear stress coefficient; p h is the pressure sustained by the oil film; \u03b7avg is the average viscosity of oil lubrication which is represented as Eq. (24) ; u s is the absolute value of the relative sliding speed between the raceway and ball; \u03c4 lim is the ultimate shear stress of oil lubrication which is represented as Eq. (28) ; h c is the central oil film thickness which is represented as Eq. (29) ; a and b denote the ellipse contact area dimensions. Fig. 4 gives the contact relationship for ball-inner raceway contact. According to three self-rotation velocity \u03c9 xj , \u03c9 yj , and \u03c9 zj , and the revolution velocity \u03c9 cj , the relative sliding speed for ball-inner raceway contact in the minor ( x\u201d, which is the ball rolling direction) and major axes ( y\u201d) are given by [30] u ix = 0 . 5 d b [ \u03c9 y j sin \u03b1i j \u2212 \u03c9 z j cos \u03b1i j ] \u2212 ( \u03c9 i \u2212 \u03c9 c j )( 0 . 5 d m \u2212 0 . 5 d b cos \u03b1i j ) \u2212 \u03c9 si j y (16) u iy = \u22120 . 5 d b \u03c9 x j + \u03c9 si j x (17) Two components of the relative sliding speed for ball-outer raceway contact is defined by u ox = 0 "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003799_pnas.0703530104-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003799_pnas.0703530104-Figure1-1.png",
        "caption": "Fig. 1. Swimming paths r(t) in two dimensions for two different chemoattractant concentration fields. (a) In a linear concentration field c(x) c0 c1 x with constant gradient vector (blue arrow), the swimming path (black line) is a drifting circle. This drift can be described by the motion of the center of the circle R(t) (red line). This centerline encloses an angle of with the gradient direction. (b) In a radial concentration field c(x) c0/ x , the swimming path is a circle that drifts along a spiral to the center of the radial distribution (blue dot). The paths shown are numerical solutions to Eqs. 1\u20134 with parameters in (a) c1 0.1 c0/r0, 1 1/r0, 5 c0/ , v0 0.5 r0/ , and in (b) 1 1/r0, 0.5 c0/ , v0 0.5 r0/ , where c0, r0, set the concentration, length and time scales of the problem, respectively. In both cases, initial conditions were r(0) (10, 1) r0, a(0) 1, and p(0) 1/c(r(0)).",
        "texts": [
            " [4] Here, p(t) is an internal variable that governs adaptation, is a relaxation time, and has units of time per volume. For a time-independent stimulus s(t) s0, the system reaches a steady state with a 1 and p 1/s0. The system is adaptive because the steady-state output is independent of the stimulus level s0. Eqs. 1\u20134 determine a unique swimming path r(t) in a self-consistent manner for given initial conditions and the concentration field. Examples of swimming paths for linear and radial concentration fields are shown in Fig. 1. These paths can be described as circles whose centers drift along well defined trajectories. Small periodic variations of the stimulus s(t) s0 s1 cos 0t evoke a periodic response of the curvature (t) 0 s1 cos ( 0t ) (s1 2) with amplitude gain and phase shift . For the signaling system (Eq. 4), the linear response coefficient exp(i ) reads 1 s0 i 0 0 2 s0 / i 0 . [5] This linear response will play a key role to characterize swimming paths. Chemotaxis in Three-Dimensional Space. Far from any surface, sperm cells swim along helical paths if no chemoattractant is present (8)",
            " The response of the torsion to weak stimuli is characterized by a linear response coefficient exp(i ), which is defined analogously to the linear response coefficient of the curvature (Eq. 5). Motion in a Plane. Eqs. 1\u20134 for the swimming path r(t) can be solved numerically. For a linear concentration field c(x) c0 c1 x, the swimming paths are drifting circles. The overall motion of these circling paths is captured by the trajectory of the circle centers, which defines the centerline R(t). The centerline R(t) is oriented approximately at constant angle with respect to the direction of the concentration gradient c1 (see Fig. 1a). In a radial concentration field with c(x) c0 / x , the centerline R(t) is a spiral that circles toward the origin at x (0, 0) (see Fig. 1b). This choice of the radial decay is is motivated by the steady-state concentration field established in three dimensions by diffusion from a source. Note that the spiral shape of the centerline does not depend on the precise form of the radial decay. To understand these results, we consider the limit of weak gradients. We determine the swimming path in a linear concentration field c(x, y) c0 c1 x by a perturbation calculation in the small parameter r0 2c1, which describes the strength of the perturbation of the swimming path by chemotactic signaling"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure9.69-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure9.69-1.png",
        "caption": "Figure 9.69 The members for which we know the normal force are shown in bold. (a) The method of joints gets stuck at joints E and D, as more than two member forces are unknown. (b) The force in member 11 follows from the vertical equilibrium of joint H, (c) after which we can find the force in member 9 from the force equilibrium of joint G. The method of joints can now continue at D. (d) We could also switch to the method of sections to calculate the force in member 14. The method of joints can then be resumed at E.",
        "texts": [
            "5 provides a summary of all the member forces. 362 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Example 3 You are given the (Baltimore) truss beam in Figure 9.68. Question: Determine the member forces N1 to N15 using the method of joints. In which order should we handle the joint equilibrium? Solution: After first determining the support reactions from the equilibrium of the truss as a whole, we can determine the forces in members 1 to 6 from the equilibrium of joints A, B and C respectively. In the situation shown in Figure 9.69a we get stuck, as more than two member forces are unknown in both D and E. Since members 8 and 12, and 10 and 13 are in a direct line with one another, we can determine the forces in the members 11 and 9 from the equilibrium of joints H and G. The vertical equilibrium of joint H in Figure 9.70 gives N11 = 2F. We now have the situation as shown in Figure 9.69b. From the equilibrium in G in the direction normal to members 10 and 13 we find next (see Figure 9.70): N9 + 1 2N11 \u221a 2 = N9 + 1 2 \u00d7 2F \u00d7 \u221a 2 = 0 \u21d2 N9 = \u2212F \u221a 2. Now that N9 is known (see Figure 9.69c), we can find the remaining member forces by consecutively elaborating the equilibrium of joints D, E, G, H and L. The order in which we handle the joints is therefore A \u21d2 B \u21d2 C \u21d2 H \u21d2 G \u21d2 D \u21d2 E \u21d2 G \u21d2 H \u21d2 L. 9 Trusses 363 Instead of determining N9 and N11 from the equilibrium of joints H and G, it is far easier to revert to the method of sections. With the section shown in Figure 9.69d across members 12, 13 and 14, we can determine the force in member 14 from \u2211 Tz|K = 0. The other member forces are then found from the equilibrium for the successive joints E, D, G, H and L. In certain cases, it can be useful to switch from one method to the other at the right moment. Table 9.6 provides a summary of member forces N1 to N15. Table 9.6 Member forces Example 3. Mem. no. i Ni (kN) 1 +4F 2 \u22124F \u221a 2 3 +2F 4 +4F 5 \u2212F \u221a 2 6 \u22123F \u221a 2 7 +2F 8 +4F Mem. no. i Ni (kN) 9 \u2212F \u221a 2 10 +F \u221a 2 11 +2F 12 +4F 13 0 14 \u22124F 15 0 We can often shorten the calculation that needs to be done by first looking for zero-force members in a truss"
        ],
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    },
    {
        "image_filename": "designv10_0_0003486_elan.200403116-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003486_elan.200403116-Figure2-1.png",
        "caption": "Fig. 2. a) A tapping mode AFM image showing nanotube bundles assembled normal to the surface and b) a cross-section of the AFM image as indicated by the line of in part (a).",
        "texts": [
            " The wrapped eppendorf tube was shaken for 3 h at 60 RPM in a minishaker at 4 8C, washed with 0.1 M sodium acetate buffer (pH 5.6) and dried with nitrogen gas before measurement. The steps involved in forming the different electrode constructs described in this work are shown in Figure 1. Prior to attachment of the protein or FAD to the ends of the tubes the SWNT modified gold electrode can be thought of as an aligned nanoelectrode array as electrode as shown schematically in Figure 1 with AFM images of the electrode interface in Figure 2. Figure 2a shows aligned nanotubes protruding normal from the electrode surface, thus con- Electroanalysis 2005, 17, No. 1 2005 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim firming the schematic in Figure 1 is realized, and Figure 2b shows a cross-section of Figure 2a through the line shown. Note the heights represented in the cross-section are anomolously low compared with lengths determined by transmission electron microscopy which has been attributed to the inability of the AFM tip to penetrate between the tubes and reach the underlying substrate [42, 43]. Additional evidence that Figure 2 shows nanotubes protruding from the gold surface is that if the time the tubes are exposed to the cysteamine modified gold surface in the presence of DCC is increased, the higher the density of surface protrusions [44]. Furthermore, these AFM images are consistent with those presented previously for shortened SWNTs aligned in a similar manner[34, 36, 43, 45, 46]. It is evident from the AFM images that the SWNTs do not exist as individual tubes but as bundles of between 5 \u2013 20 tubes. The size of the bundles increases as the time of tube assembly increases"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure2.9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure2.9-1.png",
        "caption": "FIGURE 2.9. An accelerating car on inclined pavement.",
        "texts": [
            "117) The minimum time for 0\u2212 100 km/h on a level road for this front-wheeldrive car is t = 27.78 g\u03bcx a2 l + h\u03bcx \u2248 6. 21 s. (2.118) Now consider the same car to be four-wheel-drive. Then, the traction force is 2Fx1 + 2Fx2 = 2\u03bcx (Fz1 + Fz2) = g l m (a1 + a2) = ma. (2.119) and the minimum time for 0\u2212100 km/h on a level road for this four-wheeldrive car can theoretically be reduced to t = 27.78 g \u2248 2.83 s. (2.120) 2.4 Accelerating Car on an Inclined Road When a car is accelerating on an inclined pavement with angle \u03c6 as shown in Figure 2.9, the normal force under each of the front and rear wheels, Fz1 , Fz2 , would be: Fz1 = 1 2 mg \u00b5 a2 l cos\u03c6\u2212 h l sin\u03c6 \u00b6 \u2212 1 2 ma h l (2.121) Fz2 = 1 2 mg \u00b5 a1 l cos\u03c6+ h l sin\u03c6 \u00b6 + 1 2 ma h l (2.122) l = a1 + a2 56 2. Forward Vehicle Dynamics The dynamic parts, \u00b11 2mg hl a g , depend on acceleration a and height h of mass center C and remain unchanged, while the static parts are influenced by the slope angle \u03c6 and height h of mass center. Proof. The Newton\u2019s equation in x-direction and two static equilibrium equations, must be examined to find the equation of motion and ground reaction forces"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure4.15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure4.15-1.png",
        "caption": "Fig. 4.15: Grasping configurations and mechanical designs for two-finger grippers: a) rigid fingers for two-grasp contacts; b) articulated fingers without and with palm for four-grasp contacts; c) articulated fingers without and with palm for six-grasp contacts.",
        "texts": [
            " However, most multi-task capabilities can be devoted to a two-finger design solution whose successful operation strongly depends on a suitable programming of robot and gripper, once the grasping action has been analyzed in all its phases, as illustrated in the next section. Two-finger grippers are widely used because two-finger grasp can be used in most cases. Most of the objects that are used in human or industrial manipulations have a size that is comparable with the size of the human hand or gripping devices and they have a socalled \u2018regular geometry\u2019, which describes the shape of basic easily graspable figures. Figure 4.15 summarizes feasible mechanical designs for two-finger grippers. In general, fingers can be designed as articulated fingers to mimic the human hand, as shown in Fig. 4.15. b) and c), or as coupler links to use planar linkages as gripping mechanisms, as shown in Fig. 4.15. a). A further peculiarity can be recognized in flexible tips for the contacts, which are installed on fingertips and finger phalanges with a suitable curvature in order to be easily adapted to the shape of specific objects, as schematically shown in Fig. 4.15 by small grey tips. These tips can be of flexible material to help the shape adaptation and mainly to enlarge the contact surface so that the contact pressure can be limited to avoid surface Fundamentals of the Mechanics of Robotic Manipulation 255 damage of the grasped object. Indeed, the need for a suitable low value of contact pressure is a requirement to adopt two, three, four, and even more contacts, although static equilibrium for grasping can be ensured with two contacts only in planar cases",
            "16e) dynamic force may change the equilibrium but still the object is statically grasped by the fingers; in Fig. 4.16a, f) an external disturbance, such as an impact with another object in the Chapter 4 Fundamentals of the Mechanics of Grasp256 environment, may change the equilibrium and the object may move to another static equilibrium configuration between the fingers. The above-mentioned analysis can be outlined similarly for any other case of grasping devices, even with multiple contacts, such as those in Fig. 4.15. The analysis has been focused on the interactions among object and fingers through a purely mechanical viewpoint, but the feasibility of finger configurations strongly depends on the mechatronic gripper design and operation that guide the finger motion and action. The analysis of Fig. 4.16 refers to planar grasp but it can also be extended to threedimensional cases by looking similarly at the phases in planes of a Cartesian frame reference for a three-dimensional grasp. Figure 4.17 has been modeled to summarize all the phases of a grasping action and considers the mechanics of grasp through all the aspects shown in Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure6.30-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure6.30-1.png",
        "caption": "Fig. 6.30 The answer for the quiz of page 207: To keep the rod stationary, a person must apply the force \u2212f as well as the moment \u03c4 . If the person who is gripping the handle is on a small boat, he/she will start moving in the direction of vector f .",
        "texts": [],
        "surrounding_texts": [
            "210 6 Dynamic Simulation",
            "1. Aldebaran robotics, http://www.aldebaran-robotics.com 2. Carnegie mellon university TARTAN RESCUE,\nhttp://www.rec.ri.cmu.edu/projects/tartanrescue/\n3. DARPA Robotics Challenge, http://www.theroboticschallenge.org 4. Products of Kondo Kagaku Co. Ltd. (in Japanese),\nhttp://kondo-robot.com/product/\n5. Broad Agency Announcement DARPA Robotics Challenge, DARPA-BAA-1239 (April 2012) 6. Takanishi, A., Ishida, M., Yamazaki, Y., Kato, I.: The realization of dynamic walking by the biped walking robot. In: Proceedings of IEEE Int. Conf. on Robotics and Automation, pp. 459\u2013466 (1985) 7. Goswami, A., Espiau, B., Keramane, A.: Limit cycles in a passive compass gait biped and passivity-mimicking control laws. Journal of Autonomous Robots 4(3), 273\u2013286 (1997) 8. Harashima, A.: Mechanics I, II. Shokabo (1972) (in Japanese) 9. Herdt, A., Diedam, H., Wieber, P.-B., Dimitrov, D., Mombaur, K., Diehl,\nM.: Online walking motion generation with automatic footstep placement. Advanced Robotics 24, 719\u2013737 (2010) 10. Takanishi, A., Ishida, M., Yamazaki, Y., Kato, I.: The realization of dynamic walking by the biped walking robot WL-10RD. In: Proceedings of 1985 International Conference on Advanced Robotics (ICAR), pp. 459\u2013466 (1985) 11. Takanishi, A., Egusa, Y., Tochizawa, M., Takeya, T., Kato, I.: Realization of dynamic biped walking stabilized with trunk motion. In: Proceedings of RoManSy 7: 7th CISM-IFToMM Symposium on Theory and Practice of Robots and Manipulators, pp. 68\u201379 (1990) 12. Cho, B.K., Park, S.S., Oh, J.H.: Controllers for running in the humanoid robot, HUBO. In: Proceedings of IEEE-RAS International Conference on Humanoid Robots, pp. 385\u2013390 (2009) 13. Lim, B., Lee, J., Kim, J., Lee, M., Kwak, H., Kwon, S., Lee, H., Kwon, W., Roh, K.: Optimal gait primitives for dynamic bipedal locomotion. In: Proceedings of the IEEE/RSJ International Conference on Intelligent Robots and Systems, pp. 4013\u20134018 (2012) 14. Thuilot, B., Goswami, A., Espiau, B.: Bifurcation and chaos in a simple passive bipedal gait. In: Proceedings of the 1997 IEEE International Conference on Robotics & Automation, pp. 792\u2013798 (1997)"
        ]
    },
    {
        "image_filename": "designv10_0_0000512_027836499801701202-Figure15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000512_027836499801701202-Figure15-1.png",
        "caption": "Fig. 15. The divergence of nearby trajectories in the phase plane of a biped robot, a phenomenon normally observed in a system with chaotic dynamics. The initial 9ns between the two trajectories differs by 0.00 1 rad, while the other three state variables are exactly identical. Even after only two steps (a), the difference between the trajectories (shown in solid and dotted lines) is clearly identifiable. After 12 steps (b), the trajectories bear no resemblance to each other although they lie on the same attractor.",
        "texts": [],
        "surrounding_texts": [
            "1296\na Euclidean plane. This is a consequence of the strong phasespace volume contraction, which is discussed in Section 3.4.\nThe gradual progression of the robot gait to the chaotic regime is well depicted in the first return maps of 6ns,k+t = .f (ens,k)~ shown in Figure 13. When 0 = 5.030, the gait is eight-periodic, and its first return map consists of eight points (as was shown in Fig. 13a). At 0 = 5.04\u00b0, the first return map still consists of eight distinguishable clusters of points (Fig. 13a). Through multiple period-doubling bifurcations, this eight-period gait gives rise to a 2&dquo;-period gait with a large n. This gait still preserves some similarity with the eight-period gait from which it originates. For example, step order is still preserved, and 8ns always visits the eight clusters, as shown in Figure 13a, in the same order. In this order, a large 6ns (i.e., lens I > .4 rad) is always followed by a small 8ns (i.e., l ens I < .4 rad).\nWhen 0 = 5.08\u00b0, the eight clusters of points merge into two larger packs; see Figure 13b. Some order is still preserved, since a large 9ns is still always followed by a small one. The same property continues to hold for 0 = 5.12\u00b0, but in this case the first return map appears as a continuum of points (Fig. 13c). We are therefore very close to the broad-band frequency characteristic typical of chaotic behavior. Finally, when 0 = 5.2, we observe that predictability and periodicity have been completely destroyed, since a large 6ns can be followed by another large one. The layered structure of the strange attractor can also be guessed from the first return map.\nIt is extremely interesting to note that the first return maps of all of the robot-gait descriptors look remarkably similar. For instance, the first return map of the step period T (Fig. 14) looks like a scaled and rotated first return map of 8ns (Fig. 13d). With this we can suggest that all of the characteristics of the passive chaotic gait of our robot are somehow\nensconced in the shape of its first return map, which can be viewed as a signature of the chaotic gait.\nAn important characteristic of chaotic motion is the exponential divergence of nearby trajectories (Hilborn 1994). The phase-space trajectories of our biped robot exhibit this behavior, and this section ends with a presentation of this fact. Figures 15a and 15b show the divergence of trajectories initiating from nearby phase-space points and continuing for 2 and 12 steps, respectively. The initial states of the two trajectories are\ncorresponding to the dashed line, and\nat CORNELL UNIV on November 6, 2014ijr.sagepub.comDownloaded from",
            "1297\ncorresponding to the solid line. After just 2 steps, the difference in the trajectories is clearly identifiable; after 12 steps, the trajectories bear no semblance to each other although they lie on the same attractor.\n5. Dampers Improve Gait Stability\nTaking a cue from the connection between gait stability and energy dissipation, we studied the effect of placing passive damping elements in the robot\u2019s hip joint. A significant improvement of the gait stability and overall gait versatility was achieved by this without violating the &dquo;passive&dquo; status of the robot. The damper affects a continuous dissipation of energy in the robot in addition to the energy dissipated intermittently during ground impact. Although even a linear damper may increase the range of slopes on which steady gaits exist, we obtained more encouraging results with quadratic dampers. It is possible to consider the damper coefficient as another parameter affecting the robot gait; however, we give it a special status here because of its obvious connection to control laws.\nIndeed, the passive quadratic damper placed in the robot\u2019s hip joint can be easily replaced by a motor implementing the same physical law. The damper generates a hip-joint torque uH OC - 62 H sign(6H) = _(6n, _ 6,)2sign(6.., - 6,).\nThe presence of a damper profoundly alters the passive gait of the robot on a given slope.5 In Figure 16 we show three different limit cycles of the robot on a 4\u00b0 slope-one for the damperless gait (the solid line) and the other two corresponding to two different quadratic dampers (the dashed line for 0.08 Nm/(rad/sec)2, and the innermost dotted-line cycle for 0.15 Nm/(rad/sec)2). A robot equipped with a damper may exhibit steady gaits for a larger range of slopes. Fig 17 shows the steady stable gait (asymmetric) for a 10\u00b0 slope obtained with a quadratic hip damper with a coefficient of 0.23 Nm/(rad/sec)2. For a damperless robot, no gait cycle was found beyond a 5.2\u00b0 slope.\nFigure 18 plots the robot\u2019s kinetic energy (KE) versus its potential energy (PE) during a few cycles on 4\u00b0 slope; Figure 18a corresponds to the damperless motion, and Figure 18b corresponds to the motion with a 0.15-Nm/rad/sec damper. Let us concentrate on the first figure. As reported by Goswami, Espiau, and Keramane (1996), the KE versus PE plot of a damperless compass robot consists of two straight lines: a constant-PE horizontal line representing the ground impact, and an inclined line representing the swing stage. The conservation of total mechanical energy of the robot during the swing stage ensures that the inclined line makes a 135\u00b0 angle with the KE axis. In Figure 18a we see that the PE of the robot decreases by a constant amount at every step. For\n5. Dampers may increase the size of the limit cycle\u2019s basin of attraction for a given slope, but a better tool for estimating the size of the basin is needed to prove it.\nat CORNELL UNIV on November 6, 2014ijr.sagepub.comDownloaded from",
            "1298\na steady gait, precisely this amount of KE is absorbed during the ground impact.\nIn the presence of a damper, the situation is different. Since the damper continuously dissipates energy during the swing stage, the latter is no longer Hamiltonian, and is therefore not represented by a single inclined line. It becomes a curve, the exact nature of which depends on the robot dynamics and the damper coefficient (see Fig. 18b). To exhibit a steady gait, the PE lost by the robot during a step must be exactly equal to the sum of the energy absorbed during the ground impact and the energy absorbed by the damper during the\nat CORNELL UNIV on November 6, 2014ijr.sagepub.comDownloaded from"
        ]
    },
    {
        "image_filename": "designv10_0_0003908_978-1-84882-730-1-Figure14.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003908_978-1-84882-730-1-Figure14.6-1.png",
        "caption": "Figure 14.6 General framework of mobile robot path-following",
        "texts": [
            "1 Control Objective For the reader\u2019s convenience, we rewrite the equations of motion here, see (14.1): P D J . /!; M P!CC . P /!CD! D : (14.69) After an observer and a primary control design as in Section 14.1.3.2, we only need to consider the robot model (i.e., the dynamics (14.42)): Px D cos. / OvC cos. / Qv; Py D sin. / OvC sin. / Qv; P D OwC Qw; POv D vc C\u02ddv; POw D wc C\u02ddw : (14.70) In this section, we consider a control objective of designing the control vector to force the mobile robot to follow a specified path , see Figure 14.6. If we are able to drive the robot to closely follow a virtual robot that moves along the path with a desired speed v0, which is tangential to the path, then the control objective is fulfilled, i.e., the robot is in a tube of nonzero diameter centered on the reference path and moves along the specified path at the speed v0. Roughly speaking, the approach is to steer the robot such that it heads toward the virtual robot and eliminates the distance between itself and the virtual robot. We define the following variables to mathematically formulate the control objective: xe D xd x; ye D yd x; e D d ; ze Dp x2e Cy2e ; (14"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure8.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure8.1-1.png",
        "caption": "Fig. 8.1. Free rotating disc: (a) in laminar flow, (b) transition from laminar to turbulent flow.",
        "texts": [
            " In this section the convection heat transfer coefficients in different parts of the AFPM machine are evaluated, exploiting a number of existing models. Free rotating disc The average heat transfer coefficient at the outside surface of a rotating disc may be evaluated using the formula developed for a free rotating disc [270], i.e. h\u0304 = k R Nu (8.5) where R is the radius of the disc and the average Nusselt number Nu is given according to the different flow conditions as follows: (i) For combined effects of free convection and rotation in laminar flow (Fig. 8.1a) [270] Nu = 2 5 (Re2 +Gr) 1 4 (8.6) Gr = \u03b2gR3\u03c03/2\u2206\u03d1 \u03bd2 (8.7) where Re is the Reynolds number according to eqn (2.72), \u03b2 is the coefficient of thermal expansion, \u03bd is the kinematic viscosity of the fluid (m2/s) and \u2206\u03d1 8.2 Heat Transfer Modes 255 is the temperature difference between the disc surface and surrounding air. (ii) For a combination of laminar and turbulent flow with the transition at a radius rc (Fig. 8.1b) [270] Nu = 0.015Re 4 5 \u2212 100( rc R )2 (8.8) where rc = (2.5\u00d7 105\u03bd/\u2126)1/2 (8.9) The angular speed \u2126 = 2\u03c0n when n is the rotational speed in rev/s. It is instructive to compare the heat transfer capabilities between a rotation disc and a stationary disc. If we consider a steel disc, which has a diameter of 0.4 m and rotates at 1260 rpm., the convection heat transfer coefficient may be calculated as 41 W/(m2 oC), which is about ten times that of the same disc at standstill. Alternatively, one can say that the effective heat dissipation area of the same disc can be increased by a factor of 10 when the disc rotates at the specific speed",
            " 262 8 Cooling and Heat Transfer The loss coefficient for a pipe is given by \u03bbL/d where \u03bb is a friction factor obtained as a function of Reynolds number Re and surface roughness from a Moody diagram [190]. To facilitate numeric calculations, the Moody diagram may be represented by [61]: \u03bb = 8{(8/Re)12 + (X + Y )\u2212 3 2 } 1 12 X = {2.457 ln{(7/Re)0.9 + 0.27\u03b3/D}\u22121}16 Y = {37530/Re}16 (8.27) where Re = \u03c1DhQ \u00b5A and where \u03b3 is the equivalent sand grain roughness [61]. Characteristics It is now possible to relate the theoretical prediction obtained from the ideal flow model to the actual characteristic by accounting for the various losses discussed above. Assuming that the AFPM machine (shown in Fig. 8.1) operates at a constant speed of 1200 rpm, the ideal developed pressure characteristic for a radial channel is a function described by eqn (8.22) as shown in Fig. 8.6. After introducing the slip factor, the resultant curve is shown as a dotted line as 8.3 Cooling of AFPM Machines 263 eqn (8.24). It was not possible to obtain a suitable correlation in the literature [248] for the pressure loss due to shock and leakage as was the case for the slip. The calculated characteristic curve without considering shock and leakage losses, i"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure14.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure14.6-1.png",
        "caption": "FIGURE 14.6. A 1/8 car model moving with speed v on a wavy road.",
        "texts": [
            " So, the equivalent spring keq and damper ceq are keq = k \u00b3a b cos\u03b1 \u00b42 (14.35) ceq = c \u00b3a b cos\u03b1 \u00b42 . (14.36) 14. Suspension Optimization 889 For example assume that we have determined the following stiffness and damping as a result of optimization. keq = 9869.6N/m (14.37) ceq = 87.965N s/m. (14.38) The actual k and c for a McPherson suspension with a = 19 cm (14.39) b = 32 cm (14.40) \u03b1 = 27deg (14.41) would be k = 28489N/m (14.42) c = 253.9N s/m. (14.43) Example 499 Wavy road and excitation frequency. Figure 14.6 illustrates a 1/8 car model moving with speed v on a wavy road with length d1 and peak-to-peak height d2. Assuming a stiff tire with a small radius compared to the road waves, we may consider y as the fluctuation of the road. The required time to pass one length d1 is the period of the excitation T = d1 v (14.44) 890 14. Suspension Optimization which can be used to find the frequency of excitation \u03c9 = 2\u03c0 T = 2\u03c0v d1 . (14.45) Therefore, the excitation y = Y sin\u03c9t is y = d2 2 sin 2\u03c0v d1 t. (14.46) Example 500 Function of an isolator The function of an isolator is to reduce the magnitude of motion transmitted from a vibrating foundation to the equipment, or to reduce the magnitude of force transmitted from the equipment to its foundation, both in time and frequency domain"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure4.39-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure4.39-1.png",
        "caption": "Fig. 4.39: Force sensored fingertip: a) a scheme; b) mechanical design of a prototype.",
        "texts": [
            "38 indicating the main sub-systems is shown the composition of the test-bed for two-finger grippers. In Fig. 4.38 an additional block is included to monitor the grasping force through the same force sensor for the force control. Suitable sensor equipment can be used as based on Labview virtual instruments. A PLC has been used to program the operation of the system by implementing the control scheme of Fig. 4.36, after suitable calibration start-up. The PC of the monitoring system can also be used for the programming of the PLC. Figure 4.39 shows the design of a force sensored fingertip that can be easily changed. The control block in Fig. 4.38 is obtained by using market pneumatic components, as shown in the scheme of Fig. 4.36. Fundamentals of the Mechanics of Robotic Manipulation 293 Figure 4.40 shows some illustrative results that have been obtained in experimental tests when a small plastic bottle of water has been grasped, as also shown in Fig. 4.37. It is worthy of note in the plots of Fig. 4.40 that the grasping force increases slowly and with a suitably smooth path to the equilibrium value that has been calculated a priori and imposed through the PLC program"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002277_j.prostr.2016.02.039-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002277_j.prostr.2016.02.039-Figure5-1.png",
        "caption": "Fig. 5 \u2013 TO solution (left) and optimized component final design (right).",
        "texts": [],
        "surrounding_texts": [
            "The mesh has great influence in the final solution. Highly refined meshes give very different topologies from less refined meshes. In the final design domain, a more controlled method for meshing was used in order to ensure its high quality. The control parameters also have great influence not only in the solution convergence degree, but also in the computing time, thus a convergence study on these parameters was run. In this optimisation there were two control parameters which were studied, the Relative Convergence Criterion (RCC) and the Discreteness Parameter (DP) which is the equivalent for penalty factor in TO theory. The objective function was the weighted compliance in order to consider the three load cases in the topology optimisation [HyperWorks Guide]. This response is given by Equation 1 \ud835\udc36\ud835\udc36\ud835\udc64\ud835\udc64 = \u2211\ud835\udc64\ud835\udc64\ud835\udc56\ud835\udc56\ud835\udc36\ud835\udc36\ud835\udc56\ud835\udc56 (1) where wi is the weight and Ci is the compliance of load case i which is given by Equation 2 Ci = 1 2 ui Tfi (2) where ui and fi are the displacement and force vectors, respectively, corresponding to load case i. The objective was the minimization of the weighted compliance and each load case was given the same weight. There were two constraints defined in the optimisation. The first one was regarding the volume fraction of the design domain. The second was a symmetry constraint, forcing the optimised solution to be symmetric with respect to the component\u2019s mid vertical plane as the original component is. Figure 3 and Figure 4 illustrate the TO boundary conditions and final solution pseudo-density distribution. Fig. 3 \u2013 Final Design Domain (blue) and boundary conditions. Fig. 4 \u2013 TO solution with element pseudo-density distribution. Author name / Structural Integrity Procedia 00 (2016) 000\u2013000 5"
        ]
    },
    {
        "image_filename": "designv10_0_0002567_j.jmatprotec.2021.117117-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002567_j.jmatprotec.2021.117117-Figure7-1.png",
        "caption": "Fig. 7. (a\u2013e) Prototype design of Ti-6Al-4 V to Nb gradient for rocket (Hofmann et al., 2014). Reproduced with permission from Cambridge University Press.",
        "texts": [
            " Thus, \u03b2-isomorphous elements are subject to this section to be added to titanium for producing a gradient from \u03b1-Ti to \u03b2-Ti to pure refractory metals. This is a unique advantage for both aerospace and biomedical applications, which other metals cannot deliver. There are some research studies on biomedical applications Ti-Mo and Ti-V gradients. However, they are limited to a maximum increment of 25 wt.% Mo/V. Some researchers investigated the transition to 100 wt.% refractory metals. Table 1 summarizes various gradients from titanium-based alloys to refractory metals together with some key processing parameters. Fig. 7 shows a potential application of a gradient from titanium to pure refractory metals as a rocket-nozzle. Fig. 7a shows a schematic design of a rocket. The main body is made up of titanium, but the nozzle is made of refractory metal as the nozzle experiences the maximum heat load. The nozzle section was only fabricated as a prototype of this design, shown in Fig. 7b and c. The prototype has been manufactured by the LDED process, as shown in Fig. 7d and e. The nozzle section is made of pure Nb while it is linearly graded to Ti-6Al-4 V after the neck (Hofmann et al., 2014). Schneider-Maunoury et al. (2017) investigated functionally graded Ti-6Al-4 V to pure Mo. This steep transition was performed using Mo increment of 25 wt.%. Fig. 8 shows EBSD maps at four interfaces. In the first transition of adding 25 wt.% Mo, the \u03b2 grain was rebuild from the \u03b1 grain orientation. EBSD maps show good continuity of \u03b2 grains at the interface. On the side made of 100 wt"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure9.44-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure9.44-1.png",
        "caption": "Figure 9.44 The isolated K-truss with support reactions.",
        "texts": [
            " 346 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Example 3 The third example relates to the somewhat more complicated truss in Figure 9.43, a so-called K-truss. This type of truss is sometimes used as wind bracing in bridges. Here, the K-truss has four fields and is loaded by two vertical forces of 120 kN and a horizontal force of 240 kN. Question: Determine the forces in members 7 to 13, with the correct sign for tension and compression. In the calculations, use the coordinate system given. Solution: In Figure 9.44, the truss has been isolated and the support reactions have been shown. Using the method of sections, we now encounter the difficulty that, for most of the members, no section can be found that intersects only three members. Sometimes it is possible to determine a member force if the section passes through more than three members, but in most cases, additional information is required that has to be obtained by selecting a section in a clever way, or by considering a combination of sections. Since the top chord and bottom chord members are easiest to determine, we will start with them"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure3.49-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure3.49-1.png",
        "caption": "Fig. 3.49: The F-200iB commercialized by FANUC. (Photo taken from FANUC webpage).",
        "texts": [
            " The Hexel company sells the Hexabot for applications such as: flexible fixturing for welding, polishing, deburringg grinding; transfer lines; precision positioning; high speed pick and place for packaging, loading/unloading; programmable welding; motion simulation for animation and exhibitions; testing equipment. Chapter 3: Fundamentals of the Mechanics of Robots186 In Fig. 3.48 a commercialized 3-d.o.f.s parallel manipulator, named \u2018Tricept\u2019, is shown as a robotic arm carrying a suitable wrist. The SMT Tricept 850 that is shown in Fig. 3.48 is a 5-axis machine with a declared positioning accuracy of \u00b150\u03bcm and repeatability of \u00b110\u03bcm within the workspace. Its maximum velocity is 90 m/min and its acceleration is 2 g. The F-200iB robot in Fig. 3.49 is a 6-d.o.f.s servo-driven parallel manipulator commercialized by FANUC. It has been designed for manufacturing and automotive assembly. Fanuc F-200iB has a repeatability of \u00b1 0.1 mm. Currently it is used for several applications, primarily spot welding. Maximum declared velocities are 1,500 mm/sec on X and Y directions and 300 mm/sec on Z direction. Fundamentals of Mechanics of Robotic Manipulation 187 The Eclipse parallel manipulator in Fig. 3.50 has been conceived at Seoul National University"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000138_6.2007-6461-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000138_6.2007-6461-Figure5-1.png",
        "caption": "Figure 5. Free body diagram of the moments and forces acting on rotor i. Due to the effect of \u201cblade flapping\u201d, explained in Section V, the thrust and moments that were depicted in Figure 4 are not necessarily orthogonal to the body axes. They undergo deflection depending on the velocity of the vehicle and the direction of spin of the specific rotor. The unit vector ever is parallel to the body vertical direction, zB, elon is parallel to the velocity vector, ev, and elat is orthogonal to both. The thrust is deflected by small angles a1s,i and b1s,i. This deflection also results in reaction moments about the rotor hub.",
        "texts": [
            " Euler angles of the body axes are {\u03c6, \u03b8, \u03c8} with respect to the eN, eE and eD axes, respectively, and are referred to as roll, pitch and yaw. The current velocity direction unit vector is ev, in inertial coordinates, and defines coordinates relative to the c.g. referred to as longitudinal, lateral and vertical. The rotor plane does not necessarily align with the xB, yB plane, so for the ith rotor let {xR,i,yR,i, zR,i} denote unit vectors aligned with the plane of the rotor and oriented with respect to the lateral, longitudinal, and vertical directions as shown in Figure 5. Let r be defined as the position vector from the inertial origin to the vehicle center of gravity (c.g.), and let \u03c9B be defined as the angular velocity of the aircraft in the body frame. 6 of 20 American Institute of Aeronautics and Astronautics The rotors, numbered 1 \u2212 4, are mounted outboard on the xB, yB, \u2212xB and \u2212yB axes, respectively, with position vectors ri with respect to the c.g. Thrust is produced by each rotor through the torque applied by brushless DC motors, with the dynamics of each motor given by Q = KqI (1) V = RaI +Ke\u03c9 (2) where Q is the torque developed by the motor, V is the voltage across the motor, I is the current through the motor, and \u03c9 is the angular rate at which the motor is spinning",
            " The relation between applied voltage and thrust is found by equating the power produced with the ideal power consumed at hover, and combining equations (4) and (5), to yield Q Kq V = \u03batT Kq V = T 3/2 \u221a 2\u03c1A (6) Thus, T = 2\u03c1A\u03ba2 t K2 q V 2 (7) The thrust produced by the rotors is proportional to the square of the voltage across the motor. Each of the four rotors produces a thrust, denoted Ti, controlled through the application of a voltage Vi, which is used to actuate the vehicle. The thrust produced by the ith rotor acts perpendicularly to the rotor plane along the zR,i axis, as defined in Figure 5. Each rotor also produces an aerodynamic moment, Mi, in the body fixed frame, which is a function of the motor torque as well as various aerodynamic effects which are described in Section V. The roll, pitch and yaw angles are controlled by differential thrust. Differential thrust between opposite motors provides roll and pitch torques. Differential thrust between the two pairs of counter-rotating motors provides yaw torque. To decouple the control, motors 1 and 3 rotate in the opposite direction of rotors 2 and 4. The vehicle body drag force is defined as Db, vehicle mass is m, acceleration due to gravity is g, and the inertia matrix is Ib \u2208 R3\u00d73. A free body diagram is depicted in Figure 4, with a depiction of the rotor forces and moments in Figure 5. The total force, F, can be summed as, F = \u2212Dbev +mgeD + 4\u2211 i=1 (\u2212TiRRi,Ib zR,i) (8) where RRi,I is the rotation matrix from the plane of rotor i to inertial coordinatesa. Similarly, the total moment, M, is, M = 4\u2211 i=1 (Mi + ri \u00d7 (\u2212TiRRi,BzR,i)) (9) aThe notation RA,B shall refer to rotation matrices from coordinate system A to B throughout. 8 of 20 American Institute of Aeronautics and Astronautics where RRi,B is the rotation matrix from the plane of rotor i to body coordinates. Note that the drag force was neglected in computing the moment"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.51-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.51-1.png",
        "caption": "FIGURE 8.51. The front and side views of a wheel and its steering axis.",
        "texts": [
            " The origin of the wheel-body coordinate frame of the wheel number 1 is at BdW1 = \u23a1\u23a3 a1 \u2212b1 \u2212h+Rw \u23a4\u23a6 . (8.70) Hence the transformation between the two frames is only a displacement. Br = BIW1 W1r+ BdW1 (8.71) 8.6 F Caster Theory The steer axis may have any angle and any location with respect to the wheel-body coordinate frame. The wheel-body frame C (xc, yc, zc) is a frame at the center of the wheel at its rest position, parallel to the vehicle coordinate frame. The frame C does not follow any motion of the wheel. The steer axis is the kingpin axis of rotation. Figure 8.51 illustrates the front and side views of a wheel and its steering axis. The steering axis has angle \u03d5 with (yc, zc) plane, and angle \u03b8 with (xc, zc) plane. The angles \u03d5 and \u03b8 are measured about the yc and xc axes, 498 8. Suspension Mechanisms respectively. The angle \u03d5 is the caster angle of the wheel, while the angle \u03b8 is the lean angle. The steering axis of the wheel, as shown in Figure 8.51, is at a positive caster and lean angles. The steering axis intersect the ground plane at a point that has coordinates (sa, sb,\u2212Rw) in the wheelbody coordinate frame. If we indicate the steering axis by the unit vector u\u0302, then the components of u\u0302 are functions of the caster and lean angles. C u\u0302 = \u23a1\u23a3 u1 u2 u3 \u23a4\u23a6 = 1p cos2 \u03d5+ cos2 \u03b8 sin2 \u03d5 \u23a1\u23a3 cos \u03b8 sin\u03d5 \u2212 cos\u03d5 sin \u03b8 cos \u03b8 cos\u03d5 \u23a4\u23a6 (8.72) The position vector of the point that u\u0302 intersects the ground plane, is called the location vector s that in the wheel-body frame has the following coordinates: Cs = \u23a1\u23a3 sa sb \u2212Rw \u23a4\u23a6 (8"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure3.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure3.2-1.png",
        "caption": "Fig. 3.2: Planar examples of kinematic chains of manipulators: a) serial chain as open type; b) parallel chain as closed type.",
        "texts": [
            " This aim requires the determination of a model that can be deduced by abstraction from the mechanical design of a manipulator and by stressing the fundamental kinematic parameters. The mobility of a manipulator is due to the degrees of freedom (d.o.f.s) of the joints in the kinematic chain of the manipulator, when the links are assumed to be rigid bodies. Chapter 3: Fundamentals of the Mechanics of Robots74 A kinematic chain can be of open architecture, when referring to serial connected manipulators, or closed architecture, when referring to parallel manipulators, as in the examples shown in Fig. 3.2. Of course, it is also possible to design mixed chains for so-called hybrid manipulators. Regarding the joints, although there are several designs both from theoretical and practical viewpoints, usually the joint types in robots are related to prismatic and revolute pairs with one degree of freedom. They can be modeled as shown in Fig. 3.3. However, most of the manipulators are designed by using revolute joints, which have the advantage of simple design, long durability, and easy operation and maintenance"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001270_jproc.2020.3046112-Figure20-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001270_jproc.2020.3046112-Figure20-1.png",
        "caption": "Fig. 20. (a) Slotless-Halbach machine with integrated drive. (b) WELC concept [58].",
        "texts": [
            " during three-phase short-circuit fault are being worked on for further improvement. The outer rotor configurations also leave a hollow space in the middle where the drive can be housed to increase the system-level power density. 2) Slotless-Halbach PM Machine With Embedded Cooling: This outer rotor configuration with Halbach magnets uses a slotless topology with winding embedded liquid cooling (WELC) to reduce the frequency-dependent core loss and magnet-loss during high-speed operation [56]. This electromagnetic configuration is illustrated in Fig. 20(a), while the WELC concept is shown in Fig. 20(b). The outer rotor Halbach PM maximizes torque density and reduces the harmonic contents of airgap flux density. The stator adopts fractional slot concentrated winding to minimize the end turn extension. The winding support uses commercially available nonmagnetic thermally conductive plastic with conductivity varying between 3 and 14 W/m\u00b7K [57]. With the winding embedded cooling channels passing through the winding support structure, the heat source (winding) comes into direct contact with the heat sink (lowtemperature coolant). In addition, an axial water jacket cooling is added at the stator frame to further improve the thermal performance, as shown in Fig. 20(b). The CFD analysis has shown that a maximum current density (for 18 s) of 33 A(rms)/mm2 and a continuous current density of 23 A(rms)/mm2 are feasible [58]. This translates to an improvement of 50% over conventional housing water jacket cooling. The main disadvantage of this SlotlessHalbach topology is the very high three-phase short-circuit fault, which increases the risk of demagnetization. 3) Segmented Magnet IPM Machine: The segmented magnet IPM configuration is very similar to the state-ofthe-art configurations using inner rotor topology where each pole has two magnets arranged in V-shape"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure10.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure10.6-1.png",
        "caption": "Figure 10.6 The sign of the section forces is related to a (local) coordinate system with the x axis along the member axis and the yz plane parallel to the cross-sections.",
        "texts": [
            " \u2022 A bending moment: this is the pair of couples of 120 kNm in a plane normal to the cross-sectional plane. For normal force, shear force and bending moment1 we use the symbols N , V and M respectively. Since section forces are interaction forces, their sign convention is somewhat more complicated than that for a force F or couple T . The sign of the section forces is related to a (local) coordinate system with the x axis along the member axis and the yz plane parallel to the member cross-sections (see Figure 10.6). 1 These names used in practice in no sense reflect that we are talking about interaction forces (pair of forces). 10 Section Forces 391 After applying a section, there are two cross-sectional planes. To distinguish these from one another, we call the sectional plane positive where the x axis points outwards, and the sectional plane negative where the x axis points inwards. This is shown in Figure 10.7 where the sectional planes I are positive, and sectional planes II are negative. More formally, we describe the position of a sectional plane using a socalled unit normal vector n"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003429_pnas.94.21.11307-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003429_pnas.94.21.11307-Figure4-1.png",
        "caption": "FIG. 4. A propeller in the shape of a right-handed helix connected to a spherical cell, both moving to the right at velocity v. The propeller and cell rotate in opposite directions, at angular velocities v and V, respectively. The external force F acting on the propeller is directed to the left.",
        "texts": [
            " Because most of the flagellum is in fact relatively far from the cell, we may hope to achieve a reasonably good approximation. The cell, of course, must rotate continuously in a sense opposite to the propeller rotation, there being no external torques on the system. We denote by V the angular velocity of the cell. Both the f lagellum and the cell must translate at the same speed v. The torque on the cell must be equal and opposite to the torque on the propeller; likewise, the force on the cell must be equal and opposite to the force on the propeller. Fig. 4 will help us to make a self-consistent assignment of signs. The propeller shown has been arbitrarily chosen to be a right-handed helix. The directions have been made consistent with the fact, not predicted by this general analysis, that a helical filament does tend to move like a corkscrew in a cork when it is rotated. With our sign convention, the off-diagonal elements of the propulsion matrix of a right-handed corkscrew will then be negative: B , 0. The motor will have to drive the propeller in the sense v , 0 to propel the system to the right, that is with v "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001073_b978-0-12-384050-9.50007-6-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001073_b978-0-12-384050-9.50007-6-Figure2-1.png",
        "caption": "Fig. 2. P a d grip a n d relative support d iameter: cross section through cylindrical support . As one p a d ( 1 ) is m o v e d progressively closer ( 2 and 3 ) to another ( 4 ) on the surface of the support , the normal component ( X ) of the constant force of a d  duction ( A ) grows smaller a n d the tangential component ( Y ) grows larger. Angle \u03c6 = I ( 1 8 0 \u00b0 \u2014 \u0398). In the ideal case , the grip will sl ip when the central angle 0 falls be low the point at which tan \u03c6 equals the coefficient of friction of the p a d on the support .",
        "texts": [
            ", by Schultz, 1969); in this connection it might be pointed out that corrugated and convex pads are not likely to make very satisfactory suction cups, especially on dry tree bark. Most arboreal mammals produce the necessary normal force by adducting two or more appendages deployed across some arc of the trunk or vertical branch to which they cling; these may be the digits of a single extremity on a support of small diameter, or opposing hands and feet on large trunks. This situation is diagrammed in Fig. 2. It is evident that, except in the extreme case where the arc of deployment \u03b8 ^ 180\u00b0, the force of adduc tion A will have both a normal component X and a tangential component Y. When the apposed appendages subtend a small central angle \u0398, propor tionately more of the animal's total muscular effort will go into opposing the frictional force rather than augmenting it. The minimal friction needed to keep the animal from slipping will just equal the tangential force. Therefore, F M I N = \u03a5 = \u03bc\u03a7, and \u03bc = \u03a5 / \u03a7 = tan ( 1 8 0 \u00b0 \u2014 \u03b8) 12"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000024_41.184817-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000024_41.184817-Figure6-1.png",
        "caption": "Fig. 6 . Sliding modes and quasi-sliding modes.",
        "texts": [
            " In the steady state, chattering appears as a high-frequency oscillation about the desired equilibrium point and may also serve as a source to excite the unmodeled highfrequency dynamics of the system [61, [521, [911, 11471. Since chattering is almost always objectionable, significant research effort has been directed at eliminating or reducing its effects. Several methods are described below. 1) The Continuation Approach: In many VSC designs, the control contains terms that are relaylike in nature, as shown in Fig. 6(a). The ideal relay characteristic is practically impossible to implement, so one approach to reducing the chatter is to replace relay control by a saturating, continuous approximation as shown in Fig. 603) [1431, [148]. In state space, a boundary layer around the switching surface is introduced. Within this boundary layer, the control is chosen to be a continuous approximation of the switching function. An interpretation is that a high-gain control is used near the surface. A consequence of the continuation method is that invariance is lost",
            " The reader may find it helpful to consider a simple second-order system and switching surface such as uous. (73) .i + a i + px = - k u ( s ) s = x + c x In the phase plane, the sliding surface equation describes a straight line passing through the origin with slope - c . The ideal relay control (Fig. 5(a)): Here the control takes the form u(s )=sgn(s ) = +1 when s > 0 = -1 whens < 0 The control is ideal in the sense that it switches instantly at the value s = 0. For this case it is easy to get the exact analytical solution and to sketch the entire phase portrait as shown in Fig. 6(a), from which the following conclusions can be illustrated: An ideal sliding mode exists on the line s = 0, meaning there is no chattering. This is because the control can switch infinitely close of the line s = 0. There is no steady-state error. The invariance property holds because the control The ideal saturation control (Fig. 5(b): The most simple can keep the system on the line s = 0. saturating control is described by u ( s ) = s a t ( s ) = +1 when s > L when Is1 5 L S = - L = -1 when s < -L where L > 0 and fL defines the thresholds for entering the boundary layer. Outside the boundary layer, the control is identical to the ideal relay characteristic. Within the boundary layer, however, the control is a high-gain, linear control. As a result, the system will be driven to the boundary layer, but the trajectory cannot not be forced to follow the line s = 0. In the present case, the exact phase portrait of the system can still be obtained as shown in Fig. 6(b), illustrating the following conclusions: The sliding mode does not exist because the trajectory is not forced to stay on the s = 0 surface. 14 IEEE TRANSACTIONS O N INDUSTRIAL ELECTRONICS, VOL. 40, NO. 1, FEBRUARY 1993 +* \u2018 I (C) (4 Another type of boundary layer modification that has been proposed [ 1431, [147] involves adaptively adjusting the boundary layer width to match the degree of uncertainty in the dynamics of the system. The Practical relay control (Fig. 5c): A practical relay always exhibits hysteresis, modeled by u (s )=hys(s ) = +1 when s > A, or when S < 0 and Is1 < A = -1 w h e n s < A , or when S > 0 and Is1 < A where 2A > 0 is the amount of the hysteresis in s. The hysteresis characteristic makes it impossible to switch the control on the surface s = 0. Instead, switching occurs on the lines s = f A. The exact solution of the system can be determined with some effort and the phase portrait of the response, Fig. 6(c), illustrates the following general characteristics: A nonideal sliding mode exists, meaning there always The system has limit cycle behavior in \u201csteady state.\u201d The origin is not an equilibrium point. The practical saturation control (Fig. 5(d)): A practical saturation element also exhibits hysteresis. Analysis of a system with such a control is complicated, especially in the case of a nonlinear system. Describing function techniques can be useful. Fig. 6(d) shows the phase portrait of the VSC system. Generally speaking: is chattering in the sliding mode. It is impossible to eliminate the chatter. The sliding mode does not exist at all. The VSC system is asymptotically stable in the large, but has two equilibrium points (points P and P\u2019 in Fig. 6(d)). There is steady-state error. There is no chattering phenomenon. When the slope of the linear part of the saturation function is not sufficiently large, the system ceases to possess any VSC system properties. The invariance characteristics do not exist. In conclusion, continuation approaches eliminate the high-frequency chattering at the price of losing invariance. A high degree of robustness can still be maintained with a small boundary layer width, but significant delays in the control actuator may dictate the need for a \u201cthicker\u201d boundary layer. In the extreme case, large amplitude low-frequency oscillation may result and the system may cease to possess any variable structure behavior. The invariance and robustness properties of the system no longer exist in the latter case. It is interesting to note, in Fig. 6, that in all four cases the VSC systems are stable and their state trajectories are bounded in a strip in the state space. If the width of the strip is sufficiently narrow, the dynamics within the strip may be called the \u201cquasisliding mode.\u201d Thus one can consider that the systems shown in b, c, and d of Fig. 7 have quasi-sliding modes. 2) Tuning the Reaching Law Approach: Chattering can be reduced by tuning parameters q, and k , in the reaching law Near the switching surface, s, =: 0, so IS,I = q,"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure4.11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure4.11-1.png",
        "caption": "Figure 4.11 (a) A two-force member is a straight bar that is joined at both ends with a hinge to its surroundings and is loaded only by forces at its ends. A two-force member can transfer only forces of which the line of action passes through both hinges. (b) A bar support. (c) Model of a bar support.",
        "texts": [
            " The forces that the structure exerts on the supports (for example on the foundation) are called support forces or support actions. The support forces are equal and opposite to the support reactions (see Figure 4.10). We will look at four types of supports: \u2022 bar supports; \u2022 roller supports; \u2022 hinged supports; \u2022 (fully) fixed supports. A two-force member is a straight bar which is joined to its environment at both ends by a hinge, and is loaded only by forces at the ends. From the moment equilibrium it follows that such members can transfer forces only when the line of action passes through both hinges (see Figure 4.11a). 4 Structures 121 In a bar support the two-force member is used as a link between the structure and the immovable environment (see Figure 4.11b). Figure 4.11c is a model of the bar support: the bar support is depicted as a single line between the two hinges. The immovable environment is generally shown by means of a hatched area. In Figure 4.12, the two-force member has been isolated at hinges A and B. The position of the two-force member (the line joining both hinges) fixes the line of action of the interaction forces F . Only the magnitude of F (with its sign for the correct direction) is unknown. When the body moves, point A is forced to follow a circle with centre B by the two-force member (see Figure 4"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002269_tnnls.2015.2396044-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002269_tnnls.2015.2396044-Figure1-1.png",
        "caption": "Fig. 1. Earth-fixed and body-fixed coordinate frames.",
        "texts": [
            " In Section III, we investigate the adaptive robust neural output feedback control design for the DP system of ships with unknown ship dynamics and unknown time-variant environmental disturbances when the velocities of ships are unavailable. Section IV provides two numerical simulation examples with comparison studies to illustrate the effectiveness, robustness, and adaptive ability of the proposed control scheme. Finally, the conclusions are drawn in Section V. The reference coordinate frames of ship motion are shown in Fig. 1, where X0Y0 Z0 is the earth-fixed frame and XY Z is the body-fixed frame. The accelerations of a point on the surface of the earth are usually assumed to be negligible for dynamically positioned marine vehicles. This is a good approximation, because the motion of the earth hardly affects low-speed marine vehicles. Therefore, the earth-fixed reference frame X0Y0 Z0 can be considered to be inertial. The coordinate origin O of the earth-fixed frame is taken as any point on the earth\u2019s surface. The axis O X0 is directed to the north, OY0 is directed to the east, and the O Z0 points downward normal to the earth\u2019s surface"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003641_ichr.2007.4813931-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003641_ichr.2007.4813931-Figure1-1.png",
        "caption": "Fig. 1. The three basic balancing strategies. The green dot represents the center of mass, the magenta dot represents the center of pressure, and the blue arrow represents the ground reaction force. 1. CoP Balancing (\u201cAnkle Strategy\u201d) 2. CMP Balancing (\u201cHip Strategy\u201d) 3. Step-out",
        "texts": [
            " An explanation for their findings is they only consider the position of the CoM as the defining measure of stability. It was soon observed that the use of CoM velocity in addition to position was a better measure of stability [14] [15]. Here we explore different models and strategies used for humanoid balance. There are three basic strategies: 1) CoP Balancing (ankle strategy) 2) CMP Balancing (hip strategy) 3) Stepping (change-of-support strategy) Generally, these strategies can be employed sequentially from top to bottom, advancing to the next if the current strategy is inadequate. As shown in Fig. 1, the effective horizontal force on the center of mass, equal to the horizontal ground reaction force, can be increased (CMP Balancing) or moved (Stepping) if simple CoP Balancing is not enough. Our primary goal is to determine decision surfaces that describe when a particular strategy should be used. Real humanoid robots will have many more degrees of freedom and complicated control problems, but by describing their motion in terms of these dimensionally-reduced quantities, such as center of mass and center of pressure, we create useful approximations"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003838_7.953252-Figure25-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003838_7.953252-Figure25-1.png",
        "caption": "Fig. 25. Stator and rotor elements of built 6/4 SRM.",
        "texts": [
            " 23), as well SOARES & COSTA BRANCO: SIMULATION OF A 6/4 SWITCHED RELUCTANCE MOTOR 999 1000 IEEE TRANSACTIONS ON AEROSPACE AND ELECTRONIC SYSTEMS VOL. 37, NO. 3 JULY 2001 minimizes the variance and so torque oscillations (see Fig. 24). In this section, simulation and experimental results are presented and compared for model validation. An SR drive prototype was used, consisting of the 6/4 SR machine and a H-bridge power converter. The electrical machine, constructed based on the finite element study of [1] and whose magnetic data was used in our simulation model, is shown in Fig. 25. The H-bridge power converter is shown in Fig. 26. As indicated in the figure, the converter uses IGBTs with freewheeling diodes, and the continuous voltage Vd is obtained from a diode rectifier. The power converter was implemented limiting each phase current to 5 A. This was effectuated because the objective was to operate the machine in the corner of the magnetic characteristics where the machine efficiency is near its maximum. That stator current limitation has restricted, however, our validation tests to current values that do not saturate the machine"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001514_j.engfailanal.2015.06.004-Figure23-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001514_j.engfailanal.2015.06.004-Figure23-1.png",
        "caption": "Fig. 23. (a) 6 DOF model [66], (b) 9 DOF model [48].",
        "texts": [
            " [61,62] Combined analytical-FE approach Their approach considers a combination of the structure deformation using FEM and the local contact deformations using AM Pandya and Parey [63] Experimental method Conventional photoelasticity technique is adopted to experimentally measure the change of TVMS of a cracked gear pair Note: AM denotes the analytical method, FEM the finite element method and PEP the potential energy principle. For all analytical methods, the gear tooth is assumed as a variable cross-section cantilever beam. Fig. 22. (a) Single DOF model [65], (b) 4 DOF model [10], (c) 4 DOF model [31]. (see Fig. 23a). Taking the effects of torsional and lateral stiffness and damping of the shafts into account, Omar et al. [48] established a 9 DOF model for a gear pair (see Fig. 23b). On the basis of the TVMS model of a cracked gear pair, Chen and Shao [36] established a 6 DOF LMM of a spur gear system where the y axis is parallel to the action line of the gear pair (see Fig. 24a). Based on a similar model presented in Ref. [36] (see Fig. 24b), Mohammed et al. [37] investigated the influence of tooth root crack propagation on dynamic response of a gear system. Ma et al. [55] established a 6 DOF model of a perforated gear system, which is similar to the model in Ref. [36] except for ignoring the effect of tooth friction"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure2.12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure2.12-1.png",
        "caption": "FIGURE 2.12. Free-body-diagram of a car and the trailer when moving on an uphill road.",
        "texts": [
            " Forward Vehicle Dynamics 59 Passenger cars are usually in the range 0.4 < (a1/l) < 0.6. In this range, (afwb/g) < (arwb/g) and therefore, front-wheel-brake cars can reach better forward deceleration than rear-wheel-brake cars. Hence, front brakes are much more important than the rear brakes. Example 60 F A car with a trailer. Figure 2.11 depicts a car moving on an inclined road and pulling a trailer. To analyze the car-trailer motion, we need to separate the car and trailer to see the forces at the hinge, as shown in Figure 2.12. We assume the mass center of the trailer Ct is at distance b3 in front of the only axle of the trailer. If Ct is behind the trailer axle, then b3 should be negative in the following equations. For an ideal hinge between a car and a trailer moving in a straight path, there must be a horizontal force Fxt and a vertical force Fzt . Writing the Newton\u2019s equation in x-direction and two static equilibrium equations for both the trailer and the vehicleX Fx = mta (2.146)X Fz = 0 (2.147)X My = 0 (2.148) we find the following set of equations: Fxt \u2212mt g sin\u03c6 = mt a (2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002351_s00170-017-0426-7-Figure12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002351_s00170-017-0426-7-Figure12-1.png",
        "caption": "Fig. 12 (i) Example of box outer edge profile [measured\u2014red; simulation\u2014blue]. (ii) Typical part deflection. (iii) Test cases C1.1\u2013C1.4. Comparison of experimental and simulated deflection of the box edges along its longest length for the conformally cooled mould",
        "texts": [
            " & The conformal insert showed lower temperature magnitudes and higher temperature uniformity across all thermocouples when compared to the conventional insert. In particular, thermocouples 11\u201316 showed the largest differences due to the limited insert perimeter distribution of cooling baffles Table 8 Numerical and experimental study test cases Case Mould insert Moulding material Coolant flow rate [L/min] Mould section/ coolant temp. [\u00b0C] Aim of analysis Results Moving Fixed Moving Fixed C1.1 Conformal PA6 30% GF 4.5 5.7 80 80 Conventional and conformal cooling comparison, Warpage prediction Fig. 9(i, iii) Fig. 12(iii) C1.2 4.2 5.7 95 95 Warpage prediction Fig. 9(iv) Figs. 12(iii) and 13 C1.3 4.6 5.8 95 80 Warpage prediction Figs. 12(iii) and 13 C1.4 4.4 5.7 80 95 Warpage prediction Fig. 12(iii) and 13 B1.1 Conventional 8.3 5.4 80 80 Conventional cooling comparison Warpage prediction Figs. 9(ii, iii), 12(iii), and 13 B1.2 8.3 5.4 95 95 Warpage prediction Warpage prediction Warpage prediction Fig. 13 B1.3 8.3 5.4 95 80 B1.4 8.3 5.4 80 95 C2.1 Conformal PP 4.2 4.8 60 60 Predictive capability over a range of materials Fig. 11(i) C2.2 PP 30% GF 4.4 5.0 40 40 Fig. 11(ii) C2.3 PA6 30% GF 4.5 5.1 80 80 Fig. 11(iii) C2.4 SAN 4.4 5 45 45 Fig. 11(iv) compared to conformal channels. The results confirm the improved performance of the conformal cooling approach",
            " To quantify the effectiveness of numerical part warpage predictive capability, the part deflection of the lower part edge was measured and compared to simulated predictions both for equal mould half temperatures, and for varying mould half temperatures in order to test warpage prediction in the presence of a mould temperature differential (Table 8). For each experiment, parts were moulded until processing conditions had stabilised and then five consecutive parts were used for measurement. An Optical Gauging Product (OGP) scanner was used to accurately measure the part\u2019s outer edge profile (for example Fig. 12(ii)). The main warpage deflection occurred in part side walls (Fig. 12(i)). For the simulated and experimentally measured values the inward deflection at the longer wall section of the part was measured for each test condition. Figure 12(iii) summarises the results where the mean simulated and measured deflections (of five measured parts) are shown along with the associated variation represented by 95% confidence intervals. With the polymer volumetric shrinkage dependent on both the temperature and pressure change from its injected state to room temperature conditions, any temperature differential across the part will result in differential shrinkage, which in turn leads to part warpage. Consequently, the largest deflection occurs when the mould insert core (part inside surface) is hotter than the fixed half cavity (part outside surface) and the core side of the part shrinks more than the cavity side",
            " When both mould halves are at the same temperature (80 or 95 \u00b0C), the deflection is at similar levels. It can be seen that the numerically predicted deflection is within the measured deflections 95% confidence interval for each mould temperature condition, generally being within 0.2% of the mean measured deflection. Importantly, both the moulded parts and the predicted warpage follow the trend that the deflection magnitude is driven by the temperature difference between the mould halves. As shown in Fig. 12, the main driver for part deflection is the mould insert temperature. Therefore, with conventionally cooled moulds producing a higher variation in temperatures within the part (Figs. 8 and 9(iii)), larger part deflections can be expected. An experimental comparison of the deflection of both conformal and conventional cooled parts for the same material (PA6 30% GF) and conditions is shown in Fig. 13. For this case, it can be seen that the conformally cooled parts produced \u223c0.5% less inward deflection, which depending on the application, may be a significant quality criterion improvement"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002216_j.ijfatigue.2021.106317-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002216_j.ijfatigue.2021.106317-Figure2-1.png",
        "caption": "Fig. 2. Specimen geometry and dimensions (Units in mm) for (a) the uniaxial tensile, (b) the HCF, and (c) compact tension (C(T)) test-pieces.",
        "texts": [
            " Firstly, so that we could assess the crack propagation threshold in the X direction with the Mode I driving force applied relative to the vertical build direction (Z), often referred to the Z-X configuration, and secondly, the threshold of crack propagation in the build direction (Z) with Mode I driving force applied relative to the horizontal direction (X), often referred to as the X-Z configuration. For simplicity these are referred to as V-C(T) and H-C(T) respectively, which relates to the mode I loading direction. The tensile (T) and HCF samples are similarly referred to as V-T, H-T and V-HCF and H-HCF respectively. The geometries and dimensions of the uniaxial tensile specimens, HCF specimens, and C(T) specimens were designed according to ISO 6892\u20131:2009, ISO 1099:2017, and ISO 12108:2012 standards, respectively (as seen in Fig. 2). All specimens were also ground and polished prior to mechanical tests. To obtain tensile properties for the AM build, three uniaxial tensile tests were conducted in each direction at a strain rate of 0.0002 s-1 on an MTS-810 Axial/Torsional machine on test-pieces having the design shown in Fig. 2(a). FCG threshold tests were carried out on C(T) testpieces (Fig. 2(c)) using an MTS ACUMEN3 testing machine, and crack advance was detected using a constant-amplitude sinusoidal load with a stress ratio R = 0.1 and a frequency f = 10 Hz. The threshold stress intensity factor (SIF) range, \u0394Kth, was obtained by the load shedding method. The SIF range, \u0394K, is taken as the \u0394Kth till the FCG rate of 1 \u00d7 10\u2212 6 to 1 \u00d7 10\u2212 7 mm/cycles was reached. Uniaxial HCF testing was undertaken using the test-piece design shown in Fig. 2(b) on a QBG-50 high frequency testing machine. A constant-amplitude sinusoidal load was applied with R = 0.1 and f = 85 Hz. Since the stress level used in the HCF tests is generally lower than the yield strength, intense plastic deformation of the specimen is avoided, no significant rise in Z. Wu et al. International Journal of Fatigue 151 (2021) 106317 temperature was observed during the HCF tests. The runout limit was set to 1 \u00d7 107 cycles. After tests, post-mortem fractographic analysis of the HCF test-pieces was conducted on a TESCAN MIRA3 SEM"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure2.22-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure2.22-1.png",
        "caption": "Figure 2.22 The forces acting on joint A when unloading a container.",
        "texts": [
            "20 The force F4 that closes the force polygon from F1 to F3 \u2013 and ensures equilibrium \u2013 is equal and opposite to the resultant R from F1 to F3. The vector equation for the force equilibrium ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM 2 Statics of a Particle 37 Solution (forces in kN): Rx = \u2211 Fx = Fx;1 + Fx;2 + Fx;3 + Fx;4 = +6 \u2212 2 \u2212 1 \u2212 3 = 0, Ry = \u2211 Fy = Fy;1 + Fy;2 + Fy;3 + Fy;4 = \u22121 \u2212 4 + 6 \u2212 1 = 0. The particle is in equilibrium since the resultant is zero: the forces F1 to F4 therefore together form an equilibrium system.1 Example 2 In Figure 2.22, a container with mass 880 kg is being unloaded. Forces F1, F2 and F3, act on joint A. Here, F3 stands for the weight of the container. In the figure, the system is in equilibrium. The gravitational field strength is g = 10 N/kg. Question: How large are F1 and F2? Solution: The weight of the container is F3 = mg = (880 kg)(10 N/kg) = 8800 N. The unknown forces F1 and F2 are obtained from the two equations for the force equilibrium in the x and y directions: \u2211 Fx = F1 \u00b7 sin 3\u25e6 \u2212 F2 \u00b7 cos 20\u25e6 = 0,\u2211 Fy = F1 \u00b7 cos 3\u25e6 \u2212 F2 \u00b7 sin 20\u25e6 \u2212 (8800 N) = 0"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure10.14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure10.14-1.png",
        "caption": "Figure 10.14 (a) The resultant of the normal stresses on a small area A around a point P is a small force N . This force in P is statically equivalent to (b) a small force N in the normal force centre NC (the intersection of the member axis with the cross-sectional plane), together with (c) a small moment My in the xy plane and (d) a small moment Mz in the xz plane.",
        "texts": [
            " 1 The stresses \u03c3ij (i, j = x, y, z) are the components of a quantity (the so-called stress tensor) that in a certain point for each arbitrary plane links the components of the stress vector p and the components of the unit normal vector n. is the stress 396 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM The resultant of the normal stresses on a small area A around a point P is a small force N : N = \u03c3xx A. This small force N in P is statically equivalent to a small force N in the member axis (the origin of the yz coordinate system), together with two small moments My and Mz, acting in the xy plane and the xz plane respectively (see Figure 10.14): My = y \u00b7 N = y \u00b7 \u03c3xx A, Mz = z \u00b7 N = z \u00b7 \u03c3xx A. If we sum up the contributions of all the forces N for the entire crosssection, this gives: N = \u222b A \u03c3xx dA, My = \u222b A y\u03c3xx dA, Mz = \u222b A z\u03c3xx dA. \u2022 N is the resulting force (or rather: the resulting pair of forces) due to the normal stresses in the cross-section, and is by definition known as normal force when it acts at the normal centre NC of the cross-section (the intersection of the member axis with the cross-sectional plane). \u2022 My is a moment (or rather: a pair of moments) that acts in the xy plane"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002770_tii.2021.3089340-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002770_tii.2021.3089340-Figure1-1.png",
        "caption": "Fig. 1. The shape of the square defect in the bearing raceway.",
        "texts": [
            " In this study, to simplify the dynamic model, we ignore the contact deformation between the rolling element and the edge of the defect, and assume that the shape of the defect surface is square. When the defect is very small, it is regarded as a point, and the corresponding displacement excitation is described as [31]: 0 e e 1 mod( , 2 ) 0           d iH H otherwise (1) where e is half of the tangential size of defect; dH denotes the displacement excitation, and it is defined as H = H       d H H H H H (2) where H presents the height of the defect; H is illustrated in Fig. 1, and it is formulated as 2 2 0.5H 0.5 ((0.5 ) (0.5min( , )) )    d d L B (3) where d is the diameter of the rolling element; L and B denote the length and width of the defect respectively. When the defect becomes larger and the shape of surface is assumed to be square, the displacement excitation is represented by a half sine function, which is written as 0 2 0 sin(0.5 / (mod( , 2 ) - )) mod( , 2 ) - - 0                  d i i H H otherwise (4) where  denotes the angle of the defect, 0 is the initial angular offset of the defect of the ith rolling element"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001751_0278364913483183-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001751_0278364913483183-Figure1-1.png",
        "caption": "Fig. 1. Schematic showing the method to achieve independent orientation control of two microrobots, where the microrobot angle is controlled by the rotation lag in a viscous fluid. (a) Here, R2 has a strong magnetic moment, and nearly aligns with the rotating field B with phase lag \u03c62. Microrobot R1 has a weak magnetic moment, and thus experiences a large lag angle \u03c61. For a constant field rotation rate these angles are constant. (b) Placed in a superimposed magnetic field gradient, the resulting force directions are shown as arrows for a snapshot in time. These forces are unique for a given microrobot moment orientation and magnitude in time. As a 3D field gradient is difficult to visualize, the gradient field shown here is simply a conceptual representation. The inset illustrates that the magnetic field is treated as uniform over the microrobot workspace, although the gradient is non-zero.",
        "texts": [
            " When multiple microrobots with different orientations are placed in the same spatial magnetic gradient field they will experience forces in different directions. In this paper, we show that any arbitrary 3D force on each unconstrained microrobot can be achieved if the orientation of each microrobot is unique and known. We propose to control and differentiate the orientation by rotating the magnetic field, creating a unique phase lag in each microrobot depending on its magnetic strength and fluid drag characteristics, as shown in Figure 1(a). Thus, if each microrobot possesses a different magnetic strength or rotational fluid drag coefficient, arbitrary forces can be exerted on each independently and simultaneously when averaged over one short field cycle, as shown in Figure 1(b). While it is difficult to intuitively study the relationship between such microrobot angles and their resulting force directions in 3D, we will develop such relationships mathematically. at St Petersburg State University on January 25, 2014ijr.sagepub.comDownloaded from Similar dynamic control has been used to manage the magnetic force on a magnetic sphere propelled by a single permanent magnet actuator (Mahoney and Abbott, 2011), but was not used to exert simultaneous unique forces on independent magnets"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure3.50-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure3.50-1.png",
        "caption": "Figure 3.50 The moment of the couple about a point A is T = r \u00d7 F . This moment is independent of the location of point A.",
        "texts": [
            " Comment: For a moment about the origin O of the coordinate system or a moment about one of the coordinate axes, the point O is generally omitted in the representation of the moment. Example The curved beam AB in Figure 3.49 is loaded at B by a force of which the components are defined with respect to magnitude and direction in the figure. Question: Find the moment about the x, y and z axis respectively of the force(s) at B. Solution: Tx = +(25 kN)(3 m) \u2212 (50 kN)(1 m) = +25 kNm, Ty = +(40 kN)(1 m) \u2212 (25 kN)(2 m) = \u221210 kNm, Tz = +(50 kN)(2 m) \u2212 (40 kN)(3 m) = \u221220 kNm. Two parallel forces that are equal and opposite form a couple (see Section 3.1.4). Figure 3.50 shows two forces F1 = F and F2 = \u2212 F , forming a couple in space.1 For the moment of the couple about a point A we have T |A = r1 \u00d7 F1 + r2 \u00d7 F2 = r1 \u00d7 F + r2 \u00d7 (\u2212 F) = ( r1 \u2212 r2) \u00d7 F = r \u00d7 F . 1 There is no resultant force, for F1 + F2 = 0. Figure 3.51 (a) The components of a resultant force \u2211 F in O and a resultant couple \u2211 T . (b) The resultant force vector and the resultant moment vector need not necessarily have the same direction. (c) By shifting the resultant force \u2211 F parallel to itself one can provide that the resultant moment vector has the same direction as the force vector"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure3.13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure3.13-1.png",
        "caption": "Fig. 3.13 Definition of Variables for Calculation of ZMP by 6 Axis Force/Torque Sensor",
        "texts": [
            " This sensor is mainly used for measuring the force at the end effector of industrial robots. An example of 6 axis force/torque sensor is shown in Fig. 3.12. To measure the ZMP of a humanoid robot, the force/torque sensor must be light and must be strong enough to accept the large impulsive force applied to the sensor. To obtain the ZMP from the measured data of 6 axis force/torque sensor, we set N = 1 in (3.24) and (3.25). Let the position of the ZMP in the right and the left foot be pR and pL, respectively, as shown in Fig. 3.13. Especially when the center of measurement 80 3 ZMP and Dynamics of the sensor lies on the z axis of the reference coordinate system, the position of the ZMP of each foot can be obtained very simply. For the right foot, pRx = (\u2212\u03c41y \u2212 f1xd)/f1z (3.26) pRy = (\u03c41x \u2212 f1yd)/f1z (3.27) where pR = [pRx pRy pRz ] T p1 = [0 0 d] T . 2 Measurement of ZMP by Multiple Force Sensors Next, we explain the method to measure the ZMP by using multiple force sensors. Fig. 3.14 shows the humanoid robot H5 [70]. To make the foot light, the ZMP is measured by using twelve force sensing registers: FSR and sorbothane sandwiched by two aluminum planes (Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure3.26-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure3.26-1.png",
        "caption": "Fig. 3.26 Approximated multibody system",
        "texts": [
            " For walking on the flat ground, we have pz = 0. For example, when a robot stands still, we have P\u0307 = L\u0307 = 0 and px = x (3.75) py = y. (3.76) The ZMP coincides with the ground projection of the center of mass. Figure 3.25 shows the program calculating the ZMP by using (3.73) and (3.74). Here, dP(= P\u0307), dL(= L\u0307) can be calculated by using the numerical differentiation of the momentum and the angular momentum. 3.4 Calculation of ZMP from Robot\u2019s Motion 97 We introduce the method to calculate ZMP using the simplified models. Fig. 3.26(a) models the robot neglecting the effect of the inertia tensor of each link about its center of mass and assuming the robot as a sum of point masses. In this case, the angular momentum about the origin is given by L = N\u2211 i=1 ci \u00d7 Pi. (3.77) Substituting (3.77) into (3.73) and (3.74), the ZMP can be expressed as px = \u2211N i=1 mi{(z\u0308i + g)xi \u2212 (zi \u2212 pz)x\u0308i}\u2211N i=1 mi(z\u0308i + g) (3.78) py = \u2211N i=1 mi{(z\u0308i + g)yi \u2212 (zi \u2212 pz)y\u0308i}\u2211N i=1 mi(z\u0308i + g) (3.79) where ci = [xi yi zi] T . While this equation is an approximation, the ZMP can be calculated with enough accuracy if we model each link by using multiple point masses [6]. Next, in the model shown in Fig. 3.26(b), the whole robot is modeled by a point mass. In this case, the momentum and the angular momentum about the origin are given by P = M c\u0307 (3.80) L = c \u00d7M c\u0307, (3.81) where their elements are 98 3 ZMP and Dynamics \u23a1 \u23a3 P\u0307x P\u0307y P\u0307z \u23a4 \u23a6 = \u23a1 \u23a3 Mx\u0308 My\u0308 Mz\u0308 \u23a4 \u23a6 (3.82) \u23a1 \u23a3 L\u0307x L\u0307y L\u0307z \u23a4 \u23a6 = \u23a1 \u23a3 M(yz\u0308 \u2212 zy\u0308) M(zx\u0308\u2212 xz\u0308) M(xy\u0308 \u2212 yx\u0308) \u23a4 \u23a6 . (3.83) Substituting the above equation to (3.73) and (3.74), the ZMP can be given by px = x\u2212 (z \u2212 pz)x\u0308 z\u0308 + g (3.84) py = y \u2212 (z \u2212 pz)y\u0308 z\u0308 + g . (3.85) We will use (3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure3.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure3.6-1.png",
        "caption": "Fig. 3.6. Demagnetization curve, recoil loop, energy of a PM, and recoil magnetic permeability.",
        "texts": [
            " However, the complexity and high cost of the rotor structure discourage further commercializing development. A PM can produce magnetic flux in an air gap with no exciting winding and no dissipation of electric power. As any other ferromagnetic material, a PM can be described by its B\u2014H hysteresis loop. PMs are also called hard magnetic materials, which means ferromagnetic materials with a wide hysteresis loop. The basis for the evaluation of a PM is the portion of its hysteresis loop located in the upper left-hand quadrant, called the demagnetization curve (Fig. 3.6). If reverse magnetic field intensity is applied to a previously magnetized, say, toroidal specimen, the magnetic flux density drops down to the magnitude determined by the point K. When the reversal magnetic flux density is 90 3 Materials and Fabrication removed, the flux density returns to the point L according to a minor hysteresis loop. Thus, the application of a reverse field has reduced the remanence, or remanent magnetism. Reapplying magnetic field intensity will again reduce the flux density, completing the minor hysteresis loop by returning the core to approximately the same value of flux density at the point K as before",
            " Recoil magnetic permeability \u00b5rec is the ratio of the magnetic flux density\u2013 to\u2013magnetic field intensity at any point on the demagnetization curve, i.e. \u00b5rec = \u00b5o\u00b5rrec = \u2206B \u2206H (3.4) where the relative recoil permeability \u00b5rrec = 1 . . . 4.5. Maximum magnetic energy per unit produced by a PM in the external space is equal to the maximum magnetic energy density per volume, i.e. 92 3 Materials and Fabrication wmax = (BH)max 2 J/m3 (3.5) where the product (BH)max corresponds to the maximum energy density point on the demagnetization curve with coordinates Bmax and Hmax (Fig. 3.6). Form factor of the demagnetization curve characterizes the concave shape of the demagnetization curve, i.e. \u03b3 = (BH)max BrHc = BmaxHmax BrHc (3.6) for a square demagnetization curve \u03b3 = 1 and for a straight line (rare-earth PMs) \u03b3 = 0.25. The leakage flux causes the magnetic flux to be distributed nonuniformly along the height 2hM of a PM, where hM is the height per pole. As a result, the MMF produced by the PM is not constant. The magnetic flux is higher in the neutral cross section and lower at the ends, but the behavior of the MMF distribution is the opposite [106]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000047_5.4400-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000047_5.4400-Figure5-1.png",
        "caption": "Fig. 5. Phase plane plots for feedback structures (a) kl(xl) = -3, and (b) k,(xl) = 2.",
        "texts": [
            " It is important to note, that the behavior of under the variable structure control law u(t) = k ( x , ) xAt) where k ( x l ) can be \"-2\" or \"3\". This system illustrated in Fig. 4 has two linear structures, one each for k ( x l ) = 2 and k ( x l ) = -3. With k ( x l ) = -3, the system has complex eigenvalues and with k ( x l ) = 2, the system has real eigenvalues. With the switch in the upper position, the feedback pro- DeCARLO er al.: C O N T R O L OF NONLINEAR MULTIVARIABLE SYSTEMS 213 duces an unstable free motion satisfying p] = [ -2 O 2 '1 p] x, as shown in Fig. 5(a). With the switch in the lower position, the feedback becomes positive and the system's free motion satisfies E:] = [\" 3 2 '1 p]. x2 Switching of course i s not random. It occurs with respect to a sliding or switching surface, generically denoted as U = 0. To illustrate this notion, consider the surface defined as U = ul(xl, x,) = slxl + x, = 0 with s1 > 1. If the feedback is switched according to -3, if ul(xl, x,)xl > 0 i 2, if ul(xl, xz)xl < 0 k(x1) = a behavior illustrated in the phase plane plot of Fig. 6 results. The \"unstable\" equilibrium point (0,O) i s now a saddle point with asymptotes x2 = 3x1 and x, = -xl, as shown in Fig. 5(b). Observe from the dotted line trajectory that if the state vec- tor i s perturbed below the surface, ul(xl, x,) = slxl + x, = 0, at time to, it circles to the point tl before intercepting the surface again. On the other hand, if the switching surface is u2(xl, x,) = slxl + x, = 0 with s1 < 1, then a perturbation off the surface i s always immediatelyforced back to the surface since the phase-plane velocity vectors always point towards the surface. Fig. 7 illustrates the phenomena. As suggested by Figs"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000294_cr0680742-Figure15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000294_cr0680742-Figure15-1.png",
        "caption": "Figure 15. Geometric model used to calculate the probability density function of the distance between the surface-exposed redox relay and the electrode.",
        "texts": [
            "44 In ref 44 the distribution of orientations was assumed to be such that, within a certain range of width d0, all of the possible distances between the electrode surface and the exposed relay, which is the entry point for electrons in the enzyme, occur with a constant probability density: p(d) ) 1/d0 for d \u2208 [dmin, dmin + d0] (30) Although this is indeed the only probability density function that makes it possible to predict a linear relationship between i and E if Butler-Volmer kinetics is assumed (see below), eq 30 has not been given other justification than fruitfulness and simplicity. However, this kind of distribution can be inferred from geometrical considerations that are too simple to be concealed. Let us think of the globular enzyme as a sphere laying on a plane (the electrode surface) in a random orientation (Figure 15). On the surface of the sphere, a dot marks the position of the surface-exposed cluster and its distance d to the electrode sets the electron tunneling rate. The probability that d will belong to the interval [d, d + \u03b4d] is proportional to the area of the sphere included between these two altitudes. Using spherical coordinates, this area can be defined by \u2208 [0, 2\u03c0] and \u03b8 \u2208 [\u03b80, \u03b80 + \u03b4\u03b8]; \u03b80 is the angle with the z axis (\u03b8 \u2208 [0, \u03c0]) defined by R(1 - cos \u03b80) ) d. The area is given by A) 2\u03c0R2 sin \u03b80 \u03b4\u03b8 (31) Using R sin \u03b80 \u03b4\u03b8 ) \u03b4d, it turns out that the probability density function of d is actually independent of d: p(d) ) 1 2R (32) This justifies eq 30"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001428_tfuzz.2010.2047506-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001428_tfuzz.2010.2047506-Figure3-1.png",
        "caption": "Fig. 3. (a) Ball and beam system and its (b) schematic diagram.",
        "texts": [
            " Selecting q1 = 0.5, q2 = 1.56, q3 = 0.24, \u03b51 = 0.2, \u03b52 = 0.09, and \u03b53 = 0.48, and using the fuzzy adaptive sliding-mode staticoutput-feedback controller (39), we have the simulation results, which are shown in Fig. 2. It can be seen that the closed-loop system is uniformly ultimately bounded (UUB) under the controller (39); moreover, the adaptive parameters converge to positive constants, respectively. B. Practical Experiment In this section, we further consider the ball and beam system shown in Fig. 3 and implement the adaptive sliding-mode statefeedback-control scheme, which is established in Section III, to the ball and beam system produced by Googol Technology Limited. The system parameters are given in Example 1. By implementing the controller (19), we can get the following real-time parameters of the ball and beam system, as shown in Fig. 4, and the control input, as shown in Fig. 5, which show that the designed control approach is valid in practice. VI. CONCLUSION AND FUTURE WORKS In this paper, we have employed the T\u2013S fuzzy-systemmodel approach, the SMC technique, and adaptive control scheme to construct a class of robust controllers"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000637_978-3-642-83006-8-Figure2.12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000637_978-3-642-83006-8-Figure2.12-1.png",
        "caption": "Fig. 2.12. Terminal link of the equivalent chain",
        "texts": [],
        "surrounding_texts": [
            "Let us adopt an open kinematic chain equivalent to the consi,dered clo sed mechanism structure. The equivalent chain is illustrated in Fig. 2.11. It consists of 15 links and 15 revolute joints. Let the links 1- -14 and all 15 joints have the same parameters and geometry as the cor responding links and joints of the initial closed chain structure. Let ~ ~ ~ ~ ~ ~ the terminal, 15-th link be fictitious. Let Q1' Q2' Q3 and Q1' ~2' ~3 be the axes of the coordinate frames connected to the terminal link and the support (base), respectively. Let us determine the terminal link position in such a way that the coincidence of the equivalent and closed chain structures implies coincidence of the coordinate frames Q -+ and Q', too. and the conditions for \"closing\" the equivalent chain as 0, i 1, 2, 3 (2.5.3) Suppose the position of the equivalent chain is set up quite near to the closed chain structure, i.e., 6p and 6d. are sufficiently small. Then, ~ it can be written [6, 8]: (2.5.4) where lIdR, and lIdr are two arbitrary nonzero increments from the set 6d. (i=1,2,3); Q\" Q are two unit vectors along the corresponding axes _~ ~ \"'.... r Qi (i=1,2,3). If Qi does not coincide with Qi then at least two increments are different from zero. Accordin~ to (2.5.4) the vector 6t is perpendicular to 6d\" and 6dr and, therefore, colinear with their vector product. Let C be the unit vector along 6t Then de;- = ec; e -~ C = CI lId\"Xlldr 16dR,x6drl (2.5.5) (2.5.6) where CI may be either +1 or -1. The factor CI is determined on thebasis of the following. Let us denote the unit vector of the normal on 6di by d .\u2022 The vector 6t should have such direction that the rotation of all n~ . vectors dnR, is counterclockwise if viewed from the end of 6t. Therefore the necespary condition (dn \"x6d,,) \u00b7C>O should be fulfilled. By introdu cing C from (2.5.6) one obtains: from which follows that 78 (2.5.7) Relation (2.5.6) enables one to calculate C in all cases except when 6dfl 16dr. These vectors may be parallel only if the vector 6t lies in the plane defined by the vectors dnt and dnri then, both increments are perpendicular to that plane. In this case, however, 6t is perpendicular to the third ort denoted by d ,and thus, the increment 6d should ns . *) s be, by its module, larger than 6df and 6dr \u2022 Obviously, an opposite statement is also justifiable: if there be a maximal increment bet ween 6df and 6dr (with respect to the module), the vectors 6d t and 6dr could not be parallel. Therefore, relation (2.5.6) can be always used, provided that one of its increments (e.g., 6df ) has the maximal inten sity 16d. 1 = max ( 16d\u00b71 ) , .. (j) J j = f, r, s By introducing (2.5.5) into (2.5.4) we determine the factor 9 (2.5.8) Finally, by introducing (2.5.6), (2.5.7), and (2.5.8) into (2.5.5) we obtain the necessary angular displacement ->- 6u 16dfl. 6df x6dr ICxQfl 16dfx6drl (2.5.9)"
        ]
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure4.10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure4.10-1.png",
        "caption": "Fig. 4.10: Examples of multi two-finger end-effectors with: a) two two-finger grippers for long objects; b) three two-finger grippers for large objects.",
        "texts": [
            " Therefore this chapter is focused mainly on two-finger grippers that can be designed and operated in industrial and non-industrial applications with the aforementioned characteristics for high reliability and robustness. Fundamentals of the Mechanics of Robotic Manipulation 247 In addition, two-finger grippers are also useful for an easy application for more complicated tasks. Figures 4.9 and 4.10 show examples of such typical applications: Fig. 4.9 shows how two fingers can be properly designed and located to grasp long objects by their extremities only; Fig. 4.10 shows alternative grasps by using two or more two-finger grippers to grasp long objects and even large 3D shaped objects. Chapter 4 Fundamentals of the Mechanics of Grasp248 In Fig. 4.11 a general scheme is shown for an industrial gripper with two fingers composed of: - an actuation system, that usually is of a pneumatic type with suitable pipelines and electrovalve; - a transmission system, connecting the actuator to a driving mechanism; - a driving mechanism that transmits movement and force to the fingers; - fingers and fingertips, that can be specifically designed and shaped for contacting specific objects"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure7.23-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure7.23-1.png",
        "caption": "Figure 7.23 To calculate the resulting water pressure, all the data has been shown as symbols.",
        "texts": [
            "22, the 1-metre wide strip from the slide is modelled as a line element. The figure also shows the water pressure, increasing from 14 kN/m at top A of the slide to 40 kN/m near base B. The water pressure is acting normal to the slide everywhere. In other words, all the forces on the slide pass through C, the centre of arc AB. Therefore, the resultant R of the total water pressure on the slide also passes through C. 7 Gas Pressure and Hydrostatic Pressure 259 To determine the resultant water pressure, please refer to Figure 7.23, which shows all symbols used. The water pressure as a function of \u03d5 is q(\u03d5) = 1 + r sin \u03d5 d \u00b7 q\u0302. The resultant of the water pressure on a small part of the slide with length r d\u03d5 is a small force dF : dF = q(\u03d5) \u00b7 r d\u03d5 with components dFx = dF cos \u03d5 = q(\u03d5)r cos \u03d5 d\u03d5, dFy = dF sin \u03d5 = q(\u03d5)r sin \u03d5 d\u03d5. The components Rx and Ry of the resulting water pressure are found by summing up all the contributions dFx , respectively dFy , over the length of slide AB. This summation is done by integrating between the limits \u03d5 = 0 and \u03d5 = 60\u25e6 = \u03c0/3 rad: Rx = \u222b \u03c0/3 0 q(\u03d5)r cos \u03d5 d\u03d5, Ry = \u222b \u03c0/3 0 q(\u03d5)r sin \u03d5 d\u03d5"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure6.18-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure6.18-1.png",
        "caption": "Fig. 6.18. Phasor diagram of a coreless AFPM synchronous motor (eddy current Ie is in phase with the internal voltage Ei). Numerical example 6.2 .",
        "texts": [
            "92 \u2126 and armature reaction reactances Xad = Xaq = 0.88 \u2126. The input phase voltage is V1 = 220 V, power factor cos\u03c6 = 0.96, angle between the stator current and q axis \u03a8 = 5o, EMF constant kE = 3.363 V/(rev/s), eddy current losses in stator conductors \u2206Pe = 184 W and rotational losses \u2206Prot = 79 W. Find the internal voltage Ei excited by the resultant air gap flux and the stator d- and q-axis current adjusted for eddy current losses. Solution The phasor diagram showing the EMFs Ef and Ei is drawn in Fig. 6.18. The EMF Ef excited by the rotor magnetic flux 6.8 Characteristics of Coreless AFPM Machines 213 Ef = kEn = 3.363\u00d7 3600 60 = 201.78 V The synchronous reactance is Xsd = Xsq = Xad +X1 = Xaq +X1 = 0.88 + 0.92 = 1.8 \u2126 The angle between the stator current and input voltage \u03c6 = arccos 0.96 = 16.26o. The load angle between the voltage and EMF Ef is \u03b4 = \u03c6 \u2212 \u03a8 = 16.26 \u2212 5 = 11.26o. The stator currents according to eqns (2.87), (2.88) and (2.89) are: Iad = 220(1.8 cos 11.26o \u2212 0.24 sin 11.26o)\u2212 201.78\u00d7 1",
            "5 A Ia = \u221a 4.52 + 24.52 = 24.9 A The stator winding losses according to eqn (2.49) are \u2206P1w = 3\u00d7 24.92 \u00d7 0.24 = 445.6 W The input power is Pin = m1V1Ia cos\u03c6 = 3\u00d7 220\u00d7 24.9\u00d7 0.96 = 15 763.0 W The electromagnetic power according to eqn (5.1) is Pelm = 15 763\u2212 445.6\u2212 184 = 15 317.3 W The output power according to eqn (5.3) is Pout = 15 317.3\u2212 79 = 15 238.3 W The efficiency is \u03b7 = 15 238.3 15 763.0 = 0.967 The internal voltage Ei excited by the resultant air gap flux according to the phasor diagram (Fig. 6.18) is Ei = \u221a (Ef + IadXad)2 + (IaqXaq)2 214 6 AFPM Machines Without Stator and Rotor Cores = \u221a (201.78 + 4.5\u00d7 0.88)2 + (24.5\u00d7 0.88)2 = 206.87 V The angle \u03b4i between the EMF Ei and q-axis is \u03b4i = arctan ( IaqXaq Ef + IadXad ) = arctan ( 24.5\u00d7 0.88 201.78 + 4.5\u00d7 0.88 ) = 5.97o The eddy current shunt resistance according to eqn (5.32) is Re = 3 206.872 184 = 697.75 \u2126 The current in the vertical branch of the equivalent circuit (Fig. 5.5) found on the basis of the phasor diagram (Fig. 6.18) Ied = \u2212Ei Re sin \u03b4i = \u2212206.87 697.75 sin 5.97o = \u22120.031 A Ieq = Ei Re cos \u03b4i = 206.87 697.75 cos 5.97o = 0.295 A Ie = \u221a I2 ed + I2 eq = \u221a (\u22120.031)2 + 0.2952 = 0.296 A 6.8 Characteristics of Coreless AFPM Machines 215 or Ie = Ei Re = 206.87 697.75 cos 5.97o = 0.296 A The stator current adjusted for eddy current losses found on the basis of the phasor diagram (Fig. 6.18) I \u2032ad = Ei cos \u03b4i \u2212 Ef Xad \u2212 Ei sin \u03b4i Re = 206.87 cos 5.972 \u2212 201.78 0.88 \u2212 206.87 sin 5.972 697.75 \u2248 4.5 A I \u2032aq = Ei sin \u03b4i Xaq \u2212 Ei cos \u03b4i Re = 206.87 sin 5.972 0.88 \u2212 206.87 cos 5.972 697.75 \u2248 24.8 A I \u2032a = \u221a (I \u2032ad)2 + (I \u2032aq)2 = \u221a 4.52 + 24.82 \u2248 25.2 A 7 Control In the previous chapters it was shown that there is a direct relationship between the rotor speed and the frequency of the phase voltages and currents of the AFPM machine. It was also shown that the magnitude of the induced phase voltages is directly related to the frequency and hence the rotor-speed of the AFPM machine"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001748_j.jmatprotec.2014.06.002-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001748_j.jmatprotec.2014.06.002-Figure6-1.png",
        "caption": "Fig. 6. The force analysis diagram of MPBs: (a) produced in horizontal direction; (b) produced in vertical direction, where, ND-normal direction, SD-slipping direction, T \u2013 the angle between tensile load and \u201ctrack\u2013track\u201d MPBs slipping surface, L \u2013 the angle between tensile load and \u201clayer\u2013layer\u201d MPBs slipping surface.",
        "texts": [
            " The metal ductile deformation at room temperature for traitional castings and forgings is mainly attributed to the grain lipping. Besides grain slipping, the ductile deformation of SLM arts has an additional influencing factor, the MPBs. In the ductile eformation of SLM parts, slipping along the MPBs preferentially ccurs due to the weaker bonding force between the MPBs comared with grain boundaries, and slipping of grain boundaries imultaneously happens. The slipping generates under the action f shear stress. Fig. 6(a) and (b) demonstrates the force analysis iagrams for the horizontal and vertical tensile specimens, respecively. Assuming that the tensile loading is F, the area of stress cross ection is A, the angle between the F and the slipping direction is , he angle between the F and the normal direction of slipping suraces is , then the shear stress along the slipping direction can e expressed as the following: = Fcos A/cos = F A cos cos , (1) When the tensile stress (F/A) is up to the yield point ( s) with he increase in F, the shear stress on the slipping surface reaches critical value k, which results in the beginning of slipping",
            " (3) The critical shear stress mainly depends on the physical and hemical properties of slipping surface, and has nothing to do with he applied loads. Therefore, the yield limit s changes with the ariation of the angle between the tensile load and the slipping urface. When = 0\u25e6 or 90\u25e6 (i.e., the load F is parallel or vertical to ing Technology 214 (2014) 2660\u20132667 the slipping surface), the s goes to infinity, and the slipping surface cannot slip; when = 45\u25e6 and s = 2 k, the s has a minimum value, and thus the slipping surface slips most easily. For the horizontal specimens produced by SLM (Fig. 6(a)), the sliding surfaces consisting of \u201clayer\u2013layer\u201d MPBs are always parallel to the tensile loading ( L = 0) regardless of how the processing angle changes. Consequently, it is difficult to slip along the \u201clayer\u2013layer\u201d MPBs surfaces. In this case, the ductile deformation is mainly attributed to the slipping along \u201ctrack\u2013track\u201d MPBs surfaces. The data in Table 2 indicate that the elongation increases gradually with the decrease in the angles ( T) between the tensile loading direction and the laser scanning direction (Y-axis). In theory, the elongation should somewhat decrease when T is less than 45\u25e6, but the experimental result is not so. The main reason for this deviation is that the actual MPBs surfaces are not regular planes. For the slipping angle, there is a certain deviation between the real value and the description shown in Fig. 6(a). Therefore, the specimens show a certain ductility and possess the elongation of 9.5% when T = 90\u25e6 and L = 0\u25e6. When the SLM specimens are produced in the vertical direction (Fig. 6(b)), the angle between the tensile loading and the slipping surfaces consisting of \u201clayer\u2013layer\u201d and \u201ctrack\u2013track\u201d MPBs varies with the variation of build directions. In theory, when SLM specimens are built completely in the vertical direction ( L = 90\u25e6 or T = 0\u25e6), it is difficult for both types of slipping surfaces to slip. As discussed above, however, the actual MPBs surfaces are not strictly planar. Therefore, the SLM parts produced under this condition still exhibit a certain degree of ductility"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003657_02783649922066475-Figure10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003657_02783649922066475-Figure10-1.png",
        "caption": "Fig. 10. Tether winch with level-wind mechanism (side and top views).",
        "texts": [
            " Most notably, the cable was designed and tested to withstand high tension while crossing a sharp corner such as a rock edge (kinking). The tether is wound on a winch that is located on the inner frame and is driven by a brushless motor via a spur gearbox and chain drive. A multiturn potentiometer geared to the output of the spur gear head measures cable pay-out. From the winch drum the tether cable passes over a sheeve on a levelwind mechanism then under the winch drum, and finally exits the vehicle through an exit cone (see Fig. 10). The level-wind mechanism is a reversing ball screw that is timed via a chain to lay the cable on the drum, one layer at a time. The level-wind is placed on the opposite side of the drum as the exit cone to improve compactness of the overall winch system. At the center of the drum the tether cable passes to the inside of the drum, and into a rotary slip ring. The winch and exit cone are mounted to a rigid frame that is suspended from four small cantilever beams, or flexures. The flexures are stiff in all directions except along the robot\u2019s lon- at East Tennessee State University on June 6, 2015ijr"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001362_epe.2007.4417204-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001362_epe.2007.4417204-Figure5-1.png",
        "caption": "Fig. 5. Stator iron core and winding",
        "texts": [
            " We define the following parameters : q : number of phases Nc : number of cells per phase \u03b8s : angular width of the cell to the stator \u03b8r : angular width of the cell to the rotor Ns : number of teeth to the stator Nr : number of teeth to the rotor ws : angular width of a stator tooth wr : angular width of a rotor tooth The relations making it possible to define a polyphase structure are as follows : \u03c0=\u03b8 2N rr and \u03c0=\u03b8 2N ss c s Nq 2\u03c0 =\u03b8 ; \u239f\u239f \u23a0 \u239e \u239c\u239c \u239d \u239b \u00b1\u03c0=\u03b8 q2 k12 elecs and elecr 2\u03c0=\u03b8 with k, natural entirety Then, \u239f\u239f \u23a0 \u239e \u239c\u239c \u239d \u239b \u00b1 \u03b8 =\u03b8 q2 k1 s r We also have, for reasons of symmetries : 4 w s s \u03b8 = The angular width of the rotor teeth is defined by : rrrw \u03b8\u03b2= with ] [1;0r\u2208\u03b2 In order to balance the radial efforts and to minimize the harmonic components of flows, the numbers of teeth to the stator (Ns) and the rotor (Nr) must even beings. For a three-phase machine (q = 3) and with Nc = 4, we obtain: Ns = 12 and Nr = 10. For a diphasic machine (q = 2) and with Nc = 4, we obtain: Ns = 8 and Nr = 6. We built a three-phase machine. On figures 4 and 5 we can see the stator. On figure 4, there is only the carcass out of aluminium with ferromagnetic sheets. On figure 5, we can see windings of the three phases and excitation circuit. Also, all the active parts are arranged on the static part (stator) which is beneficial to evacuating the copper and iron losses. In Fig. 6, we show that the no-load voltage is almost sinusoidal and that it is possible to modulate their amplitude (dexc is the current density of the wound excitation in A/mm2). This amplitude modulation is useful under driving operation and also under generating operation associated a bridge of diode"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002281_j.mechmachtheory.2020.103838-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002281_j.mechmachtheory.2020.103838-Figure4-1.png",
        "caption": "Fig. 4. Deformation calculation of a gear tooth fillet-foundation [15 , 21] .",
        "texts": [
            " The analytical calculation of this deflection was derived in [15] based on the theory proposed by Muskhelishvili [42] who assumed a linear distribution of normal stress and a constant tangential stress at the gear dedendum circle. And the derived equation was then firstly adopted by Chen and Shao [21] to calculate the gear mesh stiffness together with the potential energy principle, which has been used widely by other scholars. The gear tooth fillet-foundation deflection is calculated as [15 , 21 , 40] , 1 K f = 1 W E cos 2 \u03b11 [ L ( u S )2 + M u S + P [ 1 + Q tan 2 \u03b11 ]] (5) where, u and S are given in Fig. 4 ; and the coefficients L, M, P, Q can be numerically calculated and curve-fitted by polynomial functions for 2 \u2264h f \u22645 and 0.04 \u2264 \u03b8 f \u2264 0.08 with the following forms [40] : X i ( h f , \u03b8f ) = A i \u03b8 3 f + B i /h 3 f + C i \u03b8 2 f + D i /h 2 f + E i \u03b8f /h 2 f + F i \u03b8f /h f + G i \u03b8 2 f /h f + H i /h f + I i (6) X i ( h f , \u03b8f ) = A i /\u03b8 3 f + B i h 3 f + C i /\u03b8 2 f + D i h 2 f + E i h 2 f /\u03b8f + F i h f /\u03b8f + G i h f /\u03b8 2 f + H i h f + I i (7) where X i denotes the coefficients L, M, P and Q . And when calculating Q, Eq. (6) is used, while calculating L, M, P , Eq. (7) should be used. The symbol h f = R f /r int . The symbols R f , r int and \u03b8 f are defined in Fig. 4 , and the values of A i , B , C , D , E , F , G , H , and I are given in Table A.1 in the Appendix which is extracted from Ref. [40] . i i i i i i i i It has been found that it will not only make the tooth itself deformed when a gear tooth is loaded but also make the posture change of the neighboring teeth through the gear foundation structure. How to calculate this gear body structure coupling effect between the loaded tooth and its neighboring teeth has attracted the attention of the researchers"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000824_j.jtbi.2005.04.004-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000824_j.jtbi.2005.04.004-Figure3-1.png",
        "caption": "Fig. 3. (a) Collisional pseudo-elastic version of \u2018pogo-stick\u2019 (massspring) running. (b) Downhill passive running with no springs (plastic eg \u00bc 1 collisions), energetically 4 times as costly as (a), even if (a) has no elastic recovery (r \u00bc 0), (c) Level-ground running propelled by hip torque, energetically identical to (b). (d) Hodographs show candidate trajectories for the tip of the velocity vector, during collision, as it moves from v to v\u00fe. Two cases are shown: (i) corresponding to pseudo-elastic running, and (ii) corresponding to hip-powered running.",
        "texts": [
            " As an aside note, perhaps surprisingly, a sequence of super-elastic collisions eg \u00bc 1 also tends towards an energetically neutral collision as the number of sub-collisions gets large. An estimate can be generated from the principles described above for that portion of human running cost that is derived from collisional loss. The presentation here is an extension of that given in Bekker (1956) and Rashevsky (1948). Consider running as a point mass collision followed by a parabolic flight followed by a collision, etc. We take the nominal (nearly constant) forward speed as v. Various cases of this model of running are pictured in Fig. 3. The time of flight and distance per stride are T \u00bc fv=g and d \u00bc vT (Fig. 3). From Eq. (15), assuming egp0, and Eq. (7) we have Em=\u00f0bT\u00de \u00bc f2v2m 8T \u00f01 r\u00de\u00f01 eg\u00de 2 \u00bc fvmg\u00f01 r\u00de\u00f01 eg\u00de 2=8 \u00bc dg2m\u00f01 r\u00de\u00f01 eg\u00de 2=\u00f08v\u00de, cm \u00bc b Em=T mgv \u00bc b dg\u00f01 r\u00de\u00f01 eg\u00de 2 8v2 \u00bc b\u00f01 r\u00de\u00f01 eg\u00de 2F\u0302=16. \u00f021\u00de The Froude-like number F\u0302 \u00bc \u00f02gd\u00de=v2 \u00bc mgd mv2=2 (22) is based on stride length d rather than the usual \u2018. F\u0302 is the ratio of the locomotion credit per step (weight step length \u00bc mgd) to the kinetic energy \u00f0mv2=2\u00de. The usual Froude number F ffiffiffiffiffiffiffiffiffiffiffi v2=g\u2018 p does not simplify the collisional formulas here. Now, we can compare running without and then with emulating a spring with the legs. Although apparently a silly idea, we can imagine a passive runner with no springs and no spring like behavior. This (Fig. 3b) is the running analog to the springless passive-dynamic walkers. Such a runner has a plastic collision \u00f0er \u00bc 0; eg \u00bc 1\u00de at each footfall. As mentioned above, because of the tipped leg the collision in some ways appears elastic, with the relative-to-ground angle of incidence equal to the angle of reflection. For this generation-free case there is no place to use recovery, so r \u00bc 0. From Eq. (21) the collisional cost is cm \u00bc b Em=T mgv \u00bc b dg 2v2 \u00bc bF\u0302=4. (23) Energetically equivalent models can be considered that replace gravity work with another source. For now we do not allow use of work from an axial leg force. Another possible energy source is an externally applied impulse in the direction tangent to the outgoing path, as shown in Fig. 3c. Such a force impulse is mechanically equivalent (equipollent) to an externally imposed impulsive torque applied to the rigid leg at, say, the hip. Within the context of the point-mass model, but for leg extension which we are excluding at the moment, no scheme can supply mechanical work from within the animal. But, as suggested by McGeer in the context of walking (McGeer, 1990a, 1992) and running (McGeer, 1990b, 1992), point-mass concepts might apply to bodies with extended mass. First, if the upper body is an extended mass hinged to the leg(s) at the center of mass (e",
            " hip coincident with COM), the mechanics are identical to that of the point-mass model. Next imagine that, for collisional purposes, the COM is close enough to the hip that the collisional mechanics is little effected. But imagine that the center of mass is sufficiently forward of the hip that substantial hip torque is needed to balance the torque from the axial leg force (to prevent net body angular momentum increase over a stride). That is, Eq. (23) seems to apply, approximately, to an imagined runner that uses hip torques (Fig. 3c), rather than gravitational energy (Fig. 3b), to make up the collisionally absorbed energy. ARTICLE IN PRESS A. Ruina et al. / Journal of Theoretical Biology 237 (2005) 170\u2013192180 A popular mathematical model for running has a massless foot connected to a rigid point-mass body by a perfectly elastic and massless leg-spring (called variously the mass-spring model, the Spring-Loaded Inverted Pendulum SLIP model, or the pogo-stick model). In flight the leg is positioned so the foot lands forward of the body. During the support phase the body compresses the spring, passes over the foot, and flies off again forward of the foot",
            " These gaits would correspond to n \u00bc 2 (Minetti, 1998) and according to the theory here should have an initial (rear) foot contact with a forward thrust. Immediately following this the next foot (front) should redirect the mass upward into the next flight phase while slowing the forward motion. Both these features appear to be true for human children. And, as is true for horse galloping, the phasing of kinetic energy for skipping is predicted to more resemble human walking than human running (see the local maximum in Ekx at mid double-support in Fig. 3b of Minetti, 1998). As noted, the collisional model, without other costs, is not sufficient to explain why children and astronauts on the moon like to skip and adults on earth generally do not. However, given a velocity and stride length, the collision-loss minimization of this sequential-collision gait make it preferable to hopping with both feet together. One does not need fancy equipment to find that repeated jumping is more tiring than skipping. Gaits can be classified any number of ways (e.g"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001674_j.mechmachtheory.2012.10.008-FigureA.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001674_j.mechmachtheory.2012.10.008-FigureA.1-1.png",
        "caption": "Fig. A.1. Contact Radii Rj for Involute\u2013Involute contacts.",
        "texts": [
            " This paper was developed in the framework of Project DPI2006-14348 funded by the Spanish Ministry of Science and Technology. The displacements designated as (\u03b2k i, j) in the obtaining of the global and local deflections represents the flexibility of point j at the radius Rj, located on the k flank when force is applied to the point i (at the radius Ri) in the active flank. Next, a procedure for the calculation of the radius Rr n as a function of geometric parameters and the position of the gears is shown. In the first case of involute \u2013involute contact (see Fig. A.1): Rr j \u00bc CrPr j \u00bc ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi CrQr j 2 \u00fe QrPr j 2 r \u00f0A 1\u00de QrPrj \u00bc Qr0Pr0 \u00fe \u03c1r \u03c6T\u2212\u03c60 \u00fe \u03c8j\u2212\u03b8r j ; r \u00bc 1;2 \u00f0A 2\u00de Qr0Pr0 \u00bc \u03c1rtg\u03c60; r \u00bc 1;2 \u00f0A 3\u00de CrQr j \u00bc \u03c1r ; r \u00bc 1;2 \u00f0A 4\u00de \u03c6T \u00bc arccos \u03c11 \u00fe \u03c12 dT \u00f0A 5\u00de \u03c8j \u00bc \u2212 arctan x1\u2212x2 d0 \u00fe y1\u2212y2 \u00f0A 6\u00de In the second case of involute \u2013 tip rounding contact (see Fig. A.2.): Rr j \u00bc CrPr j \u00bc ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi CrQ r2 r j 2 \u00fe Qr2 r Pr j 2 r \u00f0A 7\u00de Q1 r2P1 j \u00bc Q10P10 \u00fe \u03c11 \u03c6T r2\u2212\u03c60 \u00fe \u03c8j\u2212\u03b81 j \u00f0A 8\u00de Q2 r2P2 j \u00bc C2Cr2 sen \u03c6T r2\u2212\u03bb2 \u00fe Cr2P2 \u00f0A 9\u00de C2Cr2 \u00bc RExt2\u2212Rtip ; C1Q1 r2 j \u00bc \u03c11 \u00f0A 10\u00de C2Q2 r2 j \u00bc dT cos \u03c6T r2 \u2212\u03c11 \u00f0A 11\u00de \u03c6T r2 \u00bc \u03bc1\u2212\u03be1 \u00f0A 12\u00de where \u03bc1 \u00bc arccos \u03c11 C1Cr2 "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure13.74-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure13.74-1.png",
        "caption": "Figure 13.74 Support reactions.",
        "texts": [
            " The structure has five unknown support reactions while there are only three equilibrium equations. b. The horizontal support reaction at B follows from the shear force in BE. Bh = 5 kN (\u2192). Bar AE is a two-force member so that Av = 0. Introduce a cut at G, and investigate the moment equilibrium of ABEG about G (see Figure 13.73): \u2211 T|G = +12.5 + 23 \u00d7 1, 25 + 5 \u00d7 2.5 \u2212 Bv \u00d7 2.5 = 0 \u21d2 Bv = 21.5 kN (\u2191). The support reactions at A and C are found from the equilibrium of the structure as a whole: \u2211 Fvert = 0 \u21d2 Cv = 24.5 kN (\u2191),\u2211 T|A = 0 \u21d2 Ch = 32.5 kN (\u2192),\u2211 Fhor = 0 \u21d2 Ah = 37.5 kN (\u2190). In Figure 13.74, the support reactions are shown as they act in reality. c. By resolving the horizontal and vertical support reaction at C into components parallel to and normal to CD we find the normal force NCD and the 604 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM shear force V CD (see Figure 13.75): NCD = 1 2 \u221a 2 \u00d7 (32.5 + 24.5) = 28.5 \u221a 2 kN, V CD = 1 2 \u221a 2 \u00d7 (32.5 \u2212 24.5) = 4 \u221a 2 kN. The normal force is a tensile force; the deformation symbol for the shear force is given in Figure 13.75. d. In Figures 13"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure15.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure15.2-1.png",
        "caption": "Figure 15.2 If F is constant, the total amount of work to be performed depends only on the location of the starting point and end point, and not on the route followed.",
        "texts": [
            " If the point of application of the force moves a finite distance along path s, 15 Virtual Work 711 the total amount of work is equal to the sum of the contributions of all the infinitesimal displacements. Mathematically this corresponds to integrating over the path length s: A = \u222b s F \u00b7 d u. The magnitude and direction of the force F may depend on the location on the route. If F is constant (for a vector that means constant in magnitude and direction), then F can be excluded from the integration symbol. The total amount of work performed is then (see Figure 15.2): A = \u222b s F \u00b7 d u = F \u222b s d u = F \u00b7 u. In this case, the total amount of work A depends only on the position of the starting and end points and not on the shape of the route followed. In full A = Fxux + Fyuy + Fzyz = Fu cos \u03b1. Note that no work is performed if F and u are normal to one another (in that case \u03b1 = \u00b1\u03c0/2 and cos \u03b1 = 0). Forces that are constant in magnitude and direction include gravitational forces. The dimension of work is force multiplied by distance. The applicable SI unit is the joule, denoted as J: J = N \u00b7 m = kg \u00b7 m2/s2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure3.7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure3.7-1.png",
        "caption": "Figure 3.7 Compounding several forces using (a) a line of action figure and (b) force polygon.",
        "texts": [
            " The magnitude and direction of the resultant can also be determined in a force polygon (see Figure 3.6). The line of action is determined by realising that it has to pass through the intersection of the lines of action of the forces to be compounded. Note that here the line of action of the resultant is entirely outside the body! If several forces have to be compounded together, this can be done in phases by first determining the resultant of two forces, then compounding it with the third force, and so forth. This procedure is shown in Figure 3.7a. 56 The magnitude and direction of the resultant can also be found quickly using a force polygon, as in Figure 3.7b. The force polygon does not provide information about the location of the line of action, however. To find the line of action, one would have to revert to Figure 3.7a. This figure is referred to as the line of action figure. For more than two forces, using the line of action figure becomes laborious, and the analytical approach is clearly preferable (see Section 3.1.7). To determine the magnitude and direction of the resultant, the force polygon can still be useful. If the forces F1 and F2 are almost parallel, or parallel, one can determine the magnitude and direction of the resultant R graphically in a force polygon, although the graphical construction of its line of action (the line of action figure) becomes difficult as the intersection of the lines of action is far away or even at infinity"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-FigureC.9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-FigureC.9-1.png",
        "caption": "FIGURE C.9 Partial inductance for two rectangular conductors.",
        "texts": [
            " Test results show that the errors become large for very large values of u and for small values of w as considered in Section 5.5.2. A possible solution for this is the computation of Lp11 by breaking the length into several segments as is illustrated in Section 5.6.1. This is required only for very long conductors. The accuracy problem for this case has also been observed in Ref. [10]. It is also pointed out that neighboring partial mutual inductances can be computed by using only partial self-inductance computations. Two neighboring block cells are shown in Fig. C.9. The partial self-inductance of the two neighbors cells shown can be computed using formula (C.36). If we also compute the partial self-Lp33 for a combined block 1,2, then the result is Lp33 = Lp11 + Lp22 + 2 Lp12. (C.38) Hence, we can compute Lp12 totally based on partial self- inductance only, or Lp12 = 0.5 (Lp33 \u2212 Lp11 \u2212 Lp22). (C.39) 394 COMPUTATION OF PARTIAL INDUCTANCES This can be very helpful for skin-effect models where the accuracy is an issue due to the closely located neighbors. Hence, three partial self-inductance computations are required. This helps, since we can more accurately compute a partial mutual inductance Lp12 for some geometries. We want to point out that this technique can also be applied to conductors in parallel as shown in Fig. C.9 (b). We leave the computation of Lp12 as an exercise to the reader. C.1.8 Lp12 for Two Rectangular Parallel Bars Lp12 = \ud835\udf070 4\ud835\udf0b 1 12 4\u2211 k=1 4\u2211 \ud835\udcc1=1 4\u2211 m=1 (\u22121)k+\ud835\udcc1+m+1 [ \u2212 b\ud835\udcc1 cm a3 k 6 tan\u22121 ( b\ud835\udcc1 cm ak R ) PARTIAL INDUCTANCE FORMULAS FOR ORTHOGONAL GEOMETRIES 395 \u2212 b\ud835\udcc1 c3 m ak 6 tan\u22121 ( b\ud835\udcc1 ak cm R ) \u2212 b3 \ud835\udcc1 cm ak 6 tan\u22121 ( cm ak b\ud835\udcc1 R ) + ak ( b2 \ud835\udcc1c2 m 4 \u2212 b4 \ud835\udcc1 24 \u2212 c4 m 24 ) log \u239b\u239c\u239c\u239c\u239d ak + R\u221a b2 \ud835\udcc1 + c2 m \u239e\u239f\u239f\u239f\u23a0 + b\ud835\udcc1 ( c2 m a2 k 4 \u2212 c4 m 24 \u2212 a4 k 24 ) log \u239b\u239c\u239c\u239c\u239d b\ud835\udcc1 + R\u221a c2 m + a2 k \u239e\u239f\u239f\u239f\u23a0 + cm ( b2 \ud835\udcc1a2 k 4 \u2212 b4 \ud835\udcc1 24 \u2212 a4 k 24 ) log \u239b\u239c\u239c\u239c\u239d cm + R\u221a a2 k + b2 \ud835\udcc1 \u239e\u239f\u239f\u239f\u23a0 + 1 60 (b4 \ud835\udcc1 + c4 m + a4 k \u2212 3b2 \ud835\udcc1c2 m \u2212 3c2 ma2 k \u2212 3a2 kb2 \ud835\udcc1) R ] (C"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure6.31-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure6.31-1.png",
        "caption": "FIGURE 6.31. A rigid body in a planar motion.",
        "texts": [
            " The coordinates of point I are a function of the parameter \u03b8 as follows: xI = h cot \u03b8 (6.219) yI = h+ xI cot \u03b8 = h \u00a1 1 + cot2 \u03b8 \u00a2 . (6.220) Eliminating \u03b8 between x and y, generates the path of motion for I. yI = h \u00b5 1 + x2I h2 \u00b6 (6.221) Example 247 F Plane motion of a rigid body. The plane motion of a rigid body is such that all points of the body move only in parallel planes. So, to study the motion of the body, it is enough to examine the motion of points in just one plane. 352 6. Applied Mechanisms 6. Applied Mechanisms 353 Figure 6.31 illustrates a rigid body in a planar motion and the corresponding coordinate frames. The position, velocity, and acceleration of body point P are: GrP = GdB + GRB BrP = GdB + G BrP (6.222) GvP = Gd\u0307B + G\u03c9B \u00d7 \u00a1 GrP \u2212 GdB \u00a2 = Gd\u0307B + G\u03c9B \u00d7 G BrP (6.223) where, GdB indicates the position of the moving origin o relative to the fixed origin O. The termGd\u0307B is the velocity of point o and, G\u03c9B \u00d7 G BrP is the velocity of point P relative to o. GvP/o = G\u03c9B \u00d7 G BrP (6.224) Although it is not a correct view, it might sometimes help if we interpret Gd\u0307B as the translational velocity and G\u03c9B\u00d7 G BrP as the rotational velocity components of GvP "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000579_tcst.2012.2200104-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000579_tcst.2012.2200104-Figure1-1.png",
        "caption": "Fig. 1. Draganflyer V Ti four rotor helicopter UAV. The distance from the center of mass to the rotor shaft is 24 cm.",
        "texts": [
            " The first known quadrotor flight was in the 1920s [1], but due to the difficulty of controlling four motors simultaneously with sufficient bandwidth, the project was abandoned. In 1963, the Curtiss-Wright X-19A became the first manned quadrotor to leave the ground effect. Due to a lack of stability augmentation system, stationary hover was nearly impossible [2]. Smaller, unmanned quadrotor helicopters typically consist of two pairs of counter-rotating blades mounted on a carbon fiber frame, as shown in Fig. 1. The dynamics of quadrotor helicopters have been studied in detail by several groups [3]\u2013[6]. In designing a controller for these aircraft, there are several important vehicle-specific considerations. The dynamics of quadrotors are nonlinear and multivariate. There are also Manuscript received July 25, 2011; revised May 10, 2012; accepted May 13, 2012. Manuscript received in final form May 13, 2012. This work was supported in part by the Boeing Company Strategic University Initiative. Recommended by Associate Editor A"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure9.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure9.6-1.png",
        "caption": "Figure 9.6 We assume that the dead weight of a truss applies in the joints. The total dead weight Fdw of a truss member is equally distributed over both adjacent joints.",
        "texts": [
            " (e) to (h) The same structures, but now all the rigid joints are replaced by hinged joints. With hinged joints (a) ad (b) are kinematically determinate and can be considered to be trusses. For (c) and (d), the use of hinged joints generates a mechanism; they cannot be considered trusses. The force flow in (c) and (d) occurs mainly by bending. 324 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM One also often assumes that the dead weight of a truss applies at the joints. The total dead weight Fdw of a truss member is split up into two equal forces in both adjacent joints (see Figure 9.6). This is a rough model of reality, but since the dead weight is generally small with respect to the other loads that the truss has to bear, the deviations that occur are relatively small. Figure 9.7 shows part of a truss. The letters show the names of the members in the truss. The members along the chord or perimeter of the truss are called chord members (ch), the others are referred to as bracing members (br). Chord members can be divided into top chord members (t) and bottom chord members (b)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.13-1.png",
        "caption": "FIGURE 8.13. A Robert suspension mechanism with a Panhard arm.",
        "texts": [],
        "surrounding_texts": [
            "The suspension is what links the wheels to the vehicle body and allows relative motion. This chapter covers the suspension mechanisms, and discusses the possible relative motions between the wheel and the vehicle body. The wheels, through the suspension linkage, must propel, steer, and stop the vehicle, and support the associated forces. 8.1 Solid Axle Suspension The simplest way to attach a pair of wheels to a vehicle is to mount them at opposite ends of a solid axle, such as the one that is shown in Figure 8.1. The solid axle must be attached to the body such that an up and down motion in the z-direction, as well as a roll rotation about the x-axis, is possible. So, no forward and lateral translation, and also no rotation about the axle and the z-axis, is allowed. There are many combinations of links and springs that can provide the kinematic and dynamic requirements. The simplest design is to clamp the axle to the middle of two leaf springs with their ends tied or shackled to the vehicle frame as shown schematically in Figure 8.1. A side view of a multi-leaf spring and solid axle is shown in Figure 8.2. A suspension with a solid connection between the left and right wheels is called dependent suspension. 456 8. Suspension Mechanisms The performance of a solid axle with leaf springs suspension can be improved by adding a linkage to guide the axle kinematically and provide dynamic support to carry the non z-direction forces. The solid axle with leaf spring combination came to vehicle industry from horse-drawn vehicles. Example 299 Hotchkiss drive. When a live solid axle is connected to the body with nothing but two leaf springs, it is called the Hotchkiss drive, which is the name of the car that used it first. The main problems of a Hotchkiss drive, which is shown in Figure 8.2, are locating the axle under lateral and longitudinal forces, and having a low mass ratio \u03b5 = ms/mu, where ms is the sprung mass and mu is the unsprung mass. Sprung mass refers to all masses that are supported by the spring, such as vehicle body. Unsprung mass refers to all masses that are attached to and not supported by the spring, such as wheel, axle, or brakes. Example 300 Leaf spring suspension and flexibility problem. The solid axle suspension systems with longitudinal leaf springs have many drawbacks. The main problem lies in the fact that springs themselves act as locating members. Springs are supposed to flex under load, but their flexibility is needed in only one direction. However, it is the nature of leaf springs to twist and bend laterally and hence, flex also in planes other than the tireplane. Leaf springs are not suited for taking up the driving and braking traction forces. These forces tend to push the springs into an S-shaped profile, as shown in Figure 8.3. The driving and braking flexibility of leaf springs, generates a negative caster and increases instability. Long springs provide better ride. However, long sprigs exaggerate their bending and twisting under different load conditions. Example 301 Leaf spring suspension and flexibility solution. To reduce the effect of a horizontal force and S-shaped profile appearance in a solid axle with leaf springs, the axle may be attached to the chassis by a longitudinal bar as Figure 8.4(a) shows. Such a bar is called an anti-tramp 8. Suspension Mechanisms 457 bar, and the suspension is the simplest cure for longitudinal problems of a Hotchkiss drive. A solid axle with an anti-tramp bar may be kinematically approximated by a four-bar linkage, as shown in Figure 8.4(b). Although an anti-tramp bar may control the shape of the leaf spring, it introduces a twisting angle problem when the axle is moving up and down, as shown in Figure 8.5. Twisting the axle and the wheel about the axle is called caster. The solid axle is frequently used to help keeping the wheels perpendicular to the road. Example 302 Leaf spring location problem. The front wheels need room to steer left and right. Therefore, leaf springs cannot be attached close to the wheel hubs, and must be placed closer to the middle of the axle. That gives a narrow spring-base, which means that a small side force can sway or tilt the body relative to the axle through a considerable roll angle due to weight transfer. This is uncomfortable for the vehicle passengers, and may also produce unwanted steering. The solid axle positively prevents the camber change by body roll. The wheels remain upright and hence, do not roll on a side. However, a solid axle shifts laterally from its static plane and its center does not remain on the vehicle\u2019s longitudinal axis under a lateral force. 458 8. Suspension Mechanisms A solid axle produces bump-camber when single-wheel bump occurs. If the right wheel goes over a bump, the axle is raised at its right end, and that tilts the left wheel hub, putting the left wheel at a camber angle for the duration of deflection. Example 303 Triangular linkage. A triangulated linkage, as shown in Figure 8.6, may be attached to a solid axle to provide lateral and twist resistance during acceleration and braking. Example 304 Panhard arm. High spring rate is a problem of leaf springs. Reducing their stiffness by narrowing them and using fewer leaves, reduces the lateral stiffness and increases the directional stability of the suspension significantly. A Panhard arm is a bar that attaches a solid axle suspension to the chassis laterally. Figure 8.7 illustrates a solid axle and a Panhard arm to guide the axle. Figure 8.8 shows a triangular linkage and a Panhard arm combination for guiding a solid axle. A double triangle mechanism, as shown in Figure 8.9, is an alternative design to guide the axle and support it laterally. Example 305 Straight line linkages. There are many mechanisms that can provide a straight line motion. The simplest mechanisms are four-bar linkages with a coupler point moving straight. Some of the most applied and famous linkages are shown in Figure 8.10. By having proper lengths, the Watt, Robert, Chebyshev, and Evance linkages can make the coupler point C move on a straight line vertically. Such a mechanism and straight motion may be used to guide a solid axle. Two Watt suspension mechanisms with a Panhard arm are shown in Figures 8.11 and 8.12. 460 8. Suspension Mechanisms 8. Suspension Mechanisms 461 Figures 8.13, 8.14, and 8.15 illustrate three combinations of Robert suspension linkages equipped with a Panhard arm. Example 306 Solid axle suspension and unsprung mass problem. A solid axle is counted as an unsprung member, and hence, the unsprung mass is increased where using solid axle suspension. A heavy unsprung mass ruins both, the ride and handling of a vehicle. Lightening the solid axle makes it weaker and increases the most dangerous problem in vehicles: axle breakage. The solid axle must be strong enough to make sure it will not break under any loading conditions at any age. As a rough estimate, 90% of the leaf spring mass may also be counted as unsprung mass, which makes the problem worse. The unsprung mass problem is worse in front, and it is the main reason that they are no longer used in street cars. However, front solid axles are still common on trucks and buses. These are heavy vehicles and solid axle 8. Suspension Mechanisms 463 464 8. Suspension Mechanisms suspension does not reduce the mass ratio \u03b5 = ms/mu very much. When a vehicle is rear-wheel-drive and a solid axle suspension is used in the back, the suspension is called live axle. A live axle is a casing that contains a differential, and two drive shafts. The drive shafts are connected to the wheel hubs. A live axle can be three to four times heavier than a dead I-beam axle. It is called live axle because of rotating gears and shafts inside the axle. Example 307 Solid axle and coil spring. To decrease the unsprung mass and increase vertical flexibility of solid axle suspensions, it is possible to equip them with coil springs. A sample of a solid axle suspension with coil spring is shown in Figure 8.16. The suspension mechanism is made of four longitudinal bars between the axle and chassis. The springs may have some lateral or longitudinal angle to introduce some lateral or longitudinal compliance. Example 308 De Dion axle. When a solid axle is a dead axle with no driving wheels, the connecting beam between the left and right wheels may have different shapes to do different jobs, usually to give the wheels independent flexibility. We may also modify the shape of a live axle to attach the differential to the chassis and reduce the unsprung mass. De Dion design is a modification of a beam axle that may be used as a dead axle or to attach the differential to the chassis and transfer the driving power to the drive wheels by employing universal joints and split shafts. Figure 8.17 illustrates a De Dion suspension. 8.2 Independent Suspension Independent suspensions is introduced to let a wheel to move up and down without affecting the opposite wheel. There are many forms and designs of independent suspensions. However, double A-arm and McPherson strut suspensions are the simplest and the most common designs. Figure 8.18 illustrates a sample of a double A-arm and Figure 8.19 shows a McPherson suspension. Kinematically, a double A-arm suspension mechanism is a four-bar linkage with the chassis as the ground link, and coupler as the wheel carrying link. A McPherson suspension is an inverted slider mechanism that has the chassis as the ground link and the coupler as the wheel carrying link. A double A-arm and a McPherson suspension mechanism on the left and right wheels are schematically shown in Figures 8.20 and 8.21 respectively. Double A-arm, is also called double wishbone, or short/long arm suspension. McPherson also may be written as MacPherson. Example 309 Double A-arm suspension and spring position. Consider a double A-arm suspension mechanism. The coil spring may be between the lower arm and the chassis, as shown in Figure 8.18. It is also possible to install the spring between the upper arm and the chassis, or between the upper and lower arms. In either case, the lower or the upper arm, which supports the spring, is made stronger and the other arm acts as a connecting arm. Example 310 Multi-link suspension mechanism. When the two side bars of an A-arm are attached to each other with a joint, as shown in Figure 8.22, then the double A-arm is called a multi-link mechanism. A multi-link mechanism is a six-bar mechanism that may have a better coupler motion than a double A-arm mechanism. However, multi- 466 8. Suspension Mechanisms 8. Suspension Mechanisms 467 link suspensions are more expensive, less reliable, and more complicated compare to a double A-arm four-bar linkage. There are vehicles with more than six-link suspension with possibly better kinematic performance. Example 311 Swing arm suspension. An independent suspension may be as simple as a triangle shown in Figure 8.23. The base of the triangle is jointed to the chassis and the wheel to the tip point. The base of the triangle is aligned with the longitudinal axis of the vehicle. Such a suspension mechanism is called a swing axle or swing arm. The variation in camber angle for a swing arm suspension is maximum, compared to the other suspension mechanisms. Example 312 Trailing arm suspension. Figure 8.24 illustrates a trailing arm suspension that is a longitudinal arm with a lateral axis of rotation. The camber angle of the wheel, supported by a trailing arm, will not change during the up and down motion. Trailing arm suspension has been successfully using in a variety of frontwheel-drive vehicles, to suspend their rear wheels. Example 313 Semi-trailing arm Semi-trailing arm suspension, as shown in Figure 8.25, is a compromise between the swing arm and trailing arm suspensions. The joint axis may have any angle, however an angle not too far from 45 deg is more applied. Such suspensions have acceptable camber angle change, while they can handle both, the lateral and longitudinal forces. Semi-trailing design has successfully applied to a series of rear-wheel-drive cars for several decades. Example 314 Antiroll bar and roll stiffness. Coil springs are used in vehicles because they are less stiff with better 470 8. Suspension Mechanisms ride comfort compared to leaf springs. Therefore, the roll stiffness of the vehicle with coil springs is usually less than in vehicles with leaf springs. To increase the roll stiffness of such suspensions, an antiroll bar must be used. Leaf springs with reduced layers, uni-leaf, trapezoidal, or nonuniform thickness may also need an antiroll bar to compensate for their reduced roll stiffness. The antiroll bar is also called a stabilizer. Figure 8.26 illustrates an anti-roll bare attached to a solid axle with coil springs. Example 315 Need for longitudinal compliance. A bump is an obstacle on the road that opposes the forward motion of a wheel. When a vehicle goes over a bump, the first action is a force that tends to push the wheel backward relative to the rest of the vehicle. So, the lifting force has a longitudinal component, which will be felt inside the vehicle unless the suspension system has horizontal compliance. There are situations in which the horizontal component of the force is even higher than the vertical component. Leaf springs can somewhat absorb this horizontal force by flattening out and stretching the distance from the forward spring anchor and the axle. Such a stretch is usually less than 1/2 in \u2248 1 cm. 8.3 Roll Center and Roll Axis The roll axis is the instantaneous line about which the body of a vehicle rolls. Roll axis is found by connecting the roll center of the front and rear suspensions of the vehicle. Assume we cut a vehicle laterally to disconnect the front and rear half of the vehicle. Then, the roll center of the front or rear suspension is the instantaneous center of rotation of the body with respect to the ground. 8. Suspension Mechanisms 471 472 8. Suspension Mechanisms Figure 8.27 illustrates a sample of the front suspensions of a car with a double A-arm mechanism. To find the roll center of the body with respect to the ground, we analyze the two-dimensional kinematically equivalent mechanism shown in Figure 8.28. The center of tireprint is the instant center of rotation of the wheel with respect to the ground, so the wheels are jointed links to the ground at their center of tireprints. The instant center I18 is the roll center of the body with respect to the ground. To find I18, we apply the Kennedy theorem and find the intersection of the line I12I28 and I13I38 as shown in Figure 8.29. The point I28 and I38 are the instant center of rotation for the wheels with respect to the body. The instant center of rotation of a wheel with respect to the body is called suspension roll center. So, to find the roll center of the front or rear half of a car, we should determine the suspension roll centers, and find the intersection of the lines connecting the suspension roll centers to the center of their associated tireprints. The Kennedy theorem states that the instant center of every three relatively moving objects are colinear. Example 316 McPherson suspension roll center. A McPherson suspension is an inverted slider crank mechanism. The instant centers of an example of inverted slider crank mechanism are shown in Figure 8.30. In this figure, the point I12 is the suspension roll center, which is the instant center of rotation for the wheel link number 2 with respect to the chassis link number 1. A car with a McPherson suspension system is shown in Figure 8.31. The kinematic equivalent mechanism is depicted in Figure 8.32. Suspension roll centers along with the body roll center are shown in Figure 8.33. To 8. Suspension Mechanisms 473 find the roll center of the front or rear half of a car, we determine each suspension roll center and then find the intersection of the lines connecting the suspension roll centers to the center of the associated tireprint. Example 317 Roll center of double A-arm suspension. The roll center of an independent suspension such as a double A-arm can be internal or external. The kinematic model of a double A-arm suspension for the front left wheel of a car is illustrated in Figure 8.34. The suspension roll center in Figure 8.34(a) is internal, and in Figure 8.34(b) is external. An internal suspension roll center is toward the vehicle body, while an external suspension roll center goes away from the vehicle body. A suspension roll center may be on, above, or below the road surface, as shown in Figure 8.35(a)-(c) for an external suspension roll center. When the suspension roll center is on the ground, above the ground, or below the ground, the vehicle roll center would be on the ground, below the ground, and above the ground, respectively. Example 318 F Camber variation of double A-arm suspension. When a wheel moves up and down with respect to the vehicle body, de- 474 8. Suspension Mechanisms 8. Suspension Mechanisms 475 pending on the suspension mechanism, the wheel may camber. Figure 8.36 illustrates the kinematic model for a double A-arm suspension mechanism. The mechanism is equivalent to a four-bar linkage with the ground link as the vehicle chasis. The wheel is always attached to a coupler point C of the mechanism. We set a local suspension coordinate frame (x, y) with the x-axis indicating the ground link MN . The x-axis makes a constant angle \u03b80 with the vertical direction. The suspension machanism has a length a for the upper A-arm, b for the coupler link, c for the lower A-arm, and d for the ground link. The configuration of the suspension is determined by the angles \u03b82, \u03b83, and \u03b84, all measured from the positive direction of the x-axis. When the suspension is at its equilibrium position, the links of the double A-arm suspension make initial angles \u03b820 \u03b830, and \u03b840 with the x-axis. The equilibrium position of a suspension is called the rest position. To determine the camber angle during the fluctuation of the wheel, we should determine the variation of the coupler angle \u03b83, as a function of vertical motion z of the coupler point C. Using \u03b82 as a parameter, we can find the coordinates (xC , yC) of the 476 8. Suspension Mechanisms coupler point C in the suspension coordinate frame (x, y) as xC = a cos \u03b82 + e cos (p\u2212 q + \u03b1) (8.1) yC = a sin \u03b82 + e sin (p\u2212 q + \u03b1) (8.2) where, q = tan\u22121 a sin \u03b82 d\u2212 a cos \u03b82 (8.3) p = tan\u22121 q 4b2f2 \u2212 (b2 + f2 \u2212 c2) 2 b2 + f2 \u2212 c2 (8.4) f = p a2 + d2 \u2212 2ad cos \u03b82. (8.5) The position vector of the coupler point is uC uC = xC \u0131\u0302+ yC j\u0302 (8.6) and the unit vector in the z-direction is u\u0302z = \u2212 cos \u03b80\u0131\u0302\u2212 sin \u03b80j\u0302. (8.7) Therefore, the displacement z in terms of xC and yC is: z = uC \u00b7 u\u0302z = \u2212xC cos \u03b80 \u2212 yC sin \u03b80 (8.8) 8. Suspension Mechanisms 477 The initial coordinates of the coupler point C and the initial value of z are: xC0 = a cos \u03b820 + e cos (p0 \u2212 q0 + \u03b1) (8.9) yC0 = a sin \u03b820 + e sin (p0 \u2212 q0 + \u03b1) (8.10) z0 = \u2212xC0 cos \u03b80 \u2212 yC0 sin \u03b80 (8.11) and hence, the vertical displacement of the wheel center can be calculated by h = z \u2212 z0 (8.12) The initial angle of the coupler link with the vertical direction is \u03b80 \u2212 \u03b830. Therefore, the camber angle of the wheel would be \u03b3 = (\u03b80 \u2212 \u03b83)\u2212 (\u03b80 \u2212 \u03b830) = \u03b830 \u2212 \u03b83 (8.13) The angle of the coupler link with the x-direction is equal to \u03b83 = 2 tan \u22121 \u00c3 \u2212E \u00b1 \u221a E2 \u2212 4DF 2D ! (8.14) where, D = J5 \u2212 J1 + (1 + J4) cos \u03b82 (8.15) E = \u22122 sin \u03b82 (8.16) F = J5 + J1 \u2212 (1\u2212 J4) cos \u03b82 (8.17) and J1 = d a (8.18) J2 = d c (8.19) J3 = a2 \u2212 b2 + c2 + d2 2ac (8.20) J4 = d b (8.21) J5 = c2 \u2212 d2 \u2212 a2 \u2212 b2 2ab . (8.22) Substituting (8.14) and (8.13), and then, eliminating \u03b82 between (8.13) and (8.8) provides the relationship between the vertical motion of the wheel, z, and the camber angle \u03b3. 478 8. Suspension Mechanisms Example 319 F Camber angle and wheel fluctuations. Consider the double A-arm suspension that is shown in Figure 8.36. The dimensions of the equivalent kinematic model are: a = 22.4 cm b = 22.1 cm c = 27.3 cm d = 17.4 cm \u03b80 = 24.3 deg (8.23) The coupler point C is at: e = 14.8 cm \u03b1 = 54.8 deg (8.24) If the angle \u03b82 at the rest position is at \u03b820 = 121.5 deg (8.25) then the initial angle of the other links are: \u03b830 = 18.36 deg \u03b840 = 107.32 deg (8.26) At the rest position, the coupler point is at: xC0 = \u221222.73 cm yC0 = 9.23 cm z0 = 16.92 cm (8.27) We may calculate h and \u03b3 by varying the parameter \u03b82. Figure 8.37 illustrates h as a function of the camber angle \u03b3. For this suspension mechanism, the wheel gains a positive camber when the wheel moves up, and gains a negative camber when the it moves down. The mechanism is shown in Figure 8.38, when the wheel is at the rest position and has a positive or a negative displacement. 8.4 F Car Tire Relative Angles There are four major wheel alignment parameters that affect vehicle dynamics: toe, camber, caster, and trust angle. 8. Suspension Mechanisms 479 8.4.1 F Toe When a pair of wheels is set so that their leading edges are pointed toward each other, the wheel pair is said to have toe-in. If the leading edges point away from each other, the pair is said to have toe-out. Toe-in and toe-out front wheel configurations of a car are illustrated in Figure 8.39. The amount of toe can be expressed in degrees of the angle to which the wheels are not parallel. However, it is more common to express the toein and toe-out as the difference between the track widths as measured at the leading and trailing edges of the tires. Toe settings affect three major performances: tire wear, straight-line stability, and corner entry handling. For minimum tire wear and power loss, the wheels on a given axle of a car should point directly ahead when the car is running in a straight line. Excessive toe-in causes accelerated wear at the outboard edges of the tires, while too much toe-out causes wear at the inboard edges. Toe-in increases the directional stability of the vehicle, and toe-out increases the steering response. Hence, a toe-in setting makes the steering function lazy, while a toe-out makes the vehicle unstable. With four wheel independent suspensions, the toe may also be set at the rear of the car. Toe settings at the rear have the same effect on wear, directional stability, and turn-in as they do on the front. However, we usually do not set up a rear-drive race car toed out in the rear, because of excessive instability. When driving torque is applied to the wheels, they pull themselves forward and try to create toe-in. Furthermore, when pushed down the road, a non-driven wheel or a braking wheel will tend to toe-out. Example 320 Toe-in and directional stability. Toe settings have an impact on directional stability. When the steering 480 8. Suspension Mechanisms 8. Suspension Mechanisms 481 wheel is centered, toe-in causes the wheels to tent to move along paths that intersect each other in front of the vehicle. However, the wheels are in balance and no turn results. Toe-in setup can increase the directional stability caused by little steering fluctuations and keep the car moving straight. Steering fluctuations may be a result of road disturbances. If a car is set up with toe-out, the front wheels are aligned so that slight disturbances cause the wheel pair to assume rolling directions that approach a turn. Therefore, toe-out encourages the initiation of a turn, while toe-in discourages it. Toe-out makes the steering quicker. So, it may be used in vehicles for a faster response. The toe setting on a particular car becomes a trade-off between the straight-line stability afforded by toe-in and the quick steering response by toe-out. Toe-out is not desirable for street cars, however, race car drivers are willing to drive a car with a little directional instability, for sharper turn-in to the corners. So street cars are generally set up with toe-in, while race cars are often set up with toe-out. Example 321 Toe-in and toe-out in the front and rear axles. Front toe-in: slower steering response, more straight-line stability, greater wear at the outboard edges of the tires. Front toe-zero: medium steering response, minimum power loss, minimum tire wear. Front toe-out: quicker steering response, less straight-line stability, greater wear at the inboard edges of the tires. Rear toe-in: straight-line stability, traction out of the corner, more steerability, higher top speed. 482 8. Suspension Mechanisms 8.4.2 F Caster Angle Caster is the angle to which the steering pivot axis is tilted forward or rearward from vertical, as viewed from the side. Assume the wheel is straight to have the body frame and the wheel frame coincident. If the steering axis is turned about the wheel yw-axis then the wheel has positive caster. If the steering axis is turned about the wheel \u2212yw-axis, then the wheel has negative caster. Positive and negative caster configurations on the front wheel of a car are shown in Figure 8.40. Negative caster aids in centering the steering wheel after a turn and makes the front tires straighten quicker. Most street cars are made with 4\u22126 deg negative caster. Negative caster tends to straighten the wheel when the vehicle is traveling forward, and thus is used to enhance straight-line stability. Example 322 Negative caster of shopping carts. The steering axis of a shopping cart wheel is set forward of where the wheel contacts the ground. As the cart is pushed forward, the steering axis pulls the wheel along, and because the wheel drags along the ground, it falls directly in line behind the steering axis. The force that causes the wheel to follow the steering axis is proportional to the distance between the steering axis and the wheel-to-ground contact point, if the caster is small. This distance is referred to as trail. The cars\u2019 steering axis intersects the ground at a point in front of the tireprint, and thus the same effect as seen in the shopping cart casters is achieved. While greater caster angles improves straight-line stability, they also cause an increase in steering effort. Example 323 Characteristics of caster in front axle. Zero castor provides: easy steering into the corner, low steering out of the corner, low straight-line stability. 8. Suspension Mechanisms 483 Negative caster provides: low steering into the corner, easy steering out of the corner, more straight-line stability, high tireprint area during turn, good turn-in response, good directional stability, good steering feel. When a castered wheel rotates about the steering axis, the wheel gains camber. This camber is generally favorable for cornering. 8.4.3 F Camber Camber is the angle of the wheel relative to vertical line to the road, as viewed from the front or the rear of the car. Figure 8.41 illustrates the wheel number 1 of a vehicle. If the wheel leans in toward the chassis, it is called negative camber and if it leans away from the car, it is called positive camber. The cornering force that a tire can develop is highly dependent on its angle relative to the road surface, and so wheel camber has a major effect on the road holding of a car. A tire develops its maximum lateral force at a small camber angle. This fact is due to the contribution of camber thrust, which is an additional lateral force generated by elastic deformation as the tread rubber pulls through the tire/road interface. To optimize a tire\u2019s performance in a turn, the suspension should provide a slight camber angle in the direction of rotation. As the body rolls in a turn, the suspension deflects vertically. The wheel is connected to the chassis by suspension mechanism, which must rotate to allow for the wheel deflection. Therefore, the wheel can be subject to large camber changes as the suspension moves up and down. So, the more the wheel must deflect from its static position, the more difficult it is to maintain an ideal camber angle. Thus, the relatively large wheel travel and soft roll stiffness needed to provide a smooth ride in passenger cars presents a difficult design challenge, while the small wheel travel and high roll stiffness inherent in racing cars reduces the problem. Example 324 Castor versus camber. Camber doesn\u2019t improve turn-in as the positive caster does. Camber is not generally good for tire wear. Camber in one wheel does not improve directional stability. Camber adversely affects braking and acceleration efforts. 8.4.4 F Trust Angle The trust angle \u03c5 is the angle between vehicle\u2019s centerline and perpendicular to the rear axle. It compares the direction that the rear axle is aimed with the centerline of the vehicle. A nonzero angle configuration is shown in Figure 8.42. Zero angle confirms that the rear axle is parallel to the front axle, and the wheelbase on both sides of the vehicle are the same. A reason for nonzero 484 8. Suspension Mechanisms 8. Suspension Mechanisms 485 trust angle would have unequal toe-in or toe-out on both sides of the axle. Example 325 Torque reaction. There are two kinds of torque reactions in rear-whel-drive: 1\u2212 the reaction of the axle housing to rotate in the opposite direction of the crown wheel rotation, and 2\u2212 the reaction of axle housing to spin about its own center, opposite to the direction of pinion\u2019s rotation. The first reaction leads to a lifting force in the differential causing a wind-up in springs. The second reaction leads to a lifting force on the right wheels. 8.5 Suspension Requirements and Coordinate Frames The suspension mechanism should allow a relative motion between the wheel and the vehicle body. The relative motions are needed to pass the road irregularities and steering. To function properly, a suspension mechanism should have some kinematic and dynamics requirements. 8.5.1 Kinematic Requirements To express the motions of a wheel, we attach a wheel coordinate system W (oxwywzw) to the center of the wheel. A wheel, as a rigid body, has six degrees-of-freedom with respect to the vehicle body: three translations and three rotations, as shown in Figure 8.43. 486 8. Suspension Mechanisms The axes xw, yw, and zw indicate the direction of forward, lateral, and vertical translations and rotations. In the position shown in the figure, the rotation about the xw-axis is the camber angle, about the yw-axis is the spin, and about the zw-axis is the steer angle. Consider a non-steerable wheel. Translation in zw-direction and spin about the yw-axis are the only two DOF allowed for such a wheel. So, we need to take four DOF. If the wheel is steerable, then translation in the zw-direction, spin about the yw-axis, and steer rotation about the zw-axis are the three DOF allowed. So, we must take three DOF of a steerable wheel. Kinematically, non-steerable and steerable wheels should be supported as shown in Figures 8.44 and 8.45 respectively. Providing the required freedom, as well as eliminating the taken DOF, are the kinematic requirements of a suspension mechanism. 8.5.2 Dynamic Requirements Wheels should be able to propel, steer, and stop the vehicle. So, the suspension system must transmit the driving traction and deceleration braking forces between the vehicle body and the ground. The suspension members must also resist lateral forces acting on the vehicle. Hence, the wheel suspension system must make the wheel rigid for the taken DOF. However, there must also be some compliance members to limit the untaken DOF. The most important compliant members are spring and dampers to provide returning and resistance forces in the z-direction. 8. Suspension Mechanisms 487 8.5.3 Wheel, wheel-body, and tire Coordinate Frames Three coordinate frames are employed to express the orientation of a tire and wheel with respect to the vehicle: the wheel frame W , wheel-body frame C, and tire frame T . A wheel coordinate frame W (xw, yw, zw) is attached to the center of a wheel. It follows every translation and rotation of the wheel except the spin. Hence, the xw and zw axes are always in the tire-plane, while the yw-axis is always along the spin axis. A wheel coordinate frame is shown in Figure 8.43. When the wheel is straight and the W frame is parallel to the vehicle coordinate frame, we attach a wheel-body coordinate frame C (xc, yc, zc) at the center of the wheel parallel to the vehicle coordinate axes. The wheelbody frame C is motionless with respect to the vehicle coordinate and does not follow any motion of the wheel. The tire coordinate frame T (xt, yt, zt) is set at the center of the tireprint. The zt-axis is always perpendicular to the ground. The xt-axis is along the intersection line of the tire-plane and the ground. The tire frame does not follows the spin and camber rotations of the tire however, it follows the steer angle rotation about the zc-axis. Figure 8.46 illustrates a tire and a wheel coordinate frames. Example 326 Visualization of the wheel, tire, and wheel-body frames. Figure 8.47 illustrates the relative configuration of a wheel-body frame C, a tire frame T , and a wheel frame W . If the steering axis is along the zc-axis then, the rotation of the wheel about the zc-axis is the steer angle \u03b4. Rotation about the xt-axis is the camber angle \u03b3. 488 8. Suspension Mechanisms Generally speaking, the steering axis may have any angle and may go through any point of the ground plane. Example 327 Wheel to tire coordinate frame transformation. If TdW indicates the T -expression of the position vector of the wheel frame origin relative to the tire frame origin, then having the coordinates of a point P in the wheel frame, we can find its coordinates in the tire frame using the following equation. T rP = TRW W rP + TdW (8.28) If W rP indicates the position vector of a point P in the wheel frame, W rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 (8.29) then the coordinates of the point P in the tire frame T rP are T rP = TRW W rP + Td = TRW W rP + TRW W T dW = \u23a1\u23a3 xP yP cos \u03b3 \u2212Rw sin \u03b3 \u2212 zP sin \u03b3 Rw cos \u03b3 + zP cos \u03b3 + yP sin \u03b3 \u23a4\u23a6 (8.30) 8. Suspension Mechanisms 489 where, WT dW is the W -expression of the position vector of the wheel frame in the tire frame, Rw is the radius of the tire, and TRW is the rotation matrix to go from the wheel frame W to the tire frame T . TRW = \u23a1\u23a3 1 0 0 0 cos \u03b3 \u2212 sin \u03b3 0 sin \u03b3 cos \u03b3 \u23a4\u23a6 (8.31) W T dW = \u23a1\u23a3 0 0 Rw \u23a4\u23a6 . (8.32) As an example, the center of the wheel W rP = W ro = 0 is the origin of the wheel frame W , that is at T ro = TdW = TRW W T dW = \u23a1\u23a3 0 \u2212Rw sin \u03b3 Rw cos \u03b3 \u23a4\u23a6 (8.33) in the tire coordinate frame T . 490 8. Suspension Mechanisms Example 328 F Tire to wheel coordinate frame transformation. If rP indicates the position vector of a point P in the tire coordinate frame, T rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 (8.34) then the position vector W rP of the point P in the wheel coordinate frame is W rP = WRT T rP \u2212 W T dW (8.35) = \u23a1\u23a3 xP yP cos \u03b3 + zP sin \u03b3 zP cos \u03b3 \u2212Rw \u2212 yP sin \u03b3 \u23a4\u23a6 because WRT = \u23a1\u23a3 1 0 0 0 cos \u03b3 sin \u03b3 0 \u2212 sin \u03b3 cos \u03b3 \u23a4\u23a6 (8.36) WdT = \u23a1\u23a3 0 0 Rw \u23a4\u23a6 (8.37) and we may multiply both sides of Equation (8.28) by TRT W to get TRT W T rP = W rP + TRT W TdW (8.38) = W rP + W T dW W rP = WRT T rP \u2212 W T dW . (8.39) As an example, the center of tireprint in the wheel frame is at W rP = \u23a1\u23a3 1 0 0 0 cos \u03b3 \u2212 sin \u03b3 0 sin \u03b3 cos \u03b3 \u23a4\u23a6T \u23a1\u23a3 0 0 0 \u23a4\u23a6\u2212 \u23a1\u23a3 0 0 Rw \u23a4\u23a6 = \u23a1\u23a3 0 0 \u2212Rw \u23a4\u23a6 . (8.40) Example 329 F Wheel to tire homogeneous transformation matrices. The transformation from the wheel to tire coordinate frame may also be expressed by a 4\u00d7 4 homogeneous transformation matrix TTW , T rP = TTW W rP = \u2219 TRW TdW 0 1 \u00b8 W rP (8.41) where TTW = \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 cos \u03b3 \u2212 sin \u03b3 \u2212Rw sin \u03b3 0 sin \u03b3 cos \u03b3 Rw cos \u03b3 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 . (8.42) 8. Suspension Mechanisms 491 The corresponding homogeneous transformation matrix WTT from the tire to wheel frame would be WTT = \u2219 WRT WdT 0 1 \u00b8 = \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 cos \u03b3 \u2212 sin \u03b3 0 0 sin \u03b3 cos \u03b3 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 . (8.43) It can be checked that WTT = TT\u22121W , using the inverse of a homogeneous transformation matrix rule. TT\u22121W = \u2219 TRW TdW 0 1 \u00b8\u22121 = \u2219 TRT W \u2212 TRT W TdW 0 1 \u00b8 = \u2219 WRT \u2212WRT TdW 0 1 \u00b8 (8.44) Example 330 F Tire to wheel-body frame transformation. The origin of the tire frame is at CdT in the wheel-body frame. CdT = \u23a1\u23a3 0 0 \u2212Rw \u23a4\u23a6 (8.45) The tire frame can steer about the zc-axis with respect to the wheel-body frame. The associated rotation matrix is CRT = \u23a1\u23a3 cos \u03b4 \u2212 sin \u03b4 0 sin \u03b4 cos \u03b4 0 0 0 1 \u23a4\u23a6 (8.46) Therefore, the transformation between the tire and wheel-body frames can be expressed by Cr = CRT T r+ CdT (8.47) or equivalently, by a homogeneous transformation matrix CTT . CTT = \u2219 CRT CdT 0 1 \u00b8 = \u23a1\u23a2\u23a2\u23a3 cos \u03b4 \u2212 sin \u03b4 0 0 sin \u03b4 cos \u03b4 0 0 0 0 1 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 (8.48) As an example, the wheel-body coordinates of the point P on the tread of a 492 8. Suspension Mechanisms negatively steered tire at the position shown in Figure 8.48, are: Cr = CTT T rP = \u23a1\u23a2\u23a2\u23a3 cos\u2212\u03b4 \u2212 sin\u2212\u03b4 0 0 sin\u2212\u03b4 cos\u2212\u03b4 0 0 0 0 1 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 \u23a1\u23a2\u23a2\u23a3 Rw 0 Rw 1 \u23a4\u23a5\u23a5\u23a6 = \u23a1\u23a2\u23a2\u23a3 Rw cos \u03b4 \u2212Rw sin \u03b4 0 1 \u23a4\u23a5\u23a5\u23a6 (8.49) The homogeneous transformation matrix for tire to wheel-body frame TTC is: TTC = CT\u22121T = \u2219 CRT CdT 0 1 \u00b8\u22121 = \u2219 CRT T \u2212CRT T CdT 0 1 \u00b8 = \u2219 CRT T \u2212 T CdT 0 1 \u00b8 = \u23a1\u23a2\u23a2\u23a3 cos \u03b4 sin \u03b4 0 0 \u2212 sin \u03b4 cos \u03b4 0 0 0 0 1 Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 (8.50) Example 331 F Cycloid. Assume that the wheel in Figure 8.48 is turning with angular velocity \u03c9 and has no slip on the ground. If the point P is at the center of the tireprint 8. Suspension Mechanisms 493 when t = 0, MrP = \u23a1\u23a3 0 0 \u2212Rw \u23a4\u23a6 (8.51) then we can find its position in the wheel frame at a time t by employing another coordinate frame M . The frame M is called the rim frame and is stuck to the wheel at its center. Because of spin, the M frame turns about the yw-axis, and therefore, the rotation matrix to go from the rim frame to the wheel frame is: WRM = \u23a1\u23a3 cos\u03c9t 0 sin\u03c9t 0 1 0 \u2212 sin\u03c9t 0 cos\u03c9t \u23a4\u23a6 (8.52) So the coordinates of P in the wheel frame are: W rP = WRM MrP = \u23a1\u23a3 \u2212Rw sin t\u03c9 0 \u2212Rw cos t\u03c9 \u23a4\u23a6 (8.53) The center of the wheel is moving with speed vx = Rw\u03c9 and it is at Gr =\u00a3 vxt 0 Rw \u00a4 in the global coordinate frame G on the ground. Hence, the coordinates of point P in the global frame G, would be GrP = W rP + \u23a1\u23a3 vxt 0 Rw \u23a4\u23a6 = \u23a1\u23a3 Rw (\u03c9t\u2212 sin t\u03c9) 0 Rw (1\u2212 cos t\u03c9) \u23a4\u23a6 . (8.54) The path of motion of point P in the (X,Z)-plane can be found by eliminating t between X and Z coordinates. However, it is easier to expressed the path by using \u03c9t as a parameter. Such a path is called cycloid. In general case, point P can be at any distance from the center of the rim frame. If the point is at a distance d 6= Rw, then its path of motion is called the trochoid. A trochoid is called a curtate cycloid when d < Rw and a prolate cycloid when d > Rw. Figure 8.49(a)-(c) illustrate a cycloid, curtate cycloid, and prolate cycloid respectively. Example 332 F Wheel to wheel-body frame transformation. The homogeneous transformation matrix CTW to go from the wheel frame 494 8. Suspension Mechanisms to the wheel-body frame can be found by combined transformation. CTW = CTT TTW (8.55) = \u23a1\u23a2\u23a2\u23a3 c\u03b4 \u2212s\u03b4 0 0 s\u03b4 c\u03b4 0 0 0 0 1 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 c\u03b3 \u2212s\u03b3 \u2212Rw sin \u03b3 0 s\u03b3 c\u03b3 Rw cos \u03b3 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 = \u23a1\u23a2\u23a2\u23a3 cos \u03b4 \u2212 cos \u03b3 sin \u03b4 sin \u03b3 sin \u03b4 Rw sin \u03b3 sin \u03b4 sin \u03b4 cos \u03b3 cos \u03b4 \u2212 cos \u03b4 sin \u03b3 \u2212Rw cos \u03b4 sin \u03b3 0 sin \u03b3 cos \u03b3 Rw cos \u03b3 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 If rP indicates the position vector of a point P in the wheel coordinate frame, W rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 (8.56) then the homogeneous position vector CrP of the point P in the wheel-body 8. Suspension Mechanisms 495 coordinate frame is: CrP = CTW W rP = \u23a1\u23a2\u23a2\u23a3 xP cos \u03b4 \u2212 yP cos \u03b3 sin \u03b4 + (Rw + zP ) sin \u03b3 sin \u03b4 xP sin \u03b4 + yP cos \u03b3 cos \u03b4 \u2212 (Rw + zP ) cos \u03b4 sin \u03b3 \u2212Rw + (Rw + zP ) cos \u03b3 + yP sin \u03b3 1 \u23a4\u23a5\u23a5\u23a6 (8.57) The position of the wheel center W r = 0, for a cambered and steered wheel is at Cr = CTW W r = \u23a1\u23a2\u23a2\u23a3 Rw sin \u03b3 sin \u03b4 \u2212Rw cos \u03b4 sin \u03b3 \u2212Rw(1\u2212 cos \u03b3) 1 \u23a4\u23a5\u23a5\u23a6 (8.58) The zc = Rw (cos \u03b3 \u2212 1) indicates how much the center of the wheel comes down when the wheel cambers. If the wheel is not steerable, then \u03b4 = 0 and the transformation matrix CTW reduces to CTW = \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 cos \u03b3 \u2212 sin \u03b3 \u2212Rw sin \u03b3 0 sin \u03b3 cos \u03b3 Rw (cos \u03b3 \u2212 1) 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 (8.59) that shows CrP = CTW W rP = \u23a1\u23a2\u23a2\u23a3 xP yP cos \u03b3 \u2212Rw sin \u03b3 \u2212 zP sin \u03b3 zP cos \u03b3 + yP sin \u03b3 +Rw (cos \u03b3 \u2212 1) 1 \u23a4\u23a5\u23a5\u23a6 (8.60) Example 333 F Tire to vehicle coordinate frame transformation. Figure 8.50 illustrates the first and fourth tires of a 4-wheel vehicle. There is a body coordinate frame B (x, y, z) attached to the mass center C of the vehicle. There are also two tire coordinate frames T1 (xt1 , yt1 , zt1) and T4 (xt4 , yt4 , zt4) attached to the tires 1 and 4 at the center of their tireprints. The origin of the tire coordinate frame T1 is at Bd1 BdT1 = \u23a1\u23a3 a1 \u2212b1 \u2212h \u23a4\u23a6 (8.61)"
        ]
    },
    {
        "image_filename": "designv10_0_0001630_j.msea.2017.02.038-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001630_j.msea.2017.02.038-Figure3-1.png",
        "caption": "Fig. 3. (a) Schematic representation of the 760 \u00b0C tensile testing setup; (b) thermocouples positioning.",
        "texts": [
            " Inverse pole figures were calculated using a 1\u00d71 mm2 area to eliminate local heterogeneities of the specimens. The dumbbell-shaped specimens (Fig. 1b) with a 2\u00d74 mm2 gauge section were tested in tension at room temperature (RT) and at 760 \u00b0C (1400 \u00b0F). The tensile tests were performed under a constant strain rate of 10\u22123 s\u22121 using an MTS 810 system. For RT testing, this system was coupled with an MTS 634.12 contact extensometer (initial gauge length of 20 mm). For high-temperature testing, the specimens were heated up to 760 \u00b0C with a 1 \u00b0C/s rate, using an infrared radiant heating chamber (Fig. 3a), and then maintained for 10 min at this temperature before testing. The thermal field uniformity was assessed using three K-type thermocouples distributed along the specimen gauge length and placed in contact with the specimen, the central thermocouple being used for temperature control (Fig. 3b). The hightemperature testing was performed under Ar atmosphere with a flow rate of 5\u201318 L/h. Three specimens were tested for each build orientation. The relative orientation dependence of mechanical properties as a function of the build direction (L-PBF anisotropy) was calculated as follows: \u0394x x x x = ( \u2212 ) \u22c5100%max min max where x represents either the yield stress (YS), the ultimate tensile strength (UTS), or the elongation (\u03b4). For the as-built alloy, Fig. 4a and b show overlapped, bowl-shaped melt pool contours in the vertical plane and stripe patterns in the horizontal plane, with the latter reflecting the EOS laser scanning strategy involving a 67\u00b0 hatch rotation between two successively built layers"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure6.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure6.2-1.png",
        "caption": "Figure 6.2 (a) When a block is suspended by three wires, and one of the wires suddenly breaks, there is a sudden change in the loading on the remaining wires. (b) Model of the cableway used to close the Haringvliet. Discarding a concrete block causes a sudden change in the loading on the cable gondola.",
        "texts": [
            " In an earthquake, one refers to a prescribed displacement: the earth starts to move and the structure is forced to follow the movement of the earth via its foundations. In general, one can distinguish between four different types of dynamic loading: \u2022 Periodic loads (Figure 6.1b) This type of load is caused, for example, by rotating machines, ringing bells, eddies in a stream, or people jumping on a floor. \u2022 Suddenly-applied loads (Figure 6.1c) This could include a load resulting from a snapping wire (see Figure 6.2a). Another example is the cableway in Figure 6.2b, which was used at the Haringvliet1 dam to unload concrete blocks. \u2022 Loads of short duration and collision phenomena (impact loads) (Figure 6.1d) 1 A see arm. The enclosure of the Haringvliet is one of the Delta Works in the Netherlands. 208 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Examples include explosions, wave impact, gusts of wind, or a falling pile hammer on a pile. \u2022 Stochastic loads (Figure 6.1e) This includes loads of a variable and unpredictable character, such as those resulting from wind, waves, traffic or earthquakes"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure3.12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure3.12-1.png",
        "caption": "FIGURE 3.12. Side view of a normal force Fz and stress \u03c3z applied on a stationary tire.",
        "texts": [
            " Due to equilibrium conditions, the overall integral of the normal stress over the tireprint area AP must be equal to the normal load Fz, and the integral of shear stresses must be equal to zero. Z AP \u03c3z(x, y) dA = Fz (3.20)Z AP \u03c4x(x, y) dA = 0 (3.21)Z AP \u03c4y(x, y) dA = 0 (3.22) 3.3.1 Static Tire, Normal Stress Figure 3.11 illustrates a stationary tire under a normal load Fz along with the generated normal stress \u03c3z applied on the ground. The applied loads on the tire are illustrated in the side view shown in Figure 3.12. For a stationary tire, the shape of normal stress \u03c3z(x, y) over the tireprint area depends on tire and load conditions, however its distribution over the tireprint is generally in the shape shown in Figure 3.13. 3. Tire Dynamics 105 106 3. Tire Dynamics The normal stress \u03c3z(x, y) may be approximated by the function \u03c3z(x, y) = \u03c3zM \u00b5 1\u2212 x6 a6 \u2212 y6 b6 \u00b6 (3.23) where a and b indicate the dimensions of the tireprint, as shown in Figure 3.14. The tireprints may approximately be modeled by a mathematical function x2n a2n + y2n b2n = 1 n \u2208 N"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure3.57-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure3.57-1.png",
        "caption": "Figure 3.57 The forces Av, Bv and Cv as they are actually acting on the structure.",
        "texts": [
            " By choosing the equilibrium equations carefully, and by applying them in a carefully chosen order, it is sometimes possible to cut back on the amount of calculation needed. Cv is derived directly from \u2211 Ty = 0: \u2211 Ty = \u2212(30 kNm) \u2212 (40 kN) \u00d7 (1 m) + Cv \u00d7 2 = 0 \u21d2 Cv = +35 kN. ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM 3 Statics of a Rigid Body 95 Next, we find Av directly from \u2211 Tx = 0: \u2211 Tx = \u2212Av \u00d7 (5 m) + (60 kN) \u00d7 (3 m) + (50 kNm) \u2212Cv \u00d7 (3 m) + (100 kN) \u00d7 (1 m) + (40 kN) \u00d7 (1 m) = 0 \u21d2 Av = +53 kN, after which Bv follows directly from \u2211 Fz = 0: \u2211 Fz = +{(100 + 40 + 60) kN} \u2212 Av \u2212 Bv \u2212 Cv = 0 \u21d2 Bv = +112 kN. Figure 3.57 shows the forces Av, Bv and Cv as they act on the structure in reality. To check, one could also have a look at the moment equilibrium at a point other than the origin, such as point A: \u2211 Tx |A = \u2212{(60 \u2212 35) kN} \u00d7 (2 m) + (50 kNm) \u2212 \u2212{(100 + 40) kN} \u00d7 (4 m) + (112 kN) \u00d7 (5 m) = 0. The moment equilibrium is also met about a line through A parallel to the x axis. Example 2 In Figure 3.58, a cube with edge length a and weight G is kept in equilibrium by the six forces F1 to F6. For the angle \u03b1 between the lines of action of the forces applies tan \u03b1 = 3/4"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001049_j.pmatsci.2020.100703-Figure4-36-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001049_j.pmatsci.2020.100703-Figure4-36-1.png",
        "caption": "Fig. 4-36 Residual stress during the (a) 2nd, (b) 4th, (c) 6th, (d) 8th, and (e) 10th layers during the DED-L processing of an IN718 thin-wall part. Simulation is for 300W laser power at 11 mm/s scanning speed. Reproduced from [27].",
        "texts": [
            " These models require temperature dependent mechanical properties of alloys that are difficult to obtain and are often estimated using commercial software packages. For PBF-L and PBF-EB, rapid scanning of a small beam makes a myriad of complex scanning strategies and part geometries feasible, leading to differences in the heating and cooling of the material both spatially and temporally. Fig. 4-35 shows how different scanning strategies lead to different residual stress states for the same material and processing parameters [211, 471]. Similar models have been developed for DED-L processes [27, 111, 260, 472-476], an example of which is shown in Fig. 4-36. Notably, most of the models show that the edges of the material where the part joins the base plate are 105 subject to the highest residual stresses. If these stresses surpass the yield strength of the material, it is possible that delamination or cracking would cause a build failure during AM process. Fig. 4-37 shows that the highest accumulation of residual stresses occurs near the top of the deposit for DED-GMA due to the cooling of bulk deposition [260]. However, these stresses are partially alleviated when the component and the substrate are released from the clamps resulting in high compressive stresses in the substrate [260]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure11.7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure11.7-1.png",
        "caption": "Figure 11.7 (a) A beam fixed at A and loaded at its free end B by a force F normal to the beam axis. (b) The boundary conditions V = +F and M = 0 are found from the force and moment equilibrium of a small boundary segment at B (with x \u2192 0).",
        "texts": [
            " This is illustrated using the following four examples: 1. A fixed beam, loaded at its free end by a concentrated load. 2. A simply supported beam and a beam fixed at one of its ends, both with a uniformly distributed load along its entire length. 3. A simply supported beam with a triangular load. 4. A simply supported beam with overhang (cantilever beam) and a uniformly distributed load along its entire length. 11 Mathematical Description of the Relationship between Section Forces and Loading 443 Example 1 Figure 11.7a shows a beam AB fixed at A and of length . At its free end B the beam is loaded normal to its axis by a force F . Question: Determine the V and M diagrams using the differential equations for equilibrium. Solution: For 0<x< it holds that d2M dx2 = \u2212qz = 0. Repeated integration gives dM dx = V = C1, (a) M = C1x + C2. (b) In an unloaded field the shear force is constant and the bending moment is linear. The integration constants C1 and C2 are found from the boundary conditions at the free end B. Here both V and M have a prescribed value: the shear force is equal to F (pay attention to the sign), and the bending moment is zero: x = : V = +F, (c) x = : M = 0. (d) The boundary conditions can also be derived from the force and moment equilibrium of a small member segment with length x ( x \u2192 0) at the 444 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM free end B (see Figure 11.7b). The section forces V and M , which here are acting on a negative section plane, have to be drawn in accordance with their positive directions. The equilibrium of the member segment gives \u2211 Fz = \u2212V + F = 0 \u21d2 V = +F,\u2211 Tz|B = \u2212M \u2212 V x = 0 (with x \u2192 0) \u21d2 M = 0. Substitute (c) in (a) and (d) in (b); elaboration of the boundary conditions leads to C1 = +F, C2 = \u2212F . This gives the variation of shear force V and bending moment M for beam AB: V = F, M = Fx \u2212 F = \u2212F( \u2212 x). The V and M diagrams are shown in Figure 11"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001236_j.euromechsol.2007.11.005-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001236_j.euromechsol.2007.11.005-Figure5-1.png",
        "caption": "Fig. 5. Schematic graph of spalling.",
        "texts": [
            " In the absence of the damage on a tooth\u2019s surface, the contact between teeth in contact is in a straight line along the width W . However, when there is spalling damage, the width of contact changes at the defect location because of the removal of material from the surface of the tooth. It is assumed here that no tooth contact occurs in the spalled area. On the other hand, the area moment of inertia and the surface of the cross sectional areas of the tooth in the affected area are diminished. Spalling is modelled as a rectangular indentation having the dimensions as \u00d7 ws \u00d7 hs as shown in Fig. 5. W is the width of contact of two meshing teeth for the healthy case. In the presence of the spalling this width becomes: W1 = W \u2212 ws (13) {t1, . . . , t4} are the reduced tooth thickness in the affected zone (sections Ss in Fig. 5). All these parameters are introduced in Eqs. (1), (6) and (9) by modifying area moment of inertia, width and cross sectional areas. The tooth width W is updated to its new value W1 in Eqs. (6) and (9). In Eq. (1) describing the bending deflection expression, the computation of the summation have to be done taking into account the changed shape of the elementary cross sectional areas Ai and area moment of inertia Ii as shown in Fig. 7(A). Tooth deflections and gearmesh stiffness are then computed"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure4.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure4.1-1.png",
        "caption": "Fig. 4.1. Double-sided AFPM brushless servo motor with the stator ferromagnetic core and built-in brake. Courtesy of Mavilor , Barcelona, Spain.",
        "texts": [],
        "surrounding_texts": [
            "An example of a commercial double-sided AFPM servo motor with ferromagnetic core is shown in Figs. 4.1 and 4.2. External stators have slotted ring-shaped cores made of nonoriented electrotechnical steel ribbon. The inner rotor does not have any ferromagnetic material. PMs are mounted on a nonmagnetic rotating disc. Fig. 4.3 shows a double-sided AFPM synchronous generator with the stator core wound of amorphous alloy ribbon manufactured by LE Incorporated , Indianapolis, IN, U.S.A. The volume of LE AFPM generators is approximately 60% lower than that of classical synchronous generators of the same rating. 4.3 Some Features of Iron-Cored AFPM Machines 125"
        ]
    },
    {
        "image_filename": "designv10_0_0002797_j.jsv.2021.116029-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002797_j.jsv.2021.116029-Figure7-1.png",
        "caption": "Fig. 7. Deformation of spherical asperity on the rough surface.",
        "texts": [
            " Greenwood and Williamson [31] developed a classic method to depict the contact between real roughness surfaces. In Greenwood-Williamson model (GW model), the summits of asperities on the roughness surfaces are all assumed as spheres with the same diameter. The summit heights of asperities are assumed to have Gaussian (normal) distribution. The asperities are uniformly distributed over the rough surface considering a certain density in each unit zone. When a roughness surface makes contact with a smooth surface, the deformation of spherical asperity on the rough surface is shown in Fig. 7 . In GW model, the mean height for summits is higher than the mean height for rough surface with Zs . The separation of summits mean plane and mean plane of smooth surface is d a , the summit height of asperity is Z s ; then, when Z s > d a , the asperity can be deformed. Thus, the probability of an asperity deform is represented as [32] P [ Z s > d a ] = \u222b \u221e d a f ( Z s ) d Z s = \u222b \u221e d a \u03c3s \u03c6( x ) dx = F 0 ( d a \u03c3s ) (46) where, f ( Z s ) describes the probability density function; \u03c6( x ) denotes the standard normal probability density function; \u03c3 s descibes the standard deviation of Z s ; and the value of F 0 ( x ) can be obtained by lookup table [32] "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure7.22-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure7.22-1.png",
        "caption": "Figure 7.22 The distribution of the water pressure on a 1-metre wide strip from the slide.",
        "texts": [
            " The support reaction Br in B is found from the force equilibrium: Br = R1 + R2 + R3 \u2212 Ar = 123.7 kN. Example 2 The moveable dam in Figure 7.21 consists of a circular cylindrical slide AB hinged at C and joining a rigid vertical partition wall at A. There is no water to the right of the dam. The specific weight of water is \u03b3w = 10 kN/m3. All other information required can be found in the figure. Question: Find the magnitude and direction of the resultant water pressure on a 1- metre strip from the circular cylindrical slide. Solution: In Figure 7.22, the 1-metre wide strip from the slide is modelled as a line element. The figure also shows the water pressure, increasing from 14 kN/m at top A of the slide to 40 kN/m near base B. The water pressure is acting normal to the slide everywhere. In other words, all the forces on the slide pass through C, the centre of arc AB. Therefore, the resultant R of the total water pressure on the slide also passes through C. 7 Gas Pressure and Hydrostatic Pressure 259 To determine the resultant water pressure, please refer to Figure 7"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure2.4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure2.4-1.png",
        "caption": "FIGURE 2.4. A motorcycle with different front and rear tires.",
        "texts": [
            " To eliminate the suspension deflection, we should lock the suspension, usually by replacing the shock absorbers with rigid rods to keep the vehicle at its ride height. Example 44 Different front and rear tires. Depending on the application, it is sometimes necessary to use different type of tires and wheels for front and rear axles. When the longitudinal position of C for a symmetric vehicle is determined, we can find the height of C by measuring the load on only one axle. As an example, consider the motorcycle in Figure 2.4. It has different front and rear tires. Assume the load under the rear wheel of the motorcycle Fz is known. The height h of C can be found by taking a moment of the forces about the tireprint of the front tire. h = Fz2 (a1 + a2) mg \u2212 a1 cos \u00b5 sin\u22121 H a1 + a2 \u00b6 + Rf +Rr 2 (2.35) Example 45 Statically indeterminate. A vehicle with more than three wheels is statically indeterminate. To determine the vertical force under each tire, we need to know the mechanical properties and conditions of the tires, such as the value of deflection at the center of the tire, and its vertical stiffness"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.34-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.34-1.png",
        "caption": "FIGURE 7.34. Illustration of a negative four-wheel steering vehicle in a left turn.",
        "texts": [
            "73) expresses the kinematic condition for both, positive and negative 4WS systems. Employing the wheel coordinate frame (xw, yw, zw), we define the steer angle as the angle between the vehicle x-axis and the wheel xw-axis, measured about the z-axis. Therefore, a steer angle is positive when the wheel is turned to the left, and it is negative when the wheel is turned to the right. Proof. The slip-free condition for wheels of a 4WS in a turn requires that the normal lines to the center of each tire-plane intersect at a common point. This is the kinematic steering condition. Figure 7.34 illustrates a positive 4WS vehicle in a left turn. The turning center O is on the left, and the inner wheels are the left wheels that are closer to the turning center. The longitudinal distance between point O and the axles of the car are indicated by c1, and c2 measured in the body 7. Steering Dynamics 411 coordinate frame. The front inner and outer steer angles \u03b4if , \u03b4of may be calculated from the triangles 4OAE and 4OBF , while the rear inner and outer steer angles \u03b4ir, \u03b4or may be calculated from the triangles 4ODG and 4OCH as follows"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003443_0278364903022001005-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003443_0278364903022001005-Figure8-1.png",
        "caption": "Fig. 8. 3RITz 4-DoF 4 \u2212 xRuPxiUN jRN mechanism.",
        "texts": [
            " When all the limb constraint forces are coplanar and passing through a common point, they form a desired 2-system as shown in eq (28). Based on the above analysis, we can obtain the characteristics of constraint and structure of such 4-DoF PMs, as shown in Table A4 in the Appendix. The enumeration of such 4-DoF PMs is shown in Table 4, in which 2 \u2264 p \u2264 4. Actually, the 4 \u2212 xRxR(iRjRkR)N PM was first proposed by Zaltnov and Gosselin (2001). Consider a xRuPxR(iRjR)N limb. xRiR can form a universal joint xiUN. Figure 8 shows a 4 \u2212 xRuPxiUN jRN PM. The constraint force $r i1 passes through the mechanism central point and is parallel to the first revolute axis$i1. The four limbs exert four constraint forces on the moving platform. These forces Table 4. Symmetrical 3RITz 4-DoF PMs g/p PMs 4 p \u2212 zP(iRjRkR)N 5 p \u2212 xRxRxR(iRjR)N p \u2212 xRxR(iRjRkR)N p \u2212 uPxRxR(iRjR)N p \u2212 xRuPxR(iRjR)N p \u2212 xRxRuP(iRjR)N p \u2212 uPxR(iRjRkR)N p \u2212 xRuP(iRjRkR)N are coplanar and always parallel to the base plane, intersecting at the mechanism central point"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure1.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure1.3-1.png",
        "caption": "Fig. 1.3: A modern reconstruction of the Chinese Wooden Cow that was designed and built in Vth century B.C. (from the Ancient Chinese Machines Foundation at National Cheng Kung University in Tainan).",
        "texts": [
            " In the liber X Vitruvius treats in detail mechanisms and mechanical design of machines. But technical evolution in machine design was important not only in the western world. For example, in China the culture reached heights that also needed mature technology. In the field of automation Chinese designers developed brilliant solutions. A very significant example is the Wooden Cow that was used for transportation purposes of heavy loads by using a one degree of freedom (d.o.f.) walking machine, as shown in Fig. 1.3. Another very significant example can be selected from the Arabic world which achieved great technical designs during the Middle Ages. An example is reported in Fig. 1.4, where an intriguing mechanisms is shown to operate as an automaton that provides water for washing hands and then a towel for drying hands, and finally it goes back to a start configuration. The mechanical design of the Fundamentals of Mechanics of Robotic Manipulation 9 automaton is based on a great knowledge of mechanics and hydraulics, that was only partially a heritage from the Greek culture, but it was evolved in practical applications to consciously interact with the environment"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003858_tbme.2005.846734-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003858_tbme.2005.846734-Figure6-1.png",
        "caption": "Fig. 6. Examples of repulsive potential field shapes: (a) circular, (b) elliptical, and (c) hemi-elliptical. In (d), two distinct points near the needle tip experience different repulsion forces, and a moment is applied to the line segment connecting these two points. This is the repulsive torque at the tip.",
        "texts": [
            " Obstacle volumes can be discretised (for example, the finite element analysis nodes may be used) and repulsive fields corresponding to each of the nodes in the discretization can be superimposed by straight addition as above. Alternatively, techiques from [21] can be used to construct potential functions. The scaling factors in (4) and in (6) are chosen such that the attraction and repulsion potentials are of the same order of magnitude. For a point obstacle, the resulting repulsion force is isotropic within a circular region of influence. Such a field is not feasible for needle motion planning because of the large lateral deviation that may be required at the needle tip (illustrated in Fig. 6(a)). Needle motion becomes increasingly constrained as the needle penetrates further into the tissue (i.e., the condition number of increases with needle depth). Therefore, needle trajectory compensation must be made early, and is therefore facilitated by selecting an elliptical or hemielliptical repulsive potential field shape, as shown in Fig. 6(b) and (c). An elliptical region of repulsion is defined in Cartesian coordinates by , and by in polar coordinates (7) where and are the radii at the major and minor axes of the ellipse. The normal at a point on is determined by finding the gradient of at (8) where is the orientation angle of the major axis, chosen to be . The anisotropic, elliptical repulsive force field is (9) where is the Euclidean distance , and the radius that defines the region of repulsion is . is , and is . and are the Cartesian coordinates of the obstacle, target and needle tip, respectively. In practice, it is necessary to match the field gradient along the major and minor ellipse axes by scaling both and by in (9) (10) This defines the translational case, where lies in . The effect of a repulsive potential field on needle tip heading is included by considering the repulsion force at two points (at the needle tip), and (on the needle shaft, a short distance from ). This is shown in Fig. 6(d). The force \u201cexerted\u201d by the field at each end of the line segment that connects and results in an effective moment that is used to guide angular motion at the needle tip. The choice of is somewhat arbitrary, since it simply allows for the computation of an orienting torque (11) where and are the positions of and , while and are the potentials computed at and , respectively. Examples of repulsive potentials and forces for the environment shown in Fig. 4 are plotted in Fig. 7. Angular potentials and forces are not shown"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002208_j.ymssp.2019.05.014-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002208_j.ymssp.2019.05.014-Figure6-1.png",
        "caption": "Fig. 6. Experimental equipment: (a) power flow enclosed gear rotary test machine [41]; (b) crack paths under different running cycles.",
        "texts": [
            " Considering the complex gear foundation structure and the real crack propagation paths of spur gears, the calculation method of the mesh stiffness at a certain contact position is described above and the flow chart is shown in Fig. 5. Furthermore, the time-varying mesh stiffness of spur gears can be obtained. Based on the fracture mechanics theory, the plane crack propagation paths of spur gears are simulated and the results are verified by the results of the extended finite element method and the experimental method. Firstly, the experimental crack paths are obtained using the power flow enclosed gear rotary test machine (see Fig. 6a and Ref. [41]). Then, using the parameters of the gear pair 1 (see Table 1), under the torque T = 800 N m and the rotational speed r = 1460 r/min, the crack propagation process of the experimental gear is finished after running for 120 h continuously. The crack propagation process under different running stages is shown in Fig. 6b. In order to make the proposed finite element method and the extended finite element method have the same initial crack parameters as the experimental results, the starting point of the initial crack coincides with the position of the maximum tooth root stress. The initial crack length lI and the initial crack propagation angle a are 0.2 mm and 45 , respectively (see Fig. 4b). The crack propagation paths of the proposed method, the extended finite element method and the experimental method are shown in Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001270_jproc.2020.3046112-Figure19-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001270_jproc.2020.3046112-Figure19-1.png",
        "caption": "Fig. 19. Outer rotor Halbach configuration (18-slot/12-pole).",
        "texts": [
            " Therefore, the power density, efficiency, and power factor of the electric machines can be improved significantly using this material. The main impediment for the use of Hyperco in automotive traction applications is its high cost. The advanced lamination materials are summarized in Table 3. A couple of promising HRE-free machine designs to meet the DOE 2025 goals, including relative advantages and disadvantages, are discussed in this section. 1) Outer Rotor Halbach PM Machine With Slotted Stator: This configuration uses outer rotor topology to maximize the torque density (as opposed to inner rotor topology), as shown in Fig. 19; the outer rotor also naturally provides natural retention against centrifugal force. The Halbach configuration maximizes the airgap flux density due to flux concentration. In addition, the stator is based on fractional slot concentrated winding. The concentrated winding has a shorter end-turn length and high-fill factor and can have a segmented stator. Three different HRE-free configurations with a maximum speed of 20 000 rpm have been compared and reported in [55]. This configuration can have the maximum power density compared with other HRE-free topologies (IPM machine)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000586_978-3-540-73503-8-Figure5.9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000586_978-3-540-73503-8-Figure5.9-1.png",
        "caption": "Fig. 5.9. Schematic view of a tilting train. The additional control-induced inclinaison of the compartment allows for a higher speed on existing rails while keeping the same comfort level (by maintaining the resulting felt acceleration normal to the compartment floor).",
        "texts": [
            " Another successful application of the differential moving-horizon observer can be found in [6] where this observation scheme has been applied to activated sludge processes used for waste-water treatment. In this section, an industrial patented application [5] of the differential form of moving-horizon observer presented in this section is proposed4. The problem is first stated in the context of the control of tilting trains, then the need for an observer is explained and the performance of the proposed moving-horizon observer is shown. Tilting Trains: The Control Problem The problem of controlling tilting trains is schematically depicted on figure 5.9. Typically, when the train goes into a bend of curvature \u03c1(r) at some curvilinear abscissa r on the rails, a passenger feels a centrifugal acceleration V 2\u03c1(r). This acceleration when combined with the gravity gives a resulting acceleration that is not perpendicular to the compartment floor unless the rails present a byconstruction inclinaison \u03b4(r). Consequently, the rails are inclined in accordance with some nominal optimal velocity Vnom by an angle \u03b4\u0304(r) satisfying: \u03b4\u0304(r) = tan\u22121 (1 g \u00b7 V 2 nom \u00b7 \u03c1(r) ) (5",
            " Nonlinear Moving Horizon Observers 167 Obviously, the curvature \u03c1(\u00b7) and the rails inclinaison \u03b4(\u00b7) become constant characteristics (profiles) of the rails. Now if the train follows these rails with a velocity that is significantly higher than the nominal velocity Vnom that has been used in the computation and the construction of the rails inclinaison profiles \u03b4(\u00b7), passenger would feel uncomfortable. The aim of the tilting train control is therefore to compensate for the lack of rails inclinaison by tilting the compartment using the dedicated jacks (see figure 5.9). Ideally, the additional inclinaison angle \u03b1d is clearly given at instant t by: \u03b1d(V (t), r(t)) := tan\u22121 (1 g \u00b7 V 2(t) \u00b7 \u03c1(r(t)) ) \u2212 \u03b4\u0304(r(t)) (5.73) Therefore, from a control point of view, the problem is to track a reference trajectory that depends on: \u2022 The train\u2019s velocity V (t) \u2022 The curvilinear abscissa of the train on the rails r(t) \u2022 The geometric characteristics of the rails \u03c1(\u00b7) and \u03b4\u0304(\u00b7) Remember that the origin of the control problem is related to the high velocities one aims to use that are higher than the nominal velocity Vnom"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.52-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.52-1.png",
        "caption": "FIGURE 8.52. A 3D illustration of a steered wheel with a steer axis coincident with zw.",
        "texts": [
            " Substituting \u03b4 = 0 simplifies the rotation matrix CRW and the position vector CdW to I and 0 CRW = \u23a1\u23a3 1 0 0 0 1 0 0 0 1 \u23a4\u23a6 (8.89) CdW= \u23a1\u23a3 0 0 0 \u23a4\u23a6 (8.90) indicating that at zero steering, the wheel frameW and wheel-body frame C are coincident. Example 336 F Steer angle transformation for zero lean and caster. Consider a wheel with a steer axis coincident with zw. Such a wheel has no lean or caster angle. When the wheel is steered by the angle \u03b4, we can find the coordinates of a wheel point P in the wheel-body coordinate frame using transformation method. Figure 8.52 illustrates a 3D view, and Figure 8.53 a top view of such a wheel. 8. Suspension Mechanisms 501 502 8. Suspension Mechanisms Assume W rP = [xw, yw, zw] T is the position vector of a wheel point, then its position vector in the wheel-body coordinate frame C is CrP = CRW W rP = Rz,\u03b4 W rP = \u23a1\u23a3 cos \u03b4 \u2212 sin \u03b4 0 sin \u03b4 cos \u03b4 0 0 0 1 \u23a4\u23a6\u23a1\u23a3 xw yw zw \u23a4\u23a6 = \u23a1\u23a3 xw cos \u03b4 \u2212 yw sin \u03b4 yw cos \u03b4 + xw sin \u03b4 zw \u23a4\u23a6 . (8.91) We assumed that the wheel-body coordinate is installed at the center of the wheel and is parallel to the vehicle coordinate frame"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure1.10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure1.10-1.png",
        "caption": "FIGURE 1.10. Examples of a non-radial tire\u2019s interior components and arrangement.",
        "texts": [
            "7 shows the interior structure and the carcass arrangement of a radial tire. The non-radial tires are also called bias-ply and cross-ply tires. The plies are layered diagonal from one bead to the other bead at about a 30 deg angle, although any other angles may also be applied. One ply is set on a bias in one direction as succeeding plies are set alternately in opposing directions as they cross each other. The ends of the plies are wrapped around the bead wires, anchoring them to the rim of the wheel. Figure 1.10 shows the interior structure and the carcass arrangement of a nonradial tire. The most important difference in the dynamics of radial and non-radial tires is their different ground sticking behavior when a lateral force is applied on the wheel. This behavior is shown in Figure 1.11. The radial tire, shown in Figure 1.11(a), flexes mostly in the sidewall and keeps the tread flat on the road. The bias-ply tire, shown in Figure 1.11(b) has less contact with the road as both tread and sidewalls distort under a lateral load"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000522_j.automatica.2003.10.021-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000522_j.automatica.2003.10.021-Figure1-1.png",
        "caption": "Fig. 1. Dead-zone model.",
        "texts": [
            " Based only on the intuitive concept and piece-wise description of dead-zones (Section 2), in this paper, a new approach for adaptive control of linear or nonlinear systems with dead-zones is introduced without constructing the inverse of the dead-zone. The new control law ensures a global stability of the entire adaptive system and asymptotical tracking (Section 4). Computer simulations were carried out to illustrate the e:ectiveness of the approach (Section 5). The dead-zone with input v(t) and output w(t), as shown in Fig. 1, is described by w(t) = D(v(t)) =   mr(v(t)\u2212 br) for v(t)\u00bf br; 0 for bl \u00a1v(t)\u00a1br; ml(v(t)\u2212 bl) for v(t)6 bl: (1) As stated by Tao and Kokotovic (1994), this dead-zone model is a static simpli;cation of diverse physical phenomena with negligible fast dynamics. Eq. (1) is a good model for a hydraulic servo valve or a servo motor. The key features of the dead-zone in the control problems investigated in this paper are (A1) The dead-zone output w(t) is not available for measurement. (A2) The dead-zone slopes in positive and negative region are same, i"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000028_0005-1098(94)90209-7-Figure37-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000028_0005-1098(94)90209-7-Figure37-1.png",
        "caption": "FIG. 37. Parallel trajectories inside the sticking band S have zero acceleration until they reach the border ]from Radcliffe",
        "texts": [
            " A survey of friction and controls 1107 and Southward (1990), courtesy of the publisher]. for p equals 0.01, 0.7, 2.0 and 4.0 1 500.0 , , , 200 1 o o o . o c 5 0 0 . 0 0 . 0 ' ' ' 0.0 2.0 4.0 6.0 8.0 0.0 Dimens;onless Velocity FI6. 38. Plot of boundary above which there is no stick-slip, determined by 8000 simulation trials, p is the dimensionless damping coefficient, equation (21) ]from Armstrong- H~louvry (1991)]. This is done by investigating the properties of a twodimensional slice through the three-dimensional phase space. The slice yields the x - y plane, as shown in Fig. 37; y is the integral error state. The phase space is partitioned into six regions, trajectories in each region are shown to enter another region; and the map from one region onto itself, similar to a Poincar6 map, is established. Specifically, the position-error coordinate at breakaway is mapped to the position-error coordinate at the next breakaway event. Convergence, limit cycling or divergence are determined by establishing contraction, stability or expansion in this one-dimensional map. Integrations in the method are performed numerically, and so only for specific combinations of parameters"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure3.38-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure3.38-1.png",
        "caption": "FIGURE 3.38. The stress distribution \u03c4y, the resultant lateral force Fy, and the pneumatic trail ay for a turning tire going on a positive slip angle \u03b1.",
        "texts": [
            " As the wheel turns forward, undeflected treads enter the tireprint region and deflect laterally as well as longitudinally. When a tread moves toward the end of the tireprint, its lateral deflection increases until it approaches the tailing edge of the tireprint. The normal load decreases at the tail of the tireprint, so the friction force is lessened and the tread can slide back to its original position when leaving the tireprint region. The point where the laterally deflected tread slides back is called sliding line. A turning tire under lateral force and the associated sideslip angle \u03b1 are shown in Figure 3.38. Lateral distortion of the tire treads is a result of a tangential stress distribution \u03c4y over the tireprint. Assuming that the tangential stress \u03c4y is proportional to the distortion, the resultant lateral force Fy Fy = Z AP \u03c4y dAp (3.137) is at a distance ax\u03b1 behind the center line. ax\u03b1 = 1 Fy Z AP x \u03c4y dAp (3.138) The distance ax\u03b1 is called the pneumatic trail, and the resultant moment 138 3. Tire Dynamics Mz is called the aligning moment. Mz = Mz k\u0302 (3.139) Mz = Fy ax\u03b1 (3.140) The aligning moment tends to turn the tire about the z-axis and make it align with the direction of tire velocity vector v. A stress distribution \u03c4y, the resultant lateral force Fy, and the pneumatic trail ax\u03b1 are illustrated in Figure 3.38. There is also a lateral shift in the tire vertical force Fz because of slip angle \u03b1, which generates a slip moment Mx about the forward x-axis. Mx = \u2212Mx \u0131\u0302 (3.141) Mx = Fz ay\u03b1 (3.142) The slip angle \u03b1 always increases by increasing the lateral force Fy. However, the sliding line moves toward the tail at first and then moves forward by increasing the lateral force Fy. Slip angle \u03b1 and lateral force Fy work as action and reaction. A lateral force generates a slip angle, and a slip angle generates a lateral force"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure2.8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure2.8-1.png",
        "caption": "Fig. 2.8 Definition of Cross Product. v = \u03c9 \u00d7 p represents the velocity of the a point on the circle. It is at right angles to both \u03c9 and p.",
        "texts": [
            "17) 2 \u03c9 describes the velocity of all the points on the rotating object Let p be a vector which represents a given point p on the surface of the object. Let us also define the base of this vector to be somewhere on the rotation axis. In this case the velocity of point p will be described by \u03c9 \u00d7 p. The \u00d7 operator is the cross product or outer product of two vectors and is defined as follows. 30 2 Kinematics By using the two vectors, \u03c9 and p we obtain a new vector v which has the following characteristics (Fig. 2.8) |v| = |\u03c9||p| sin \u03b8 (v\u22a5\u03c9) \u2229 (v\u22a5p). However, there are two vectors that satisfy this condition so we use the right screw rule and choose the one which matches the direction a right handed screw will move when turned. This can be represented as v = \u03c9 \u00d7 p (2.18) and this is called the \u201ccross product of \u03c9 and p\u201d. When given the elements of \u03c9 and p, the cross product can be calculated as follows. As an exercise we can also check to see that the equations below can be derived from the above. \u03c9 \u00d7 p = \u23a1 \u23a3 \u03c9x \u03c9y \u03c9z \u23a4 \u23a6\u00d7 \u23a1 \u23a3 px py pz \u23a4 \u23a6 \u2261 \u23a1 \u23a3 \u03c9ypz \u2212 \u03c9zpy \u03c9zpx \u2212 \u03c9xpz \u03c9xpy \u2212 \u03c9ypx \u23a4 \u23a6 (2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003786_1045389x10369718-Figure17-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003786_1045389x10369718-Figure17-1.png",
        "caption": "Figure 17. The dielectric elastomers actuate by means of electrostatic forces applied via compliant electrodes on the elastomer film. Reproduced with permission from Elsevier, http://dx.doi.org/ 10.1016/j.sna.2009.01.002.",
        "texts": [
            " For practical applications, incorporation of defects into the polymers can significantly lower the transition temperature near to the room temperature, broaden the transition region, and eliminate the energy barrier in the transformation (Zhang et al., 1998; Bohannan et al., 2000; Xia et al., 2002; Huang et al., 2004b). A dielectric electroactive actuator is made of a dielectric elastomer with compliant electrodes coated on both sides. The modulus of the dielectric elastomer, such as silicone elastomer, acrylic elastomer, and polyurethane elastomer, is very low, while the dielectric constant is very high. The schematic operation principle of the dielectric electroactive actuator is shown in Figure 17 (Shankar et al., 2009). When the voltage is applied between the electrodes, the sandwiched elastomer is squeezed in the thickness direction which causes expansion in the plane direction. The induced strain is proportional to the square of the electric field, dielectric constant and inversely proportional to the polymer elastic modulus (Pelrine et al., 1998). The performance of the dielectric electroactive elastomer actuators can be markedly improved by prestraining the elastomer. SRI International researchers (Pelrine et al"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001394_tro.2010.2043757-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001394_tro.2010.2043757-Figure8-1.png",
        "caption": "Fig. 8. Compliant multicontact behavior. A four-contact stance is shown here. Contact CoP\u2019s are controlled to stay at the center of the bodies in contact. The CoM is controlled to remain at a fixed location over the area defined by the horizontal projection of the four contacts. The table is pivoted rapidly by an external user to challenge the pose. To add additional difficulty, the internal tension between the left hand and the right foot is commanded to track a sinusoidal trajectory. All other tensions and normal moments are commanded to become null. Overall, the robot was able to track efficiently the commanded internal force and moment references with maximum errors around 0.3 N. We show the data corresponding to the left hand, including the tangential moments on the y-direction, the internal tension between the right hand and the left hand, and the normal moment with respect to the moving surface. As we can see, the tangential moment is maintained close to zero, which means that the desired anchor point of the virtual linkage, which, in our case, is the center of the hand, becomes the actual CoP. The tensions are also tracked efficiently as well as the normal moment. We also show data of the robot\u2019s CoM and the normal component of the reaction forces in the hands produced by the moving table. The desired CoM position was tracked with offset values below 0.1 mm, despite the large variation on the movement of the table that oscillated to angles up to 75\u25e6 from the resting position.",
        "texts": [
            " The motion-control law is a velocity-saturation law defined by the position-feedback rule given by vdes = \u2212Kpm K\u22121 vm (xhand \u2212 xdes) (61) with \u03bd = min { 1, vmax \u2016 vdes \u2016 } . (62) This ensures that we track user-defined trajectories without violating predefined velocity limits. To respond compliantly to impacts, we use moderate gains in the above motion-control law. In Fig. 6, we show the response to impact with the panel and with the table when establishing the contact interactions. We conduct a second simulation shown in Fig. 8 that studies a compliant behavior that emerges when the environment is dynamically moving. The robot is commanded to establish a four-contact stance leaning against a pivoting table. The tasks used to implement this behavior are similar to the ones used in the previous example; however, the robot is commanded to stay in the four-contact stance all the time. To demonstrate force tracking at the internal level, the tension between the left hand and the right foot is commanded to follow a sinusoidal trajectory"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure9.56-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure9.56-1.png",
        "caption": "Figure 9.56 (a) The forces in members 1 and 2 follow from the equilibrium of joint A. (b) The closed force polygon for the equilibrium of joint A. FA;1 and FA;2 are the forces that members 1 and 2 exert on joint A. (c) Joint A with all the forces acting on it. From this figure we can see that N1 is a compressive force and N2 is a tensile force.",
        "texts": [
            " Member 1 is a compression member and exerts a compressive force on joint A. Member 2 is a tension member. Figure 9.55b shows the forces as they really act on both the joint and on the two members. 354 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Graphical solution for the equilibrium of joint A: FA;1 and FA;2 are the forces that members 1 and 2 exert on joint A. The forces FA;1 and FA;2 have their line of action along the members 1 and 2, but we do not know their magnitudes, nor their directions (see Figure 9.56a). Joint A is in equilibrium if all forces acting on joint A form a closed force polygon. Figure 9.56b shows the closed force polygon for the equilibrium of joint A. From here, we can read off the magnitude of FA;1 and FA;2 (or calculate it): FA;1 = 8F \u221a 2, FA;2 = 4F \u221a 5. From the force polygon, we can also find the directions of FA;1 and FA;2, but we cannot see whether they are tensile or compressive forces. To do so, we first have to draw the forces found as they act on joint A, see Figure 9.56c. Only then we can see that FA;1 is a compressive force, and FA;2 is a tensile force, so that N(1) = \u2212FA;1 = \u22128F \u221a 2, N(2) = +FA;2 = +4F \u221a 5. Note that the forces in the force polygon have not been denoted as N . The force polygon provides information only on the magnitude of the member forces, and not on the sign for tension or compression. The order in which one writes down the forces in a force polygon does not influence the result (vector addition is associative and commutative). Figure 9.57 shows two equivalent force polygons"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure7.11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure7.11-1.png",
        "caption": "Fig. 7.11. Phasor diagrams for motoring action of overexcited AFPM machine: (a) for iad = 0; (b) at unity power factor. Note that \u03b4 is negative for motoring action and positive for generation action.",
        "texts": [
            " The developed torque of the machine is expressed by eqn (5.13). The first term in eqn (5.13) expresses the torque component due to PM excitation, while the second term expresses the reluctance torque component. For surface mounted AFPM machines it can be assumed that Lsd = Lsq, hence the torque is directly proportional to iaq. In the case of an embedded AFPM machine like in Fig. 7.10a it is clear that Lsd < Lsq, which implies that iad in eqn (5.13) must be negative in order to generate a positive reluctance torque. Fig. 7.11a shows the phasor diagrams of the motor for iad = 0 while Fig. 7.11b shows the phasor diagram of the motor at unity power factor. The inductances in the d and q\u2212axis windings are expressed by eqns (5.17) and the magnet flux linkage \u03c8f is further expressed as \u03c8f = Lad if (7.13) where Lad is the armature reaction (mutual) inductance in the d-axis and if is a fictitious magnetising current representing the PM excitation. 7.2 Control of Sinusoidal AFPM Machine 231 The flux linkage and mutual inductance between the d\u2212axis winding and the fictitious permanent magnet (field) winding need further explanation",
            " It is always good in terms of efficiency to control the drive at maximum torque per ampere. For the surface mounted (nonsalient) AFPM machine, where it can be taken that Lsd = Lsq, it is clear from eqn (5.13) that maximum torque per ampere for this machine is obtained by keeping iad = 0. However, for 234 7 Control embedded (salient-pole) AFPM machines where Lsd 6= Lsq, maximum torque per ampere is obtained by making iad negative, that is by controlling the current phasor at some current angle \u03a8 as shown in Fig. 7.15 (see also Fig. 7.11b). Note from Fig. 7.15 that negative torque is obtained by keeping both iad and iaq negative, or in the case of the nonsalient machine, only iaq is negative. In the sub base speed region (\u03c9 < \u03c9b) of the drive, where there is enough voltage available so that the current regulators do not saturate, the machine is controlled to operate at maximum torque per ampere. For the nonsalient AFPM machine this implies that the current angle \u03a8 = 0, while for the salient-pole machine \u03a8 = \u03a80 (see Fig. 7.15), which is an (average) optimum angle for all load conditions. However, in the high speed region (\u03c9 > \u03c9b), the EMF voltage \u03c9Ladif = \u03c9\u03c8f of eqn (7.23) (see also Fig. 7.11b) becomes larger than the possible output voltage of the inverter. The current regulators therefore start to saturate, current control is lost and the torque of the machine rapidly drops as the speed increases. To avoid this torque decrease and to ensure a wider high-speed region one has to deviate from the maximum torque per ampere control and start to inject a larger negative d\u2212axis current to reduce the q\u2212axis terminal voltage (Fig. 7.11b) as the speed increases. This implies that the current angle \u03a8 must be increased as shown in Fig. 7.15 such that always 2\u03c9 \u221a (Lsdiad + Ladif + iaqR1)2 + (Lsqiaq + iadR1)2 < Vd where Vd is is the dc-link voltage of the inverter, and keeping the current amplitude Iam = \u221a i2ad + i2aq equal or less than the rated current amplitude 7.2 Control of Sinusoidal AFPM Machine 235 depending on the load. At a maximum current angle, \u03a8m, the speed range can be extended much more if the current angle is held constant at \u03a8 = \u03a8m and the amplitude of the phase current is reduced as shown in Fig. 7.15. Note from Fig. 7.11b that as the current angle \u03a8 increases the apparent power of the machine decreases and the power factor improves. The above current control strategy can be used for both salient and nonsalient AFPM machines. It must be noted, however, that the speed range obtained with the above control method depends very much on the value of the stator phase inductance of the machine (see Numerical example 7.2 ). If the phase inductance is relatively small (this implies a low per unit stator phase reactance), as is the case with slotless and air-cored AFPM machines, the high speed range of the machine is very limited",
            " Solution (a) Maximum speed and electromagnetic power With a surface-mounted AFPM machine maximum torque per ampere is obtained with iad = 0 A. Thus, at rated stator current iaq = \u221a 2\u00d7105 = 148.5 A. To obtain maximum speed the output voltage of the inverter must be at a maximum, i.e. from eqn (7.27) the rms phase voltage V1 = 0.612\u00d7 755/ \u221a 3 = 266.8 V. Thus, the peak phase voltage, V1m = \u221a 2 \u00d7 266.8 = 377.3 V or 7.3 Sensorless Position Control 245 simply 1/2Vd, and note that V1m = \u221a v1d 2 + v1q 2. By considering the drive in the steady state and ignoring the resistive voltage drop, from eqn (7.22), eqn (7.23) and Fig. 7.11 we have v1d = V1m sin \u03b4 \u2248 \u2212\u03c9Lsqiaq (7.34) v1q = V1m cos \u03b4 \u2248 \u03c9\u03c8f (7.35) or else tan \u03b4 \u2248 \u2212Lsqiaq \u03a8f = \u22122.13\u00d7 10\u22123 \u00d7 148.5 0.538 and \u03b4 \u2248 \u221230.45o Using eqn (7.34) the angular frequency is \u03c9 \u2248 V1m sin \u03b4 \u2212Lsqiaq = 377.2\u00d7 sin(\u221230.45) \u22122.13\u00d7 10\u22123 \u00d7 148.5 = 604.5 rad/s and the mechanical speed n = 30\u03c9/p\u03c0 = 1443 rpm. The maximum speed with maximum torque per ampere control is thus just less than the rated speed of 1500 r/min of the machine. Using eqn (7.34), an approximate equation for the steady-state electromagnetic power (7.12) can be found as follows: Pelm = \u2126Td = 3 2 \u03c8f \u03c9iaq \u2248 3 2 ( \u03c8fV1m \u2212Lsq ) sin \u03b4 (7.36) Thus, = 3 2 \u00d7 0.538\u00d7 377.3 \u22122.13\u00d7 10\u22123 \u00d7 sin(\u221230.45) \u2248 72.45 kW which is also just less than the rated power of 75 kW of the machine. (b) Per unit high speed. The speed of the AFPM brushless motor can be increased beyond base speed by increasing the current angle \u03a8 , as shown in Fig. 7.11(b), keeping the peak value of the sinusoidal phase current Iam = constant. Note that Iam = \u221a iad 2 + iaq 2. At rated conditions Iam = 148.5 A and V1m = 377.3 V. The same power as in (a), i.e. 72.45 kW must be obtained. Hence, it is clear from eqn (7.36) that \u03b4 will be the same as in (a), i.e. \u03b4 = \u221230.45o. With the machine considered as 100% efficient, the electromagnetic power is equal to the input power Pelm \u2248 3 2 V1mIam cos\u03c6 where 246 7 Control cos\u03c6 \u2248 72 450 (3/2)\u00d7 377.3\u00d7 148.5 = 0.862 and \u03c6 \u2248 30"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003443_0278364903022001005-Figure11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003443_0278364903022001005-Figure11-1.png",
        "caption": "Fig. 11. 2RITz 3-DoF 3 \u2212 xRuPxR(iRjR)N PM.",
        "texts": [
            " The standard base of the mechanism twist system is given by $m1 = (1 0 0 ; 0 0 0) $m2 = (0 1 0 ; 0 0 0) $m3 = (0 0 0 ; 0 0 1). (42) The standard base of the mechanism constraint system is given by $r m1 = (1 0 0 ; 0 0 0) $r m2 = (0 1 0 ; 0 0 0) $r m3 = (0 0 0 ; 0 0 1). (43) When the limb kinematic chain consists of five kinematic pairs, it only exerts one constraint. If the constraint is a couple, no translations can be constrained. So the constraint must be a force and the combined effect of all the p \u00b7 q forces must equal one couple not parallel to the XY plane and two forces parallel to the XY plane. Figure 11 shows such a 3-DoF 3 \u2212 xRuPxR(iRjR)N PM. Note that the three limb central points form a triangle, ABC, which is fixed relative to the moving platform. In the initial configuration, as shown in Figure 11, the triangle ABC is parallel to the moving platform plane. Note that each limb constraint force is parallel to the first revolute axis and passes through the corresponding limb central point. It is obvious that the three limb constraint forces lie in the triangle ABC and do not intersect at a common point. Thus, the standard base of the mechanism constraint system is the same as eq (43) and the moving platform loses two translational DoFs in the base plane and one rotational DoF about the normal of the base plane"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001692_1.3659004-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001692_1.3659004-Figure7-1.png",
        "caption": "Fig . 7 R e p r e s e n t a t i o n o f b o u n d a r y l a y e r s a r o u n d r o t a t i n g c y l i n d e r fo r a = 0 . 2 a n d R X I 0 4",
        "texts": [
            " At this point it is helpful to introduce the concept of a \"boundary-layer origin.'' The positions of the boundary-layer instability and separation points are a function of a boundary-layer length Reynolds number. For a stationary surface such as a flat plate, airfoil, or nonrotating cylinder, the boundary-layer length is measured from the front stagnation point. The shear (or rotation) in the boundary layer has opposite direction (or is of opposite sign) on opposite sides of the body from this point (Fig. 7). It seems logical to define the beginning of the boundary layer, or a boundary-layer origin, with respect to such a condition if we accept the definition and idea of a boundary layer as a shear layer. For a stationary surface, this boundary-layer origin is coincident with the stagnation point; however, for a moving surface this boundary-layer origin will, in general, not coincide with the stagnation point. On the rotating cylinder, the forward stagnation point is translated in a direction opposite to the direction of rotation. (In addition, it is interesting to note that the stagnation point as a point of zero velocity can no longer lie on the surface of the body.) If the boundary or shear layers are now investigated, we find that the upper and lower layers conforming to the required conditions for a boundary layer begin at a point where the cylinder surface velocity is equal to the fluid velocity (i.e., where the relative velocity is zero). The boundary layers are then as shown in Fig. 7. It is seen that the boundary-layer origin is translated in the same direction as the direction of motion of the surface (direction of rotation) but opposite to that of the stagnation point. Boundary-layer profiles obtained for the cylinder rotating at two different velocity ratios are presented in Figs. 8 and 9. The outlines of the boundary layers are shown. It is interesting to note that, for a = 1, both boundary layers are about the same length. In order to calculate characteristics of the boundary layer, it is necessary to have a set of initial conditions"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001620_j.addma.2017.11.001-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001620_j.addma.2017.11.001-Figure8-1.png",
        "caption": "Fig. 8. Model configuration and boundary conditions as presented by [98] on the left and by [95] on the right.",
        "texts": [
            " he initial and boundary conditions are expressed as: T (x1, x2, x3, 0) = Tpreheat with (x1, x2, x3) \u2208 D T (x1, x2, x3, 0) = Tr with (x1, x2, x3) /\u2208 D T (x1, x2, x3, \u221e) = Tr with (x1, x2, x3) /\u2208 D \u2212 \u2202T \u2202n | beam = q \u2212 qrad \u2212 \u2202T \u2202n | topsurface = \u2212qrad here D is the union between the substrate and the layer domains. he heat flux q represents the energy source. Tpreheat and Tr are he preheating temperature and the build chamber temperature, espectively. The heat loss, qrad, due to radiation can be expressed s: rad = \u03b5 ( T4 \u2212 Tr 4 ) (5) here \u03b5 = \u03b5 (T) is the emissivity and is the Stefan-Boltzmann onstant, whose value is 5.67\u00b710\u22128 Wm\u22122K\u22124. Fig. 8 shows the model configuration and the boundary condiions presented by Shen and Chou [95] and by Galati, Iuliano, Salmi nd Atzeni [98]. .1.2. Modelling assumptions and material properties Qi, Yan, Lin and Zhang [89] assumed that the powder density of 16L stainless steel was equal to 64 % of the bulk density. The solid hermal conductivity was written as a linear function of porosity nd the emissivity was ignored, thus the heat losses due to radiation ere neglected. Moreover, the material state change was not taken nto consideration"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000613_978-94-015-9261-1-Figure8.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000613_978-94-015-9261-1-Figure8.6-1.png",
        "caption": "Figure 8.6. Robot in Cartesian space",
        "texts": [
            " Let a multi-link robot be presented by an m-degree-of-freedom (d .o.f.) spatial kinematic rnechanisrn with rotational joints described by the equa tions q B(q) w+ b(q,w) w u, (8.45) (8.46) where q = {qj} E Q c IRm and w = {Wj} are the vectors of joint (gen eralized) coordinates and their rares , and u = {Uj} E IRm is the vector of the generalized torques associated with control variables . Consider the external (absolute) Cartesian space 1R3 with the coordinate frame TO = I . The position of the jth link with respect to TO is characterized by the pair (Figure 8.6) (yJ,TJ), j = 1,2, .. . ,m, (8.47) where yj = col(Yi , Y~, Y~) E Y C 1R3 is the vector of Cartesian coordinates of a principal point of the link and Tj E 50(3) is the orthogonal matrix associated with the frame fixed to the link. The matrix Ti .can be expressed through the Euler angles ai, or the vector ai = col(a{, a~ , a~), and then (8.48) where Ai are canonical rotation matrices defined by (8.6). F\u00fcr a robot having joints with one d.o.I, only, the angular attitude of the links can be found recursively from the equations [279, 184J (8"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure4.13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure4.13-1.png",
        "caption": "Figure 4.13 (a) If the body moves, the bar AB forces point A to follow a circular path with centre B. (b) If the motion remains small with respect to the length of the bar, the arc can be approximated by the tangent at A to the circle.",
        "texts": [
            "11c is a model of the bar support: the bar support is depicted as a single line between the two hinges. The immovable environment is generally shown by means of a hatched area. In Figure 4.12, the two-force member has been isolated at hinges A and B. The position of the two-force member (the line joining both hinges) fixes the line of action of the interaction forces F . Only the magnitude of F (with its sign for the correct direction) is unknown. When the body moves, point A is forced to follow a circle with centre B by the two-force member (see Figure 4.13a). If the displacement remains very small with respect to the length of the two-force member (which is generally the case), then the arc is almost the same as the tangent at A to the circle (Figure 4.13b). Note that the diagram of the structure is much smaller than the actual structure, and also the displacement is strongly magnified in the diagram. The bar support at A prevents displacement in the direction of the bar. Displacement in the direction perpendicular to the bar is free (Figure 4.13b), as is a rotation of the body about A. Imagine that in the free displacement of the body, ux;A and uy;A are displacements of A in the x and y directions, and that \u03d5z;A is the rotation of the body about A. Generalising, the rotation is called a motion. For a bar support at A (with the bar in the y direction) the generalised motions are: ux;A = unknown (free motion), uy;A = 0 (prescribed motion), \u03d5z;A = unknown (free motion). The bar support prevents free motion of point A by exerting forces on it"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure9.19-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure9.19-1.png",
        "caption": "FIGURE 9.19. A trebuchet in motion.",
        "texts": [
            " The origin of a global coordinate frame is set at the pivot point. The counterweight m1 is at (x1, y1) and is hinged at the shorter arm of the beam at a distance c from the end B. The mass of the projectile is m2 and it is at the end of a massless sling with a length l attached to the end of the longer arm of the beam. The three independent variable angles \u03b1, \u03b8, and \u03b3 describe the motion of the device. We consider the parameters a, b, c, d, l, m1, and m2 constant, and determine the equations of motion by the Lagrange method. Figure 9.19 illustrates the trebuchet when it is in motion. The position coordinates of masses m1 and m2 are x1 = b sin \u03b8 \u2212 c sin (\u03b8 + \u03b3) (9.360) y1 = \u2212b cos \u03b8 + c cos (\u03b8 + \u03b3) (9.361) 570 9. Applied Dynamics and x2 = \u2212a sin \u03b8 \u2212 l sin (\u2212\u03b8 + \u03b1) (9.362) y2 = \u2212a cos \u03b8 \u2212 l cos (\u2212\u03b8 + \u03b1) . (9.363) Taking a time derivative provides the velocity components x\u03071 = b\u03b8\u0307 cos \u03b8 \u2212 c \u00b3 \u03b8\u0307 + \u03b3\u0307 \u00b4 cos (\u03b8 + \u03b3) (9.364) y\u03071 = b\u03b8\u0307 sin \u03b8 \u2212 c \u00b3 \u03b8\u0307 + \u03b3\u0307 \u00b4 sin (\u03b8 + \u03b3) (9.365) x\u03072 = l (c\u2212 \u03b1\u0307) cos (\u03b1\u2212 \u03b8)\u2212 a\u03b8\u0307 cos (\u03b8) (9.366) y\u03072 = a\u03b8\u0307 sin \u03b8 \u2212 l \u00b3 \u03b8\u0307 \u2212 \u03b1\u0307 \u00b4 sin (\u03b1\u2212 \u03b8) "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001516_s00170-012-4271-4-Figure13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001516_s00170-012-4271-4-Figure13-1.png",
        "caption": "Fig. 13 Effects of part\u2019s orientation on SLM fabrication critical position. a Fabrication direction. b Fabrication direction. c Fabrication direction",
        "texts": [
            " The experiments above provided theoretical basis for adjusting part\u2019s orientation: that is the relationships among energy input and critical building angle were mutually constrained. According to the conclusion shown in Fig. 9, the important overhanging surface\u2019s inclined angle should be adjusted to be bigger than the reliable building angle, by means of being at the expense of the quality of unimportant surfaces, where the inclined angle will be decreased and the amount of manual-added supports are increased. Figure 13 represents a typical example of adjusting overhanging surface\u2019s orientation to change the inclined angle and manualadded support\u2019s style. In Fig. 13, placing style (a) has the smallest z-directional height, so the fabrication time will be the shortest. However, placing style (a) needs the most of manual-added supports, so it will have the worst forming quality at the downward surface. Placing style (b) needs the least amount of manualadded supports, but it has the biggest z-directional height, so longest forming time will be needed. Placing style (c) is the desired one, where the inclined angle \u03b1 is reliable building angle according to Fig. 9"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003578_mech-34238-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003578_mech-34238-Figure1-1.png",
        "caption": "Figure 1. Schematic of a 3\u2013UPU Parallel Manipulator.",
        "texts": [
            " A coordinate frame A : xyz is attached to the fixed base at point O and another coordinate frame B : uvw is attached to the moving platform at point P. Furthermore, we assume that points A1, A2, and A3 lie on the x y plane whereas points B1, B2, and B3 lie on the u v plane. We also define an instantaneous reference frame C : x y z with its origin located at point P and the x , y , and z axes parallel to the x, y, and z axes of the fixed coordinate frame, respectively. All joint screws are defined with respect to this instantaneous reference frame. THE 3\u2013UPU PARALLEL MANIPULATOR Figure 1 shows the 3 UPU parallel manipulator which consists of a moving platform, a fixed base, and three limbs of identical kinematic structure. Each limb connects the fixed base to the moving platform by a universal joint followed by a pris- Copyright  2002 by ASME s of Use: http://www.asme.org/about-asme/terms-of-use Downl matic joint and another universal joint. A linear actuator drives each of the three prismatic joints [10]. The connectivity of each limb is equal to 5. Therefore, the instantaneous twist, $p, of the moving platform can be expressed as a linear combination of 5 instantaneous twists: $p \u03b8\u03071 i $\u03021 i \u03b8\u03072 i $\u03022 i d\u03073 i $\u03023 i \u03b8\u03074 i $\u03024 i \u03b8\u03075 i $\u03025 i for i 1 2 3 (8) where $\u03021 i s1 i bi di s1 i $\u03022 i s2 i bi di s2 i $\u03023 i 0 s3 i $\u03024 i s4 i bi s4 i $\u03025 i s5 i bi s5 i where s j i is a unit vector along the jth joint axis of the ith limb, bi PBi, and di AiBi d3 i s3 i"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002097_978-1-4939-3017-3-Figure5.8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002097_978-1-4939-3017-3-Figure5.8-1.png",
        "caption": "Figure 5.8.1. The geometry of the snakeboard.",
        "texts": [
            " Spinning the momentum wheel causes a countertorque to be exerted on the board. The configuration of the board is given by the position and orientation of the board in the plane, the angle of the momentum wheel, and the angles of the back and front wheels. Let (\u03b8, x, y) represent the orientation and position of the center of the board, \u03c8 the angle of the momentum wheel relative to the board, and \u03c61 and \u03c62 the angles of the back and front wheels, also relative to the board. Take the distance between the center of the board and the wheels to be r. See Figure 5.8.1. Following Bloch, Krishnaprasad, Marsden, and Murray [1996], a simplification is made that we shall also assume here, namely that \u03c61 = \u2212\u03c62, J1 = J2. The parameters are also chosen such that J +J0+J1+J2 = mr2, where m is the total mass of the board, J is the inertia of the board, J0 308 5. Nonholonomic Mechanics is the inertia of the rotor, and J1, J2 are the inertias of the wheels. This simplification eliminates some terms in the derivation but does not affect the essential geometry of the problem"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000047_5.4400-Figure9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000047_5.4400-Figure9-1.png",
        "caption": "Fig. 9. Two-dimensional illustration of domain of sliding mode.",
        "texts": [
            " Specifically, stability to the switching surface requires selecting a generalized Lyapunov function V(t, x) which i s positive definite and has a negative time derivative in the region of attraction. Formally stated: Definition I [2]: A domain D in the manifold U = 0 i s a sliding mode domain if for each E > 0, there i s 6 > 0, such that any motion starting within a n-dimensional &vicinity of D may leavethe n-dimensional e-vicinity of Donlythrough the n-dimensional e-vicinity of the boundary of D. See Fig. 9. Since the region D lies on the surface ~ ( x ) = 0, dimension [D] = n - m. Hence: Theorem 1: For the n - m-dimensional domain D to be the domain of a sliding mode, it i s sufficient that in some n-dimensional domain Q 3 D, there exists a function V(t, x, a) continuously differentiable with respect to all of its arguments, satisfying the following conditions: 1) V(t, x, U) is positive definite with respect to U , i.e., V(t, x, U) > 0 with U # 0 and arbitrary t, x, and V(t, x, 0) = 0; and on the sphere 11u11 = p for all x E Q and any t the relations i) inf V(t, x, a) = h, h, > 0 (3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002799_j.nanoen.2021.106169-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002799_j.nanoen.2021.106169-Figure4-1.png",
        "caption": "Fig. 4. Magnetic soft robot with a six-arm configuration for object transport. (A) Illustration of the testing platform for object transport. Yellow arrows represent the magnetization direction of magnetic soft robot. (B) Side view images of the rolling locomotion of the soft robot when transporting a pill under an 80-mT rotating Bxz generated by a permanent magnet. (C) Climbing over a 4-mm high obstacle. (D) Climbing up and down a 16\u00ba slope. (E) Making a right-angle turn and a U-turn. (F) Crossing an S-shaped glass tube with an inner diameter of 17.7 mm. (G) Top view images of the robot folding in half under an 80-mT Bxy aligned along different field directions. \u03b8 is a clockwise angle from the x-axis for describing the direction of Bxz and Bxy. Only one soft robot was used in these tests and images of the same robot at different times were superimposed into one image in (C)-(F) to better show the motion of the robot during the transport process. Scale bar: 10 mm.",
        "texts": [
            " Considering that the weight of soft robots is influencing their locomotion speed considerably [42], taking the weight of our robot as a reference, soft robots within the range of 20 times larger and 20 times smaller (which is considered as the same weight level in this work) are sorted out for comparison. It can be seen that the speed of our soft robot is at a high level (top-3) among robots in Fig. 3E. In this section, we show another soft robot for object transport with potential practical application in complex and unstructured environments. The soft robot was programmed to have a radial magnetization profile along its six arms, as shown in Fig. 4A. Although the soft robot has multiple arms, it only needs one step to complete its magnetization (within milliseconds), showing great potential in the high-efficiency production of such robots. To demonstrate the functionality and performance of the soft robot, we designed a multi-configuration testing platform that consisted of a 4 mm high obstacle, a 16\u00ba slope, a right-angle turn, a U-turn and an Sshaped glass tube (Fig. 4A). The soft robot first lies prone on the ground with a cylindrical pill placed on its back (Fig. 4B). When an actuating field along the z-axis is applied, the robot folds its six arms. As the magnetic field rotates to the x-axis, the robot tilts its arms towards the field direction until three arms touch the ground. As the field direction turns from the negative z-axis to negative x-axis, the robot stands upside down and somersaults forward, holding the pill inside. This locomotion is enabled by a relatively high rotating magnetic field (about 80 mT) generated by a permanent magnet, which allows the soft robot to grasp the pill tight enough. Controlled rolling locomotion of the soft robot on the platform was achieved by continuously changing the magnetic field direction (Movie S4). To better demonstrate the motion of the robot during the transport process, we used the vector graphics software Adobe Illustrator to create composite pictures shown in Fig. 4C\u2013F. Postures of the same robot at different moments were extracted from their original photos and superimposed onto the same background photo while keeping their relative positions on the testing platform unchanged. The flexible movement demonstrated in these complex environments is enabled by the simplicity of the steering of the soft robot, under magnetic fields. Since the soft robot tends to deform and align its arms with the field direction, we can alter the direction of Bxy to steer the robot along a desired direction (Fig. 4G, and Movie S5). In short, the motion mechanism of our six-arm soft robot is to achieve forward rolling under a rotating magnetic field in the x-z plane (Fig. 4B) and controlled steering under a magnetic field in the x-y plane (Fig. 4G). Supplementary material related to this article can be found online at doi:10.1016/j.nanoen.2021.106169. To summarize, we developed a direct magnetization method based on magnetic field focusing technology, where the local magnetization distribution can be programmed flexibly and reconfigurably within milliseconds and millimeter-scale resolution. The magnetization process of soft magnetic composites is inherently decoupled from their fabrication process, which implies that after fabrication, the magnetization Y"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001218_bfb0109984-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001218_bfb0109984-Figure8-1.png",
        "caption": "Fig. 8. Twisting algorithm phase trajectory",
        "texts": [
            " It must be pointed out that, given the system to be controlled and the desired sliding manifold, it is possiblc to define the above constants by uncertainty maximization. Nevertheless this evaluation procedure usually defines very large parameters, and, as a consequence, very large control signals as well, which are not really needed for controlling the real system. In practice, the convergence conditions for the control algorithms are only sufficient but not necessary and the tuning of the controller is often better made heuristically. This algorithm is characterized by a twisting around the origin of the ylOy~ 2-sliding plane (Fig.8). The finite time convergence to the origin of the plane is due to the switching of the control amplitude between two different values such that the abscissas and ordinates axes are crossed nearer and nearer to the origin. The control amplitude commutes at each axis crossing, and the sign of the sliding variable time derivative y~ is needed. The control algorithm is defined by the following control law [20], in which the condition on [ul must be taken into account when considering the chattering avoidance problem, - u i f lul > 1 v(t) = --Vmsign(yl) i f yly2 <_ O; [ul < 1 (19) --VMsign(yl) i f YlY2 > O; lur < 1 and the corresponding sufficient conditions for the finite time convergence to the sliding manifold are [20] { VM ~> ~n - 4I'M Vrn ~ so vm > (20) ImVM - - r > FM~,, + ~ By taking into account the different limit trajectories arising from the uncertain dynamics (14) and evaluating the time intervals between subsequent crossings of the abscissa axis, it is possible to define the following upper bound for the convergence time 1 ttwoo tMa + (~tw\" 1 __--0t w ~/lYlM, I (21) where YlM~ is the value of the Yl variable at the first abscissa crossing in the y l O y 2 plane, tM1 the corresponding time instant and ~ , F~VM+FMV~ Ot~ = ~ - r ~ - ~ ) ~ / r ~ v ~ + ~ 0t w __ / l~Vm+~ V FmVM-~ In many practical cases the Y2 variable is completely unmeasurable, then its sign can be estimated by the sign of the first difference of the available sliding variable Yl in a time interval 5, i"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001467_j.mechmachtheory.2010.05.001-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001467_j.mechmachtheory.2010.05.001-Figure6-1.png",
        "caption": "Fig. 6. A 3-DOF spherical parallel manipulator.",
        "texts": [
            " In this study, the definition of GTW is applied to the analysis of parallel manipulators. Within the GTW, the parallel manipulator has good motion/force transmissibility at every configuration and is far away from its singularity. It is noteworthy that the limit of LTI may be different for different design requirements. To introduce the analysis method and the proposed frame-free index in detail, the motion/force transmission performance analysis of two parallel manipulators will be presented in this section. Fig. 6 shows a 3-DOF spherical parallel manipulator (SPM), which has a moving platform connected to the base by three identical spherical RRR legs. Such an SPM is usually applied to a wrist robot or a camera-orienting device [24] because of its high rotational workspace, high speed, and high load capacity. What is special about this SPM is that all axes of the nine R joints intersect at the origin O of the global coordinate system O\u2212xyz. In the i-th leg, the axial directions of three R joints Ri,1, Ri,2, and Ri,3 are denoted by three unit vectors Ui,1, Ui,2, and Ui,2, respectively. The R joint attached to the base in each leg is actuated. Since the symmetric SPM has three rotational DOFs, the orientation of the moving platform with respect to the coordinate system O\u2212xyz can be described by the Tilt-and-Torsion (T&T) angles (\u03d5, \u03b8, \u03c3) [25], where \u03d5 is the azimuth, \u03b8 is the tilt angle, and \u03c3 is the torsion angle. When the T&T angles are (0\u00b0, 0\u00b0, 0\u00b0), the SPM is considered to be at its initial orientation, and the vectors Ui,1 and Ui,3 lie in the same vertical plane. In Fig. 6, angles \u03b11, \u03b12, \u03b21, and \u03b22 denote the four geometric parameters of the SPM. Based on the analysis in Section 3.1.3, the transmissionwrench of the i-th leg is a pure torque, the axis of which is perpendicular to the plane ORi,2 Ri,3 and passes through the origin O. Thus, the unit TWS of the i-th leg can be expressed as $T i = 0;Ui; 2 \u00d7 Ui; 3 = jUi; 2 \u00d7 Ui; 3j \u00f025\u00de According to Eq. (16), the input transmission index of the i-th leg can be represented as \u03bbi = jsin\u2220Ri; 1 ORi; 2 Ri; 3j \u00f026\u00de \u2220Ri,1\u2212ORi,2\u2212ORi,3 represents the angle between the planes Ri,1ORi,2 and Ri,2ORi,3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002567_j.jmatprotec.2021.117117-Figure24-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002567_j.jmatprotec.2021.117117-Figure24-1.png",
        "caption": "Fig. 24. Conceptualization of a thermal protection system employing attachment clips to join a hot-service IN718 aeroshell to a Ti-6Al-4 V integrally stiffened tank structure (Domack and Baughman, 2005). Reproduced with permission Emerald Publishing Limited.",
        "texts": [
            " Due to the incompatibility of titanium and nickel during the welding process, AM-FGM has increasingly attracted attention in joining titanium-based alloys to nickel-based alloys. One potential application of a tailored graded composition component from titanium to nickel is in aerospace launch vehicles thermal protection system to reduce the stress concentrations (due to the thermal gradients during flight) of the Inconel clips that join thermal protection systems to launch vehicle structure. Fig. 24 shows the concept design of an attachment bracket joining a hot-service aeroshell to an integrally stiffened tank structure. Another potential application is for the fabrication of reliable joints in the oil and gas industry (Domack and Baughman, 2005). Table 4 summarizes various gradients from titanium-based to nickel-based alloys together with some key processing parameters. Domack and Baughman (2005) employed different manufacturing techniques for creating a gradient from Ti-6Al-4 V to Inconel 718"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000538_acs.chemrev.8b00172-Figure22-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000538_acs.chemrev.8b00172-Figure22-1.png",
        "caption": "Figure 22. (1) Schematic representation of a crystal of highly ordered pyrolytic graphite where the layers of graphite have an interlayer spacing of 3.35 \u00c5. (2) The electrochemistry of carbon nanotubes resembles that of graphite. (3) This corresponds to a single MWNT in which edge-plane-like sites are shown at the end of the tube and along the tube axis. These similarities result in identical electrochemical responses for CNT-modified electrodes when compared with the edge-plane pyrolytic graphite electrode. Reproduced with permission from ref 289. Copyright 2012 WileyVCH Verlag GmbH & Co. (4) A conceptual schematic model of the structure of graphene, indicating its basal and edge-plane-like sites. (5) SEM image of a single atomic layer of graphite, known as graphene. Note, in reality, the graphene utilized in the majority of the work is from one to more than four layers thick. Reproduced with permission from ref 290. Copyright 2010 The Royal Society of Chemistry.",
        "texts": [
            " CNFs should not be confused with CNTs, as CNFs have cylindrical nanostructures with graphene layers arranged in various forms as illustrated in Figure 20-1. Even though CNTs have higher electrical conductivity than CNFs,265 the arrangement of graphene sheets in CNFs offers remarkable benefits that make CNFs more attractive than CNTs in the development of electrochemical biosensors.266 First, the stacking of graphene sheets with various shapes generates more edge sites (electrochemically active parts; see also Figure 22) exposed on the exterior of CNFs, facilitating better electron transfer of electroactive analytes than CNTs.267 In addition, compared with CNTs, the entire surface area of CNFs can be activated without damaging the structural integrity of their backbones. Thus, CNFs are an excellent candidate to serve as immobilization matrixes for bioreceptors while still preserving the highly efficient electron transfer along the CNF structure (Figure 20-2).268 5.1.3. Two-Dimensional (2-D) Nanomaterials. In 2-D nanomaterials, two of the dimensions are not confined to the nanoscale",
            " 2019, 119, 120\u2212194 139 Thus, using graphene prepared by a top-down-based technique in electrochemical sensing can alleviate misinterpretation when detecting analytes that can be catalyzed by metals, such as H2O2 (as discussed above, bottom-up graphene synthesis through CVD can encounter the same metal impurity challenges). Graphene also offers better analytical performance than CNTs because graphene sheets have a larger number of edges, where the active sites are located, and thus enables a faster heterogeneous electron transfer than CNTs (Figure 22).287 Further comparisons of electrochemical sensing properties for CNTs and graphene are given in many excellent review articles.288,289 5.1.3.2. Graphene-like 2-D Nanomaterials. Graphene-like 2-D nanomaterials have been widely investigated for electrochemical sensing due to their similar structural properties to graphene and the tunable band gap capability that is not present in graphene. The band gap is of interest as it allows electrical conduction through a material to be switchable (on DOI: 10"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000233_978-1-4614-0460-6-Figure2.36-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000233_978-1-4614-0460-6-Figure2.36-1.png",
        "caption": "Fig. 2.36 Inverted pendulum",
        "texts": [
            "9*clim; % Define simulation variables dt = tau/20; % sec/step steps = round(50*tau/dt); % Euler integration initial conditions x = 1e-3; % m dx = 0; % m/s % Initialize final position and time vectors y = zeros(1, steps); time = zeros(1, steps); for cnt = 1:steps ddx = (-(c - gamma)*dx - k*x)/m; % m/s^2 dx = dx + ddx*dt; % m/s x = x + dx*dt; % m % Write results to vectors y(cnt) = x; % m time(cnt) = cnt*dt; % s end figure(1) plot(time, y*1e3, 'k') xlim([0 max(time)]) ylim([-5 5]) set(gca,'FontSize', 14) xlabel('t (s)') ylabel('x (mm)') In order to model and simulate divergent instability, we will use the inverted pendulum, composed of a mass, m, supported by a massless rod of length, l, that rotates about the frictionless pivot, O. The rod is held in its vertical equilibrium position by springs and dampers as shown in Fig. 2.36. The free body diagram for 2.5 Unstable Behavior 69 a rotation, y, of the mass/rod counterclockwise from the equilibrium position is provided in Fig. 2.37. The forces that need to be considered include: \u2022 the gravity force, mg \u2022 the two spring forces, kd, where d \u00bc l 2 sin y\u00f0 \u00de is the horizontal deflection due to rotation of the rod as shown in Fig. 2.38 \u2022 the two viscous damping forces, c _d, where _d \u00bc dd dt \u00bc l 2 cos y\u00f0 \u00de _y is the horizontal velocity \u2022 the reaction forces at the pivot in the horizontal, Rx, and vertical, Ry, directions",
            "5 Unstable Behavior 71 means that the spring force is insufficient to counteract gravity and the pendulum simply falls over for nonzero initial conditions. However, if k l2 2 mgl> 0, the pendulum oscillates around its equilibrium position when disturbed from equilibrium and the response eventually decays to zero due to the viscous dampers. The limiting spring stiffness is found using klim l2 2 mgl \u00bc 0. Rewriting gives klim \u00bc 2mg l . If k< klim, divergent instability occurs. Consider the inverted pendulum in Fig. 2.36 with m \u00bc 0.5 kg, c \u00bc 1 N-s/m, and l \u00bc 0.3 m. The limiting spring stiffness is klim \u00bc 2 0:5\u00f0 \u00de9:81 0:3 \u00bc 32:7 N/m. For initial conditions of y\u00f00\u00de \u00bc 5 and _y\u00f00\u00de \u00bc 0, let\u2019s determine the response y\u00f0t\u00de using Euler integration. As with the aircraft wing example, the first step is to solve the equation of motion for the acceleration term. \u20acy \u00bc c l2 2 _y\u00fe k l2 2 mgl y ml2 The angular velocity of the pendulum for the current time step is then determined by: _y \u00bc _y\u00fe \u20acy dt; where the angular velocity on the right-hand side of the equation is the value from previous time step"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000533_j.humov.2007.04.003-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000533_j.humov.2007.04.003-Figure2-1.png",
        "caption": "Fig. 2. The principles of dynamic walking yield a periodic walking gait for both unpowered and powered robots. The single support phase can be produced entirely by passive dynamics, with the legs acting like pendulums as in the inverted pendulum theory. However, a key feature of dynamic walking is that there is a collision between the swing leg and ground, dissipating energy. Energy may be restored passively by descending a gentle slope as in passive dynamic walking (McGeer, 1990a), or actively through actuation such as push-off (Kuo, 2002), as demonstrated by a recent robot (Collins et al., 2005). The conditions of push-off and collision can produce initial conditions for the next step, with no need for active stabilization nor for control of prescribed trajectories. With a knee that passively prevents hyperextension (McGeer, 1990b), body weight can also be supported with no actuation. Dynamic walking extends the inverted pendulum theory, indicating that there is an energetic consequence due to the collision, with the transition between pendulum-like steps termed the step-to-step transition (Donelan et al., 2002a).",
        "texts": [
            " This is because of the many variables that must be orchestrated in the human. The many variables must either be reduced to a smaller number, or specified through a simplifying context such as an optimization principle. Even if these many variables are correctly specified, they may be too numerous to yield testable predictions or contribute to cogent understanding. The rigorous testing of theories requires an appropriate approach that provides both simplification and prediction. The principles of dynamic walking (Fig. 2) provide just such an approach. Here we use the term \u2018dynamic walking\u2019 to refer to systems in which the passive dynamics of the limbs dominate the motion, with minimal actuation applied to sustain periodic behavior. These principles were originally developed by McGeer (1990a), who also demonstrated passive dynamic walking machines that could descend a gentle slope under gravity power alone. These same principles apply to powered walking on level ground (Kuo, 2002), as demonstrated by several recent walking machines (Collins, Ruina, Tedrake, & Wisse, 2005; Wisse, 2005)",
            " Most of these features have been added to relatively modest dynamic walking models, and have yielded simulations amenable to comparison with human walking, with each feature implying at least one comparison. The challenge of simultaneously incorporating many features (e.g., multiple muscles) in a model is the greatly increased number of comparisons or experiments needed to test the model. Several new features may be incorporated in dynamic walking models while adding only modest complexity. Articulating ankles would allow the feet, presently fixed to the leg in most dynamic walking models (see Fig. 2), to produce more human-like push-off and collision. The impulsive behavior of the model is energetically favorable, because the step-to-step transition requires the least work if the duration and displacement is minimized. This cannot, however, explain why humans perform the transition over the considerable duration of double support, or about 10% of a stride. One explanation is simply that neither push-off nor collision can realistically be performed perfectly impulsively, because of the limited forces that the body structure can sustain"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001205_s10846-016-0442-0-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001205_s10846-016-0442-0-Figure7-1.png",
        "caption": "Fig. 7 Half vehicle (Bicycle) representation in dynamic model",
        "texts": [],
        "surrounding_texts": [
            "Going through past publications in path tracking control for autonomous vehicle, one can notice that a good portion of the studies employed linear handling model also known as bicycle model in order to develop steering controller. To put things into perspective, vehicle model usually plays a crucial role in two aspects. One aspect is for the vehicle system simulation. In developing any controller, simulation stage is one of the most crucial stage. During this stage, the controller properties are investigated and controller parameters are tuned to yield the best performance for the controller. For this, a vehicle model is usually used to simulate vehicle behaviours under the influence of the proposed controller. Second aspect, control law in trajectory tracking control is sometimes derived from the mathematical representation of the vehicle system. Usually, a linearized vehicle model is used for this purpose. The equations for vehicle model are manipulated to form state space representation before further developed into a control law for the controller. In this section, both aspects will be addressed and existing approaches for the vehicle modelling will be reviewed from the previous researches. In general, car-like robot can be classified as a nonholonomic system. Holonomic properties of a system refers to the relationship between the controllable DOF and total number of DOF for the system. A system is holonomic if the controllable DOF if equal to the total number of DOF and non-holonomic if the controllable DOF is less than the total DOF. As stated in Katrakazas et al. [7], normal vehicle is considered non-holonomic due to its total of four DOF which are motion in the two Cartesian coordinates direction, orientation, and heading but only two controllable DOF which are the longitudinal direction (forward and backward) and lateral direction (bounded steering input). In this section, only non-holonomic model of a vehicle will be addressed. It is worth to note that common models for small mobile robot will not be reviewed here. Vehicle model usually assumed the vehicle body as a rigid body with concentrated sprung mass at the centre of gravity. Path tracking control is similar to any handling control where the behaviour of the vehicle is observed in longitudinal plane. Usually known as handling model, important properties usually observed are the lateral and yaw motions. For handling model, vertical motions are always considered constant by assuming flat road surface where ride properties and vertical forces from suspension is negligible. Most common vehicle model adopt bicycle concept where the vehicle is reduced to a two-tire configuration at the front and rear by assuming same behaviour for left and right tires. With this assumption, the right and left tire is merged and represented by one tire at front and rear, each. Figure 2 shows the common representation of vehicle model usually used in handling studies depicting the full vehicle model. Usual configurations consists of geometric, kinematic and dynamic model. Geometrical dimensions of the vehicle are considered in geometric model with no regard on the kinematic (acceleration and its derivatives) and dynamic (forces, inertia, and energy) properties. Similarly, kinematic model consider the motion of the vehicle in terms of its acceleration, velocity and position related to the geometric of the whole vehicle. Dynamic model, on the other hand, considers the motion of the vehicle in terms of its internal forces, inertia and energy properties. Each of the model configuration has its own benefits and purpose depending on usage and which properties do we want to study. Therefore, this section will be split into three subsection of these 3 model configurations. 2.1 Geometric Vehicle Model Geometric vehicle model is particularly important in order to relate the vehicle\u2019s dimensions, radius of turn and radius of curvature of road undertaken by the vehicle during turning. As stated earlier, this model only consider the dimension and positions of the vehicle during manoeuvring with no regard to its velocity and acceleration. It is developed based on the Ackerman steering configurations that the line perpendicular to each of the vehicle wheel should intersect at the centre point of the vehicle cornering arc where the radius of turn is R. The importance of this model is in developing one of the most popular path tracking controller, Pure Pursuit [5, 10\u201316]. This model is rarely used to simulate any of the vehicle states. From Fig. 3, angle of turn, \u03b4 is given by the following equation. \u03b4 = tan\u22121 ( L R ) (1) Another geometric model relates the vehicle position with respect to the trajectory and its subsequent error. This model is particularly important in order to estimate the tracking error between the actual vehicle position and the desired trajectory which is the basis of another geometric-type Stanley controller as well as one of the most common quantification method to evaluate a path tracking performance. Development of both the model and quantification method will be described in details within its respective sections. 2.2 Kinematic Vehicle Model In mechanics field, kinematic is defined as the study of motion of a body without regards of the body\u2019s internal forces, inertia and energy. Therefore, kinematic vehicle model can be defined as vehicle model that is describing the motion of the vehicle in terms of its position, velocity and acceleration without any regard of its internal dynamics. The motion will be described solely based on its geometry. Kinematic vehicle modelling can be found in most of the studies on path tracking controller due to its simplicity and important relationship to describe the vehicle motion. This model is usually used to describe the vehicle velocity and acceleration in lateral direction as well as its yaw motion with respect to the vehicle fixed axes (also known as local coordinates) and global axes (also known as global coordinates). While it is always used together with geometric model to develop controller algorithm as well as to formulate the tracking error for quantification method, this model is also often used to simulate the vehicle\u2019s position, velocity, and acceleration during manoeuvring. There are several approaches used previously in order to develop and represent the vehicle kinematics. A simple kinematic model is commonly used [17\u201319] for this purpose as depicted in Fig. 4. The vehicle is described with 2 sets of Cartesian coordinates, local (x-y) and global (X-Y ). Local coordinates are set to be fixed on the vehicle body. Global coordinates, on the other hand is set to be fixed on the earth coordinates. Local coordinates are always regarded as moving coordinates with respect to the global coordinates since the vehicle is constantly moving. Based on Fig. 4, kinematic equation of the vehicle describing the velocity in global coordinates, vX and vY as well as local coordinates, vx and vy can be deduced as in Eq. 2. Here, v is the total velocity for the vehicle, \u03b8 is the vehicle heading with respect to the local coordinates and \u03c8 is the vehicle\u2019s orientation with respect to the global coordinates as depicted in Fig. 4. [ vX vY ] = [ cos \u03c8 \u2212 sin \u03c8 sin \u03c8 cos \u03c8 ] [ vx vy ] where vx = v cos \u03b8 vy = v sin \u03b8 (2) A simpler kinematic model can be found in which the vehicle is modelled a bicycle. With this, responses from each wheel is simplified to be only one wheel per axle [10, 20\u201323]. By assuming only one wheel per axle, this model neglect the effect of slip from each wheel to the direction of the vehicle. Therefore, for this model, direction of vehicle\u2019s velocity is the same as vehicle heading direction. Referring to notation in Fig. 5, \u03b8 is the vehicle heading with respect to the local coordinated and \u03c8 is the vehicle\u2019s orientation with respect to the global coordinates and r is the vehicle\u2019s yaw rate or \u03c8\u0307 as depicted kinematic model for the bicycle model is shown in Eq. 3.\u23a1 \u23a3 vX vY r \u23a4 \u23a6 = \u23a1 \u23a3 cos \u03c8 0 sin \u03c8 0 0 1 \u23a4 \u23a6 [ v \u03c8\u0307 ] (3) Another studies [24, 25] extended the kinematic bicycle model by including side slip angle of both front and rear wheel to account for slippery terrain the study needs to address. The equation is as given in Eq. 4 [25] where VLRis rear wheel longitudinal slip velocities, VSRis rear wheel side slip velocities, \u03b2R is the sideslip angle for rear wheel and \u03b2F is the sideslip angle for front wheel. vX = v cos \u03c8 \u2212 VLR cos \u03c8 \u2212 VSR sin \u03c8 vY = v sin \u03c8 \u2212 VLR sin \u03c8 + VSR cos \u03c8 \u03c8\u0307 = v\u2212VLR L [tan \u03b2R + tan (\u03b4 \u2212 \u03b2F )] (4) 2.3 Dynamic Vehicle Model Opposite to kinematic model, dynamic vehicle model is the vehicle model that is describing the motion of the vehicle in terms of its position, velocity and acceleration by considering the internal forces, energy or momentum within the system. In this model, forces from tire and mass of the vehicle body are considered and may include the overall geometric and kinematic relationships as described in previous sections. Common method in deriving the mathematical model of the vehicle handling dynamics is using Newtonian equation of motions. In longitudinal plane, handling model usually contains the translational motion in lateral, x, longitudinal, y, and rotational motion about z\u2212axis also known as yaw, \u03c8 , and yaw rate, r Figs. 6 and 7 show the schematic representation of the handling model of full vehicle and half vehicle (bicycle) respectively. Equation of motion for dynamic vehicle handling model can be shown in Eq. 5 for full vehicle [17] and Eq. 6 for half vehicle model (Bicycle) [26]. This is considering the vehicle with sprung mass mb, moment of inertia of the sprung mass about z-axis, ICG, acceleration ax of longitudinal motion in x-direction, ay of lateral motion in y-direction, and yaw motion with yaw angle, \u03c8 , about z-axis. Position of each vehicle\u2019s wheel is denoted with subscripts i, j , where i =front/rear, j =left/right.\u2211 Fx = mbax\u2211 Fy = mbay\u2211 Mz = Iz\u03c8\u0308 For Full Vehicle Model Fxrr + Fxrl + Fxf l cos \u03b4 + Fyf l sin \u03b4 + Fxf r cos \u03b4 + Fyf r sin \u03b4 = mbax Fyrr + Fyrl \u2212 Fxf l sin \u03b4 \u2212 Fxf r sin \u03b4 + Fyf l cos \u03b4 + Fxf r cos \u03b4 + Fyf r cos \u03b4 = mbay( \u2211 Mzij + [\u2212Fyrr \u2212 Fyrl ] lr + [ Fxf l sin \u03b4 + Fyf l cos \u03b4 + Fxf r sin \u03b4 + Fyf r cos \u03b4 ] lf + [ Fxf r cos \u03b4 \u2212 Fxf l sin \u03b4 + Fyf l sin \u03b4 \u2212 Fyf r sin \u03b4 \u2212 Fxrl + Fxrr ] t 2 ) = Iz,CG\u03c8\u0308 (5) For Bicycle Vehicle Model Fxr + Fxf cos \u03b4 + Fyf sin \u03b4 = mbax Fyr \u2212 Fxf sin \u03b4 + Fyf cos \u03b4 = mbay Fyr lr + [ Fxf sin \u03b4 + Fyf cos \u03b4 ] lf = Iz,CG\u03c8\u0308 (6) Looking at above equations, one can notice that the main contributing external factor in this model is the tire forces which is the main source of external disturbance as well as the traction governing the vehicle motion. Tire forces come from the interaction between the tire and road surfaces mainly caused by the deformation of the tire during different manoeuvring in both the longitudinal and lateral directions. While the forces are non-linear function, linearization of the dynamic model is commonly employed by assuming linear forces acting on each of the tires. This is suitable if the wheel angle, \u03b4,is small which correspond to small lateral slip angle. Such linearization has been carried out on a full vehicle dynamic model [17, 27, 28] and bicycle model [14, 20, 22\u201324, 26, 29, 30] previously. Linearization of lateral and longitudinal tire forces, Fxand Fy respectively is done by assuming proportionality between tire forces and tire slips. This model is valid for small value of lateral and longitudinal slips. Previous studies outlined the region as below 0.5gacceleration for low longitudinal slip [27] and lateral slip angle lower than 5\u00b0[31]. Linearization of lateral and longitudinal forces, Fx and Fy are shown respectively in Eq. 7. Here, CxandCy are defined as the longitudinal and lateral (cornering) stiffness of the tire, sis the longitudinal slip and \u03b1 is the lateral slip angle. Fx \u2248 Cxs Fy \u2248 Cy\u03b1 (7) Non-linear vehicle model, on the other hand, utilised tire model to simulate the lateral and longitudinal forces generated on each wheel during manoeuvring. Consideration of tire non-linear dynamics will better simulate the vehicle response especially in high speed and large steering angle. This approach can be seen in several studies previously used in both full vehicle model [19] and the simplified bicycle model [32]. Most of these studies are using the well-known Pacejka tire model also known as the Magic Formula. In Eq. 8, function P may represent either Fx , Fy , or Mzwhere B, C, D, \u03c6, Sh, and Sv represent the stiffness factor, shape factor, peak value, curvature factor, horizontal shift, and vertical shift properties of the function, respectively. These are parametersdependent functions in terms of various constants where for each set of functions, there will be a different set of B, C, D, \u03c6, Sh, and Sv . Another study can be found using Calspan tire formula [19]. This model estimate the normalised lateral and longitudinal forces, Fx and Fy respectively using a polynomial expression of a saturation function f (\u03c3). In order to use this model, one must have the knowledge of tyre vertical force, Fz, coefficient of friction for the road surface, \u03bc, lateral stiffness coefficient, Ks , modified longitudinal stiffness coefficient, K \u2032 c,constant for tyre camber angle, Y\u03b3 , and the tire camber angle, \u03b3 and the normalised lateral and longitudinal forces can be evaluated using Eq. 9. P (Fz, \u03b1, \u03c3 ) = D sin (C arctan (B\u03c6)) + Sv (8) Fy \u03bcFz = f (\u03c3)Ks tan \u03b1\u221a K2 s tan2 \u03b1+K \u2032 cs 2 + Y\u03b3 \u03b3 Fx \u03bcFz = f (\u03c3)K \u2032 cs\u221a K2 s tan2 \u03b1+K \u2032 cs 2 (9) Both of these tire models are depending on several inputs; most importantly, the tire vertical force, Fz. Usually, vertical force acting on each tire was considered constant [32]. Nevertheless, studies by Ping et al. [19], Amer et al. [33] and Wang et al. [17] extended the nonlinearity of the model by including a load distribution model to estimate the vertical tire forces especially during braking and acceleration where the vehicle will experience significant pitch and hence, load transfer between rear and front part of the vehicle. Also, most of the full vehicle models incorporate the effect of yaw rate to the total acceleration for the vehicle [17, 18, 30]. Referring to Fig. 6 and Eq. 5, total lateral and longitudinal acceleration of the vehicle, ayand ax,respectively, can be written as follows. ax = v\u0307x \u2212 rvy ay = v\u0307y + rvx (10) Another dynamic approach usually employed in vehicle modelling is Euler-Lagrange method. Different from Newtonian method that focus on analysis of forces and moments acting within the vehicle system, Euler-Lagrange method focuses on the changes in potential and kinetic energies due to outside disturbances into the system. For ground vehicles, Lagrangian approach usually used for models that include ride models, i.e. vehicle model for motions in vertical plane [34\u201336]. In path tracking studies, this method was seldom used for ground vehicles due to the complex derivations that may entail, according to the various subsystems in lateral dynamics. However, it can be found used in various studies [37\u2013 40] to derive mathematical model of smaller scales autonomous mobile robots. Also, Lagrangian method is used in the multi-body dynamics software MSC ADAMS in deriving the mathematical models of the vehicle system. Users can also obtain the Lagrange equations used in modelling the system [41]. In general, the Euler-Lagrange formula can be generalised in Eq. 11. d dt ( \u2202L \u2202q\u0307i ) \u2212 \u2202L \u2202qi = Qx i (11) Where qi are generalized coordinates and Qx i are corresponding external forces. The Lagrangian, L is given in Eq.12, where T and Uare the kinetic and potential energy of the system, respectively. In vehicle handling model application for motions in lateral direction, a set of local coordinates based on vehiclefixed frame usually used, where q \u2032 = [ vx vy \u03c8\u0307 \u03d5 ] . Here, vx and vy are vehicle velocity in lateral and longitudinal direction respectively, \u03c8\u0307 is the vehicle yaw rate about vertical axis and \u03d5 is the vehicle roll angle about the longitudinal axis [34]. Detail derivations of the Euler-Lagrange approach in vehicle modelling can be referred to [34, 36, 42]."
        ]
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure1.8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure1.8-1.png",
        "caption": "Fig. 1.8. Solid cylindrical shaft loaded with two masses m1 and m2.",
        "texts": [
            " Coreless stator configurations eliminate any ferromagnetic material from the stator (armature) system, thus making the associated eddy current and hysteresis core losses nonexisting. This type of configuration also eliminates axial magnetic attraction forces between the stator and rotor at zero-current state. It is interesting that slotless AFPM machines are often classified according to their winding 10 1 Introduction arrangements and coil shapes, namely, toroidal, trapezoidal and rhomboidal forms [38, 49, 87]. All spinning shafts, even in the absence of external load, deflect during rotation. Fig. 1.8 shows a shaft with two rotating masses m1 and m2. The mass m1 can represent a disc-type rotor of an AFPM machine while the mass m2 can represent a load. The mass of the shaft is msh. The combined mass of the rotor, load and shaft can cause deflection of the shaft that will create resonant vibration at certain speed called whirling or critical speed . The frequency when the shaft reaches its critical speed can be found by calculating the frequency at which transverse vibration occurs. The critical speed in rev/s of the ith rotating mass can be found as [227] ni = 1 2\u03c0 \u221a g \u03c3i (1.4) where g = 9.81 m/s2 is the acceleration due to gravity and \u03b4i is the static deflection at the ith position of the rotor due to ith rotor only, i.e., \u03c3i = miga 2 i (L\u2212 ai)2 3EiIiL (1.5) 1.6 Rotor Dynamics 11 In the above equation Ei is the modulus of elasticity (for steel E = 200\u00d7 109 Pa), Ii is the area moment of inertia of cross-sectional area, L is the length of the shaft and ai is the location of the ith rotor from the left end of the shaft (Fig. 1.8). The area moment of inertia can be found as Ii = \u03c0D4 i 64 (1.6) The resultant angular critical speed \u2126cr = 2\u03c0ncr for the shaft loaded with i number of rotors (\u2126i = 2\u03c0ni) can be found on the basis of Dunkerley\u2019s equation [80] as 1 \u21262 cr = \u2211 i 1 \u21262 i (1.7) or Rayleigh\u2019s equation [227] \u2126cr = \u221a g \u2211 i(mi\u03c3i)\u2211 i(mi\u03c32 i ) (1.8) The shaft is also considered a rotor with mass msh concentrated at 0.5L where L is the length of the shaft (bearing\u2013to\u2013bearing). Dunkerley\u2019s empirical method uses the frequencies that each individual load creates when each load acts alone and then combines them to give an approximation for the whole system [80]",
            " Moduli of elasticity, specific mass densities, diameters and widths (lengths) are as follows: (a) E1 = 200 \u00d7 109 Pa, \u03c11 = 7600 kg/m3, D1 = 0.3 m, w1 = 0.01 m for the disc-type rotor; (b)E2 = 200 \u00d7 109 Pa, \u03c13 = 7650 kg/m3, D2 = 0.15, w2 = 0.1 m for the driven wheel; (c) Esh = 210 \u00d7 109 Pa, \u03c1sh = 7700 kg/m3, Dsh = 0.0245 m, L = 0.6 m for the shaft. 20 1 Introduction The location of the rotor from the left end of the shaft is a1 = 0.2 m and the location of the driven wheel from the same end of the shaft is a2 = 0.35 m (Fig. 1.8). The acceleration of gravity is 9.81 m/s2. Solution Mass of rotor m1 = \u03c11 \u03c0D2 1 4 w1 = 7600 \u03c00.32 4 \u00d7 0.01 = 5.372 kg Mass of driven wheel m2 = \u03c12 \u03c0D2 2 4 w2 = 7650 \u03c00.152 4 \u00d7 0.1 = 13.519 kg Mass of shaft msh = \u03c1sh \u03c0D2 sh 4 L = 7700 \u03c00.02452 4 \u00d7 0.6 = 2.178 kg Area moment of inertia of the rotor according to eqn (1.6) I1 = \u03c00.34 64 = 3.976\u00d7 10\u22124 m4 Area moment of inertia of the driven wheel to eqn (1.6) I2 = \u03c00.154 64 = 2.485\u00d7 10\u22125 m4 Area moment of inertia of the shaft according to eqn (1.6) Ish = \u03c00"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure4.27-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure4.27-1.png",
        "caption": "Figure 4.27 (a) A trough bridge, composed of surface elements; (b) the same bridge with the walls replaced by trusses.",
        "texts": [
            " In mechanics, a structure is a three-dimensional cohesive whole of structural elements that has to be able to resist external influences (the loads). In many cases, structures appear to have been designed and built in such a way that the loads are transferred to the foundation via certain planes. In such cases, the three-dimensional structure can be modelled as a system of so-called planar structures (or two-dimensional structures). This is illustrated using two examples. Planar structures 4 Structures 127 The first example is the bridge in Figure 4.27a. The loading by the traffic is transferred from the plane of the road deck to the vertical walls. These walls are in practice the spanning structure and transfer the load via the supports to the abutments, which subsequently transfer it to the foundation. Surface elements (plates) can be used for the road deck and the walls; and together they form a so-called trough bridge. If the transverse measurements of the bridge are small compared to the span, the bridge can be modelled as a line element, or in other words, a bar with a U-section. In order to limit the use of material and thereby reduce the self-weight that has to be carried, the surface elements can be replaced by planar structures made of line elements, as has been done in Figure 4.27b for the vertical walls. The second example is the apartment building in Figure 4.28a. The structure consists of only surface elements. The vertical floor loading is transferred to the vertical walls and from there is transferred to the foundation. The horizontal wind loading is also distributed across the floors via the walls to the foundation. Figure 4.28b represents the same building, but now all the horizontal and vertical surface elements in the main load-bearing structure have been replaced by planar structures made up of beams and columns",
            " It is therefore certainly worth the effort of further investigating these types of planar structures. 128 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Planar trusses and frames are planar structures that are loaded in their plane (see Figure 4.29). The difference between a truss and a frame is determined by the nature of the joints in the connections. \u2022 in a truss, the bars are joined together by hinges at all the connections;1 \u2022 in frames, all the joints are fixed and entirely stiff. The truss in Figure 4.29a appeared in the bridge in Figure 4.27b. The open circles, which represent the hinged joints, are generally omitted as in a truss all the joints are by definition hinged. The structure in Figure 4.29b is a frame. You will recognise part of the building in Figure 4.28b here, with the vertical floor loading and the horizontal wind loading. Sometimes the stiffnesses of the joints are accentuated by thickenings in the connections, but generally they are omitted. If there are also hinged joints in a frame, they have to be clearly depicted by means of open circles"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure2.21-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure2.21-1.png",
        "caption": "FIGURE 2.21. A cresting vehicle at a point where the hill has a radius of curvature Rh.",
        "texts": [
            " Forward Vehicle Dynamics 2.8 F Vehicles on a Crest and Dip When a road has an outward or inward curvature, we call the road is a crest or a dip. The curvature can decrease or increase the normal forces under the wheels. 2.8.1 F Vehicles on a Crest Moving on the convex curve of a hill is called cresting. The normal force under the wheels of a cresting vehicle is less than the force on a flat inclined road with the same slope, because of the developed centrifugal force mv2/RH in the \u2212z-direction. Figure 2.21 illustrates a cresting vehicle at the point on the hill with a radius of curvature RH . The traction and normal forces under its tires are approximately equal to Fx1 + Fx2 \u2248 1 2 m (a+ g sin\u03c6) (2.279) Fz1 \u2248 1 2 mg \u2219\u00b5 a2 l cos\u03c6+ h l sin\u03c6 \u00b6\u00b8 \u22121 2 ma h l \u2212 1 2 m v2 RH a2 l (2.280) 2. Forward Vehicle Dynamics 79 Fz2 \u2248 1 2 mg \u2219\u00b5 a1 l cos\u03c6\u2212 h l sin\u03c6 \u00b6\u00b8 + 1 2 ma h l \u2212 1 2 m v2 RH a1 l (2.281) l = a1 + a2. (2.282) Proof. For the cresting car shown in Figure 2.21, the normal and tangential directions are equivalent to the \u2212z and x directions respectively. Hence, the governing equation of motion for the car isX Fx = ma (2.283) \u2212 X Fz = m v2 RH (2.284)X My = 0. (2.285) Expanding these equations produces the following equations: 2Fx1 cos \u03b8 + 2Fx2 cos \u03b8 \u2212mg sin\u03c6 = ma (2.286) \u22122Fz1 cos \u03b8 \u2212 2Fz2 cos \u03b8 +mg cos\u03c6 = m v2 RH (2.287) 2Fz1a1 cos \u03b8 \u2212 2Fz2a2 cos \u03b8 + 2 (Fx1 + Fx2)h cos \u03b8 +2Fz1a1 sin \u03b8 \u2212 2Fz2a2 sin \u03b8 \u2212 2 (Fx1 + Fx2)h sin \u03b8 = 0. (2.288) We may eliminate (Fx1 + Fx2) between the first and third equations, and solve for the total traction force Fx1 + Fx2 and wheel normal forces Fz1 , Fz2 ",
            " The traction and normal forces under the tires of the vehicle are approximately equal to Fx1 + Fx2 \u2248 1 2 m (a+ g sin\u03c6) (2.306) Fz1 \u2248 1 2 mg \u2219\u00b5 a2 l cos\u03c6+ h l sin\u03c6 \u00b6\u00b8 \u22121 2 ma h l + 1 2 m v2 RH a2 l (2.307) Fz2 \u2248 1 2 mg \u2219\u00b5 a1 l cos\u03c6\u2212 h l sin\u03c6 \u00b6\u00b8 + 1 2 ma h l + 1 2 m v2 RH a1 l (2.308) l = a1 + a2. (2.309) Proof. To develop the equations for the traction and normal forces under the tires of a dipping car, we follow the same procedure as a cresting car. The normal and tangential directions of a dipping car, shown in Figure 2.21, are equivalent to the z and x directions respectively. Hence, the governing 2. Forward Vehicle Dynamics 85 equations of motion for the car are X Fx = ma (2.310)X Fz = m v2 RH (2.311)X My = 0. (2.312) Expanding these equations produces the following equations: 2Fx1 cos \u03b8 + 2Fx2 cos \u03b8 \u2212mg sin\u03c6 = ma (2.313) \u22122Fz1 cos \u03b8 \u2212 2Fz2 cos \u03b8 +mg cos\u03c6 = m v2 RH (2.314) 2Fz1a1 cos \u03b8 \u2212 2Fz2a2 cos \u03b8 + 2 (Fx1 + Fx2)h cos \u03b8 +2Fz1a1 sin \u03b8 \u2212 2Fz2a2 sin \u03b8 \u2212 2 (Fx1 + Fx2)h sin \u03b8 = 0. (2.315) The total traction force (Fx1 + Fx2) may be eliminated between the first and third equations"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003728_1.3099008-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003728_1.3099008-Figure8-1.png",
        "caption": "Fig 8. Copper model of a chariot from Tell Agrab, third millennium BC (above), and a clay model of a cart from the Indus Valley, before 2000 BC (below). (From Childe, 1956b).",
        "texts": [
            " The pictograph is inconclusive as to whether the wheels were Feeny et al: Historical review of dry friction and stick-slip phenomena 323 simple rollers or something more sophisticated. It seems that sledges were initially placed on rollers. Sumerian remains from around 3100 BC include an actual captive roller, which would work as long as the friction between the wheels and the ground was greater than that between the wheels and the circumferential bearing. Drawings of a Tell-Agrab chariot dated 2800 BC (Fig 8) clearly show axled wheels (Sleeswyk, 1982). By 2700 BC, Sumerians were using two- and four-wheeled chariots. The wheel has also been found to occur during the third millennium in the Indus valley, Crete, and Ur (Childe, 1956b). The wheel created jobs, such as that of Xi Zhong of the Xue Clan: Officer of Carriages in the period of Xia, 21st to 16th centuries BC (Zhou, 1983). Egyptian carriage axles had leather hub caps and were packed with grease, typically animal fat. A chariot recovered from the tomb of Yuaa and Thuiu (around 1400 BC) still had some of the original lubricant on its axle (Davison, 1957/58)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003459_0278364905058363-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003459_0278364905058363-Figure2-1.png",
        "caption": "Fig. 2. A legged posture is shown in which the ZMP is located just beneath the CM, the horizontal ground reaction force field is tangent to a circle centered about the ZMP, and the horizontal ground reaction force magnitude is a function of only radial distance. In this case, the net horizontal force is zero, but the net moment is non-zero. Thus, both the vertical component of moment and the CM work performed by the ground reaction force cannot be computed solely on the bases of the ZMP trajectory and the resulting ground reaction force vector.",
        "texts": [
            " For example, the resultant horizontal ground reaction force could be zero while the vertical component of moment and/or the work performed by the ground reaction force could be non-zero. Consider a legged posture in which the following conditions are satisfied: (1) the ZMP is located just beneath the CM; (2) the horizontal ground reaction force field is tangent to a circle centered about the ZMP; (3) the horizontal ground reaction force magnitude is a function of only radial distance. In this situation, shown in Figure 2, the net horizontal force is zero, but the net moment is non-zero. Another example is two particles of equal mass subject to two at NORTH CAROLINA STATE UNIV on April 28, 2015ijr.sagepub.comDownloaded from forces equal in magnitude but acting in opposite directions; while the net force is zero and the CM is at rest, the particles are moving and the work conducted by the external forces is non-zero. In other words, neither \u03b4WG.R. = FG.R,\u03b4 rZMP nor \u03b4WG.R. = FG.R,\u03b4 rCM is a permissible expression for the work performed by the ground reaction force"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure1.22-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure1.22-1.png",
        "caption": "FIGURE 1.22. A three-wheel motorcycle.",
        "texts": [
            " Tire and Rim Fundamentals 12\u2212 Six-axle multi-trailers 13\u2212 Seven or more axle multi-trailers Figure 1.21 illustrates the FHWA classification. The definition of FHWA classes follow. Motorcycles: Any motorvehicle having a seat or saddle and no more than three wheels that touch the ground is a motorcycle. Motorcycles, motor scooters, mopeds, motor-powered or motor-assisted bicycles, and three-wheel motorcycles are in this class. Motorcycles are usually, but not necessarily, steered by handlebars. Figure 1.22 depicts a three-wheel motorcycle. Passenger Cars : Street cars, including sedans, coupes, and station wagons manufactured primarily for carrying passengers, are in this class. Figure 1.23 illustrates a two-door passenger car. Passenger cars are also called street cars, automobiles, or autos. Other Two-Axle, Four-Tire Single-Unit Vehicles: All two-axle, four-tire vehicles other than passenger cars make up this class. This class includes pickups, panels, vans, campers, motor homes, ambulances, hearses, carryalls, and minibuses"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002595_978-3-030-19416-1-Figure2.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002595_978-3-030-19416-1-Figure2.1-1.png",
        "caption": "Fig. 2.1 Different encapsulation methods",
        "texts": [
            " These formulations provide slow-release properties and have excellent water-retention properties. They can also be used as compost after their degradation. 2 Polymer Based Micro- and\u00a0Nanoencapsulation of\u00a0Agrochemicals 14 Natural polymers are gaining more popularity and approval over synthetic polymers as materials for encapsulation due to their eco-friendly nature, cost- effectiveness, easy availability, and biodegradability (Campos et\u00a0al. 2014b). There are many different types of encapsulation methods formulated to encapsulate agrochemicals in recent years (Fig.\u00a02.1). A few common methods are: Interfacial polymerization is a process of microencapsulation in which polycondensation reaction takes place at the phase interface of two different solvents having one reactive monomer each (Liu et\u00a0al. 2012). For example, natural pyrethinnanocapsules (Guo et\u00a0al. 2014). Microcapsule formation takes place by precipitation of the solution containing the dispersion of encapsulating material and core material in the organic phase along with a solution immiscible with the encapsulating material (Zhou et\u00a0al"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure3.8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure3.8-1.png",
        "caption": "Figure 3.8 Compounding two parallel forces F1 and F2 graphically by adding the equilibrium system P1 = P and P2 = P . Here we use for the forces the visual notation.",
        "texts": [
            " For more than two forces, using the line of action figure becomes laborious, and the analytical approach is clearly preferable (see Section 3.1.7). To determine the magnitude and direction of the resultant, the force polygon can still be useful. If the forces F1 and F2 are almost parallel, or parallel, one can determine the magnitude and direction of the resultant R graphically in a force polygon, although the graphical construction of its line of action (the line of action figure) becomes difficult as the intersection of the lines of action is far away or even at infinity. In Figure 3.8, F1 and F2 are two parallel forces. The body on which the forces act is not shown. A graphical construction of the line of action is possible by having two equal yet opposite forces P1 = P and P2 = P apply to point A on the line of action of F1, and to point B on the line of action of F2, with AB as their common line of action. The magnitude of P can be chosen arbitrarily. Since P1 and P2 together form an equilibrium system, the combined effect of the forces F1, F2, P1 and P2 is equal to that of only F1 and F2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000453_j.engfracmech.2014.03.008-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000453_j.engfracmech.2014.03.008-Figure3-1.png",
        "caption": "Fig. 3. Residual stresses obtained by sin2 W-method for as-built and at 650 C heat treated specimens depicted for (a) the x-direction, (b) the 45 -direction and (c) the y-direction. The schematic depicted in (d) displays the reference coordinate system and the measuring position (point 1). Analyses were conducted on the {220} plane under tilt angles ranging from 0 to \u00b145 .",
        "texts": [
            " Consequently, it can be concluded, that (in comparison to other alloys tested so far) the 316L does not suffer severely from defects induced by SLM processing due to its well-balanced properties featuring high ductility even in the as-built condition (c.f. Table 1). Such properties eventually lead to fairly low notch sensitivity. Still, the results indicate that there is the possibility of slight enhancement of the fatigue performance of SLM processed and post-treated 316L. A reduction of residual stresses (c.f. Fig. 3), achieved by the 650 C annealing, resulted in a small increase in the fatigue limit of about 30 MPa. The role of residual stress will be further discussed in the next section. An increase of 50 MPa compared to the initial state could be achieved through the HIP processing, which expectedly proved to be the most effective means for increasing fatigue strength in the case of Ti\u2013 6Al\u20134V [17]. The results for fatigue crack growth (FCG) obtained using potential drop technique are shown in Figs. 1 and 2",
            " crack growth parallel and perpendicular to building direction, respectively. The mean threshold values are equal to 4.7 \u00b1 0.6 MPa m1/2 for the orientation corresponding to Fig. 1 and 4.6 \u00b1 0.7 MPa m1/2 for the orientation corresponding to Fig. 2. Consequently, HIP processing of SLM parts leads to an isotropic material behavior with respect to crack growth. In order to determine residual stresses in the SLM processed material XRD analyses were performed on both as-built and heat treated (650 C) specimens. Results from these measurements are given in Fig. 3. The average error of data concerning internal stresses is about \u00b125 MPa. The x- and the y-directions of the residual stresses as well as the building direction during SLM are depicted schematically in Fig. 3d. Measurements on samples treated at 650 C were conducted on fatigued specimens and untested specimens as reference. Based on the reference it could be shown that after fatigue a similar condition prevails. Consequently, no significant stress redistribution takes place during fatigue loading in the elastically loaded part of the samples. In order to characterize residual stresses inside the sample, measurements in a depth of 100 lm and 200 lm were performed on both as-built and heat treated (650 C) material. Measurements on the surface were taken at a minimum of two randomly selected points. All measurements showed no significant internal stresses for the x-direction and the highest stresses for the y-direction. Stress values obtained for an angle of 45 are in between. The in-depth evolution of the internal stress is dependent on the direction. In x-direction (Fig. 3a), the absolute value of the stress increases steadily, but still remains on a low level compared to the y-direction. In the 45 -direction (Fig. 3b) and the y-direction (Fig. 3c) the in-depth evolution of the internal stresses is qualitatively fairly similar. Highest values for the internal stress are depicted in a depth of 100 lm followed by a slight decrease of the values towards the sample interior. This kind of residual stress distribution seems to be affected by the process strategy during SLM fabrication. Shell and core of the samples are manufactured based on two different sets of parameters and the final step of SLM processing on each layer is the re-melting of the cohesive zone between shell and core. Consequently, in a depth of 100 lm between shell and core the highest values for internal stress have to be expected. By heat treating at 650 C residual stresses could be halved as shown for the y-direction in Fig. 3c. The high residual tensile stresses found in building direction in combination with the superimposed testing load according to Fig. 2 lead to even higher total loads compared to the orientation depicted in Fig. 1 where internal residual compressive stresses (x-direction corresponds to the loading direction for this case) have to be considered. In consequence, the internal stresses seem to play a minor role in terms of FCG performance in the current study contrary to the findings in [17] for Ti\u20136Al\u20134V",
            " 1\u20133 for the as-built and the at 650 C heat treated material it can be concluded that residual stresses have hardly any impact on the fracture mechanics behavior of 316L. This finding can be concluded based on the following argumentation: The stress relief heat treatment does not influence microstructural characteristics other than residual stress as will be shown in the remainder of this section. The as-built and the 650 C heat treated condition both show the same crack growth characteristics (c.f. Fig. 1). As those samples differ in residual stress state (Fig. 3), residual stress hardly influences the fracture mechanics behavior. Consequently, the differences in the crack growth behavior between samples with different building direction (c.f. Figs. 1 and 2) and HIP treatment have to be explained on a microstructural basis emphasizing the role of grain size and shape as well as texture. The EBSD micrographs shown in Fig. 4 depict for as-built and at 650 C heat treated material a very similar microstructure. The grains are elongated along the building direction, c"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure9.9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure9.9-1.png",
        "caption": "FIGURE 9.9. A homogeneous rectangular link.",
        "texts": [
            "62 \u23a4\u23a6\u23a1\u23a3 cos\u03b12 cos\u03b22 cos \u03b32 \u23a4\u23a6 = \u23a1\u23a3 0 0 0 \u23a4\u23a6 (9.189) we obtain \u03b12 = 10.9 deg (9.190) \u03b22 = 79.1 deg (9.191) \u03b32 = 90.0 deg . (9.192) The third principal axis is for I3 = 40\u23a1\u23a3 20\u2212 40 \u22122 0 \u22122 30\u2212 40 0 0 0 40\u2212 40 \u23a4\u23a6\u23a1\u23a3 cos\u03b13 cos\u03b23 cos \u03b33 \u23a4\u23a6 = \u23a1\u23a3 0 0 0 \u23a4\u23a6 (9.193) which leads to \u03b13 = 90.0 deg (9.194) \u03b23 = 90.0 deg (9.195) \u03b33 = 0.0 deg . (9.196) 548 9. Applied Dynamics Example 356 Moment of inertia of a rigid rectangular bar. Consider a homogeneous rectangular link with mass m, length l, width w, and height h, as shown in Figure 9.9. The local central coordinate frame is attached to the link at its mass center. The moments of inertia matrix of the link can be found by the integral method. We begin with calculating Ixx Ixx = Z B \u00a1 y2 + z2 \u00a2 dm = Z v \u00a1 y2 + z2 \u00a2 \u03c1dv = m lwh Z v \u00a1 y2 + z2 \u00a2 dv = m lwh Z h/2 \u2212h/2 Z w/2 \u2212w/2 Z l/2 \u2212l/2 \u00a1 y2 + z2 \u00a2 dx dy dz = m 12 \u00a1 w2 + h2 \u00a2 (9.197) which shows Iyy and Izz can be calculated similarly Iyy = m 12 \u00a1 h2 + l2 \u00a2 (9.198) Izz = m 12 \u00a1 l2 + w2 \u00a2 . (9.199) Since the coordinate frame is central, the products of inertia must be zero"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure4.17-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure4.17-1.png",
        "caption": "Figure 4.17 (a) An example of a simple roller support. (b) If the roller is large enough and the movements and rotations remain small, a large part of the roller can be omitted. The roller support changes into a bar support.",
        "texts": [],
        "surrounding_texts": [
            "Figure 4.15 is a schematic representation of a roller support. For a roller support at A, the body can move parallel to the so-called roller track, and can also rotate freely about A. Only motion of A perpendicular to the roller track is prevented; this is the direction in which the interaction force is exerted. For the roller support in Figure 4.16, with the roller track parallel to the x axis, the following applies for the motion at A: ux;A = unknown (free motion), uy;A = 0 (prescribed motion), \u03d5z;A = unknown (free motion). ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM 4 Structures 123 The following applies for the forces in A in the coordinate system given: Fx;A = 0 (prescribed force), Fy;A = unknown (free force), Tz;A = 0 (prescribed force). The roller support therefore has two degrees of freedom and generates one support reaction. Note the parallel with a bar support! Figure 4.18 shows a steel roller support used in older bridge structures. Due to the continuous sideways movement, the roller can end up askew after a while. To prevent this happening, the roller is provided with a tooth structure on its sides (comparable to a cogwheel). In order to prevent displacement in the z direction, a groove is sometimes cut into the roller that fits over an open ridge in the rail and bearing pedestal. This example of a steel roller support provides a good picture of how it works. Roller supports can be made of materials other than steel, but then as sliding supports. Examples include supports made of rubber or plastics (neoprene), occasionally in combination with Teflon to reduce friction. The roller support shown can transfer only compressive forces and no tensile forces. This is not a problem as long as the loading generates only compressive forces in the support. Such a load could be, for example, the ever-present weight of the structure. Generally speaking, the weight of a structure, such as a bridge, is sufficiently large to ensure that the roller support is continuously loaded by compressive forces. If a roller support also has to be able to transfer tensile forces, special structural provisions have to be made. It is assumed here that a roller support can transfer both tensile and compressive forces. 124 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM"
        ]
    },
    {
        "image_filename": "designv10_0_0000352_j.matdes.2014.07.006-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000352_j.matdes.2014.07.006-Figure3-1.png",
        "caption": "Fig. 3. Temperature distributions during SLM process at P = 250 W and v = 200 mm/s: (a) on the top surface of the molten pool and (b) on the cross-section of the molten pool as the laser beam reached the center of the first layer (point 1); (c) on the top surface of the molten pool and (d) on the cross-section of the molten pool as the laser beam reached the center of the second layer (point 2).",
        "texts": [
            " The processing parameters in the experiments were same as those used in the simulations (Table 2). Samples for metallographic examinations were cut, ground and polished according to standard procedures. The cross-sectional microstructures of the SLM-fabricated AlSi10Mg parts were characterized using a PMG3 optical microscopy (Olympus Corporation, Japan). The characteristic surface morphologies of the SLM-fabricated AlSi10Mg parts were characterized by a Quanta 200 scanning electron microscopy (SEM) (FEI Company, The Netherlands) in secondary electron mode at 20 kV. Fig. 3 shows the transient temperature distribution on the top surface and cross-section of the molten pool as the laser beam reaches different layers during the SLM process, using P of 250 W and v of 200 mm/s. At the center of the first powder layer (point 1, Fig. 2), the isotherm curves on the top surface of the molten pool were similar to a series of ellipses. Moreover, the isotherm curves were intensive at the fore part of the ellipses compared with those at the back-end of it. The dashed line circle presented the isotherm of the melting temperature of AlSi10Mg (600 C). The area inside the black isotherm line possessed higher temperature than that of the melting temperature of AlSi10Mg, which induced a small molten pool within this area. The predicted working temperature of the molten pool decreased gradually from 1482 C in the center of the molten pool to 535 C at the edge of the pool. The width of the molten pool was approximately 94.2 lm (Fig. 3a), while the length and depth of which were approximately 129.1 lm and 61.7 lm, respectively (Fig. 3b). As the laser beam reached the center of the second layer (point 2, Fig. 2), the operative SLM temperature of the molten pool ranged from 1548 C in the center to 664 C at the edge of the pool. The width of the molten pool (111.4 lm) increased by 15.4% compared to that in the first layer (Fig. 3c). The length (148.3 lm) and depth (67.5 lm) of the molten pool were also slightly larger than those in the first layer, increasing by 12.9% and 8.5%, respectively (Fig. 3d). From the simulation results, it can be seen that the isotherm curves arrange more intensive at the fore part of the ellipses (unscanned zone) than that at the back-end of the ellipses (scanned zone). This is mainly due to the change of thermal conductivity caused by the transition from powder to solid, which in turn contributes to heat transmission in the powder layer. During the SLM process, as the laser beam moves from the first layer to the second layer, the average temperature and the size of the molten pool increase gradually"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure8.4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure8.4-1.png",
        "caption": "Fig. 8.4. The relative eddy in the PM channel.",
        "texts": [
            " 260 8 Cooling and Heat Transfer The Actual Radial Channel The actual characteristics of a hydraulic machine differ from the ideal case owing to two reasons: (i) the uneven spatial distribution of velocities in the blade passages, and (ii) the leakage and recirculation of flow and hydraulic losses such as friction and shock losses. These are completely different issues [78] and shall be dealt with separately. Slip factor As a result of the unbalanced velocity distribution of the leading and trailing edges of a PM channel and the rotation effects [232], there exists, according to Stodola [198, 232], a relative eddy within the blade passage shown in Fig. 8.4. This results in the reduction of the tangential velocity components and is called slip, which is usually accounted for using a slip factor. For approximately radial blades, the Stanitz slip factor ks (80o < \u03b22 < 90o) is ks = 1\u2212 0.63\u03c0/nb (8.23) where \u03b22 is the blade angle at exit and nb is the number of the blades. When applying a slip factor, the pressure relation (8.22) becomes \u2206p = \u03c1\u21262(ksR 2 out \u2212R2 in) + \u03c1 2 ( 1 A2 1 \u2212 1 A2 2 )Q2 (8.24) 8.3 Cooling of AFPM Machines 261 Shock, leakage and friction Energy losses due to friction, separation of the boundary layer (shock loss) and leakage should also be considered in the flow analysis"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003429_pnas.94.21.11307-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003429_pnas.94.21.11307-Figure3-1.png",
        "caption": "FIG. 3. A special propeller with a symmetrical propulsion matrix (B 5 C). See the text.",
        "texts": [
            ", the removal of an external torque N1 that was doing no work. The argument can be made by invoking the minimum dissipation theorem for inertialess f lows (16). If F2 were greater than F1, the rotationally unconstrained system of case ii could reduce its energy dissipation by reverting spontaneously to the kinematics of case i. We conclude that F2 # F1. [In view of Eq. 8, BCyAD $ 0. Because A . 0 and D . 0, AD . 0; therefore,] BC $ 0. [9] The next step in the proof is to exhibit [a special] propeller for which, manifestly, B 5 C. As shown in Fig. 3, it consists of two short rods mounted obliquely at the end of stiff radial wires of infinitesimal thickness [and of length a] attached to a stiff axial wire likewise of infinitesimal thickness, i.e., of negligible flow resistance. Let the rods be short compared with their distance from the axis and let the symmetry axis of each rod make an angle of 45\u00b0 with the axial direction. It follows then from the symmetry of the rods with respect to the orthogonal directions, axial and circumferential, that the longitudinal force F associated with a circumferential speed av will be the same as the circumferential force Nya associated with a longitudinal speed v"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002414_s11837-018-3207-3-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002414_s11837-018-3207-3-Figure1-1.png",
        "caption": "Fig. 1. Schematic of metal 3D printing process using Ultrafuse 316LX filament.",
        "texts": [
            " The usage of Ultrafuse metal filament is the same as using a plastic filament. The filament is first heated to its glass-transition temperature, then extruded from the nozzle to print a \u2018\u2018green\u2019\u2019 part. Users can customize the parameters and develop new process schemes to achieve a desired density or efficiency. Thereafter, the green part is subjected to a debinding process to remove the plastic content and form a \u2018\u2018brown\u2019\u2019 part, followed by a sintering process to finalize the metal part, as shown in Fig. 1. The debinding and sintering process is the same as the industry-standard process for injection-molded metal parts.11 The objective of this study is to investigate the material properties of a SS 316L part obtained from a green part printed using a low-cost desktop-size 3D printer with Ultrafuse filament, in comparison with SLM SS 316L alloy. The cost efficiency of using Ultrafuse filament for SS 316L parts is evaluated. Preliminary guideline on how to select SLM or FDM for fabrication of metal parts is also discussed"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure5.18-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure5.18-1.png",
        "caption": "FIGURE 5.18. A rigid body with coordinate frame B (oxyz) moving freely in a fixed global coordinate frame G(OXY Z).",
        "texts": [
            "390) = \u23a1\u23a3 c\u03d5c\u03c8 \u2212 c\u03b8s\u03d5s\u03c8 c\u03c8s\u03d5+ c\u03b8c\u03d5s\u03c8 s\u03b8s\u03c8 \u2212c\u03d5s\u03c8 \u2212 c\u03b8c\u03c8s\u03d5 \u2212s\u03d5s\u03c8 + c\u03b8c\u03d5c\u03c8 s\u03b8c\u03c8 s\u03b8s\u03d5 \u2212c\u03d5s\u03b8 c\u03b8 \u23a4\u23a6 G G\u03b1B = \u23a1\u23a2\u23a2\u23a3 cos\u03c8 \u00b3 \u03b8\u0308 + \u03d5\u0307\u03c8\u0307 sin \u03b8 \u00b4 + sin\u03c8 \u00b3 \u03d500 sin \u03b8 + \u03b8\u0307\u03d5\u0307 cos \u03b8 \u2212 \u03b8\u0307\u03c8\u0307 \u00b4 cos\u03c8 \u00b3 \u03d5\u0308 sin \u03b8 + \u03b8\u0307\u03d5\u0307 cos \u03b8 \u2212 \u03b8\u0307\u03c8\u0307 \u00b4 \u2212 sin\u03c8 \u00b3 \u03b8\u0308 + \u03d5\u0307\u03c8\u0307 sin \u03b8 \u00b4 \u03d5\u0308 cos \u03b8 \u2212 \u03c8\u0308 \u2212 \u03b8\u0307\u03d5\u0307 sin \u03b8 \u23a4\u23a5\u23a5\u23a6 . 5. Applied Kinematics 279 5.11 Rigid Body Acceleration Consider a rigid body with an attached local coordinate frame B (oxyz) moving freely in a fixed global coordinate frame G(OXY Z). The rigid body can rotate in the global frame, while the origin of the body frame B can translate relative to the origin of G. The coordinates of a body point P in local and global frames, as shown in Figure 5.18, are related by the equation GrP = GRB BrP + GdB (5.391) where GdB indicates the position of the moving origin o relative to the fixed origin O. The acceleration of point P in G is GaP = Gv\u0307P = Gr\u0308P = G\u03b1B \u00d7 \u00a1 GrP \u2212 GdB \u00a2 +G\u03c9B \u00d7 \u00a1 G\u03c9B \u00d7 \u00a1 GrP \u2212 GdB \u00a2\u00a2 + Gd\u0308B . (5.392) Proof. The acceleration of point P is a consequence of differentiating the 280 5. Applied Kinematics velocity equation (5.326) or (5.327). GaP = Gd dt GvP = G\u03b1B \u00d7 G BrP + G\u03c9B \u00d7 G B r\u0307P + Gd\u0308B = G\u03b1B \u00d7 G BrP + G\u03c9B \u00d7 \u00a1 G\u03c9B \u00d7 G BrP \u00a2 + Gd\u0308B = G\u03b1B \u00d7 \u00a1 GrP \u2212 GdB \u00a2 +G\u03c9B \u00d7 \u00a1 G\u03c9B \u00d7 \u00a1 GrP \u2212 GdB \u00a2\u00a2 + Gd\u0308B",
            " Example 199 Acceleration of joint 2 of a 2R planar manipulator. A 2R planar manipulator is illustrated in Figure 5.19. The elbow joint has a circular motion about the base joint. Knowing that 0\u03c91 = \u03b8\u03071 0k\u03020 (5.394) we can write 0\u03b11 = 0\u03c9\u03071 = \u03b8\u03081 0k\u03020 (5.395) 0\u03c9\u03071 \u00d7 0r1 = \u03b8\u03081 0k\u03020 \u00d7 0r1 = \u03b8\u03081RZ,\u03b8+90 0r1 (5.396) 0\u03c91 \u00d7 \u00a1 0\u03c91 \u00d7 0r1 \u00a2 = \u2212\u03b8\u030721 0r1 (5.397) and calculate the acceleration of the elbow joint 0r\u03081 = \u03b8\u03081RZ,\u03b8+90 0r1 \u2212 \u03b8\u0307 2 1 0r1. (5.398) Example 200 Acceleration of a moving point in a moving body frame. Assume the point P in Figure 5.18 is indicated by a time varying local position vector BrP (t). Then, the velocity and acceleration of P can be found by applying the derivative transformation formula (5.287). GvP = Gd\u0307B + B r\u0307P + B G\u03c9B \u00d7 BrP = Gd\u0307B + BvP + B G\u03c9B \u00d7 BrP (5.399) GaP = Gd\u0308B + B r\u0308P + B G\u03c9B \u00d7 B r\u0307P + B G\u03c9\u0307B \u00d7 BrP +B G\u03c9B \u00d7 \u00a1 B r\u0307P + B G\u03c9B \u00d7 BrP \u00a2 = Gd\u0308B + BaP + 2 B G\u03c9B \u00d7 BvP + B G\u03c9\u0307B \u00d7 BrP +B G\u03c9B \u00d7 \u00a1 B G\u03c9B \u00d7 BrP \u00a2 . (5.400) 5. Applied Kinematics 281 It is also possible to take the derivative from Equation (5.323) with the assumption B r\u0307P 6= 0 and find the acceleration of P "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure14.29-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure14.29-1.png",
        "caption": "Figure 14.29 For taut cables (pk/ 1) the distributed load along the cable can be approximated by an equal distributed load along the chord.",
        "texts": [
            " In that case the cable shape is a parabola. In Table 14.2, the results are compared for a parabola and a catenary, both with = 200 m and pk = 40 m. The differences are relatively minor. In the example, the ratio between the sag and span is pk = 40 m 200 m = 0.2. The differences decrease sharply as the ratio pk/ decreases. For taut cables (pk/ 1), the catenary can be approximated by a parabola, for which the distributed load along the cable is replaced by an equal distributed load along the chord (see Figure 14.29). 662 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Example 1 Figure 14.30 represents the longitudinal section of a shelter. The vertical load on the ridge is q = 125 N/m. The horizontal component of the force that the guys exert on the ridge is H = 500 N. Question: Determine the sag of the ridge in the middle of the shelter. Solution: The following applies for the sag: pk = 1 8q 2 H = 1 8 (125 N/m)(2 m)2 500 N = 125 mm. Example 2 The dimensions for the suspension bridge in Figure 14.31 are derived from the bridge over the Storebaelt (Large Belt) in Denmark"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure14.26-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure14.26-1.png",
        "caption": "Figure 14.26 Replacing g (force per length measured along the cable) by q (force per horizontally measured length): q = g ds/dx.",
        "texts": [
            " The tensile force is a maximum at the supports A and B, with x = \u00b1100 m Nmax = \u221a (15 kN)2 + {(0.12 kN/m)(\u00b1100 m)}2 = 19.2 kN. Check: NA = Nmax \u221a A2 h + A2 v = \u221a (15 kN)2 + (12 kN)2 = 19.2 kN. Of course the same applies at B. Figure 14.25 shows a cable under its uniformly distributed dead weight g. In the cable equation H d2z dx2 = \u2212q. 656 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM q is a vertical force per horizontally measured length. The dead weight g of the cable is a vertical force per length measured along the cable.1 In order to replace the dead weight g by the load q , Figure 14.26 shows an infinitesimally small cable element with length ds. From the figure we find g ds = q dx and so q = g ds dx = g \u221a (dx)2 + (dz)2 dx = g \u221a 1 + ( dz dx )2 . The cable equation is now: H d2z dx2 = \u2212g \u221a 1 + ( dz dx )2 . (25) To solve this second degree differential equation we assume:2 dz dx = sinh u (26) 1 The symbol g is used for the dead weight of the cable, instead of the formal qdw. By doing so, we avoid the recurring index \u201cdw\u201d and maintain the distinction with q (force per horizontally measured length)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure12.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure12.2-1.png",
        "caption": "FIGURE 12.2 PEEC model with voltage sources for an incident electric field.",
        "texts": [
            " Essentially, we need to add the source voltage along the inductive branch of the cell due to the incident electric field Einc(r, t). We integrate the external fields over the cell shown in Fig. 12.1, or Ve(t) = 1 \u222b\u222b\ud835\udcc1 Einc(r, t) \u22c5 d\ud835\udcc1 d , (12.2) where Einc(r, t) = Einc x (r, t) x\u0302 + Einc y (r, t) y\u0302 + Einc z (r, t) z\u0302. (12.3) These sources are simply added to the basic PEEC loop, which is an addition to the circuit based on part of the Kirchhoff\u2019s voltage law (KVL). Hence, it is obvious that the extension of a PEEC model with the electric field is straightforward (Fig. 12.2). The right hand side basic loop in Fig. 12.2 represents the PEEC model for a dielectric with the incident field applied. In the modified nodal analysis (MNA) equations, the independent field sources will appear in the right-hand-side source vector of the MNA equations as shown in Fig. 6.10 and (6.55). As an example, we experiment a simple problem that was solved using both PEEC [4] and an approximate transmission line solution [11]. The example problem consists of a two-wire transmission line, which is 1 m long and the conductors are spaced by 5 cm"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003908_978-1-84882-730-1-Figure8.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003908_978-1-84882-730-1-Figure8.1-1.png",
        "caption": "Figure 8.1 Interpretation of tracking errors",
        "texts": [
            "1) can be transformed to a diagonal form to overcome difficulties caused by nonzero off-diagonal terms in the system matrices. 2. Interpret tracking errors in a frame attached to the path \u02dd such that the tracking error dynamics are of a triangular form to which the backstepping technique can be applied. 3. Use the orientation tracking error as a virtual control to stabilize the cross-track error. 8.1.2 Coordinate Transformations 8.1.2.1 Transform Ship Dynamics to a \u201cDiagonal Form\u201d Introduce the following coordinate transformation (changing the ship position, see Figure 8.1): Nx D xC \"cos. /; Ny D yC \"sin. /; Nv D vC \"r; (8.12) where \" D m23=m22. Using the above change of coordinates, the ship dynamics (8.1) can be written as PNx D ucos. / Nv sin. /; PNy D usin. /C Nv cos. /; P D r; (8.13) PuD N u; PNv D \u02dbur \u02c7 NvC r; Pr D N r ; where we have chosen the primary controls u and r as: u Dm11 N u m22vr m23Cm32 2 r2Cd11u; r D m22m33 m23m32 m22 N r 1 m22 .m11m22 m222/uvC m11m32 m22 m23Cm32 2 urC .m32d22 m22d32/v .m22d33 m32d23/r ; (8.14) with N u and N r being considered as new controls to be designed later. 8.1 Full-State Feedback 169 Remark 8.2. After the ship system (8.1) is transformed to the diagonal form (8.13), the methods mentioned in the preceding three chapters can be used to obtain controllers for specific tasks such as stabilization and model reference trajectorytracking if . Nx; Ny/ are considered as the ship position instead of .x;y/. 8.1.2.2 Transform Path-tracking Errors We now interpret the path-tracking errors in a frame attached to the reference path \u02dd as follows (see Figure 8.1): 2 64 xe ye N e 3 75D J T . / 2 64 Nx xd Ny yd d 3 75 ; (8.15) where d is the angle between the path and the X -axis defined by d D arctan y 0 d .s/ x 0 d .s/ ! ; (8.16) with x 0 d .s/ and y 0 d .s/ being defined in (8.7). In Figure 8.1, OEXEYE is the earth-fixed frame, OdXdYd is a frame attached to the path \u02dd such that OdXd and OdYd are parallel to the surge and sway axes of the ship, Nud is tangential to the path, CG Ob is the center of gravity, and Os is referred to as the center of oscillation of the ship. Therefore xe , ye , and N e can be 170 8 Path-tracking Control of Underactuated Ships referred to as the tangential tracking error, cross-tracking error, and heading error, respectively. Differentiating (8.15) along the solutions of the first three equations of (8"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000024_41.184817-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000024_41.184817-Figure1-1.png",
        "caption": "Fig. 1. A simple VSC example. (a) System model. (b) Regions defined by the switching logic.",
        "texts": [],
        "surrounding_texts": [
            "CONVENTIONS FOR SYMBOLS\nCapital, italic letters represent matrices, e.g., A and K. Boldface, roman lower-case letters represent vectors,\nbrief discussions about its historical development are presented.\nA. The Basic Notion of VSC\nsecond-order system [431, [ 1541 similar to the following The basic idea of VSC was originally illustrated by a\nX = y y = 2 y -x + U U = -*x ( l a )\nwhere\n(1b) * = 4 when s ( x , y ) > 0\n= -4 when s ( x , y ) < 0 e.g., a and k. and\nk .\nrespectively.\ns ( x , y ) = X U , u = 0 . 5 x + y . (IC) Lower-case, italic letters represent scalars, e.g., a and\ndim-x and d i m 4 stand for dimensions of x and B, A block diagram of the system is shown in Fig. l(a). The variable s(x, y) in (1.14 is the product of two functions\nI. INTRODUCTION x = 0 and U = 0 . 5 ~ + y = 0.\nARIABLE structure control (VSC) with sliding mode V control was first proposed and elaborated in the early 1950\u2019s in the Soviet Union by Emelyanov and several coresearchers [43], [79], [ 1541. In their pioneer works, the plant considered was a linear second-order system modeled in phase variable form. Since then, VSC has developed into a general design method being examined for a wide spectrum of system types including nonlinear systems, multi-input/multi-output systems, discrete-time models, large-scale and infinite-dimensional systems, and stochastic systems. In addition, the objectives of VSC has been greatly extended from stabilization to other control functions. The most distinguished feature of VSC is its ability to result in very robust control systems; in many cases invariant control systems result. Loosely speaking, the term \u201cinvariant\u201d means that the system is completely insensitive to parametric uncertainty and external disturbances. Today, research and development continue to apply VSC to a wide variety of engineering systems. In this introductory section, the basic notion of VSC and The functions describe lines dividing the phase plane (xy plane) into regions where s ( x , y ) has different sign as shown in Fig. l(b). As such, the lines (2) are often called switching lines and s(x, y ) is called a switching function. The lines also define the set of points in the phase plane where s ( x , y ) = 0. This set of points is known as the switching su$uce, despite the fact that the set composed of two lines is not a surface in the strict sense. All of these terms are more carefully defined and used later. The feedback gain q9 is switched according to (lb), i.e., the sign of s ( x , y). Therefore, the system (la, b) is analytically defined in two regions of the phase plane by two different mathematical models: In region I where s(x, y) = x u > 0, model is x = y y = 2 y - x - (3) 4x = 2 y - 5 x In region I1 where s ( x , y ) = x u < 0, model is\n(4) x = y 3 = 2 y - x + 4x = 2y + 3x.\nManuscript received Nov. 26, 1991; revised December 28, 1991, April 30. 1992. and Julv 17. 1992. The phase plane trajectories for (3) and (4) are shown as\nJ. Y. Hung is\u2019with the Electrical Engineering Department, Auburn port&ts in-Fig. 2(a)\u201dand (b). The equilibrium point of (3) University, Auburn, AL 36849.\nAeronautics and Astronautics, Beijing 100083, China. J. C. Hung is with the Electrical and Computer Engineering Department, The University of Tennessee, Knoxville, TN 37996, and is the author to whom correspondence should be addressed.\nW. Gao is with The Seventh Research Division, Beijing University of is an unstable focus at the origin. The equilibrium point of (4) is a saddle at the origin; the saddle point is also unstable. The phase portrait for the system (la, b) is formed by drawing the portrait for (2) in region I of the phase plane IEEE Log Number 9204072.\n0278-0046/93$03.00 0 1993 IEEE",
            "HUNG et al.: VARIABLE STRUCTURE CONTROL: A SURVEY 3\nI\u2019\nand drawing the portrait for (3) in region 11. The resultant portrait is shown in Fig. 2(c). To obtain the complete phase portrait, the trajectory of the system on the set s ( x , y ) = 0 must be described. On the line x = 0, the phase trajectories of regions I and I1 are just joined together without any ambiguity. On the line\nThe complete phase portrait of the system shows that there are no unusual motion characteristics on the line x = 0 other than possible discontinuities on motion direction. However, the line U = 0 contains only endpoints of those trajectories coming from both sides of the line. These points constitute a special trajectory along the\nU = 0 line, representing motion called a sliding mode. Thus, a phase trajectory of this system generally consists of two parts, representing two modes of the system. The first part is the reaching mode, also called nonsliding mode, in which the trajectory starting from anywhere on the phase plane moves toward a switching line and reaches the line in finite time. The second part is the sliding mode in which the trajectory asymptotically tends to the origin of the phase plane, as defined by the differential equation (5). Four basic notions of this example VSC system should be observed:\n1) Since the origin of the phase plane represents the equilibrium state of the system, the sliding mode represents the behavior of the system during the transient period. In other words, the line that describes U = 0 defines the transient response of the system during the sliding mode. 2) During the sliding mode, trajectory dynamics ( 5 ) are of a lower order than the original model (1). 3) During the sliding mode, system dynamics are solely governed by the parameters that describe the line U = 0. 4) The sliding mode is a trajectory that is not inherent in either of the two structures defined by (2) or (3).\nDuring the control process, the structure of the control system (la, b) varies from one structure (2) to another (31, thus earning the name variable structure control. To emphasize the important role of the sliding mode, the control is also often called sliding mode control. It should be noted that a variable structure control system can be devised without a sliding mode, but such a system does not possess the associated merits. In the next section, the general VSC problem is stated.\nB. Statement of the VSC Problem\nequation For a given control system represented by the state\nx = A(x, t ) + B ( x , t ) u (6)\nwhere dim-x = n and dim-u = m , find:\nm switching functions, represented in vector form as s(x), and a variable structure control\nu(x, t ) = u+(x , t ) when s(x) > 0\n= U-(x , t ) when s(x) < 0\nsuch that the reaching modes satisfy the reaching condition, namely, reach the set s(x) = 0 (switching surface) in finite time. The physical meaning of above statement is as follows:\n1) Design a switching surface s(x) = 0 to represent a desired system dynamics, which is of lower order than the given plant. 2) Design a variable structure control u(x, t ) such that any state x outside the switching surface is driven to",
            "4 IEEE TRANSACTIONS O N INDUSTRIAL ELECTRONICS, VOL. 40, NO. I , FEBRUARY 1993\nreach the surface in finite time. On the switching surface, the sliding mode takes place, following the desired system dynamics. In this way, the overall VSC system is globally asymptotically stable. Then, a function s(x) = c l x l + c2x2 + ... +x, defines a surface (8) C , X l + c2x2 + c3x3 + ... +x, = 0\nC. Brief Theoretical Background Before the emergence of the early stages of VSC development, its foundation had already been laid. Elements of the foundation consist of the theory of oscillation and the qualitative theory of differential equations. Brief discussions of these elements are given below.\n(a) Phase Plane Method: As a powerful graphical tool for studying second-order dynamic systems, the phase plane method was well established in the realm encompassing the qualitative (geometric) theory of differential equations and oscillation theory. The classical literature of Andronov et al. 161 and Flugge-Lotz [52] cited many early works in these areas. In their outstanding works, two contributions provided the foundation for the emergence of vsc:\n1) Regionwise linearization of nonlinear dynamic systems in which linearization of nonlinear systems was applied in partitioned regions of the phase plane. This gave the initial prototype VSC systems. 2) The sliding mode motion, a term first used by Nikolski [ l l l l . This was the first concept of sliding mode control.\n(b) Theory of DiSferential Equations with a Nonanalytic Right-Hand Side: Two kinds of nonanalyticity are of importance with respect to VSC:\n1) Finite discontinuous right-hand side, which is the\n2) Double-valued right-hand side, which is the relay\nThe problem is that a differential equation is not defined at the point where the right-hand side of the equation is not analytic because the existence and uniqueness of the solutions at these points are not guaranteed. Hence, the phase plane method cannot give a complete solution without defining an auxiliary equation at these points. The auxiliary equation is the model of switching that occurs in VSC systems with discontinuous control. Five methods have been suggested to define the differential equation for the system at points of discontinuous dynamics. They are summarized as follows.\nMethod 1: First, transform the system model to controllable canonic form. For example, the controllable canonic form for a time-invariant, single-input linear system is\nrelay type discontinuity, and\ntype discontinuity with hysteresis.\nx, = x 2\nx,-1 =x, , (7)\nin the n-dimensional phase space. If switching is constrained to occur on this ( n - 1)-dimensional surface, then all points of discontinuity lie on the surface. By solving (8) for x , and then substituting the result in (71, the differential equation of the sliding mode is found:\nx1 = x 2\n(9)\nX n p i = - c l x l - c2x2 - *- . - Cnp I X , 1\nIn summary, the coefficients in the switching function (8) define the characteristic equation of the sliding mode if the system model is described in controllable canonic form.\nMethod 2: This method obtains the differential equation of sliding mode by means of two coordinate transformations. Consider a linear, time-invariant plant and a switching function described by\nX = A x + B u s(x) = c x\nwhere dim-x = n and dim-u = dim-s = m. The dynamics of the sliding mode could be more easily described if the state vector was composed with s as m of the state variables. Therefore, the objective is to transform the model (10) into a form that contains s as m of the state variables. The plant is first transformed by y = Tlx to the form [1581\nY , =A,lYl + & Y 2 - Y 2 =Az,y , +&y2 + B u\nwhere dim-y, = n - m, dim-y2 = m, and det second transformation\n# 0. A\n[ ;] = 7$:]\nbrings (11) to the form I601\nY, = & y , + & S\ns = & y l + A22s + B u\nWhen the system is in the sliding mode, the dynamics satisfy s = 0 and S = 0. From (121, the differential equation for the sliding mode is easily solved by setting s = 0:\nY l = A l l y , . (13)\nMethod 3: Fillipov established a systematic mathematical theory for differential equations with discontinuities [%I, [51], [110]. Consider a general plant\nx = f(x,u) (14)\nand switching function s(x) where dim-x = n, dim-u = m."
        ]
    },
    {
        "image_filename": "designv10_0_0001972_978-3-319-15171-7-FigureD.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001972_978-3-319-15171-7-FigureD.2-1.png",
        "caption": "Fig. D.2 Avariety of digraphs: a balanced graph, b strongly connected graph, c a directed spanning tree (with root 1)",
        "texts": [
            "1 A digraph G = (V, E) is said to be: (i) balanced, if every vertex in V has the same in- and out-degrees, (ii) strongly connected, if every vertex in V is globally reachable, (iii) connected, if the graph is undirected and strongly connected, (iv) a directed spanning tree or simply spanning tree if it is acyclic and there exists a vertex i \u2208 V , which is called root, such that, for every other vertex j = i , there is only one path starting at i and ending at j . In addition, it is said to have a spanning tree if there exists a spanning subgraph G \u2032 of G such that G \u2032 is a spanning tree. An example of the balanced graph is illustrated in Fig. D.2a, where vertices 1, 2 and 4 have degree 1, while vertices 3 and 5 have degree 2. It is easy to confirm that the graph is also strongly connected. Replacing the edge (1, 5) of Fig. D.2a by (1, 3) makes the degree of vertices 3 and 5 unbalanced but the resulting graph in Fig. D.2b is still strongly connected. An example of the directed tree is illustrated in Fig. D.2c, whose root is vertex 1. It is also useful to introduce a matrix L(G) called graph Laplacian associated with a digraph G. Appendix D: Basic Graph Theory 293 Definition D.2 Consider a digraph G = (V, E)with vertex set V := {1, \u00b7 \u00b7 \u00b7 , n} and edge set E \u2282 V \u00d7 V . Then, the graph Laplacian L(G) associated with the graph G is defined by L(G) := [Li j ] = \u23a7 \u23a8 \u23a9 di if j = i \u22121 if j \u2208 Ni 0 otherwise . For example, the graph Laplacian L(G) associated with Fig. D.1a and Fig.D.2a, b, c, which are denoted by G1, \u00b7 \u00b7 \u00b7 , G4 respectively, are given as L(G1) = \u23a1 \u23a2 \u23a2 \u23a2 \u23a2 \u23a3 2 \u22121 \u22121 0 0 0 1 \u22121 0 0 \u22121 0 1 0 0 \u22121 0 \u22121 2 0 0 0 0 0 0 \u23a4 \u23a5 \u23a5 \u23a5 \u23a5 \u23a6 , L(G2) = \u23a1 \u23a2 \u23a2 \u23a2 \u23a2 \u23a3 1 \u22121 0 0 0 0 1 \u22121 0 0 0 0 2 \u22121 \u22121 0 0 0 1 \u22121 \u22121 0 \u22121 0 2 \u23a4 \u23a5 \u23a5 \u23a5 \u23a5 \u23a6 , L(G3) = \u23a1 \u23a2 \u23a2 \u23a2 \u23a2 \u23a3 1 \u22121 0 0 0 0 1 \u22121 0 0 \u22121 0 3 \u22121 \u22121 0 0 0 1 \u22121 0 0 \u22121 0 1 \u23a4 \u23a5 \u23a5 \u23a5 \u23a5 \u23a6 , L(G4) = \u23a1 \u23a2 \u23a2 \u23a2 \u23a2 \u23a3 0 0 0 0 0 \u22121 1 0 0 0 \u22121 0 1 0 0 0 0 0 1 \u22121 0 0 \u22121 0 1 \u23a4 \u23a5 \u23a5 \u23a5 \u23a5 \u23a6 , respectively. By the definition, the graphLaplacian L(G)must have a zero right eigenvaluewith eigenvector 1n , i"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure6.40-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure6.40-1.png",
        "caption": "FIGURE 6.40. A universal joint.",
        "texts": [
            "280) and from quadrilateral \u00a4EFNB, we find \u03b23 +\u2220EFN + \u03b22 + \u03b21 +\u2220EBN = 2\u03c0. (6.281) However, \u2220EBN = \u03c0 2 (6.282) \u2220EFN = \u03c0 2 (6.283) and therefore, \u03b23 + \u03b22 + \u03b21 = \u03c0. (6.284) The output angle \u03b84 is equal to \u03b84 = \u03c0 \u2212 (\u03b22 + \u03b21) , (6.285) and thus, \u03b84 = \u03b23. (6.286) Now the angle \u03b3 may be written as \u03b3 = \u03c0 \u2212 \u03b84 \u2212 \u03b1 (6.287) where \u03b84 is the output angle, found in Equation (6.173). Therefore, the coordinates xC and yC can be calculated as two parametric functions of \u03b82 for a given set of a, d, c, e, and \u03b1. 6.6 F Universal Joint Dynamics The universal joint shown in Figure 6.40 is a mechanism used to connect rotating shafts that intersect in an angle \u03d5. The universal joint is also known as Hook\u2019s coupling, Hook joint, Cardan joint, or yoke joint. Figure 6.41 illustrates a universal joint. There are four links in a universal joint: link number 1 is the ground, which has a revolute joint with the input link 2 and the output link 4. The input and the output links are connected with a cross-link 3. The universal joint is a three-dimensional four-bar linkage for which the cross-link acts as a coupler link"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure3.22-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure3.22-1.png",
        "caption": "Figure 3.22 (a) A triangular block subject to three forces and a couple; (b) the resultant force R on the block, determined using a force polygon.",
        "texts": [
            " To summarise, with respect to the resultant of a system of forces and couples, one can distinguish the following cases: \u2022 R = 0 and \u2211 Tz|O = 0 There is a resultant force, and the line of action does not pass through O. \u2022 R = 0 and \u2211 Tz|O = 0 There is a resultant force of which the line of action passes through O. \u2022 R = 0 and \u2211 Tz|O = 0 There is no resultant force, but there is a resultant couple. \u2022 R = 0 and \u2211 Tz|O = 0 The forces and couples together form an equilibrium system. Example Three forces and a couple are exerted on the triangular block in Figure 3.22a. The magnitude and the direction of the forces can be found in the diagram, as can the direction of couple T . The magnitude of the couple is 30 kNm. Figure 3.21 (a) The resultant force R at O and the associated couple \u2211 Tz|O are statically equivalent to (b) a force R at a distance a = ( \u2211 Tz|O)/R from O. ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM 3 Statics of a Rigid Body 67 Question: Determine the magnitude, direction, and line of action of the resultant force on the block. Solution: For convenience sake, the units (kN and/or m) are not always shown in the interim calculations. For the components of the resultant force R applies Rx = 3\u2211 i=1 Fx;i = \u221210 + 30 + 0 = +20 kN, Ry = 3\u2211 i=1 Fy;i = 0 + 20 \u2212 40 = \u221220 kN so that R = \u221a R2 x + R2 y = \u221a 202 + (\u221220)2 = 20 \u221a 2 kN. The magnitude and direction of R and of its components can of course also be determined graphically by using a force polygon (see Figure 3.22b). The moment about O of the three forces and the couple is \u2211 Tz|O = +10 \u00d7 6 (for F1) +(20 \u00d7 6 \u2212 30 \u00d7 3) (for F2, resolved into its components) \u221240 \u00d7 4 (for F3) +30 (for T ) = \u221240 kNm. 68 The resultant R must have the same moment about O as the three forces and the couple. Imagine (x, y) is a point on the line of action of R, with components Rx and Ry (see Figure 3.23a). Then \u2211 Tz|O = \u221240 kNm = Ryx \u2212 Rxy. With Rx = +20 kN and Ry = \u221220 kN, this gives the following equation for the line of action of the resultant R: \u221240 kNm = (\u221220 kN)x \u2212 (+20 kN)y \u21d2 x + y = 2 m"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001770_0954405415619883-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001770_0954405415619883-Figure3-1.png",
        "caption": "Figure 3. (a) Laser scanning strategy (Zigzag pattern, 90 )21 and (b) building strategy of the four series tensile test samples.42",
        "texts": [
            " Therefore, it is important to understand the effects of the parameters during the building process: spot size, power, layer thickness, scanning speed, scanning pattern, and so on. In the study of Wang et al.,21 the SLM-processed Inconel 718 tensile samples were fabricated by a continuous wave IPG YLR-200 fiber laser with laser power of 170W and beam spot size of 100mm. The powder layer thickness was 20mm and the laser beam was scanning at a speed of 416.7mm/s in the strategy of zigzag scanning pattern with 90 rotating after each layer, which also have been mostly used in SLM process, as shown in Figure 3(a).21 The overlap rate of the scanning lines was 30% at the beam sport size of 100mm, which means the distance between scanning lines (hatch spacing) was 70mm. A fine dendritic structure was found on both cross section and vertical section, as shown in Figure 5(a) and (b), which also has been found by Amato et al.39 who reported that the dimensions of the fine equiaxed/columnar microstructure are 0.5\u20131mm observed in samples from cylindrical components produced with the EOS M270 SLM system containing a 200W Yb:YAG fiber laser with a diameter laser beam of 100mm scanning at 800 or 1200mm/s",
            " Then at UNIV CALIFORNIA SAN DIEGO on March 7, 2016pib.sagepub.comDownloaded from the surface became relatively smooth with few scattered metallic globules to a sound surface with the near-full density of 98.4% at the recommended laser energy density of 330 J/m. In order to know the effect of orientation on the tensile strength and microstructures, Chlebus et al.42 have built four series of tensile specimens with differing orientation of their axes with respect to the build direction, as shown in Figure 3. The computer numerical control (CNC) finishing process was used after the buildup process to remove residues of support structures and to achieve a smooth, notch-free surface. Generally, the LRF, DLD, and LNSM processes will have a bigger beam spot size and layer thickness. The detail of the parameters in the studies can be seen in Table 2. The base phase in Inconel 718 alloy is g phase, also called g matrix, and the major precipitates are diskshaped g$ phase and spheroidal g# phase. There are also some needle/plate-like d phase, discrete metalcarbide (MC) particles, and round, island-like Laves phase"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003656_j.bios.2008.09.026-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003656_j.bios.2008.09.026-Figure2-1.png",
        "caption": "Fig. 2. A typical diagram of an enzymatic biofuel cell.",
        "texts": [
            " Here they are transferred to a high otential electron acceptor such as oxygen. As current now flows ver a potential difference, power is generated directly from the uel/analyte by the catalytic activity of the enzyme. Thus enzyme lectrodes are widely used both in biosensors as well as biofuel cell pplications. Enzymatic biofuel cells are one of the sub-classes of iofuel cell, while the other sub-class is microbial biofuel cell, which se a complete living cell as the biological element. The general onfiguration of an enzymatic biofuel cell is shown in Fig. 2. l H e u s n c ( w o h r d o t e s f b t e a T enzymatic biosensor. The performance of an enzyme electrode is largely governed by he materials used and the mechanism of their assembly on the urface of the electrode. Since the invention of the first enzyme lectrode was disclosed (Clark and Lyons, 1962; Charlton et al., 963; Updike and Hicks, 1967) about half-a-century ago, research n this area has intensified and there is a large base of literature in his area. Instead of covering the detailed inventions and develop- ent on the subject, we are attempting here to highlight the recent evelopments in material science that have made an impact on the evelopment of biosensors and enzymatic biofuel cells"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure9.7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure9.7-1.png",
        "caption": "FIGURE 9.7. Two coordinate frames with a common origin at the mass center of a rigid body.",
        "texts": [
            "157) If the local coordinate frame Oxyz is located such that the products of inertia vanish, the local coordinate frame is called the principal coordinate frame and the associated moments of inertia are called principal moments of inertia. Principal axes and principal moments of inertia can be found by solving the following equation for I:\u00af\u0304\u0304\u0304 \u00af\u0304 Ixx \u2212 I Ixy Ixz Iyx Iyy \u2212 I Iyz Izx Izy Izz \u2212 I \u00af\u0304\u0304\u0304 \u00af\u0304 = 0 (9.158) det ([Iij ]\u2212 I [\u03b4ij ]) = 0. (9.159) Since Equation (9.159) is a cubic equation in I, we obtain three eigenvalues I1 = Ix I2 = Iy I3 = Iz (9.160) 544 9. Applied Dynamics that are the principal moments of inertia. Proof. Two coordinate frames with a common origin at the mass center of a rigid body are shown in Figure 9.7. The angular velocity and angular momentum of a rigid body transform from the frame B1 to the frame B2 by vector transformation rule B2\u03c9 = B2RB1 B1\u03c9 (9.161) B2L = B2RB1 B1L. (9.162) However, L and \u03c9 are related according to Equation (9.52) B1L = B1I B1\u03c9 (9.163) and therefore, B2L = B2RB1 B1I B2RT B1 B2\u03c9 = B2I B2\u03c9 (9.164) which shows how to transfer the moment of inertia from the coordinate frame B1 to a rotated frame B2 B2I = B2RB1 B1I B2RT B1 . (9.165) Now consider a central frame B1, shown in Figure 9"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001795_tfuzz.2014.2312026-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001795_tfuzz.2014.2312026-Figure5-1.png",
        "caption": "Fig. 5. One-link Manipulator.",
        "texts": [
            " The simulation results are shown in Figs. 2-3. The H\u221e control performance is illustrated in Fig. 4, under different attenuation levels (\u03c1 = 0.1, 0.2, 0.3). It is shown in Fig. 4 that the integral of error signals under different prescribed attenuation levels decreases obviously with the effects of both fuzzy approximation errors and external disturbances (d1(t)=0.5sint,d2(t)=0.5cost). Example 2: Consider a one-link manipulator with the inclu- sion of motor dynamics, and assume that there is no stochastic disturbance (Fig. 5). Then the dynamic equation of such 1063-6706 (c) 2013 IEEE. Personal use is permitted, but republication/redistribution requires IEEE permission. See http://www.ieee.org/publications_standards/publications/rights/index.html for more information. system is given by{ Dq\u0308 +Bq\u0307 +N sin(q) = \u03c4 M\u03c4\u0307 +H\u03c4 = u\u2212Kmq\u0307 (74) where q, q\u0307 and q\u0308 denote the link position, velocity, and acceleration, respectively. \u03c4 is the torque produced by the electrical subsystem, u is the input of the system used to represent the electromechanical torque"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002502_tfuzz.2020.2985638-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002502_tfuzz.2020.2985638-Figure1-1.png",
        "caption": "Fig. 1: A one-link manipulator with motor dynamics system.",
        "texts": [
            " The Zeno behavior is that countless events may occur in a limited time interval in the system, which leads to the instability of the system. It is easy to see that if tk+1 \u2212 tk > 0, for any k, then the Zeno behavior will not occur. According to the proof of Theorem 1, we can get tk+1 \u2212 tk \u2265 (\u03b3|u(t)|+ d1)/\u03ba. Then, it is easy to conclude tk+1 \u2212 tk > 0, and therefore, the Zeno behavior can be successfully avoided. IV. SIMULATION EXAMPLES To validate the feasibility and effectiveness of the design scheme, a practical one-link manipulator with its motor dynamics system depicted in Fig. 1 is considered. The similar example has been used frequently for simulation in the literature [14] and [31]. Furthermore, compared with the simulation example in [14] and [31], the influence of the stochastic factor is also considered in the system to make the simulation close to a practical system. The system considered is  Aw\u0308 +Dw\u0307 +M sin(w) = \u03c1, N\u03c1\u0307+ F\u03c1 = u\u2212Rmw\u0307, y = \u03c21, (48) where w, w\u0307 and w\u0308 denote the position, velocity and acceleration of the link, respectively. \u03c1 is the motor shaft angle"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.27-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.27-1.png",
        "caption": "FIGURE 8.27. An example of a double A-arm front suspensions.",
        "texts": [
            " Such a stretch is usually less than 1/2 in \u2248 1 cm. 8.3 Roll Center and Roll Axis The roll axis is the instantaneous line about which the body of a vehicle rolls. Roll axis is found by connecting the roll center of the front and rear suspensions of the vehicle. Assume we cut a vehicle laterally to disconnect the front and rear half of the vehicle. Then, the roll center of the front or rear suspension is the instantaneous center of rotation of the body with respect to the ground. 8. Suspension Mechanisms 471 472 8. Suspension Mechanisms Figure 8.27 illustrates a sample of the front suspensions of a car with a double A-arm mechanism. To find the roll center of the body with respect to the ground, we analyze the two-dimensional kinematically equivalent mechanism shown in Figure 8.28. The center of tireprint is the instant center of rotation of the wheel with respect to the ground, so the wheels are jointed links to the ground at their center of tireprints. The instant center I18 is the roll center of the body with respect to the ground. To find I18, we apply the Kennedy theorem and find the intersection of the line I12I28 and I13I38 as shown in Figure 8"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure5.26-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure5.26-1.png",
        "caption": "Figure 5.26 (a) The interaction forces between joint D and members SD and BD are found from the equilibrium of the separate members; (b) the interaction forces as they really act.",
        "texts": [
            " VOLUME 1: EQUILIBRIUM If one selects the left-hand part AS, this gives \u2211 T (AS) z |S = Ah \u00d7 4 \u2212 Av \u00d7 4 = 0 (d) or, if one assumes the right-hand part BS \u2211 T (BS) z |S = \u221260 \u00d7 2 \u2212 Bh \u00d7 4 + Bv \u00d7 4 = 0. (e) Both equations are equivalent. The solution is Ah = Bh = 15 kN. All the support reactions are shown in Figure 5.24c. b. The interaction forces in hinge S follow from the force equilibrium of AS or BS (see Figure 5.25). The equilibrium of the left-hand part AS gives Sh = Sv = 15 kN. Check: For these forces, the left-hand part BS is also in equilibrium. c. To find the forces acting on joint D, the joint is isolated (see Figure 5.26a). There are three interaction forces acting between joint D and member SD. The magnitude of these forces is found from the equilibrium of member SD. In the same way, one can use the equilibrium of BD to find the magnitude of the three interaction forces between joint D and member BD. The result is shown in Figure 5.26b. Check: Joint D has to meet the conditions of force and moment equilibrium. 5 Calculating Support Reactions and Interaction Forces 173 The previous section shows that a vertical load on a three-hinged frame generates not only vertical, but also horizontal support reactions (see Figures 5.27a and 5.27b). Horizontal forces on foundations in soft soil often cause problems. To reduce the horizontal forces on the foundation, one can decide to link the bearings A and B of the three-hinged frame by means of a so-called tie-rod"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002049_j.ymssp.2021.107945-Figure16-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002049_j.ymssp.2021.107945-Figure16-1.png",
        "caption": "Fig. 16. Hinge and bending models of creases in an origami base structure: (a) torsional spring (top) and improved three-spring compliant crease model (bottom), and (b) the pseudo-rigid-body bending model.",
        "texts": [
            " Facets are joined by creases to form a unit cell (element). The crease design determines how the facets would be folded and deployed. A crease made of elastic materials provides torsional actuation for the facets to rotate about the crease during folding and deploying processes. Normally the crease is straight for a rigid origami pattern and thus the facets can be folded and deployed without deformation. There are two common crease models used in designing origami structures, namely, the hinge model and the bending model [76], as illustrated in Fig. 16. The mechanism of the hinge model is the use of active materials as the torsional and bending stiffness at the crease to drive the connected facets. In this model, the crease is a straight line with zeroth-order geometric continuity, which has been employed in the majority of the current rigid origami structures [229,230]. However, the bending stiffness at the crease of the hinge model would require revaluation if the crease range or material is changed. An evaluation on mechanical response of the different active crease materials has been conducted by using more compliant models in addition to the single torsional spring in [215,231\u2013234]",
            " In contrast to a traditional hinge model, a compliant joint could be more appropriate to approximate the fold of origami mechanisms [235,236]. An improved three-spring compliant crease model can be used to capture the torsional deformations of the creases and the extensional stretching energy of the origami structures [237]. The other modelling method, the bending model, is based on the bending deformation of the materials within the crease range. The advantage of the hinge model type is the massive foldability. However, the driving field for folding can be difficult to be defined, unless it is mechanically constrained. Fig. 16(b) illustrates a pseudo-rigid-body model to design origami structures. The rotational stiffness kcrease of the single compliant crease can be expressed as: kcrease \u00bc EcreaseLt3 12W \u00f034\u00de where parameters L, W and t represent the length, width and thickness of the crease model, and Ecrease is Young\u2019s modulus of the crease material. Fig. 17 shows the geometric continuity and non-zero thickness solution for the origami structure with non-negligible fold thickness. Modelling of an origami structure is usually conducted by neglecting facet thickness as shown in Fig",
            " [248] combined the quaternion rotation sequence and dual quaternion method to examine the rigid foldability of the foldable structure. Cai et al. [249] found that although the folding angles of the Kreslingpattern could meet with the requirement of the rigid foldability, its coordinate relationship would not meet with the requirement, meaning that the Kresling-pattern is not rigidly foldable. In order to study the compliance of the facets in the rigid origami structures, Rommers et al. [250] used the pseudo-rigidbody model, as shown in Fig. 16(a), to incorporate stiffness performance of a single vertex compliant facet origami mechanism, where the compliant facets were divided into two rigid facets using virtual hinge connections with torsional stiffness. Different joint methods were tested to manipulate the moment curve of the single vertex compliant facet mechanism for the origami design in [251]. Yu et al. [252] used the thin plate bending theory to study the deflections of the facets in the Miuraori base structure and obtained the deflection curves of the facets under different stress conditions"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure10.7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure10.7-1.png",
        "caption": "Figure 10.7 The sectional planes I are positive because the x axis points out of the matter and the unit normal vector n points in the positive x direction. The sectional planes II are negative because the x axis points into the matter and the unit normal vector n points in the negative x direction.",
        "texts": [
            " The sign of the section forces is related to a (local) coordinate system with the x axis along the member axis and the yz plane parallel to the member cross-sections (see Figure 10.6). 1 These names used in practice in no sense reflect that we are talking about interaction forces (pair of forces). 10 Section Forces 391 After applying a section, there are two cross-sectional planes. To distinguish these from one another, we call the sectional plane positive where the x axis points outwards, and the sectional plane negative where the x axis points inwards. This is shown in Figure 10.7 where the sectional planes I are positive, and sectional planes II are negative. More formally, we describe the position of a sectional plane using a socalled unit normal vector n. This is a unit vector (a vector with length 1) pointing outwards from the matter and normal to the sectional plane that is considered. The position of a sectional plane in space is fully determined by the scalar components nx ; ny ; nz of the unit normal vector n. Since the cross-section is normal to the x axis as chosen along the member axis, ny = nz = 0. The sectional plane is now said to be positive if the unit normal vector n points in the positive x direction (nx = +1), and negative if n is pointing in the negative x direction (nx = \u22121). Again, see Figure 10.7. Figure 10.8 shows the positive directions in the given xz coordinate system of the normal force N , the shear force V and the bending moment M . The sign conventions are as follows: \u2022 A normal force N is positive when it acts on a positive sectional plane in the positive x direction and on a negative sectional plane in the negative x direction. To simplify: a normal force N is positive as a tensile force and negative as a compressive force. This sign convention has already been used in trusses (see Section 9"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000448_s11740-009-0192-y-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000448_s11740-009-0192-y-Figure7-1.png",
        "caption": "Fig. 7 Final temperature and residual stresses",
        "texts": [
            " Pertaining to the equivalent stresses and single stress peaks, noticeably larger values are observed after the cooling period of the 8th layer. While the support shows certain flexibility, the largest stresses of about 490 N/mm2 are observed within the intersection between the cantilever and the structural plane respectively at the upper shells of the part. Analogue to the precedent layers, the whole structure still shows a bending of the externals into the positive vertical direction. This effect disappears for additional layers of the cantilever as described in the following. Subsequently, Fig. 7 represents the temperature and the residual stresses within the structure for an entire cooling cycle at the ambient temperature of 20 C. Essentially, it is observed that the structural plane with the cantilever is deformed 0.55 mm into the negative z-direction. This phenomenon is related with the growth of the solidified volume especially in the upper regions, whereby the deformations are inverted. It has to be mentioned, that the resulting strains are a fundamental result of the mechanical boundary conditions (BCs), i"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002816_tii.2021.3057832-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002816_tii.2021.3057832-Figure1-1.png",
        "caption": "Fig. 1. Experimental setup of an LM positioning system.",
        "texts": [
            " In addition, compared to [18]\u2013[20], it ensures the resultant tracking error to be bounded by an explicitly prespecified region whose size is independent on the disturbance bound. This paper is organized as follows. Section II describes the experimental setup of the LM positioning system and its dynamic model. The BFASM control design method and the new MBF are presented in Section III, where the stability analysis and parameter selections are also addressed. Section IV discusses the experimental results on the LM positioner. Finally, conclusion is given in Section V. The experimental setup of LM positioning system (by Baldor Electric Company) is shown in Fig. 1. The positioning stage is of a travel range of 500 mm and is directly driven by an LM. The axial position of the stage is detected by a position encoder (by Renishaw PLC) of a resolution of 1 \u00b5m. To proceed with the controller design, the plant model of the LM is given by my\u0308 = u\u2212 f \u2212 d (1) f = kv y\u0307 + kc sgn(y\u0307) (2) where y denotes the position of the stage, m is the moving mass of the positioner that may include any payload, u is the control input, f is the friction force, and d is the lumped uncertainties including unmodeled system dynamics, disturbance and measurement noise"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure16.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure16.2-1.png",
        "caption": "Figure 16.2 (a) Simply supported beam with (b) the isolated left-hand segment and (c) the influence line for the bending moment at C.",
        "texts": [
            " For example, for the case in Figure 16.1c: Av = +0.75 \u00d7 (10 kN) + 0.5 \u00d7 (20 kN) + 0.25 \u00d7 (30 kN) = +25 kN. From the influence line we can see that at the position of the force of 10 kN the influence is 0.75, or in other words: the contribution of this force to Av is: +0.75 \u00d7 (10 kN) = +7.5 kN. The total support reaction is found by superposing all the individual contributions. Example 2 \u2013 Influence line for a bending moment We will now determine the bending moment MC at midspan C for the beam in Example 1 (see Figure 16.2a). Solution: Assume the bending moment MC is positive if it causes tension at C at the underside of the beam. The magnitude of MC follows from the moment equilibrium of AC (or CB) about C. It should be remembered that it makes a difference in the equilibrium equations whether the load is to the left or right of the cross-section under consideration. 746 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM If the load is to the left of C (0 \u2264 x < 1 2 ) then the bending moment at C equals (see Figure 16.2b) MC = +F \u2212 x \u00b7 1 2 \u2212 F \u00b7 ( 1 2 \u2212 x ) = + 1 2Fx. If the load is to the right of C ( 1 2 <x \u2264 ) then the bending moment at C equals MC = +F \u2212 x \u00b7 1 2 = + 1 2F( \u2212 x). The variation of MC/F as a function of x, as shown graphically in Figure 16.2c, is referred to as the influence line for the bending moment at C. The influence line has its maximum in the middle. This means that the bending moment at C is a maximum when the point load F is at midspan: MC = ( + 1 4 ) \u00d7 F = + 1 4F . Example 3 \u2013 Influence line for a shear force For the same beam we will now determine the influence line for the shear force VC at C (see Figure 16.3a). Solution: The direction assumed for VC is shown in Figure 16.3b. The shear force VC can be found from the vertical force equilibrium of AC (or CB)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000554_978-3-540-85629-0-Figure8.15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000554_978-3-540-85629-0-Figure8.15-1.png",
        "caption": "Fig. 8.15. Example of motion primitive: circular arc.",
        "texts": [
            "9) Although the trajectory composed by a set of linear segments is continuous, it is characterized by discontinuous derivatives at the intermediate points. Then, undesired discontinuities in the velocity and acceleration profiles of the motion are present and, as discussed in more details in Sec. 8.11, this technique is often accompanied by the use of blending functions which guarantee a smooth transition between two consecutive segments. Another typical motion primitive is the circular arc, starting from a given point p0 and with the center located on a desired axis, univocally determined by a unitary vector z1 and a generic point d, see Fig. 8.15. Given these data, it is possible to compute the circular path. Firstly, it is necessary to determine the exact location of the center, provided that the trajectory does not degenerate in a point12. Let us define r = p0 \u2212d, then the position of the 12 In this case the point p0 is located on the axis through the center of the circle. 8.3 Definition of the Geometric Path Through Motion Primitives 357 center is o1 = d + (rT z1) z1 while the radius is \u03c1 = |p0 \u2212 o1|. The parametric representation of the circular arc in the frame F1 (defined by o1\u2212 x1y1z1, see Fig. 8.15) is simply p1(u) = \u23a1 \u23a3\u03c1 cos(u) \u03c1 sin(u) 0 \u23a4 \u23a6 , with 0 \u2264 u \u2264 \u03b8 that, expressed in the base frame F0, has the form p(u) = o1 + R p1(u) (8.10) where R is the rotation matrix of the frame F1 with respect to the frame F0, given by R = [x1 y1 z1]. Example 8.7 A two-dimensional trajectory composed by the following motion primitives 358 8 Multidimensional Trajectories and Geometric Path Planning p0 p1 z1 d \u03b8 line [0 0 0]T [1 0 0]T circle [1 0 0]T [0 0 1]T [1 1 0]T \u03c0/2 circle [2 1 0]T [0 0 1]T [3 1 0]T \u2212 \u03c0/2 line [3 2 0]T [5 2 0]T circle [5 3 0]T [0 0 1]T [5 3 0]T \u03c0 line [5 4 0]T [4 5 0]T is represented in Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002349_access.2017.2661967-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002349_access.2017.2661967-Figure3-1.png",
        "caption": "Fig. 3: PT 500 experimental test rig and corresponding layout graph.",
        "texts": [
            " Primarily, comparison study is carried out for evaluating the different performance of statistical parameters and vector distances, which are critical for generating discriminatory multi-speed fault signatures. Afterwards, the experimental results of roller bearing diagnosis and the generation of fault signatures are investigated and presented using the proposed approach in this paper. In this study, PT 500 machinery diagnostic system [26] is used to collect vibration signals of four conditions of roller bearing, as shown in Fig. 3 (a) and (b). Roller bearing faults kit is composed of motor assembly, motor control unit, shaft, four types of bearings, belt drive kit, and computerised vibration analyser. The control unit is used to collect speed and horsepower data. The piezo-electric sensor and measurement amplifier are used for acceleration measurement. During the tests, vibration data were captured at a sampling frequency of 8kHz for different bearing conditions. Roller bearings used in this paper are illustrated in Fig. 4, which are bearing A without damage, bearing B with outer race damage, bearing C with inner race damage, and bearing D with rolling element damage"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure8.18-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure8.18-1.png",
        "caption": "FIGURE 8.18 Zero thickness approximation of an L-shaped 3D object.",
        "texts": [
            " On the other hand, the connection between the upper and lower conductors can be established in the circuit domain by replacing the connection with a short circuit or a small macromodel such as a resistor in series with a partial inductance connected between node points A and B. It is clear that the meshing of plane in the middle is simplified considerably by approach. Physically thin conductor can be represented by a zero thickness meshing approximation. The key issue clearly is the joining of the thin planes as shown in Fig. 8.18. Further approximations can be made later by adding some thickness in the PEEC plane models. For the meshing, the planes are replaced by zero thickness planes located at the center of the actual conductor thickness as shown in Fig. 8.18. The planes need to be connected at the intersection where the nodes from both planes are connected. Importantly, the dominant physics and circuit properties are preserved by this simplification. AUTOMATIC GENERATION OF GEOMETRICAL OBJECTS 203 Today, different techniques exist for the meshing of complex geometries. Among the techniques for mesh generation, a distinction has been made between Indirect Methods and Direct Methods. The first indirect approach generates the quadrilateral surfaces by starting from an initial triangular mesh (e",
            " Some of the direct methods of interest are the looping algorithm, the domain decomposition through geometric optimization, and bubble mesh. We only briefly consider the principles of the looping algorithm. To illustrate some of the techniques used, we give a short introduction into the so-called looping algorithm. Sluiter and Hansen [32], as well as Talbert and Parkinson [33], have developed a recursive algorithm for the decomposition of a planar domain into quadrilateral shapes. It starts out with a planar domain with a fixed contour, which consists of piecewise linear segments as shown in Fig. 8.18. The required mesh density is prespecified according 204 GEOMETRICAL DESCRIPTION AND MESHING to some local requirements. Hence, it starts out with the closed polygonal made of n nodes and n edges. At each step, the algorithm subdivides the oval polygon through a segment (known as split line) of length Lsplit, where the ends connect to two of the n nodes as shown in Fig. 8.21. Other quantities needed in the algorithm are the length of the polynomial diagonal Dpol as well as the two areas of the subdomains A1 and A2 and the four angles 1\u20134 shown in Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001252_j.ymssp.2007.12.001-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001252_j.ymssp.2007.12.001-Figure2-1.png",
        "caption": "Fig. 2. (a) Rolling element bearing components and load distribution, (b) two degree-of-freedom model [10].",
        "texts": [
            " For instance, Lim and Singh [20,21] studied the vibration transmitted through rolling element bearings in rotor systems and in geared rotors. ARTICLE IN PRESS N. Sawalhi, R.B. Randall / Mechanical Systems and Signal Processing 22 (2008) 1924\u201319511928 The main fundamental components of a rolling element bearing are the outer race, the inner race, the cage and the rolling elements. The important geometrical quantities are the number of rolling elements nb, the element diameter Db, the pitch diameter Dp and the contact angle a. Fig. 2a shows these parameters and components along with the load zone associated with a unidirectional vertical load. The model discussed ahead was originally developed by Fukata et al. [17]. It provides the basis for the following developments. The non-linearity in a rolling bearing, arising from the non-linear forces between the different elements (Hertzian contact), the time-varying stiffness (load transmission dependency on the positioning of the supporting elements) and the clearance between the rolling elements and the bearing races, have been incorporated into this model. This gives a much-improved simulation of the dynamics of the REBs. The bearing was modelled as a two-DOF system, which provides the load\u2013deflection relationships, while ignoring the mass and the inertia of the rolling elements. The two orthogonal DOF are related to the inner race (rotor). Contact forces are summed over each of the rolling elements to give the overall forces on the shaft in the load direction and at right angles to it. Fig. 2b explains the essentials of this model as presented by Liew et al. [22], who gave an interesting summary presentation of four different ways to model the bearing, and used them in a transient software (developed in FORTRAN) to study the dynamics of a rotor system. All these models (two DOF with and without inertia and five DOF with and without inertia) are based upon the Hertzian contact relationship [10]. The overall contact deformation (contact compression) for the jth rolling element dj is a function of the inner race displacement relative to the outer race in the x- and y-directions (xs, ys), the element position fj (time varying) and the clearance (c)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001538_j.jmatprotec.2011.09.020-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001538_j.jmatprotec.2011.09.020-Figure7-1.png",
        "caption": "Fig. 7. Channelled gas rail design (ref: compl",
        "texts": [
            " This initial analysis is shown in Fig. 5 where the edges of he build area are denoted by gridlines and the gas rail with 9 flow utlets is the rectangular feature to the right edge of the figure. A number of design iterations were made to the gas rail, with the verall goal of producing an even gas flow over the entire process rea. Typically these iterations involved the segmentation of the as rail to provide individual channels to each individual gas outlet which can be seen in the cross-sectional view in Fig. 7), the opti- isation of the channel width to ensure similar flow rates through ach channel and the optimisation of the height of the outlet above he powder bed. All iterations were analysed to verify the efficacy f the design solution by CFD in the first instance. Two of these terations are described in Section 3.2. .2. Gas flow design developments The first design iteration manufactured (overall dimensions: = 37 mm, L = 430 mm, D = 13.5 mm) consisted of eight square utlets (L = 10 mm by H = 3 mm) each fed by a separate channel D = 11 mm by H = 3",
            " Tests showed that the samples at the back of the substrate (A1, B1, C1, D1 and E1, as shown in Fig. 2) were under poor flow conditions. This was later confirmed by the examination of a statistically significant number of samples built using this design gas rail. As uniformity of gas flow was shown to significantly affect the strength of the parts another iteration of gas rail design was undertaken to further improve the gas flow, this became the gas rail for the completed condition and can be seen in Fig. 7. This rail design includes a series of individual adjustable nozzles and a diffuser (offset diamond formation, four diamonds \u00d7 four rows) at the inlet of the gas flow rail to ensure even distribution to the channels. In addition to this the development of the individual feeds for each gas outlet are visible on the Figure. These additions to the gas rail were analysed using CFD with the results indicating a further improvement in the uniformity of the flow as shown in Fig. 8. It can be seen from this result that a stable and uniform flow across the build area was achieved"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001218_bfb0109984-Figure10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001218_bfb0109984-Figure10-1.png",
        "caption": "Fig. 10. Super-Twisting ~lgorithm phase trajectory",
        "texts": [
            "~m, VM,(5, ylM)) , VM2(5;ylM) ) (25) where VM, < VM~ are the solutions of thc second order algebraic equation [Im + r ~ (2 -- ~*)] ~:M--~--- + = (3Fro - o~*/'M) T -- 4 ~ - ~ 8 4 0 In accordance with the definition of real higher order sliding mode, in the case of approximated evaluation of the YtM values the size of the boundary layer of the sliding manifold is A -~ 0 ( 5 ' ) , and it can be minimized by choosing ~ as follows [5] 44 [ i l 3Fmzo~*FM] V M - 3 F m - a * F M 1 + + 4FM ] An extension of the real sub-opt imal 2-sliding control algorithm to a class of sampled data systems characterized by a constant gain flmction in (14), i.e., 7 = i, was recently presented in [9]. A development of the cited digital controller is presented in the next chapter. This algorithm has been developed for the case of systems with relative degree one in order to avoid chattering in VSS. Also in this case the trajectories on the 2-sliding plane are characterized by twistings around the origin (Fig.10), but the continuous control law u(t) is consti tuted by two terms. The first is defined by means of its discontinuous time derivative, while the other, which is present during the reaching phase only, is a continuous function of the available sliding variable. The control algorithm is defined by the following control law [20] u(t) = u,(t) + uz(t) I - ~ i f I~1 > 1 ul(t) = - W s i g n ( y l ) i f lul <_ 1 -AIs01%ign(yl ) i f lYll > ~o u2(t) = -Alyl lPsign(yl) i f iYll _< So (26) and the corresponding sufficient conditions for the finite t ime convergence to the sliding manifold are [20] / \"- W > ~, rM(w+~) A: >_ r-~ rm(w-~) (27) 0 < p < _ 0 "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure14.22-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure14.22-1.png",
        "caption": "Figure 14.22 With uniformly distributed loads, the cable assumes the shape of a parabola. At A and B the tangents to the parabola are shown.",
        "texts": [
            " This can be denoted as z = zk + h x in which zk is the distance from the chord to the cable (see Figure 14.21): zk = 1 2qx( \u2212 x) H = M H . 652 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM M = 1 2qx( \u2212 x) is the bending moment in a simply supported beam with a uniformly distributed load (see Section 10.2.2, Example 1). The cable has the same parabolic shape under the chord as the M diagram; the scale factor is H . Assume pk is the sag of the parabola under the chord, that is the distance between the parabola and the chord at the middle of the span (see Figure 14.22): pk = zk ( x = 1 2 ) = 1 8q 2 H . (17a) If the sag pk of the parabola under the chord is given, the horizontal component of the cable force follows from H = 1 8q 2 pk . (17b) The slope of the cable is then dz dx = 1 2q H \u2212 qx H + h . (18) At the supports A (x = 0) and B (x = ) the slope is ( dz dx ) A = h + 1 2q H , ( dz dx ) B = h \u2212 1 2q H . 14 Cables, Lines of Force and Structural Shapes 653 Check: These expressions can also be determined directly from Figure 14.22, where the tangents to the parabola at A and B are shown. Using (17a) we find tan \u03b1 = + ( dz dx ) A = 2pk + 1 2h 1 2 = 1 2q H + h , (19a) tan \u03b2 = \u2212 ( dz dx ) B = 2pk \u2212 1 2h 1 2 = 1 2q H \u2212 h , (19b) The maxixum sag in the cable appears where dz/dx = 0. Here the parabola has its vertex. Equation (18) gives xvertex = 1 2 + Hh q (20a) or, using (17b) xvertex = 1 2 + 1 8hl pk . (20b) If we select the coordinate system at the vertex of the parabola, as in Figure 14.23, the formulas are far easier. With the boundary conditions (x = 0; z = 0) and (x = 0; dz/dx = 0) the cable shape is z = \u2212 1 2qx2 H "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure16.13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure16.13-1.png",
        "caption": "Figure 16.13 (a) Hinged beam with uniformly distributed full load and the influence lines for (b) the bending moment at D and (c) the shear force at D.",
        "texts": [
            " The influence quantity for the bending moment (MD/F ) has the dimension of length. Figure 16.12b therefore includes the values in metres. The influence quantity for the shear force (VD/F ) is dimensionless. The influence lines give the following for the bending moment: MD = \u2212 ( 5 4 m ) \u00d7 (15 kN) \u2212 ( 5 6 m) ) \u00d7 (30 kN) \u2212 ( 5 12 m ) \u00d7 (45 kN) = \u221262.5 kNm and for the shear force VD = \u2212 1 4 \u00d7 (15 kN) \u2212 1 6 \u00d7 (30 kN) \u2212 1 12 \u00d7 (45 kN) = \u221212.5 kN. The correctness of these values can be checked by considering the equilibrium equations. Example 2 \u2013 Uniformly distributed load In Figure 16.13a, the beam in the previous example is loaded along its entire length by a uniformly distributed load q = 80 kN/m. Here too we will determine the bending moment and the shear force at D. 16 Influence Lines 757 First we calculate the bending moment. The influence quantity MD/F , which is a function of x, is hereafter for simplicity denoted by f (x). The contribution dMD to the bending moment MD of the distributed load q over a small length dx is found by multiplying the small resulting force q dx by the associated value of f (x) of the influence line: dMD = f (x) \u00b7 q dx. The bending moment at D due to the distributed load between x = x1 and x = x2 is found by summing up all the contributions, or in other words, by integrating: MD = \u222b x2 x1 f (x)q dx. Since the distributed load is constant here, q can be taken outside the integration symbol: MD = q \u222b x2 x1 f (x) dx. The integral represents the area of the influence line between x1 and x2 (see Figure 16.13b). The bending moment due to a uniformly distributed load q is therefore equal to the load q , multiplied by the area of the influence line for the part where the load is acting. The signs have to be taken into account when determining the magnitude of the areas. The bending moment at D due to the uniformly distributed full load is found from the influence line in Figure 16.13b: 758 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM MD = { + 1 2 \u00d7 (10 m) \u00d7 ( 5 2 m ) \u2212 1 2 \u00d7 (10 m) \u00d7 ( 5 4 m )} \u00d7 (80 kN/m) = 500 kNm. In the same way, the shear force at D is found from the area of the influence line in Figure 16.13c. Since it is immediately clear that the total area of the influence line over field AB is zero, we have only to determine the area over field BC: VD = \u2212 1 2 \u00d7 (10 m) \u00d7 1 4 \u00d7 (80 kN/m) = \u2212100 kN. If the load consists of a single point load, the influence line shows directly where the force has the maximum effect. Also for a uniformly distributed load the most unfavourable placement is rather easy to find. With a set of loads, however, this is no longer the case, and several positions will have to be investigated"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure7.10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure7.10-1.png",
        "caption": "FIGURE 7.10 Example for numerical solution of c direction layered representation.",
        "texts": [
            " Issues of importance are both the accuracy and speed for the partial element evaluation. A general evaluation of partial elements for nonorthogonal structures consists of the combination of analytic and numerical techniques. Some of the formulations are presented in Appendices C and D. 170 NONORTHOGONAL PEEC MODELS We consider an example for partial element evaluation using a mixed numerical\u2013 analytic combination. The example involves a quadrilateral cell problem where the thickness c is represented by a set of layers or sheets as shown in Fig. 7.10. In this example, the thickness of the 3D hexahedral structure is assumed to be thin compared to the other cell dimensions. Hence, the numerical integration is applied in the thin c direction as shown with the sheets in Fig. 7.10. If we use local coordinates, the kth sheet is specified by a \u2208 [\u22121,+1], b \u2208 [\u22121,+1], and c \u2208 [ck\u22121, ck]. The values of the {ck} are the values obtained from the Gaussian quadrature rule for the Lvl levels. Hence, each inductive cell is represented by four sheets as is shown in Fig. 7.10. There are thus a total of four (Lvl) \u2212 1 sheets for each inductive cell shown in Fig. 7.10. The value of Lvl determines the accuracy for a given aspect ratio of the cell and the accuracy required. For the inductance computation, the partial inductance of each sheet cell is required with respect to all the other sheet cells. Importantly, the integral (7.17) is reduced to the evaluation Nc fourfold integrals of the following form Lpaa\u2032 = \ud835\udf070\u222ba\u222bb \u222ba\u2032\u222bb\u2032 (a\u0302 \u22c5 a\u0302\u2032) ha\u2032 g[r(a, b, c), r(a\u2032, b\u2032, c\u2032)] ha da\u2032 db\u2032 da db, (7.29) where Nc is the product of the number of layers represented by each of the two cells"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002797_j.jsv.2021.116029-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002797_j.jsv.2021.116029-Figure2-1.png",
        "caption": "Fig. 2. Time dependent displacement excitations for different defects. (a) A half-sine and (b) a piecewise function.",
        "texts": [
            " At the position \u03c6cj , the axial and radial distances between the centers of curvatures of two raceways are expressed as { D a j = ( r i + r o \u2212 d b ) sin \u03b10 + z D r j = ( r i + r o \u2212 d b ) cos \u03b10 + x cos \u03c6c j + y sin \u03c6c j (1) where r i and r o denote the curvature radius for inner and outer raceways; \u03b10 denotes the initial contact angle for two raceways; and d b is the ball diameter; The axial and radial distances between the curvature centers for ball and outer raceway are given as { X a j = ( r o \u2212 0 . 5 d b ) sin \u03b10 + z j X r j = ( r o \u2212 0 . 5 d b ) cos \u03b10 + y j (2) The elastic deformations between two raceways and ball are written as { \u03b4i j = \u221a ( D a j \u2212 X a j )2 + ( D r j \u2212 X r j )2 \u2212 ( r i \u2212 0 . 5 d b ) \u03b4oj = \u221a X a j 2 + X r j 2 \u2212 ( r o \u2212 0 . 5 d b + H d ) (3) where, H d is the time dependent displacement excitation (TDDE) caused by different defects. As plotted in Fig. 2 , the TDDE caused by a rectangular defect can be described by two different functions [17] . The TDDE H d considering a half-sine function is given as H d = { d sin ( \u03c0 \u03b8d ( mod ( \u03b8di , 2 \u03c0) \u2212\u03b8d0 ) ) 0 \u2264 mod ( \u03b8di , 2 \u03c0) \u2212\u03b8d0 \u2264 \u03b8d 0 other (4) The TDDE H d considering a piecewise function is given as H d = \u23a7 \u23aa \u23a8 \u23aa \u23a9 d sin ( \u03c0 2 \u03b8dc ( mod ( \u03b8di , 2 \u03c0) \u2212 \u03b8d0 ) ) 0 \u2264 mod ( \u03b8di , 2 \u03c0) \u2212 \u03b8d0 \u2264 \u03b8dc d \u03b8dc < mod ( \u03b8di , 2 \u03c0) \u2212 \u03b8d0 \u2264 \u03b8d \u2212\u03b8dc d sin ( \u03c0 2 \u03b8dc ( mod ( \u03b8di , 2 \u03c0) \u2212 \u03b8d0 ) ) \u03b8d \u2212\u03b8dc < mod ( \u03b8di , 2 \u03c0) \u2212 \u03b8d0 \u2264 \u03b8d 0 other (5) where, d denotes the maximum excitation amplitude; \u03b8d denotes the tangential angular distance for defect; mod is the remainder function"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure4.5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure4.5-1.png",
        "caption": "Fig. 4.5: A scheme for an electromagnetic end-effector with a releasing mechanism.",
        "texts": [
            "3 examples of suction-cups are shown as those that are used in robot pneumatic end-effectors. Examples of pneumatic vacuum end-effectors are shown for the case of a multi suction-cup end-effector in Fig. 4.4a) and for a palm end-effector with several suction-cups in Fig. 4.4b). Chapter 4 Fundamentals of the Mechanics of Grasp244 The electrostatic and magnetic systems can be used for objects with ferromagnetic properties so that they can be grasped through the action of electromagnetic forces. In Fig. 4.5 a scheme for a typical electromagnetic end-effector is shown together with a slider-crank mechanism, which is useful for smooth releasing. However, all grasping devices should fulfil the following requirements, mainly for installation on robots: low-cost, robust design and operation, simplicity in the mechanical design and operation, and easy integration. In particular, robotic systems require low-cost grasping devices in order to confirm the economical advantage of a robotization. The gripper design can be considered of Fundamentals of the Mechanics of Robotic Manipulation 245 fundamental importance since a grasping device is the robot extremity that interacts with the environment and performs tasks"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000376_j.jallcom.2014.06.172-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000376_j.jallcom.2014.06.172-Figure2-1.png",
        "caption": "Fig. 2. Definitions of the axes and planes discussed within SLM fabrication with respect to the build layers.",
        "texts": [
            " As with all FCC alloys, nickel alloys exhibit preferential solidification in the {001} direction [11]; this been exploited in turbine blade manufacture with the production of DS castings and subsequently in single crystal castings where very low-angle grain boundaries significantly improve mechanical properties [11]. It has been reported that due to the remelting of the previous layers (or substrate in the case of the first layer) there exists a strong trend of epitaxial growth of grains from one layer to the next [12]. In addition to this tendency for epitaxy between layers, the nature of the heat flow via conduction through the previous build layers and substrate in the \u2018 Z\u2019 direction (as defined in Fig. 2) encourages the formation of large elongated or \u2018columnar\u2019 grains similar to those observed in the DS casting. These columnar grains have been observed in both SLM powder-bed [13] and blown-powder laser fabricated [14\u201318] materials. The results presented by Liu et al. [15] are of particular interest as they show not only the columnar grains within the blown-powder laser fabricated material, but also fine grained regions formed due to the scan spacing. It is known that the material CM247LC is particularly susceptible to weld-cracking in the as-fabricated state"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure3.38-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure3.38-1.png",
        "caption": "Fig. 3.38: A scheme for a spatial 2R manipulator and its mechanical parameters.",
        "texts": [
            "26) From this description of the dynamics the influence of the Jacobian is evident and the effect of a singularity can be interpreted as an increase to infinity for the Cartesian inertia matrix MX, and Coriolis and centrifugal matrix CX. Consequently, in the direction of the singularity the end-effector should exert large action F to obtain small kinematic motion by feeling large inertia effects. In this section an analysis on the specific system of a general spatial 2R manipulator, whose mechanical and kinematical models are shown in Fig. 3.38, is carried out to show the direct application of Newton\u2013Euler equations with computer simulation and mechanical design purposes. It is also useful to show from a teaching viewpoint the dynamical phenomena and the computational difficulties in deriving a closed-form formulation with direct understanding purposes. Referring to the scheme of Fig. 3.38 for a two-link spatial manipulator, links 1 and 2 can be identified with the rigid bodies with G1 and G2 as centers of mass. Assuming no forces are acting on the link 2, when the payload has been included in its mass characteristics, the dynamic equations can be expressed in the following form - link 1 of the chain: in 12211 FfRf \u2212= (3.6.27) in 122121 in i112211 R x x R TfOOFGOtt \u2212++= - link 2 of the chain: in 22 Ff \u2212= (3.6.28) in 2 in 2222 x TFGOt \u2212= Fundamentals of Mechanics of Robotic Manipulation 161 Evaluation of inertia actions on the links requires a previous kinematic analysis"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure10.5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure10.5-1.png",
        "caption": "FIGURE 10.5. A car moving on a road with sideslip angle \u03b2 and heading angle \u03c8.",
        "texts": [
            "15) Gr3 = Gd+ GRB Br3 = \u23a1\u23a2\u23a3 XC + 1 2 w sin\u03c8 \u2212 a2 cos\u03c8 YC \u2212 1 2 w cos\u03c8 \u2212 a2 sin\u03c8 \u23a4\u23a5\u23a6 (10.16) 588 10. Vehicle Planar Dynamics Gr4 = Gd+ GRB Br4 = \u23a1\u23a2\u23a3 XC + 1 2 w sin\u03c8 \u2212 a3 cos\u03c8 YC \u2212 1 2 w cos\u03c8 \u2212 a3 sin\u03c8 \u23a4\u23a5\u23a6 (10.17) Gr5 = Gd+ GRB Br5 = \u23a1\u23a2\u23a3 XC \u2212 1 2 w sin\u03c8 \u2212 a3 cos\u03c8 YC + 1 2 w cos\u03c8 \u2212 a3 sin\u03c8 \u23a4\u23a5\u23a6 (10.18) Gr6 = Gd+ GRB Br6 = \u23a1\u23a2\u23a3 XC \u2212 1 2 w sin\u03c8 \u2212 a2 cos\u03c8 YC + 1 2 w cos\u03c8 \u2212 a2 sin\u03c8 \u23a4\u23a5\u23a6 . (10.19) The rotation matrix between the global G and body coordinate B is GRB = \u2219 cos\u03c8 \u2212 sin\u03c8 sin\u03c8 cos\u03c8 \u00b8 . (10.20) Example 376 Crouse angle, attitude angle, and heading angle. Figure 10.5 illustrates a car moving on a road with the angles \u03c8 = 15deg (10.21) \u03b2 = 16deg . (10.22) The heading angle of the car is Heading angle = \u03c8 = 15deg (10.23) which is the angle between the car\u2019s longitudinal x-axis and a reference X-axis on the road. The attitude angle of the car is Attitude angle = \u03b2 = 16deg (10.24) which is the angle between the direction of the car\u2019s motion and its longitudinal axis. The cruise angle of the car is Heading angle = \u03b2 + \u03c8 = 31deg (10.25) which is the angle between the car\u2019s direction of motion and the reference X-axis on the road"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000406_j.actamat.2015.06.036-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000406_j.actamat.2015.06.036-Figure1-1.png",
        "caption": "Fig. 1. (a) Schematic of EBM system. (b), (c), and (d) SEM images showing Ti\u20136Al\u20134V ELI powder with different magnifications.",
        "texts": [
            " It is worth noting that the a/b duplex microstructure of EBM-built Ti\u20136Al\u20134V has not yet been characterized in detail, particularly on the structure of a/b interface and the elemental segregation behavior at a/b interface. In order to elucidate the microstructure\u2013property relationship from the viewpoint of atomic scale, APT was employed for a/b interface analysis in this work. Furthermore, the paper provides a quantitative composition and crystallography analysis of the microstructure of EBM-built Ti\u20136Al\u20134V, and discusses its microstructural evolution under such a complex process in detail. All test samples were fabricated by an Arcam A2XX system (as schematically shown in Fig. 1a) using the standard processing themes provided by Arcam AB. Pre-alloyed Ti\u20136Al\u20134V ELI (Grade 23) powder supplied by Arcam AB was used for the evaluation of graded microstructure and mechanical properties. The perfectly spherical morphology and clean surface of each powder are revealed in Fig. 1b\u2013d, indicating good flowability and no oxidation. The powder size ranges from 45 to 105 lm. The nominal composition of as-supplied powders is 6Al\u20134V\u20130.03C\u20130.1Fe\u20130.15O\u20130.01N\u2013 0.003H and Ti Bal. (wt.%). Two build themes were employed in this work, i.e. Ti6Al4V-PreHeat-50 lm and Ti6Al4V-melt-50 lm. A 10 mm-thick stainless steel start plate is heated by the electron beam when a pressure of 5.0e 4 mBar within the build chamber is achieved. Once a bottom temperature of 730 C is reached, parts are built directly onto the preheated start plate by selectively melting layers of 50 lm under a controlled vacuum in the temperature range of 600\u2013650 C"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure9.4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure9.4-1.png",
        "caption": "FIGURE 9.4. A disc with mass m and radius r, mounted on a massless turning shaft.",
        "texts": [
            " K = 1 2 Z B v\u03072dm = 1 2 Z B (\u03c9 \u00d7 r) \u00b7 (\u03c9 \u00d7 r) dm = \u03c92x 2 Z B \u00a1 y2 + z2 \u00a2 dm+ \u03c92y 2 Z B \u00a1 z2 + x2 \u00a2 dm+ \u03c92z 2 Z B \u00a1 x2 + y2 \u00a2 dm \u2212\u03c9x\u03c9y Z B xy dm\u2212 \u03c9y\u03c9z Z B yz dm\u2212 \u03c9z\u03c9x Z B zx dm = 1 2 \u00a1 Ixx\u03c9 2 x + Iyy\u03c9 2 y + Izz\u03c9 2 z \u00a2 \u2212Ixy\u03c9x\u03c9y \u2212 Iyz\u03c9y\u03c9z \u2212 Izx\u03c9z\u03c9x (9.93) The kinetic energy can be rearranged to a matrix multiplication form K = 1 2 \u03c9T I \u03c9 (9.94) = 1 2 \u03c9 \u00b7 L. (9.95) When the body frame is principal, the kinetic energy will simplify to K = 1 2 \u00a1 I1\u03c9 2 1 + I2\u03c9 2 2 + I3\u03c9 2 3 \u00a2 . (9.96) Example 349 A tilted disc on a massless shaft. Figure 9.4 illustrates a disc with mass m and radius r, mounted on a massless shaft. The shaft is turning with a constant angular speed \u03c9. The disc is attached to the shaft at an angle \u03b8. Because of \u03b8, the bearings at A and B must support a rotating force. We attach a principal body coordinate frame at the disc center as shown in the figure. The angular velocity vector in the body frame is B G\u03c9B = \u03c9 cos \u03b8 \u0131\u0302+ \u03c9 sin \u03b8 j\u0302 (9.97) 536 9. Applied Dynamics and the mass moment of inertia matrix is BI = \u23a1\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3 mr2 2 0 0 0 mr2 4 0 0 0 mr2 4 \u23a4\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6 "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000372_jcb.41.2.600-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000372_jcb.41.2.600-Figure8-1.png",
        "caption": "FIGURE 8 Effect of cycloheximide on arginine- 14C incorporation into TCAinsoluble protein of Chlamydomonas (pf 16) . Arginine 14C (2 .5 \u00b5c/ml, 172 me/mmole) added at time zero and cycloheximide added at 50 min . Closed circles, no cycloheximide : open circles, 1 .0\u00b5g/ml; crosses, 10 .0 \u00b5g/ml .",
        "texts": [],
        "surrounding_texts": [
            "When both flagella of the wild-type strain are amputated in the presence of cycloheximide, the flagella elongate to about one-half normal length and maintain this length (Fig . 12) . The regeneration kinetics in the presence of cycloheximide are similar to the kinetics of the controls until just prior to the inhibition of elongation . These results indicate that the cells have a limiting amount of some protein, possibly flagellar precursor, which is necessary for regeneration . The following experiment was performed to determine if the limiting protein suggested in the experiment above was actually being used for flagella formation and, if so, at what stage of regeneration. The flagella of a population of cells ROSENBAUM, MOULDER, AND RINGO Flagellai Elongation in Chlamydomonas 605 on M arch 9, 2015 jcb.rupress.org D ow nloaded from FIGURE 6 Photomicrographs chosen from the experiment described in Fig . 5 to illustrate the simultaneous shortening and elongation processes in a single cell. The time (minutes) after amputation of the right flagellum is shown in the lower right corner of each micrograph . X ca. 1100. TIME-(MINUTES) were amputated, and the cells were allowed to regenerate flagella to various lengths . Re-amputation was then carried out in the presence of cycloheximide, and the amount of elongation was assessed . Fig . 13 shows that the flagella regenerate to about one-half normal length when cycloheximide is added at the time of amputation (control) . When the inhibitor is added to cells which have already regenerated flagella to about 34' length and are then re-amputated, regeneration 606 THE JOURNAL OF CELL BIOLOGY . VOLUME 41, 1969 FIGURE 7 Two single flagellum amputations carried out on a wild-type Chlamydonwnas (21 gr), successively removing opposite flagella. Open circles, amputated flagellum in the first regeneration, intact flagellum in \u2022 the second regeneration ; closed circles, intact flagellum in the first regeneration, amputated flagellum in the second regeneration . The arrow indicates the time of the second amputation . 110 120 130 140 is much less . If the cells are allowed to regenerate flagella of almost normal length and are re-amputated, the amount of regeneration in cycloheximide is almost similar to that obtained in the controls . The Rate of Incorporation of Arginine- 1H into Protein of Regenerating Flagella The previous result indicated that there might be some use of flagellar precursor proteins during on M arch 9, 2015 jcb.rupress.org D ow nloaded from the first half of elongation. Pulse-labeling experiments were carried out to determine the actual rate of incorporation into the proteins of elongating flagella. Arginine-requiring cells were shown to have regeneration kinetics similar to those of the wild-type cells and to incorporate labeled arginine into their protein at a rate which proceeded linearly after about 2 min in the presence of the isotope . The cells were deflagellated, pulsed with arginine 3H for six 10-min intervals during the course of regeneration, and radioautographed . The rate of incorporation (grains/A/min) was determined by assessing the number of silver grains over the flagella for each time interval . Cells with nonregenerating flagella (interphase cells) were also found to incorporate some amino acid into their flagellar proteins .? This turnover 7 Cells with nonregenerating flagella will incorporate incorporation was corrected for by pulsing cells which had nonregenerating flagella for 10 min and then subtracting this incorporation from that obtained for each 10-min pulse in the regenerating cells. The results in Fig . 14 show that the greatest rate of incorporation occurs at the time when the flagella are at about one-half length . The rate then decreases as the flagella attain their normal length. Thus, the highest rate of incorporation into flagellar protein occurs at about the time when, the cycloheximide studies indicate, the cells have the lowest amount of flagellar precursors . amino acids into their flagellar proteins . The protein turnover in Chlamydomonas flagellar axonemes, membranes, and matrix components will be the subject of another report (Rosenbaum, J ., K . Carlson, and G. Witman. In preparation) . ROSENRAUM, MOULDER, AND RINGO Flagellar Elongation in Chlamydonwnas 607 on M arch 9, 2015 jcb.rupress.org D ow nloaded from The Ef ect of Colchicine on Flagellar Regeneration in Chlamydomonas Colchicine is a well known inhibitor of the formation of microtubular systems (see references 9, 10 for reviews). Its mode of action has been investigated, and it has been shown to bind to the subunits of microtubules isolated from various organelles, including the mitotic apparatus and flagella (3, 4, 29, 35) . It was therefore used in studying flagellar regeneration in Chlamydomonas . 608 THE JOURNAL OF CELL BIOLOGY \u2022 VOLUME 41, 1969 10 9 8 N72 \u00db_ 6 5 =4F C9z 3 w 2 l I I I 1 1 I 1 10 20 30 40 50 60 70 80 TIME-(MINUTES) FIGURE 12 Effect of cycloheximide (10 .0 \u00b5g/ml) on flagellar regeneration in a population of wild-type Chlamydomonas (21 gr) . Open circles, control ; closed circles, cycloheximide added at time of amputation . EFFECT OF COLCHICINE ON FLAGELLAR REGENERATION WHEN BOTH FLAGELLA ARE AMPUTATED : When used at the proper concentration, colchicine was found to prevent flagellar elongation in Chlamydomonas . s Fig. 15 shows s The potency of colchicine as an inhibitor of flagellar regeneration will vary somewhat with different lots of colchicine . Dose response curves run on each new batch of colchicine show that 1 .0-1 .5 mg/ml will usually completely inhibit flagellar regeneration . on M arch 9, 2015 jcb.rupress.org D ow nloaded from the effect of varying concentrations of colchicine on flagellar regeneration . At 0.1 mg/ml, normal regeneration occurs ; at 0 .4 mg/ml, there is partial regeneration ; and at 1 .0 mg/ml, flagellar regeneration is completely inhibited as long as colchicine is present. These results are shown more clearly in Fig . 16 in which the complete kinetics of regeneration in the presence of varying amounts of colchicine are described . It can be seen that under conditions of partial inhibition (0 .4 mg/ml) the rate of elongation is decreased from the very beginning of regeneration . This is in contrast to the partial inhibition observed with cycloheximide (Fig. 12) where regeneration kinetics are normal until the time of inhibition at half-length . Since cycloheximide will partially inhibit flagellar formation by inhibiting protein synthesis, it was necessary to determine whether colchicine had a similar effect . In Fig. 17, it can be seen that a concentration of 1 mg/ml colchicine has little or no effect on arginine-14C incorporation into total Chlamydomonas TCA-precipitable protein . Therefore, the inhibition by colchicine of flagellar regeneration in Chlamydomonas occurs by a mechanism other than inhibition of protein synthesis . The results also indicate that colchicine acts rapidly to inhibit flagellar regeneration . When colchicine is added to cells which have partially regenerated their flagella, inhibition of further elongation occurs almost immediately (Fig . 18) . Once this inhibition has occurred, the flagellar length is maintained, and little or no shortening has been observed in the continued presence of the drug s EFFECT OF COLCHICINE ON REGENERA-"
        ]
    },
    {
        "image_filename": "designv10_0_0002379_j.ijheatmasstransfer.2017.04.055-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002379_j.ijheatmasstransfer.2017.04.055-Figure3-1.png",
        "caption": "Fig. 3. Computed temperature and velocity distribution at 1500 ms (at th",
        "texts": [
            " Table 1 Material compositions of compacted graphite cast iron and Ni-based alloy powder (wt%). Fe C Ni Al Cr Co S Cast iron Bal 3.6 \u2013 \u2013 \u2013 \u2013 0.02 Ni-based alloy \u2013 \u2013 Bal 3.4 31 2.6 \u2013 parallelized using OPENMP tools in a computing platform with 12 2.5 GHz CPU and 24 GB memory. The total time of actual computation is about 30 h. Table 1 presents the compositions of powder and base metal. Table 2 shows the thermophysical properties of materials. Table 3 shows the calculation data used in this simulation. Fig. 3 shows the computed temperature field and velocity field within the molten pool at the time of 1500 ms (i.e. at the midlength of first layer). In Fig. 3(a), the temperature field is shown by the contour. The velocity in the molten pool is indicated by the black arrows. The liquidus and solidus are present by the black isotherms. In Fig. 3(b), the contour indicates the velocity magnitude. A reference vector of 1 m/s is shown in the figure. The relative importance of conduction and convection in the molten pool is evaluated by Peclet number [32], PeT \u00bc uLR ah \u00f016\u00de where u represents the characteristic velocity (1 m/s in this case), LR represents the characteristic length (laser beam radius, 10 3 m, in this case), ah represents the thermal diffusivity (10 5 m2/s in this case). Thus, the PeT is on the order of 102, which means the heat transfer in the molten pool is dominated by convection (liquid metal flow)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure1.11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure1.11-1.png",
        "caption": "Fig. 1.11. Topologies of (a) RFPM machine (b) AFPM machine.",
        "texts": [
            "35) In pace with the application of new materials, innovation in manufacturing technology and improvements in cooling techniques, further increase in the power density (output power per mass or volume) of the electrical machine has been made possible. There is an inherent limit to this increase for conventional radial flux PM (RFPM) machines because of [30, 53, 106, 165, 192]: \u2022 the bottle-neck feature for the flux path at the root of the rotor tooth in the case of induction and d.c. commutator machines or brushless machines with external rotors (Fig. 1.11); \u2022 much of the rotor core around the shaft (rotor yoke) is hardly utilised as a magnetic circuit; \u2022 heat from the stator winding is transferred to the stator core and then to the frame \u2014 there is poor heat removal through the stator air gap, rotor and shaft without forced cooling arrangements. These limitations are inherently bound with radial flux structures and cannot be removed easily unless a new topology is adopted. The AFPM machine, recognised as having a higher power density than the RFPM machine, is more compact than its radial flux counterpart [29, 53, 106, 165]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure2.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure2.6-1.png",
        "caption": "Fig. 2.6. Various types of kinematic chains",
        "texts": [
            " Generally speaking, these parameters are the length of the manipulator link, the angle between the axes lying at two ends of the link, etc. Precise definitions of link parameters for the Denavit-Hartenberg kinematic notation will be presented in Sect. 2.3. Dynamic parameters describing the link are its mass, the location of its centre of mass and the inertia tensor. These parameters are used in dynamic modelling of the mechanism. We will consider only kinematic pairs that have single degree-of-freedom rotation or translation. A chain in which no link enters more than two kinematic pairs is said to be a simple kinematic pair (Fig. 2.6). On the other hand, a complex kinematic chain contains at least one link that enters more than two kinematic pairs. An open kinematic chain has at least one link that belongs to one kinematic pair only. If each link enters into at least two kinematic pairs, the chain is said to be closed. 22 2 Manipulator Kinematic Model Here we will consider only simple, open kinematic chains. The closed kinematic chains will be discussed in Chap. 3 which is concerned only with constrained end-effector motion"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.50-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.50-1.png",
        "caption": "FIGURE 7.50. The optimal multi-link steering mechanism along with the length of its links.",
        "texts": [
            "824m, which is the best length for the base of the triangle PBC. The behavior of the multi-link steering mechanism for different values of x, is shown in Figure 7.48. The Ackerman condition is also plotted to 7. Steering Dynamics 433 434 7. Steering Dynamics compare with the optimal multi-link mechanism. The optimality of x = \u22120.824m may be more clear in Figure 7.49 that shows the difference \u2206 = \u03b42 \u2212 \u03b4Ac for different values of x. The optimal multi-link steering mechanism along with the length of its links is shown in Figure 7.50. The mechanism and the meaning of negative value for x are shown in Figure 7.51 where the mechanism is in a positive turning position. 7.7 F Trailer-Truck Kinematics Consider a car pulling a one-axle trailer, as shown in Figure 7.52. We may normalize the dimensions such that the length of the trailer is 1. The 7. Steering Dynamics 435 positions of the car at the hinge point and the trailer at the center of its axle are shown by vectors r and s. Assuming r is a given differentiable function of time t, we would like to examine the behavior of the trailer by calculating s, and predict jackknifing"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003702_s0043-1648(96)07467-4-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003702_s0043-1648(96)07467-4-Figure7-1.png",
        "caption": "Fig. 7. Point placement on gear wheel tooth flank at different pivot angles.",
        "texts": [
            " Since a tooth pair can be considered as two springs serially connected, the combined stiffness of the tooth pair will be C8T sE/19. Wear simulations of spur gears are made for an FZG gear set running at low speed, which also have been used by Andersson and Eriksson [3]. The data used in the simulations are summarized in Journal: WEA (Wear) Article: 7467 val. The interaction between the teeth are assumed to be quasi static. The Points on the flanks are aligned in such a way that they, when in mesh, can be considered to be on a straight line perpendicular to the line of action, according to Fig. 7. The first point is at the base diameter, the last point at the tip diameter and the points in between on an involute curve. The points can in a coordinate system according to Fig. 6 be defined as: d b \u00df s (cos b qb sin b ) pi i i i2 d b c s (sin b yb cos b ) pi i i i2 b s(see Fig. 6)i The flanks are then mirrored and rotated to fit the desired configuration. The simulatedwear of the pinion and the gear are presented in the Fig. 8(a) and (b). The wear depths are varying over the teeth flanks with the maximum wear at the root of the pinion and gear respectively"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure6.48-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure6.48-1.png",
        "caption": "FIGURE 6.48. A 6-bar linkage.",
        "texts": [
            " a = 10 cm b = 45 cm e = 3 cm Determine the difference time between go and return half cycle of the slider motion if \u03c92 = 2\u03c0 rad/ s. 376 6. Applied Mechanisms 10. Two possible configurations for an inverted slider-crank mechanism. Consider an inverted slider-crank mechanism with the following links. a = 10 cm d = 45 cm e = 5 cm If \u03b82 = 30deg, what would be the angle \u03b83 and position of the slides b? 11. Instant center of rotation. Find the instant center of rotations for the 6-bar linkage shown in Figure 6.48. 12. A coupler point of a four-bar linkage. Consider a four-bar linkage with the following links a = 10 cm b = 25 cm c = 30 cm d = 25 cm and a coupler point with the following parameters. e = 10 cm \u03b1 = 30deg Determine the coordinates of the coupler point if \u03b82 = 30deg. 13. A coupler point of a slider-crank mechanism. 6. Applied Mechanisms 377 Consider a slider-crank mechanism with the following parameters. a = 10 cm b = 45 cm e = 3 cm c = 10 cm \u03b1 = 30deg Determine the coordinates of the coupler point"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.16-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.16-1.png",
        "caption": "FIGURE 7.16. Langensperger invention for the steering geometry condition.",
        "texts": [
            " Correct steering geometry was a major problem in the early days of carriages, horse-drawn vehicles, and cars. Four- or six-wheel cars and carriages always left rubber marks behind. This is why there were so many three-wheeled cars and carriages in the past. The problem was making a mechanism to give the inner wheel a smaller turning radius than the outside wheel when the vehicle was driven in a circle. The required geometric condition for a front-wheel-steering four-wheelcarriage was introduced in 1816 by George Langensperger in Munich, Germany. Langensperger\u2019s mechanism is illustrated in Figure 7.16. Rudolf Ackerman met Langensperger and saw his invention. Ackerman 7. Steering Dynamics 395 acted as Langensperger\u2019s patent agent in London and introduced the invention to British carriage builders. Car manufacturers have been adopting and improving the Ackerman geometry for their steering mechanisms since 1881. The basic design of vehicle steering systems has changed little since the invention of the steering mechanism. The driver\u2019s steering input is transmitted by a shaft through some type of gear reduction mechanism to generate steering motion at the front wheels"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000790_j.engfailanal.2011.07.006-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000790_j.engfailanal.2011.07.006-Figure2-1.png",
        "caption": "Fig. 2. Geometrical parameters for the fillet-foundation deflection [3,4].",
        "texts": [
            " The Hertzian contact stiffness Kh is given by 1 Kh \u00bc 4\u00f01 v2\u00de pEW \u00f010\u00de Besides the tooth deformation, the fillet-foundation deflection also influences the stiffness of gear tooth. Sainsot et al. [26] derived the fillet-foundation deflection of the gear based on the theory of Muskhelishvili [27]. And then, they applied it to circular elastic rings to derive an analytical formula reflecting the gear body-induced tooth deflections by assuming linear and constant stress variations at root circle. It can be calculated as [3,4,7,26], df \u00bc F cos2 am WE L uf Sf 2 \u00feM uf Sf \u00fe P 1\u00fe Q tan2 am ( ) \u00f011\u00de where W is the tooth width. uf and Sf are given in Fig. 2. The coefficients L , M , P , Q can be approached by polynomial functions [26]: X i \u00f0hfi; hf \u00de \u00bc Ai=h 2 f \u00fe Bih 2 fi \u00fe Cihfi=hf \u00fe Di=hf \u00fe Eihfi \u00fe Fi \u00f012\u00de X i denotes the coefficients L\u2044, M\u2044, P\u2044 and Q\u2044. hfi \u00bc rf =rint; rf ; rint and hf are defined in Fig. 2, the values of Ai, Bi, Ci, Di, Ei and Fi are given in Table 1. The stiffness with consideration of gear fillet-foundation deflection can be obtained by 1 Kf \u00bc df F \u00f013\u00de of the coefficients of Eq. (12) [22]. The total equivalent mesh stiffness of one tooth pair in mesh can be obtained by Ke \u00bc 1 1 Kb1 \u00fe 1 Ks1 \u00fe 1 Ka1 \u00fe 1 Kf 1 \u00fe 1 Kb2 \u00fe 1 Ks2 \u00fe 1 Ka2 \u00fe 1 Kf 2 \u00fe 1 Kh \u00f014\u00de Here, the subscripts 1, 2 mean the pinion and gear, respectively. At very early stage of crack propagation, it usually starts from some local positions where the stress concentrations are observed"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001996_j.rcim.2015.12.004-Figure22-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001996_j.rcim.2015.12.004-Figure22-1.png",
        "caption": "Fig. 22. Examination of the internal cross sections. (a) Cut parts of the case one; the cross section A is parallel to the building direction and the cross section B is perpendicular to the building direction. (b) Close up of the cross section A. (c) Close up of the cross section B.",
        "texts": [
            " It can be seen that with the changing of stepover distance, both wire-feed rate and travel speed are changed to produce the desired bead geometry suitable for the given range of step-over distance. Using the generated paths and the welding process parameters, programme code for the robotic manipulator and the welding power source is automatically generated. The near-net shape component was deposited with 15 layers, as shown in Fig. 21a. Fig. 21b shows the finished component after surface milling. In order to check the potential internal defects, the finished component was cut into two pieces as shown in Fig. 22a. The cross section A is parallel to the building direction and the cross section perimental WAAM system [21]. B is perpendicular to the building direction. Close up of the cross sections are shown in Fig. 22b and c. It can be seen that the deposited part is 100% solid and free of porosity. Fig. 23 shows an example of a solid structure without holes. The geometry has a length of 100 mmwhile the width ranges from 25 mm to 50 mm as described in Fig. 23. Table 3 provides the geometrical parameters that were determined by the adaptive MAT path planning algorithm. The deposition path comprised of three closed loops with varied step-over distances. The red arrows in Fig. 23 indicate the start points and the travel directions"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure2.16-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure2.16-1.png",
        "caption": "Fig. 2.16. Time history of joint coordinate qi",
        "texts": [
            " If external sensors are to be included in order to modify the trajectory, the motion generation has to be 46 2 Manipulator Kinematic Model done in real time. Then, the time required for obtaining the inverse kinematics solution is critical, especially for high-speed robots (e.g. 5 m/s). If this time is several times greater than the time required by the servosystem, the cubic spline interpolation is used to generate continuous joint positions and velocities [13]. Very often it is not necessary to stop at each user-supplied point. Such motion may be realized in various ways. One of them is shown in Fig. 2.16 [3]. Here, the quadratic change of acceleration in the vicinity of the transition points provides for continuity of velocity and acceleration. It is applied to the tasks where it is sufficient to pass close enough to the user-supplied points. According to the notation introduced in Fig. 2.16, the time history of joint position qJt), velocity IMt) and acceleration qi(t) in time interval -<1 <t\u00abl' can be ex pressed as while for the intermediate part of the trajectory the velocity is constant t qi(t)=JqCiy+qBi, <1 <t< T1 -<2 1 (2.80) (2.81) The time function q(t), -<1 <t< T1 -<2 depends solely on the variables related only to the three points A, Band C on the trajectory and does not depend on the other trajectory segments. This feature is convenient in real time motion synthesis. 2.5 Manipulator Path Generation 47 One of the classical methods of manipulator path generation which provides for continuity of position, velocity and acceleration on the entire trajectory is the use of cubic spline functions [16,18]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.17-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.17-1.png",
        "caption": "FIGURE 7.17. Steering of a three-axle vehicle.",
        "texts": [
            " The basic design of vehicle steering systems has changed little since the invention of the steering mechanism. The driver\u2019s steering input is transmitted by a shaft through some type of gear reduction mechanism to generate steering motion at the front wheels. 7.2 Vehicles with More Than Two Axles If a vehicle has more than two axles, all the axles, except one, must be steerable to provide slip-free turning at zero velocity. When an n-axle vehicle has only one non-steerable axle, there are n \u2212 1 geometric steering conditions. A three-axle vehicle with two steerable axles is shown in Figure 7.17. To indicate the geometry of a multi-axle vehicle, we start from the front axle and measure the longitudinal distance ai between axle i and the mass center C. Hence, a1 is the distance between the front axle and C, and a2 is the distance between the second axle and C. Furthermore, we number the wheels in a clockwise rotation starting from the driver\u2019s wheel as number 1. For the three-axle vehicle shown in Figure 7.17, there are two indepen- 396 7. Steering Dynamics dent Ackerman conditions: cot \u03b42 \u2212 cot \u03b41 = w a1 + a3 (7.44) cot \u03b43 \u2212 cot \u03b46 = w a2 + a3 . (7.45) Example 269 A six-wheel vehicle with one steerable axle. When a multi-axle vehicle has only one steerable axle, slip-free rotation is impossible for the non-steering wheels. The kinematic length or wheelbase of the vehicle is not clear, and it is not possible to define an Ackerman condition. Strong wear occurs for the tires, especially at low speeds and large steer angles",
            " Consider a two-axle vehicle with the following dimensions. l = 4m w = 1.3m g = 1.2m Determine Rmin, RMax, and \u2206R for \u03b4 = 30deg. 4. Rear wheel steering lift truck. A battery powered lift truck has the following dimensions. l = 55 in w = 30 in Calculate the radius of rotations if \u03b4 = 55deg for a1 = a2. 452 7. Steering Dynamics 5. Wheel angular velocity. Consider a two-axle vehicle with the following dimensions. l = 2.7m w = 1.36m What is the angular velocity ratio of \u03c9o/\u03c9i? 6. A three-axle vehicle. A three-axle vehicle is shown in Figure 7.17. Find the relationship between \u03b42 and \u03b43, and also between \u03b41 and \u03b46. 7. A three-axle truck. Consider a three-axle truck that has only one steerable axle in front. The dimensions of the truck are a1 = 5300mm a2 = 300mm a3 = 1500mm w = 1800mm. Determine maximum steer angles of the front wheels if the truck is supposed to be able to turn with R = 11m. 8. A vehicle with a one-axle trailer. Determine the angle between the trailer and vehicle with the following dimensions. a1 = 1000mm a2 = 1300mm wv = 1500mm b1 = 1200mm b2 = 1800mm wt = 1100mm g = 800mm \u03b4i = 12deg "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000736_s00170-010-2631-5-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000736_s00170-010-2631-5-Figure8-1.png",
        "caption": "Fig. 8 Some samples and detail of a support structure, case D",
        "texts": [
            " Case D: Samples made with base geometry in the Z\u2013X to Z\u2013Y plane (with different rotation angle) and main axis of the sample bowed with respect to Z direction. In order to obtain better properties on components made in vertical position (case C), a strategy of growing the components/samples with some inclination with respect to Z axis has been proposed. The need of supports depends on the angle of fabrication (with respect to the horizontal plane): as much as this angle is larger (sample closer to vertical axis), the less is the need for these supports. In the pictures corresponding to Fig. 8, some of these manufactured samples are shown; the supports may be observed as well as a detail of the surface of these supports. In the case of 30\u00ba and 45\u00ba, the supports are needed in all the length of the samples, while for the case of 60\u00ba, these supports is only needed at the beginning of the angles. 5.2 Notch impact (Charpy) tests The same criteria have been followed for the notch impact tests (Charpy). In this case, samples have been manufactured by SLM, too. Batches of three standardized samples of 10\u00d710\u00d755 mm have been produced in two orientations corresponding to the ones with best and worst tensile results (case A, 30\u00ba in the XY plane and case C, 0\u00ba with z axis)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003766_j.ymssp.2008.03.016-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003766_j.ymssp.2008.03.016-Figure1-1.png",
        "caption": "Fig. 1. (a) Part of driving system for a bucket wheel with planetary gearbox (gears: z1-sun, z2-planet, z3-stationary rim, z4\u2013z9 three-stage cylindrical gearbox) and (b) three independent drives of bucket wheel.",
        "texts": [
            " In real practice, local faults of planetary gearboxes have not occurred, but heavy destruction of planetary gearboxes have been noticed, which are caused by a prolonged run of a planetary gearbox at the condition of the arm run perturbation. It may be stated that the paper gives a new approach to the condition monitoring of planetary gearboxes. It has been shown that only a root cause analysis based on factors having an influence on the vibration solves the problem of planetary gearbox condition monitoring. & 2008 Elsevier Ltd. All rights reserved. Fig. 1 shows a complex gearbox system for which the first stage is a planetary gearbox with a stationary rim. The planetary gearboxes considered consists of a sun gear z1 \u00bc 39 teeth, planetary gear z2 \u00bc 27 and rim gear z3 \u00bc 93. . All rights reserved. ax: +48 71 344 8123. W. Bartelmus). Considering the system given in Fig. 1a and using notation f12 as the meshing frequency for a pair of gear wheels marked in Fig. 1 as z1, z2, z3 f 12 \u00bc f 23 \u00bc n1z1z3 60\u00f0z1 \u00fe z3\u00de \u00bc 950 39 93 60\u00f039\u00fe 93\u00de \u00bc 435:067 Hz; (1) where n1 is the input rotation velocity RPM. The arm frequency is f a \u00bc n1z1 60\u00f0z1 \u00fe z3\u00de \u00bc 950 39 60\u00f039\u00fe 93\u00de \u00bc 4:67 Hz: (2) A planet passing frequency is fp \u00bc sfa \u00bc 4 4.67 \u00bc 18.68 Hz, where s is the number of planets. The spectrum of vibration presented by frequencies (1) and their harmonics together with inter-stage modulation components [1] and modulation caused by the arm rotation and passing planet/satellite frequency against the point of sensor attachment give the gearbox vibration spectrum",
            " So the classical method of condition monitoring given in [7,8] is not always suitable. The influence of varying load should be stressed for which some aspects are also considered in [14\u201319]. Considering the external load variation for bucket wheels excavators and keeping in mind the variation in digging ground properties and bucket wheel features generated vibration will be cyclostationary. The described whole driving system of a bucket wheel excavator consists of three subsystems as shown in Fig. 1b, and such a whole system is used for driving a bucket wheel. The bucket wheel diameter is 17.5 m and has 11 evenly distributed buckets of capacity 4.6 m3. The rotation speed of the bucket wheel is about 3 RPM, which gives the rotation cycle of 20 s. The gearbox system is monitored by a commercial diagnostic system, which makes it possible to estimate the properties of the vibration signal. The acceleration vibration signal from one sensor attached radial on the stiff flange on each planetary gearbox is received",
            " 9b) to the load variation caused by design factors, such as its tooth flexibility, is also seen. This should especially be so for spur gears such as are used in the planetary gearbox considered. For a better identification of arm perturbation, which is seen in Fig. 4, further identification on a perturbation period using Wigner-Ville distribution is given in Fig. 10. Besides the above considerations, signal analysis has been made for another two gearboxes that run in the complete bucket drive system Fig. 1. For condition assessment, measures of load susceptibility, as shown in Fig. 9, are chosen to be the mean (Lmean) and standard deviation (Lstd) of the load variation estimated from the acceleration vibration signals. This means that the load is proportional to the estimated values of Lmean, and Lstd of the vibration time course given in Fig. 9 represents its variation. Fig. 11 shows the values of relative changes of gearbox condition parameters. Besides the above-mentioned load susceptibility measures, which are also measures of planetary gearbox condition, further two condition measures are taken into consideration, these being the RMS values of the signals filtered around the first mesh harmonic f1270.5f12 (1) and first and second harmonic amplitudes of the arm frequency (2). To evaluate the condition change, relative values of the condition measures are considered. The considered relative values are given in Fig. 11. In the figure, the three planetary gearboxes are marked as G1, G2 and G3 and compose a part of the bucket wheel drive system, Fig. 1. The signals were received from four planetary gearboxes some time before replacement (time t1) of gearbox G1 and some time after replacement, time marked as t2. The decision to replace gearbox G1 was made on subjective grounds, not on the basis of measurements from the commercial monitoring system installed on the drive system, because the monitoring system did not fulfil its task. As one can see for the planetary gearbox G1, the RMS value dropped about 14% (Fig. 11a) when comparing the measurements for two times t1 and t2 (measurements for two gearboxes marked as G1, one before replacement and one after replacement), and for the PG marked G2 and G3 there were increases of about 9% and 8% of RMS values, respectively (measurements at times t1 and t2 for two gearboxes G2 and G3)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure9.11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure9.11-1.png",
        "caption": "FIGURE 9.11. A simple pendulum.",
        "texts": [
            " We thus have n second-order ordinary differential equations in which q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn are the dependent variables and t is the independent variable. The coordinates q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn are called generalized coordinates and can be any measurable parameters to provide the configuration of the system. The number of equations and the number of dependent variables are equal, therefore, the equations are theoretically sufficient to determine the motion of all mi. Example 364 Equation of motion for a simple pendulum. A pendulum is shown in Figure 9.11. Using x and y for the Cartesian position of m, and using \u03b8 = q as the generalized coordinate, we have x = f(\u03b8) = l sin \u03b8 (9.268) y = g(\u03b8) = l cos \u03b8 (9.269) K = 1 2 m \u00a1 x\u03072 + y\u03072 \u00a2 = 1 2 ml2\u03b8\u0307 2 (9.270) and therefore, d dt \u00b5 \u2202K \u2202\u03b8\u0307 \u00b6 \u2212 \u2202K \u2202\u03b8 = d dt (ml2\u03b8\u0307) = ml2\u03b8\u0308. (9.271) The external force components, acting on m, are Fx = 0 (9.272) Fy = mg (9.273) 558 9. Applied Dynamics and therefore, F\u03b8 = Fx \u2202f \u2202\u03b8 + Fy \u2202g \u2202\u03b8 = \u2212mgl sin \u03b8. (9.274) Hence, the equation of motion for the pendulum is ml2\u03b8\u0308 = \u2212mgl sin \u03b8"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001265_j.triboint.2005.11.014-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001265_j.triboint.2005.11.014-Figure1-1.png",
        "caption": "Fig. 1. Structuring of the mechanical subsystem components and interfaces of the lifting unit [4].",
        "texts": [
            " His approach is modular, and lays down strict guidelines for behavioral models of machine elements, modules, and interfaces. Sellgren [2] defined an interface as an attachment relation between two mating faces. That definition is elaborated on by Andersson and Sellgren [3] in terms of an interaction relation between two functional surfaces. A functional surface is a carrier of a function, and in this paper a contact surface also considered. The modeling principle proposed and elaborated on in Ref. [4] is illustrated by the link mechanism of the lifting system shown in Fig. 1. The different components of the mechanism are represented by CAD models, and the interfaces are represented by symbols from i1 to i6. An interface can be both a mechanical contact and an interface element, as when it is a bearing or a system component such as a coupling. This paper considers only mechanical contacts. ARTICLE IN PRESS Nomenclature A amplitude F friction force Fapp external applied tangential force Fc Coulomb sliding friction force, Fc \u00bc mN Fs maximal static friction force in the Stribeck friction model f frequency i exponent in the Stribeck friction model and in the Dahl friction model k spring constant in the 1 DOF system ksat constant in combined Coulomb and viscous friction model ktanh constant in combined Coulomb and tanh friction model kv viscous friction coefficient m mass of the sliding body in the 1 DOF system N normal load, for the 1 DOF system: N \u00bc mg x position of the sliding body in the 1 DOF system x0 position of the wall in the 1 DOF system v sliding speed vs sliding speed coefficient in the Stribeck friction model d maximal micro-slip before macro-sliding occurs m coefficient of friction S"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001383_025403-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001383_025403-Figure5-1.png",
        "caption": "Figure 5. (a) Steady state numerical calculation of the melt-pool shape and (b) 2D representation of the melt pool on an (x, y) plane discretized with 25 \u00b5m \u00d7 25 \u00b5m cells.",
        "texts": [
            " The main issue here is to calculate the interaction area between different meltpool sizes and a unique powder distribution area (figure 4). When the melt-pool size is totally included into the powder distribution (melt-pool 1 on figure 4), the melt-pool area forms the interaction area, whereas when the melt-pool length is longer than the powder stream radius (melt-pool 2), one has to consider a specific interaction area, smaller than the melt-pool and restricted by the powder area. The melt-pool geometry was calculated using steady state calculations on COMSOL Multiphysics\u2122 software (figure 5(a)), using the Eulerian formalism where matter flows through a fixed mesh matrix (Lagrange linear tetrahedrons elements), simultaneously irradiated by a stationary laser source. The laser irradiation \u03c6 (W m\u22122) = f (x, y) was considered as quasi-spherical (equation (1)) in our description, due to a defocusing that deteriorates its homogeneous \u2018top-hat\u2019 spatial distribution near the focal point (figure 1(b)). Using a finite element code allowed us to consider non-linear physical properties for TA6V (k = f (T ), CP = f (T ), table 1), together with realistic boundary conditions (thermal loss, see equation (14), and with very short calculation times (1 min)",
            " Also, in the COMSOL code, the latent heat of fusion HF was addressed through a modification of the specific heat on the interval of solidification (Cp = C0 p + HF/ Tm). In turn, COMSOL solves the steady state heat equation and provides a 3D description of the fusion isotherm, which can be depicted in the (O, x, y) interaction plane by two halfellipses: one for the front part of the melt pool (prior to the heat deposit), and the second and longer half-ellipse for the rear part of the melt pool (planar representation in figure 5(b)). The corresponding geometrical equations for both fusion isotherms is thus given by x2 L2 front + y2 (e/2)2 + z2 p2 = 1 (5) x2 L2 rear + y2 (e/2)2 + z2 p2 = 1, with Lfront = front part of the melt-pool length, Lrear = rear part of the melt-pool length, e = melt-pool width and p = melt-pool depth. A simplified description of temperature values in the molten pool is thus provided by equation (6), using geometrical features of the melt pool and the maximum temperature Tmax obtained by finite element simulations",
            " However, this Tmax value remains questionable, due to the strong assumptions concerning thermal conductivity of TA6V above the melting point Tm. This description allows us to define local melt-pool depths pij on elemental cells. T = Tm + 1 2 (Tmax \u2212 Tm) \u00d7 ( 1 + cos ( \u03c0 ( x2 (e/2)2 + y2 L2 front + z2 p2 ))) , for y > 0 (front semi-ellipse), and T = Tm + 1 2 (Tmax \u2212 Tm) \u00d7 ( 1 + cos ( \u03c0 ( x2 (e/2)2 + y2 L2 rear + z2 p2 ))) , (6) for y < 0 (rear semi-ellipse). Following this, we could verify for different laser conditions that the simulated melt-pool width (figure 5) coincides rather well with the width e1 of the first manufactured layer near the substrate. Once the melt pool is obtained, the physical domain of powder\u2013melt-pool interaction was discretized to calculate the cell by cell contribution to wall manufacturing: we considered the interaction between a 25 \u00b5m length melt-pool cell and the powder, by introducing a local surface powder feeding parameter D\u2217 m (g s\u22121 m\u22122) in the description of powder feeding (figure 6). The elemental contribution hij to the layer growth, provided that all the incident powder melts, becomes hij = D\u2217 m \u00b7 l \u03c1 \u00b7 V , (7) with D\u2217 m = local feed rate (g s\u22121 m\u22122), l = cell size (= 2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002053_tro.2014.2361937-Figure10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002053_tro.2014.2361937-Figure10-1.png",
        "caption": "Fig. 10. Computer-aided design of custom instrumented foot: an adapter with two 6-axis load cells (model: Mini45, ATI Industrial Automation, Apex, NC) mounted onto a prosthetic foot plate, as described in [20].",
        "texts": [
            " Because of experimental time constraints, we did not examine variations in foot geometry or body weight other than differences across subjects. We implemented the approximate controller (9) on the Vanderbilt leg (see Fig. 9), a powered knee\u2013ankle prosthesis developed at Vanderbilt University (see [5] for design details). This device had encoders to measure joint angles/velocities and two brushless dc actuators to provide control of the knee and ankle joints. The leg did not have sensors to measure the foot orientation or COP, requiring us to design and integrate the custom instrumented foot in Fig. 10 (see [20]). An onboard microcontroller computed4 the desired control torques, which were 4Note that the microcontroller was preprogrammed with an impedance control law of the form (13), so during the stance period an off-board computer sent the microcontroller impedance commands that inverted the impedance loop and inserted desired torques from virtual constraint control law (9). converted into open-loop current inputs to each actuator\u2014these control loops are depicted in Fig. 11. The integrated leg-and-foot system provided the feedback needed to implement the output PD control law (9) as in Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001513_acs.chemrev.0c00999-Figure16-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001513_acs.chemrev.0c00999-Figure16-1.png",
        "caption": "Figure 16. Visualization of ideal features of next-generation programmable nano/microrobots.",
        "texts": [
            "462 Not only can we create more novel and diverse geometries due to the latest advances in additive manufacturing techniques,463 but we can also exploit the smooth gradient transition in materials,464 analogous to what is seen in the animal kingdom. Such tools create a problem of having infinite possibilities of design and complexity which requires a deterministic structure where the whole unit is needed to be designed before fabrication, resulting in rigid and slow prototyping developments while still requiring specialized machinery and code instructions. For example, different parts of the microrobot should be responsible for different parts of its behavior (Figure 16). The engine is responsible for motion, while a sensing center is responsible for probing its environment. Coupled to this center is the decision making \u201cbrain\u201d. Thus, the development of organs or in robotic terms \u201cmodules\u201d will become paramount. So, at the end of this Review, as we look forward to the future, we introduce the most vital modules which must be developed for the next generation of intelligent microrobots. First, a portion of any suchmicrorobotmust have a propulsion capability. As was mentioned earlier, this could be based on a reaction between a fuel and catalyst or direct decomposition, or instead converting electromagnetic energy into mechanical energy"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002492_j.msea.2019.138395-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002492_j.msea.2019.138395-Figure3-1.png",
        "caption": "Fig. 3. (a) LPBF-fabricated SS316L blocks; (b) The dimension of the dog-bone-shaped tensile sample; (c) The relative positions of the three sets of tensile samples in the LPBF-fabricated blocks.",
        "texts": [
            " Three distinct configurations were adopted for each set of tensile samples so that their tensile axes (TAs) are aligned with the three crystallographic directions (< 100> ,< 110>and<111>) of the LPBF-fabricated SS316L. The three sets of samples will be called hereafter as< 100> ,< 110> and<111> samples. Note that the angular error of the 35\u00b0 angle for the< 111> samples was within 1\u00b0 after machining. The underlying correlations between the alignment of TAs and the corresponding crystallographic directions will be explained in detail in Section 3.2.1. Fig. 3 shows the three block specimens on the build plate, the dimension of the dog-bone-shaped tensile sample and the three different sample configurations inside the block specimens. Prior to tensile testing, the gauge sections of the dog-bone-shaped samples were ground up to 1200 grit SiC paper to remove the machining grooves perpendicular to the TAs. Tensile tests were conducted at ambient temperature by a Satec tensile machine (Model 60UD) with a strain rate of 8.33\u00d710\u22124 s\u22121. The strain was measured by an Instron AVE 2 noncontacting video extensometer",
            " The detailed mechanism will be elaborated in Section 3.2.2. In principle, the single-crystalline-like texture makes it possible to align certain crystallographic directions following specific angular relations with respect to the specimen coordinate system. In the present case, specifically, the [001], [1\u203e10] and [1\u203e11\u203e] crystallographic directions of the SS316L FCC matrix were aligned with the specimen X axis, specimen Z axis and the direction about 35\u00b0 away from the specimen Z axis, respectively. Therefore, the extracting strategy in Fig. 3 could result in tensile samples with their TAs parallel to the<100> ,< 110> and<111>directions. To validate the correlation between the TAs and the crystallographic directions, EBSD analysis was conducted on the XZ plane of the block samples, which is the common plane where the TAs of the three tensile samples lie in. The IPF orientation maps and the corresponding pole figures are shown in Fig. 7. It can be observed that the TAs of the three types of tensile samples matched well with the< 100> ,<110> and<111> crystallographic directions"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002021_j.addma.2017.08.014-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002021_j.addma.2017.08.014-Figure7-1.png",
        "caption": "Fig. 7 Macroscale part model: (a) mesh design and dimensions, (b) boundary conditions of the part model.",
        "texts": [
            " Very similar heating and cooling thermal cycles calculated from the microscale model and mesoscale model were observed except the melt pool in the microscale model cools faster than that in the mesoscale due to the smaller melt pool volume which leads to a faster cooling rate. The calculated body heat flux was applied to the mesoscale hatch model which has several bundles of single scans with the heating time of 0.4 milliseconds, and it is given by Eq. 6: \ud835\udc5e = \ud835\udc34 \u2219 \ud835\udc43 \ud835\udc51\ud835\udc60 \u2219 \ud835\udc51\ud835\udc5a \u2219 \ud835\udc3b (6) Page 14 of 27 2.6 Macroscale part model Model dimensions and mesh: In the macroscale part model, one quarter of a twin cantilever with a support structure was modeled (Fig. 7) as the part is symmetrical. The dimensions of the cantilever (quarter model) are 55 mm (length) 5 mm (width) \u00d7 12 mm (height). The elements of the cantilever are hexahedral elements with 333 \u00b5m (length) \u00d7 500 \u00b5m (width) \u00d7 225 \u00b5m (height). The support structure (marked by yellow color in Fig. 7) is the same material as the cantilever. In the actual SLM process, support structures (usually porous) are widely applied to mitigate distortion during the build especially for parts with complex geometries and overhang structures [33]. In this study, same mesh density was created for the support structure and the cantilever. The initial temperatures of both the powders and the substrate were set to 20 \u00b0C. The bottom surface of the substrate was fixed during the SLM process. Page 15 of 27 Thermal-mechanical coupled analysis: The macroscale part model is a thermal-mechanical coupled analysis which has two analysis steps",
            " [34] initiated a novel experiment setup for in situ monitoring displacement and temperature of the substrate for SLM process. This monitoring experimental setup could be applied to determine the percentages of heat dissipation to the substrate, to the powder bed, and to the surroundings. A future numerical study may focus on the effects of different convection coefficient values of different surfaces on the final residual stress distribution and distortion of a SLMed part. Details of the heat transfer modeling approach applied in this study are shown in Fig. 7. Part build-up modeling: In this study, the part along with the support was divided into 12 layers in Y direction (part build-up direction in Fig. 7b). At the very beginning of the simulation, all the elements of the part and support were deactivated. When the first powder layer was placed, the corresponding elements set was activated and heated up by the equivalent body flux for 0.4 milliseconds (exposure time calculated from the microscale scan model). A cooling cycle for the whole model of 10 seconds was applied immediately after the heating cycle. Then a new powder layer was deposited, heated up, and cooled down to room temperature. This depositing-heatingcooling cycle continues until the whole part was built. In order to predict the cantilever distortion after the SLM process, the removal of support structure by cutting was achieved by deactivating the associated elements in the cutting plane (Fig. 7b) to release the internal residual stress after the whole part (including the support) was build. The cantilever distortion can be predicted and compared with the experimental measurement. 3.1 Material state (powder-liquid-solid) transition The material state contours and the corresponding temperature fields during the cantilever building process are shown in Fig. 8. The newly deposited powder materials (in green) were placed on the top surface of the cantilever and heated by the equivalent body heat flux"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001670_amp2.10021-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001670_amp2.10021-Figure2-1.png",
        "caption": "FIGURE 2 Effect of surface finishing on additively manufactured gear hobs (275 mm long, 100 mm diameter) produced by VBN Components: (A) before grinding and (B) after grinding (Reproduced with permission from VBN Components)",
        "texts": [
            " The surface roughness has been reported to be controlled and improved by selecting a proper feedstock feed rate, deposition speed, layer thickness and laser power.[8] However, due to the intrinsic characteristics of the AM process, optimizing these in-process parameters could only have limited effects. Multiple post-process surface treatment methods have also been used to improve the geometrical accuracy and surface roughness of the final parts, including machining, mechanical grinding and polishing, abrasive flow polishing, chemical milling and electroplating.[9] Figure 2 shows an AM gear hob before and after mechanical grinding. These hobs are made of hard steel and are currently produced by VBN Components in Sweden. Powder bed fusion (PBF) is one of the most commonly used AM techniques. PBF requires powdered feedstock that is sequentially processed in thin layers and solidified either by a laser beam in laser-based powder bed fusion (L-PBF), or by an electron beam in electron beam melting (EBM).[10] As far as L-PBF is concerned, the powder consolidation occurs through a liquid phase sintering mechanism in direct metal laser sintering (DMLS), or via a complete meltingsolidification mechanism in selective laser melting (SLM)",
            " As a consequence, the debate is now open in the literature on two basic requirements, namely further reduction of the residual porosity and development of appropriate means to detect the formation of pores and microstructural defects in situ. The achievement of these two goals can significantly contribute to the future industrial development and advancement of PBF. The authors would like to acknowledge Oerlikon AM GmbH (Feldkirchen, Germany) for providing the images of AM parts in Figure 1 and PBF processing in Figure 4. The authors are grateful to VBN Components AB (Uppsala, Sweden) for the images of AM parts before and after surface finishing, shown in Figure 2. The authors would also like to express their gratitude to Mr Mark Kirby at Renishaw Canada Solutions Centre for the images of process-induced defects in Figure 7. CONFLICT OF INTEREST The authors have no conflicts of interest to declare. AUTHOR CONTRIBUTIONS A.S. and A.N. made substantial contributions to conception and design, and to search, analysis and interpretation of literature information; A.S. was involved in drafting the manuscript and A.N. in revising it critically for important intellectual content; they gave final approval of the version to be published"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001786_j.mechatronics.2011.02.007-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001786_j.mechatronics.2011.02.007-Figure5-1.png",
        "caption": "Fig. 5. No payload (disturbance-free) configuration.",
        "texts": [
            " The neural-adaptive control adjusts to payloads nearly as quickly as the LP adaptive control, and improves performance significantly the next time a payload is added since it has learned how to react. The neural-adaptive control does not overshoot as much as the LP adaptive control when the payload removed. Moreover, performance may keep improving every time the same payload is added, with convergence level depending on the approximation ability (size) of the CMAC. During the experiments, the quadrotor remains mounted to the bearing allowing motion in only roll, pitch, and yaw (Fig. 5). Since the quadrotor is constrained in Z, we provided a constant control u1 = 162,977. The linear control gains for the other control signals are K \u00bc 4 K2 \u00bc K3 \u00bc 75;000 K4 \u00bc 50;000 The number of CMAC layers and cell sizes are the same as used in the simulation. Testing adaptation ability, we add a payload after the quadrotor has found a stable equilibrium without payload. Specifically, a mass of 90 gm is hung from the X-axis arm (Fig. 6), equivalent to a step-disturbance of 0.021 Nm in pitch torque"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000743_70.370511-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000743_70.370511-Figure5-1.png",
        "caption": "Fig. 5. Robot trajectory, joint 2 plot, and cost curves-Example 1",
        "texts": [
            " Therefore, the two schemes should produce different joint trajectories for the same end-effector path using the same cost function. When the end-effector is not moving, WLN scheme does not change its configuration, but GPM algorithm continues to change its configuration until it reaches an optimum. For better comparison, a constant end-effector speed is used for LN, WLN, and GPM control schemes in the following examples that show the advantages of the WLN solution. A. Example I Figs. 4 and 5 show the 3-D maps corresponding to the straight line end-effector trajectory shown in Fig. 5. The performance criterion W ( H ) defined in (12), is plotted as a function of the end-effector trajectory and self-motion. Since a higher value of this performance criterion represents proximity to the joint limits, we refer to it as \u201ccost\u201d which needs to be minimized. The \u201cend-effector trajectory\u201d axis represents the moving distance of the end-effector. The \u201cselfmotion\u201d axis indicates the self motion of the manipulator. On these maps, we use joint 1 value for self-motion since it has a one-to-one correspondence with the self-motion along the end-effector trajectory",
            " These curves continue to climb toward the high-cost area and hit the joint limit. It may be noted that the projection of the gradient onto the null space is zero at the maximum or minimum. Since i t is very small near the local maximum, it does not produce enough self-motion to change the direction toward a low cost area. Therefore, the GPM using (10) provides very close to the LN solution. To compare the proposed WLN scheme with LN and GPM solutions, we choose a starting configuration slightly away from the local maximum or the middle peak on the 3-D map in Fig. 5. The cost curves for the three schemes are shown on this 3-D map for the same starting configuration. Until Point A, WLN and GPM trajectories are similar to the LN trajectory. There after, WLN scheme changes its direction toward the low cost area. Since, the magnitude of the self 290 IEEE TRANFACTIONS ON ROBOTICS AND AUTOMATION. VOL. I I , NO. 2, APRIL 1995 motion in the WLN scheme is affected by the overall magnitude of the gradient (and not just on the projection in the null space direction), the deviation from the LN path is significant in this case",
            " to reduce the self-motion after C will make the situation undesirable between Ed-effector lrajectory Joint 2 limit 140 130 120 110 et deg 100 90 BO 70 0 2 4 6 8 10 12 14 16 18 Time (second) Fig. 7. Robot trajectory and joint 2 plot-Example 2 B and C. In that case, the trajectory will not deviate much from the LN trajectory, and it may be too late to avoid the joint limits. Thus it is not possible to find a I. which would work well throughout the trajectory. After point A, the cost function of WLN is the same as GPM since none of the joints changes the direction of motion, and the desired end-effector trajectory requires the joints to move toward their limits. However, as seen in Fig. 5 , the two joint trajectories are quite different. Fig. 5 also shows the joint 2 angle plots for the three schemes. As shown LN runs into the joint limit. WLN slows down the speed of joint 2 and stops it from moving toward its limit, and GPM moves joint 2 away from its limit. The GPM effect on the trajectory is much later and more severe compared to the WLN scheme. The joint velocity plots for the three schemes are shown in Fig. 6. Since the projection of the gradient onto the null space is too large during the time t = 3.3 to 4 seconds, the joint velocities go over their limits for the GPM trajectory"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure5.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure5.1-1.png",
        "caption": "FIGURE 5.1 System of coupled conductor loops.",
        "texts": [
            " We also show that the computation of partial inductances is an important part of this process and we treat the general case for partial inductance computations for nonorthogonal structures in Chapter 7 and Appendix C. Circuit Oriented Electromagnetic Modeling Using the PEEC Techniques, First Edition. Albert E. Ruehli, Giulio Antonini, and Lijun Jiang. \u00a9 2017 John Wiley & Sons, Inc. Published 2017 by John Wiley & Sons, Inc. Included in Appendix C is a collection of formulas for partial inductances and some other integrals. As presented in Ref. [3], we start out with the basic concept of the computation of loop inductances for N loops. Figure 5.1 shows an example of such a loop structure for which we want to compute the inductances. We assume that they all have a small gap where we can inject a current and measure the induced voltage. In a realistic example, we can ignore the missed coupling due to the gaps since they are small compared to the loop size. The example in Fig. 5.1 does not have multibranch connected loops, a subject which is considered in this chapter. In this example, we show N coupled loops. The desired inductance equivalent circuit is shown in Fig. 5.2. We give the definition of inductance for this simple case. LOOP INDUCTANCE COMPUTATIONS 91 Definition 1 (Inductance for N loops) The N2 inductances for a system of N loop are defined by Lkm = \ud835\udf13km Im = Vk s Im . (5.1) Where Im is the current flowing in the mth loop and VK is the voltage induced in the kth loop",
            "14), we get Lloop = 4\u2211 i=1 4\u2211 j=1 1 ck cm \ud835\udf070 4\ud835\udf0b\u222bck \u222bcm \u222blk,i \u222blm,j t\u0302m,j \u22c5 t\u0302k,i Rrm d\ud835\udcc1m d\ud835\udcc1k dcm dck. (5.15) We observe that we compute the partial self-inductance if we apply this result to only one part where m = k. With this, we have a form of the equations that leads to the partial inductances in the following section. We use the loop inductance derived in the previous section as a starting point for a much more powerful circuit-oriented general approach for inductance computations using partial inductances. By considering Fig. 5.1, we see that we approximate each section of the loops by a rectangular bar. As a first example to the general case for inductance computations, we consider the loop in Fig. 5.3. We recognize that we can simplify the loop inductance equation in (5.15) into Lloop = 4\u2211 i=1 4\u2211 j=1 Lpij, (5.16) where Lpij = 1 k m \ud835\udf070 4\ud835\udf0b\u222bk \u222bm \u222b\ud835\udcc1k,i \u222b\ud835\udcc1m,j t\u0302k,i \u22c5 t\u0302m,j Rkm d\ud835\udcc1m d\ud835\udcc1k dm dk. (5.17) From this example, we find the following definition of a partial inductance Definition 2 (Partial inductance) A partial self-inductance Lpii is defined for a single piece of conductor as Vi = Lpii sIi, where Vi is the voltage along the conductor i and Ii is the current in the conductor i"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000378_s0888-3270(03)00077-3-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000378_s0888-3270(03)00077-3-Figure1-1.png",
        "caption": "Fig. 1. A schematic of the experimental system.",
        "texts": [
            " By using selected segments from the available experimental data, typical sample feature vectors are generated for both normal bearings and bearings with different types of faults under different load conditions. Then different pattern classification methods have been studied in the decision making stage, including the neural-fuzzy inference system, which is believed to be most suitable for complex situations due to its adaptability and the capability of the network to realise a non-linear approximation. The ball bearings are installed in a motor driven mechanical system, as shown in Fig. 1. A 2 hp, three-phase induction motor (Reliance Electric 2HP IQPreAlert motor), was connected to a dynamometer and a torque sensor by a self-aligning coupling. The dynamometer is controlled so that desired torque load levels can be achieved. An accelerometer with a bandwidth up to 5000Hz and a 1V/g output is mounted on the motor housing at the drive-end of the motor to acquire the vibration signals from the bearing. The data collection system consists of a high bandwidth amplifier particularly designed for vibration signals and a data recorder with a sampling frequency of 12,000Hz per channel. The data recorder is equipped with low-pass filters at the input stage for anti-aliasing. On the other hand, the frequency content of interest in the vibration signals of the system under study did not exceed 5000Hz, for which the sampling rate is ample. To develop the new diagnostic technique, four sets of data were obtained from the experimental system (shown in Fig. 1): (i) under normal conditions; (ii) with inner race faults; (iii) with a ball fault (iv) with outer race faults. Faults were introduced into the drive-end bearing of the motor using the EDM method. The bearings used in this work are deep grove ball bearing manufactured by NTN. Some parameters are listed below: Bearing specs, NTN p/n 6205c3: Basic dynamic load rating: 14000N Basic static load rating: 7850N Radial internal clearance: 0.013\u20130.028 Pitch diameter(Pd): 1.535 in Ball diameter(Bd): 0.312 in Ball pass frequency at outer ring(OR): 3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001803_icra.2015.7139759-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001803_icra.2015.7139759-Figure2-1.png",
        "caption": "Fig. 2: (a): terminal part of the i-th hexarotor arm showing the body frame FPi and the generated thrust Tthrusti and drag Tdragi ; (b) and (c): Visualization of the possible reorientation of the propeller around XPi (case (b)) and YPi (case (c)). The angle of reorientation is denoted with \u03b1i in (b) and \u03b2i in (c)",
        "texts": [
            " 6 (2) where the angular parameters \u03b1i and \u03b2i represent the tilt angles that uniquely define the rotation plane of the i-th propeller, XPiY Pi or, equivalently, the direction of ZPi in FB . The angles \u03b1i and \u03b2i have a clear geometrical interpretation, in fact the i-th propeller plane XPi Y Pi is obtained from XBY B by first applying a rotation of \u03b1i about the line OBOPi and then a rotation of \u03b2i about Y Pi , which lies on XBY B and is perpendicular to OBOPi . The \u03b1i and \u03b2i rotation is pictorially represented in Fig. 2. For convenience, we group the following parameters into four 6-tuples: \u03b1 = (\u03b11, \u03b12, \u03b13, \u03b14, \u03b15, \u03b16), \u03b2 = (\u03b21, \u03b22, \u03b23, \u03b24, \u03b25, \u03b26), \u03bb = (\u03bb1, \u03bb2, \u03bb3, \u03bb4, \u03bb5, \u03bb6) and Lx = (Lx1 , Lx2 , Lx3 , Lx4 , Lx5 , Lx6). In this paper we consider the case in which \u03bbi, Lxi , \u03b1i, \u03b2i, for i = 1 . . . 6, are constant during flight. Nevertheless, we allow \u03b1i, \u03b2i to be changed during a pre-flight setup, in order, e.g., to minimize the sum of the overall control effort for a specific task, as shown in Sec. IV-B. Utilizing the standard Newton-Euler approach for dynamic systems, it is possible to derive the complete dynamic equations of the hexarotor by considering the forces and torques that are generated by each propeller rotation together with the significant gyroscopic and inertial effects"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001515_j.trac.2008.05.009-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001515_j.trac.2008.05.009-Figure1-1.png",
        "caption": "Figure 1. Glucose sensor based on carbon-nanotube (CNT) nanoelectrode ensembles.",
        "texts": [
            " CNTs have thus been examined as a new electrode material to reduce the overpotential [27]. It is found that the CNT offers remarkably decreased overvoltage for NADH oxidation as well as reducing the surface-fouling effects of the electrodes. These characteristics indicate great promise for CNTs in developing highly-sensitive, low-potential, stable amperometric biosensors based on dehydrogenase enzymes [27]. Lin and co-workers have demonstrated a novel glucose biosensor using CNT-nanoelectrode ensembles (Fig. 1) [28]. The CNT array was fabricated by growing CNTs directly on the patterned catalysts [24]. The density of the CNT array can be adjusted by changing the density of the catalysts. The GOx molecules were attached to the open-ended tips of the CNTs by forming amide linkages between their amine residues and carboxylic-acid groups on the CNT tips via carbodiimide chemistry. For this CNT-array sensor, each CNT electrode acts as nanoelectrode. Because the distance between two nanoelectrodes is much larger than the diameter of each nanoelectrode, there is negligible overlap of diffusion layers between nanoelectrodes, so this kind of CNT-array biosensor has faster response time, lower background current, and higher signal/noise ratio than bulk-CNT-modified electrodes"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003690_1.1802311-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003690_1.1802311-Figure3-1.png",
        "caption": "Fig. 3 Spindle dynamics model",
        "texts": [
            " The forces are transferred to inner rings B3 to B5 through bearing balls, then to the spindle shaft through inner ring B5, which is fixed to the spindle shaft. Finally, the forces are transmitted to the front bearings by inner ring B1 which is also fixed to the spindle shaft, then to the housing by outer ring A2, which is fixed to the housing. The whole spindle is self-balanced in the axial direction under the preload. Equations of Motion for the Rigid Disk. A diagram of the spindle shaft with the pulley is shown in Fig. 3. O2xyz is a fixed coordinate system, where the x-axis is coincident with the centerline of the shaft before the shaft is deformed. The coordinates of point P on the disk are affected by the displacements u, v , w, uy , and uz. Assume that the disk is fixed, the sequence of rotation is not important considering the small deformations. By rotating the coordinate system O2xyz at an angle 2uy about axis y, then 2uz about axis z , the coordinates of point P in the O2xyz system are changed due to the rotations uy and uz , and can be expressed as: H x y z J 5H x0 0 0 J 1@T#H 0 r cos f r sin f J (1) where x0 is the coordinate of the disk center, f5Vt , and transformation matrix @T# is: @T#5F cos uz 0 sin uz 0 1 0 2sin uz 0 cos uz GF 1 0 0 0 cos uy 2sin uy 0 sin uy cos uy G (2) The coordinates of point P become: Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure9.24-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure9.24-1.png",
        "caption": "FIGURE 9.24 Geometry of U-shaped test problem called HShoe.",
        "texts": [
            " Finally, a GIBC surface formulation solver [50] was used in the comparison. All computations are done on a single processor with 8 GB of random access memory. So, this limited some of the very high frequency accuracy. For very high frequency results, a larger number of small cells and computer memory are required. The examples given are interesting wide frequency range problems where the 3D current path is not predetermined. Specifically, the horseshoe (HShoe) problem consists of one layer shown in Fig. 9.24, but it has wide flat corners that lead to a current redistribution with frequency. The geometrical data for this example is shown in Table 9.1. The additional data required is w= 10 \u03bcm and the gap width is wc = 0.2 \u03bcm. We used two different models for the contacts area at the inside of the gap. In one case, we use a 0.4 \u03bcm section at the contact surface with a conductivity that is 104 times larger than copper. This leads to contacts at the entire cross-sectional surface. Alternatively, point contacts are used at the front of the gap as shown in Fig. 9.24. The second problem in Fig. 9.25 consists of an L-shaped conductor with a close ground plane. At high frequencies, the current will take the path that minimizes the inductance right 240 SKIN EFFECT MODELING under the L-shaped wire. However, at low frequencies, the current takes the path of least resistance that is across the plane. The lengths of the L conductors are relatively small for this redistribution to be very large. Hence, the change in inductance is relatively modest. Much larger changes in L and R have been obtained for longer lines [45]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003926_0167-6911(91)90050-o-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003926_0167-6911(91)90050-o-Figure7-1.png",
        "caption": "Fig. 7. Phase-space behaviour of uncontrolled R~ssler type system (12).",
        "texts": [
            " Suppose now that n = 3 and ~\u00a2'(Y,, Y2, Y 3 ) = Y ~ Y ( Y ) ~, Y(Y) : = 1 + I lYll 2 Choosing c 1 = 1 and c 2 = 2, then, by Theorem 2, the adaptive strategy u(t) ~ p ( x ( t ) ) [ z ( t ) + 2 z ' ( t ) + z \" ( t ) + y ( z ( t ) , z ' ( t ) , z \" ( t ) ) ~ ( z ( t ) + 2 z ' ( t ) + z \" ( t ) ) ] , ~(t) = ( z ( t ) + 2 z ' ( t ) + z ' ( t ) ) 2 + y ( z ( t ) , z ' ( t ) , z \" ( t ) ) l z ( t ) + 2 z ' ( t ) + z \" ( t ) [, renders the zero state of (1) a global attractor. As a specific example of a system subsumed by (1), we select the following system of R6ssler type (see [15,19]) with control: z '\" (t) + a l z\" ( t ) + z ' ( t ) - ( z \" ( t ) + a lz ' ( t ) + z ( t ) ) ( z ' ( t ) + alz( t ) + a2) + a 3 = bu(t) , (12) with unknown parameters a i and b ~ 0, and which, in the absence of control, is known to exhibit chaotic behaviour for particular parameter values. For the values a 1 = -0 .398 , a 2 = - 4 , a 3 = 2, b = 1, Figure 7 depicts the phase-space behaviour of the uncontrolled system (specifically, the solution with initial data (z(0), z'(0), z\"(0)) = (1, 0, 0) is shown). 2 l 6 E.P. Rvan / Uniuersal adaptit~e stabilizer E.P. Ryan / Universal adaptive stabilizer 217 1 0 -I -5 -2 -10 0 t . 5 0 Fig. 8. Controlled state evolution. K(.) v(\u00a2( - ) ) t \u2022 5 Fig. 9. Evolution of control parameters. Figures 8 and 9 depict the behaviour of the adaptively-controlled (z(0), z'(0), z\"(0), K(0)) = (1, 0, 0, 0) and r : o ~ o 2 cos o"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000961_j.euromechsol.2008.07.007-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000961_j.euromechsol.2008.07.007-Figure2-1.png",
        "caption": "Fig. 2. Geometrical parameters for the fillet-foundation deflection.",
        "texts": [
            " (2) The computation of the fillet-foundation gear body deflection of the gear is based on the theory of Muskhelishvili (1975) applied to circular elastic rings (Sainsot et al., 2004). Assumed linear and constant stress variations at root circle, this theory makes it possible to derive an analytical formula for gear body-induced tooth deflections. This analytical expression is given by the following equation (Wang, 2003; Sainsot et al., 2004): \u03b4 f = F cos2 \u03b1m W E { L\u2217 ( u f S f )2 + M\u2217 ( u f S f ) + P\u2217(1 + Q\u2217 tg2 \u03b1m) } (3) where W is the tooth width. u f and S f are given in Fig. 2. The coefficients L\u2217 , M\u2217 , P\u2217 , Q \u2217 can be approached by polynomial functions (Sainsot et al., 2004): X\u2217 i (h f i, \u03b8 f ) = Ai/\u03b8 2 f + Bih 2 f i + Cih f i/\u03b8 f + Di/\u03b8 f + Eih f i + Fi (4) X\u2217 i denotes the coefficients L\u2217 , M\u2217 , P\u2217 and Q \u2217 . h f i = r f /rint , h f i and \u03b8 f are defined in Fig. 2, the values of Ai , Bi , Ci , Di , Ei and Fi are given in Table 1. The corresponding fillet-foundation stiffness can be obtained by: k f = F \u03b4 f . (5) From the results derived by Yang and Sun (1985), the stiffness of Hertzian contact of two meshing teeth (commonly nonlinear) is practically a constant along the entire line of action independent to both the position of contact and the depth of interpenetration. kh can be given by: kh = \u03c0 EW 4(1 \u2212 \u03bd2) (6) where \u03bd is the Poisson\u2019s ratio. The local deformation is then expressed by: \u03b4h = F kh "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003403_978-1-4615-5633-6-Figure8.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003403_978-1-4615-5633-6-Figure8.6-1.png",
        "caption": "Figure 8.6. Control Characteristics",
        "texts": [
            " This is equivalent to maintaining the an gular difference across the line a constant and has been termed as Constant Angle (CA) control [Padiyar and Uma Rao (1995)]. Assuming the voltage magnitudes at the two ends of the line are regulated, maintaining constant an gle is equivalent to maintaining constant voltage difference between two ends of the line. Both CC (Constant Current) and CA controllers can be of PI type with dy namic compensation for improving the response. The steady state control characteristics of both CC and CA control are shown in Fig.8.6 (a) and (b) respectively. Assuming VTCSC to be positive in the capacitive region, the char acteristics have three segments OA, AB and BC. The control range is AB. OA INTERACTIONS WITH SERIES COMPENSATORS 215 and BC correspond to the limits on XTcsc. In Fig.5.23 (b), the control range AB is described by the equation (8.18) where h is the magnitude of the line current, XLR is the net line reactance ( taking into account the fixed series compensation if any), VLO is the constant (regulated) voltage drop across the line (including TCSC)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure3.32-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure3.32-1.png",
        "caption": "FIGURE 3.32. A turning tire on the ground to show the no slip travel distance dF , and the actual travel distance dA.",
        "texts": [
            "1, before dropping to an almost steady-state value \u03bcds. The friction coefficient \u03bcx (s) may be assumed proportional to s when s is very small \u03bcx (s) = Cs s s << 1 (3.98) where Cs is called the longitudinal slip coefficient. The tire will spin when s & 0.1 and the friction coefficient remains almost constant. The same phenomena happens in braking at the values \u03bcbp and \u03bcbs. Proof. Slip ratio, or simply slip, is defined as the difference between the actual speed of the tire vx and the equivalent tire speeds Rw\u03c9w. Figure 3.32 illustrates a turning tire on the ground. The ideal distance that the tire would freely travel with no slip is denoted by dF , while the actual distance the tire travels is denoted by dA. Thus, for a slipping tire, dA > dF , and for a spinning tire, dA < dF . The difference dF \u2212 dA is the tire slip and therefore, the slip ratio of the tire is s = dF \u2212 dA dA . (3.99) To have the instant value of s, we must measure the travel distances in an infinitesimal time length, and therefore, s \u2261 d\u0307F \u2212 d\u0307A d\u0307A "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001786_j.mechatronics.2011.02.007-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001786_j.mechatronics.2011.02.007-Figure1-1.png",
        "caption": "Fig. 1. Movement due to rotor speeds: thicker arrowed lines indicate increased rotor speed.",
        "texts": [
            " This work substantially modifies the previous method, achieving robustness to disturbances, and tests the method experimentally with a quadrotor mounted on a test-bed moving in roll, pitch, and yaw. In the early 1900s, the Breguet Brothers built their first human carrying quadrotor helicopter called the Breguet\u2013Richet Gyroplane No. 1. The Breguet Brothers found that such machine exhibited poor stability characteristics [21]. Although proving difficult to control, a quadrotor exhibits numerous advantages over other rotary wing UAVs such as helicopters. Control actuation consists of changing motor speeds rather than changing blade pitch (Fig. 1). Thus, quadrotors compare favorably to traditional helicopter design in the case of small, inexpensive electrically actuated UAVs where mechanical complexity is a disadvantage [22]. Recent advances in technology, including sensors and microcontrollers, now allow small UAV\u2019s to be built relatively easily and cheaply. Practical applications of UAV quadrotors will require a high level of controllability and flying capabilities. Successful autonomous operation of quadrotors in windy conditions while carrying an unknown payload remains an open problem"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002054_tie.2016.2593683-Figure9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002054_tie.2016.2593683-Figure9-1.png",
        "caption": "Fig. 9 Schematic plot of vibration and noise test",
        "texts": [
            " After the nodal force in the 1/3 2-D model is calculated, nodal force in 3-D space is obtained according to the motor periodicity and assuming that the electromagnetic force distributes uniformly in the axial direction. Then, the nodal force in electromagnetic mesh is transferred to structural mesh and serves as the excitation for vibration calculation, as shown in Fig. 8. Fig. 7 Comparison of BEMF from simulation and test at 3000 rpm. (a) (b) Fig. 8 Nodal force transfer. (a) force in electromagnetic mesh. (b) force in structural mesh. Based on mode superposition method [17], the stator surface vibration is calculated and compared with the test result. Fig. 9 shows the schematic plot of vibration and noise test. The motor shaft is connected with the dynamometer by a coupling and the motor is located in a ring by two clamping blocks. The ring is fixed to the foundation bed. A noise enclosure is used to isolate the dynamometer noise and a microphone is placed 35 cm above the motor to measure the noise at near distance. Fig. 10 shows the comparison of radial vibration from simulation and test. It can be seen that the calculated results agree well with those from experimental test either under noload or under the rated load and main test vibration peaks can almost be found in the simulation results"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure10.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure10.2-1.png",
        "caption": "FIGURE 10.2. Illustration of a moving vehicle, indicated by its body coordinate frame B in a global coordinate frame G.",
        "texts": [
            " Vehicle Planar Dynamics dynamics, we usually show them by a special character and call them roll rate, pitch rate, and yaw rate respectively. \u03d5\u0307 = p (10.1) \u03b8\u0307 = q (10.2) \u03c8\u0307 = r (10.3) The resultant of external forces and moments, that the vehicle receives from the ground and environment, makes the vehicle force system (F,M). This force system will be expressed in the body coordinate frame. BF = Fx\u0131\u0302+ Fy j\u0302+ Fzk\u0302 (10.4) BM = Mx\u0131\u0302+My j\u0302+Mzk\u0302 (10.5) The individual components of the 3D vehicle force system are shown in Figure 10.2. These components have special names and importance. 1. Longitudinal force Fx. It is a force acting along the x-axis. The resultant Fx > 0 if the vehicle is accelerating, and Fx < 0 if the vehicle is braking. Longitudinal force is also called forward force, or traction force. 2. Lateral force Fy. It is an orthogonal force to both Fx and Fz. The resultant Fy > 0 if it is leftward from the driver\u2019s viewpoint. Lateral force is usually a result of steering and is the main reason to generate a yaw moment and turn a vehicle",
            " The resultant Mz > 0 if the tire tends to turn about the z-axis. The yaw moment is also called the aligning moment. The position and orientation of the vehicle coordinate frame B(Cxyz) is measured with respect to a grounded fixed coordinate frame G(OXY Z). 10. Vehicle Planar Dynamics 585 The vehicle coordinate frame is called the body frame or vehicle frame, and the grounded frame is called the global coordinate frame. Analysis of the vehicle motion is equivalent to expressing the position and orientation of B(Cxyz) in G(OXY Z). Figure 10.2 shows how a moving vehicle is indicated by a body frame B in a global frame G. The angle between the x and X axes is the yaw angle \u03c8 and is called the heading angle. The velocity vector v of the vehicle makes an angle \u03b2 with the body x-axis which is called sideslip angle or attitude angle. The vehicle\u2019s velocity vector v makes an angle \u03b2 + \u03c8 with the global X-axis that is called the cruise angle. These angles are shown in the top view of a moving vehicle in Figure 10.3. There are many situations in which we need to number the wheels of a vehicle"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure2.12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure2.12-1.png",
        "caption": "Fig. 2.12: A scheme for the analysis of manipulations for intelligent packing of batteries.",
        "texts": [
            " The intelligent packing has been designed for batteries including a test of correct production in order to pack a battery in a given box with a proper polarity direction. The process has been examined by starting from a feed on a conveyor with a loading station at which a battery can be at a vertical or horizontal posture with undetermined polarity direction. The required manipulation can be considered as intelligent when the manipulative task includes the determination of battery orientation so that the robot can pack the battery into a suitable box with a proper battery polarity. The manipulation has been properly analyzed as shown in Fig. 2.12 where precision points have been determined together with alternative paths depending on the battery polarity and posture. In Fig. 2.12 a layout for a robotized work-cell is also indicated as a function of the feeding conveyor in A and packing station in B. The corresponding flowchart for battery manipulation is reported in Fig. 2.13. Fundamentals of the Mechanics of Robotic Manipulation54 A scheme for a specific robotization has been designed as in Fig. 2.14 and the laboratory practising layout is shown in Fig. 2.15 by using the available robot. The three alternative paths in Fig. 2.12 can be performed by a 5 d.o.f.s robot since three translations are required for the transfer motion from A to B, and two rotations (about vertical and horizontal axes) are necessary to rotate a battery from any posture in A to the horizontal deposition in the pack box in B with the proper polarity orientation. The feeding conveyor has been sensored with two proximity sensors to detect the posture of a battery as in a vertical or horizontal posture. Thus, the robot can move to grasp a battery by means of a suitable mode corresponding to the current posture and quote of a battery"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000846_j.ymssp.2017.05.024-Figure13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000846_j.ymssp.2017.05.024-Figure13-1.png",
        "caption": "Fig. 13. A single tooth pit modeling [58].",
        "texts": [
            " Chaari et al. [61] compared healthy planetary gear dynamic response and the response of planetary gears containing tooth defects (tooth pitting and cracking) in both the time and frequency domains and in the joint time frequency domains using the Wigner-Ville distribution. The faults were approximated by giving a gear mesh stiffness reduction using the square waveform method. Chaari et al. [5], Zhe et al. [96] and Abouel-seoud et al. [142] modeled a single gear tooth pit in rectangular shape as shown in Fig. 13(a). Chaari et al. [5] investigated a single tooth pit or tooth breakage effect on gear mesh stiffness and gear vibration frequency spectrum. Zhe et al. [96] studied a single tooth pit effect on the vibration of a planetary gear set and used the grey relational analysis to estimate pitting damage levels. Abouel-seoud et al. [142] presented a single rectangular tooth pit effect on the mesh stiffness of a wind turbine gearbox and analyzed vibration signal fault signatures in the time and frequency domain",
            " [70] investigated fault mechanism of a spur gear pair with spalling defect. Later, Ma and Chen [120] used a four-degree of freedom gear system with local crack or spalling fault to investigate the failure mechanism and characteristics. The two types of failures were compared using time history vibration, phase contrail, Poincar\u00e9 section, spectrum analysis and fractal dimension. Experimental results agreed with the theoretical results to some extent. Rincon et al. [152] used the finite element methods to analyze a single elliptical pit (see Fig. 13(b)) effect on mesh stiffness. Parey et al. [178] investigated the dynamic responses of a 6 DOF lumped parameter model and concluded that empirical mode decomposition pre-processed kurtosis and crest factor give early detection of pitting as compared to raw signal. The gear tooth profile error, eccentricities and defect were incorporated in the dynamic model by modifying the dynamic mesh force profile. Chaari et al. [179] investigated the effect of gear tooth pitting and crack on dynamic response of a planetary gear set, respectively"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure5.16-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure5.16-1.png",
        "caption": "FIGURE 5.16. Illustration of a simple pendulum.",
        "texts": [
            " Applied Kinematics The angular acceleration of B in G can always be expressed in the form G\u03b1B = G\u03b1B u\u0302\u03b1 (5.357) where u\u0302\u03b1 is a unit vector parallel to G\u03b1B. The angular velocity and angular acceleration vectors are not parallel in general, and therefore, u\u0302\u03b1 6= u\u0302\u03c9 (5.358) G\u03b1B 6= G\u03c9\u0307B. (5.359) However, the only special case is when the axis of rotation is fixed in both G and B frames. In this case G\u03b1B = \u03b1 u\u0302 = \u03c9\u0307 u\u0302 = \u03c6\u0308 u\u0302. (5.360) Example 195 Velocity and acceleration of a simple pendulum. A point mass attached to a massless rod and hanging from a revolute joint is called a simple pendulum. Figure 5.16 illustrates a simple pendulum. A local coordinate frame B is attached to the pendulum that rotates in a global frame G. The position vector of the bob and the angular velocity vector G\u03c9B are Br = l\u0131\u0302 (5.361) Gr = GRB Br = \u23a1\u23a3 l sin\u03c6 \u2212l cos\u03c6 0 \u23a4\u23a6 (5.362) B G\u03c9B = \u03c6\u0307k\u0302 (5.363) G\u03c9B = GRT B B G\u03c9B = \u03c6\u0307 K\u0302. (5.364) 5. Applied Kinematics 275 GRB = \u23a1\u23a3 cos \u00a1 3 2\u03c0 + \u03c6 \u00a2 \u2212 sin \u00a1 3 2\u03c0 + \u03c6 \u00a2 0 sin \u00a1 3 2\u03c0 + \u03c6 \u00a2 cos \u00a1 3 2\u03c0 + \u03c6 \u00a2 0 0 0 1 \u23a4\u23a6 = \u23a1\u23a3 sin\u03c6 cos\u03c6 0 \u2212 cos\u03c6 sin\u03c6 0 0 0 1 \u23a4\u23a6 (5.365) Its velocity is therefore given by B Gv = B r\u0307+ B G\u03c9B \u00d7 B Gr = 0 + \u03c6\u0307k\u0302 \u00d7 l\u0131\u0302 = l \u03c6\u0307j\u0302 (5"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002492_j.msea.2019.138395-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002492_j.msea.2019.138395-Figure2-1.png",
        "caption": "Fig. 2. Block and cube specimens used in the study and the associated coordinate system.",
        "texts": [
            " Standard laser power for SS316L was adopted, resulting in a volumetric energy density of 106 J/mm3, which is close to the values mentioned in the previous studies where great densification was achieved [29,30]. No contour parameters were used at the border of the specimens. The LPBF process was conducted under an argon atmosphere with an oxygen level below 300 ppm. Specimens with two different geometries (cubes and blocks) were built on regular steel plates. The cubes (10mm\u00d710mm\u00d710mm) were used for density analysis and microstructure characterization. The blocks (50mm\u00d750mm\u00d710mm) were prepared for tensile testing and the subsequent deformation behaviour analysis. Fig. 2 presents the two geometries and the associated coordinate system with respect to the scanning directions (SD1 and SD2) and the building direction (BD). The cube specimens were sectioned, mounted and then ground up to 800 grit SiC paper. The samples were polished with diamond suspensions of 9 \u03bcm, 3 \u03bcm and 1 \u03bcm, followed by a final polishing using a 0.05 \u03bcm colloidal silica suspension. The density of the SS316L samples was measured using Archimedes\u2019 method, and was also checked by a Keyence VHX-S550E optical microscope"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000233_978-1-4614-0460-6-Figure6.42-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000233_978-1-4614-0460-6-Figure6.42-1.png",
        "caption": "Fig. 6.42 Force balance to determine k21",
        "texts": [
            "31) Dividing by 2l/3 and solving for fk11 give the stiffness k11. 2 3x2m2 2 x2m1 fm22 x1 x2 2 3x2m2 2 x2m1 fm22 x1 x2 fm21 Fig. 6.38 Force balance to determine m21 The other on-diagonal term, k22, is determined from the force required to give a unit displacement to x2 while holding x1 fixed. The forces are displayed in Fig. 6.41. Summing the moments about x1 gives: X M \u00bc fk22 2l 3 k2x2 2l 3 \u00bc fk22 2l 3 k2 2l 3 \u00bc 0: (6.35) fk22 fk11 fk11 6.6 Shortcut Method for Determining Mass, Stiffness, and Damping Matrices 233 the forces in Fig. 6.42 and substituting for fk22 provide the desired result. X f \u00bc fk21 \u00fe fk22 k2x2 \u00bc fk21 \u00fe fk22 k2 \u00bc 0 fk21 \u00bc fk22 \u00fe k2 \u00bc k2 \u00fe k2 \u00bc 0 (6.37) The final stiffness matrix is: k \u00bc k11 k12 k21 k22 \u00bc k1 0 0 k2 : (6.38) Because the dampers appear at the same locations as the springs, the damping matrix is: c \u00bc c11 c12 c21 c22 \u00bc c1 0 0 c2 : (6.39) Chapter Summary \u2022 In practice, it is commonly required that we have an actual dynamic system and would like to build a model that we can use to represent its vibratory behavior"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001446_j.oceaneng.2016.06.041-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001446_j.oceaneng.2016.06.041-Figure1-1.png",
        "caption": "Fig. 1. Body-referenced coordinate system on an AUV.",
        "texts": [
            " In Section 3, we design an adaptive controller for AUVs based on sliding mode only considering the unknown model parameters and disturbances. In Section 4, we design the controls for the unknown dead zone and saturation. The simulation results are presented in Section 5, followed by experimental results in Section 6. Conclusions are drawn in Section 7. 2. Problem formulation 2.1. Motion equations In this section, the dynamical model of a slender body AUV is briefly introduced. There two commonly used frames, namely, inertia referenced frame and body referenced frame, to describe the motion of an AUV. As shown in Fig. 1, the AUV used in this work is centered at the center of buoyancy (CB) and has the bodyreferenced x axis forward, z axis to port (left), and y axis up. Looking forward from the bridge of the body, roll (\u03c6) about the x axis is positive counterclockwise, yaw (\u03c8) about the y axis is positive turning left, and pitch (\u03b8) about the z axis is positive bow up. In general, the model in D3 can be divided into three lightly interacting subsystems for the horizontal plane, vertical plane and roll. We also provide a simple view in the horizontal plane in Fig. 1, where the body axis and inertia axis frames are related through rotations about the y axis with the yaw angle \u03c8. The sideslip angle is defined by \u03b2 = ( + )v v varc tan /z x y 2 2 , and the angle of attack is defined by \u03b1 = \u2212 ( )v varc tan /y x . Through the coordinate frame transition, the kinematics of the AUV is described as ( ) \u03b8 \u03c8 \u03c8 \u03c6 \u03b8 \u03c8 \u03c6 \u03c8 \u03c6 \u03b8 \u03c8 \u03c6 \u03b8 \u03b8 \u03c6 \u03b8 \u03c6 \u03b8 \u03c8 \u03c8 \u03c6 \u03b8 \u03c8 \u03c6 \u03c8 \u03c6 \u03b8 \u03c8 \u03c6 \u0307 \u0307 \u0307 = \u2212 + \u2212 \u2212 + \u2212 \u23a1 \u23a3 \u23a2\u23a2\u23a2 \u23a4 \u23a6 \u23a5\u23a5\u23a5 \u23a1 \u23a3 \u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2 \u23a4 \u23a6 \u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5 \u23a1 \u23a3 \u23a2\u23a2 \u23a4 \u23a6 \u23a5\u23a5 1 x y z v v v cos cos sin sin sin cos cos sin cos sin cos sin sin cos cos cos sin cos sin cos sin sin sin cos cos cos sin sin sin x y z \u03c8 \u03b8 \u03c6 \u03b8 \u03c6 \u03b8 \u03c6 \u03c6 \u03c6 \u03b8 \u03c6 \u03b8 \u03c6 \u03c9 \u03c9 \u03c9 \u0307 \u0307 \u0307 = \u2212 \u2212 ( ) \u23a1 \u23a3 \u23a2 \u23a2 \u23a2 \u23a4 \u23a6 \u23a5 \u23a5 \u23a5 \u23a1 \u23a3 \u23a2 \u23a2 \u23a2 \u23a4 \u23a6 \u23a5 \u23a5 \u23a5 \u23a1 \u23a3 \u23a2 \u23a2 \u23a2 \u23a4 \u23a6 \u23a5 \u23a5 \u23a5 0 sec cos sec sin 0 sin cos 1 tan cos tan sin 2 x y z where \u03c9 \u03c9 \u03c9, ,x y z are the angle velocity in the body-referenced frame"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000846_j.ymssp.2017.05.024-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000846_j.ymssp.2017.05.024-Figure5-1.png",
        "caption": "Fig. 5. Dynamic modeling of a one-stage gearbox [39].",
        "texts": [
            " [34]: \u2018\u2018Both methods are equally accurate if proper attention is paid to defining the boundary conditions and the degree of discretization, which may be different for the two methods. Solution costs will differ depending on the discretization characteristics of the LPM and various FMM derivations and, of course, on the efficiency of the programmer.\u201d Many lumped parameter models were reviewed in [26,27,30\u201332] for fixed-axis and planetary gearbox transmission systems. We will not repeat reviewing these lumped parameter models again. Instead, two examples are given for a better understanding. One is for a fixed axis gearbox and the other one is for a planetary gear set. Fig. 5 gives a model for a one-stage fixed axis gearbox. This model was originally proposed by Bartelmus [35]. Later, it was used for dynamic analysis of gears with tooth damages [36\u201339]. This model has 8 degrees of freedom: driving motor rotation, driven motor rotation, pinion rotation, gear rotation, x- and y-direction translational motions of the pinion and the gear, respectively. Each gear is considered as a rigid plate with a central hole. Gear mesh interface and each bearing are modeled using a spring-damper system, respectively"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001598_tpel.2012.2227808-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001598_tpel.2012.2227808-Figure6-1.png",
        "caption": "Fig. 6. Phasor/vector diagram for one phase of PMSM",
        "texts": [
            " Influence of temperature on the magnetic remanence is approximately linear below Curie temperature, which is expressed in Equation (9) 0 0( ) ( )[1 \u0394 ( )]r r BB T B T T T   (9) where T is magnet\u2019s operation temperature, T0 is the preferred temperature, Br(T0) is the remanence at T0, and \u0394B is the reversible temperature coefficient, which is a negative number. What each search coil measures is a vector summation of flux due to permanent magnets and armature coils induced flux, based on an assumption that no saturation occurs. To analyze the reason of flux unbalance, it is required to decouple these two components. The phasor diagram for the Copyright (c) 2011 IEEE. Personal use is permitted. For any other purposes, permission must be obtained from the IEEE by emailing pubs-permissions@ieee.org. 4 operation of the PMSM as a generator is shown in Fig 6, in which the subscripts f, a, and fa represent field, armature and the combination of field and armature respectively. Under no-load condition when the rotor is revolving at the synchronous speed, voltage Ef is generated by field MMF Ff in each phase of the search coil. The MMF distribution can be described as space vectors, while the back EMF are time phasors. Superposition of the field MMF and the armature MMF, known as armature reaction, produces combined MMF Ffa, which is the vector sum of Ff and Fa. Additionally, this MMF is responsible for the resultant flux, which induces a back EMF in the search coil under load, denominated as Efa in Fig. 6. The PMSM used in the experiment has surface mounted NdFeB permanent magnets, resulting equal d-q axis inductance. Therefore phase current is controlled in the same direction of q axis. For an interior permanent magnet machine, phase current has a d axis component. In either case, a search coil measured voltage can be decoupled as: ( ) ( ) ( )fa f aE t E t E t  (10) Since the PMSM is always controlled by an inverter module, its phase current is never a pure sinusoidal waveform. To get the amplitude of each harmonic, Fourier series transformation is applied into both side of Equation (10), one can get _ _ _ jk t jk t jk t fa k f k a k k k k a e a e a e             (11) where k represents harmonic order; \u03c9 represents fundamental frequency; and a represents amplitude of each harmonic"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure1.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure1.6-1.png",
        "caption": "FIGURE 1.6. The plus one (+1) concept is a rule to find the tire to a rim with a 1 inch increase in diameter.",
        "texts": [
            " Tire and Rim Fundamentals SL \u2261 Standard load; Tire for normal usage and loads XL \u2261 Extra load; Tire for heavy loads rf \u2261 Reinforced tires Arrow \u2261 Direction of rotation Some tread patterns are designed to perform better when driven in a specific direction. Such tires will have an arrow showing which way the tire should rotate when the vehicle is moving forwards. Example 17 F Plus one (+1) concept. The plus one (+1) concept describes the sizing up of a rim and matching it to a proper tire. Generally speaking, each time we add 1 in to the rim diameter, we should add 20mm to the tire width and subtract 10% from the aspect ratio. This compensates the increases in rim width and diameter, and provides the same overall tire radius. Figure 1.6 illustrates the idea. By using a tire with a shorter sidewall, we get a quicker steering response and better lateral stability. However, we will have a stiffer ride. Example 18 F Under- and over-inflated tire. Overheat caused by improper inflation of tires is a common tire failure. An under-inflated tire will support less of the vehicle weight with the air pressure in the tire; therefore, more of the vehicle weight will be supported by the tire. This tire load increase causes the tire to have a larger tireprint that creates more friction and more heat"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003862_s0143-8166(03)00063-0-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003862_s0143-8166(03)00063-0-Figure2-1.png",
        "caption": "Fig. 2. The proposed geometrical characteristics.",
        "texts": [
            " * The effect of fluid motion due to the thermocapilary phenomena can be taken into account using a modified thermal conductivity for calculating the melt pool boundaries. Experimental work and estimations in the literature [15] suggest that the effective thermal conductivity in the presence of thermocapilary flow is at least twice the stationary melt conductivity. This increase can be generally presented by K \u00f0T\u00de \u00bc aK\u00f0Tm\u00de if T > Tm; \u00f08\u00de where a is the correction factor and K is modified thermal conductivity \u00f0W=m K\u00de: * Power attenuation is considered using the method developed by Picasso et al. [8] with some minor modifications. Fig. 2 shows the proposed geometrical characteristics in the process zone which is used in the development of the following equations. Based on their work P1 \u00bc Plbw\u00f0y\u00de 1 Pat Pl ; \u00f09\u00de P2 \u00bc PlZpbp Pat Pl 1\u00fe \u00f01 bw\u00f0y\u00de\u00de 1 Pat Pl ; \u00f010\u00de ARTICLE IN PRESS E. Toyserkani et al. / Optics and Lasers in Engineering 41 (2004) 849\u2013867 853 where P1 is total energy directly absorbed by the substrate (W), P2 is energy that is carried into the melt pool by powder particles (W), and Pat is attenuated laser power by powder particles (W)",
            " The ratio between the attenuated and average laser power can be obtained by [8] Pat Pl \u00bc \u2019m 2rcrlrpvp cos\u00f0yjet\u00de if rjetorl; \u2019m 2rcrjetrpvp cos\u00f0yjet\u00de if rjetXrl: 8>>< >: \u00f012\u00de In these equations, \u2019m is powder feedrate \u00f0kg=s\u00de; rc is powder density \u00f0kg=m 3\u00de; rl is radius of the laser beam on the substrate (m), rp is radius of powder particles (m), vp is powder particles velocity (m/s), yjet is the angle between powder jet and substrate (deg), rjet is radius of powder spray jet (m), bw\u00f0y\u00de is workpiece absorption factor, bp is particle absorption factor, and y is angle of top surface of melt pool with respect to horizontal line as shown in Fig. 2 (deg). The powder efficiency Zp can be considered as the ratio between the melt pool surface and the powder stream\u2019s area (Fig. 2) as Zp \u00bc A liq jet Ajet ; \u00f013\u00de where A liq jet is the intersection between the melt pool area on the workpiece and powder stream, and Ajet is the cross-sectional area of the powder stream on the workpiece. If we assume, the absorption of a flat plane inclined to a circular laser beam depends linearly on the angle of inclination y as shown in Fig. 2 and bw\u00f00\u00de is the workpiece absorption of a flat surface, bw\u00f0y\u00de can be calculated from bw\u00f0y\u00de \u00bc bw\u00f00\u00de\u00f01\u00fe awy\u00de; \u00f014\u00de where y is the angle shown in Fig. 2 and aw is a constant coefficient obtained experimentally for each material [8,12]. * The temperature dependency of material properties and absorption coefficients on the temperature are taken into account in the model. * In order to reduce the computational time, a combined heat transfer coefficient for the radiative and convective boundary conditions is calculated based on the relationship given by Goldak [16] and Yang et al. [17]: hc \u00bc 24:1 10 4etT 1:61: \u00f015\u00de ARTICLE IN PRESS E. Toyserkani et al",
            " The first one is the time between two deposition steps and the second one is the time step for calculating the melt pool area. After performing Step 2 and before repeating Step 1, the following corrections are applied: * All thermo-physical properties and absorption factor bw\u00f00\u00de are updated based on the new temperature distribution. * The new bw\u00f0y\u00de is calculated based on the updated y and Eq. (14). y is obtained from the developed geometry at any time based on the proposed triangle, which connects the tip of melt pool to the highest melted point on the clad as shown in Fig. 2. * The new Zp is obtained based on the new melt pool geometry using Eq. (13). * The new Pw is calculated using Eq. (11). Many numerical methods for solving Eq. (1) have been reported since 1940. Finite element methods (FEM) are one of the reliable and efficient numerical techniques which have been used for many years. The FEM can solve different form of partial differential equations with different boundary conditions. In this work, the governing PDE Eq. (1) is highly nonlinear due to material properties with dependency on temperature and a moving heat source with a Gaussian distribution"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure3.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure3.2-1.png",
        "caption": "FIGURE 3.2. Tire coordinate system.",
        "texts": [
            " To understand its importance, it is enough to remember that a vehicle can maneuver only by longitudinal, vertical, and lateral force systems generated under the tires. Figure 3.1 illustrates a model of a vertically loaded stationary tire. To model the tire-road interactions, we determine the tireprint and describe the forces distributed on the tireprint. 3.1 Tire Coordinate Frame and Tire Force System To describe the tire-road interaction and force system, we attach a Cartesian coordinate frame at the center of the tireprint, as shown in Figure 3.2, assuming a flat and horizontal ground. The x-axis is along the intersection line of the tire-plane and the ground. Tire plane is the plane made by narrowing the tire to a flat disk. The z-axis is perpendicular to the ground, opposite to the gravitational acceleration g, and the y-axis makes the coordinate system a right-hand triad. To show the tire orientation, we use two angles: camber angle \u03b3 and sideslip angle \u03b1. The camber angle is the angle between the tire-plane and 96 3. Tire Dynamics the vertical plane measured about the x-axis",
            " The camber angle can be recognized better in a front view as shown in Figure 3.3. The sideslip angle \u03b1, or simply sideslip, is the angle between the velocity vector v and the x-axis measured about the z-axis. The sideslip can be recognized better in a top view, as shown in Figure 3.4. The force system that a tire receives from the ground is assumed to be located at the center of the tireprint and can be decomposed along x, y, and z axes. Therefore, the interaction of a tire with the road generates a 3D force system including three forces and three moments, as shown in Figure 3.2. 1. Longitudinal force Fx. It is a force acting along the x-axis. The resultant longitudinal force Fx > 0 if the car is accelerating, and Fx < 0 if the car is braking. Longitudinal force is also called forward force. 2. Normal force Fz. It is a vertical force, normal to the ground plane. The resultant normal force Fz > 0 if it is upward. Normal force is also called vertical force or wheel load. 3. Lateral force Fy. It is a force, tangent to the ground and orthogonal to both Fx and Fz. The resultant lateral force Fy > 0 if it is in the y-direction",
            " The origin of the coordinate system is at the center of the tireprint when the tire is standing stationary. The x-axis is at the intersection of the tire-plane and the ground plane. The z-axis is downward and perpendicular to the tireprint. The y-axis is on the ground plane and goes to the right to make the coordinate frame a righthand frame. The sideslip angle \u03b1 is considered positive if the tire is slipping to the right, and the camber angle \u03b3 is positive when the tire leans to the right. The SAE coordinate system is as good as the coordinate system in Figure 3.2 and may be used alternatively. However, having the z-axis directed downward is sometimes inefficient and confusing. Furthermore, in SAE convention, the camber angle for the left and right tires of a vehicle have opposite signs. So, the camber angle of the left tire is positive when the tire leans to the right and the camber angle of the right tire is positive when the tire leans to the left. 3.2 Tire Stiffness As an applied approximation, the vertical tire force Fz can be calculated as a linear function of the normal tire deflection 4z measured at the tire center"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure1.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure1.2-1.png",
        "caption": "Figure 1.2 The forces exerted on a swimming shark. Source: Natuur en Techniek 90/02, p. 136.",
        "texts": [
            "1 Examples from the field of mechanics Mechanics is the subdivision of physics which addresses equilibrium and the motion of matter. Mechanics therefore includes for example: \u2022 The description of the movement of natural and artificial heavenly bod- ies. Figure 1.1 is a schematic representation of our Galaxy. The vast majority of all stars are in a flat disk. Above and below this disk, there are some 200 globular star clusters that revolve in ellipsoidal orbits 2 around the centre of the Galaxy. \u2022 A calculation of the forces exerted on a swimming shark. In Figure 1.2a, in which the shark is at rest, the shark is subject to forces resultant from its weight W and the upward force A caused by the water pressure. As a result, the animal tips over (see Figure 1.2b). If the shark\u2019s tail generates a thrust T , vertical forces are generated that keep the shark in vertical equilibrium (see Figure 1.2c). \u2022 Balancing on a surfboard (see Figure 1.3). \u2022 A calculation as to the deformation of an oil platform at sea subject to wave action. Figure 1.4 shows a concrete platform designed for the Norwegian Troll field with a water depth of 340 metres. The seabed consists of extremely weak clay. The sea conditions are extremely rough with waves over 10 metres in height. The mass of the deck is 60,000 tons (60 \u00d7 106 kg). \u2022 The description of water currents in a river, estuary, or sea. Figure 1.5 represents a current model for the North Sea"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000848_s40192-016-0045-4-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000848_s40192-016-0045-4-Figure1-1.png",
        "caption": "Fig. 1 Illustration of a typical SLM or SLS process",
        "texts": [
            " The energy beam may be either an electron beam or laser, the latter of which is of interest in this work. These laser powder bed fusion processes are referred to by many different names, the most common of which are selective laser sintering (SLS) and selective laser melting (SLM), depending on the nature of the powder fusion process. When the powder to be fused is metal, the terms direct metal laser sintering (DMLS) and direct metal laser melting (DMLM) are commonly used. A typical SLS or SLM process is illustrated in Fig. 1. The powder feed is contained in a hopper or dispenser bed. In the embodiment illustrated in Fig. 1, an elevator in the powder reservoir lifts a prescribed dose of powder above the level of a build plate which is then spread in a thin even layer over the build surface by a recoater mechanism. The recoater mechanism may consist of a hard scraper, soft squeegee, or roller [9]. Powder may also be supplied by a hopper from above the build surface. The powder layer thickness is typically between 10 and 100 \u03bcm. Selective portions of the powder layer corresponding with a slice of the part to be manufactured are then sintered or melted by a focused laser scanning across the surface"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001218_bfb0109984-Figure11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001218_bfb0109984-Figure11-1.png",
        "caption": "Fig. 11. Drift algorithm phase trajectories",
        "texts": [],
        "surrounding_texts": [
            "When using the drift algorithm the phase trajectories on the 2-sliding plane are characterized by loops with constant sign of the sliding variable yl (Fig.l 1), furthermore it is characterized by the use of sampled values of the available signal Yl with sampling period 6. The control algorithm is defined by tile following control law, in which the condition on lul must be considered when dealing with the chattering avoidance problem. [14, 20] - u i f [u I > 1 v(t) = - -Vmsign(Ayl i ) i f y l A y l , < 0; lu[ < 1 (28) --VMsign(Ayli) i f y l A y l i > 0; lul < 1 where Vm and VM are suitable positive constants such that Vm < VM and y_~ Vm sufficiently large, and Ayl i = Yl (t,) --Yl (ti --6), t E [t i, t i+1). The corresponding sufficient conditions for the convergence to the sliding manifold are rather cumbersome [14] and are omitted here for the sake of simplicity. After substituting Y2 for Ayl i a first order sliding mode on y2 = 0 would be achieved. This implies Yl = const., but, since an artificial switching time delay appears, we ensure a real sliding on Y2 with most of time spent in the set Yly'~ < 0, and therefore, Yl ---+ 0. The accuracy of the real sliding on Y2 ---- 0 is proportional to the sampling time interval 6; hence the duration of the transient process is proportional to 6 -1. Such an algorithm does not satisfy the definition of a real sliding algorithm [20] requiring the convergence time to be uniformly bounded with respect to 6. Let us consider a variable sampling time 6i+l[yl(ti)] = t,+l - t i , i = O, 1, 2 , . . . with 6 = max(6M,min(6m,Tl[yl(ti)]P)) himself where 0.5 < p _< 1, 6M > 6m > 0, > 0. Then with 7/, V~M sufficiently small and Vm sufficiently large the drift algorithm constitutes a second order real sliding algorithm with respect to 6 -* 0. This algorithm has no overshoot if parameters are chosen properly [14]."
        ]
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure13.15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure13.15-1.png",
        "caption": "Figure 13.15 (a) The distribution of the water pressure and (b) the resulting water pressure on the walls. (c) The isolated bottom with all the forces and couples acting on it.",
        "texts": [
            " From the equilibrium of the slice modelled as a line element, determine the water pressure on the bottom AB. Draw the water pressure on both the bottom and the walls. Include the values. b. Isolate bottom AB, and draw all the forces acting on it. Include the values. c. For the entire slice, draw the M , V and N diagrams, with the deformation symbols. Include relevant values. d. Determine the maximum bending moment. Where does it occur? 560 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Solution: a. Figure 13.15a shows the water pressure on the slice modelled as a line element. The water pressure on the bottom is constant. Let the water pressure there be qw. The water pressure on the walls varies linearly from zero at the water level to qw at the bottom. The upward water pressure qw on the bottom must be in equilibrium with the dead weight of the bottom and walls of the strip: (6 m) \u00d7 qw (\u2191) = {(6 m) + 2 \u00d7 (3 m)} \u00d7 (12 kN/m) (\u2193) = 144 kN (\u2193). This gives qw = 24 kN/m. b. The resulting water pressure R on the walls is (see Figure 13.15b): R = 1 2 (24 kN/m)(2.5 m) = 30 kN. The forces R, which pass through the centroid of the load diagram and therefore act (2.5 m)/3 above bottom AB, exert horizontal forces of 30 kN on AB and couples of (30 kN)(2.5 m)/3 = 25 kNm. The bottom AB can be seen as an eccentrically compressed beam. In addition, at A and B, the vertical forces due to the dead weight of the walls are (3 m)(12 kN/m) = 36 kN. In Figure 13.15c, the base AB has been isolated, and all the forces are shown. The resulting (uniformly) distributed load q on base AB is equal to the difference between the upward water pressure qw = 24 kN/m (\u2191) and the dead weight qw = 12 kN/m (\u2193): q = qw \u2212 qdw = (24 kN/m) \u2212 (12 kN/m) = 12 kN/m) (\u2191). c. Figures 13.16a to 13.16c shows the M , V and N diagrams for the entire structure. 13 Calculating M, V and N Diagrams 561 Walls: Due to the linearly distributed water pressure, the M diagram is a cubic, and the V diagram is a parabola"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure11.17-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure11.17-1.png",
        "caption": "Figure 11.17 (a) The magnitude and direction of the support reactions at A and B follow from (b) the shear force diagram. (c) The shear forces directly to the left and right of joint B. The support reaction at B is the same magnitude as the step change in the shear force diagram.",
        "texts": [
            " Note that p1 = 1 8 \u00d740 \u00d752 = 125 kNm and p2 = 1 8 \u00d740 \u00d722 = 20 kNm, or in other words, for each field: \u201cp = 1 8q 2\u201d. 11 Mathematical Description of the Relationship between Section Forces and Loading 455 The bending moment in field (1) is a maximum where the tangent to the M diagram is horizontal, or where dM dx = V = \u221240x + 84 = 0 \u21d2 x = 2.1 m. The maximum bending moment therefore occurs to the left of the middle of AB. Substituting x = 2.1 in the expression for M(1) gives the value of the maximum bending moment: Mmax = \u221220 \u00d7 2.12 + 84 \u00d7 2.1 = 88.2 kNm. The support reactions at A and B are shown in Figure 11.17a. Their magnitude and direction can be found directly from the shear force diagram (see Figure 11.17b). This is shown below for the support reaction at B. Figure 11.17c shows (only) the shear forces directly to the left and right of joint B. The vertical force equilibrium of joint B gives Bv = 116 + 80 = 196 kN. The support at B is carrying 116 kN from the left-hand field and 80 kN from the right-hand field. The support reaction at B is exactly the same magnitude as the \u201cstep change\u201d in the shear force diagram. The support reactions derived from the shear force diagram can be checked using the equilibrium of the beam as a whole. With R = 7 \u00d7 40 = 280 kN (see Figure 11"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001996_j.rcim.2015.12.004-Figure9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001996_j.rcim.2015.12.004-Figure9-1.png",
        "caption": "Fig. 9. Overlapped surfaces at different step-over distances. (a) No overlapping when dZw, where w is single-bead width. (b) Less overlapping when d*rdrw, where d* is the critical step-over distance. (c) Ideal overlapping when d\u00bcd*. (d) Excessive overlapping when dod*.",
        "texts": [
            " In order to optimise the welding parameters based on the desired bead geometry, a database of 3721 combinations (61 by 61 input matrix) of welding parameters has been built, with the wirefeed rate ranging from 2 to 8 m/min in the steps of 0.1 m/min and the travel speed ranging from 0.4 to 1 m/min in the steps of 0.01 m/min. An output matrix (61 by 61) of the predicted bead height and width is then generated through the trained neural network as shown in Fig. 8. This database is used to select the optimal welding parameters that will produce a given bead geometry. The step-over distance between paths plays an important role in achieving high surface quality and geometrical accuracy of the sliced layer, as shown in Fig. 9. Some preliminary investigations on multi-bead overlapping models have been made in recent years [22, 23]. Referring to our previous work [21], the tangent overlapping model (TOM) is described as follows. As shown in Fig. 10a, for two identical beads with individual width w and height h, Bead 1 is first deposited on the substrate, and Bead 2 is deposited next to Bead 1 with a step-over distance of d. When the area of the critical valley is equal to the overlapping area, the critical step-over d* is obtained"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001514_j.engfailanal.2015.06.004-Figure9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001514_j.engfailanal.2015.06.004-Figure9-1.png",
        "caption": "Fig. 9. Limiting lines related to the tooth thickness [42]: (a) parabolic curve under small crack sizes, (b) straight line, (c) parabolic curve under large crack sizes.",
        "texts": [
            " Based on Chen\u2019s model [36], Chen and Shao [39] developed a coupled planetary gear dynamic model with the tooth crack in which tooth cracks with different sizes and inclination angles were assumed in one tooth of the sun gear and planet gear, respectively. Chen and Shao [40] derived the TVMS of the internal gear pair on the basis of the PEP, and evaluated the ring gear crack on the reduction of the mesh stiffness. Considering the effects of the flexible ring gear rim and the tooth root crack on the mesh stiffness, Chen et al. [41] developed an improved TVMS calculation model for a planetary gear pair. By revising limiting lines related to the tooth thickness (see Fig. 9), Mohammed et al. [42] presented an improved method to calculate the TVMS of the gear with tooth root crack, and compared the effects of different crack sizes on the vibration responses of a cracked gear system. Their results show that the improved method calculates the TVMS more accurately than the traditional method does. On the foundation of Ref. [35], Ma et al. [43] proposed an improved mesh stiffness model for the cracked spur gear pair where two parabolic curves are used to simulate the crack propagation path and limiting line, respectively [42] (see Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure16.11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure16.11-1.png",
        "caption": "Figure 16.11 (a) Hinged beam with influence lines for (b) the vertical support reaction at B, (c) the bending moment at E, (d) the bending moment at G and (e) the shear force at G; (f) positive directions for bending moment and shear force.",
        "texts": [
            "10b) the paths through S1C and CS2 are parallel. After all, in the mechanism, segments S1C and CS2 can displace only with respect to one another, and cannot rotate with respect to one another. In the influence line for VS1 (Figure 16.10c) the paths through ABS1 and S1C are not parallel, as in accordance with the mechanism the segments ABS1 and S1C can rotate with respect to one another due to the hinge at S1. Example 3 \u2013 Various influence lines Figures 16.11b to 16.11e show various influence lines for the hinged beam in Figure 16.11a. The positive direction of the support reaction Bv at B is shown in Figure 16.11a. For the bending moment and the shear force, the positive directions are related to the xz coordinate system in Figure 16.11f. These influence lines are also shown in Section 16.1.2. There, we did not 16 Influence Lines 755 address the amount of arithmetic needed for a direct calculation from the equilibrium equation. The method of virtual work gives the same result with far less effort. If the correct mechanism is selected, and the virtual displacement is applied in such a way that the required quantity performs negative work over a unit displacement or unit rotation, the deformed mechanism is the influence line we are looking for"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003685_9780470692059-Figure3.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003685_9780470692059-Figure3.6-1.png",
        "caption": "Figure 3.6 Determination of the kinetic constant of a first-order kinetics reaction.",
        "texts": [
            "11) k = v/[A], which means that the dimensions of k are: M s\u22121/M = s\u22121. From the definition of the first-order kinetics, one can write: \u2212dS dt = kS (3.12) \u2212dS S = k dt (3.13) \u222b \u2212(dS/S) = \u222b k dt (3.14) \u2212 ln S = kt + cte (3.15) At t = 0, cte = \u2212 ln S0. Thus, Equation (3.15) can be written as \u2212 ln S = kt \u2212 ln S0 (3.16) Paul Sensecall Albani: \u201cchapter03\u201d \u2014 2007/6/18 \u2014 12:27 \u2014 page 29 \u2014 #9 or ln S = \u2212kt + ln S0 (3.17) Plotting ln S as a function of t yields a straight line with a slope equal to\u2212k and a y-intercept equal to ln S0 (Figure 3.6). Equation (3.17) can be modified and written as: ln S S0 = \u2212kt (3.18) S S0 = e\u2212kt (3.19) S = S0e\u2212kt (3.20) Thus, S decreases exponentially with time. When S = S0/2, we obtain ln 1 2 = \u2212kt1/2 (3.21) or ln 2 = kt1/2 (3.22) t1/2 = 0.693/k (3.23) t1/2 is the half-life of the kinetic reaction. The role of the enzyme in these types of reactions is to accelerate the rate of product formation. However, the substrate possesses a binding site on the enzyme, and so in order Paul Sensecall Albani: \u201cchapter03\u201d \u2014 2007/6/18 \u2014 12:27 \u2014 page 30 \u2014 #10 30 Principles and Applications of Fluorescence Spectroscopy to transform the substrate to a product, an enzyme\u2013substrate complex is formed"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure2.16-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure2.16-1.png",
        "caption": "Figure 2.16 The components of F .",
        "texts": [
            "15 shows that dx = \u22128 m, dy = +24 m, dz = +12 m. The x component of d is negative as it is pointing in the negative x 1 Remember the d of direction. ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM 2 Statics of a Particle 33 direction. The magnitude (length) d of d is d = \u221a d2 x + d2 y + d2 z = \u221a (\u22128 m)2 + (24 m)2 + (12 m)2 = 28 m. Using this information, we can calculate the components of F : Fx = F dx d = (35 kN) \u00d7 \u22128 m 28 m = \u221210 kN, Fy = F dy d = (35 kN) \u00d7 24 m 28 m = +30 kN, Fz = F dz d = (35 kN) \u00d7 12 m 28 m = +15 kN. Figure 2.16 shows the components of F as they are working on the foundation block. In order to determine the resultant of the forces on a particle in space, we first resolve all the forces into their x, y and z component, and then add all the associated components together. This is illustrated in an example. Example Figure 2.17 shows the schematised situation in a salvage operation. A shows the wreckage of a crashed lorry on a slope. People are trying to salvage the wreckage using cables AB and AC and winches in B and C"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001975_tro.2013.2281564-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001975_tro.2013.2281564-Figure1-1.png",
        "caption": "Fig. 1. Cable-driven continuum robotic structure under consideration with rigid-body discretization illustrated.",
        "texts": [
            " To find the net external moment M i,ex and force F i,ex on each body, the inertial and active forces and moments are added together. A key benefit of the method of virtual power is its ability to directly include moments and forces in the mechanics calculation: as long as a moment or force vector can be calculated, it can be added to the net external moment or force terms. In addition, this model may also be used to formulate a time-invariant model of the continuum robot static equilibrium by neglecting the inertial effects. The cable-driven continuum robot under consideration is illustrated in Fig. 1. An elastic core is the robot\u2019s backbone, along which are rigidly mounted disks. Three cables actuate the robot. This structure leads to a natural choice for the discretization into subsegments with each disk modeled as a rigid body. The mass and inertia of each rigid body are determined by the mass and inertia of the disk, and elastic core surrounding the disk, which is illustrated in Fig. 1. The kinematics assumes circular subsegment arcs separate each rigid body. Based on the elastic core properties, each subsegment\u2019s modeled elastic effects (bending and torsion) apply moments to the subsegment\u2019s two adjacent disks. Compressive and shear loads are neglected due to the incompressibility of the modeled elastic core compared with its bending and twist. Gravitational loading will be applied at each disk\u2019s center of mass. Actuation and cable-disk friction will be calculated as a force and moment at each disk\u2019s center of mass",
            " (51) As discussed in Section IV-D, the friction model uses an iterative solver to implement the simultaneous calculation of cable-disk contact forces and the tension along the length of the continuum robot. In order to determine the number of iterations for the cable tension/friction force convergence, the virtual power static equilibrium script is used to track the change in subsegment cable tension at each iteration. Fig. 5 shows the convergence of the contact forces for an eight disk robot with a prescribed tension of 10 N in cable 1 (see Fig. 1) at static equilibrium. This 10 N tension is the maximum tension applied in the subsequent analyses, and the maximum number of iterations will be needed for the maximum actuating tension. Table II shows the maximum percent error at each iteration for the cable subsegments (in each case, the maximum error was in subsegment 8). Three iterations were chosen\u2014the maximum subsegment tension error will be less than 1% in the highest actuation case. Fig. 6 illustrates the dynamic response of the \u03b2 curvatures of the continuum robot\u2019s second, fourth, sixth, and eighth subsegments with zero tension in the three cables",
            " Because the cable tensions are all zero, there is no contact force between the cabling and the disks, resulting in zero friction along the arm. Because the friction is the only dissipative force in the model, the energy of the continuum robot in the zero actuation simulation is constant. Fig. 7 shows the first oscillation of the continuum robot after release: the robot initially drops over 0.1342 s, then springs up over 0.1036 s. C. In-Plane Actuation Simulation Fig. 8 illustrates the dynamic responses of the selected \u03b2 curvatures of the virtual power dynamic model in response to a step input of a 5 N tension in cable 1 (see Fig. 1) from the initial condition of zero actuation static equilibrium. Because the ac- tuation remains in the x\u2013z plane due to the purely x-component of the hole radius rlcl,1 defined in (28), the dynamic response will remain in the x\u2013z plane and the \u03b3 curvature and the \u03b5 twist angle will remain zero. Unlike Fig. 6, the nonzero actuation will cause a contact force between the cabling and disk, leading to friction that dampens the oscillations. In addition, because of the use of step functions to actuate the virtual systems, preliminary simulations demonstrated the need to add numerical dampening terms to the simulation to reduce the speed of oscillations and improve their stability",
            " are defined in (52) and (53), shown below, where Mi,bD is the bending dampening moment, M i:(i\u22121),tD and M i:i,tD are the torsional dampening moments, and Cb and Ct are the bending and torsional mode dampening parameters. The subsegment\u2019s dampening moments may be formulated into a resultant moment for each disk\u2019s M i,Dmp using (54), shown below. This term is added to the other moment terms in (47). Suitable values for these terms were found to be Cb = 10\u22126N\u00b7m2 \u00b7s and Ct = 10\u22125N\u00b7m\u00b7s based on preliminary simulations Fig. 10 illustrates the dynamic responses of the \u03b2 and \u03b3 curvatures of a dynamic model in response to a step input of a 5 N tension in cable 2 (see Fig. 1) from the initial condition of zero actuation static equilibrium. Fig. 11 illustrates the dynamic response of the torsional twist angle \u03b5. Unlike the first two case studies, because of the out-of-plane actuation, the gravitational loading will cause twist along the length of the subsegments. For the scaling of the current continuum robot, the magnitude and impact of these torsional vibrations is relatively small compared with the impact of variation in curvatures. However, as macro-scale robots are considered and the distributed mass along the continuum arm increases, this effect will significantly impact the continuum robot shape"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure2.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure2.2-1.png",
        "caption": "Fig. 2.2. Revolute joint",
        "texts": [
            " In this section we will introduce some basic notations and definitions relevant to manipulator kinematics formulation. We will be concerned with the manipu lator structure, link, kinematic pair, kinematic chain, joint coordinates, world coordinates, direct and inverse kinematic problems and redundancy. Let us consider the manipulator shown in Fig. 2.1. It consists of n rigid bodies interconnected by joints. The joints are either revolute or prismatic 20 2 Manipulator Kinematic Model (sliding). Revolute joints enable rotational motion of one link with respect to the other (Figure 2.2). In prismatic joints the motion is translational (Figure 2.3). The mechanical structure of the mechanism depends on the type of the joints that are included in the given robot and the disposition of the joints. For example, the robot shown in Fig. 2.1 has six revolute degrees of freedom, with the second and the third always being in parallel (anthropomor phic manipulator). 2.2 Definitions 21 Different schematic representations have been introduced in order to de scribe manipulator configurations simply"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure9.24-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure9.24-1.png",
        "caption": "Fig. 9.24. Double-disc AFPM brushless motor for gearless elevators. Courtesy of Kone, Hyvinka\u0308a\u0308, Finland.",
        "texts": [
            " Specifications of single-sided AFPM brushless motors for gearless passenger elevators are given in Table 9.10 [113]. Laminated stators have from 96 to 120 slots with three phase short pitch winding and class F insulation. For example, the MX05 motor rated at 2.8 kW, 280 V, 18.7 Hz has the stator 306 9 Applications 9.7 Elevators 307 winding resistance R1 = 3.5 \u2126, stator winding reactance X1 = 10 \u2126, 2p = 20, sheave diameter 340 mm and weighs 180 kg. A double disc AFPM brushless motor for gearless elevators is shown in Fig. 9.24 [113]. Table 9.11 lists specifications data of double disc AFPM brushless motors rated from 58 to 315 kW [113]. The double disc configuration offers high output torque and compensation of axial forces [P105]. 308 9 Applications Ultra-flat PM micromotor, the so called penny-motor is shown in Figs 9.25 and 9.26 [154]. The thickness is 1.4 to 3.0 mm, outer diameter about 12 mm, torque constant up to 0.4 \u00b5Nm/mA and speed up to 60,000 rpm. A 400-\u00b5m eight pole PM and a three-strand, 110-\u00b5m disc shaped, lithographically produced stator winding have been used [154]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003774_j.1469-185x.1974.tb01299.x-FigureI-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003774_j.1469-185x.1974.tb01299.x-FigureI-1.png",
        "caption": "Fig. I . Illustrates the movement of fluid relative to a wave passing along the organelle in flagella and cilia. In the cilium we have only shown the recovery stroke, which is the phase of the beating cycle when the bending wave is prominent. The effective stroke, which is a near rigid clockwise gyration, causes the fluid movement to the right.",
        "texts": [
            " The bend in the cilium produced by these two angular movements at the base is then propagated up to the tip of the cilium during the recovery stroke, in such a way that the cilium moves back close to the cell surface in preparation for the beginning of the next effective stroke. Generally the recovery stroke takes longer than the effective stroke ; both strokes are active processes. Flagella (including most sperm tails) have the same internal structure as cilia, but are functionally different in that, as a broad generalization, flagella tend to move fluid along the flagellar axis and usually perpendicular to the surface of the organism, while cilia tend to move fluid parallel to the cell surface (Fig. I). T h e movements of both cilia and flagella are associated with the propagation of bending waves along the organelle ; while the bending pattern of cilia is highly asymmetric, the movement of a flagellum is associated with more symmetrical waves. Internally these organelles contain an \u2018 axoneme\u2019 array of nine double peripheral microtubules and two central microtubules (the 9+2 fibril complex); changes in shape of the organelles are believed to result from forces generated by interaction between the peripheral doublets that cause sliding of these fibrils relative to one another, and, since the fibrils are anchored together at the base and do not change in length, this fibril sliding deforms the 9 + 2 complex (Satir, 1968; Summers & Gibbons, 1971)",
            " Thus Gray & Hancock (1955) obtained the following relationship between the normal and tangential coefficients of resistance, which follows from the discussion in (u) : where (4) c\u2018 = log (2h/r0) - 4 \u2019 Fig. 10. Illustrates the Gray and Hancock theory. An element 6s of the organelle is moving with velocity V with components V, in the normal and V, in the tangential directions. To obtain the direction of the force exerted on the fluid, we double the V, coordinate, while keeping VT the same, then join up the rectangle to obtain the direction as shown. Fig. I I. Comparison between the velocity profile for an inert and self-propelling body. The penetration depth of noticeable fluid motion is S for the self-propelling body. where h is the flagellar wavelength and r,, is the radius of the flagellum. These two coefficients (slightly modified) have been used by Holwill & Burge (1963), Holwill (1966b), Holwill & Sleigh (1967), Brokaw (1970, 1971, 1972), Chwang & Wu (1971) and Schreiner (1971) to study movements of spermatozoa and flagella which differ from those of Gray & Hancock"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-FigureD.8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-FigureD.8-1.png",
        "caption": "FIGURE D.8 Partial mutual inductance for two wires in arbitrary relative directions.",
        "texts": [
            "28) where a1 = xs2 \u2212 xe1, a2 = xe2 \u2212 xe1 (D.29a) a3 = xe2 \u2212 xe1, a4 = xs2 \u2212 xs1 (D.29b) b1 = y2 \u2212 ys1, b2 = y2 \u2212 ye1 (D.29c) c1 = ze2 \u2212 z1, c2 = zs2 \u2212 z1 (D.29d) rkm\ud835\udcc1 = \u221a a2 k + b2 m + c2 \ud835\udcc1 . (D.29e) Again, \ud835\udf16 = 10\u221237 is a very small number to prevent singularities. 418 COMPUTATION OF PARTIAL COEFFICIENTS OF POTENTIAL D.2 PARTIAL POTENTIAL COEFFICIENT FORMULAS FOR NONORTHOGONAL GEOMETRIES D.2.1 Pp for Wire Filaments with an Arbitrary Direction In this section, the filaments are oriented in any mutual orientation as shown in Fig. D.8. This allows an arbitrary relative orientation of the filaments. Using the rotation and translation operations in (C.2) to (C.5), we can place the two wires in any location in the global rectangular coordinate system. Hence, this result will be ideally suited for the filament representation of non-orthogonal quadrilateral and hexahedral shapes. It is important to minimize the number of divisions and multiplications. For this reason, we present the formulation in a more compute friendly form Pp12 = 1 4\ud835\udf0b\ud835\udf00\ud835\udcc11\ud835\udcc12 [ 2(T2 + \ud835\udcc12)tanh\u22121 ( \ud835\udcc11 R1 + R2 ) + 2(T1 + \ud835\udcc11)tanh\u22121 ( \ud835\udcc12 R1 + R4 ) \u2212 2T2tanh\u22121 ( \ud835\udcc11 R3 + R4 ) \u22122T1tanh\u22121 ( \ud835\udcc12 R2 + R3 ) \u2212 Wd SE ] (D"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001491_0278364919842269-Figure16-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001491_0278364919842269-Figure16-1.png",
        "caption": "Fig. 16. A soft elastic rod has a hollow chamber offset from the central axis which is subject to some quasi-static and uniform pressure P(t). The chamber is offset from the central axis by a vector r(s) in the local frame. The pressure results in a force at the cap of the chamber, which also generates a moment due to the offset. The outer curve of the chamber has more surface area than the inner curve, resulting in a net distributed force and moment.",
        "texts": [
            " We consider a soft robot with one or more hollow actuation chambers offset from the neutral axis. There is a vector ri from the cross-section center of mass to the center of the ith chamber, which is similar to the tendon robot variable ri. We restrict our attention to cases with rsi = 0 so that the channel has a constant offset from the cross-section centroid. Fluid pressure is applied in the chamber, which results in a bending motion of the robot. We assume quasistatic fluid dynamics in the chamber so that there is a single uniform pressure Pi(t). The situation is illustrated in Figure 16. We note that robots with multiple sections connected in serial can be modeled by piecewise integration with breaks at each section transition, but we only consider a single section here. The distributed loading and point wrenches caused by pressurizing the chamber are similar to the effects of a tendon robot, but the two are not quite equivalent because tendon forces are transmitted along the tendon tangent line, while pressure forces act normal to a cross-sectional plane. We assume all chambers extend to the distal end of the robot, with flat chamber caps of area Ai so that the magnitude of the force on the cap is PiAi"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure5.4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure5.4-1.png",
        "caption": "Fig. 5.4. Stator winding patterns for 8-pole microgenerator: (a) 2 turns per pole; (b) 3-turns pole. Photo courtesy of Georgia Institute of Technology, Atlanta, GA, U.S.A.",
        "texts": [],
        "surrounding_texts": [
            "Fig. 5.3 shows an AFPM microgenerator developed at Georgia Institute of Technology, Atlanta, GA, U.S.A. [12, 13]. The stator uses interleaved, electroplated copper windings that are dielectrically isolated from a 1-mm thick NiFeMo substrate by a 5 \u00b5m polyimide layer. The rotor consists of a 2 to 12 pole, 500 \u00b5m thick, annular PM (OD = 10 mm, ID = 5 mm) and a 500 \u00b5m thick Hiperco 50 (FeCoV alloy) ring as a return path for the magnetic flux. SmCo PMs and Hiperco alloy have been used because microgenerators will be operating in high temperature environment, i.e., powered by gas-fueled turbine engines. Specifications are given in Table 5.1 [12]. A 16 W of mechanical\u2013to\u2013electrical power conversion and delivery of 8 W of d.c. power to a resistive load at a rotational speed of 305 krpm has been demonstrated [12]. The power density of the AFPM microgenerator was 59 W/cm3."
        ]
    },
    {
        "image_filename": "designv10_0_0002277_j.prostr.2016.02.039-Figure9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002277_j.prostr.2016.02.039-Figure9-1.png",
        "caption": "Fig. 9 \u2013 Metrological test result. Deviations in mm.",
        "texts": [
            " In this section, four manufacturing issues were defined:  Process accuracy [Vandenbroucke et al. (2007), Wang et al. (2013)]  Supports [Vandenbroucke et al. (2007), Hussein et al. (2013), Jhabvala et al. (2012), Wang et al. (2013)]  Surface Roughness [Vandenbroucke et al. (2007)]  Geometrical feasibility and possibilities [Ponche et al. (2012), Vayre et al. (2012)] The machine SLM\u00ae 125 HL from SLM Solutions GmbH was used to produce the final component. Figure 8 illustrates the manufactured component. The metrological test was done using 3D scanning, see Figure 9. The machine used was the Comet L3D from Steinbichler. The scanned image, saved in STL format, was then compared with the original STL used for the component\u2019s production. The deviation goes up to 0.8 mm in some areas of the component. This deviation is related with the SLM process itself. The high temperature gradients present during the manufacturing of the component leave residual stresses in the material which, in worst cases, can even rip some supports away from the platform. Miguel Seabra et al",
            " (2007), Wang et al. (2013)]  Supports [Vandenbroucke et al. (2007), Hussein et al. (2013), Jhabvala et al. (2012), Wang et al. (2013)]  Surface Roughness [Vandenbroucke et al. (2007)]  Geometrical feasibility and possibilities [Ponche et al. (2012), Vayre et al. (2012)] The machine SLM\u00ae 125 HL from SLM Solutions GmbH was used to produce the final component. Figure 8 illustrates the manufactured component. Fig. 8 \u2013 Optimised component manufactured. The metrological test was done using 3D scanning, see Figure 9. The machine used was the Comet L3D from Steinbichler. The scanned image, saved in STL format, was then compared with the original STL used for the component\u2019s production. The deviation goes up to 0.8 mm in some areas of the component. This deviation is related with the SLM process itself. The high temperature gradients present during the manufacturing of the component leave residual stresses in the material which, in worst cases, can even rip some supports away from the platform. Fig. 9 \u2013 Metrological test result. Deviations in mm. Author name / Structural Integrity Procedia 00 (2016) 000\u2013000 7 The mechanical tests were done with an Instron 3669 machine. In order to replicate the load cases in the machine, an interface set-up was designed and manufactured. Figure 10 illustrates the tests set-up for Load Case 2. Table 2 shows the comparison between the results obtained with FEM and experimental tests. The TO was successfully implemented and proved to be an effective way of taking advantage of the manufacturing freedom provided by SLM"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure16.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure16.3-1.png",
        "caption": "Figure 16.3 (a) Simply supported beam with (b) the positive direction assumed for shear force VC and (c) the influence line for the shear force at C.",
        "texts": [
            " If the load is to the right of C ( 1 2 <x \u2264 ) then the bending moment at C equals MC = +F \u2212 x \u00b7 1 2 = + 1 2F( \u2212 x). The variation of MC/F as a function of x, as shown graphically in Figure 16.2c, is referred to as the influence line for the bending moment at C. The influence line has its maximum in the middle. This means that the bending moment at C is a maximum when the point load F is at midspan: MC = ( + 1 4 ) \u00d7 F = + 1 4F . Example 3 \u2013 Influence line for a shear force For the same beam we will now determine the influence line for the shear force VC at C (see Figure 16.3a). Solution: The direction assumed for VC is shown in Figure 16.3b. The shear force VC can be found from the vertical force equilibrium of AC (or CB). If the load is to the left of C ( 1 2 <x \u2264 ) the equilibrium of AC gives VC = Av \u2212 F = F \u2212 x \u2212 F = \u2212F x . 16 Influence Lines 747 If the load is to the right of C ( 1 2 <x \u2264 ) then the shear force at C is equal to the support reaction Av at A: VC = Av = F \u2212 x . This determines the influence line for the shear force at C. The influence line is shown in Figure 16.3c. If the point load is to the left of C the influence line is negative. The shear force acts in the direction opposite to the one assumed in Figure 16.3b. The shear force at C is a maximum if the point load F is directly to the left or right of C: VC = (\u00b10.5) \u00d7 F = \u00b1 1 2F. A change of sign occurs at C. Figure 16.4 shows the influence lines for a hinged beam. The positive direction of the support reaction Bv at B is shown in Figure 16.4a. For the bending moment and the shear force, the positive directions according to the xz coordinate system are given in Figure 16.4f. We do not address the calculation of each of these influence lines in detail here",
            " Determine this minimum value. d. Using the influence line, determine the value of X when the uniformly distributed load q acts over the entire length of the beam. e. Where must the uniformly distributed load q be placed so that X is a maximum? Determine this maximum value. f. Where must the uniformly distributed load q be placed so that X is a minimum? Determine this minimum value. 16.3: 1\u20139 A point load F , a set of loads F1; F2; F3 and a uniformly distributed load q can move across the hinged beam in Figure 16.3. Use F = 40 kN, F1 = 20 kN, F2 = 50 kN, F3 = 30 kN and q = 8 kN/m. The same questions are asked for nine different quantities X: 1. X = vertical support reaction at B. 2. X = vertical support reaction at C. 3. X = vertical support reaction at D. 4. X = bending moment at B. 5. X = bending moment at C. 6. X = bending moment at E. 7. X = shear force directly to the left of C. 8. X = shear force directly to the right of C. 9. X = shear force at S2. Questions: a. Draw the influence line for X. b. Where must the load F be placed so that X is a maximum"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.53-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.53-1.png",
        "caption": "FIGURE 7.53. A jackknifed configuration of a car pulling a one-axle trailer.",
        "texts": [
            " We may normalize the dimensions such that the length of the trailer is 1. The 7. Steering Dynamics 435 positions of the car at the hinge point and the trailer at the center of its axle are shown by vectors r and s. Assuming r is a given differentiable function of time t, we would like to examine the behavior of the trailer by calculating s, and predict jackknifing. When the car is moving forward, we say the car and trailer are jackknifed if r\u0307 \u00b7 z < 0 (7.185) where z = r\u2212 s. (7.186) A jackknifed configuration is shown in Figure 7.53, while Figure 7.52 is showing an unjackknifed configuration. Mathematically, we want to know if the truck-trailer will jackknife for a 436 7. Steering Dynamics given path of motion r = r(t) and what conditions we must impose on r(t) to prevent jackknifing. The velocity of the trailer can be expressed by s\u0307 = c (r\u2212 s) (7.187) where c = r\u0307 \u00b7 z (7.188) and the unjackknifing condition is c > 0. (7.189) Assume the twice continuously differentiable function r is the path of car motion. If |z| = 1, and r has a radius of curvature R(t) > 1, and r\u0307(0) \u00b7 z(0) > 0 (7"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000376_j.jallcom.2014.06.172-Figure10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000376_j.jallcom.2014.06.172-Figure10-1.png",
        "caption": "Fig. 10. Diagram showing how the grain structure in the transverse (X\u2013Y) and longitudinal (X\u2013Z) planes relate to each other; (a) is a 3D representation constructed from micrographs and (b) shows this relationship schematically.",
        "texts": [
            " Overall the material consists of regions of long elongated grains in the build direction which are observed as almost equiaxed in the X\u2013Y plane separated by bands of fine grains which form a macro 1 mm \u2018diamond\u2019 pattern when viewed in the X\u2013Y plane as discussed earlier. The fine grains display some elongation, but in directions perpendicular to the build direction suggesting a different heat flow direction during material solidification. The overall relationship between these two regimes within the material can be seen in Fig. 10(a) and is shown schematically in Fig. 10(b). It is obvious that this 1 mm diamond pattern is not a naturally occurring feature of this material and can be traced back to the \u2018island\u2019 scan-strategy outlined in the experimental section; it is however less obvious how an area made up of 5 mm 5 mm islands could form a 1 mm 1 mm repeating pattern. In order to appreciate this, a series of 6 layers must be considered. On the first of these layers, each square of the island pattern is selectively melted. On the 2nd build layer this pattern is shifted by 1 mm in both its \u2018X\u2019 and \u2018Y\u2019 direction and then the island pattern is selectively melted"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure5.9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure5.9-1.png",
        "caption": "Fig. 5.9. Arthropoidal six-degree-of freedom robot",
        "texts": [
            " Numerical efficiency of robot kinematics and dynamics depends on the number of joints and relative dis position of the joint axes, etc. We pointed out in Sect. 5.2 that the floating-point arithmetics, including some trigonometric calculations and square roots, are carried out within specialized mathematical coprocessors. In order to have a clear insight into the numerical complexity of robot kinematics, let us consider two typical industrial robots: The first, shown in Fig. 2.1 (p. 19), has three degrees of freedom, while the second is a six degree-of freedom robot (Fig. 5.9). Both of them have kinematic chains for which the inverse kinematic solution in analytical form can be found. The number of numerical operations for both robots are given in Table 5.1 (see p. 173). We see 172 5 Microprocessor Implementation of Control Algorithms that the complexity is practically the same for the direct and inverse kinematic problem. Either the direct or inverse kinematic problem for this six-degree-of freedom robot can be solved within 5 to 10 ms by the use of an up-to-date 16-bit microcomputer"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000554_978-3-540-85629-0-Figure9.34-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000554_978-3-540-85629-0-Figure9.34-1.png",
        "caption": "Fig. 9.34. Trajectory based on motion primitives, approximated by a B-spline curve with 12 control points.",
        "texts": [
            " An alternative approach consists in approximating the path with an intrinsically continuous trajectory, for instance based on cubic (or higher degree) B-spline curves. In the example described below, the trajectory to be approximated is the one shown in Fig. 8.17, where the orientation is not taken into account. This trajectory can be approximated by interpolating the points\u23a1 \u23a3 qx qy qz \u23a4 \u23a6 = \u23a1 \u23a30 1.00 1.70 2.00 2.00 2.00 2.00 2.00 2.00 2.00 0 0.00 0.29 0.99 1.70 2.00 1.70 1.00 0.01 0.00 0 0.00 0.00 0.00 0.29 0.99 1.70 2.00 2.00 2.00 \u23a4 \u23a6 obtained by \u201csampling\u201d the trajectory itself, see Fig. 9.34 where the original trajectory, and the approximating cubic B-spline are reported. In particular, the knots (chosen according to a cord-length distribution in the interval [0, 1]) and the control points (computed with the algorithm of Sec. 8.4) which define the B-spline curve are u = [ 0, 0, 0, 0, 0.15, 0.26, 0.38, 0.49, 0.61, 0.73, 0.84, 0.99, 1, 1, 1, 1 ] and P = \u23a1 \u23a30 0.33 0.91 1.79 2.05 1.98 2.00 1.99 2.00 2.00 2.00 2.00 0 0 \u22120.07 0.20 1.00 1.77 2.10 1.78 0.92 0.33 0.00 0 0 0 \u22120.00 0.01 \u22120.05 0"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure6.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure6.1-1.png",
        "caption": "FIGURE 6.1. A revolute and a prismatic joint.",
        "texts": [
            " In this chapter, we review the analysis and design methods for such mechanisms. 6.1 Four-Bar Linkage An individual rigid member that can have relative motion with respect to all other members is called a link. A link may also be called a bar, body, arm, or a member. Any two or more links connected together, such that no relative motion can occur among them, are considered a single link. Two links are connected by a joint where their relative motion can be expressed by a single coordinate. Joints are typically revolute (rotary) or prismatic (translatory). Figure 6.1 illustrates a geometric form for a revolute and a prismatic joint. A revolute joint (R), is like a hinge that allows relative rotation between the two connected links. A prismatic joint (P), allows a relative translation between the two connected links. Relative rotation or translation, between two connected links by a revolute or prismatic joint, occurs about a line called axis of joint. The value of the single variable describing the relative position of two connected links at a joint is called the joint coordinate or joint variable"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001837_tro.2014.2309194-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001837_tro.2014.2309194-Figure3-1.png",
        "caption": "Fig. 3. Two-dimensional planar robot driven by two co-activated tendon servos for the curvature degree of freedom and a servo for insertion/retraction of the robot. The insertion and the curvature degrees of freedom are mechanically decoupled.",
        "texts": [
            " 3) Estimate Jacobian J\u0302k+1: Optimization in the previous step relies on continuous estimations of the Jacobian, and is estimated by solving the following optimization problem: minimize J\u0302k + 1 \u2016\u0394J\u0302\u20162 subject to \u0394xk = J\u0302k+1W\u0394yk J\u0302k+1 = J\u0302k + \u0394J\u0302 (10) where \u0394J\u0302 is the optimization variable, J\u0302k is the Jacobian matrix at time k, J\u0302k+1 is the new Jacobian estimate, and {\u0394xk ,\u0394yk} are measured displacements from actuator and end-effector sensors between the last Jacobian estimate at time k and the present time. We chose to minimize the Frobenius norm (element-wise L2-norm) of \u0394J\u0302 such that the column vectors of the Jacobian transition smoothly. A continuum manipulator was constructed to evaluate the model-less control algorithm (see Fig. 3). The robot is a 2-D planar manipulator, with a flexible backbone constructed of a polypropylene beam (length: 280 mm, cross-section: 0.8 mm \u00d7 12 mm, 2500\u20135400 psi, McMaster Carr, Santa Fe Springs, CA, USA). The backbone is segmented into eight equal subsections by 3-D printed plates acting as tendon guides. The curvature of the backbone is driven by two 0.6-mm diameter stranded steel cables (Sava Industries, Riverdale, NJ, USA) on either side and are equidistant from the central backbone. Each cable terminates on a tension sensor (LSP-5, range: 0\u20135N, Transducer Techniques, Temecula, CA, USA)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000717_0278364912452673-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000717_0278364912452673-Figure3-1.png",
        "caption": "Fig. 3. Schematic representation of the 3D-LIPM with point foot. The model comprises a point foot at position rankle, a point mass at position r with mass m and a massless telescoping leg link with an actuator that exerts a force f on the point mass that keeps it at constant height z0. The projection matrix P projects the point mass location onto the xy-plane. The gravitational acceleration vector is g.",
        "texts": [
            " The second model (Section 6) is obtained by adding a finite-sized foot and ankle actuation to the first model, enabling modulation of the center of pressure (CoP). The third model (Section 7) extends the second by the addition of a reaction mass and hip actuation, enabling the human-like use of rapid trunk (Horak and Nashner 1986; van der Burg et al. 2005) or arm motions (Roos et al. 2008; Pijnappels et al. 2010). The 3D-LIPM, described by Kajita and Tanie (1991); Kajita et al. (2001) and depicted in Figure 3, comprises a point mass with position r at the end of a telescoping massless mechanism (representing the leg), which is in contact with the flat ground. The point mass is kept on a horizontal plane by suitable generalized forces in the mechanism. at UCSF LIBRARY & CKM on August 26, 2014ijr.sagepub.comDownloaded from Torques may be exerted at the base of the pendulum. For this first model, however, we set all torques at the base to zero. Hence, the base of the pendulum can be seen as a point foot, with position rankle"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure3.30-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure3.30-1.png",
        "caption": "Fig. 3.30. External coordinates",
        "texts": [
            "64\u00bb can be written in a more suitable form as (3.68) where: [ FA(3 x 1) 1 RA (6 x 1) = ------ M A (3 X 1) (3.69) In order to define a functional robot motion we introduce the generalized position vector x \u2022. For a robot with six degrees offreedom we adopt the position 94 3 Dynamics and Dynamic Analysis of Manipulation Robots vector in the form (3.70) where XA, YA' ZA are Cartesian coordinates of the gripper point which determine its position and e, cp,!/J are Euler's angles which determine its orientation (Fig. 3.30). Let us consider a manipulation task in which the gripper cannot move freely, its motion being imposed by this constraint. Here we have a closed chain. Imposed gripper motion reduces the number of degrees of freedom (d.o.f). Let nr be this reduced number of d.oJ. If n is the number of d.o.f, it holds that nr ~ n. The equality holds when there is no constraint. We now introduce nr free and independent parameters U 1, ... , Un which define the constrained position of the gripper. The reduced position vector X r is introduced (3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003786_1045389x10369718-Figure15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003786_1045389x10369718-Figure15-1.png",
        "caption": "Figure 15. Structure of the grafted elastomer. Reproduced with permission from Institute of Physics Publishing Ltd, http://dx.doi.org/ 10.1088/0964-1726/17/2/025007.",
        "texts": [
            " Electronic electroactive polymers include electrostrictive graft elastomers, electrostrictive cellulose papers, electro-viscoelastic elastomers, ferroelectric electroactive polymers, dielectric electroactive elastomers, all-organic composites, liquid crystal elastomers, and electroactive conducting polymers. Ionic electroactive polymers consist of ionic polymer-metal composites, ionic polymeric gels, and CNTs. at University of Sussex Library on February 20, 2013jim.sagepub.comDownloaded from Electrostrictive graft elastomer consists of two components, a flexible backbone macromolecule and a grafted polymer. Metal electrodes are fabricated on the opposite sides of the elastomer film (Su et al., 1999, 2003). The molecular structure of the electrostrictive graft elastomer is shown in Figure 15. Both the backbone and grafted unit have polarized monomers containing atoms with partial electric charges. The grafted polar phase, which is crystalline, provides moieties to respond to an applied electric field and the cross-linking sites provide the elastomer system. The electric charges generate dipole moments. As a result, an electric field applied to the polymer may therefore cause the rotation of the dipole moment. With increasing intensity of the electric field, the monomer dipole moments of the crystal graft unit accumulate, and finally the electrical field can rotate the whole unit"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure15.13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure15.13-1.png",
        "caption": "Figure 15.13 The displacement of point i due to a translation ux0; uy0 and a large rotation \u03d5z0.",
        "texts": [
            " The body is in equilibrium when the following condition is satisfied: G = \u03bb1 \u2211 i Fxi + \u03bb2 \u2211 i Fyi + \u03bb3 \u2211 i (xiFyi \u2212 yiFxi) = 0 for each arbitrary choice of \u03bb1; \u03bb2; \u03bb3 (not all concurrently zero). The displacement of a rigid body in a plane is defined by the displacement of, for example, point O to O\u2032 and a rotation about O. Assume that the components of the translation (displacement) are ux0; uy0 and the rotation is \u03d5z0. Instead of using its coordinates xi ; yi , we can define the location of an arbitrary point i also by its angle \u03b1i and the radius ri . The displacement of point i is (see Figure 15.13) uxi = ux0 \u2212 a, uyi = uy0 + b. a and b are the result of the rotation \u03d5z0. Due to the rotation, point i moves through an arc length \u03d5z0ri along the circle with radius ri and centre O\u2032. For a and b it applies that a = ri { cos \u03b1i \u2212 cos(\u03b1i + \u03d5z0) } , b = ri { sin(\u03b1i + \u03d5z0) \u2212 sin \u03b1i } . The expressions are much simplified when the rotation is small. If \u03d5z0 1 722 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM the circle can be replaced by its tangent (see Figure 15.14). The displacement of point i due to the rotation is then \u03d5z0ri with the following components: a = \u03d5z0ri sin \u03b1i = \u03d5z0yi, b = \u03d5z0ri cos \u03b1i = \u03d5z0xi",
            " Conclusion: Due to a translation ux0; uy0 and a small rotation \u03d5z0 of the uxi = ux0 \u2212 a = ux0 \u2212 \u03d5z0yi, uyi = ux0 + b = uy0 + \u03d5z0xi. Note that the geometric relationships between the various displacement quantities are linear for small rotations. When a body is given a virtual displacement, the virtual work performed by all forces acting on it is \u03b4A = \u2211 i Fxi\u03b4uxi + \u2211 i Fyi\u03b4uyi. The virtual displacements \u03b4uxi ; \u03b4uyi of point i can be expressed by three independent virtual displacements \u03b4ux0; \u03b4uy0; \u03b4\u03d5z0 of the body. The equa- Figure 15.13 The displacement of point i due to a translation ux0; uy0 and a large rotation \u03d5z0. body the displacement of point i is 15 Virtual Work 723 tion for \u03b4A assumes the form of expression G (see Section 15.3.1) only if the relationships between \u03b4uxi; \u03b4uyi and \u03b4ux0; \u03b4uy0; \u03b4\u03d5z0 are linear. The geometric relationships appeared to be linear only for bodies subject to minor rotations. We can also say that the virtual displacements have to be very small. In that case, the virtual work performed by the force at point i is Fxi\u03b4uxi + Fyi\u03b4uyi = Fxi(\u03b4ux0 \u2212 yi\u03b4\u03d5z0) + Fyi(\u03b4uy0 + xi\u03b4\u03d5z0)",
            " VOLUME 1: EQUILIBRIUM the pair of forces in Figure 15.15 we find \u03b4A = \u2212F \u00b7 b\u03b4\u03d5z0 + F \u00b7 (a + b)\u03b4\u03d5z0 = Fa \u00b7 \u03b4\u03d5z0 = Tz \u00b7 \u03b4\u03d5z0. The work performed by a couple is equal to the product of couple and rotation. The work is positive when the couple and rotation are in the same direction. In deriving the virtual work equation, we found that the geometric relationships had to be linear in the virtual displacements. Mathematically, this means that the first-order variation has to be assumed for these virtual displacements. We previously deduced (see Figure 15.13) that the following applies for the displacement of an arbitrary point i, due to a translation ux0; uy0 and a rotation \u03d5z0: uxi = ux0 \u2212 ri { cos \u03b1i \u2212 cos(\u03b1i + \u03d5z0) } , uyi = uy0 + ri { sin(\u03b1i + \u03d5z0) \u2212 sin \u03b1i } . The first-order variation of uxi is defined as \u03b4uxi = \u2202uxi \u2202ux0 \u03b4ux0 + \u2202uxi \u2202\u03d5z0 \u03b4\u03d5z0. In the same way, the first-order variation of uyi is defined as \u03b4uyi = \u2202uyi \u2202uy0 \u03b4uy0 + \u2202uyi \u2202\u03d5z0 \u03b4\u03d5z0. 15 Virtual Work 725 Elaborating these expressions (for \u03d5z0 = 0) indeed leads to \u03b4uxi = \u03b4ux0 \u2212 ri sin \u03b1i\u03b4\u03d5z0 = \u03b4ux0 \u2212 yi\u03b4\u03d5z0, \u03b4uyi = \u03b4uy0 + ri cos \u03b1i\u03b4\u03d5z0 = \u03b4uy0 + xi\u03b4\u03d5z0"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001236_j.euromechsol.2007.11.005-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001236_j.euromechsol.2007.11.005-Figure8-1.png",
        "caption": "Fig. 8. Model of the spur gear train.",
        "texts": [
            " , etc.} of the tooth. The width of contact changes as the line of contact moves and is evaluated instantaneously by: W1 = W \u2212 wb (14) The tooth and gearmesh stiffness of the gear pair is also calculated according to Eqs. (1), (6), (9) and (11) by taking into account the geometric changes due to the tooth breakage and updating the values of width, cross sectional area and area moment of inertia (Fig. 7(B)). A one-stage spur gear set is considered. It is divided into two rigid blocks as presented in Fig. 8. Each block has four degrees of freedom (two translations and two rotations). Pinion (12) and gear (21) are assumed to be rigid bodies and the shafts with torsional rigidity. Shafts are supported by bearings modelled each one by two linear spring. The gearmesh stiffness is modelled by linear spring acting on the line of action of the meshing teeth. The expression of the displacement on the line of action is expressed by (Chaari et al., 2006a): \u03b4(t) = (x1 \u2212 x2) sin(\u03b1) + (y1 \u2212 y2) cos(\u03b1) + \u03b812rb12 + \u03b821rb21 (15) xi and yi are the translations of block i (i = 1,2)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure13.9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure13.9-1.png",
        "caption": "FIGURE 13.9. A bicycle vibrating model of a vehicle.",
        "texts": [
            "200) u2 = \u2219 1 0.157 58 \u00b8 (13.201) Therefore, the free vibrations of the quarter car is x = nX i=1 ui (Ai sin\u03c9it+Bi cos\u03c9it) i = 1, 2 (13.202)\u2219 xs xu \u00b8 = \u2219 1 \u22123.1729\u00d7 10\u22123 \u00b8 (A1 sin 8.8671t+B1 cos 8.8671t) + \u2219 1 0.157 58 \u00b8 (A2 sin 55.269t+B2 cos 55.269t) (13.203) 13.4 Bicycle Car and Body Pitch Mode Quarter car model is excellent to examine and optimize the body bounce mode of vibrations. However, we may expand the vibrating model of a vehicle to include pitch and other modes of vibrations as well. Figure 13.9 illustrates a bicycle vibrating model of a vehicle. This model includes the body bounce x, body pitch \u03b8, wheels hop x1 and x2 and independent road excitations y1 and y2. The equations of motion for the bicycle vibrating model of a vehicle are 854 13. Vehicle Vibrations as follow. mx\u0308+ c1 \u00b3 x\u0307\u2212 x\u03071 \u2212 a1\u03b8\u0307 \u00b4 + c2 \u00b3 x\u0307\u2212 x\u03072 + a2\u03b8\u0307 \u00b4 +k1 (x\u2212 x1 \u2212 a1\u03b8) + k2 (x\u2212 x2 + a2\u03b8) = 0 (13.204) Iz \u03b8\u0308 \u2212 a1c1 \u00b3 x\u0307\u2212 x\u03071 \u2212 a1\u03b8\u0307 \u00b4 + a2c2 \u00b3 x\u0307\u2212 x\u03072 + a2\u03b8\u0307 \u00b4 \u2212a1k1 (x\u2212 x1 \u2212 a1\u03b8) + a2k2 (x\u2212 x2 + a2\u03b8) = 0 (13.205) m1x\u03081 \u2212 c1 \u00b3 x\u0307\u2212 x\u03071 \u2212 a1\u03b8\u0307 \u00b4 + kt1 (x1 \u2212 y1) \u2212k1 (x\u2212 x1 \u2212 a1\u03b8) = 0 (13"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000554_978-3-540-85629-0-Figure5.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000554_978-3-540-85629-0-Figure5.3-1.png",
        "caption": "Fig. 5.3. Translation operations on a modified trapezoidal.",
        "texts": [
            "7 has been determined analytically for t \u2208 [0, T/2], being T the duration of the trajectory, and h the desired displacement: q(t) = \u23a7\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23a8 \u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23a9 h 2 + \u03c0 [ 2t T \u2212 1 2\u03c0 sin ( 4\u03c0t T )] , 0 \u2264 t < T 8 h 2 + \u03c0 [ 1 4 \u2212 1 2\u03c0 + 2 T ( t \u2212 T 8 ) + 4\u03c0 T 2 ( t \u2212 T 8 )2 ] , T 8 \u2264 t < 3 8T h 2 + \u03c0 [ \u2212\u03c0 2 + 2(1 + \u03c0) t T \u2212 1 2\u03c0 sin ( 4\u03c0 T ( t \u2212 T 4 ))] , 3 8T \u2264 t \u2264 T 2 . The second part of the trajectory can be obtained by exploiting its symmetry and the rules stated above. In particular, since the trajectory is symmetric with respect to the point (T/2, h/2), it is convenient to translate q(t) so that the entire trajectory will be centered on (0, 0) (and accordingly t \u2208 [\u2212T/2, T/2] and q(t) \u2208 [\u2212h/2, h/2]). The new trajectory obtained by translation, that is q\u2032(t) = q ( t \u2212 (\u2212T/2) ) \u2212 h/2, (5.1) is shown in Fig. 5.3. At this point, the second half of the trajectory can be easily deduced by reflecting q\u2032(t) through the origin (by applying a reflection about the t-axis and then one about the q-axis), i.e. q\u2032\u2032(t) = \u2212q\u2032(\u2212t) as shown in Fig. 5.4. Now, it is necessary to translate again q\u2032\u2032(t), with a dual operation with respect to (5.1): q\u2032\u2032\u2032(t) = q\u2032\u2032 ( t \u2212 (T/2) ) + h/2. 226 5 Operations on Trajectories Following this procedure, the expression of the modified trapezoidal trajectory for t \u2208 [T/2, T ] results q(t)= \u23a7\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23a8 \u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23a9 h + h 2 + \u03c0 [ \u03c0 2 + 2(1+\u03c0) t \u2212 T T \u2212 1 2\u03c0 sin ( 4\u03c0 T ( t \u2212 3T 4 ))] , 1 2T \u2264 t < 5 8T h + h 2 + \u03c0 [ \u22121 4 + 1 2\u03c0 + 2 T ( t \u2212 7T 8 ) \u2212 4\u03c0 T 2 ( t \u2212 7T 8 )2 ] , 5 8T \u2264 t < 7 8T h + h 2 + \u03c0 [ 2(t \u2212 T ) T \u2212 1 2\u03c0 sin ( 4\u03c0 T (t \u2212 T ) )] , 7 8T \u2264 t \u2264 T"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002658_j.addma.2019.101034-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002658_j.addma.2019.101034-Figure2-1.png",
        "caption": "Fig. 2. (a) Overview of the scanning strategy with the optimized parameters during SLM. (b) Schematic representation of sample sectioning from the SLMed parts, with the coordinate system showing the plane orientations.",
        "texts": [
            " AlSi10Mg and NI-AlSi10Mg alloys were produced in a SLM Solution SLM125HL selective laser melting system, which employed an ytterbium fibre laser with a wavelength of 1.06 \u03bcm and a maximum power of 400W. Prior to the SLM processing, the powder was dried at 60 \u2103 in a vacuum furnace for 12 h and then sieved to remove agglomerations with the size over 75 \u03bcm. Based on a series of preliminary trials, the building parameters were optimized as follows: laser power (P)= 300W, scanning speed (V) =1.65m/s, hatching spacing (H)= 130 \u03bcm, layer thickness (L)= 30 \u03bcm, and a zigzag scan pattern with 67\u00b0 rotation between the adjacent layers (Fig. 2a). Samples were built onto an Al platform preheated to 200 \u2103 and the building process was performed under an inert and high purity (\u2265 99.99 %) argon atmosphere with the oxygen concentration below 0.05 vol.%. Chemical compositions of the SLMed alloys were analysed by Spectrometer Services PTY. Ltd. using inductively coupled plasma (ICP) emission spectroscopy. The densities of the SLMed alloys were measured by the Archimedes method. The measurement used the theoretical density of 2.67 g/cm3 for both alloys. For each alloy, the measurement was repeated at least 3 times to ensure reproducibility. Table 1 lists the composition and relative densities of both SLMed alloys. Both alloys exhibited high relative densities (\u2265 99.6 %) after SLM. For anisotropy examination, samples were sectioned along both longitudinal (YZ or XZ) and transverse (XY) directions from the SLMed parts as schematically shown in Fig. 2b. A Reichert-Jung POLYVAR MET optical microscope and a JEOL 7100 F field emission scanning electron microscope (FESEM) were used to characterize the microstructures of (i) the AlSi10Mg and NIAlSi10Mg powders and (ii) the SLMed samples. Block SLMed samples were cut across both transverse (XY plane) and longitudinal (YZ plane) planes from the SLMed parts, then mechanically ground and polished using standard metallographic methods down to 1 \u03bcm diamond suspension. Samples were subsequently etched with the Keller\u2019s reagent (2 ml HF, 5ml HNO3, 3ml HCL and 190ml H2O) to reveal the microstructures"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure14.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure14.1-1.png",
        "caption": "Figure 14.1 (a) Cable AB loaded by two vertical forces. (b) The isolated cable AB. (c) The isolated cable part AE.",
        "texts": [
            "5, we calculate the cable shape due to its dead weight. The dead weight is a force measured along the length of the cable. The cable shape resulting from the dead weight is a catenary. If the (vertical) sag of the cable with respect to its chord is small compared with the (horizontal) span, then the catenary can be approximated by the simpler parabola. Finally, in Section 14.1.6, we present a number of examples. Calculating the cable shape and cable forces from the equilibrium is illustrated using the cable in Figure 14.1a, supported at the fixed points A and B, and loaded by the vertical forces FC = 75 kN and FD = 30 kN. The cable has a (horizontal) span = 60 m and a difference in elevation between supports A and B of h = 9 m. The distances between the supports and the lines of action of the forces are shown in the figure. The z coordinate of the cable of point E is also given: zE = 22 m. The dead weight of the cable is so small compared to the load that it can be ignored. 634 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Questions: a. Determine the cable shape, or in other words, the z coordinates of C and D where kinks occur in the cable. b. Determine the maximum and minimum cable force. Solution: a. With fully flexible cables, no bending moments can be transferred, and the cable remains straight between the places where forces are applied. Each straight part of the cable can be seen as a line element subject to a tensile force N , the cable force. In Figure 14.1b the cable has been isolated. There are four unknown support reactions: Ah, Av, Bh and Bv. There are three equilibrium equations: \u2211 Fx = \u2212Ah + Bh = 0, (1) \u2211 Fz = \u2212Av \u2212 Bv + (75 kN) + (30 kN) = 0, (2) \u2211 Ty |B = +Ah(9 m) \u2212 Av(60 m) + (75 kN)(40 m) + (30 kN)(20 m) = 0. (3) For a unique solution to these three equations with four unknowns, we need a fourth equation. This is found from the moment equilibrium of the part of the cable to the right or left of E, the point where the z coordinate of the cable is given. Here, we select the part to the left of E, as this equilibrium equation contains only the unknowns Ah and Av and in combination with Equation (3) leads to the quicker result (see Figure 14.1c): \u2211 Ty |E = +Ah(22 m) \u2212 Av(30 m) + (75 kN)(10 m) = 0. (4) From (3) and (4) we find Ah = 60 kN, Figure 14.1 (a) Cable AB loaded by two vertical forces. (b) The isolated cable AB. (c) The isolated cable part AE. 14 Cables, Lines of Force and Structural Shapes 635 Av = 69 kN. From (1) and (2) we then find Bh = 60 kN, Bv = 36 kN. If the z coordinate of E (or of another point on the cable) is not given, the result remains undetermined, and the cable can assume various shapes, such as the two dotted shapes in Figure 14.2. The final shape of the cable is determined by the length of the cable. The approach via a given cable length is considerably more complicated than that with a locally given z coordinate of the cable, such as that at point E"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure2.9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure2.9-1.png",
        "caption": "Fig. 2.9. Denavit-Hartenberg kinematic parameters for a revolute kinematic pair",
        "texts": [
            " Let us now consider how the link coordinate frames are assigned to the links according to the Denavit-Hartenberg approach, together with kinematic par ameters which are introduced here. Consider a simple open kinematic chain with n links. Each link is characterized by two dimensions: the common normal distance ai (along the common normal between axes of joint i and (i + 1)), and the twist angle (Xi between these axes in the plane perpendicular to ai \u2022 Each joint axis has two normals to it: ai -1 and ai (Fig. 2.9). The relative position of these normals along the axis of joint i is given by di \u2022 2.3 Direct Kinematic Problem 27 Denote by OjXjYjZj the local coordinate system assigned to link i. We will first consider revolute joints (Fig. 2.9). The origin of the coordinate frame of link i is set to be at the intersection of the common normal between the axis of joint (i+ 1) and i, and the axis of joint i. In the case of intersecting joint axes, the origin is set to be at the point of intersection of the joint axes. If the axes are parallel, the origin is chosen to make the joint distance zero for the next link whose coordinate origin is defined. The axes of the link coordinate system OjXjYjZj are to be selected in the following way. The Zj axis of system i should coincide with the axis of joint (i+ 1), about which rotation qj+l is performed. The Xj axis will be aligned with any common normal which exists (usually the normal between axes of joint i and (i + 1)) and is directed from joint i to joint (i + 1). In the case of intersecting joint axes, the Xj axis is chosen to be parallel or anti parallel to the vector-cross product Zj-l x Zj' The Yj axis satisfies Xi x Yi =Zj' The joint coordinate qj for a revolute joint is now defined as the angle between axes Xi _ 1 and Xj (Fig. 2.9). It is zero when these axes are parallel and have the same direction. The twist angle lXi is measured from axis Zj-l to Zj' i.e. as a rotation about Xj axis. Let us now consider the prismatic joint (Fig. 2.10). Here, the distance dj becomes joint variable qj, while parameter aj has no meaning and it is set to zero. The origin of the coordinate system corresponding to the sliding joint i is chosen to coincide with the next defined link origin. The Zj axis is aligned with the axis of joint (i+ 1)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002149_j.matdes.2015.06.063-Figure14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002149_j.matdes.2015.06.063-Figure14-1.png",
        "caption": "Fig. 14. Strategy of regional re-melting for fatigue applications.",
        "texts": [
            " For a reliable fatigue life, residual stresses must be relieved. The role of residual stresses in this study can be neglected as a stress-relieving post-process heat treatment was performed on all the specimens. Based on the effect of near-surface pores on the stress concentration and, therefore, on the fatigue crack initiation, a change in scanning strategy in the SLM process is proposed. The current scanning strategy is a combination of hatch-islands for scanning of core and contour scanning for scanning at the boundaries (Fig. 14 left). An additional regional re-melting like the one shown in Fig. 14 (right) would be helpful for decreasing the pores in the 200 lm re-melted region. It would also smoothen the top surface before application of powder in the next layer which results in higher density. Such a combination of scanning strategy will help reduce the fatigue scatter with a slight increase in energy density. Capability of computed tomography was investigated for characterization of the internal structure of additive manufactured parts. Materials with two different levels of material defects were studied"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure6.8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure6.8-1.png",
        "caption": "Fig. 6.8 Motion of a rigid body with constant spatial velocity of vo = [0.3 0 1]T (m/s), \u03c9 = [1 0 0]T (rad/s). State of every 0.3 second is displayed during the motion of 10 seconds. A trajectory of a rigid body with constant spatial velocity generally becomes a spiral.",
        "texts": [
            " However, such case gives a simple translation without rotation, thus we get e\u039et = [ E vot 0 0 0 1 ] . (6.13) When the position and orientation of a rigid body was (p(t),R(t)) at time t and its spatial velocity was given as \u039e, its configuration at t+\u0394t is given by [ R(t+\u0394t) p(t+\u0394t) 0 0 0 1 ] = e\u039e\u0394t [ R(t) p(t) 0 0 0 1 ] . (6.14) Equations (6.12),(6.13) and (6.14) can be coded as a Matlab program shown in Fig. 6.7. In this program a new variable uLINK.vo is introduced to hold the linear part of the spatial velocity vo. Figure 6.8 shows the motion of a rigid body under a constant spatial velocity calculated by the program of Fig. 6.7. In this section, we discuss the dynamics and the simulation method of a rigid body in 3D space. It is well known that the dynamics of rigid body can be separated into the translation of the center of mass (CoM) and the rotation around it. Thus the rigid body dynamics in 3D space is given by the Newton\u2019s equation for 2 For more detailed derivation, see the textbook of Murray, Li and Sastry [101](pp"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure14.18-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure14.18-1.png",
        "caption": "Figure 14.18 Cable with distributed load (force per horizontally measured length).",
        "texts": [
            "14a: Av = 24 kN (\u2191). Check: Figure 14.16 shows the forces acting on the pulley at B. The forces NCB and G, both 20 kN, are known. The force equilibrium can be used to find the support reactions at B: Bh = H = 16 kN (\u2192), Bv = (12 kN) + (20 kN) = 32 kN) (\u2191). Check: The horizontal support reaction at B is equal to H . The vertical support reaction is equal to the support reaction at B of the beam in Figure 14.14a, increased with the vertical cable force G. The support reactions at A and B are shown in Figure 14.17. Figure 14.18 shows a cable subject to a distributed load qz = qz(x). The cable shape is z = z(x). In Figure 14.19, a small cable element of length x has been isolated from the deformed cable and blown up. As x \u2192 0 the distributed load qz on the cable element can be considered to be uniformly distributed. Assume the cable force at the left-hand section is a tensile force N , with a horizontal component H and a vertical component V .1 The cable force at the right-hand section could have changed with respect to magnitude and 1 V is not the transverse force in the cable, but the vertical component of the tensile force N "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure8.11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure8.11-1.png",
        "caption": "Fig. 8.11. Configuration of the AFPM machine with shaft-integral fan.",
        "texts": [
            " Depending on the operation conditions obtained on site, either an air-blast or a suction fan may be used as shown in Fig. 8.10. In both cases, intake and/or discharge ducts are needed to direct and condition the air flow. Since the inlet air temperature, for a given volumetric flow rate, has a significant effect on the machine temperature, this cooling arrangement can also help prevent recirculation of hot air should the machine operate in a confined space (e.g. small machine room). For high speed AFPM machines, a shaft-integral fan may be 266 8 Cooling and Heat Transfer a good option. Fig. 8.11 shows the assembly of a large power AFPM machine developed in the Department of Electrical and Electronic Engineering at the University of Stellenbosch, South Africa, in which the rotor hub part serves as both cooling fan and supporting structure for the rotor discs. It can be seen that the \u201cblades\u201d of the hub are not curved as the machine may operate in both directions of rotation. Heat Pipes The concept of a passive two-phase heat transfer device capable of transferring large amount of heat with a minimal temperature drop was introduced by R"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure7.13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure7.13-1.png",
        "caption": "Fig. 7.13. The main components of IRCC",
        "texts": [
            " The disadvantages are that the mechanical parts being assembled must usually be specially prepared, resulting in the impossibility of reprogramming some other task which thereby limits the use of the robot. RCC (Figure 7.11) absorbs the errors of the mutual position of the as sembling parts independently compensating both lateral and angular errors (Figure 7.12). 7.2 Environment Sensors 219 Lateral reaction forces act on RCC so that it only translates the gripper, and when the tip of the shaft comes to the hole contact forces produce the torque that acts only upon the angular part of RCC thus assisting the insertion. Further development of RCC led to IRCC (Instrumented RCC, Fig. 7.13), with the main purpose of incorporating passive compliance, but equipped with both force and torque sensors that provide useful information for the control system, enhancing the operating capabilities of the robot [6]. The principle of measurement is similar to that of the sensor from the Figure 7.9, differing only in the method of measuring deformations: instead of inductive, optical sensors are used. Sources of infrared light are fixed to one plate in the vicinity of the elastic elements, and two-dimensional sensors detect the beam"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure9.22-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure9.22-1.png",
        "caption": "FIGURE 9.22. A pendulum with a vibrating pivot.",
        "texts": [],
        "surrounding_texts": [
            "Dynamics of a rigid vehicle may be considered as the motion of a rigid body with respect to a fixed global coordinate frame. The principles of Newton and Euler equations of motion that describe the translational and rotational motion of the rigid body are reviewed in this chapter. 9.1 Force and Moment In Newtonian dynamics, the forces acting on a system of connected rigid bodied can be divided into internal and external forces. Internal forces are acting between connected bodies, and external forces are acting from outside of the system. An external force can be a contact force, such as the traction force at the tireprint of a driving wheel, or a body force, such as the gravitational force on the vehicle\u2019s body. External forces and moments are called load, and a set of forces and moments acting on a rigid body, such as forces and moments on the vehicle shown in Figure 9.1, is called a force system. The resultant or total force F is the sum of all the external forces acting on a body, and the resultant or 522 9. Applied Dynamics total moment M is the sum of all the moments of the external forces. F = X i Fi (9.1) M = X i Mi (9.2) Consider a force F acting on a point P at rP . The moment of the force about a directional line l passing through the origin is Ml = lu\u0302 \u00b7 (rP \u00d7F) (9.3) where u\u0302 is a unit vector on l. The moment of the force F, about a point Q at rQ is MQ = (rP \u2212 rQ)\u00d7F (9.4) so, the moment of F about the origin is M = rP \u00d7F. (9.5) The moment of a force may also be called torque or moment. The effect of a force system is equivalent to the effect of the resultant force and resultant moment of the force system. Any two force systems are equivalent if their resultant forces and resultant moments are equal. If the resultant force of a force system is zero, the resultant moment of the force system is independent of the origin of the coordinate frame. Such a resultant moment is called couple. When a force system is reduced to a resultant FP andMP with respect to a reference point P , we may change the reference point to another point Q and find the new resultants as FQ = FP (9.6) MQ = MP + (rP \u2212 rQ)\u00d7FP = MP + QrP \u00d7FP . (9.7) The momentum of a moving rigid body is a vector quantity equal to the total mass of the body times the translational velocity of the mass center of the body. p = mv (9.8) The momentum p is also called translational momentum or linear momentum. Consider a rigid body with momentum p. The moment of momentum, L, about a directional line l passing through the origin is Ll = lu\u0302 \u00b7 (rC \u00d7 p) (9.9) 9. Applied Dynamics 523 where u\u0302 is a unit vector indicating the direction of the line, and rC is the position vector of the mass center C. The moment of momentum about the origin is L = rC \u00d7 p. (9.10) The moment of momentum L is also called angular momentum. A bounded vector is a vector fixed at a point in space. A sliding or line vector is a vector free to slide on its line of action. A free vector is a vector that may move to any point as long as it keeps its direction. Force is a sliding vector and couple is a free vector. However, the moment of a force is dependent on the distance between the origin of the coordinate frame and the line of action. The application of a force system is emphasized by Newton\u2019s second and third laws of motion. The second law of motion, also called the Newton\u2019s equation of motion, states that the global rate of change of linear momentum is proportional to the global applied force. GF = Gd dt Gp = Gd dt \u00a1 mGv \u00a2 (9.11) The third law of motion states that the action and reaction forces acting between two bodies are equal and opposite. The second law of motion can be expanded to include rotational motions. Hence, the second law of motion also states that the global rate of change of angular momentum is proportional to the global applied moment. GM = Gd dt GL (9.12) Proof. Differentiating from angular momentum (9.10) shows that Gd dt GL = Gd dt (rC \u00d7 p) = \u00b5 GdrC dt \u00d7 p+ rC \u00d7 Gdp dt \u00b6 = GrC \u00d7 Gdp dt = GrC \u00d7 GF = GM. (9.13) Kinetic energy K of a moving body point P with mass m at a position GrP , and having a velocity GvP , is K = 1 2 mGv2P = 1 2 m \u00b3 Gd\u0307B + BvP + B G\u03c9B \u00d7 BrP \u00b42 . (9.14) 524 9. Applied Dynamics The work done by the applied force GF on m in moving from point 1 to point 2 on a path, indicated by a vector Gr, is 1W2 = Z 2 1 GF \u00b7 dGr. (9.15) However, Z 2 1 GF \u00b7 dGr = m Z 2 1 Gd dt Gv \u00b7 Gvdt = 1 2 m Z 2 1 d dt v2dt = 1 2 m \u00a1 v22 \u2212 v21 \u00a2 = K2 \u2212K1 (9.16) that shows 1W2 is equal to the difference of the kinetic energy between terminal and initial points. 1W2 = K2 \u2212K1 (9.17) Equation (9.17) is called principle of work and energy. Example 342 Position of center of mass. The position of the mass center of a rigid body in a coordinate frame is indicated by BrC and is usually measured in the body coordinate frame. BrC = 1 m Z B Br dm (9.18) \u23a1\u23a3 xC yC zC \u23a4\u23a6 = \u23a1\u23a2\u23a2\u23a3 1 m R B x dm 1 m R B y dm 1 m R B z dm \u23a4\u23a5\u23a5\u23a6 (9.19) Applying the mass center integral on the symmetric and uniform L-section rigid body with \u03c1 = 1 shown in Figure 9.2 provides the position of mass center C of the section. The x position of C is xC = 1 m Z B xdm = 1 A Z B x dA = \u2212b 2 + ab\u2212 a2 4ab+ 2a2 (9.20) and because of symmetry, we have yC = \u2212xC = b2 + ab\u2212 a2 4ab+ 2a2 . (9.21) 9. Applied Dynamics 525 When a = b, the position of C reduces to yC = \u2212xC = 1 2 b. (9.22) Example 343 F Every force system is equivalent to a wrench. The Poinsot theorem states: Every force system is equivalent to a single force, plus a moment parallel to the force. Let F andM be the resultant force and moment of a force system. We decompose the moment into parallel and perpendicular components,Mk andM\u22a5, to the force axis. The force F and the perpendicular moment M\u22a5 can be replaced by a single force F0 parallel to F. Therefore, the force system is reduced to a force F0 and a moment Mk parallel to each other. A force and a moment about the force axis is called a wrench. The Poinsot theorem is similar to the Chasles theorem that states: Every rigid body motion is equivalent to a screw, which is a translation plus a rotation about the axis of translation. Example 344 F Motion of a moving point in a moving body frame. The velocity and acceleration of a moving point P as shown in Figure 5.12 are found in Example 200. GvP = Gd\u0307B + GRB \u00a1 BvP + B G\u03c9B \u00d7 BrP \u00a2 (9.23) GaP = Gd\u0308B + GRB \u00a1 BaP + 2 B G\u03c9B \u00d7 BvP + B G\u03c9\u0307B \u00d7 BrP \u00a2 +GRB \u00a1 B G\u03c9B \u00d7 \u00a1 B G\u03c9B \u00d7 BrP \u00a2\u00a2 (9.24) Therefore, the equation of motion for the point mass P is GF = mGaP = m \u00b3 Gd\u0308B + GRB \u00a1 BaP + 2 B G\u03c9B \u00d7 BvP + B G\u03c9\u0307B \u00d7 BrP \u00a2\u00b4 +m GRB \u00a1 B G\u03c9B \u00d7 \u00a1 B G\u03c9B \u00d7 BrP \u00a2\u00a2 . (9.25) 526 9. Applied Dynamics Example 345 Newton\u2019s equation in a rotating frame. Consider a spherical rigid body, such as Earth, with a fixed point that is rotating with a constant angular velocity. The equation of motion for a moving point vehicle P on the rigid body is found by setting Gd\u0308B = B G\u03c9\u0307B = 0 in the equation of motion of a moving point in a moving body frame (9.25) BF = mBaP +mB G\u03c9B \u00d7 \u00a1 B G\u03c9B \u00d7 BrP \u00a2 + 2mB G\u03c9B \u00d7 B r\u0307P (9.26) 6= mBaP which shows that the Newton\u2019s equation of motion F = ma must be modified for rotating frames. Example 346 Coriolis force. The equation of motion of a moving vehicle point on the surface of the Earth is BF = mBaP +mB G\u03c9B \u00d7 \u00a1 B G\u03c9B \u00d7 BrP \u00a2 + 2mB G\u03c9B \u00d7 BvP (9.27) which can be rearranged to BF\u2212mB G\u03c9B \u00d7 \u00a1 B G\u03c9B \u00d7 BrP \u00a2 \u2212 2mB G\u03c9B \u00d7 BvP = mBaP . (9.28) Equation (9.28) is the equation of motion for an observer in the rotating frame, which in this case is an observer on the Earth. The left-hand side of this equation is called the effective force Feff , Feff = BF\u2212mB G\u03c9B \u00d7 \u00a1 B G\u03c9B \u00d7 BrP \u00a2 \u2212 2mB G\u03c9B \u00d7 BvP (9.29) because it seems that the particle is moving under the influence of this force. The second term is negative of the centrifugal force and pointing outward. The maximum value of this force on the Earth is on the equator r\u03c92 = 6378.388\u00d7 103 \u00d7 \u00b5 2\u03c0 24\u00d7 3600 366.25 365.25 \u00b62 = 3.3917\u00d7 10\u22122m/ s2 (9.30) which is about 0.3% of the acceleration of gravity. If we add the variation of the gravitational acceleration because of a change of radius from R = 6356912m at the pole to R = 6378388m on the equator, then the variation of the acceleration of gravity becomes 0.53%. So, generally speaking, a sportsman such as a pole-vaulter who has practiced in the north pole can show a better record in a competition held on the equator. The third term is called the Coriolis force or Coriolis effect, FC, which is perpendicular to both \u03c9 and BvP . For a mass m moving on the north hemisphere at a latitude \u03b8 towards the equator, we should provide a 9. Applied Dynamics 527 lateral eastward force equal to the Coriolis effect to force the mass, keeping its direction relative to the ground. FC = 2mB G\u03c9B \u00d7 Bvm = 1.4584\u00d7 10\u22124 Bpm cos \u03b8 kgm/ s2 (9.31) The Coriolis effect is the reason why the west side of railways, roads, and rivers wears. The lack of providing the Coriolis force is the reason for turning the direction of winds, projectiles, flood, and falling objects westward. Example 347 Work, force, and kinetic energy in a unidirectional motion. A vehicle with mass m = 1200 kg has an initial kinetic energy K = 6000 J. The mass is under a constant force F = F I\u0302 = 4000I\u0302 and moves from X(0) = 0 to X(tf ) = 1000m at a terminal time tf . The work done by the force during this motion is W = Z r(tf ) r(0) F \u00b7 dr = Z 1000 0 4000 dX = 4\u00d7 106Nm = 4MJ (9.32) The kinetic energy at the terminal time is K(tf ) =W +K(0) = 4006000 J (9.33) which shows that the terminal speed of the mass is v2 = r 2K(tf ) m \u2248 81.7m/ s. (9.34) Example 348 Direct dynamics. When the applied force is time varying and is a known function, then, F(t) = m r\u0308. (9.35) The general solution for the equation of motion can be found by integration. r\u0307(t) = r\u0307(t0) + 1 m Z t t0 F(t)dt (9.36) r(t) = r(t0) + r\u0307(t0)(t\u2212 t0) + 1 m Z t t0 Z t t0 F(t)dt dt (9.37) This kind of problem is called direct or forward dynamics. 528 9. Applied Dynamics 9.2 Rigid Body Translational Dynamics Figure 9.3 depicts a moving body B in a global coordinate frame G. Assume that the body frame is attached at the mass center of the body. Point P indicates an infinitesimal sphere of the body, which has a very small mass dm. The point mass dm is acted on by an infinitesimal force df and has a global velocity GvP . According to Newton\u2019s law of motion we have df = GaP dm. (9.38) However, the equation of motion for the whole body in a global coordinate frame is GF = mGaB (9.39) which can be expressed in the body coordinate frame as BF = mB GaB +m B G\u03c9B \u00d7 BvB (9.40)\u23a1\u23a3 Fx Fy Fz \u23a4\u23a6 = \u23a1\u23a3 max +m (\u03c9yvz \u2212 \u03c9zvy) may \u2212m (\u03c9xvz \u2212 \u03c9zvx) maz +m (\u03c9xvy \u2212 \u03c9yvx) \u23a4\u23a6 . (9.41) In these equations, GaB is the acceleration vector of the body mass center C in the global frame,m is the total mass of the body, and F is the resultant of the external forces acted on the body at C. 9. Applied Dynamics 529 Proof. A body coordinate frame at the mass center is called a central frame. If frame B is a central frame, then the center of mass, C, is defined such that Z B Brdm dm = 0. (9.42) The global position vector of dm is related to its local position vector by Grdm = GdB + GRB Brdm (9.43) where GdB is the global position vector of the central body frame, and therefore, Z B Grdm dm = Z B GdB dm+ GRB Z m Brdm dm = Z B GdB dm = GdB Z B dm = mGdB. (9.44) A time derivative of both sides shows that mGd\u0307B = mGvB = Z B Gr\u0307dm dm = Z B Gvdm dm (9.45) and another derivative is mGv\u0307B = mGaB = Z B Gv\u0307dm dm. (9.46) However, we have df = Gv\u0307P dm and therefore, mGaB = Z B df . (9.47) The integral on the right-hand side accounts for all the forces acting on the body. The internal forces cancel one another out, so the net result is the vector sum of all the externally applied forces, F, and therefore, GF = m GaB = m Gv\u0307B . (9.48) In the body coordinate frame we have BF = BRG GF = m BRG GaB = m B GaB = m BaB +m B G\u03c9B \u00d7 BvB. (9.49) 530 9. Applied Dynamics The expanded form of the Newton\u2019s equation in the body coordinate frame is then equal to BF = m BaB +m B G\u03c9B \u00d7 BvB\u23a1\u23a3 Fx Fy Fz \u23a4\u23a6 = m \u23a1\u23a3 ax ay az \u23a4\u23a6+m \u23a1\u23a3 \u03c9x \u03c9y \u03c9z \u23a4\u23a6\u00d7 \u23a1\u23a3 vx vy vz \u23a4\u23a6 = \u23a1\u23a3 max +m (\u03c9yvz \u2212 \u03c9zvy) may \u2212m (\u03c9xvz \u2212 \u03c9zvx) maz +m (\u03c9xvy \u2212 \u03c9yvx) \u23a4\u23a6 . (9.50) 9.3 Rigid Body Rotational Dynamics The rigid body rotational equation of motion is the Euler equation BM = Gd dt BL = BL\u0307+B G\u03c9B \u00d7 BL = BI B G\u03c9\u0307B + B G\u03c9B \u00d7 \u00a1 BI B G\u03c9B \u00a2 (9.51) where L is the angular momentum BL = BI B G\u03c9B (9.52) and I is the moment of inertia of the rigid body. I = \u23a1\u23a3 Ixx Ixy Ixz Iyx Iyy Iyz Izx Izy Izz \u23a4\u23a6 (9.53) The elements of I are functions of the mass distribution of the rigid body and may be defined by Iij = Z B \u00a1 r2i \u03b4mn \u2212 ximxjn \u00a2 dm , i, j = 1, 2, 3 (9.54) where \u03b4ij is Kronecker\u2019s delta. \u03b4mn = \u00bd 1 if m = n 0 if m 6= n (9.55) The expanded form of the Euler equation (9.51) is Mx = Ixx\u03c9\u0307x + Ixy\u03c9\u0307y + Ixz\u03c9\u0307z \u2212 (Iyy \u2212 Izz)\u03c9y\u03c9z \u2212Iyz \u00a1 \u03c92z \u2212 \u03c92y \u00a2 \u2212 \u03c9x (\u03c9zIxy \u2212 \u03c9yIxz) (9.56) 9. Applied Dynamics 531 My = Iyx\u03c9\u0307x + Iyy\u03c9\u0307y + Iyz\u03c9\u0307z \u2212 (Izz \u2212 Ixx)\u03c9z\u03c9x \u2212Ixz \u00a1 \u03c92x \u2212 \u03c92z \u00a2 \u2212 \u03c9y (\u03c9xIyz \u2212 \u03c9zIxy) (9.57) Mz = Izx\u03c9\u0307x + Izy\u03c9\u0307y + Izz\u03c9\u0307z \u2212 (Ixx \u2212 Iyy)\u03c9x\u03c9y \u2212Ixy \u00a1 \u03c92y \u2212 \u03c92x \u00a2 \u2212 \u03c9z (\u03c9yIxz \u2212 \u03c9xIyz) . (9.58) which can be reduced to M1 = I1\u03c9\u03071 \u2212 (I2 \u2212 I2)\u03c92\u03c93 M2 = I2\u03c9\u03072 \u2212 (I3 \u2212 I1)\u03c93\u03c91 (9.59) M3 = I3\u03c9\u03073 \u2212 (I1 \u2212 I2)\u03c91\u03c92 in a special Cartesian coordinate frame called the principal coordinate frame. The principal coordinate frame is denoted by numbers 123 to indicate the first, second, and third principal axes. The parameters Iij , i 6= j are zero in the principal frame. The body and principal coordinate frame sit at the mass center C. Kinetic energy of a rotating rigid body is K = 1 2 \u00a1 Ixx\u03c9 2 x + Iyy\u03c9 2 y + Izz\u03c9 2 z \u00a2 \u2212Ixy\u03c9x\u03c9y \u2212 Iyz\u03c9y\u03c9z \u2212 Izx\u03c9z\u03c9x (9.60) = 1 2 \u03c9 \u00b7 L (9.61) = 1 2 \u03c9T I \u03c9 (9.62) that in the principal coordinate frame reduces to K = 1 2 \u00a1 I1\u03c9 2 1 + I2\u03c9 2 2 + I3\u03c9 2 3 \u00a2 . (9.63) Proof. Let mi be the mass of the ith particle of a rigid body B, which is made of n particles and let ri = Bri = \u00a3 xi yi zi \u00a4T (9.64) be the Cartesian position vector of mi in a central body fixed coordinate frame Oxyz. Assume that \u03c9 = B G\u03c9B = \u00a3 \u03c9x \u03c9y \u03c9z \u00a4T (9.65) is the angular velocity of the rigid body with respect to the ground, expressed in the body coordinate frame. 532 9. Applied Dynamics The angular momentum of mi is Li = ri \u00d7mir\u0307i = mi [ri \u00d7 (\u03c9 \u00d7 ri)] = mi [(ri \u00b7 ri)\u03c9 \u2212 (ri \u00b7 \u03c9) ri] = mir 2 i\u03c9 \u2212mi (ri \u00b7 \u03c9) ri. (9.66) Hence, the angular momentum of the rigid body would be L = \u03c9 nX i=1 mir 2 i \u2212 nX i=1 mi (ri \u00b7 \u03c9) ri. (9.67) Substitution for ri and \u03c9 gives us L = \u00b3 \u03c9x \u0131\u0302+ \u03c9y j\u0302+ \u03c9z k\u0302 \u00b4 nX i=1 mi \u00a1 x2i + y2i + z2i \u00a2 \u2212 nX i=1 mi (xi\u03c9x + yi\u03c9y + zi\u03c9z) \u00b7 \u00b3 xi\u0131\u0302+ yij\u0302+ zik\u0302 \u00b4 (9.68) and therefore, L = nX i=1 mi \u00a1 x2i + y2i + z2i \u00a2 \u03c9x \u0131\u0302 + nX i=1 mi \u00a1 x2i + y2i + z2i \u00a2 \u03c9y j\u0302 + nX i=1 mi \u00a1 x2i + y2i + z2i \u00a2 \u03c9z k\u0302 \u2212 nX i=1 mi (xi\u03c9x + yi\u03c9y + zi\u03c9z)xi\u0131\u0302 \u2212 nX i=1 mi (xi\u03c9x + yi\u03c9y + zi\u03c9z) yij\u0302 \u2212 nX i=1 mi (xi\u03c9x + yi\u03c9y + zi\u03c9z) zik\u0302 (9.69) or L = nX i=1 mi \u00a3\u00a1 x2i + y2i + z2i \u00a2 \u03c9x \u2212 (xi\u03c9x + yi\u03c9y + zi\u03c9z)xi \u00a4 \u0131\u0302 + nX i=1 mi \u00a3\u00a1 x2i + y2i + z2i \u00a2 \u03c9y \u2212 (xi\u03c9x + yi\u03c9y + zi\u03c9z) yi \u00a4 j\u0302 + nX i=1 mi \u00a3\u00a1 x2i + y2i + z2i \u00a2 \u03c9z \u2212 (xi\u03c9x + yi\u03c9y + zi\u03c9z) zi \u00a4 k\u0302 (9.70) 9. Applied Dynamics 533 which can be rearranged as L = nX i=1 \u00a3 mi \u00a1 y2i + z2i \u00a2\u00a4 \u03c9x\u0131\u0302 + nX i=1 \u00a3 mi \u00a1 z2i + x2i \u00a2\u00a4 \u03c9y j\u0302 + nX i=1 \u00a3 mi \u00a1 x2i + y2i \u00a2\u00a4 \u03c9zk\u0302 \u2212 \u00c3 nX i=1 (mixiyi)\u03c9y + nX i=1 (mixizi)\u03c9z ! \u0131\u0302 \u2212 \u00c3 nX i=1 (miyizi)\u03c9z + nX i=1 (miyixi)\u03c9x ! j\u0302 \u2212 \u00c3 nX i=1 (mizixi)\u03c9x + nX i=1 (miziyi)\u03c9y ! k\u0302. (9.71) By introducing the moment of inertia matrix I with the following elements, Ixx = nX i=1 \u00a3 mi \u00a1 y2i + z2i \u00a2\u00a4 (9.72) Iyy = nX i=1 \u00a3 mi \u00a1 z2i + x2i \u00a2\u00a4 (9.73) Izz = nX i=1 \u00a3 mi \u00a1 x2i + y2i \u00a2\u00a4 (9.74) Ixy = Iyx = \u2212 nX i=1 (mixiyi) (9.75) Iyz = Izy = \u2212 nX i=1 (miyizi) (9.76) Izx = Ixz = \u2212 nX i=1 (mizixi) . (9.77) we may write the angular momentum L in a concise form Lx = Ixx\u03c9x + Ixy\u03c9y + Ixz\u03c9z (9.78) Ly = Iyx\u03c9x + Iyy\u03c9y + Iyz\u03c9z (9.79) Lz = Izx\u03c9x + Izy\u03c9y + Izz\u03c9z (9.80) 534 9. Applied Dynamics or in a matrix form L = I \u00b7 \u03c9 (9.81)\u23a1\u23a3 Lx Ly Lz \u23a4\u23a6 = \u23a1\u23a3 Ixx Ixy Ixz Iyx Iyy Iyz Izx Izy Izz \u23a4\u23a6\u23a1\u23a3 \u03c9x \u03c9y \u03c9z \u23a4\u23a6 . (9.82) For a rigid body that is a continuous solid, the summations must be replaced by integrations over the volume of the body as in Equation (9.54). The Euler equation of motion for a rigid body is BM = Gd dt BL (9.83) where BM is the resultant of the external moments applied on the rigid body. The angular momentum BL is a vector defined in the body coordinate frame. Hence, its time derivative in the global coordinate frame is GdBL dt = BL\u0307+B G\u03c9B \u00d7 BL. (9.84) Therefore, BM = dL dt = L\u0307+ \u03c9 \u00d7 L = I\u03c9\u0307 + \u03c9\u00d7 (I\u03c9) (9.85) or in expanded form BM = (Ixx\u03c9\u0307x + Ixy\u03c9\u0307y + Ixz\u03c9\u0307z) \u0131\u0302 +(Iyx\u03c9\u0307x + Iyy\u03c9\u0307y + Iyz\u03c9\u0307z) j\u0302 +(Izx\u03c9\u0307x + Izy\u03c9\u0307y + Izz\u03c9\u0307z) k\u0302 +\u03c9y (Ixz\u03c9x + Iyz\u03c9y + Izz\u03c9z) \u0131\u0302 \u2212\u03c9z (Ixy\u03c9x + Iyy\u03c9y + Iyz\u03c9z) \u0131\u0302 +\u03c9z (Ixx\u03c9x + Ixy\u03c9y + Ixz\u03c9z) j\u0302 \u2212\u03c9x (Ixz\u03c9x + Iyz\u03c9y + Izz\u03c9z) j\u0302 +\u03c9x (Ixy\u03c9x + Iyy\u03c9y + Iyz\u03c9z) k\u0302 \u2212\u03c9y (Ixx\u03c9x + Ixy\u03c9y + Ixz\u03c9z) k\u0302 (9.86) and therefore, the most general form of the Euler equations of motion for a rigid body in a body frame attached to C are Mx = Ixx\u03c9\u0307x + Ixy\u03c9\u0307y + Ixz\u03c9\u0307z \u2212 (Iyy \u2212 Izz)\u03c9y\u03c9z \u2212Iyz \u00a1 \u03c92z \u2212 \u03c92y \u00a2 \u2212 \u03c9x (\u03c9zIxy \u2212 \u03c9yIxz) (9.87) My = Iyx\u03c9\u0307x + Iyy\u03c9\u0307y + Iyz\u03c9\u0307z \u2212 (Izz \u2212 Ixx)\u03c9z\u03c9x \u2212Ixz \u00a1 \u03c92x \u2212 \u03c92z \u00a2 \u2212 \u03c9y (\u03c9xIyz \u2212 \u03c9zIxy) (9.88) Mz = Izx\u03c9\u0307x + Izy\u03c9\u0307y + Izz\u03c9\u0307z \u2212 (Ixx \u2212 Iyy)\u03c9x\u03c9y \u2212Ixy \u00a1 \u03c92y \u2212 \u03c92x \u00a2 \u2212 \u03c9z (\u03c9yIxz \u2212 \u03c9xIyz) . (9.89) 9. Applied Dynamics 535 Assume that we are able to rotate the body frame about its origin to find an orientation that makes Iij = 0, for i 6= j. In such a coordinate frame, which is called a principal frame, the Euler equations reduce to M1 = I1\u03c9\u03071 \u2212 (I2 \u2212 I2)\u03c92\u03c93 (9.90) M2 = I2\u03c9\u03072 \u2212 (I3 \u2212 I1)\u03c93\u03c91 (9.91) M3 = I3\u03c9\u03073 \u2212 (I1 \u2212 I2)\u03c91\u03c92. (9.92) The kinetic energy of a rigid body may be found by the integral of the kinetic energy of a mass element dm, over the whole body. K = 1 2 Z B v\u03072dm = 1 2 Z B (\u03c9 \u00d7 r) \u00b7 (\u03c9 \u00d7 r) dm = \u03c92x 2 Z B \u00a1 y2 + z2 \u00a2 dm+ \u03c92y 2 Z B \u00a1 z2 + x2 \u00a2 dm+ \u03c92z 2 Z B \u00a1 x2 + y2 \u00a2 dm \u2212\u03c9x\u03c9y Z B xy dm\u2212 \u03c9y\u03c9z Z B yz dm\u2212 \u03c9z\u03c9x Z B zx dm = 1 2 \u00a1 Ixx\u03c9 2 x + Iyy\u03c9 2 y + Izz\u03c9 2 z \u00a2 \u2212Ixy\u03c9x\u03c9y \u2212 Iyz\u03c9y\u03c9z \u2212 Izx\u03c9z\u03c9x (9.93) The kinetic energy can be rearranged to a matrix multiplication form K = 1 2 \u03c9T I \u03c9 (9.94) = 1 2 \u03c9 \u00b7 L. (9.95) When the body frame is principal, the kinetic energy will simplify to K = 1 2 \u00a1 I1\u03c9 2 1 + I2\u03c9 2 2 + I3\u03c9 2 3 \u00a2 . (9.96) Example 349 A tilted disc on a massless shaft. Figure 9.4 illustrates a disc with mass m and radius r, mounted on a massless shaft. The shaft is turning with a constant angular speed \u03c9. The disc is attached to the shaft at an angle \u03b8. Because of \u03b8, the bearings at A and B must support a rotating force. We attach a principal body coordinate frame at the disc center as shown in the figure. The angular velocity vector in the body frame is B G\u03c9B = \u03c9 cos \u03b8 \u0131\u0302+ \u03c9 sin \u03b8 j\u0302 (9.97) 536 9. Applied Dynamics and the mass moment of inertia matrix is BI = \u23a1\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3 mr2 2 0 0 0 mr2 4 0 0 0 mr2 4 \u23a4\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6 . (9.98) Substituting (9.97) and (9.98) in (9.90)-(9.92), with 1 \u2261 x, 2 \u2261 y, 3 \u2261 z, provides that Mx = 0 (9.99) My = 0 (9.100) Mz = mr2 4 \u03c9 cos \u03b8 sin \u03b8. (9.101) Therefore, the bearing reaction forces FA and FB are FA = \u2212FB = \u2212Mz l = \u2212mr2 4l \u03c9 cos \u03b8 sin \u03b8. (9.102) Example 350 Steady rotation of a freely rotating rigid body. The Newton-Euler equations of motion for a rigid body are GF = mGv\u0307 (9.103) BM = I B G\u03c9\u0307B + B G\u03c9B \u00d7 BL. (9.104) 9. Applied Dynamics 537 Consider a situation in which the resultant applied force and moment on the body are zero. GF = BF = 0 (9.105) GM = BM = 0 (9.106) Based on the Newton\u2019s equation, the velocity of the mass center will be constant in the global coordinate frame. However, the Euler equation reduces to \u03c9\u03071 = I2 \u2212 I3 I1 \u03c92\u03c93 (9.107) \u03c9\u03072 = I3 \u2212 I1 I22 \u03c93\u03c91 (9.108) \u03c9\u03073 = I1 \u2212 I2 I3 \u03c91\u03c92 (9.109) that show the angular velocity can be constant if I1 = I2 = I3 (9.110) or if two principal moments of inertia, say I1 and I2, are zero and the third angular velocity, in this case \u03c93, is initially zero, or if the angular velocity vector is initially parallel to a principal axis. Example 351 Angular momentum of a two-link manipulator. A two-link manipulator is shown in Figure 9.5. Link A rotates with angular velocity \u03d5\u0307 about the z-axis of its local coordinate frame. Link B is attached to link A and has angular velocity \u03c8\u0307 with respect to A about the xA-axis. We assume that A and G were coincident at \u03d5 = 0, therefore, the rotation matrix between A and G is GRA = \u23a1\u23a3 cos\u03d5(t) \u2212 sin\u03d5(t) 0 sin\u03d5(t) cos\u03d5(t) 0 0 0 1 \u23a4\u23a6 . (9.111) Frame B is related to frame A by Euler angles \u03d5 = 90deg, \u03b8 = 90deg, and \u03c8, hence, ARB = \u23a1\u23a3 c\u03c0c\u03c8 \u2212 c\u03c0s\u03c0s\u03c8 \u2212c\u03c0s\u03c8 \u2212 c\u03c0c\u03c8s\u03c0 s\u03c0s\u03c0 c\u03c8s\u03c0 + c\u03c0c\u03c0s\u03c8 \u2212s\u03c0s\u03c8 + c\u03c0c\u03c0c\u03c8 \u2212c\u03c0s\u03c0 s\u03c0s\u03c8 s\u03c0c\u03c8 c\u03c0 \u23a4\u23a6 \u23a1\u23a3 \u2212 cos\u03c8 sin\u03c8 0 sin\u03c8 cos\u03c8 0 0 0 \u22121 \u23a4\u23a6 (9.112) 538 9. Applied Dynamics and therefore, GRB = GRA ARB (9.113) = \u23a1\u23a3 \u2212 cos\u03d5 cos\u03c8 \u2212 sin\u03d5 sin\u03c8 cos\u03d5 sin\u03c8 \u2212 cos\u03c8 sin\u03d5 0 cos\u03d5 sin\u03c8 \u2212 cos\u03c8 sin\u03d5 cos\u03d5 cos\u03c8 + sin\u03d5 sin\u03c8 0 0 0 \u22121 \u23a4\u23a6 . The angular velocity of A in G, and B in A are G\u03c9A = \u03d5\u0307K\u0302 (9.114) A\u03c9B = \u03c8\u0307\u0131\u0302A. (9.115) Moment of inertia matrices for the arms A and B can be defined as AIA = \u23a1\u23a3 IA1 0 0 0 IA2 0 0 0 IA3 \u23a4\u23a6 (9.116) BIB = \u23a1\u23a3 IB1 0 0 0 IB2 0 0 0 IB3 \u23a4\u23a6 . (9.117) These moments of inertia must be transformed to the global frame GIA = GRB AIA GRT A (9.118) GIB = GRB BIB GRT B . (9.119) 9. Applied Dynamics 539 The total angular momentum of the manipulator is GL = GLA + GLB (9.120) where GLA = GIA G\u03c9A (9.121) GLB = GIB G\u03c9B = GIB \u00a1 G A\u03c9B + G\u03c9A \u00a2 . (9.122) Example 352 Poinsot\u2019s construction. Consider a freely rotating rigid body with an attached principal coordinate frame. HavingM = 0 provides a motion under constant angular momentum and constant kinetic energy L = I \u03c9 = cte (9.123) K = 1 2 \u03c9T I \u03c9 = cte. (9.124) Because the length of the angular momentum L is constant, the equation L2 = L \u00b7 L = L2x + L2y + L2z = I21\u03c9 2 1 + I22\u03c9 2 2 + I23\u03c9 2 3 (9.125) introduces an ellipsoid in the (\u03c91, \u03c92, \u03c93) coordinate frame, called the momentum ellipsoid. The tip of all possible angular velocity vectors must lie on the surface of the momentum ellipsoid. The kinetic energy also defines an energy ellipsoid in the same coordinate frame so that the tip of the angular velocity vectors must also lie on its surface. K = 1 2 \u00a1 I1\u03c9 2 1 + I2\u03c9 2 2 + I3\u03c9 2 3 \u00a2 (9.126) In other words, the dynamics of moment-free motion of a rigid body requires that the corresponding angular velocity \u03c9(t) satisfy both Equations (9.125) and (9.126) and therefore lie on the intersection of the momentum and energy ellipsoids. For clarity, we may define the ellipsoids in the (Lx, Ly, Lz) coordinate system as L2x + L2y + L2z = L2 (9.127) L2x 2I1K + L2y 2I2K + L2z 2I3K = 1. (9.128) Equation (9.127) is a sphere and Equation (9.128) defines an ellipsoid with\u221a 2IiK as semi-axes. To have a meaningful motion, these two shapes must intersect. The intersection may form a trajectory, as shown in Figure 9.6. 540 9. Applied Dynamics It can be deduced that for a certain value of angular momentum there are maximum and minimum limit values for acceptable kinetic energy. Assuming I1 > I3 > I3 (9.129) the limits of possible kinetic energy are Kmin = L2 2I1 (9.130) Kmax = L2 2I3 (9.131) and the corresponding motions are turning about the axes I1 and I3 respectively. Example 353 F Alternative derivation of Euler equations of motion. Assume that the moment of the small force df is shown by dm and a mass element is shown by dm, then, dm = Grdm \u00d7 df = Grdm \u00d7 Gv\u0307dm dm. (9.132) The global angular momentum dl of dm is equal to dl = Grdm \u00d7 Gvdm dm (9.133) and according to (9.12) we have dm = Gd dt dl (9.134) Grdm \u00d7 df = Gd dt \u00a1 Grdm \u00d7 Gvdm dm \u00a2 . (9.135) 9. Applied Dynamics 541 Integrating over the body results inZ B Grdm \u00d7 df = Z B Gd dt \u00a1 Grdm \u00d7 Gvdm dm \u00a2 = Gd dt Z B \u00a1 Grdm \u00d7 Gvdm dm \u00a2 . (9.136) However, utilizing Grdm = GdB + GRB Brdm (9.137) where GdB is the global position vector of the central body frame, can simplify the left-hand side of the integral toZ B Grdm \u00d7 df = Z B \u00a1 GdB + GRB Brdm \u00a2 \u00d7 df = Z B GdB \u00d7 df + Z B G Brdm \u00d7 df = GdB \u00d7 GF+ GMC (9.138) where MC is the resultant external moment about the body mass center C. The right-hand side of Equation (9.136) is Gd dt Z B \u00a1 Grdm \u00d7 Gvdm dm \u00a2 = Gd dt Z B \u00a1\u00a1 GdB + GRB Brdm \u00a2 \u00d7 Gvdm dm \u00a2 = Gd dt Z B \u00a1 GdB \u00d7 Gvdm \u00a2 dm+ Gd dt Z B \u00a1 G Brdm \u00d7 Gvdm \u00a2 dm = Gd dt \u00b5 GdB \u00d7 Z B Gvdmdm \u00b6 + Gd dt LC = Gd\u0307B \u00d7 Z B Gvdmdm+ GdB \u00d7 Z B Gv\u0307dmdm+ d dt LC . (9.139) We use LC for angular momentum about the body mass center. Because the body frame is at the mass center, we haveZ B Grdm dm = mGdB = mGrC (9.140)Z B Gvdmdm = mGd\u0307B = mGvC (9.141)Z B Gv\u0307dmdm = mGd\u0308B = mGaC (9.142) and therefore, Gd dt Z B \u00a1 Grdm \u00d7 Gvdm dm \u00a2 = GdB \u00d7 GF+ Gd dt GLC . (9.143) 542 9. Applied Dynamics Substituting (9.138) and (9.143) in (9.136) provides the Euler equation of motion in the global frame, indicating that the resultant of externally applied moments about C is equal to the global derivative of angular momentum about C. GMC = Gd dt GLC . (9.144) The Euler equation in the body coordinate can be found by transforming (9.144) BMC = GRT B GMC = GRT B Gd dt LC = Gd dt GRT B LC = Gd dt BLC = BL\u0307C + B G\u03c9B \u00d7 BLC . (9.145) 9.4 Mass Moment of Inertia Matrix In analyzing the motion of rigid bodies, two types of integrals arise that belong to the geometry of the body. The first type defines the center of mass and is important when the translation motion of the body is considered. The second is the moment of inertia that appears when the rotational motion of the body is considered. The moment of inertia is also called centrifugal moments, or deviation moments. Every rigid body has a 3 \u00d7 3 moment of inertia matrix I, which is denoted by I = \u23a1\u23a3 Ixx Ixy Ixz Iyx Iyy Iyz Izx Izy Izz \u23a4\u23a6 . (9.146) The diagonal elements Iij , i = j are called polar moments of inertia Ixx = Ix = Z B \u00a1 y2 + z2 \u00a2 dm (9.147) Iyy = Iy = Z B \u00a1 z2 + x2 \u00a2 dm (9.148) Izz = Iz = Z B \u00a1 x2 + y2 \u00a2 dm (9.149) and the off-diagonal elements Iij , i 6= j are called products of inertia Ixy = Iyx = \u2212 Z B xy dm (9.150) 9. Applied Dynamics 543 Iyz = Izy = \u2212 Z B yz dm (9.151) Izx = Ixz = \u2212 Z B zx dm. (9.152) The elements of I for a rigid body, made of discrete point masses, are defined in Equation (9.54). The elements of I are calculated about a body coordinate frame attached to the mass center C of the body. Therefore, I is a frame-dependent quantity and must be written like BI to show the frame it is computed in. BI = Z B \u23a1\u23a3 y2 + z2 \u2212xy \u2212zx \u2212xy z2 + x2 \u2212yz \u2212zx \u2212yz x2 + y2 \u23a4\u23a6 dm (9.153) = Z B \u00a1 r2I\u2212 r rT \u00a2 dm (9.154) = Z B \u2212r\u0303 r\u0303 dm. (9.155) Moments of inertia can be transformed from a coordinate frame B1 to another coordinate frame B2, both installed at the mass center of the body, according to the rule of the rotated-axes theorem B2I = B2RB1 B1I B2RT B1 . (9.156) Transformation of the moment of inertia from a central frame B1 located at B2rC to another frame B2, which is parallel to B1, is, according to the rule of parallel-axes theorem, B2I = B1I +mr\u0303C r\u0303TC . (9.157) If the local coordinate frame Oxyz is located such that the products of inertia vanish, the local coordinate frame is called the principal coordinate frame and the associated moments of inertia are called principal moments of inertia. Principal axes and principal moments of inertia can be found by solving the following equation for I:\u00af\u0304\u0304\u0304 \u00af\u0304 Ixx \u2212 I Ixy Ixz Iyx Iyy \u2212 I Iyz Izx Izy Izz \u2212 I \u00af\u0304\u0304\u0304 \u00af\u0304 = 0 (9.158) det ([Iij ]\u2212 I [\u03b4ij ]) = 0. (9.159) Since Equation (9.159) is a cubic equation in I, we obtain three eigenvalues I1 = Ix I2 = Iy I3 = Iz (9.160) 544 9. Applied Dynamics that are the principal moments of inertia. Proof. Two coordinate frames with a common origin at the mass center of a rigid body are shown in Figure 9.7. The angular velocity and angular momentum of a rigid body transform from the frame B1 to the frame B2 by vector transformation rule B2\u03c9 = B2RB1 B1\u03c9 (9.161) B2L = B2RB1 B1L. (9.162) However, L and \u03c9 are related according to Equation (9.52) B1L = B1I B1\u03c9 (9.163) and therefore, B2L = B2RB1 B1I B2RT B1 B2\u03c9 = B2I B2\u03c9 (9.164) which shows how to transfer the moment of inertia from the coordinate frame B1 to a rotated frame B2 B2I = B2RB1 B1I B2RT B1 . (9.165) Now consider a central frame B1, shown in Figure 9.8, at B2rC , which rotates about the origin of a fixed frame B2 such that their axes remain parallel. The angular velocity and angular momentum of the rigid body transform from frame B1 to frame B2 by B2\u03c9 = B1\u03c9 (9.166) B2L = B1L+ (rC \u00d7mvC) . (9.167) 9. Applied Dynamics 545 Therefore, B2L = B1L+mB2rC \u00d7 \u00a1 B2\u03c9\u00d7B2rC \u00a2 = B1L+ \u00a1 m B2 r\u0303C B2 r\u0303TC \u00a2 B2\u03c9 = \u00a1 B1I +m B2 r\u0303C B2 r\u0303TC \u00a2 B2\u03c9 (9.168) which shows how to transfer the moment of inertia from frame B1 to a parallel frame B2 B2I = B1I +mr\u0303C r\u0303TC . (9.169) The parallel-axes theorem is also called the Huygens-Steiner theorem. Referring to Equation (9.165) for transformation of the moment of inertia to a rotated frame, we can always find a frame in which B2I is diagonal. In such a frame, we have B2RB1 B1I = B2I B2RB1 (9.170) or \u23a1\u23a3 r11 r12 r13 r21 r22 r23 r31 r32 r33 \u23a4\u23a6\u23a1\u23a3 Ixx Ixy Ixz Iyx Iyy Iyz Izx Izy Izz \u23a4\u23a6 = \u23a1\u23a3 I1 0 0 0 I2 0 0 0 I3 \u23a4\u23a6\u23a1\u23a3 r11 r12 r13 r21 r22 r23 r31 r32 r33 \u23a4\u23a6 (9.171) which shows that I1, I2, and I3 are eigenvalues of B1I. These eigenvalues can be found by solving the following equation for \u03bb:\u00af\u0304\u0304\u0304 \u00af\u0304 Ixx \u2212 \u03bb Ixy Ixz Iyx Iyy \u2212 \u03bb Iyz Izx Izy Izz \u2212 \u03bb \u00af\u0304\u0304\u0304 \u00af\u0304 = 0. (9.172) 546 9. Applied Dynamics The eigenvalues I1, I2, and I3 are principal moments of inertia, and their associated eigenvectors are called principal directions. The coordinate frame made by the eigenvectors is the principal body coordinate frame. In the principal coordinate frame, the rigid body angular momentum is\u23a1\u23a3 L1 L2 L3 \u23a4\u23a6 = \u23a1\u23a3 I1 0 0 0 I2 0 0 0 I3 \u23a4\u23a6\u23a1\u23a3 \u03c91 \u03c92 \u03c93 \u23a4\u23a6 . (9.173) Example 354 Principal moments of inertia. Consider the inertia matrix I I = \u23a1\u23a3 20 \u22122 0 \u22122 30 0 0 0 40 \u23a4\u23a6 . (9.174) We set up the determinant (9.159)\u00af\u0304\u0304\u0304 \u00af\u0304 20\u2212 \u03bb \u22122 0 \u22122 30\u2212 \u03bb 0 0 0 40\u2212 \u03bb \u00af\u0304\u0304\u0304 \u00af\u0304 = 0 (9.175) which leads to the following characteristic equation. (20\u2212 \u03bb) (30\u2212 \u03bb) (40\u2212 \u03bb)\u2212 4 (40\u2212 \u03bb) = 0 (9.176) Three roots of Equation (9.176) are I1 = 30.385, I2 = 19.615, I3 = 40 (9.177) and therefore, the principal moment of inertia matrix is I = \u23a1\u23a3 30.385 0 0 0 19.615 0 0 0 40 \u23a4\u23a6 . (9.178) Example 355 Principal coordinate frame. Consider the inertia matrix I I = \u23a1\u23a3 20 \u22122 0 \u22122 30 0 0 0 40 \u23a4\u23a6 (9.179) the direction of a principal axis xi is established by solving\u23a1\u23a3 Ixx \u2212 Ii Ixy Ixz Iyx Iyy \u2212 Ii Iyz Izx Izy Izz \u2212 Ii \u23a4\u23a6\u23a1\u23a3 cos\u03b1i cos\u03b2i cos \u03b3i \u23a4\u23a6 = \u23a1\u23a3 0 0 0 \u23a4\u23a6 (9.180) 9. Applied Dynamics 547 for direction cosines, which must also satisfy cos2 \u03b1i + cos 2 \u03b2i + cos 2 \u03b3i = 1. (9.181) For the first principal moment of inertia I1 = 30.385 we have\u23a1\u23a3 20\u2212 30.385 \u22122 0 \u22122 30\u2212 30.385 0 0 0 40\u2212 30.385 \u23a4\u23a6\u23a1\u23a3 cos\u03b11 cos\u03b21 cos \u03b31 \u23a4\u23a6 = \u23a1\u23a3 0 0 0 \u23a4\u23a6 (9.182) or \u221210.385 cos\u03b11 \u2212 2 cos\u03b21 + 0 = 0 (9.183) \u22122 cos\u03b11 \u2212 0.385 cos\u03b21 + 0 = 0 (9.184) 0 + 0 + 9.615 cos \u03b31 = 0 (9.185) and we obtain \u03b11 = 79.1 deg (9.186) \u03b21 = 169.1 deg (9.187) \u03b31 = 90.0 deg . (9.188) Using I2 = 19.615 for the second principal axis\u23a1\u23a3 20\u2212 19.62 \u22122 0 \u22122 30\u2212 19.62 0 0 0 40\u2212 19.62 \u23a4\u23a6\u23a1\u23a3 cos\u03b12 cos\u03b22 cos \u03b32 \u23a4\u23a6 = \u23a1\u23a3 0 0 0 \u23a4\u23a6 (9.189) we obtain \u03b12 = 10.9 deg (9.190) \u03b22 = 79.1 deg (9.191) \u03b32 = 90.0 deg . (9.192) The third principal axis is for I3 = 40\u23a1\u23a3 20\u2212 40 \u22122 0 \u22122 30\u2212 40 0 0 0 40\u2212 40 \u23a4\u23a6\u23a1\u23a3 cos\u03b13 cos\u03b23 cos \u03b33 \u23a4\u23a6 = \u23a1\u23a3 0 0 0 \u23a4\u23a6 (9.193) which leads to \u03b13 = 90.0 deg (9.194) \u03b23 = 90.0 deg (9.195) \u03b33 = 0.0 deg . (9.196) 548 9. Applied Dynamics Example 356 Moment of inertia of a rigid rectangular bar. Consider a homogeneous rectangular link with mass m, length l, width w, and height h, as shown in Figure 9.9. The local central coordinate frame is attached to the link at its mass center. The moments of inertia matrix of the link can be found by the integral method. We begin with calculating Ixx Ixx = Z B \u00a1 y2 + z2 \u00a2 dm = Z v \u00a1 y2 + z2 \u00a2 \u03c1dv = m lwh Z v \u00a1 y2 + z2 \u00a2 dv = m lwh Z h/2 \u2212h/2 Z w/2 \u2212w/2 Z l/2 \u2212l/2 \u00a1 y2 + z2 \u00a2 dx dy dz = m 12 \u00a1 w2 + h2 \u00a2 (9.197) which shows Iyy and Izz can be calculated similarly Iyy = m 12 \u00a1 h2 + l2 \u00a2 (9.198) Izz = m 12 \u00a1 l2 + w2 \u00a2 . (9.199) Since the coordinate frame is central, the products of inertia must be zero. To show this, we examine Ixy. Ixy = Iyx = \u2212 Z B xy dm = Z v xy\u03c1dv = m lwh Z h/2 \u2212h/2 Z w/2 \u2212w/2 Z l/2 \u2212l/2 xy dxdy dz = 0 (9.200) 9. Applied Dynamics 549 Therefore, the moment of inertia matrix for the rigid rectangular bar in its central frame is I = \u23a1\u23a3 m 12 \u00a1 w2 + h2 \u00a2 0 0 0 m 12 \u00a1 h2 + l2 \u00a2 0 0 0 m 12 \u00a1 l2 + w2 \u00a2 \u23a4\u23a6 . (9.201) Example 357 Translation of the inertia matrix. The moment of inertia matrix of the rigid body shown in Figure 9.10, in the principal frame B(oxyz) is given in Equation (9.201). The moment of inertia matrix in the non-principal frame B0(ox0y0z0) can be found by applying the parallel-axes transformation formula (9.169). B0 I = BI +m B0 r\u0303C B0 r\u0303TC (9.202) The mass center is at B0 rC = 1 2 \u23a1\u23a3 l w h \u23a4\u23a6 (9.203) and therefore, B0 r\u0303C = 1 2 \u23a1\u23a3 0 \u2212h w h 0 \u2212l \u2212w l 0 \u23a4\u23a6 (9.204) that provides B0 I = \u23a1\u23a3 1 3h 2m+ 1 3mw2 \u221214 lmw \u221214hlm \u221214 lmw 1 3h 2m+ 1 3 l 2m \u221214hmw \u221214hlm \u2212 14hmw 1 3 l 2m+ 1 3mw2 \u23a4\u23a6 . (9.205) 550 9. Applied Dynamics Example 358 Principal rotation matrix. Consider a body inertia matrix as I = \u23a1\u23a3 2/3 \u22121/2 \u22121/2 \u22121/2 5/3 \u22121/4 \u22121/2 \u22121/4 5/3 \u23a4\u23a6 . (9.206) The eigenvalues and eigenvectors of I are I1 = 0.2413 , \u23a1\u23a3 2.351 1 1 \u23a4\u23a6 (9.207) I2 = 1.8421 , \u23a1\u23a3 \u22120.8511 1 \u23a4\u23a6 (9.208) I3 = 1.9167 , \u23a1\u23a3 0 \u22121 1 \u23a4\u23a6 . (9.209) The normalized eigenvector matrix W is equal to the transpose of the required transformation matrix to make the inertia matrix diagonal W = \u23a1\u23a3 | | | w1 w 2 w 3 | | | \u23a4\u23a6 = 2RT 1 = \u23a1\u23a3 0.856 9 \u22120.515 6 0.0 0.364 48 0.605 88 \u22120.707 11 0.364 48 0.605 88 0.707 11 \u23a4\u23a6 . (9.210) We may verify that 2I \u2248 2R1 1I 2RT 1 =WT 1I W = \u23a1\u23a3 0.2413 \u22121\u00d7 10\u22124 0.0 \u22121\u00d7 10\u22124 1.842 1 \u22121\u00d7 10\u221219 0.0 0.0 1.916 7 \u23a4\u23a6 . (9.211) Example 359 F Relative diagonal moments of inertia. Using the definitions for moments of inertia (9.147), (9.148), and (9.149) it is seen that the inertia matrix is symmetric, andZ B \u00a1 x2 + y2 + z2 \u00a2 dm = 1 2 (Ixx + Iyy + Izz) (9.212) and also Ixx + Iyy \u2265 Izz (9.213) Iyy + Izz \u2265 Ixx (9.214) Izz + Ixx \u2265 Iyy. (9.215) 9. Applied Dynamics 551 Noting that (y \u2212 z) 2 \u2265 0 it is evident that \u00a1 y2 + z2 \u00a2 \u2265 2yz and therefore Ixx \u2265 2Iyz (9.216) and similarly Iyy \u2265 2Izx (9.217) Izz \u2265 2Ixy. (9.218) Example 360 F Coefficients of the characteristic equation. The determinant (9.172)\u00af\u0304\u0304\u0304 \u00af\u0304 Ixx \u2212 \u03bb Ixy Ixz Iyx Iyy \u2212 \u03bb Iyz Izx Izy Izz \u2212 \u03bb \u00af\u0304\u0304\u0304 \u00af\u0304 = 0 (9.219) for calculating the principal moments of inertia, leads to a third-degree equation for \u03bb, called the characteristic equation. \u03bb3 \u2212 a1\u03bb 2 + a2\u03bb\u2212 a3 = 0 (9.220) The coefficients of the characteristic equation are called the principal invariants of [I]. The coefficients of the characteristic equation can directly be found from the following equations: a1 = Ixx + Iyy + Izz = tr [I] (9.221) a2 = IxxIyy + IyyIzz + IzzIxx \u2212 I2xy \u2212 I2yz \u2212 I2zx = \u00af\u0304\u0304\u0304 Ixx Ixy Iyx Iyy \u00af\u0304\u0304\u0304 + \u00af\u0304\u0304\u0304 Iyy Iyz Izy Izz \u00af\u0304\u0304\u0304 + \u00af\u0304\u0304\u0304 Ixx Ixz Izx Izz \u00af\u0304\u0304\u0304 = 1 2 \u00a1 a21 \u2212 tr \u00a3 I2 \u00a4\u00a2 (9.222) a3 = IxxIyyIzz + IxyIyzIzx + IzyIyxIxz \u2212 (IxxIyzIzy + IyyIzxIxz + IzzIxyIyx) = IxxIyyIzz + 2IxyIyzIzx \u2212 \u00a1 IxxI 2 yz + IyyI 2 zx + IzzI 2 xy \u00a2 = det [I] (9.223) 552 9. Applied Dynamics Example 361 F The principal moments of inertia are coordinate invariants. The roots of the inertia characteristic equation are the principal moments of inertia. They are all real but not necessarily different. The principal moments of inertia are extreme. That is, the principal moments of inertia determine the smallest and the largest values of Iii. Since the smallest and largest values of Iii do not depend on the choice of the body coordinate frame, the solution of the characteristic equation is not dependent of the coordinate frame. In other words, if I1, I2, and I3 are the principal moments of inertia for B1I, the principal moments of inertia for B2I are also I1, I2, and I3 when B2I = B2RB1 B1I B2RT B1 . We conclude that I1, I2, and I3 are coordinate invariants of the matrix [I], and therefore any quantity that depends on I1, I2, and I3 is also coordinate invariant. The matrix [I] has only three independent invariants and every other invariant can be expressed in terms of I1, I2, and I3. Since I1, I2, and I3 are the solutions of the characteristic equation of [I] given in (9.220), we may write the determinant (9.172) in the form (\u03bb\u2212 I1) (\u03bb\u2212 I2) (\u03bb\u2212 I3) = 0. (9.224) The expanded form of this equation is \u03bb3 \u2212 (I1 + I2 + I3)\u03bb 2 + (I1I2 + I2I3 + I3I1) a2\u03bb\u2212 I1I2I3 = 0. (9.225) By comparing (9.225) and (9.220) we conclude that a1 = Ixx + Iyy + Izz = I1 + I2 + I3 (9.226) a2 = IxxIyy + IyyIzz + IzzIxx \u2212 I2xy \u2212 I2yz \u2212 I2zx = I1I2 + I2I3 + I3I1 (9.227) a3 = IxxIyyIzz + 2IxyIyzIzx \u2212 \u00a1 IxxI 2 yz + IyyI 2 zx + IzzI 2 xy \u00a2 = I1I2I3. (9.228) Being able to express the coefficients a1, a2, and a3 as functions of I1, I2, and I3 determines that the coefficients of the characteristic equation are coordinate-invariant. Example 362 F Short notation for the elements of inertia matrix. Taking advantage of the Kronecker\u2019s delta (5.138) we may write the el- 9. Applied Dynamics 553 ements of the moment of inertia matrix Iij in short notation forms Iij = Z B \u00a1\u00a1 x21 + x22 + x23 \u00a2 \u03b4ij \u2212 xixj \u00a2 dm (9.229) Iij = Z B \u00a1 r2\u03b4ij \u2212 xixj \u00a2 dm (9.230) Iij = Z B \u00c3 3X k=1 xkxk\u03b4ij \u2212 xixj ! dm (9.231) where we utilized the following notations: x1 = x x2 = y x3 = z. (9.232) Example 363 F Moment of inertia with respect to a plane, a line, and a point. The moment of inertia of a system of particles may be defined with respect to a plane, a line, or a point as the sum of the products of the mass of the particles into the square of the perpendicular distance from the particle to the plane, line, or point. For a continuous body, the sum would be definite integral over the volume of the body. The moments of inertia with respect to the xy, yz, and zx-plane are Iz2 = Z B z2dm (9.233) Iy2 = Z B y2dm (9.234) Ix2 = Z B x2dm. (9.235) The moments of inertia with respect to the x, y, and z axes are Ix = Z B \u00a1 y2 + z2 \u00a2 dm (9.236) Iy = Z B \u00a1 z2 + x2 \u00a2 dm (9.237) Iz = Z B \u00a1 x2 + y2 \u00a2 dm (9.238) and therefore, Ix = Iy2 + Iz2 (9.239) Iy = Iz2 + Ix2 (9.240) Iz = Ix2 + Iy2 . (9.241) 554 9. Applied Dynamics The moment of inertia with respect to the origin is Io = Z B \u00a1 x2 + y2 + z2 \u00a2 dm = Ix2 + Iy2 + Iz2 = 1 2 (Ix + Iy + Iz) . (9.242) Because the choice of the coordinate frame is arbitrary, we can say that the moment of inertia with respect to a line is the sum of the moments of inertia with respect to any two mutually orthogonal planes that pass through the line. The moment of inertia with respect to a point has similar meaning for three mutually orthogonal planes intersecting at the point. 9.5 Lagrange\u2019s Form of Newton\u2019s Equations of Motion Newton\u2019s equation of motion can be transformed to d dt \u00b5 \u2202K \u2202q\u0307r \u00b6 \u2212 \u2202K \u2202qr = Fr r = 1, 2, \u00b7 \u00b7 \u00b7n (9.243) where Fr = nX i=1 \u00b5 Fix \u2202fi \u2202q1 + Fiy \u2202gi \u2202q2 + Fiz \u2202hi \u2202qn \u00b6 . (9.244) Equation (9.243) is called the Lagrange equation of motion, where K is the kinetic energy of the n degree-of-freedom (DOF ) system, qr, r = 1, 2, \u00b7 \u00b7 \u00b7 , n are the generalized coordinates of the system, F = \u00a3 Fix Fiy Fiz \u00a4T is the external force acting on the ith particle of the system, and Fr is the generalized force associated to qr. Proof. Let mi be the mass of one of the particles of a system and let (xi, yi, zi) be its Cartesian coordinates in a globally fixed coordinate frame. Assume that the coordinates of every individual particle are functions of another set of coordinates q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, and possibly time t. xi = fi(q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, t) (9.245) yi = gi(q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, t) (9.246) zi = hi(q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, t) (9.247) If Fxi, Fyi, Fzi are components of the total force acting on the particle mi, then the Newton equations of motion for the particle would be Fxi = mix\u0308i (9.248) Fyi = miy\u0308i (9.249) Fzi = miz\u0308i. (9.250) 9. Applied Dynamics 555 We multiply both sides of these equations by \u2202fi \u2202qr \u2202gi \u2202qr \u2202hi \u2202qr respectively, and add them up for all the particles to have nX i=1 mi \u00b5 x\u0308i \u2202fi \u2202qr + y\u0308i \u2202gi \u2202qr + z\u0308i \u2202hi \u2202qr \u00b6 = nX i=1 \u00b5 Fxi \u2202fi \u2202qr + Fyi \u2202gi \u2202qr + Fzi \u2202hi \u2202qr \u00b6 (9.251) where n is the total number of particles. Taking a time derivative of Equation (9.245), x\u0307i = \u2202fi \u2202q1 q\u03071 + \u2202fi \u2202q2 q\u03072 + \u2202fi \u2202q3 q\u03073 + \u00b7 \u00b7 \u00b7+ \u2202fi \u2202qn q\u0307n + \u2202fi \u2202t (9.252) we find \u2202x\u0307i \u2202q\u0307r = \u2202 \u2202q\u0307r \u00b5 \u2202fi \u2202q1 q\u03071 + \u2202fi \u2202q2 q\u03072 + \u00b7 \u00b7 \u00b7+ \u2202fi \u2202qn q\u0307n + \u2202fi \u2202t \u00b6 = \u2202fi \u2202qr . (9.253) and therefore, x\u0308i \u2202fi \u2202qr = x\u0308i \u2202x\u0307i \u2202q\u0307r = d dt \u00b5 x\u0307i \u2202x\u0307i \u2202q\u0307r \u00b6 \u2212 x\u0307i d dt \u00b5 \u2202x\u0307i \u2202q\u0307r \u00b6 . (9.254) However, x\u0307i d dt \u00b5 \u2202x\u0307i \u2202q\u0307r \u00b6 = x\u0307i d dt \u00b5 \u2202fi \u2202qr \u00b6 = x\u0307i \u00b5 \u22022fi \u2202q1\u2202qr q\u03071 + \u00b7 \u00b7 \u00b7+ \u22022fi \u2202qn\u2202qr q\u0307n + \u22022fi \u2202t\u2202qr \u00b6 = x\u0307i \u2202 \u2202qr \u00b5 \u2202fi \u2202q1 q\u03071 + \u2202fi \u2202q2 q\u03072 + \u00b7 \u00b7 \u00b7+ \u2202fi \u2202qn q\u0307n + \u2202fi \u2202t \u00b6 = x\u0307i \u2202x\u0307i \u2202qr (9.255) and we have x\u0308i \u2202x\u0307i \u2202q\u0307r = d dt \u00b5 x\u0307i \u2202x\u0307i \u2202q\u0307r \u00b6 \u2212 x\u0307i \u2202x\u0307i \u2202qr (9.256) 556 9. Applied Dynamics which is equal to x\u0308i x\u0307i q\u0307r = d dt \u2219 \u2202 \u2202q\u0307r \u00b5 1 2 x\u03072i \u00b6\u00b8 \u2212 \u2202 \u2202qr \u00b5 1 2 x\u03072i \u00b6 . (9.257) Now substituting (9.254) and (9.257) in the left-hand side of (9.251) leads to nX i=1 mi \u00b5 x\u0308i \u2202fi \u2202qr + y\u0308i \u2202gi \u2202qr + z\u0308i \u2202hi \u2202qr \u00b6 = nX i=1 mi d dt \u2219 \u2202 \u2202q\u0307r \u00b5 1 2 x\u03072i + 1 2 y\u03072i + 1 2 z\u03072i \u00b6\u00b8 \u2212 nX i=1 mi \u2202 \u2202qr \u00b5 1 2 x\u03072i + 1 2 y\u03072i + 1 2 z\u03072i \u00b6 = 1 2 nX i=1 mi d dt \u2219 \u2202 \u2202q\u0307r \u00a1 x\u03072i + y\u03072i + z\u03072i \u00a2\u00b8 \u22121 2 nX i=1 mi \u2202 \u2202qr \u00a1 x\u03072i + y\u03072i + z\u03072i \u00a2 . (9.258) where 1 2 nX i=1 mi \u00a1 x\u03072i + y\u03072i + z\u03072i \u00a2 = K (9.259) is the kinetic energy of the system. Therefore, the Newton equations of motion (9.248), (9.249), and (9.250) are converted to d dt \u00b5 \u2202K \u2202q\u0307r \u00b6 \u2212 \u2202K \u2202qr = nX i=1 \u00b5 Fxi \u2202fi \u2202qr + Fyi \u2202gi \u2202qr + Fzi \u2202hi \u2202qr \u00b6 . (9.260) Because of (9.245), (9.246), and (9.247), the kinetic energy is a function of q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn and time t. The left-hand side of Equation (9.260) includes the kinetic energy of the whole system and the right-hand side is a generalized force and shows the effect of changing coordinates from xi to qj on the external forces. Let us assume that the coordinate qr alters to qr+ \u03b4qr while the other coordinates q1, q2, q3, \u00b7 \u00b7 \u00b7 , qr\u22121, qr+1, \u00b7 \u00b7 \u00b7 , qn and time t are unaltered. So, the coordinates of mi are changed to xi + \u2202fi \u2202qr \u03b4qr (9.261) yi + \u2202gi \u2202qr \u03b4qr (9.262) zi + \u2202hi \u2202qr \u03b4qr (9.263) 9. Applied Dynamics 557 Such a displacement is called virtual displacement. The work done in this virtual displacement by all forces acting on the particles of the system is \u03b4W = nX i=1 \u00b5 Fxi \u2202fi \u2202qr + Fyi \u2202gi \u2202qr + Fzi \u2202hi \u2202qr \u00b6 \u03b4qr. (9.264) Because the work done by internal forces appears in opposite pairs, only the work done by external forces remains in Equation (9.264). Let\u2019s denote the virtual work by \u03b4W = Fr (q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, t) \u03b4qr. (9.265) Then we have d dt \u00b5 \u2202K \u2202q\u0307r \u00b6 \u2212 \u2202K \u2202qr = Fr (9.266) where Fr = nX i=1 \u00b5 Fxi \u2202fi \u2202qr + Fyi \u2202gi \u2202qr + Fzi \u2202hi \u2202qr \u00b6 . (9.267) Equation (9.266) is the Lagrange form of equations of motion. This equation is true for all values of r from 1 to n. We thus have n second-order ordinary differential equations in which q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn are the dependent variables and t is the independent variable. The coordinates q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn are called generalized coordinates and can be any measurable parameters to provide the configuration of the system. The number of equations and the number of dependent variables are equal, therefore, the equations are theoretically sufficient to determine the motion of all mi. Example 364 Equation of motion for a simple pendulum. A pendulum is shown in Figure 9.11. Using x and y for the Cartesian position of m, and using \u03b8 = q as the generalized coordinate, we have x = f(\u03b8) = l sin \u03b8 (9.268) y = g(\u03b8) = l cos \u03b8 (9.269) K = 1 2 m \u00a1 x\u03072 + y\u03072 \u00a2 = 1 2 ml2\u03b8\u0307 2 (9.270) and therefore, d dt \u00b5 \u2202K \u2202\u03b8\u0307 \u00b6 \u2212 \u2202K \u2202\u03b8 = d dt (ml2\u03b8\u0307) = ml2\u03b8\u0308. (9.271) The external force components, acting on m, are Fx = 0 (9.272) Fy = mg (9.273) 558 9. Applied Dynamics and therefore, F\u03b8 = Fx \u2202f \u2202\u03b8 + Fy \u2202g \u2202\u03b8 = \u2212mgl sin \u03b8. (9.274) Hence, the equation of motion for the pendulum is ml2\u03b8\u0308 = \u2212mgl sin \u03b8. (9.275) Example 365 A pendulum attached to an oscillating mass. Figure 9.12 illustrates a vibrating mass with a hanging pendulum. The pendulum can act as a vibration absorber if designed properly. Starting with coordinate relationships xM = fM = x (9.276) yM = gM = 0 (9.277) xm = fm = x+ l sin \u03b8 (9.278) ym = gm = l cos \u03b8 (9.279) we may find the kinetic energy in terms of the generalized coordinates x and \u03b8. K = 1 2 M \u00a1 x\u03072M + y\u03072M \u00a2 + 1 2 m \u00a1 x\u03072m + y\u03072m \u00a2 = 1 2 Mx\u03072 + 1 2 m \u00b3 x\u03072 + l2\u03b8\u0307 2 + 2lx\u0307\u03b8\u0307 cos \u03b8 \u00b4 (9.280) Then, the left-hand side of the Lagrange equations are d dt \u00b5 \u2202K \u2202x\u0307 \u00b6 \u2212 \u2202K \u2202x = (M +m)x\u0308+ml\u03b8\u0308 cos \u03b8 \u2212ml\u03b8\u0307 2 sin \u03b8 (9.281) d dt \u00b5 \u2202K \u2202\u03b8\u0307 \u00b6 \u2212 \u2202K \u2202\u03b8 = ml2\u03b8\u0308 +mlx\u0308 cos \u03b8. (9.282) 9. Applied Dynamics 559 The external forces acting on M and m are FxM = \u2212kx (9.283) FyM = 0 (9.284) Fxm = 0 (9.285) Fym = mg. (9.286) Therefore, the generalized forces are Fx = FxM \u2202fM \u2202x + FyM \u2202gM \u2202x + Fxm \u2202fm \u2202x + Fym \u2202gm \u2202x = \u2212kx (9.287) F\u03b8 = FxM \u2202fM \u2202\u03b8 + FyM \u2202gM \u2202\u03b8 + Fxm \u2202fm \u2202\u03b8 + Fym \u2202gm \u2202\u03b8 = \u2212mgl sin \u03b8 (9.288) and finally the Lagrange equations of motion are (M +m)x\u0308+ml\u03b8\u0308 cos \u03b8 \u2212ml\u03b8\u0307 2 sin \u03b8 = \u2212kx (9.289) ml2\u03b8\u0308 +mlx\u0308 cos \u03b8 = \u2212mgl sin \u03b8. (9.290) Example 366 Kinetic energy of the Earth. Earth is approximately a rotating rigid body about a fixed axis. The two motions of the Earth are called revolution about the sun, and rotation about an axis approximately fixed in the Earth. The kinetic energy of the Earth due to its rotation is K1 = 1 2 I\u03c921 = 1 2 2 5 \u00a1 5.9742\u00d7 1024 \u00a2\u00b56356912 + 6378388 2 \u00b62\u00b5 2\u03c0 24\u00d7 3600 366.25 365.25 \u00b62 = 2.5762\u00d7 1029 J 560 9. Applied Dynamics and the kinetic energy of the Earth due to its revolution is K2 = 1 2 Mr2\u03c922 = 1 2 \u00a1 5.9742\u00d7 1024 \u00a2 \u00a1 1.49475\u00d7 1011 \u00a22\u00b5 2\u03c0 24\u00d7 3600 1 365.25 \u00b62 = 2.6457\u00d7 1033 J where r is the distance from the sun and \u03c92 is the angular speed about the sun. The total kinetic energy of the Earth is K = K1 +K2. However, the ratio of the revolutionary to rotational kinetic energies is K2 K1 = 2.6457\u00d7 1033 2.5762\u00d7 1029 \u2248 10000. Example 367 F Explicit form of Lagrange equations. Assume the coordinates of every particle are functions of the coordinates q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn but not the time t. The kinetic energy of the system made of n massive particles can be written as K = 1 2 nX i=1 mi \u00a1 x\u03072i + y\u03072i + z\u03072i \u00a2 = 1 2 nX j=1 nX k=1 ajkq\u0307j q\u0307k (9.291) where the coefficients ajk are functions of q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn and ajk = akj . (9.292) The Lagrange equations of motion d dt \u00b5 \u2202K \u2202q\u0307r \u00b6 \u2212 \u2202K \u2202qr = Fr r = 1, 2, \u00b7 \u00b7 \u00b7n (9.293) are then equal to d dt nX m=1 amrq\u0307m \u2212 1 2 nX j=1 nX k=1 ajk \u2202qr q\u0307j q\u0307k = Fr (9.294) or nX m=1 amrq\u0308m + nX k=1 nX n=1 \u0393rk,nq\u0307kq\u0307n = Fr (9.295) where \u0393ij,k is called the Christoffel operator \u0393ij,k = 1 2 \u00b5 \u2202aij \u2202qk + \u2202aik \u2202qj \u2212 \u2202akj \u2202qi \u00b6 . (9.296) 9. Applied Dynamics 561 9.6 Lagrangian Mechanics Assume for some forces F = \u00a3 Fix Fiy Fiz \u00a4T there is a function V , called potential energy, such that the force is derivable from V F = \u2212\u2207V. (9.297) Such a force is called potential or conservative force. Then, the Lagrange equation of motion can be written as d dt \u00b5 \u2202L \u2202q\u0307r \u00b6 \u2212 \u2202L \u2202qr = Qr r = 1, 2, \u00b7 \u00b7 \u00b7n (9.298) where L = K \u2212 V (9.299) is the Lagrangean of the system and Qr is the nonpotential generalized force. Proof. Assume the external forces F = \u00a3 Fxi Fyi Fzi \u00a4T acting on the system are conservative. F = \u2212\u2207V (9.300) The work done by these forces in an arbitrary virtual displacement \u03b4q1, \u03b4q2, \u03b4q3, \u00b7 \u00b7 \u00b7 , \u03b4qn is \u2202W = \u2212\u2202V \u2202q1 \u03b4q1 \u2212 \u2202V \u2202q2 \u03b4q2 \u2212 \u00b7 \u00b7 \u00b7 \u2202V \u2202qn \u03b4qn (9.301) then the Lagrange equation becomes d dt \u00b5 \u2202K \u2202q\u0307r \u00b6 \u2212 \u2202K \u2202qr = \u2212 \u2202V \u2202q1 r = 1, 2, \u00b7 \u00b7 \u00b7n. (9.302) Introducing the Lagrangean function L = K \u2212 V converts the Lagrange equation to d dt \u00b5 \u2202L \u2202q\u0307r \u00b6 \u2212 \u2202L \u2202qr = 0 r = 1, 2, \u00b7 \u00b7 \u00b7n (9.303) for a conservative system. The Lagrangean is also called kinetic potential. If a force is not conservative, then the virtual work done by the force is \u03b4W = nX i=1 \u00b5 Fxi \u2202fi \u2202qr + Fyi \u2202gi \u2202qr + Fzi \u2202hi \u2202qr \u00b6 \u03b4qr = Qr \u03b4qr (9.304) and the equation of motion would be d dt \u00b5 \u2202L \u2202q\u0307r \u00b6 \u2212 \u2202L \u2202qr = Qr r = 1, 2, \u00b7 \u00b7 \u00b7n (9.305) where Qr is the nonpotential generalized force doing work in a virtual displacement of the rth generalized coordinate qr. 562 9. Applied Dynamics Example 368 Spherical pendulum. A pendulum analogy is utilized in modeling of many dynamical problems. Figure 9.13 illustrates a spherical pendulum with mass m and length l. The angles \u03d5 and \u03b8 may be used as describing coordinates of the system. The Cartesian coordinates of the mass as a function of the generalized coordinates are \u23a1\u23a3 X Y Z \u23a4\u23a6 = \u23a1\u23a3 r cos\u03d5 sin \u03b8 r sin \u03b8 sin\u03d5 \u2212r cos \u03b8 \u23a4\u23a6 (9.306) and therefore, the kinetic and potential energies of the pendulum are K = 1 2 m \u00b3 l2\u03b8\u0307 2 + l2\u03d5\u03072 sin2 \u03b8 \u00b4 (9.307) V = \u2212mgl cos \u03b8. (9.308) The kinetic potential function of this system is then equal to L = 1 2 m \u00b3 l2\u03b8\u0307 2 + l2\u03d5\u03072 sin2 \u03b8 \u00b4 +mgl cos \u03b8 (9.309) which leads to the following equations of motion: \u03b8\u0308 \u2212 \u03d5\u03072 sin \u03b8 cos \u03b8 + g l sin \u03b8 = 0 (9.310) \u03d5\u0308 sin2 \u03b8 + 2\u03d5\u0307\u03b8\u0307 sin \u03b8 cos \u03b8 = 0. (9.311) Example 369 Controlled compound pendulum. A massive arm is attached to a ceiling at a pin joint O as illustrated in Figure 9.14. Assume that there is viscous friction in the joint where an ideal motor can apply a torque Q to move the arm. The rotor of an ideal motor has no moment of inertia by assumption. 9. Applied Dynamics 563 The kinetic and potential energies of the manipulator are K = 1 2 I\u03b8\u0307 2 = 1 2 \u00a1 IC +ml2 \u00a2 \u03b8\u0307 2 (9.312) V = \u2212mg cos \u03b8 (9.313) where m is the mass and I is the moment of inertia of the pendulum about O. The Lagrangean of the manipulator is L = K \u2212 V = 1 2 I\u03b8\u0307 2 +mg cos \u03b8 (9.314) and therefore, the equation of motion of the pendulum is M = d dt \u00b5 \u2202L \u2202\u03b8\u0307 \u00b6 \u2212 \u2202L \u2202\u03b8 = I \u03b8\u0308 +mgl sin \u03b8. (9.315) The generalized force M is the contribution of the motor torque Q and the viscous friction torque \u2212c\u03b8\u0307. Hence, the equation of motion of the manipulator is Q = I \u03b8\u0308 + c\u03b8\u0307 +mgl sin \u03b8. (9.316) Example 370 An ideal 2R planar manipulator dynamics. An ideal model of a 2R planar manipulator is illustrated in Figure 9.15. It is called ideal because we assume the links are massless and there is no friction. The masses m1 and m2 are the mass of the second motor to run 564 9. Applied Dynamics the second link and the load at the endpoint. We take the absolute angle \u03b81 and the relative angle \u03b82 as the generalized coordinates to express the configuration of the manipulator. The global positions of m1 and m2 are\u2219 X1 Y2 \u00b8 = \u2219 l1 cos \u03b81 l1 sin \u03b81 \u00b8 (9.317)\u2219 X2 Y2 \u00b8 = \u2219 l1 cos \u03b81 + l2 cos (\u03b81 + \u03b82) l1 sin \u03b81 + l2 sin (\u03b81 + \u03b82) \u00b8 (9.318) and therefore, the global velocity of the masses are\u2219 X\u03071 Y\u03071 \u00b8 = \u2219 \u2212l1\u03b8\u03071 sin \u03b81 l1\u03b8\u03071 cos \u03b81 \u00b8 (9.319) \u2219 X\u03072 Y\u03072 \u00b8 = \u23a1\u23a3 \u2212l1\u03b8\u03071 sin \u03b81 \u2212 l2 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 sin (\u03b81 + \u03b82) l1\u03b8\u03071 cos \u03b81 + l2 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 cos (\u03b81 + \u03b82) \u23a4\u23a6 . (9.320) The kinetic energy of this manipulator is made of kinetic energy of the masses and is equal to K = K1 +K2 = 1 2 m1 \u00b3 X\u03072 1 + Y\u0307 2 1 \u00b4 + 1 2 m2 \u00b3 X\u03072 2 + Y\u0307 2 2 \u00b4 = 1 2 m1l 2 1\u03b8\u0307 2 1 + 1 2 m2 \u00b5 l21\u03b8\u0307 2 1 + l22 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b42 + 2l1l2\u03b8\u03071 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 cos \u03b82 \u00b6 . (9.321) 9. Applied Dynamics 565 The potential energy of the manipulator is V = V1 + V2 = m1gY1 +m2gY2 = m1gl1 sin \u03b81 +m2g (l1 sin \u03b81 + l2 sin (\u03b81 + \u03b82)) . (9.322) The Lagrangean is then obtained from Equations (9.321) and (9.322) L = K \u2212 V (9.323) = 1 2 m1l 2 1\u03b8\u0307 2 1 + 1 2 m2 \u00b5 l21\u03b8\u0307 2 1 + l22 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b42 + 2l1l2\u03b8\u03071 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 cos \u03b82 \u00b6 \u2212 (m1gl1 sin \u03b81 +m2g (l1 sin \u03b81 + l2 sin (\u03b81 + \u03b82))) . which provides the required partial derivatives as follows: \u2202L \u2202\u03b81 = \u2212 (m1 +m2) gl1 cos \u03b81 \u2212m2gl2 cos (\u03b81 + \u03b82) (9.324) \u2202L \u2202\u03b8\u03071 = (m1 +m2) l 2 1\u03b8\u03071 +m2l 2 2 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 +m2l1l2 \u00b3 2\u03b8\u03071 + \u03b8\u03072 \u00b4 cos \u03b82 (9.325) d dt \u00b5 \u2202L \u2202\u03b8\u03071 \u00b6 = (m1 +m2) l 2 1\u03b8\u03081 +m2l 2 2 \u00b3 \u03b8\u03081 + \u03b8\u03082 \u00b4 +m2l1l2 \u00b3 2 \u03b8\u03081 + \u03b8\u03082 \u00b4 cos \u03b82 \u2212m2l1l2\u03b8\u03072 \u00b3 2\u03b8\u03071 + \u03b8\u03072 \u00b4 sin \u03b82 (9.326) \u2202L \u2202\u03b82 = \u2212m2l1l2\u03b8\u03071 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 sin \u03b82 \u2212m2gl2 cos (\u03b81 + \u03b82) (9.327) \u2202L \u2202\u03b8\u03072 = m2l 2 2 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 +m2l1l2\u03b8\u03071 cos \u03b82 (9.328) d dt \u00b5 \u2202L \u2202\u03b8\u03072 \u00b6 = m2l 2 2 \u00b3 \u03b8\u03081 + \u03b8\u03082 \u00b4 +m2l1l2\u03b8\u03081 cos \u03b82 \u2212m2l1l2\u03b8\u03071\u03b8\u03072 sin \u03b82 (9.329) Therefore, the equations of motion for the 2R manipulator are Q1 = d dt \u00b5 \u2202L \u2202\u03b8\u03071 \u00b6 \u2212 \u2202L \u2202\u03b81 = (m1 +m2) l 2 1\u03b8\u03081 +m2l 2 2 \u00b3 \u03b8\u03081 + \u03b8\u03082 \u00b4 +m2l1l2 \u00b3 2 \u03b8\u03081 + \u03b8\u03082 \u00b4 cos \u03b82 \u2212m2l1l2\u03b8\u03072 \u00b3 2\u03b8\u03071 + \u03b8\u03072 \u00b4 sin \u03b82 +(m1 +m2) gl1 cos \u03b81 +m2gl2 cos (\u03b81 + \u03b82) (9.330) 566 9. Applied Dynamics Q2 = d dt \u00b5 \u2202L \u2202\u03b8\u03072 \u00b6 \u2212 \u2202L \u2202\u03b82 = m2l 2 2 \u00b3 \u03b8\u03081 + \u03b8\u03082 \u00b4 +m2l1l2\u03b8\u03081 cos \u03b82 \u2212m2l1l2\u03b8\u03071\u03b8\u03072 sin \u03b82 +m2l1l2\u03b8\u03071 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 sin \u03b82 +m2gl2 cos (\u03b81 + \u03b82) . (9.331) The generalized forces Q1 and Q2 are the required forces to drive the generalized coordinates. In this case, Q1 is the torque at the base motor and Q2 is the torque of the motor at m1. The equations of motion can be rearranged to have a more systematic form Q1 = \u00a1 (m1 +m2) l 2 1 +m2l2 (l2 + 2l1 cos \u03b82) \u00a2 \u03b8\u03081 +m2l2 (l2 + l1 cos \u03b82) \u03b8\u03082 \u22122m2l1l2 sin \u03b82 \u03b8\u03071\u03b8\u03072 \u2212m2l1l2 sin \u03b82 \u03b8\u0307 2 2 +(m1 +m2) gl1 cos \u03b81 +m2gl2 cos (\u03b81 + \u03b82) (9.332) Q2 = m2l2 (l2 + l1 cos \u03b82) \u03b8\u03081 +m2l 2 2\u03b8\u03082 +m2l1l2 sin \u03b82 \u03b8\u0307 2 1 +m2gl2 cos (\u03b81 + \u03b82) . (9.333) Example 371 Mechanical energy. If a system of masses mi are moving in a potential force field Fmi = \u2212\u2207iV (9.334) their Newton equations of motion will be mir\u0308i = \u2212\u2207iV i = 1, 2, \u00b7 \u00b7 \u00b7n. (9.335) The inner product of equations of motion with r\u0307i and adding the equations nX i=1 mir\u0307i \u00b7 r\u0308i = \u2212 nX i=1 r\u0307i \u00b7\u2207iV (9.336) and then, integrating over time 1 2 nX i=1 mir\u0307i \u00b7 r\u0307i = \u2212 Z nX i=1 ri \u00b7\u2207iV (9.337) shows that K = \u2212 Z nX i=1 \u00b5 \u2202V \u2202xi xi + \u2202V \u2202yi yi + \u2202V \u2202zi zi \u00b6 = \u2212V +E (9.338) 9. Applied Dynamics 567 where E is the constant of integration. E is called the mechanical energy of the system and is equal to kinetic plus potential energies. E = K + V (9.339) Example 372 Falling wheel. Figure 9.16 illustrates a wheel turning, without slip, over a cylindrical hill. We may use the conservation of mechanical energy to find the angle at which the wheel leaves the hill. At the initial instant of time, the wheel is at point A. We assume the initial kinetic and potential, and hence, the mechanical energies are zero. When the wheel is turning over the hill, its angular velocity, \u03c9, is \u03c9 = v r (9.340) where v is the speed at the center of the wheel. At any other point B, the wheel achieves some kinetic energy and loses some potential energy. At a certain angle, where the normal component of the weight cannot provide more centripetal force, mg cos \u03b8 = mv2 R+ r . (9.341) the wheel separates from the surface. Employing the conservation of energy, we have EA = EB (9.342) KA + VA = KB + VB. (9.343) The kinetic and potential energy at the separation point B are KB = 1 2 mv2 + 1 2 IC\u03c9 2 (9.344) VB = \u2212mg (R+ r) (1\u2212 cos \u03b8) (9.345) 568 9. Applied Dynamics where IC is the mass moment of inertia for the wheel about its center. Therefore, 1 2 mv2 + 1 2 IC\u03c9 2 = mg (R+ r) (1\u2212 cos \u03b8) (9.346) and substituting (9.340) and (9.341) provides\u00b5 1 + IC mr2 \u00b6 (R+ r) g cos \u03b8 = 2g (R+ r) (1\u2212 cos \u03b8) (9.347) and therefore, the separation angle is \u03b8 = cos\u22121 2mr2 IC + 3mr2 . (9.348) Let\u2019s examine the equation for a disc wheel with IC = 1 2 mr2. (9.349) and find the separation angle. \u03b8 = cos\u22121 4 7 (9.350) \u2248 0.96 rad \u2248 55.15 deg Example 373 Turning wheel over a step. Figure 9.17 illustrates a wheel of radius R turning with speed v to go over a step with height H < R. We may use the principle of energy conservation and find the speed of the wheel after getting across the step. Employing the conservation of energy, 9. Applied Dynamics 569 we have EA = EB (9.351) KA + VA = KB + VB (9.352) 1 2 mv21 + 1 2 IC\u03c9 2 1 + 0 = 1 2 mv22 + 1 2 IC\u03c9 2 2 +mgH (9.353)\u00b5 m+ IC R2 \u00b6 v21 = \u00b5 m+ IC R2 \u00b6 v22 + 2mgH (9.354) and therefore, v2 = vuutv21 \u2212 2gH 1 + IC mR2 . (9.355) The condition for having a real v2 is v1 > vuut 2gH 1 + IC mR2 . (9.356) The second speed (9.355) and the condition (9.356) for a solid disc are v2 = r v21 \u2212 4 3 Hg (9.357) v1 > r 4 3 Hg (9.358) because we assumed that IC = 1 2 mR2. (9.359) Example 374 Trebuchet. A trebuchet, shown schematically in Figure 9.18, is a shooting weapon of war powered by a falling massive counterweight m1. A beam AB is pivoted to the chassis with two unequal sections a and b. The figure shows a trebuchet at its initial configuration. The origin of a global coordinate frame is set at the pivot point. The counterweight m1 is at (x1, y1) and is hinged at the shorter arm of the beam at a distance c from the end B. The mass of the projectile is m2 and it is at the end of a massless sling with a length l attached to the end of the longer arm of the beam. The three independent variable angles \u03b1, \u03b8, and \u03b3 describe the motion of the device. We consider the parameters a, b, c, d, l, m1, and m2 constant, and determine the equations of motion by the Lagrange method. Figure 9.19 illustrates the trebuchet when it is in motion. The position coordinates of masses m1 and m2 are x1 = b sin \u03b8 \u2212 c sin (\u03b8 + \u03b3) (9.360) y1 = \u2212b cos \u03b8 + c cos (\u03b8 + \u03b3) (9.361) 570 9. Applied Dynamics and x2 = \u2212a sin \u03b8 \u2212 l sin (\u2212\u03b8 + \u03b1) (9.362) y2 = \u2212a cos \u03b8 \u2212 l cos (\u2212\u03b8 + \u03b1) . (9.363) Taking a time derivative provides the velocity components x\u03071 = b\u03b8\u0307 cos \u03b8 \u2212 c \u00b3 \u03b8\u0307 + \u03b3\u0307 \u00b4 cos (\u03b8 + \u03b3) (9.364) y\u03071 = b\u03b8\u0307 sin \u03b8 \u2212 c \u00b3 \u03b8\u0307 + \u03b3\u0307 \u00b4 sin (\u03b8 + \u03b3) (9.365) x\u03072 = l (c\u2212 \u03b1\u0307) cos (\u03b1\u2212 \u03b8)\u2212 a\u03b8\u0307 cos (\u03b8) (9.366) y\u03072 = a\u03b8\u0307 sin \u03b8 \u2212 l \u00b3 \u03b8\u0307 \u2212 \u03b1\u0307 \u00b4 sin (\u03b1\u2212 \u03b8) . (9.367) 9. Applied Dynamics 571 which shows that the kinetic energy of the system is K = 1 2 m1v 2 1 + 1 2 m2v 2 2 = 1 2 m1 \u00a1 x\u030721 + y\u030721 \u00a2 + 1 2 m2 \u00a1 x\u030722 + y\u030722 \u00a2 = 1 2 m1 \u00b3\u00a1 b2 + c2 \u00a2 \u03b8\u0307 2 + c2\u03b3\u03072 + 2c2\u03b8\u0307\u03b3\u0307 \u00b4 \u2212m1bc\u03b8\u0307 \u00b3 \u03b8\u0307 + \u03b3\u0307 \u00b4 cos \u03b3 + 1 2 m2 \u00b3\u00a1 a2 + l2 \u00a2 \u03b8\u0307 2 + l2\u03b1\u03072 \u2212 2l2\u03b8\u0307\u03b1\u0307 \u00b4 \u2212m2al\u03b8\u0307 \u00b3 \u03b8\u0307 \u2212 \u03b1\u0307 \u00b4 cos (2\u03b8 \u2212 \u03b1) . (9.368) The potential energy of the system can be calculated by y-position of the masses. V = m1gy1 +m2gy2 = m1g (\u2212b cos \u03b8 + c cos (\u03b8 + \u03b3)) +m2g (\u2212a cos \u03b8 \u2212 l cos (\u2212\u03b8 + \u03b1)) (9.369) Having the energies K and V , we can set up the Lagrangean L. L = K \u2212 V (9.370) Using the Lagrangean, we are able to find the three equations of motion. d dt \u00b5 \u2202L \u2202\u03b8\u0307 \u00b6 \u2212 \u2202L \u2202\u03b8 = 0 (9.371) d dt \u00b5 \u2202L \u2202\u03b1\u0307 \u00b6 \u2212 \u2202L \u2202\u03b1 = 0 (9.372) d dt \u00b5 \u2202L \u2202\u03b3\u0307 \u00b6 \u2212 \u2202L \u2202\u03b3 = 0. (9.373) The trebuchet appeared in 500 to 400 B.C. China and was developed by Persian armies around 300 B.C. It was used by the Arabs against the Romans during 600 to 1200 A.D. The trebuchet is also called the manjaniq, catapults, or onager. The \"Manjaniq\" is the root of the words \"machine\" and \"mechanic\". 9.7 Summary The translational and rotational equations of motion for a rigid body, expressed in the global coordinate frame, are GF = Gd dt Gp (9.374) GM = Gd dt GL (9.375) 572 9. Applied Dynamics where GF and GM indicate the resultant of the external forces and moments applied on the rigid body, measured at the mass center C. The vector Gp is the momentum and GL is the moment of momentum for the rigid body at C p = mv (9.376) L = rC \u00d7 p. (9.377) The expression of the equations of motion in the body coordinate frame are BF = Gp\u0307+ B G\u03c9B \u00d7 Bp = m BaB +m B G\u03c9B \u00d7 BvB (9.378) BM = BL\u0307+B G\u03c9B \u00d7 BL = BI B G\u03c9\u0307B + B G\u03c9B \u00d7 \u00a1 BI B G\u03c9B \u00a2 (9.379) where I is the moment of inertia for the rigid body. I = \u23a1\u23a3 Ixx Ixy Ixz Iyx Iyy Iyz Izx Izy Izz \u23a4\u23a6 . (9.380) 9. Applied Dynamics 573 The elements of I are functions of the mass distribution of the rigid body and are defined by Iij = Z B \u00a1 r2i \u03b4mn \u2212 ximxjn \u00a2 dm , i, j = 1, 2, 3 (9.381) where \u03b4ij is Kronecker\u2019s delta. Every rigid body has a principal body coordinate frame in which the moment of inertia is in the form BI = \u23a1\u23a3 I1 0 0 0 I2 0 0 0 I3 \u23a4\u23a6 . (9.382) The rotational equation of motion in the principal coordinate frame simplifies to M1 = I1\u03c9\u03071 \u2212 (I2 \u2212 I2)\u03c92\u03c93 M2 = I2\u03c9\u03072 \u2212 (I3 \u2212 I1)\u03c93\u03c91 (9.383) M3 = I3\u03c9\u03073 \u2212 (I1 \u2212 I2)\u03c91\u03c92. The equations of motion for a mechanical system having n DOF can also be found by the Lagrange equation d dt \u00b5 \u2202L \u2202q\u0307r \u00b6 \u2212 \u2202L \u2202qr = Qr r = 1, 2, \u00b7 \u00b7 \u00b7n (9.384) L = K \u2212 V (9.385) where L is the Lagrangean of the system, K is the kinetic energy, V is the potential energy, and Qr is the nonpotential generalized force. Qr = nX i=1 \u00b5 Qix \u2202fi \u2202q1 +Qiy \u2202gi \u2202q2 +Qiz \u2202hi \u2202qn \u00b6 (9.386) The parameters qr, r = 1, 2, \u00b7 \u00b7 \u00b7 , n are the generalized coordinates of the system, Q = \u00a3 Qix Qiy Qiz \u00a4T is the external force acting on the ith particle of the system, and Qr is the generalized force associated to qr. When (xi, yi, zi) are Cartesian coordinates in a globally fixed coordinate frame for the particle mi, then its coordinates may be functions of another set of generalized coordinates q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn and possibly time t. xi = fi(q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, t) (9.387) yi = gi(q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, t) (9.388) zi = hi(q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, t) (9.389) 574 9. Applied Dynamics 9.8 Key Symbols a, b, w, h length a acceleration C mass center d position vector of the body coordinate frame df infinitesimal force dm infinitesimal mass dm infinitesimal moment E mechanical energy F force FC Coriolis force g gravitational acceleration H height I moment of inertia matrix I1, I2, I3 principal moment of inertia K kinetic energy l directional line L moment of momentum L = K \u2212 V Lagrangean m mass M moment p momentum P,Q points in rigid body r radius of disc r position vector R radius R rotation matrix t time u\u0302 unit vector to show the directional line v \u2261 x\u0307, v velocity V potential energy w eigenvector W work W eigenvector matrix x, y, z, x displacement \u03b4ij Kronecker\u2019s delta \u0393ij,k Christoffel operator \u03bb eigenvalue \u03d5, \u03b8, \u03c8 Euler angles \u03c9,\u03c9 angular velocity k parallel \u22a5 orthogonal 9. Applied Dynamics 575 Exercises 1. Kinetic energy of a rigid link. Consider a straight and uniform bar as a rigid bar. The bar has a mass m. Show that the kinetic energy of the bar can be expressed as K = 1 6 m (v1 \u00b7 v1 + v1 \u00b7 v2 + v2 \u00b7 v2) where v1 and v2 are the velocity vectors of the endpoints of the bar. 2. Discrete particles. There are three particles m1 = 10kg, m2 = 20 kg, m3 = 30 kg, at r1 = \u23a1\u23a3 1 \u22121 1 \u23a4\u23a6 r1 = \u23a1\u23a3 \u22121\u22123 2 \u23a4\u23a6 r1 = \u23a1\u23a3 2 \u22121 \u22123 \u23a4\u23a6 . Their velocities are v1 = \u23a1\u23a3 2 1 1 \u23a4\u23a6 v1 = \u23a1\u23a3 \u221210 2 \u23a4\u23a6 v1 = \u23a1\u23a3 3 \u22122 \u22121 \u23a4\u23a6 . Find the position and velocity of the system at C. Calculate the system\u2019s momentum and moment of momentum. Calculate the system\u2019s kinetic energy and determine the rotational and translational parts of the kinetic energy. 3. Newton\u2019s equation of motion in the body frame. Show that Newton\u2019s equation of motion in the body frame is\u23a1\u23a3 Fx Fy Fz \u23a4\u23a6 = m \u23a1\u23a3 ax ay az \u23a4\u23a6+ \u23a1\u23a3 0 \u2212\u03c9z \u03c9y \u03c9z 0 \u2212\u03c9x \u2212\u03c9y \u03c9x 0 \u23a4\u23a6\u23a1\u23a3 vx vy vz \u23a4\u23a6 . 4. Work on a curved path. A particle of mass m is moving on a circular path given by GrP = cos \u03b8 I\u0302 + sin \u03b8 J\u0302 + 4 K\u0302. Calculate the work done by a force GF when the particle moves from \u03b8 = 0 to \u03b8 = \u03c0 2 . (a) GF = z2 \u2212 y2 (x+ y) 2 I\u0302 + y2 \u2212 x2 (x+ y) 2 J\u0302 + x2 \u2212 y2 (x+ z) 2 K\u0302 576 9. Applied Dynamics (b) GF = z2 \u2212 y2 (x+ y) 2 I\u0302 + 2y x+ y J\u0302 + x2 \u2212 y2 (x+ z) 2 K\u0302 5. Principal moments of inertia. Find the principal moments of inertia and directions for the following inertia matrices: (a) [I] = \u23a1\u23a3 3 2 2 2 2 0 2 0 4 \u23a4\u23a6 (b) [I] = \u23a1\u23a3 3 2 4 2 0 2 4 2 3 \u23a4\u23a6 (c) [I] = \u23a1\u23a3 100 20 \u221a 3 0 20 \u221a 3 60 0 0 0 10 \u23a4\u23a6 6. Rotated moment of inertia matrix. A principal moment of inertia matrix B2I is given as [I] = \u23a1\u23a3 3 0 0 0 5 0 0 0 4 \u23a4\u23a6 . The principal frame was achieved by rotating the initial body coordinate frame 30 deg about the x-axis, followed by 45 deg about the z-axis. Find the initial moment of inertia matrix B1I. 7. Rotation of moment of inertia matrix. Find the required rotation matrix that transforms the moment of inertia matrix [I] to an diagonal matrix. [I] = \u23a1\u23a3 3 2 2 2 2 0.1 2 0.1 4 \u23a4\u23a6 8. F Cubic equations. The solution of a cubic equation ax3 + bx2 + cx+ d = 0 9. Applied Dynamics 577 where a 6= 0, can be found in a systematic way. Transform the equation to a new form with discriminant 4p3 + q2, y3 + 3py + q = 0 using the transformation x = y \u2212 b 3a , where, p = 3ac\u2212 b2 9a2 q = 2b3 \u2212 9abc+ 27a2d 27a3 . The solutions are then y1 = 3 \u221a \u03b1\u2212 3 p \u03b2 y2 = e 2\u03c0i 3 3 \u221a \u03b1\u2212 e 4\u03c0i 3 3 p \u03b2 y3 = e 4\u03c0i 3 3 \u221a \u03b1\u2212 e 2\u03c0i 3 3 p \u03b2 where, \u03b1 = \u2212q + p q2 + 4p3 2 \u03b2 = \u2212q + p q2 + 4p3 2 . For real values of p and q, if the discriminant is positive, then one root is real, and two roots are complex conjugates. If the discriminant is zero, then there are three real roots, of which at least two are equal. If the discriminant is negative, then there are three unequal real roots. Apply this theory for the characteristic equation of the matrix [I] and show that the principal moments of inertia are real. 9. Kinematics of a moving car on the Earth. The location of a vehicle on the Earth is described by its longitude \u03d5 from a fixed meridian, say, the Greenwich meridian, and its latitude \u03b8 from the equator, as shown in Figure 9.20. We attach a coordinate frame B at the center of the Earth with the x-axis on the equator\u2019s plane and the y-axis pointing to the vehicle. There are also two coordinate frames E and G where E is attached to the Earth and G is the global coordinate frame. Show that the angular velocity of B and the velocity of the vehicle are B G\u03c9B = \u03b8\u0307 \u0131\u0302B + (\u03c9E + \u03d5\u0307) sin \u03b8 j\u0302B + (\u03c9E + \u03d5\u0307) cos \u03b8 k\u0302 B GvP = \u2212r (\u03c9E + \u03d5\u0307) cos \u03b8 \u0131\u0302B + r\u03b8\u0307 k\u0302. Calculate the acceleration of the vehicle. 578 9. Applied Dynamics \u03b8 X Y Z x y z r \u03c9E Z (a) (b) G B Y z x y E \u03d5 \u03b8 y z \u03c9E P P 9. Applied Dynamics 579 (c) the pivot O has a uniform motion on a circle rO = R cos\u03c9t I\u0302 +R sin\u03c9t J\u0302. 13. Equations of motion from Lagrangean. Consider a physical system with a Lagrangean as L = 1 2 m (ax\u0307+ by\u0307) 2 \u2212 1 2 k (ax+ by) 2 . and find the equations of motion. The coefficients m, k, a, and b are constant. 14. Lagrangean from equation of motion. Find the Lagrangean associated to the following equations of motions: (a) mr2\u03b8\u0308 + k1l1\u03b8 + k2l2\u03b8 +mgl = 0 580 9. Applied Dynamics (b) r\u0308 \u2212 r \u03b8\u0307 2 = 0 r2 \u03b8\u0308 + 2r r\u0307 \u03b8\u0307 = 0 15. Trebuchet. Derive the equations of motion for the trebuchet shown in Figure 9.18. 16. Simplified trebuchet. Three simplified models of a trebuchet are shown in Figures 9.23 to 9.25. Derive and compare their equations of motion. 9. Applied Dynamics 581 10"
        ]
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        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure11.4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure11.4-1.png",
        "caption": "Figure 11.4 (a) Model of a column and its load. (b) The boundary condition (end condition) N = \u221246.4 kN follows from the force equilibrium in the x direction of the small end segment in A (with x \u2192 0). (c) N diagram.",
        "texts": [
            " They can be derived from the equilibrium of a small member segment with length x ( x \u2192 0) at the boundaries of the field (an end and/or a joining). For a statically determinate member, there are always sufficient boundary conditions to find the normal force variation without previously determining the support reactions. We will illustrate this by means of two examples previously covered in Section 10.2.3: \u2022 A column subject to its dead weight. \u2022 A simply supported member that is loaded over two-thirds of its length by a uniformly distributed axial load along the member axis. Example 1 Figure 11.4a gives the model of a column and its load. Question: Determine the variation of the normal force (the N diagram) from the differential equations for the equilibrium. Solution: The units used are m and kN; they are omitted hereafter from the calculation. 438 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM In the given coordinate system: qx = 1.92 kN/m so that N = \u2212 \u222b qx dx = \u2212 \u222b 1.92 dx = (\u22121.92x + C) kN. (a) The integration constant C is found from the fact that there is a compressive force of 46.4 kN at the top of the column. The boundary condition (end condition) is therefore x = 0 : N = \u221246.6 kN. (b) This boundary condition can also be derived more formally from the force equilibrium in the x direction of a small member segment with length x ( x \u2192 0) at the top of the column (see Figure 11.4b): \u2211 Fx = 46.4 + N = 0 \u2192 N = \u221246.4 kN. With x \u2192 0 the contribution of qx x, due to the distributed load, disappears. Substitute the values of x and N from (b) in (a) and we find C = \u221246.6 kN. This gives the variation of the normal force N : N = (\u22121.92x \u2212 46.4) kN. The normal force diagram is shown in Figure 11.4c. The results agree with what we found earlier in Section 10.2.3, Example 1. 11 Mathematical Description of the Relationship between Section Forces and Loading 439 Figure 11.5 (a) A simply supported member with a uniformly distributed axial load on section BC. (b) The boundary condition (joining condition) at B, N(1) = N(2), follows from the force equilibrium in the x direction of a small member segment at the joining at B (with x \u2192 0). (c) The boundary condition (end condition) at C, N(2) = 0, follows from the force equilibrium in the x direction of a small end segment at C (with x \u2192 0)"
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        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure3.1-1.png",
        "caption": "Fig. 3.1: A scheme for the manipulator architecture of a robot with the arm, wrist, and end-effector.",
        "texts": [
            " This case of study is an example of how a proper manipulation analysis through elementary actions can give useful results and suggestions to enhance and optimize a robotized manipulation in terms of both manipulation robot programming and robotic work-cell design. Chapter 3: Fundamentals of the mechanics of robots The manipulator architecture of a robot is composed of an arm mostly for movements of translation, a wrist for movement of orientation, and an end-effector for interaction with the environment and/or external objects, as shown in Fig. 3.1 Generally, the term manipulator refers specifically to the arm design, but it can also include the wrist when attention is addressed to the overall manipulation characteristics of a robot. A kinematic study of robots deals with the determination of configuration and motion of manipulators by looking at the geometry during the motion, but without considering the actions that generate or limit the manipulator motion. Therefore, a kinematic study makes it possible to determine and design the motion characteristics of a manipulator but independently from the mechanical design details and actuator\u2019s capability",
            " Secondary workspaces can be defined also as regions of points that can be reached several times and it is convenient to determine regions with the same number of reaches. Thus, a measure of the reachability for each rectangle Pij of a workspace grid can be introduced as a function of the number nq of times that a rectangle has been reached during the scanning process of the joints\u2019 mobility. Because of this numerical definition, a frequency matrix with entries fqij can be generated during the generation of Pij itself by giving to fqij the values of the number of times that Pij has been reached. A numerical example has been reported in Fig. 3.1.5 for a COMAU robot SMART 6.100A by using the aforementioned binary matrix procedure for workspace determination and evaluation. Chapter 3: Fundamentals of the Mechanics of Robots98 The boundary of an (N-j+1)R hyper-ring can be expressed algebraically when it is thought to be generated by enveloping the torus family traced by the parallel circles in the boundary of the revolving (N-j)R hyper-ring, according to Eq.(3.1.32). An equation for a torus family can be expressed with respect to the j-th link frame, assuming C j \u2260 0 and cos\u03b1 j \u2260 0, as a function of the radial r j and axial z j reaches, in the form B+)D+zC(+)A-z+r( 0=j 2 jjj 2 j 2 j 2 j (3"
        ],
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    {
        "image_filename": "designv10_0_0003835_j.actamat.2005.05.003-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003835_j.actamat.2005.05.003-Figure1-1.png",
        "caption": "Fig. 1. During laser powder deposition a material in powder form is injected into a laser beam and melted simultaneously with a thin layer of the substrate to form a continuous track of material. Partial overlapping of individual tracks in a suitable pattern produces a continuous layer of material. By overlapping such layers, threedimensional objects are generated. Alternative schematics, showing additional details of this rapid manufacturing process, can be found in [3,8,13].",
        "texts": [
            " This model was applied to the study of the influence of substrate size and idle time between the deposition of consecutive layers on the microstructure and hardness of a ten-layer AISI 420 steel wall built by LPD. The results show that the thermal history and, hence, the microstructure and properties of the final part, depend significantly on these parameters. 2005 Acta Materialia Inc. Published by Elsevier Ltd. All rights reserved. Keywords: Solid freeform processes; Laser deposition; Finite element analysis; Phase transformation kinetics; Martensitic steels Laser powder deposition (LPD) (Fig. 1) [1\u201310] is a very promising technique for the rapid manufacture, repair and modification of metallic components. Although the advantages of the process are widely recognized, several authors [11\u201315] have reported that parts built by LPD often present non-uniform microstructure and properties and this may restrain the wide acceptance of this process in industry. The microstructure and 1359-6454/$30.00 2005 Acta Materialia Inc. Published by Elsevier Ltd. A doi:10.1016/j.actamat.2005.05.003 * Corresponding author"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002349_access.2017.2661967-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002349_access.2017.2661967-Figure4-1.png",
        "caption": "Fig. 4: Four fault conditions of the experimental bearings.",
        "texts": [
            " In this study, PT 500 machinery diagnostic system [26] is used to collect vibration signals of four conditions of roller bearing, as shown in Fig. 3 (a) and (b). Roller bearing faults kit is composed of motor assembly, motor control unit, shaft, four types of bearings, belt drive kit, and computerised vibration analyser. The control unit is used to collect speed and horsepower data. The piezo-electric sensor and measurement amplifier are used for acceleration measurement. During the tests, vibration data were captured at a sampling frequency of 8kHz for different bearing conditions. Roller bearings used in this paper are illustrated in Fig. 4, which are bearing A without damage, bearing B with outer race damage, bearing C with inner race damage, and bearing D with rolling element damage. To study and evaluate the performance of the proposed approach, four conditions of roller bearing were monitored and respectively recorded under five rotating speeds. That is, 1000, 1500, 2000, 2500 and 3000 r.p.m. The experimental data set contains four conditions of roller bearing, for each condition in one speed, 200 data sets are used, and therefore the total number of data set corresponding to four roller bearings is 800"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure5.20-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure5.20-1.png",
        "caption": "FIGURE 5.20. Axis of rotation u\u0302 when it is coincident with the local z-axis.",
        "texts": [
            "413) coincides with a global coordinate axis, then the Equations (5.20), (5.21), or (5.22) will be reproduced. Proof. Interestingly, the effect of rotation \u03c6 about an axis u\u0302 is equivalent to a sequence of rotations about the axes of a local frame in which the 284 5. Applied Kinematics local frame is first rotated to bring one of its axes, say the z-axis, into coincidence with the rotation axis u\u0302, followed by a rotation \u03c6 about that local axis, then the reverse of the first sequence of rotations. Figure 5.20 illustrates an axis of rotation u\u0302 = u1I\u0302+u2J\u0302+u3K\u0302 , the global frame G (OXY Z), and the rotated local frame B (Oxyz) when the local z-axis is coincident with u\u0302. Based on Figure 5.20, the local frame B (Oxyz) undergoes a sequence of rotations \u03d5 about the z-axis and \u03b8 about the yaxis to bring the local z-axis into coincidence with the rotation axis u\u0302, followed by rotation \u03c6 about u\u0302, and then perform the sequence backward. Therefore, the rotation matrix GRB to map coordinates in local frame to their coordinates in global frame after rotation \u03c6 about u\u0302 is GRB = BR\u22121G = BRT G = Ru\u0302,\u03c6 = [Rz,\u2212\u03d5Ry,\u2212\u03b8 Rz,\u03c6Ry,\u03b8 Rz,\u03d5] T = RT z,\u03d5R T y,\u03b8 R T z,\u03c6R T y,\u2212\u03b8 R T z,\u2212\u03d5 (5.421) but sin\u03d5 = u2p u21 + u22 (5"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003669_ivs.2007.4290230-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003669_ivs.2007.4290230-Figure2-1.png",
        "caption": "Fig. 2 3D Quadrotor model.",
        "texts": [
            ", proposed a controller which is based upon the compensation of the Coriolis and gyroscopic torques and the use of PD 2 feedback structure [25]. II. DYNAMIC MODELING A quadrotor is an under actuated aircraft with fixed pitch angle four rotors as shown in Figure 1. Modeling a vehicle such as a quadrotor is not an easy task because of its complex structure. Our aim is to develop a model of the vehicle as realistically as possible. A 1-4244-1068-1/07/$25.00 \u00a92007 IEEE. 894 Having four rotors with fixed angles makes quadrotor has four input forces which are basically the thrust provided by each propellers as shown in Figure 2. Forward (backward) motion is maintained by increasing (decreasing) speed of front (rear) rotor speed while decreasing (increasing) rear (front) rotor speed simultaneously which means changing the pitch angle. Left and right motion is accomplished by changing roll angle by the same way. The front and rear motors rotate counter-clockwise while other motors rotate clockwise so yaw command is derived by increasing (decreasing) counter-clockwise motors speed while decreasing (increasing) clockwise motor speeds. The dynamical model and expressions in [25] used with a minor alteration. Let I = {ex, ey, ez} signify an inertial frame, and A = {ex, ey, ez} express a frame which is rigidly attached to the aircraft as shown in Fig.2. Model of the quadrotor derived as following, { } 1 Re ( ) , 1, 2,3, 4 z z f f a a r i i i v v ge T m R RS I I G I Q i \u03be \u03c4 \u03c9 \u03c4 = = \u2212 = \u03a9 \u03a9 = \u2212\u03a9\u00d7 \u03a9 \u2212 + = \u2212 \u2208 where the vector [ ]Tx y z\u03be = denotes the position of the origin of the body fixed frame A with respect to the inertial frame I , the vector denotes the linear velocity of the origin of A expressed in I, denotes the angular velocity of the airframe expressed in the body fixed frame A. T \u23a4\u23a6 ]T\u03a9 = \u03a9 \u03a9 \u03a9 x y zv v v v\u23a1= \u23a3 1 2 3[ \u00d7 denotes the vector cross product, m represents the mass of the airframe"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001943_tim.2020.3031194-Figure15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001943_tim.2020.3031194-Figure15-1.png",
        "caption": "Fig. 15: The clamping mechanism for Hybrid inspection UAV [81]. (a)UAV system; (b)Clamping mechanism;",
        "texts": [
            " designed a guiding wheel-based hybrid UAV [80], as shown in Fig.14(a). To realize the high-efficiency power inspection, the flying mode was used for the crossing obstacles. And the translation mode was for the movement along power lines through the guiding wheel. Fig.14(b) shows the diagram of hybrid operation mode by the simulation experiments. To improve the stability and safety of the UAVs, Wang et al. designed a clamping mechanism for the hybrid inspection UAV [81]. And the light-weight and rigid material was adopted to keep the weight of UAVs. Fig.15 shows the special structures of the UAV system and the clamping mechanism. To realize longterm and close-proximity power line inspection, inspired by the cooperative idea, Bian et al. proposed a novel hybrid inspection scheme based on multi robots for autonomous power line inspection [82]. As shown in Fig.16(a), it could be divided into two parts, UAV system and a climbing robot. The UAV system was in charge of transportation, loading and unloading of the climbing robot. And the climbing robot was designed for sensor data collection which was used for the following defect detection"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000206_978-0-387-74315-8-Figure9-3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000206_978-0-387-74315-8-Figure9-3-1.png",
        "caption": "Figure 9-3: Part of an articulated body",
        "texts": [
            ") Let Akl be the matrix sa sks, pll Pk\"l . s\u00b7 . . . . Akl = Rk (4' - 4' - 4' + 4' ) R, ' (9.27) \"\" pll IIp pp then Equation (9.26) is given by 170 L I: Akl fi + {Jk = O , (9.28) 1=1 and combining the equations for allloops gives Au AlL fI {Jl + =0 (9.29) ALI ALL fL {JL Equation (9.29) can be solved for the loop-closure forces (assuming the coefficient matrix is not singular), and once the forces are known they can be substituted into Equation (9.18) to find the accelerations of the handles. 171 Articulated Body Referring to Figure 9-3, let bl and b2 be the handles of the loop-free articulated bodies Bl and B2 which form the articulated body B when connected as shown. We are interested in calculating the articulated-body inertias of several handles in the one articulated body B. Let bl and b2 have articulated-body inertias \u00ee1 and \u00ee: in Bl and B2 respectively, and \u00ee~ and \u00ee: in B. From Equations (6.17 and 7.19) we have that (9.30) and (9.31) where S is the motion sub-space of the joint connecting bl and b2\u2022 Making -B- -A--B -B -A use of the relation 11\u00b7 S = 11 S, 12 can be expressed in terms of 11 and 12 as follows The corresponding equation for bias forces is Equations (9",
            " The simplest way to calculate cross-inertias is to calculate them from acceleration propagators. The acceleration propagator A.. gives the lJ acceleration of link i (ignoring bias accelerations) in terms of the acceleration of link j which was induced by a test force on link J', or on a link connected to link i via link j, according to . _. a.=A .. a .. l 'J J (9.34) The inverse cross-inertia is expressed in terms of the acceleration propagator as . .. iP .. = A .. iP .. lJ 'J J (9.35) Referring to Figure 9-3, if an acceleration al is induced in bl by a force acting on some link in BI' then the acceleration a 2 induced in b2 is given by (9.36) The force f which produces this acceleration is given by (9.37) and satisfies \u2022 S' S f=O. (9.38) The acceleration propagator is then calculated as follows: 173 \"S\"A\" _lAS\"A\" a = - (S 12 S) S 12 al' A \"\"\" S-A;. -1 \" S-A - a 2 = (1- S( S 12 ~) S 12 ) al' so (9.39) \u00ee: may be substituted for \u00ee: in Equation (9.39) since \u00ee: s= ~ s. Acceleration propagators across several joints can be constructed as products of acceleration propagators across single joints"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000832_j.addma.2016.05.007-Figure9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000832_j.addma.2016.05.007-Figure9-1.png",
        "caption": "Figure 9. Residual stress information in Y direction (cooled down to 20 oC).",
        "texts": [
            " The maximum temperatures of different scanning strategies were also different from each other at this time point. Generally speaking, the island scanning case has the highest maximum temperature which is caused by residual heat effect from short scanning path in a single island. In addition, steady state maximum temperature can be very hard to reach in these cases due to frequent change of scanning direction, path angle and length. Residual stress contours in X and Y directions for the substrate and deposited layers at eight scanning strategies have been summarized in Figure 8 and Figure 9. Compressive stress has been observed to be concentrated at the regions close to four corners of deposited layers in the substrate for all cases. High S11 (or S22) stresses of all cases have been shown at the two edges of the deposited layers along X (or Y) direction and gradually reduced when the location approached the center. The directional stress distribution phenomenon indicates that all the tested scanning strategies may not be able to solve the edge/interface stress concentration problem; potential part crack still exists at these regions"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002672_j.mechmachtheory.2020.103786-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002672_j.mechmachtheory.2020.103786-Figure7-1.png",
        "caption": "Fig. 7. Dynamic model of 6-DOF gear system.",
        "texts": [
            " 6 (b) and (c) are the stiffness of spalling defect in the double\u2013single tooth engagement region and double-tooth engagement region when L = 8 mm and w = 0.65 mm, respec- tively. By comparing Fig. 6 (b) with (c), it can be found that the stiffness of the single-tooth engagement region decreases more than that of the double-tooth engagement region. This is because when the gear meshes, the stress of a single tooth is larger; thus, the spalling fault is usually located in the single-tooth engagement region. A 6-DOF lumped-parameter model is established. As shown in Fig. 7 , the dynamic system can be described as a onestage gearbox consisting of a motor, driver (pinion), driven gear, and load. Here, m, I, T, \u03c9, k x , k y , c x , and c y represent the mass, moment of inertia, torque applied, rotating speed, radial support stiffness in the x direction, radial support stiffness in the y direction, radial support damping in the x direction, and radial support damping in the y direction, respectively, while the subscripts 1 and 2 denote the pinion and driven gear, respectively"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001205_s10846-016-0442-0-Figure9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001205_s10846-016-0442-0-Figure9-1.png",
        "caption": "Fig. 9 Pure Pursuit Strategy",
        "texts": [
            " Similarly, heading error is calculated based on Eq. 13. Pure pursuit method extends this approach at this stage by fitting a smooth circular curve from vehicle to the carrot point for the vehicle to follow. This circular curve is defined such as its line passes through two points; the vehicle position and the carrot points. This curvature will ensure smooth steering for the vehicle and reduce the oscillatory nature of the basic follow-the-carrot method as reported by Wit [15]. The overall method is as depictedin Fig. 9. Given that the circular arc goes through the carrot point and control point on the vehicle with centre at O; a local coordinate axes with origin on the control point; and intended steering angle \u03b4 as shown in Fig. 9. One should notice that the triangle produced is an isosceles triangle with common length R. Using the rule of sine involving look-ahead length, ld and the angle \u03b1 between the vehicle\u2019s heading vector and the look-ahead vector, a simple mathematical equation can be derived in Eq. 17 [14]. ld sin 2\u03b1 = R sin( \u03c0 2 \u2212\u03b1) ld sin \u03b1 cos \u03b1 = R sin \u03c0 2 cos \u03b1\u2212cos \u03c0 2 sin \u03b1 (15) Simplifying, R = ld 2 sin \u03b1 (16) Thus, steering angle can be derived by substituting Eq. 16 into Eq. 1 previously such that: \u03b4 = tan\u22121 ( L R ) = tan\u22121 ( 2L sin \u03b1 ld ) = tan\u22121 \u239b \u239d2L ( ey ld ) ld \u239e \u23a0 = tan\u22121 ( 2Ley l2 d ) (17) It can be seen that the steering angle generated in Eq"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure13.67-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure13.67-1.png",
        "caption": "Figure 13.67 The normal force NCD in member CD follows from the vertical force equilibrium of joint C.",
        "texts": [
            " At 13 Calculating M, V and N Diagrams 599 the middle of AC: M = Mmax = p = 1 8 \u00d7 20 \u00d7 42 = 40 kNm. The V diagram can be determined from the M diagram. In field BC, the shear force is zero, in field AC it varies linearly. The shear forces at A and to the left of C are 2p 1 2 AC = 2 \u00d7 40 1 2 \u00d7 4 = 40 kN. Their deformation symbols follow from the slope of the M diagram (see Figure 13.66c). Check: The shear forces found must agree with the support reactions of the simply supported beam AC. The vertical force equilibrium of joint C gives (see Figure 13.67) NCD = +40 kN. Using the equilibrium of joint D, we can now find the normal forces in AD and BD. They turn out to be compressive forces: NAD = NBD = \u2212100/3 kN. The calculation is left to the reader. In Figure 13.66a, beam ACB has been isolated, and all forces acting on it are shown. At A and B, there are acting support reactions and (components of the) compressive forces exerted by members AD and BD. Check: By reducing the support reactions (pointed upwards) at A and B by the vertical component (pointed downwards) of these member forces we 600 ENGINEERING MECHANICS"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure5.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure5.2-1.png",
        "caption": "Fig. 5.2. Exploded view of the AFPM brushless motor with film coil coreless stator winding and single-sided rotor PM excitation system. Courtesy of Embest , Soeul, South Korea.",
        "texts": [
            " The 356-mm diameter e-TORQTM motors have been used successfully by students of North Dakota State University for direct propulsion of a solar powered car participating in 2003 American Solar Challenge (Fig. 5.1b). A well designed solar-powered vehicle needs a very efficient and very light electric motor to convert the maximum amount of solar energy into mechanical energy at minimum rolling resistance. Coreless AFPM brushless motors satisfy these requirements. Small ironless motors may have printed circuit stator windings or film coil windings. The film coil stator winding has many coil layers while the printed circuit winding has one or two coil layers. Fig. 5.2 shows an ironless brushless motor with film coil stator winding manufactured by EmBest , Soeul, South Korea. This motor has single-sided PM excitation system at one side of the stator and backing steel disc at the other side of the stator. Small film coil motors can be used in computer peripherals, computer hard disc drives (HDDs) [139, 140], mobile phones, pagers, flight recorders, card readers, copiers, printers, plotters, micrometers, labeling machines, video recorders and medical equipment. 5"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001199_b978-0-08-100433-3.00013-0-Figure13.9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001199_b978-0-08-100433-3.00013-0-Figure13.9-1.png",
        "caption": "Figure 13.9 Polycarbonate wiring conduits made by fused deposition modeling. Courtesy Stratasys and Bell helicopter.",
        "texts": [
            " By contrast, FDM tooling reduces machining time by about 68%: from 14 to 4.5 days. High-performance thermoplastic can be adopted for hydroforming applications. For instance, ULTEM 9085, a flame-retardant resin, can withstand pressures up to 10,000 psi and can be produced by Stratasys\u2019s Fortus machine. The FDM process can also be used to directly manufacture aircraft parts. Bell Helicopter [30] builds tough polycarbonate wiring conduits for their heavy-lift \u201ctiltrotor\u201d Osprey; these conduits are shown in Fig. 13.9. Technicians installed the branching conduits in six mating sections inside the Osprey\u2019s twin vertical stabilizers for on-the-ground testing of its wiring. After some minor variations, the team requested five more complete sets for a total of 42 conduit models. It took only 2.5 days to make them all. In the past it would have taken take 6 weeks if the prototypes were made of cast aluminum. NASA [31] also adopted Stratasys\u2019s 3D printer for the Mars rover; about 70 of the parts that make up the rover were built digitally, directly from computer designs, in a production-grade heated chamber"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001270_jproc.2020.3046112-Figure16-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001270_jproc.2020.3046112-Figure16-1.png",
        "caption": "Fig. 16. 3-D-printed (a) liquid cooled [47] and (b) air cooled [48] heat sinks developed by ORNL.",
        "texts": [
            " A variety of coolants and fin designs have been theoretically and experimentally studied in the literature to achieve the best possible cooling of the system. Advanced cooling concepts, such as microchannel cooling, jet impingement, two-phase cold plates, double-sided packaging, and direct substrate cooling, are also being implemented for such high-powerdensity power electronic applications. With advancements in AI, another area of interest is to use such techniques to develop complex thermal management structures. In this area, ORNL has developed complex heat sink structures using AI, as shown in Fig. 16, which were 3-D-printed using Additive Manufacturing Technology and have been demonstrated to outperform their conventional counterparts [47], [48]. These structures were optimized for minimum junction temperatures for highpower-density applications. In a similar direction, new cooling structures are being developed, which alleviates intrinsic thermal management issues in power electronics. For example, a thermal imbalance mitigation scheme proposed by Sahu et al. [68] mitigates the intrinsic heat spreading issues in insulated metal-based substrates by developing suitable liquidcooled heat sinks using AI and Multiphysics finite element simulations (see Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure7.16-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure7.16-1.png",
        "caption": "Fig. 7.16. Mechanical structure of the tactile sensor",
        "texts": [
            " A sensor must be accompanied by an analog signal conditioning device and a digital circuit that controls the scanning of tactile array and transmits the information to the control system. Typical scanning rate for the 16 x 16 taxel matrix is 4 kHz (taxels are scanned at 1 MHz). Tactile sensor based on magnetostriction (the effect of changing the magnetic permeability of the material being deformed) was developed at North Carolina 100 ms 2 3 p- 4 bar 5 7.2 Environment Sensors 221 Fig. 7.15. Dependence of material resistivity for applied pressure State University [7]. Since there are no moving parts (Fig. 7.16), the sensor is reliable and robust. It consists of 16 x 16 taxels, each of which represents a miniature transformer. An A.c. current is applied to the primary coils, and the output is measured after the multiplexer. In this experiment only two levels of the output voltage were used, so the binary picture was produced. The tactile picture was scanned at the rate of 1 kHz. Circuit board The magnetoelastic material that was used showed promising results when the temperature sensitivity (a common problem with resistive elastomers), dynamic range, sensitivity, linearity and small hysteresis were considered"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001205_s10846-016-0442-0-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001205_s10846-016-0442-0-Figure3-1.png",
        "caption": "Fig. 3 Geometric model based on Ackerman Steering Configuration",
        "texts": [
            " As stated earlier, this model only consider the dimension and positions of the vehicle during manoeuvring with no regard to its velocity and acceleration. It is developed based on the Ackerman steering configurations that the line perpendicular to each of the vehicle wheel should intersect at the centre point of the vehicle cornering arc where the radius of turn is R. The importance of this model is in developing one of the most popular path tracking controller, Pure Pursuit [5, 10\u201316]. This model is rarely used to simulate any of the vehicle states. From Fig. 3, angle of turn, \u03b4 is given by the following equation. \u03b4 = tan\u22121 ( L R ) (1) Another geometric model relates the vehicle position with respect to the trajectory and its subsequent error. This model is particularly important in order to estimate the tracking error between the actual vehicle position and the desired trajectory which is the basis of another geometric-type Stanley controller as well as one of the most common quantification method to evaluate a path tracking performance. Development of both the model and quantification method will be described in details within its respective sections"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure12.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure12.3-1.png",
        "caption": "FIGURE 12.3. Three parallel springs.",
        "texts": [
            " We may substitute a set of serial dampers with only one equivalent damping ceq that produces the same velocity x\u0307 under the same force fc. For three parallel dampers, the velocity and force balance x\u0307 = x\u03071 + x\u03072 + x\u03073 (12.13) fc = \u2212c1x\u0307 = \u2212c2x\u0307 = \u2212c3x\u0307 = \u2212ceqx\u0307 (12.14) show that the equivalent damping is 1 ceq = 1 c1 + 1 c2 + 1 c3 . (12.15) We assume that displacement x has no effect on the force of a linear damper. 12. Applied Vibrations 733 Example 424 Parallel springs and dampers. Parallel springs have the same displacement x, with a resultant force, fk, equal to sum of the individual forces P fi. Figure 12.3 illustrates three parallel springs between a massless plate and the ground. The equilibrium position of the springs is the un-stretched configuration shown in Figure 12.3(a). Applying a displacement x to all the springs in Figure 12.3(b) generates the free body diagram shown in Figure 12.3(c). Each spring makes a force \u2212kx opposite to the direction of displacement. The resultant force of the springs is fk = \u2212k1x\u2212 k2x\u2212 k3x. (12.16) We may substitute parallel springs with only one equivalent stiffness keq that produces the same force fk under the same displacement. fk = \u2212keqx (12.17) Therefore, the equivalent stiffness of the parallel springs is sum of their stiffness. keq = k1 + k2 + k3 (12.18) Parallel dampers have the same speed x\u0307, and a resultant force fc equal to the sum of individual forces"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002358_j.arcontrol.2018.10.009-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002358_j.arcontrol.2018.10.009-Figure1-1.png",
        "caption": "Fig. 1. Quadrotor frame.",
        "texts": [
            " Najjaran / Annual Reviews in Control 0 0 0 (2018) 1\u201316 w e f [ o t t t t a d t s h L 4 m c t a p d d g c S m i n c m i 4 d c o l of the center of gravity ( \u03be) and the ZYX Euler angles ( ) are defined. The B-frame attached to the body of the quadrotor is a righthand reference frame denoted by ( O B , x B , y B , z B ) where ( O B ) is the axis origin and coincides with the center of the quadrotor\u2019s cross structure and ( x B , y B , z B ) pointing towards the front, left and up, respectively. The linear velocity ( V ), the angular velocity ( \u03c9), the torques ( ) are defined using the B-frame. Fig. 1 illustrates quadrotor\u2019s structure and reference frames. The linear position ( \u03be) is determined using the vector between the origins of the B-frame and E-frame represented in the E-frame. The Euler angles ( = [ \u03c6 \u03b8 \u03c8 ] T ) representing the attitude of the quadrotor are defined by the orientation of B-frame with respect to the E-frame. A rotation matrix is needed to map the orientation of a vector from B-frame to E-frame ( Gentle, 2007 ), and it is given by: R = [ c \u03b8 c \u03c8 \u2212c \u03c6s \u03c8 + c \u03c8 s \u03c6s \u03b8 s \u03c6s \u03c8 + c \u03c6c \u03c8 s \u03b8 c \u03b8 s \u03c8 c \u03c6c \u03c8 + s \u03c6s \u03b8 s \u03c8 \u2212c \u03c8 s \u03c6 + c \u03c6s \u03b8 s \u03c8 \u2212s \u03b8 c \u03c6s \u03b8 c \u03c6c \u03b8 ] (8) where c x = cos (x) and s x = sin (x) "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure2.14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure2.14-1.png",
        "caption": "Fig. 2.14. Phasor diagrams of salient-pole synchronous machine: (a) overexcited generator (generator arrow system); (b) underexcited motor (consumer arrow system).",
        "texts": [
            "84) The angle \u03a8 is between the q-axis and armature current Ia. When the current arrows are in the opposite direction, the phasors Ia, Iad, and Iaq are reversed by 180o. The same applies to the voltage drops. The location of the armature current Ia with respect to the d- and q-axis for generator and motor mode is shown in Fig. 2.13. Phasor diagrams for synchronous generators are constructed using the generator arrow system. The same system can be used for motors, however, the consumer arrow system is more convenient. An overexcited generator (Fig. 2.14a) delivers both active and reactive power to the load or utility grid. An underexcited motor (Fig. 2.14b) draws both active and reactive power from the line. For example, the load current Ia (Fig. 2.14b ) lags the voltage phasor V1 by the angle \u03c6. An overexcited motor, consequently, draws a leading current from the circuit and delivers reactive power to it. In the phasor diagrams according to Fig. 2.14 [126] the stator core losses have been neglected. This assumption is justified only for power frequency machines with unsaturated armature cores. For an underexcited synchronous motor (Fig. 2.14b) the input voltage V1 projections on the d and q axes are V1 sin \u03b4 = IaqXsq \u2212 IadR1 2.7 Phasor Diagrams 55 V1 cos \u03b4 = Ef + IadXsd + IaqR1 (2.85) where \u03b4 is the load angle between the voltage V1 and EMF Ef (q-axis). For an overexcited motor V1 sin \u03b4 = IaqXsq + IadR1 56 2 Principles of AFPM Machines V1 cos \u03b4 = Ef \u2212 IadXsd + IaqR1 (2.86) The currents of an overexcited motor Iad = V1(Xsq cos \u03b4 \u2212R1 sin \u03b4)\u2212 EfXsq XsdXsq +R2 1 (2.87) Iaq = V1(R1 cos \u03b4 +Xsd sin \u03b4)\u2212 EfR1 XsdXsq +R2 1 (2.88) are obtained by solving the set of eqns (2.85). The rms armature current as a function of V1, Ef , Xsd, Xsq, \u03b4, and R1 is Ia = \u221a I2 ad + I2 aq = V1 XsdXsq +R2 1 \u00d7 \u221a [(Xsq cos \u03b4 \u2212R1 sin \u03b4)\u2212 EfXsq]2 + [(R1 cos \u03b4 +Xsd sin \u03b4)\u2212 EfR1]2 (2.89) The phasor diagram can also be used to find the input electric power. For a motor Pin = m1V1Ia cos\u03c6 = m1V1(Iaq cos \u03b4 \u2212 Iad sin \u03b4) (2.90) Thus, the electromagnetic power for a motor mode is Pelm = Pin \u2212\u2206P1w \u2212\u2206P1Fe = m1[IaqEf + IadIaq(Xsd \u2212Xsq)]\u2212\u2206P1Fe (2.91) On the basis of phasor diagrams (Fig. 2.14), equivalent circuits of a AFPM synchronous machine can be drawn (Fig. 2.15). The main dimensions of a double-sided PM brushless motor with internal disc rotor can be determined using the following assumptions: (a) the electric and magnetic loadings are known; (b) the number of turns per phase per one stator is N1; (c) the phase armature current in one stator winding is Ia; (d) the back EMF per phase per one stator winding is Ef . The peak line current density at the average radius per one stator is expressed by eqn (2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure9.21-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure9.21-1.png",
        "caption": "FIGURE 9.21 Horizontal partitioning of one layer of a metal sheet.",
        "texts": [
            " Other improvements may apply for the modeling of a special class of problems. For example, it was suggested in Ref. [11], that for some geometries, a set of template current cell can be precomputed so that some of the computational work is done before the actual computation of the current. We give an example of the meshing and equivalent circuit for a 3D VFI model for a metal body with a finite thickness. To give details on the meshing, we show the conventional horizontal x\u2013y PEEC cells for a particular level in Fig. 9.21. For most conventional cases, the cells at all levels must be vertically aligned as is shown in Fig. 9.2. The vertical layer-to-layer connection and a top view of this connection is shown in Fig. 9.22. Hence, all three directions of currents internal to the conductors are properly represented in this conventional PEEC mesh. The PEEC equivalent circuit corresponding to a conductor node is shown in Fig. 9.23. Different approximations have been made to include the partial inductance couplings [46\u201348]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure4.8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure4.8-1.png",
        "caption": "FIGURE 4.8. A tire with radius Rw rolling on the ground and moving with velocity v and angular velocity \u03c9w.",
        "texts": [
            " Because of friction in the driveline, especially in the gearbox and torque convertor, the power at the drive wheels is always less than the power at the engine output shaft. The ratio of output power to input power is a number called efficiency \u03b7 = Pout Pin . (4.47) If we show the efficiency of transmission by \u03b7t and the efficiency of torque convertor by \u03b7c, then the overall efficiency of the driveline is \u03b7 = \u03b7c \u03b7t. The power at the wheel is the output power of driveline Pout = Pw and the engine power is the input power to the driveline Pin = Pe. Therefore, Pw = \u03b7Pe. (4.48) Figure 4.8 illustrates a driving wheel with radius Rw that is turning with angular velocity \u03c9w on the ground and moving with velocity vx. vx = Rw \u03c9w (4.49) There are two gearing devices between the engine and the drive wheel: gearbox and differential. Assigning ng for the transmission ratio of the gearbox and nd for the transmission ratio of the differential, the overall transmission ratio of the driveline is n = ng nd. (4.50) So, the angular velocity of the engine \u03c9e is n times of the angular velocity of the drive wheel \u03c9w"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000252_j.automatica.2006.10.008-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000252_j.automatica.2006.10.008-Figure5-1.png",
        "caption": "Fig. 5. Chattering attenuation.",
        "texts": [
            " In the absence of noises the controller provides in finite time for keeping + \u0307 \u2261 0. As a result the asymptotically stable 3-sliding mode =\u0307=\u0308=0 is established. Note that since (19) is quasi-continuous, the applied controller (29)\u2013(34) produces the control u, whose derivative u\u0307 remains continuous until the entrance into the 2-sliding mode =\u0307=0. The graph of u\u0307 is very similar to Fig. 3e and is omitted. The graphs of 3-sliding deviations , \u0307, \u0308 and the control u are demonstrated in Figs. 5a,c, respectively. The 2-sliding convergence in the plane , \u0307 is demonstrated in Fig. 5b. It is seen from Fig. 5d that + \u0307 \u2261 0 is kept in 2-sliding mode. Due to this equality the accuracies achieved at t = 10 are of the same order: | | = |x \u2212 xc| 4.0\u00d710\u22124, |\u0307|=|x\u0307\u2212 x\u0307c| 4.0\u00d710\u22124, |\u0308|=|x\u0308\u2212 x\u0308c| 4.0 \u00d7 10\u22124. 2-Sliding homogeneity and contractivity are shown to provide for all needed features of 2-sliding-mode controllers (Theorems 1\u20134). Construction of controllers is not difficult due to the simplicity of the plane geometry. New finite-time-stable 2-sliding controllers were obtained in such a way, significantly increasing the choice of known 2-sliding controllers"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure7.8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure7.8-1.png",
        "caption": "Figure 7.8 (a) The membrane force n = pr in the circumferential direction is independent of the aperture angle \u03b1. (b) The formula n = pr is also referred to as the boiler formula as calculating the force in the walls of a steam boiler was an important field of application.",
        "texts": [
            " If the dead weight of the gas and the membrane can be ignored, the equation for the vertical force equilibrium is 2Nv \u2212 pb \u00b7 2r sin \u03b1 = 0 so that Nv = pbr sin \u03b1. The horizontal component of N is then Nh = pbr cos \u03b1 250 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM and the resulting support reaction is N = pbr. The calculation relates to a strip with width b. The requested support reactions per length are:1 n = N b = pr. Note that the magnitude of the force n = pr is independent of the aperture angle \u03b1 (see Figure 7.8a). Obviously, in the circular cylinder pneu, the (distributed) circumferential tensile forces have the same magnitude everywhere. The formula is also applicable for a closed ring (see Figure 7.8b, where \u03b1 = 180\u25e6) and is known as the boiler formula, as calculating forces in the walls of a steam boiler was an important field of application. Example 2 The second example relates to a pneu designed by architect Frei Otto2 to cover a settlement in Antarctica (see Figure 7.9). This design is discussed in his book \u201cZugbeanspruchte Konstruktionen\u201d. The pneu is shaped like a segment of a sphere and rests on a concrete ring beam. The diameter of the sphere is r = 2200 m. The diameter of the ring beam is rbeam = 1000 m",
            " In order to prevent this, the dead weight qdw has to be larger than nv = 134 kN/m. If the ring beam is made of concrete, with a specific weight of 24 kN/m3, then the cross-section A of the beam has to obey qdw = A \u00d7 (24 kN/m3) \u2265 nv = 134 kN/m \u21d2 A \u2265 134 kN/m 24 kN/m3 = 5.6 m2. The cross-section of the ring beam has to be at least 5.6 m2. b. Due to the horizontal forces nh, the ring beam is pulled inwards from all sides (see Figure 7.13a). Here, you will recognise the loading case of the closed ring from Figure 7.8b, but now with an underpressure instead of an overpressure. 7 Gas Pressure and Hydrostatic Pressure 253 A compressive force N \u20321 is formed in the ring. This can be calculated using the boiler formula from the previous example, or directly from the equilibrium of the half ring beam in Figure 7.13b: N \u2032 = 2rbeamnh 2 = rbeamnh = (1000 m)(263 kN/m) = 263 MN. Comment: This force is relatively large for a concrete cross-section of 5.6 m2. The compressive force in the ring may therefore call for a larger cross-section"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000005_978-3-642-84379-2-Figure11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000005_978-3-642-84379-2-Figure11-1.png",
        "caption": "Fig. 11",
        "texts": [
            " Let us consider again a system with a scalar discontinuous control and show that, besides the above two types of motion, some other motions may occur on the discontinuity boundary depending on the way the function u is implemented. Let the switch of this system feature a dead zone besides a hysteresis (Fig. 10). Assume for certainty that for u = 0 (i.e. in the dead zone), the projection of vector f(x, t,O) on the gradient to the discontinuity surface is negative, i.e. s = grads'f(x,t,O) < 0 as shown in Fig. 11. In such a case a sliding mode occurs in which the control alternately takes up the values 0 and u-. Consequently, the velocity vector 1\u00b0 1 This effect was observed experimentally on an analog computer for very thin boundary layers. 1 Examples of Discontinuous Systems with Ambiguous Sliding Equations 35 in the state space may be found with the help of Filippov's procedure yielding the point of intersection of the tangential plane with the straight line which connects the ends of vectors f(x,t,u-) and f(x,t, O) (Fig. 11). If the control in the dead zone differs from zero, one may obtain some other velocity vectors whose ends lie on the intersection of the tangential plane and a straight line connecting any two points of the f(x, t, u) locus. For second order systems, the totality of such velocities lie on the segment of the line running between the ends of vectors fO and f(x, t, ueq ) (which refers to the locus presented in Fig. 11, of.course). For a system of an arbitrary order, this totality covers some area on the tangential plane. Thus, depending on the regularization technique used in treating a system nonlinear with respect to scalar control we may obtain different sliding motion equations. This proves that, generally speaking, the equations describing motion in nonlinear systems outside the discontinuity surface do not allow one to unambiguously obtain the equations of the sliding motion over that surface. Like systems with a scalar control, general-type systems (1",
            " This sets forward a problem of finding the set of feasible sliding equations and its subsets if the way control is realized in the boundary layer is known. One important regularity should be noticed in the examples of Sect. 1: in a certain point on the intersection of the discontinuity surfaces, the sliding motion velocity vector always belongs to the minimal convex set which contains all velocity vectors of the system motion in the boundary layer in the neighbourhood of that point1\u2022 If in the first example (Fig. 11) u in the boundary layer is allowed to take up any value between u + and u -, then such a minimal convex set should include a spatial curve, locus of f(x, t, u) for u - ~ u ~ u +. The intersection of this set with the plane tanential to the discontinuity surface determines the totality of righthand parts of the sliding mode differential equations. In particular, a minimal convex set for a second order system will be a segment confined between the locus and the straight line that passes through the ends of vectors f(x, t, u +) and f(x, t, u-)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003774_j.1469-185x.1974.tb01299.x-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003774_j.1469-185x.1974.tb01299.x-Figure4-1.png",
        "caption": "Fig. 4. Features of protozoon Opalina. (a) Metachronal wave patterns -viewed from above (Sleigh, 1966). (b) Beat of cilium (Sleigh, 1968).",
        "texts": [
            " (i) Opalina Opulina ana arum is a multinucleate protozoon found in the rectum of frogs (Sleigh, 1960, 1969); although it is not a true ciliate (Sleigh, 1973) it bears many ciliary organelles, sometimes called flagella. The organism has the shape of a thin flat disk, its average size being 200-300 ,um long, 200 ,um wide and only 20 ,um thick. The ciliary organelles are borne in rows about 3 ,urn apart, with cilia within the rows spaced about Q pm apart. The cilia are normally 1-15 ,um long, but may be longer at the posterior end. An animal that is swimming straight forwards is seen to be covered with transverse lines which move steadily backwards towards the posterior end (Fig. 4a). These lines are the crests of the metachronal waves, and they show up most clearly under darkground illumination. The wave crests are composed of many cilia moving together in their effective stroke (see Tamm & Horridge, 1970). The cilia that comprise these waves show a beat pattern that can be represented approximately by Fig. 4b, if one neglects the (minor) lateral movements in the three-dimensional beat, and they lie Mechanics of ciliary motion 91 close together throughout their beat, so that the wave has a continuous outline similar to that shown later in Fig. 13b. There is a strong suggestion of mechanical metachronal coordination by least mutual interference because adjacent cilia are closely packed and tend to be nearly parallel with one another. Because of the form of the waves the cilia are unlikely to exert much individual effect on the fluid, but the movement of the whole wave over the body as a nearly solid ridge must move the water quite effectively (Sleigh, 1962; Parducz, 1967)",
            " Conversely, during the effective stroke in symplectic metachronism, the cilia are closer together, so w(s, t ) is greater than I. During the recovery stroke, we need only interchange the two cases. If we are considering other types of metachronism, we would only consider the relative ciliary velocity component in the metachronal wave direction. The velocity field for the movement of a particular cilium can be obtained numerically. Calculations were made on the movements of cilia taken from the three organisms Oputina (Fig. 4), Paramecium (Fig. 6) and Pleurobrachia (Fig. 5 ) . In the case of Paramecium we have taken a three-dimensional beat pattern from Machemer (1972b). The time-averaged velocity fields are shown in Fig. 16. In solving the integral equations, non-dimensional variables y, K and T were defined as follows: where C, and CT are the normal and tangential resistance coefficients respectively (see (2)), cr is the radian frequency of the cilium, c is the metachronal wave speed, 6\" the angle of inclination of the direction of propagation of the metachronal waves to the effective stroke (i"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003835_j.actamat.2005.05.003-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003835_j.actamat.2005.05.003-Figure8-1.png",
        "caption": "Fig. 8. Finite element mesh used to simulate multilayer LPD of a 10 layer wall. The analysis was performed for L/2 equal to 4.0 and 30.5 mm. A denser mesh of finite elements was used where higher thermal gradients are expected. A symmetry plane exists if the wall is built up along the mid-plane of the base material. In this case, only one-half of the overall problem was solved since a zero heat flux was imposed along the symmetry plane.",
        "texts": [
            " A substrate of the same steel, quenched and double tempered for 1 h at 200 C, with an initial microstructure of tempered martensite and carbides and a hardness of 560 HV, was assumed. The Gaussian laser beam was focused into a spot 3 mm in diameter, measured at e 2 of maximum intensity. Based on preliminary calculations, an absorbed laser beam power P 0 = (1 R) \u00c6 P = 325 W was used to create a melt pool with a 1-mm diameter, in agreement with the desired track width. Calculations were performed for Dt = 1, 2, 3, 4, 5 and 10 s and two different substrate sizes (Fig. 8): a larger substrate (A) with L = 61.0 mm and mass of 102.8 g and a small substrate (B) with L = 8.0 mm and mass of 13.5 g. The initial temperature of the deposited material was 1450 C while the substrate was initially at 27 C. The surface emissivity, e, and the convective heat transfer coefficient, h, were set at 0.5 and 30 W/m2K [58], respectively. The results obtained are illustrated by data for points in the symmetry plane of the wall (Fig. 8). The thermal field is strongly influenced by the substrate size, as shown in Fig. 9. When the part is built on a large substrate (A, m = 102.8 g), the temperature of the substrate remains below Ms during the entire deposition process (Fig. 9). As a result, the substrate retains its heat extraction capability and the newly deposited layers cool down below the Ms temperature before the deposition of a new layer. As a result, austenite in these layers transforms to martensite, resulting in a high hardness"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure4.36-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure4.36-1.png",
        "caption": "Figure 4.36 A shored frame.",
        "texts": [
            " If the structure with hinged supports itself consists of two parts joined by a hinge, this is referred to as a three-hinged frame (see Figure 4.34). If the beam structure is not bent but arched, then the structure in Figure 4.35a is called a two-hinged arch, and the structure in Figure 4.35b a three-hinged arch. 130 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM It will be clear that a wide range of planar structures can be constructed using line elements. Two types of structure not mentioned in the earlier categories are shown here. The structure in Figure 4.36 is called a shored frame. The structures in Figure 4.37 go by the name of trussed beams. Although the structures in Figures 4.36 and 4.37 include hinges at all the connections, none of these structures are trusses. A characteristic of a truss is that all the ends of the members that merge in a connection are hinged together. This is not the case in the circled connections. Here the hinge is attached to the outside of a so-called continuous beam, and is not fitted internally in the beam. Structures are supported in such a way that all free movements are prevented"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002672_j.mechmachtheory.2020.103786-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002672_j.mechmachtheory.2020.103786-Figure1-1.png",
        "caption": "Fig. 1. Geometric model of gear profile: (a) cantilever beam model; (b) transition curve by tool.",
        "texts": [
            " The reliability of a gear model depends largely on the accuracy of the time-varying mesh stiffness [27] . Therefore, accurately calculating the gear mesh stiffness is necessary to properly conduct dynamic simulations and obtain accurate dynamic characteristics of a gear system. Based on the principles of material mechanics and elasticity mechanics, the spur gear teeth are transformed into a cantilever beam with a variable cross-section starting from the root circle. The geometric parameters of gear teeth are shown in Fig. 1 , where AB is the addendum curve, BC is the involute curve of gear teeth, and CD is the transition curve of tooth profile. According to [28] , the transition curve function can be expressed as { x = r sin \u03c6 \u2212 ( a 1 sin \u03b3 + r \u03c1 ) cos ( \u03b3 \u2212 \u03c6) y = r cos \u03c6 \u2212 ( a 1 sin \u03b3 + r \u03c1 ) sin ( \u03b3 \u2212 \u03c6) (1) where r is the radius of pitch circle, a 1 is the distance between the cutter round corner at the tip and the centerline, r \u03c1 is the radius of the cutter tip round corner, and b 1 is the distance between the center of the tip corner and the centerline of the gear groove. The corresponding values are calculated using Eqs. (2)\u2013(5) . \u03b3 is a variable parameter, \u03b10 \u2264 \u03b3 \u2264 \u03c0 / 2, which is shown in Fig. 1 (b). h a \u2217 is the addendum coefficient, c \u2217 is the addendum clearance coefficient, and \u03b10 is the pressure angle of pitch circle. r \u03c1 = c \u2217m 1 \u2212 sin \u03b10 (2) a 1 = ( h \u2217 a + c \u2217) m \u2212 r \u03c1 (3) b 1 = \u03c0m 4 + h \u2217 a m tan \u03b10 + r \u03c1cos \u03b10 (4) \u03c6 = a 1 tan \u03b3 + b 1 r (5) Yang and Lin [29] proposed that the mesh stiffness of gears is composed of the Hertzian contact stiffness k h, bending stiffness k b, and axial compression stiffness k a. Later, Tian [30] added the shear stiffness k s to Yang and Lin\u2019s formula"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001943_tim.2020.3031194-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001943_tim.2020.3031194-Figure7-1.png",
        "caption": "Fig. 7: Bi-brachiate inspection robot [59]. (a)Inspection robot; (b)Simulation experiment;",
        "texts": [
            " Through the special mechanism structure, it could realize the translational motion along power lines and crossing obstacles between power towers. Combined with different sensors, the sensor system is responsible for the environmental perception of power lines to serve the path planning of climbing robots. Meanwhile, it could also deal with the related defect detection tasks. Aimed at the high-voltage power line inspection, a new climbing robot was designed by the Institute of Automation, Chinese Academy of Sciences, Beijing, China [59]\u2013[61]. It was a Bi-brachiate inspection robot and Fig.7 shows its special structure. To get across the obstacles on the power lines, an adaptive fuzzy logic algorithm was proposed to realize the controlling of inspection robot [62]. Meanwhile, the CDMA network was built for the communication between the climbing robot and the ground maintenance center [63]. For the insulator detection of the high-voltage grid, Huang et al. designed a light climbing robots [64], as shown in Fig.8. Through the traveling of the power insulators, it could realize the insulator detection with zero resistance value"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000233_978-1-4614-0460-6-Figure6.41-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000233_978-1-4614-0460-6-Figure6.41-1.png",
        "caption": "Fig. 6.41 Force balance to determine k22",
        "texts": [
            " For the k11 on-diagonal term, we need the force required to give a unit displacement to x1 while holding x2 stationary. The forces are shown in Fig. 6.39. The moment sum about the motionless x2 is given by: X M \u00bc fk11 2l 3 \u00fe k1x1 2l 3 \u00bc fk11 2l 3 \u00fe k1 2l 3 \u00bc 0: (6.31) Dividing by 2l/3 and solving for fk11 give the stiffness k11. 2 3x2m2 2 x2m1 fm22 x1 x2 2 3x2m2 2 x2m1 fm22 x1 x2 fm21 Fig. 6.38 Force balance to determine m21 The other on-diagonal term, k22, is determined from the force required to give a unit displacement to x2 while holding x1 fixed. The forces are displayed in Fig. 6.41. Summing the moments about x1 gives: X M \u00bc fk22 2l 3 k2x2 2l 3 \u00bc fk22 2l 3 k2 2l 3 \u00bc 0: (6.35) fk22 fk11 fk11 6.6 Shortcut Method for Determining Mass, Stiffness, and Damping Matrices 233 the forces in Fig. 6.42 and substituting for fk22 provide the desired result. X f \u00bc fk21 \u00fe fk22 k2x2 \u00bc fk21 \u00fe fk22 k2 \u00bc 0 fk21 \u00bc fk22 \u00fe k2 \u00bc k2 \u00fe k2 \u00bc 0 (6.37) The final stiffness matrix is: k \u00bc k11 k12 k21 k22 \u00bc k1 0 0 k2 : (6.38) Because the dampers appear at the same locations as the springs, the damping matrix is: c \u00bc c11 c12 c21 c22 \u00bc c1 0 0 c2 : (6"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure5.48-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure5.48-1.png",
        "caption": "Figure 5.48 Graphical check of the moment equilibrium of AE, CD and BE: in all the cases, the lines of action of the three (resulting) forces pass through a single point.",
        "texts": [
            "47a, the various structural parts have been isolated, and all the interaction forces are shown. First look at the equilibrium of CD. From the moment equilibrium about C follows Dv = 10 kN. From the moment equilibrium about D follows Cv = 30 kN. The horizontal force equilibrium gives Ch = Dh. 5 Calculating Support Reactions and Interaction Forces 183 Next look at the right-hand rafter AE (see Figure 5.47b). The moment equilibrium about E gives Ch = 15 kN so that Dh = 15 kN. The force equilibrium gives Eh = \u221215 kN, Ev = 5 kN. The direction of Eh was obviously assumed falsely. In Figure 5.48, all the interaction forces are shown as they act in reality. Check: BE must also meet the conditions of the force and moment equilibrium. Figure 5.48 shows that the force equilibrium conditions are satisfied. Only the moment equilibrium has to be checked. c. If three forces act on a body, there is moment equilibrium only if the lines of action of the forces intersect at one point. In Figure 5.48, this check for moment equilibrium has been performed for each of the structural parts. 184 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM The bearing capacity of the beam in Figure 5.49a can be increased by introducing intermediate supports. These structures are referred to as trussed beams when these intermediate supports are realised by a bar system applied directly to the beam (see Figure 5.49b). Trussed beams are used in simple bearing structures and for auxiliary structures in the construction industry (formwork bearers)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000546_j.matdes.2018.02.018-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000546_j.matdes.2018.02.018-Figure2-1.png",
        "caption": "Figure 2: Schematic presentation of the LPF technique (reproduced with permission from Elsevier) [14].",
        "texts": [
            " It is also known as laser consolidation, laser direct metal fabrication, laser cladding with blown powder, etc. This technique is based on a laser beam, focused on a specific area of a part where the powder is simultaneously supplied through a nozzle pointing to the same area, layer by layer. In contrast to the LPB process, which employs a bed of powder metal that is \u2018selectively\u2019 melted by a laser, LPF is accomplished by simultaneously delivering metallic powder and focused laser energy, as schematically shown in Figure 2. AC CEP TE D M AN USC RIP T In LPF, a high power laser (e.g., Fiber, Disk, Nd:YAG or CO2) is used to fabricate a melt zone on a substrate within an inert atmosphere; simultaneously, powder is injected to the melt through the laser beam. Subsequently, parts are made layer by layer using a 3D CAD file with each layer assembled track-upon-track via a controlled tool-path. Advantages of the LPF process include the production of near-net-shape, full density parts with good mechanical properties and minimal imperfections (e"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000637_978-3-642-83006-8-Figure3.19-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000637_978-3-642-83006-8-Figure3.19-1.png",
        "caption": "Fig. 3.19. Compensation of ZMP displacement by: a) hip joint, b) ankle joint",
        "texts": [
            " The primary task of the feedback with respect to ZMP position is to prevent its excursion out of the allowable region, or, to prevent the system from falling down by rotation about the foot edge. If this is fulfilled, a fUrther requirement imposed is to ensure that the actual ZMP position is as close as possible to its nominal. Ou]; further considerations will be restricted to biped motion in the sagittal plane, what means that the ground reaction force position will deviate only in the direction of x axis by t,x. Fig. 3.19. illustrates the case when the vertical ground reaction force Rz deviates from the x nominal position 0 bY t,x; thus, the moment Rz\u00b7t,x MZMP is a measure of the mechanism overall behaviour. In the same way, we can consider the mechanism motion in the frontal plane, and Rz\u00b7t,y = M2MP is a measure of the mechanism behaviour in the 250 direction of y axis. Let us assume the correction of the Rz acting point in one direction is done by the action at only one joint, arbitrarilly selected in advance. A basic assumption introduced for the purpose of simplicity is that the action at the chosen joint will not cause a change in the motion at any other joint. If we consider only this action, the system will behave as if composed of two rigid links connected at joint k, as presented in Fig. 3.19. In other words, the servo systems are supposed to be suffi ciently stiff. In Fig. 3.19, two situations are illustrated, when the joint which has to compensate the ZMP displacement is the hip (case a) and the ankle joint (case b) of supporting leg. In both cases, this joint is denoted by k, and all links above and below it are considered as a single rigid body. The upper link is of total mass m and inertia * JIIoment J k for the axis of joint k. Of course, numerical values aredifferent for case a) and case b). Furthermore, the ~istance from the ground surface to k is denoted by L, from k to c (c is the mass centre of the upper link,) bY Ro, whereas t.P~MP stands for the correctional 251 additional actuator torque, applied at joint k. In Fig. 3.19. the up per (compensating) link is presented as a single link above joint k. In fact, in both cases presented, the compensating link includes also the other leg which is in swing phase, and which is not drawn in the 'I< figure. The inertia moment J k has to be calculated in such a way to include all the links which are found further onward with respect to the selected compensating joint. In this analysis, all the joints ex cept the k-th one are considered \"frozen\", and, as the consequence, the lower link, representing the sum of all the links below the k-th joint, is also considered as rigid body, which is standing on the gro und surface and does not move",
            " Assume the /lIechanism performs the gait such that a displacement of the ground reaction force R in the x direction occurs, so M~MP Rz \u00b76X. Then, the quantity 6P~MP is to be determined on the basis of the value M~MP and of the known mechanism and gait characteristics. It is suppo sed that the additional to~que 6P~MP will cause change in acceleration of the compensating link 6~, while velocities will not change due to the action of 6P~MP' 64>\"'0. From the equations of planar motion of con sidered system of two rigid bodies (Fig. 3.19) which is driven by k \u2022 \u2022 \u2022 2 , 6P ZMP ' and under assumption that terms (~6~) and (6~) from expresslon for normal component of angular acceleration of the upper link are ne glected, it follows that 1 + m.~.L.cospcosa + m.~.L.sinpsina 'I< 'I< (3.4.13) J k J k The control input to the actuator of the compensating joint which has to realize 6P~MP can be computed from the model of the actuator devia tion from the nominal. Thus, .k 0 0 6qk 0 0 6q .. k 0 k 'k \u00b7k fk k k 0 k k 6qT a 22 a 23 6q + (6Pc +6P ZMP ) + (6u +6U ZMP ) c :k 0 k k 6,k 0 bk 6 l R a 32 a 33 lR c (3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001415_978-3-319-74528-2-Figure2.17-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001415_978-3-319-74528-2-Figure2.17-1.png",
        "caption": "Fig. 2.17 The autonomous underwater vehicle MONSUN [300]",
        "texts": [
            " It has omnidirectional motion based on six linear directions and it can turn in place. Other special platforms relevant to swarm robotics are some approaches from modular robotics, where autonomous robot modules can physically attach to each other but also operate as independent entity, as done in the projects SYMBRION and REPLICATOR [240]. There are also a few projects doing underwater swarm robotics, that is, they develop autonomous underwater vehicles (AUV). Examples are the robots developed in the project CoCoRo [350] and the AUV called MONSUN (see Fig. 2.17) [300]. Especially communication between robots is challenging underwater. Radio cannot be used because electromagnetic waves do not propagate far in water. While CoCoRo uses visible light for communication, MONSUN uses an underwater acoustic modem, which has a much bigger range than light. However, also dealing with acoustic waves underwater is challenging because they are reflected at the surface, possibly at confining walls, or even at 52 2 Short Introduction to Robotics water layers of different temperatures"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001498_tie.2010.2045322-Figure9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001498_tie.2010.2045322-Figure9-1.png",
        "caption": "Fig. 9. Adjacent double-phase (phase-c and phase-d) open-circuit fault.",
        "texts": [
            " It can be noticed that the fundamental component of solution (31) presented in this paper is balanced and that the amplitudes of the third-harmonic components are almost equal. Furthermore, the maximum torque (57.5%) is calculated by constraining the phase rms current values to the normal rms current value. With 15.4% higher current, the maximum torque that can be achieved by solution (31) is about 66.4% of the normal torque. Finally, consider that two adjacent phases (phases c and d) are under fault (Fig. 9). Under this condition, the fault-axis direction is along the direction of the resultant magnetic axis of phases c and d. Therefore, the fault axis is aligned again with phase a. Phases b and e are symmetrical with respect to the fault axis. From Fig. 9, it can be seen that, under this fault, three of the remaining healthy phases are located in one half of the machine; this is a highly asymmetrical spatial distribution. It is found that, in this case, the imposition of a constraint to make the fundamental components of the fault-tolerant currents equal yields a solution with less output torque. Due to the type of this fault, the fundamental current components can be chosen to be imbalanced to obtain a solution with higher output torque. The fault-tolerant currents in the three healthy phases can be defined as ia = I11 \u00b7 cos(\u03c9t) + I31 \u00b7 cos(3\u03c9t) ib = I12 \u00b7 cos(\u03c9t \u2212 \u03b812) + I32 \u00b7 cos(3\u03c9t \u2212 \u03b832) ic = 0 id = 0 ie = I12 \u00b7 cos(\u03c9t + \u03b812) + I32 \u00b7 cos(3\u03c9t + \u03b832)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003605_027836499601500604-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003605_027836499601500604-Figure1-1.png",
        "caption": "Fig. 1. Laser tracking system with orientation measurement. (From Vincze et al. ~~ 994]).",
        "texts": [
            " When they exist, multiple closed-loop examples are discussed as well, in this case in terms of S and M. These examples are representative of recent work and are not meant to be exhaustive. 3.1. C = 6 For single open-loop calibration with full pose measurement, then Dj = 0 and SJ = 6. One of the most interesting recent devices for full pose measurement is the single-beam laser tracking system in Vincze et al. (1994). The robot is fitted with a retroreflector, and the laser beam is deflected by a mirror on a universal joint (Fig. 1). Position is measured using interferometry, as usual. The novel aspect is orientation measurement, by imaging of the diffraction pattern of the edges of the retroreflector. Orientation resolutions of arcsec are stated, and motions can be tracked that accelerate at 100 nVs 2 Masory et al. (1993) proposed the use of full pose measurement along with leg length measurements in the calibration of a Stewart platform (S = 12 and M = 6). The procedure is a simple extrapolation from single-loop calibration"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001270_jproc.2020.3046112-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001270_jproc.2020.3046112-Figure3-1.png",
        "caption": "Fig. 3. IPM rotors. (a) Double-V-shape in Chevy Bolt 2016 [3]. (b) V-shape in Tesla Model 3 2017 [4]. (c) U-shape in Toyota Prius 2017 [5].",
        "texts": [
            " Authorized licensed use limited to: Univ of Calif Santa Barbara. Downloaded on May 15,2021 at 01:41:56 UTC from IEEE Xplore. Restrictions apply. The desired performances of traction electric machines with high power density, high efficiency, and low-torque ripple with negligible cogging torque are achieved through a comprehensive rotor and stator design. On the rotor side, three variants of rotor designs are common for IPM machines used in traction applications: V-shape, doubleV-shape, and U-shape, which are shown in Fig. 3. Each of the designs has its advantages and disadvantages [12]. Generally, the double V-shape rotor has the highest torque density and efficiency but has higher magnet losses, which burdens the thermal management. The magnet utilization is the best for V-shape rotors but would have relatively lower corner speeds than the other two design types due to higher no-load voltage. The U-shape is a tradeoff between the V- and double-V designs that offer some design flexibility. On the stator side, the winding choices are between distributed versus concentrated and between stranded versus hairpin"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002463_j.mechmachtheory.2019.03.019-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002463_j.mechmachtheory.2019.03.019-Figure8-1.png",
        "caption": "Fig. 8. Rectilinear motion of a 3-DOF robot manipulator.",
        "texts": [
            " Another possible application may be yielding a fixed-time trajectory with given constraints. To assess the effectiveness of the proposed approach and make a comparison with classical trajectory planning methods, two numerical examples on the generation of trajectories for robot manipulators, one in Cartesian space and another in joint space, are described in this section. The first example represents the generation of a rectilinear motion in task space for the end-effector of a 3-DOF serial manipulator as illustrated in Fig. 8 , the coordinates of the start and end points for the straight path are given in Table 1 . The link lengths of the robot are assumed to be l 1 = 0 . 85 m , l 2 = 0 . 95 m , and l 3 = 0 . 65 m . This task has been studied in [20] and executed with the use of higher-degree polynomials (of order 9, 7 and 5) guaranteeing the continuity of the trajectory up to the third derivative (jerk) and the same motion duration time. However, the polynomial profiles described therein only support setting the acceleration limit values while other bounds cannot be specified independently since the extreme kinematic values are mutually interrelated"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002604_tie.2019.2902834-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002604_tie.2019.2902834-Figure2-1.png",
        "caption": "Fig. 2. Illustration of payload\u2019s swing angles",
        "texts": [
            " The unit direction vector from the quadrotor towards the payload is denoted by q (t) = [ sxcy sy cxcy ]T (3) 0278-0046 (c) 2018 IEEE. Personal use is permitted, but republication/redistribution requires IEEE permission. See http://www.ieee.org/publications_standards/publications/rights/index.html for more information. where sx, cx denote sin(\u03b3x (t)) and cos(\u03b3x (t)), sy , cy denote sin(\u03b3y (t)) and cos(\u03b3y (t)), respectively, \u03b3y (t) \u2208 R denotes the payload angle with respect to its projection on the xIzI plane, \u03b3x (t) \u2208 R denotes the projection of the payload angle along the zI axis, as shown in Fig. 2. According to the definition of q(t) in (3), the following equations are satisfied q (t) T q (t) = 1, q (t) T q\u0307 (t) = 0 (4) The following assumptions will be invoked in the subsequent control design and stability analysis. Assumption 1: The cable is massless, always taut, and the length of the cable is an unknown constant. This means that T (t) > 0, and L\u0307 \u2261 0, for \u2200t \u2265 0. So that following equations are satisfied [11]: PL (t) = PQ (t) + Lq (t) (5) P\u0307L (t) = P\u0307Q (t) + Lq\u0307 (t) . (6) where PL (t) donates the position of the payload"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000880_6.2004-4900-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000880_6.2004-4900-Figure4-1.png",
        "caption": "Figure 4. Relation with Proportional Navigation",
        "texts": [
            " Then, an interesting similarity is found in relation to proportional navigation missile guidance. The formula in Eq. (1) for the lateral acceleration command in the trajectory following guidance logic can be shown to be equivalent to the formula a\u22a5LOS = N \u2032VC \u03bb\u0307 for the acceleration command perpendicular to the line-of-sight in the proportional navigation with a navigation constant of N \u2032=2, under the assumption that the reference point is stationary in the computation of the line-of-sight rate and the closing velocity. This equivalence can be shown using Figure 4. First, noticing that there is an angular difference between the vehicle lateral acceleration(as) and the acceleration(a\u22a5LOS) perpendicular to the LOS a\u22a5LOS = as cos \u03b7 Using the acceleration command formula as = 2 V 2 L1 sin \u03b7 leads to a\u22a5LOS = 2 V 2 L1 sin \u03b7 \u00b7 cos \u03b7 = 2 (V cos \u03b7) ( V L1 sin \u03b7 ) where, assuming that the target point is stationary, the first bracket is the closing velocity (the relative velocity component in the direction of the LOS) and the second bracket is the LOS rate. Therefore, it can be shown that a\u22a5LOS = 2 \u00b7 VC \u00b7 \u03bb\u0307 which is the form of the proportional navigation formula with the navigation constant equal to 2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure3.68-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure3.68-1.png",
        "caption": "Fig. 3.68: Singular configurations of CaPaMan: a) with \u03d5 =90 deg.; b) with the three connecting bars in a coplanar configuration; c) with two aligned connecting bars.",
        "texts": [
            "63). This condition represents a cylinder with elliptical cross-section. The cross-section plane is perpendicular to the unit vector n=[1,1,1]t. An intersection of the elliptic cylinder with a plane x + y + z = 0 is its elliptic directrix with center in point at the origin of the reference frame and semi-major axis p = m/3; semi-minor axis q = m2/ 31/2 with m = 3 rP / 2. The elliptic cylinder divides the space into two regions free from singularities: the region inside and outside the cylinder. Figure 3.68 shows the singular configurations of CaPaMan, as they have been determined with the above-mentioned Jacobian analysis. The static behavior of CaPaMan can be investigated in term of force transmission capability through a static analysis. Force transmission is characterized by the parallelogram linkages in the sense that, because of the prismatic joint and assuming friction as negligible, the only force applied to a leg by the movable plate is the one of Rk which is contained in the plane of the mechanism, as shown in Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000908_j.jmatprotec.2013.01.020-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000908_j.jmatprotec.2013.01.020-Figure1-1.png",
        "caption": "Fig. 1. Straight solid walls or block support.",
        "texts": [
            " Thomas and Bibb (2008) investigated the SLM process and found that it is limited in its ability to build overhanging structures with angles less than 40\u201345\u25e6 from horizontal without building fixed support structures. The removal of support structure from the part/base plate is a tedious job; in particular, large amount of supports for delicate parts would increase the difficulties and time of support removal, causing small pieces of the part to break off. Furthermore, the commonly used conventional support method which consists of straight rectangular solid walls or blocks as in Fig. 1 is unsuitable for recovering the raw loose powder which are trapped inside the support structures during the build. Especially when the support is removed by EDM-wire cutting, the part along with the support is submerged in water that washes away all trapped raw powders. Since SLM is a layer wise process with relatively 1020 A. Hussein et al. / Journal of Materials Processin l r p e \u2022 \u2022 \u2022 \u2022 t r ( p w s u l c a a u n i d l f d 2 s t o a m s e to support the overhang of the cantilever part. The cantilever part ow build rate, it is important that this non-productive time is educed"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure1.5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure1.5-1.png",
        "caption": "Fig. 1.5 Concept of three dimensional linear inverted pendulum - the motions of the center of mass is constrained on a specified plane by controlling the contact force and the orientation of the plane is independent to the motions of the center of mass",
        "texts": [
            " One method is based on the dynamics of the linear inverted pendulum whose height of the center of the mass is kept by controlling the contact force and the length of the pendulum. Another one generates the patterns using the ZMP as the criterion to judge the contact stability. Chapter 4 starts from the introduction of a method based on two dimensional linear inverted pendulum, the method is extended to that based on three dimensional one, and it is applied to generate the patterns of multilink models. Fig.1.5 shows the concept of three dimensional linear inverted pendulum. The ZMP-based method is overviewd as well. The relationship between the derivatives of the joint angles and the ZMP can be given by nonlinear differential equations. It is called the ZMP equations. It is difficult to find trajectories of the joint angles which let the ZMP follow specified ZMP ones due to the nonlinearity. The ZMP equations were simplified under several assumptions and their solutions were found by a batch processing in the early stage"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001252_j.ymssp.2007.12.001-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001252_j.ymssp.2007.12.001-Figure3-1.png",
        "caption": "Fig. 3. Five-DOF bearing-pedestal model.",
        "texts": [
            " The random fluctuation of the ball positions as a result of this can be either forward or backward of the mean. This was related to the nominal motion of the cage (ocdt) and was modelled by introducing random numbers of zero average with uniformly distributed fluctuations within fj, rand \u00bc71.5ocdt (70.051). The model of Feng et al. [2] has been adopted here for studying the dynamics of rolling element bearings and has now an extra DOF (small sprung mass with relatively high damping to represent a typical high frequency). The new model has five DOF and is symbolically represented in Fig. 3. Although the effects of EHL were not added to the model when adopting it here\u2014due to excessive computational demands in the analysis\u2014the effects of the EHL are somewhat integrated in the model. The EHL stiffness modelling generally increases the stiffness of the bearing over the purely Hertzian analysis as the ARTICLE IN PRESS N. Sawalhi, R.B. Randall / Mechanical Systems and Signal Processing 22 (2008) 1924\u201319511930 EHL film takes up some of the clearance of the bearing and can even create negative clearance or preload (Wijnat et al",
            " (30) and (31), respectively. mi \u20acyi \u00fe ksiyi cm\u00f0r2 _y2 r1 _y1 _x2 \u00fe _x1 \u00fe _et\u00de kmri\u00f0r2y2 r1y1 x2 \u00fe x1 \u00fe et\u00de \u00bc 0, (30) ji \u20acy\u00fe cmri\u00f0r2 _y2 r1 _y1 _x2 \u00fe _x1 \u00fe _et\u00de \u00fe kmri\u00f0r2y2 r1y1 x2 \u00fe x1 \u00fe et\u00de \u00bc Ti, (31) where i \u00bc 1, 2 (1 for gear, 2 for pinion). These equations are solved iteratively using Matlabs/Simulinks. There are 34 DOF in the new model as opposed to the 16 DOF in the previous model [1]. This is illustrated in the schematic diagram of Fig. 8. The extra 18 DOF are due to the inclusion of the five-DOF bearing model (Fig. 3), and to the fact that translational DOF are now considered both along the Line of Action (LOA) and perpendicular to it (Figs. 3 and 7). The main assumptions, on which the new model is based, are as follows. (1) Shaft mass and inertia are lumped at the bearings or at the gears. (2) Translational DOF are considered along the LoA and perpendicular to it, with the LoA aligned vertically. (3) Two resonances of the gear casing are considered. ARTICLE IN PRESS N. Sawalhi, R.B. Randall / Mechanical Systems and Signal Processing 22 (2008) 1924\u20131951 1935 (4) The static TE of the gears (geometric and elastic) is included"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure14.35-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure14.35-1.png",
        "caption": "FIGURE 14.35. A 1/8 car model and a unit step displacement base excitation.",
        "texts": [
            "164) G2 = X\u0308B \u03c92nY = FTB kY = FTE e\u03c92me = FTR e\u03c92me \u00b3 1 + ma m \u00b4 (14.165) the optimal design curve can also be expressed as a minimization condition for any other G2-function with respect to any other S2-function, such as transmitted force to the base FTE e\u03c92me for an eccentric excited system XE e\u03b5E . This minimization is equivalent to the optimization of an engine mount. 14.4 F Time Response Optimization Transient response optimization depends on the type of transient excitation, as well as cost function definition. Figure 14.35 illustrates a 1/8 car model and a unit step displacement. y = \u00bd 1 t > 0 0 t \u2264 0 (14.166) If the transient excitation is a step function, and the optimization criteria is minimization of the peak value of acceleration versus peak value of relative displacement, then there is optimal \u03beF for any fn that provides the best transient behavior of a 1/8 car model. This behavior is shown in Figure 14.36. \u03beF = 0.4 (14.167) 14. Suspension Optimization 919 920 14. Suspension Optimization Proof. The equation of motion for the base excited one-DOF system shown in Figure 14.35, is x\u0308+ 2\u03be\u03c9n x\u0307+ \u03c92n x = 2\u03be\u03c9n y\u0307 + \u03c92n y (14.168) Substituting y = 1 in Equation (14.168) provides the following initial value problem to determine the absolute outplacement of the mass m: x\u0308+ 2\u03be\u03c9n x\u0307+ \u03c92n x = \u03c92n (14.169) y(0) = 0 (14.170) y\u0307(0) = 0 (14.171) Solution of the differential equation with zero initial conditions is x = 1\u2212 1 2 A ib e\u2212A\u03c9nt + 1 2 A ib e\u2212A\u03c9nt (14.172) where A and A are two complex conjugate numbers. A = \u03be + i q 1\u2212 \u03be2 (14.173) A = \u03be \u2212 i q 1\u2212 \u03be2 (14.174) Having x and y = 1 are enough to calculate the relative displacement z = x\u2212 y"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure6.36-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure6.36-1.png",
        "caption": "FIGURE 6.36. A small motion of the actual suspension and wheel configurations.",
        "texts": [
            " Point C is on the spindle and supposed to be the wheel center. When the wheel moves up and down, the wheel center moves on a the couple point curve shown in the figure. The wheel\u2019s center of proper suspension mechanism is supposed to move vertically, however, the wheel center of the suspension moves on a high curvature path and generates an undesired camber. A small motion of the kinematic model of suspension is shown in Figure 6.35, and the actual suspension and wheel configurations are shown in Figure 6.36. 6. Applied Mechanisms 359 360 6. Applied Mechanisms 6.5.2 Coupler Point Curve for a Slider-Crank Mechanism Figure 6.37 illustrates a slider-crank mechanism and a coupler point at C. When the mechanism moves, coupler point C will move on a coupler point curve with the following parametric equation: xC = a cos \u03b82 + c cos (\u03b1\u2212 \u03b3) (6.262) yC = a sin \u03b82 + c sin (\u03b1\u2212 \u03b3) (6.263) The angle \u03b82 is the input angle and acts as a parameter, and angle \u03b3 can be calculated from the following equation. \u03b3 = sin\u22121 a sin \u03b82 \u2212 e b (6"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001395_j.jmatprotec.2016.01.017-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001395_j.jmatprotec.2016.01.017-Figure1-1.png",
        "caption": "Fig. 1. Experimental setup in Au",
        "texts": [
            " The effective energy and powder deposition density are two arameters that describe the combined effects of the laser power, ravel speed, laser beam diameter and powder feed rate on the mount of heat input during the LAM process. The amount of heat nput defines values for the melt pool temperature and cooling rate. hus, E and PDD are directly related to the melt pool temperature nd its cooling rate, and indirectly related to the microstructure ormation. . Experimental setup A schematic view of the state of the art Laser Additive Manufacuring (LAM) apparatus with injected powder melting used in this esearch is illustrated in Fig. 1. This system is composed of several ntegrated instruments such as a high power laser (1.1kW IPG fiber aser YLR-1000-IC with 300 m feeding fiber and a Gaussian power istribution), powder feeder (Sulzer Metco TWIN 10-C), multi-axes NC machine (5-axis Fadal VMC 3016), nozzle and an intelligent ontroller with a data acquisition system and sensors. A Jenoptik IR-TCM 384 camera module with a resolution of 84 \u00d7 288 pixles and a measuring accuracy of \u00b12% is used to monior the thermal behavior of the LAM process"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000233_978-1-4614-0460-6-Figure2.7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000233_978-1-4614-0460-6-Figure2.7-1.png",
        "caption": "Fig. 2.7 Time-domain representation of x\u00f0t\u00de \u00bc 2A cos ont\u00fe b\u00f0 \u00de for: (a) _x0 \u00bc 0 and x0 6\u00bc 0; and (b) x0 \u00bc 0 and _x0 6\u00bc 0",
        "texts": [
            " Specifically, A \u00bc ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x0 2 2 \u00fe _x0 2on 2r and b \u00bc tan 1 _x0 2on x0 2 ! \u00bc tan 1 _x0 onx0 . We can plot x\u00f0t\u00de \u00bc 2A cos ont\u00fe b\u00f0 \u00de for two distinct cases: (1) _x0 \u00bc 0 and x0 6\u00bc 0; and (2) x0 \u00bc 0 and _x0 6\u00bc 0. For the first case, b \u00bc 0 and the representation of the counterclockwise rotating vector x in the complex plane at t \u00bc 0 is shown in Fig. 2.6a. For the second case, b \u00bc p 2 and x is shown at t \u00bc 0 in Fig. 2.6b. The timedomain representations of these two cases (i.e., the projection of the vectors on the real axis) are included in Fig. 2.7a (first case) and 2.7b (second case). We can express undamped free vibration (i.e., the solution to Eq. 2.4) in several forms. Using somewhat careless notation (the A in the following list is not the same A we just discussed), these forms can be generically written as: \u2022 x\u00f0t\u00de \u00bc A cos ont\u00fe B\u00f0 \u00de \u2022 x\u00f0t\u00de \u00bc C sin ont\u00fe D\u00f0 \u00de 2.1 Equation of Motion 31 \u2022 x\u00f0t\u00de \u00bc E sin ont\u00f0 \u00de \u00fe F cos ont\u00f0 \u00de \u2022 x\u00f0t\u00de \u00bc Geiont \u00fe He iont; where A through F are real-valued (no imaginary part) and G and H are complex. IN A NUTSHELL It might seem like a long (mathematical) way to go, but the end result is that free vibration of a single degree of freedom systemwith no damping is sinusoidal and occurs at the natural frequency of the system"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000591_j.phpro.2011.03.033-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000591_j.phpro.2011.03.033-Figure1-1.png",
        "caption": "Figure 1.A schematic view of the SLM process",
        "texts": [],
        "surrounding_texts": [
            "Maraging steels are well known for combining good material properties like high strength, high toughness, good weldability and dimensional stability during aging heat treatment. Mainly maraging steels are used for two application areas: the aircraft and aerospace industry in which superior mechanical properties and weldability of maraging steels are the most important features, and secondly in tooling applications which require superior machinability [4]. The nominal composition of maraging steel, grade 300, is given in Table 1. Maraging steels are a special class of high-strength steels that differ from conventional steels in that they are hardened by a metallurgical reaction that does not involve carbon [4]. The relatively soft body centered cubic martensite, that is formed upon cooling, is hardened by the precipitation of intermetallic compounds at temperatures of about 480\u00b0C."
        ]
    },
    {
        "image_filename": "designv10_0_0001075_j.actamat.2018.10.032-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001075_j.actamat.2018.10.032-Figure4-1.png",
        "caption": "Fig. 4 a) the volume fraction of \u03b3\" phases; b) the diagrammatic sketch of \u03b3\" phase; c) the distribution of the diameter of \u03b3\" phases; d) the relationship between the thickness and the diameter; e) the Time-Temperature-Transformation curves of the three kinds of samples; f) the incubation time of \u03b3\" phases in the three kinds of samples at 720\u2103.",
        "texts": [
            " Table 5 The volume fractions of \u03b3\" phase and \u03b3' phase in the three kinds of samples \u03b3\" phase (%) \u03b3' phase (%) \u03b3'-Ni3Al phase (%) \u03b3'-Ni3Ti phase (%) DA sample 13.9 5.13 3.64 1.49 S-15 sample 16.2 5.31 3.65 1.66 S-45 sample 16.9 5.37 3.65 1.72 As is well known, the \u03b3\" phase is a disc-shaped precipitation. When observing it from different directions in the coordinate system, the \u03b3\" phase looks like an ellipse. The long axis of the ellipse is equal to the actual diameter of the \u03b3\" phase, but the short axis of the ellipse is not equal to the actual thickness of \u03b3\" phase (Fig. 4(b)). Based on this understanding, only the diameters of \u03b3\" phases are counted in this study and the results are shown in Fig. 4(c). It can be seen that the average diameter of \u03b3\" phase in the DA sample is larger than that in the S-15 sample and the S-45 sample. Even so, however the S-15 sample and the S-45 sample show \u03b3\" phases with similar average diameter. In order to explain this phenomenon, JMatPro software is used to calculate the TTT curves of the three types of samples. In the DA sample, the element contents in the niobium micro-segregation region is employed for calculation due to the fact that \u03b3\" phases mainly precipitate in this region. As revealed in Fig. 4(e), the TTT curve of \u03b3\" phase of the DA sample moves to higher temperature and shorter time compared with that of the S-15 sample and the S-45 sample. This may be due to high niobium content in the niobium micro-segregation region [20]. According to the TTT curves, the incubation periods of \u03b3\" phases precipitating at 720\u2103 are different in the three kinds of samples. For the DA sample, the incubation time is only 14.4min, which is the shortest in all three types of specimens as displayed in Fig. 4(f). Therefore, the sizes of \u03b3\" phases in the DA sample is the largest. In addition, the incubation time of the S-15 sample is almost equal to that of the S-45 sample, which ANUSCRIP T results in \u03b3\" phases with similar size existing in both samples. In fact, the thicknesses of \u03b3\" phases are related to their diameters. According to the research of Han [21], there is a linear relationship between the thickness and the diameter. As shown in Fig. 4(d), the relationship can be obtained based on other researchers\u2019 works. It can be expressed as L=4.5026e-9.649. Therefore, the distribution of the thickness is similar to that of the diameter. M ANUSCRIP T ACCEPTE D 3.2 Room temperature tensile properties Fig. 5(a) shows the room temperature tensile properties of all the samples. It is interesting to note that the ultimate tensile strength and the elongation increase first and then decrease, while the yield strength only increases, with the decrease of Laves phases",
            " According to Table 5, the volume fraction of \u03b3'-Ni3Ti is only 1.49 %. Even micro-segregation is considered and its segregation degree is assumed to be the same as that of niobium, the relative volume fraction of titanium is only 2.11 % and the difference is only 0.6 %. Such a small difference can be considered to have no significant effect on the properties. Therefore, it is reasonable to assume that the \u03b3' phase is evenly distributed. Thirdly, elements micro-segregations can influence the formation of \u03b3\" phases as shown in Fig. 4(e). High niobium content in the micro-segregation region can induce \u03b3\" phase to precipitate and grow up fast. So the size of \u03b3\" phase in the DA sample is larger than that in the S-15 sample and the S-45 sample. But the formation of \u03b3' phase is mainly related to the concentrations of aluminum and titanium. High niobium content in the micro-segregation region has little impact on the incubation time of the formation of \u03b3' phase at 620\u2103 as indicated in Fig. 4(e). Therefore, it can be deduced that the niobium micro-segregation region have little effect on the formation of \u03b3' phase. Based on this, it is thought that the sizes of the \u03b3' phases in the three kinds of specimens are the same. Many models have been established to describe the quantitative relationship between the characteristic of the strengthening precipitations and the yield strength [35-37]. However, these models cannot be used for the prediction of the strength of the DA sample for the following two reasons"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure13.11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure13.11-1.png",
        "caption": "FIGURE 13.11. A half car vibrating model of a vehicle.",
        "texts": [
            " When the excitation frequency of a multiple DOF system increases, we will see that observable vibration moves from a coordinate to the others in the order of natural frequencies and associated mode shapes. When the excitation frequency is exactly at a natural frequency, the relative amplitudes of vibration are exactly similar to the associated mode shape. If the excitation frequency is not on a natural frequency, then vibration of the system is a combination of all modes of vibration. However, the weight factor of the closer modes are higher. 13.5 Half Car and Body Roll Mode To examine and optimize the roll vibration of a vehicle we may use a half car vibrating model. Figure 13.11 illustrates a half car model of a vehicle. 13. Vehicle Vibrations 859 This model includes the body bounce x, body roll \u03d5, wheels hop x1 and x2 and independent road excitations y1 and y2. The equations of motion for the half car vibrating model of a vehicle are as follow. mx\u0308+ c (x\u0307\u2212 x\u03071 + b1\u03d5\u0307) + c (x\u0307\u2212 x\u03072 \u2212 b2\u03d5\u0307) +k (x\u2212 x1 + b1\u03d5) + k (x\u2212 x2 \u2212 b2\u03d5) = 0 (13.234) Ix\u03d5\u0308+ b1c (x\u0307\u2212 x\u03071 + b1\u03d5\u0307)\u2212 b2c (x\u0307\u2212 x\u03072 \u2212 b2\u03d5\u0307) +b1k (x\u2212 x1 + b1\u03d5)\u2212 b2k (x\u2212 x2 \u2212 b2\u03d5) + kR\u03d5 = 0 (13.235) m1x\u03081 \u2212 c (x\u0307\u2212 x\u03071 + b1\u03d5\u0307) + kt (x1 \u2212 y1) \u2212k (x\u2212 x1 + b1\u03d5) = 0 (13"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001457_j.phpro.2010.08.083-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001457_j.phpro.2010.08.083-Figure1-1.png",
        "caption": "Fig. 1. Typical cross-section of a laser sintered track from metal powder on steel substrate",
        "texts": [
            " The formation of the first-layer single tracks is a focus of this study. The tracks of SS grade 904L powder on a SS grade 304L substrate were produced by the 50 W and 25 W power laser beam at different scanning speeds ranging from 0.06 to 0.24 m/s and 0.03 to 0.12 m/s correspondingly. The thickness of the deposited powder layer was 50 \u03bcm. The length of the scan line was 20 mm for all the experiments. The width of the tracks, the zone of powder consolidation and the remelted depth (penetration into substrate) were analyzed (Fig. 1). It was found that the width of the single track varied from 130 to 100 \u03bcm, the width of the zone of powder consolidation varied from 260 to 130 \u03bcm. The track width and the zone of powder consolidation decreased with the scanning speed (Fig. 2). For the laser power P=25 W, the substrate remelted depth is absent for all the range of the scanning speeds used, i.e. the laser power is sufficient to melt the powder but, under the given process parameters, is not enough to melt the substrate and to create a joint molten pool"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure2.10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure2.10-1.png",
        "caption": "Fig. 2.10. Denavit-Hartenberg parameters for a prismatic joint",
        "texts": [
            " The Xj axis will be aligned with any common normal which exists (usually the normal between axes of joint i and (i + 1)) and is directed from joint i to joint (i + 1). In the case of intersecting joint axes, the Xj axis is chosen to be parallel or anti parallel to the vector-cross product Zj-l x Zj' The Yj axis satisfies Xi x Yi =Zj' The joint coordinate qj for a revolute joint is now defined as the angle between axes Xi _ 1 and Xj (Fig. 2.9). It is zero when these axes are parallel and have the same direction. The twist angle lXi is measured from axis Zj-l to Zj' i.e. as a rotation about Xj axis. Let us now consider the prismatic joint (Fig. 2.10). Here, the distance dj becomes joint variable qj, while parameter aj has no meaning and it is set to zero. The origin of the coordinate system corresponding to the sliding joint i is chosen to coincide with the next defined link origin. The Zj axis is aligned with the axis of joint (i+ 1). The Xj axis is chosen to be parallel or antiparallel to vector Zj -1 x Zj' Zero joint coordinate qj = dj is defined when the origin of systems OJ-1Xj-l Yi-l Zj-l and OjXjYjZj coincide. For a prismatic joint, angle OJ between axis Xj _ 1 and Xj is fixed and represents a kinematic parameter together with the twist angle IXj \u2022 The origin of the reference coordinate system Oxyz is set to be coincident with the origin of the first link"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000676_s00604-014-1181-1-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000676_s00604-014-1181-1-Figure2-1.png",
        "caption": "Fig. 2 Schematic representation of the assay procedure for the ECL device. a SPCE: (a) silver ink, (b) PVC film, (c) insulating dielectric, (d) Ag/AgCl reference electrode, (e) carbon ink counter electrode, (f) two carbon ink working electrodes; b NPG modified SPCE; c after immobilization of aptamer; d capture with cells; e blocking with BSA; and f immobilization with the ZnO@CQDs labeled ConA. Reproduced from [55] with permission from Elsevier",
        "texts": [
            " Cancer biomarkers-aptasensors Since each cancer cell line has the specific intra-or extracellular biomarkers, which distinguish it from normal cell lines, therefore, methods that can enable sensitive and selective detection cancer cells through precise molecular recognition of their biomarkers are highly desired [54]. Based on this concept, Yu et al. [55] exploited modified SPCEs to develop a new electrochemiluminescence (ECL) platform for ultrasensitive detection of K562 leukemia cells. The assay principle for the ECL device is illustrated in Fig. 2. Two working electrodes were used for one determination to obtain more exact results. The SPCE was modified with nanoporous gold to provide a good pathway of electron transfer and to enhance the immobilized amount of aptamers. Then aptamers were used for cell capture and the concanavalin A conjugated ZnO@CQDs was used for selective recognition of the cell surface carbohydrate. The method showed a good analytical performance for the detection of K562cells with a detection limit of 46 cells mL\u22121. Mucin 1 (MUCl) is an integral membrane glycoprotein expressed by most if not all \u2018wet\u2019 epithelia, such as on bladder, breast, gastric, pancreas and ovary"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure5.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure5.2-1.png",
        "caption": "Figure 5.2 (b) The assumed directions of the support reactions in A and B; (c) the support reactions as they are actually acting; (d) the closed force polygon for the force equilibrium of joint B; (e) isolated joint B with all the forces acting on it.",
        "texts": [
            " The support reactions are shown in Figure 5.1b. The equilibrium equations are \u2211 Fx = Ah = 0,\u2211 Fy = \u2212(6 kN) + Av = 0,\u2211 Tz|A = \u2212(6 kN) \u00d7 (1.5 m) + Am = 0. The solution is Ah = 0, Av = 6 kN and Am = 9 kNm. 5 Calculating Support Reactions and Interaction Forces 155 The fact that the solutions found are positive confirms the correctness of the directions assumed for these support reactions. Note that the support reaction Av and the force of 6 kN at C together form a couple that is in equilibrium with the fixed-end moment Am. Example 2 In Figure 5.2a, a block with a weight of 60 kN is supported on three bars. Questions: a. Determine the support reactions at A and B. b. Determine the forces N(a), N(b) and N(c) in the bars1 with the correct sign for tension and compression, based on the convention that a force N as tensile force is positive and as compressive force is negative. Solution (units in kN and m): a. Figure 5.2b shows the support reactions. The directions of Ah and Av are such that the line of action of their resultant coincides with two-force member (a). For the others, the directions of the support reactions have been assumed arbitrarily. On the basis of the slope of bar (a), it follows that Ah/Av = 4/3, or Ah = (4/3)Av. Av can be determined using the moment equilibrium about B: \u2211 Tz|B = \u2212Av \u00d7 4 \u2212 60 \u00d7 4 = 0 \u21d2 Av = \u221260 kN so that Ah = 4 3Av = \u221280 kN. 1 The upper index indicates the relevant bar. The brackets can be omitted as they do not create any confusion. 156 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Bh is found from the horizontal force equilibrium: \u2211 Fx = Ah + Bh = \u221280 + Bh = 0 \u21d2 Bh = 80 kN. Bv follows from the vertical force equilibrium: \u2211 Fy = Av + Bv \u2212 60 = \u221260 + Bv \u2212 60 = 0 \u21d2 Bv = 120 kN. Bv can also be determined from the moment equilibrium about A. In Figure 5.2c, the support reactions are shown as they act in reality. Only the direction of the support reactions at A was falsely assumed. b. Figure 5.2c shows directly that a tensile force is acting in bar (a): N(a) = \u221a A2 h + A2 v = \u221a 802 + 602 = 100 kN. The forces in the bars (b) and (c) can be determined from the force equilibrium of joint B. The force polygon in Figure 5.2d shows that bar (b) exerts a force of 40 kN on joint B. This force \u201cpushes\u201d against the joint. Figure 5.2e shows the interaction forces between bar and joint. In bar (b), there is a compressive force N(b) = \u221240 kN. Bar (c) is exerting a force of 80 \u221a 2 kN on joint B, also a compressive force, so that N(c) = \u221280 \u221a 2 kN. Alternative solution (units in kN and m): The questions a and b are now answered in reverse order. b. In Figure 5.3a, the block has been isolated at A\u2032 and B\u2032. N(a), N(b) and N(c) are the (tensile) forces that the bars are exerting on the block. In Figure 5.3b, they have been resolved into their components"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000005_978-3-642-84379-2-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000005_978-3-642-84379-2-Figure6-1.png",
        "caption": "Fig. 6",
        "texts": [
            " The possibility of rejecting a zero measure set permits determining the velocity vector for sliding modes, too. Indeed, although in the points on the discontinuity surface vector f(x, t) in system (1.1) is uncertain, these points in an n-dimensional vicinity make a zero measure set and may be disregarded. Thus, in systems with a single discontinuity surface, the Filippov method yields the following result: (a) the minimal convex set of all vectors f(x, t) in the vicinity of some point (x, t) on the discontinuity surface (Fig. 6) is actually a straight line connecti-ng the ends of vectors f + and f - (to which vector f(x, t) tends with x tending to the point under consideration from either side of the discontinuity surface), and (b) since vector fO in sliding mode lies 10 Chapter 1 Scope of the Theory of Sliding Modes on a plane tangential to the discontinuity surface, the end of this vector may be found as a point of intersection of this plane and the straight line connecting the ends of vectors f + and f -. At first sight, the Filippov continuation method seems quite applicable to treat controls of the type (1",
            "6) which is true along the intersection of its discontinuity surfaces and may be regarded as its sliding mode equation. Before we try to substantiate the validity of the equation obtained in the described manner - via introduction of a boundary layer and a subsequent limiting procedure -let us compare this equation in a scalar control system against the equation resulting from Filippov's procedure. Shown in Fig. 8 for point (x, t) are velocity vectors fO and f(x, t, ueq ) obtained following these two procedures; the way they could be obtained have been discussed in Chap. 1, Sect. 2 (Fig. 6) and in Chap. 2, Sect. 2 (Fig. 7). Since the locus of f(x, t, u) is a spatial curve, the two vectors - vector fO obtained by Filippov's technique and vector f(x, t, ueq ) found with the use of the equivalent control technique - are, generally speaking, not identical and moreover, not even colin ear. These vectors are known to coincide only when the locus is a straight line (which is the case in a control-linear system). Of low probability is the case when the locus is not a straight line but, accidentally, crosses the tangential plane in the same point as does the straight line connecting the ends of vectors f + and f - (a dotted lineJin Fig",
            " To determine the average velocity associated with that motion, find first the shifts of the state, L1x, over two neighboring intervals, one with u = u - and 1 Examples of Discontinuous Systems with Ambiguous Sliding Equations 31 N ________ iL-____ ~---------S(X)=O J_L-_---' /J.X Fig. 9 the other with u = u +: (3.1) where L1t! and L1t2 are the durations of the intervals found from the relations 2.1 L1t! = _' grads\u00b7f 2.1 L1t2 = ----. grad s\u00b7f+ Equations (3.1) and (3.2) determine the average velocity! fO = __ ~ad s\u00b7 f - f + _ grad s\u00b7 f + fgrads\u00b7(f- - f+) grads\u00b7(f- - f+) (3.2) (3.3) for the motion equation x = fO(x, t) in the sliding mode. Let us show now that Filippov's procdure yields the same result. The velocity vector, r (Fig. 6), belongs to a straight line which connects the ends of vectors f + and f -, i.e. x = fO(x, t), fO=l1f+ +(1-I1)f-, (0~11~ 1) where 11 is a parameter depending on the mutual position and magnitudes of the column vectors f + and f - and the row vector grad s, the gradient of function s(x). Finding 11 from the condition grad s'fo = 0, we obtain the equation . grads\u00b7f- f+ grads\u00b7f+ fx = + --=---'----,- grad s . (f - - f + ) grad s . (f - - f + ) , (3.4) describing the sliding motion, the right hand part of the equation being identical to (3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001598_tpel.2012.2227808-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001598_tpel.2012.2227808-Figure2-1.png",
        "caption": "Fig. 2. Implantation of search coils in test machine and FEA model",
        "texts": [
            " Figure 1 presents the test machine utilized in this paper and its FEA model, which is built using a commercial FEA software package MagNet by Infolytica. This three phase Y-connected machine has a concentrated armature winding and a sinusoidal back EMF. Fig. 1. Test machine and its FEA model To allow maximum degrees of freedom during the testing phase, twelve search coils were wound around the stator teeth of this machine, for multi-fault detection and health monitoring. Four turns are used for each search coil to mitigate the variance between turns for better voltage measurement accuracy. Their implementation is illustrated in Fig. 2. Their voltages were recorded by a data acquisition system for further analysis. Details of the experimental implementation are described in Section VI. A. Eccentricity Eccentricity in a machine is a condition of uneven air-gap between the stator and rotor. If the condition is severe, the unbalanced magnetic pull (UMP) could cause stator and rotor contact [29]. Generally, eccentricity is classified into three types: static eccentricity, dynamic eccentricity and mixed eccentricity. Static eccentricity is the case that there is a displacement of the axis of rotation, which usually could be caused by an oval stator or misaligned mounting of bearings, rotors or stators"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001482_0278364910388677-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001482_0278364910388677-Figure7-1.png",
        "caption": "Fig. 7. (a) Teaching interface used to demonstrate footholds. Red ball at the foot indicates the chosen (dangerous) foothold, light green ball below indicates the demonstrated optimal (safe) foothold. (b) Terrain templates in multiple scales extracted from the demonstrated foothold. The white spheres indicate the position of the foot on each template.",
        "texts": [
            " This reward function can then be used to rank all of the footholds in F , and the one with the highest reward selected as the target foothold.2. The task of learning the trade-offs for foothold selection has thus been reduced to learning the weight vector w. An expert is shown logs of runs executed by the robot on different kinds of terrain. They inspect each run for suboptimal foot placement choices made by the robot, and in each of those situations, labels the correct foothold fc that they think is the best greedy choice from among the entire set of reachable footholds3 F . Figure 7(a) shows a screenshot of our teaching interface, where the red ball depicts the sub-optimal foot placement choice made by the robot, and the white ball below shows the optimal foot placement chosen by the expert. When the expert labels the foothold fc from the set F as the optimal foothold, they are implicitly providing information that the reward for the chosen foothold fc is better than the reward for all others: wTxc > wTxi \u2200i \u2208 F ; i = c. (2) This corresponds to the widely studied rank learning or preference learning problem in the machine learning literature (Cohen et al",
            " Manual creation of such a library of templates would be too time consuming, and it would be nearly impossible to attain good generalization performance, since the rewards for each template would have to be tuned carefully. Hence, we propose an algorithm that uses expertdemonstrated footholds to simultaneously learn a small set of templates, and a foothold ranking function that uses these templates. We first describe our methodology for template extraction from expert demonstrations, followed by the template learning algorithm. From each foothold demonstration \u3008F , fc\u3009 made by the expert, we extract a set of templates on multiple scales. Figure 7(b) shows the three scales of templates that were extracted from the demonstrated foothold (light green ball in Figure 7(a)). The scales that we use are dictated by the geometry of our quadruped robot, and are designed to independently capture different properties of the terrain that make up its reward function. The smallest scale (24 mm\u00d724 mm) is of the order of the size of the foot, and encodes information about surface friction properties and micro-features that define the slip characteristics of the foothold. We intend to capture information about clearance from obstacles and drop-offs using the medium scale (54 mm\u00d754 mm)",
            " The pose features included measures of progress towards the goal, stability margins, reachability of the foothold, differences in leg heights, and knee and body clearance from the terrain. The terrain features we used (on each scale) were slope and curvature along the x- and y-axes,7 and standard deviation of the terrain. This resulted in a total of nine features being used per length scale. The same three length scales were used for both feature computation and template extraction, and are depicted in Figure 7(b). The features and templates are mirrored in the y direction for the left and right legs, but independent functions were learnt for front and hind legs as their properties tend to be quite different. A foothold ranking function was learnt for each of the three feature sets mentioned above, using the algorithms described in Sections 5 and 5.4. The l1 regularization parameter \u03bb was set individually in each case by choosing the value that minimized cross-validation error on the training data. Table 1 shows the number of templates selected on each scale, with and without the use of terrain features"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure9.23-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure9.23-1.png",
        "caption": "Fig. 9.23. MonoSpaceTM elevator: (a) elevator propulsion system; (b) EcodiskTM motor. Courtesy of Kone, Hyvinka\u0308a\u0308, Finland.",
        "texts": [
            "22 and specifications are given in Table 9.8 [130]. The concept of gearless electromechanical drive for elevators was introduced in 1992 by Kone Corporation in Hyvinka\u0308a\u0308, Finland [113]. With the aid of a disc type low speed compact AFPM brushless motor (Table 9.10), the penthouse machinery room can be replaced by a space-saving direct electromechanical drive. In comparison with a low speed axial flux cage induction motor of similar diameter, the AFPM brushless motor has a doubled efficiency and a three times higher power factor. Fig. 9.23a shows the propulsion system of the elevator while Fig. 9.23b shows how the AFPM brushless motor is installed between the guide rails of the car and the hoistway wall. Table 9.9 contains key parameters for the comparison of different hoisting technologies [113]. The disc type AFPM brushless motor is a clear winner. Specifications of single-sided AFPM brushless motors for gearless passenger elevators are given in Table 9.10 [113]. Laminated stators have from 96 to 120 slots with three phase short pitch winding and class F insulation. For example, the MX05 motor rated at 2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure11.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure11.1-1.png",
        "caption": "FIGURE 11.1. The DOF of a roll model of rigid vehicles are: x, y, \u03d5, \u03c8.",
        "texts": [
            "5t for 0 < t < 2\u03c0 and \u03b4 (t) = 0 for t \u2264 0 and t \u2265 2\u03c0. (c) \u03b4 (t) = sin t for 0 < t < \u03c0 and \u03b4 (t) = 0 for t \u2264 0 and t \u2265 \u03c0. 11 F Vehicle Roll Dynamics In this chapter, we develop a dynamic model for a rigid vehicle having forward, lateral, yaw, and roll velocities. The model of a rollable rigid vehicle is more exact and more effective compared to the rigid vehicle planar model. Using this model, we are able to analyze the roll behavior of a vehicle as well as its maneuvering. 11.1 F Vehicle Coordinate and DOF Figure 11.1 illustrates a vehicle with a body coordinate B(Cxyz) at the mass center C. The x-axis is a longitudinal axis passing through C and directed forward. The y-axis goes laterally to the left from the driver\u2019s viewpoint. The z-axis makes the coordinate system a right-hand triad. When the car is parked on a flat horizontal road, the z-axis is perpendicular to the ground, opposite to the gravitational acceleration g. The equations of motion of the vehicle are expressed in B(Cxyz). Angular orientation and angular velocity of a vehicle are expressed by three angles: roll \u03d5, pitch \u03b8, yaw \u03c8, and their rates: roll rate p, pitch rate 666 11"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-FigureD.7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-FigureD.7-1.png",
        "caption": "FIGURE D.7 Two orthogonal rectangular conducting sheets.",
        "texts": [
            "23c) Then, the closed form of this potential coefficient is Pp12 = 1 4\ud835\udf0b\ud835\udf16 1 S1S2 4\u2211 k=1 4\u2211 m=1 (\u22121)m+k [ b2 m \u2212 Z2 2 ak log (ak + rkm + \ud835\udf16 ) + a2 k \u2212 Z2 2 bm log (bm + rkm + \ud835\udf16 ) \u2212 1 6 (b2 m \u2212 2Z2 + a2 k)rkm \u2212 bm Z ak tan\u22121 ( ak bm rkm Z )] , (D.24) where a1 = xs2 \u2212 xe1, a2 = xe2 \u2212 xe1 (D.25a) a3 = xe2 \u2212 xs1, a4 = xs2 \u2212 xs1 (D.25b) b1 = ys2 \u2212 ye1, b2 = ye2 \u2212 ye1 (D.25c) b3 = ye2 \u2212 ys1, b4 = ys2 \u2212 ys1 (D.25d) Z = z2 \u2212 z1 + \ud835\udf16, rkm = \u221a a2 k + b2 m + Z2 . (D.25e) D.1.8 Pp12 for Two Orthogonal Rectangular Sheet Cells Two rectangular cell sheets at an angle of 90\u2218 are shown in Fig. D.7. The partial potential is similar to (D.6) given by Pp12 = 1 4\ud835\udf0b\ud835\udf16 1 12 \u222b ye1 ys1 \u222b xe1 xs1 \u222b ze2 zs2 \u222b xe2 xs2 1 R1,2 dx2 dz2 dx1 dy1, (D.26) where S1 = (xe1 \u2212 xs1)(ye1 \u2212 ys1) (D.27a) S2 = (xe2 \u2212 xs2)(ze2 \u2212 zs2) (D.27b) R1,2 = \u221a (x1 \u2212 x2)2 + (y1 \u2212 y2)2 + (z1 \u2212 z2)2. (D.27c) PARTIAL POTENTIAL COEFFICIENTS FOR ORTHOGONAL GEOMETRIES 417 Then, the final formulation of Pp12 for two orthogonal charge sheets is Pp12 = 1 4\ud835\udf0b\ud835\udf16 1 S1S2 4\u2211 k=1 2\u2211 m=1 2\u2211 \ud835\udcc1=1 (\u22121)\ud835\udcc1+m+k+1 [ akbmc\ud835\udcc1 log (ak + rkm\ud835\udcc1 + \ud835\udf16 ) + ( a2 k 2 \u2212 c2 \ud835\udcc1 6 ) c\ud835\udcc1 log (bm + rkm\ud835\udcc1 + \ud835\udf16 ) \u2212 bmc\ud835\udcc1 3 rkm\ud835\udcc1 + bm ( a2 k 2 \u2212 b2 m 6 ) log (c\ud835\udcc1 + rkm\ud835\udcc1 + \ud835\udf16 ) \u2212 akc2 \ud835\udcc1 2 tan\u22121 ( akbm c\ud835\udcc1 rkm\ud835\udcc1 ) \u2212 a3 k 6 tan\u22121 ( bmc\ud835\udcc1 ak rkm\ud835\udcc1 ) \u2212 b2 mak 2 tan\u22121 ( akc\ud835\udcc1 bm rkm\ud835\udcc1 )] , (D"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001483_0954405414567522-Figure10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001483_0954405414567522-Figure10-1.png",
        "caption": "Figure 10. (a) Powder stacking strategy, (b) laser irradiation region and (c) meshing of a simulation model considering powder arrangements.49",
        "texts": [
            " 2513 C) that a 3D model is necessary for obtaining accurate results. In their 3D model, a higher maximum temperature was observed as the laser beam moved from the center of the first layer to the center of the second layer. In SLS, the temperature gradient under the beam spot is greater than that in the already sintered areas, according to Ren et al.63 In a semi-scanned layer, the residual heat in the powder-half is higher than that of the solid-half.65 Liu et al.49 developed a 3D finite element model, which takes into account powder arrangements as shown in Figure 10, in order to study SLS. They came to the conclusion that the temperature field of laser sintering is intermittent in the micro scale, because of the discretely distributed particles, and that the temperature distribution is remarkably inhomogeneous with the maximum temperature to be observed in the top layer. This approach offers enhanced accuracy. However, it would not be recommended for real parts, as modeling in the micro- at Gazi University on February 4, 2015pib.sagepub.comDownloaded from scale, would mean an enormous number of elements and a huge computation cost"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003657_02783649922066475-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003657_02783649922066475-Figure5-1.png",
        "caption": "Fig. 5. The pantograph-style leg mechanism and actuator.",
        "texts": [
            " This section describes the legs, the body frames, and the rappelling winch system. at East Tennessee State University on June 6, 2015ijr.sagepub.comDownloaded from The design philosophy for the leg was to make a tough appendage that could withstand repeated falls and collisions with obstacles, and support its load on any type of terrain material ranging from mud to hard rock. Each leg is a pantograph mechanism that amplifies the motion of a linear drive actuator to generate a straight-line foot motion (see Fig. 5). The actuator moves the upper control point, while the lower control point is fixed to the robot\u2019s body. The linkage geometry amplifies actuator motion by four times at the foot, and amplifies axial foot forces by the same factor at the actuator. The actuator is compact and located far away from the foot and leg shank area, which greatly reduces the chance of damage during walking and rappelling. The pantograph also eliminates so-called geometric work energy losses during body lifting and lowering, the former of which is typically the peak energy demand for a slow-moving walking robot (Hirose 1983)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure4.26-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure4.26-1.png",
        "caption": "Fig. 4.26: A gripping mechanism and its parameters for a two-finger gripper.",
        "texts": [
            " Making use of the F computation and G.I. formulation it is possible to deduce design charts in order to derive the optimal linkage proportions for a given gripping mechanism kinematic chain, and to operate dimensional modifications in an optimization procedure. In Fig. 4.25 examples of analytical expressions for the above-mentioned indices are shown as referring to illustrative gripping mechanisms. The differences among these grippers can be remarked by and through the formulas of the proposed performance indices. In Fig. 4.26 the kinematic chain of a gripping mechanism for a widely-used two-finger gripper is shown as referring to one finger only, due to the symmetry of the mechanical Chapter 4 Fundamentals of the Mechanics of Grasp276 design. The plots in Fig. 4.27 show the numerical results of the kinematic analysis of the gripping mechanism of Fig. 4.26 by using the formulation and indices that are outlined in section 4.6. It is worth noting how the design and operation charts for C.I. and G.I. indicate optimal design sizes and operation angles for optimum performances. Fundamentals of the Mechanics of Robotic Manipulation 277 Fig. 4.26: a) Capability Index (C.I.) versus mechanism sizes; b) Capability Index (C.I.) versus input crank angle at the open configuration; c) Grasping Index (G.I.) versus mechanism sizes; d) Grasping Index (G.I.) versus input crank angle at the open configuration. Chapter 4 Fundamentals of the Mechanics of Grasp278 The design problem for grippers consists of sizing all the components of the gripper to ensure suitable grasping performances for a given class of objects with a certain weight and dimension"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003906_robot.2010.5509646-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003906_robot.2010.5509646-Figure1-1.png",
        "caption": "Fig. 1. The base frame attached to the robot is connected to the inertial frame via 6 unactuated virtual DOFs",
        "texts": [
            " The floating base framework provides the most general representation of a rigid-body system unattached to the world, and is necessary to describe the complete dynamics of the system with respect to an inertial frame. The system configuration is represented as: q = [ qT r xT b ]T (1) where qr \u2208 R n is the joint configuration of the rigid-body robot with n joints and xb \u2208 SE(3) is the position and orientation of the coordinate system attached to the robot 978-1-4244-5040-4/10/$26.00 \u00a92010 IEEE 3406 base, and measured with respect to an inertial frame. Figure 1 illustrates this representation by showing the 6 virtual degrees of freedom attached from inertial frame to the robot base frame. When the robot is in contact with the environment, the equations of motion with respect to an inertial frame are given by: M(q)q\u0308 + h(q, q\u0307) = ST \u03c4 + JT C(q)\u03bb (2) with variables defined as follows: \u2022 M(q) \u2208 R n+6\u00d7n+6: the floating base inertia matrix \u2022 h(q, q\u0307) \u2208 R n+6: the floating base centripetal, Coriolis, and gravity forces. \u2022 S = [ In\u00d7n 0n\u00d76 ] : the actuated joint selection matrix \u2022 \u03c4 \u2208 R n: the vector of actuated joint torques \u2022 JC \u2208 R k\u00d7n+6: the Jacobian of k linearly independent constraints \u2022 \u03bb \u2208 R k: the vector of k linearly independent constraint forces By constraints, we mean locations on the robot in environmental contact, where external forces or torques are applied, and no motion is observed with respect to the inertial frame"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure9.40-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure9.40-1.png",
        "caption": "Fig. 9.40. Computer created 3D image of the axial flux 12.5 kW HTS motor for pod propulsor: 1 \u2014 HTS field excitation coils, 2 \u2014 armature excitation coil, 3 \u2014 cores, 4 \u2014 back yoke, 5 \u2014 fiberglass reinforced plate of cryostat, 6 \u2014 cryostat.",
        "texts": [
            " The pod propulsor has outer diameter of 0.8 m and is 2 m long. The diameter of the propeller is 1 m. When a large conventional motor is used with a pod propulsor, the outside diameter of the propulsor becomes also large. This HTS motor is an axial flux (disc type) three phase, eight-pole machine. The stationary field excitation system and rotating armature simplify the cooling system. Technical specifications of the HTS motor are given in Table 9.12. The expanded view of the motor is shown in Fig. 9.40. The main features include: \u2022 machine is cooled by low-cost, easy-to-handle liquid nitrogen; \u2022 uses the bismuth-based 1G BSCCO HTS wires that were fabricated in the lengths of several kilometers with the aid of SEI innovative manufacturing process; \u2022 very low noise; \u2022 very low leakage magnetic flux; \u2022 the surface of the machine is kept at a room temperature and, therefore, can be installed in used in any place. 2 Sumitomo Electric Industries, Ltd., Fuji Electric Systems Co., Ltd., Hitachi, Ltd., Ishikawajima-Harima Heavy Industries Co"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002277_j.prostr.2016.02.039-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002277_j.prostr.2016.02.039-Figure6-1.png",
        "caption": "Fig. 6 \u2013 Mesh of the original component (left) and the optimized component (right).",
        "texts": [],
        "surrounding_texts": [
            "The mesh has great influence in the final solution. Highly refined meshes give very different topologies from less refined meshes. In the final design domain, a more controlled method for meshing was used in order to ensure its high quality. The control parameters also have great influence not only in the solution convergence degree, but also in the computing time, thus a convergence study on these parameters was run. In this optimisation there were two control parameters which were studied, the Relative Convergence Criterion (RCC) and the Discreteness Parameter (DP) which is the equivalent for penalty factor in TO theory. The objective function was the weighted compliance in order to consider the three load cases in the topology optimisation [HyperWorks Guide]. This response is given by Equation 1 \ud835\udc36\ud835\udc36\ud835\udc64\ud835\udc64 = \u2211\ud835\udc64\ud835\udc64\ud835\udc56\ud835\udc56\ud835\udc36\ud835\udc36\ud835\udc56\ud835\udc56 (1) where wi is the weight and Ci is the compliance of load case i which is given by Equation 2 Ci = 1 2 ui Tfi (2) where ui and fi are the displacement and force vectors, respectively, corresponding to load case i. The objective was the minimization of the weighted compliance and each load case was given the same weight. There were two constraints defined in the optimisation. The first one was regarding the volume fraction of the design domain. The second was a symmetry constraint, forcing the optimised solution to be symmetric with respect to the component\u2019s mid vertical plane as the original component is. Figure 3 and Figure 4 illustrate the TO boundary conditions and final solution pseudo-density distribution. Fig. 3 \u2013 Final Design Domain (blue) and boundary conditions. Fig. 4 \u2013 TO solution with element pseudo-density distribution. Author name / Structural Integrity Procedia 00 (2016) 000\u2013000 5"
        ]
    },
    {
        "image_filename": "designv10_0_0000233_978-1-4614-0460-6-Figure2.14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000233_978-1-4614-0460-6-Figure2.14-1.png",
        "caption": "Fig. 2.14 Natural frequency dependence on R for the rolling cylinder example",
        "texts": [
            " For small angles, we can approximate sin y\u00f0 \u00de as y and rewrite this equation as: \u20acy\u00fe 2 3 g R r\u00f0 \u00de y \u00bc 0: (2.42) h1 \u03b8 h2r cos(q )(R\u2212r) R R\u2212r This is the system equation of motion written in the same form as Eq. 1.10, which is the standard form for the differential equation of harmonic motion. Using this form, d2x dt2 \u00fe o2x \u00bc 0, we can immediately identify the natural frequency for the system: on \u00bc ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 3 g R r\u00f0 \u00de s : (2.43) This equation tells us that a larger radius, R, of the concave surface should give a lower natural frequency. Using Fig. 2.14, we see that this makes intuitive sense. Let\u2019s return to the spring\u2013mass system and show how to quickly identify the natural frequency using the energy expressions. First, we can recognize that the kinetic energy is maximum when the potential energy is zero since their sum is a constant. In equation form, we have Kmax + 0 \u00bc constant. For the undamped, single degree of freedom, lumped parameter model, this maximum kinetic energy is identified when the oscillating mass is passing through its x \u00bc 0 position (where the velocity is maximum, _xmax)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000145_j.biotechadv.2011.09.003-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000145_j.biotechadv.2011.09.003-Figure1-1.png",
        "caption": "Fig. 1. Scheme of",
        "texts": [
            " S, 66860 Perpignan Cedex, France. Environmentalmonitoring, food industry or clinical analyses require the development of selective tools for biomolecule detection. Biosensors can be regarded as complementary tools to classical analytical methods (e.g. high performance liquid chromatography) due to their unique feature, their inherent simplicity, relative low cost, rapid response and proneness to miniaturization, thereby allowing continuous monitoring. A biosensor is a device composed of two intimately associated elements (Fig. 1): \u25cb a bioreceptor, that is an immobilized sensitive biological element (e.g. enzyme, DNA probe, antibody) recognizing the analyte (e.g. enzyme substrate, complementary DNA, antigen). Although antibodies and oligonucleotides are widely employed, enzymes are by far the most commonly used biosensing elements in biosensors. \u25cb a transducer, that is used to convert the (bio)chemical signal resulting from the interaction of the analyte with the bioreceptor into an electronic one. The intensity of generated signal is directly or inversely proportional to the analyte concentration"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003721_s0924-0136(03)00283-8-Figure13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003721_s0924-0136(03)00283-8-Figure13-1.png",
        "caption": "Fig. 13. The CAD picture (a) and the laser-sintered part (b) made from LaserTool.",
        "texts": [
            " Table 1 depicts the technical and economical features of the laser-sintered and the post-sintered parts. One can use both parts (as laser-sintered or post-sintered) dependent on the application and the required properties. Potential applications are mold inserts for plastic/powder injection molding and pressure die-casting. Fig. 12 shows a general view of the production process for rapid tooling using the developed material. To verify the applicability of this procedure, a case study was performed. Fig. 13a depicts the CAD picture of the selected part. It is a mold insert for plastic injection molding. To increase the productivity of the manufacturing process, spiral-cooling channels were designed inside the mold and incorporated in the CAD data. Fig. 13b shows the corresponding part processed from LaserTool\u00ae. Liquid phase laser sintering of a multi-component iron base powder blend was investigated. It was found that the powder characteristics such as size, shape, and distribution of the particles have a significant influence on the sintered density. The chemical composition of the powder blend is also of special concern. Optimization of the carbon content of the mixture in order to decrease the surface tension and the viscosity of the Fe melt is important"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001252_j.ymssp.2007.12.001-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001252_j.ymssp.2007.12.001-Figure1-1.png",
        "caption": "Fig. 1. University Of New South Wales (UNSW) spur gear test rig: test rig photo (a), schematic diagram (b).",
        "texts": [
            " The comparison between the simulated results and the experimental results is presented in Section 6 for the three types of localised faults in the bearing, viz. inner/outer races and rolling elements. While this part of the paper (part I) is concerned with the modelling of localised faults, the second part (part II) discusses the modelling of extended faults and compares the simulated results with those experimentally obtained for extended inner and outer race faults. Some of the theoretical development of this paper was presented in conference papers [3,4]. The gearbox test rig (Fig. 1) under investigation was built by Sweeney [5] to investigate the effect of gear profile errors on transmission error (TE). In this test rig, the single stage gearbox (in this case a spur gear set with 1:1 ratio and 32 teeth on each gear) is driven primarily by a three-phase electric motor, but with circulating power via a hydraulic pump/motor set. The input and output shafts of the gearbox are arranged in parallel and each shaft is supported by two double row ball bearings (Koyo 1205). The flywheels are used to reduce the fluctuations of the input and output shaft speeds"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001674_j.mechmachtheory.2012.10.008-Figure12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001674_j.mechmachtheory.2012.10.008-Figure12-1.png",
        "caption": "Fig. 12. Meshing forces normal direction.",
        "texts": [
            " The relative velocity between surfaces for each point in contact is obtained as the difference between absolute velocities v \u2192 Pi 1=2\u00f0 \u00de \u00bc v \u2192 Pi 1=0\u00f0 \u00de\u2212v \u2192 Pi 2=0\u00f0 \u00de \u00f023\u00de the absolute velocity for the contact points can be obtained from the rigid solid motion field as v \u2192 Pi 1=0\u00f0 \u00de \u00bc v \u2192 C1 1=0\u00f0 \u00de \u00fe\u03a9 \u2192 1=0\u00f0 \u00de C1Pi ; v \u2192 Pi 2=0\u00f0 \u00de \u00bc v \u2192 C2 2=0\u00f0 \u00de \u00fe\u03a9 \u2192 2=0\u00f0 \u00de C2Pi \u00f024\u00de 1(1/0), vC2(2/0) are the absolute velocities for each gear centre and \u03a9(1/0) y \u03a9(2/0) their rotational velocity. CiPi is the vector he gear centre Ci to the contact point Pi that according to Fig. 12 can be obtained from CjQ i j \u00fe Qi jPi ; j \u00bc 1;2 \u00f025\u00de CjQ i j is the base radius for the contact point Pi on the gear j (or equivalent base circle for the Involute\u2013Tip rounding contact ) and Qi jPi is the distance between the tangent point and the contact point defined in the reference OXY (see Fig. 12) by the ing expressions \u00bc \u03c11 sin \u03c6T \u00fe \u03c8\u00f0 \u00de \u2212 cos \u03c6T \u00fe \u03c8\u00f0 \u00de ; Q1Pi \u00bc Q1Pi \u2212 cos \u03c6T \u00fe \u03c8\u00f0 \u00de \u2212 sin \u03c6T \u00fe \u03c8\u00f0 \u00de ; \u00bc \u03c12 \u2212 sin \u03c6T \u00fe \u03c8\u00f0 \u00de cos \u03c6T \u00fe \u03c8\u00f0 \u00de ; Q2Pi \u00bc Q2Pi cos \u03c6T \u00fe \u03c8\u00f0 \u00de sin \u03c6T \u00fe \u03c8\u00f0 \u00de ; \u00f026\u00de The direction of the resultant friction force is defined by the relative velocity vector projected onto the tangential direction, which can be obtained from (27) for direct (i=1,\u2026,N/2) and reverse contacts (i=N/2+1,\u2026,N) t \u2192 1i \u00bc sin \u03c8\u00fe \u03c6i\u00f0 \u00de \u2212 cos \u03c8\u00fe \u03c6i\u00f0 \u00de ; t \u2192 2i \u00bc \u2212 t \u2192 1i; i \u00bc 1;\u2026; N 2 t \u2192 1i \u00bc sin \u03c6i\u2212\u03c8\u00f0 \u00de cos \u03c6i\u2212\u03c8\u00f0 \u00de ; t \u2192 2i \u00bc \u2212 t \u2192 1i; i \u00bc N 2 \u00fe 1;\u2026;N \u00f027\u00de Finally, the resultant friction forces translated to the geometric centre of each gear are defined by where result the an obtain LTE \u03b81\u00f0 F1x F1y T1 F2x F2y T2 8>>>< >>>: 9>>>= >>>; Friction \u00bc \u2212 XN i\u00bc1 Fif tanh v \u2192 Pi 1=2\u00f0 \u00de\u22c5 t \u2192 1i v0 0 @ 1 Asgn v \u2192 Pi 1=2\u00f0 \u00de\u22c5 t \u2192 1i \u22c5 t \u2192 1i XN i\u00bc1 \u2212Fif tanh v \u2192 Pi 1=2\u00f0 \u00de\u22c5 t \u2192 1i v0 0 @ 1 Asgn v \u2192 Pi 1=2\u00f0 \u00de\u22c5 t \u2192 1i \u22c5 t \u2192 1i 0 @ 1 A C1Pi XN i\u00bc1 Fif tanh v \u2192 Pi 1=2\u00f0 \u00de\u22c5 t \u2192 1i v0 0 @ 1 Asgn v \u2192 Pi 1=2\u00f0 \u00de\u22c5 t \u2192 1i \u22c5 t \u2192 1i XN i\u00bc1 Fif tanh v \u2192 Pi 1=2\u00f0 \u00de\u22c5 t \u2192 1i v0 0 @ 1 Asgn v \u2192 Pi 1=2\u00f0 \u00de\u22c5 t \u2192 1i \u22c5 t \u2192 1i 0 @ 1 A C2Pi 8>>>>>>>>>>< >>>>>>>>>>: 9>>>>>>>>>>= >>>>>>>>>>; \u00f028\u00de In Section 2 the procedure adopted to obtain the meshing forces has been presented"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure8.20-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure8.20-1.png",
        "caption": "Figure 8.20 (a) The horizontal earth pressure \u03c3e;h on the sheet piling, composed of the horizontal water pressure \u03c3w and the horizontal grain pressure \u03c3g;h. (b) The horizontal load on a 1-metre wide strip of sheet piling modelled as a line element, together with the shear force V and the bending moment M at A.",
        "texts": [
            "19a shows the distribution of the vertical earth pressure \u03c3e;v, including the contribution of the water pressure \u03c3w. If the water pressure is subtracted from the vertical earth pressure, it gives the vertical grain pressure \u03c3g;v (see Figure 8.19b). The horizontal (active) grain pressure \u03c3g;h is equal to the vertical grain pressure \u03c3g;v multiplied by Ka = 1/3 (see Figure 8.19c). \u00d7 (4.5 m)(24 kN/m) = 54 kN. 300 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM The sheet piling is subjected not only to the horizontal grain pressure, but also to water pressure. Figure 8.20a shows the distribution of the horizontal earth pressure \u03c3e;h, composed of the horizontal water pressure \u03c3w and the horizontal grain pressure \u03c3g;h. Figure 8.20b shows the horizontal load acting on a 1-metre strip isolated from the sheet piling and modelled as a line element. The load diagram can be split into a rectangle and two triangles, of which the resultants R and their distances a to A are easy to calculate: R1 = 0.5 \u00d7 (1.5 m)(8 kN/m) = 6 kN, a1 = 3.5 m, R2 = (3 m)(8 kN/m) = 24 kN, a2 = 1.5 m, R3 = 0.5 \u00d7 (3 m)(40 kN/m) = 60 kN, a1 = 1.0 m. From the equilibrium of part AB of the sheet piling strip, we find with the directions of V and M as shown in Figure 8.20b: V = R1 + R2 + R3 = (6 + 24 + 60) kN = 90 kN, M = R1a1 + R2a2 + R3a3 = (6 kN)(3.5 m) + (24 kN)(1.5 m) + (60 kN)(1 m) = 117 kNm. These are the requested forces acting in A on the 1-metre wide vertical strip AB from the sheet piling. In practice, one often uses a method developed by Coulomb1 based on flat slide planes. Here, one assumes that the pressure on the wall is caused by a 1 Charles Auguste de Coulomb (1736\u20131806), French scientist, known for his experiments in friction and electrostatic forces"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.7-1.png",
        "caption": "FIGURE 8.7. A solid axle and a Panhard arm to guide the axle.",
        "texts": [
            " Example 303 Triangular linkage. A triangulated linkage, as shown in Figure 8.6, may be attached to a solid axle to provide lateral and twist resistance during acceleration and braking. Example 304 Panhard arm. High spring rate is a problem of leaf springs. Reducing their stiffness by narrowing them and using fewer leaves, reduces the lateral stiffness and increases the directional stability of the suspension significantly. A Panhard arm is a bar that attaches a solid axle suspension to the chassis laterally. Figure 8.7 illustrates a solid axle and a Panhard arm to guide the axle. Figure 8.8 shows a triangular linkage and a Panhard arm combination for guiding a solid axle. A double triangle mechanism, as shown in Figure 8.9, is an alternative design to guide the axle and support it laterally. Example 305 Straight line linkages. There are many mechanisms that can provide a straight line motion. The simplest mechanisms are four-bar linkages with a coupler point moving straight. Some of the most applied and famous linkages are shown in Figure 8"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure3.32-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure3.32-1.png",
        "caption": "Fig. 3.32. Scheme of manipulation grinding task",
        "texts": [
            "86) together with the condition it A = \u00b0 can be written in the form of eq. (3.75) where T61 : 0(1 x 3) -- --1----- I E= T62 I 0(1 x3) I (3.87) ----1---- I 0(3 x 3) I 1(3 x 3) I (5 x 6) Now, all the elements of the dynamic model (Eq. (3.77)) are determined and the model can be solved. For a manipulator with six degrees offreedom the nominal dynamics are calculated and the results are presented in the next Section. Numerical Example Let us consider the manipulation robot during grinding of the working object which is moved with constant velocity of 0.1 m/s (Fig. 3.32). During the period towards the working object. Let us assume that the contact between the gripper 3.4 Dynamics of Robots Under Action of External Reaction Forces 99 (cutting tool) and the working object is impactless and that the end of period Tl is, at the same time, the start of the interval T2 (point Al in Fig. 3.32). The motion from Al to A2 is constrained, i.e. the manipulator is considered as a closed chain. The relative manipulator motion with respect to the surface, as mentioned before, is defined by means of two parameters, U 1 and u2 \u2022 The parameter U 1 is constant during the grinding process and the parameter U2 can be defined in the form: (3.88) where L[m] is the cutting length and T[s] is the total grinding time. Following the theory explained above, the reduced Jacobian form and the reduced associated vector are afx afx I -- 1 0 I 0 0 0 aUI aU2 I afy afy I 0(3 x (n-3\u00bb 0 1 I 0 0 0 aUI aU2 I 0 0 I 0 0 0 J= I --+---r afz afz -- I 0 0 I 1 0 0 aUI aU2 I 0 0 I 0 1 0 ---1----- 0 0 I 0 0 0\u00abn-3)x2) I I\u00abn-3)x(n-3)) 0 ax 0 ay 0 A= r az ------ 0 0\u00abn-3) x i) 0 0 The tangents 1'1 and 1'2' defined with Eq"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001609_j.ijmecsci.2010.05.005-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001609_j.ijmecsci.2010.05.005-Figure4-1.png",
        "caption": "Fig. 4. (a) Defect on outer race and (b) defect on inner race.",
        "texts": [
            " In other words the respective spring force comes into play when the instantaneous spring length is shorter than its unstressed length (the term in the bracket should be positive); otherwise the separation between ball and race takes place and the resulting force is set to zero. The total restoring force is the sum of the restoring forces from each of the rolling elements. Resolving the total restoring force along the X- and Y-axis we obtain FX \u00bc XZ i \u00bc 1 K \u00f0xcosyi\u00feysinyi\u00de Cr 3=2 cosyi \u00f05\u00de FY \u00bc XZ i \u00bc 1 K \u00f0xcosyi\u00feysinyi\u00de Cr 3=2 sinyi \u00f06\u00de Eqs. (5) and (6) are modified to consider the effect of the defect on the bearing surface. The defect is modeled as a circumferential half sinusoidal wave. Fig. 4(a) and (b) shows the defect on the outer raceway and inner raceway, respectively. The total deflection of the paths of bearing is the sum of the characteristic of the defect and that of the static deflection considered. Radial displacement is obtained by considering the resulting distortion. The restoring force for the presence of defect on the bearing race is given as FXD \u00bc XZ i \u00bc 1 K \u00bd\u00f0xcosyi\u00feysinyi\u00de \u00f0Cr\u00feHD sin\u00f0p=j\u00f0yt yi\u00de\u00de\u00de 3=2 cosyi \u00f07\u00de Fig. 6. (a) Time domain and power spectrum of vibrations due to ou FXD \u00bc XZ i \u00bc 1 K\u00bd\u00f0xcosyi\u00feysinyi\u00de \u00f0Cr\u00feHD sin\u00f0p=j\u00f0yt yi\u00de\u00de\u00de 3=2 sinyi \u00f08\u00de j\u00bc Defect size Raceway radius If the defect is on the outer race yt \u00bcoct\u00fe2p=Z\u00f0Z i\u00de \u00f09\u00de where i\u00bcZ to 1"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003872_0891-5849(95)02224-4-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003872_0891-5849(95)02224-4-Figure7-1.png",
        "caption": "Fig. 7. Standard curves of various ROOH. Absorbance of ferric thiocyanate complex at 500 nm was generated by various concentrations of various ROOH as determined by iodometric spectrophotometric method.",
        "texts": [
            " This was done by extracting the lipid phase with chlo- Owing to a 1:1 stoichiometry of the iodometric method, the response of the ferric thiocyanate assay (mol Fe 3\u00f7 per mol LOOH) could be evaluated. Figure 6 shows the response of the ferric thiocyanate method as a function of LOOH concentration based on iodometry for MeO1OOH. Dividing this response with the molar absorptivity of the ferric thiocyanate complex, the number of Ferric thiocyanate assay 59 60 B. MmALmVI~ et al. equivalents of Fe 3+ formed per 1 mol of LOOH was calculated (4.30 in the actual example). Figure 7 shows the stoichiometries for several hydroperoxides obtained in this work. It is seen that hydroperoxy compounds can be grouped into three distinct groups with respect to the stoichiometry of their reactions with ferrous ions: I-I202 stands alone with a 1:2 stoichiometry; tertiary hydroperoxides (t-BuOOH and CuOOH) both oxidize about three ferrous ions per one ROOH molecule; whereas fatty acid hydroperoxides oxidize more than four ferrous ions each. There are no significant differences within the last group among primary (CeOOH) and secondary hydroperoxides, as well as among positional isomers (9- and 13- hydroperoxylinoleic acid ) or among methyl esters (MeOIOOH) and free acids (O1OOH) (Table 1 )"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003482_rob.4620120607-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003482_rob.4620120607-Figure7-1.png",
        "caption": "Figure 7. A simple six-legged insect robot.",
        "texts": [
            " If the robot executes the Kelly and Murray: Geometric Phases and Robotic Locomotion 429 cyclic shape change shown in Figure 6, the x-component of the associated holonomy is and the y-component Y(1) = exp (-, dwh A dl,) 116 2;3 3 = exp (lo lo 5 dbdwh) = exp(U6) = 1/6. Because the group action is just addition in the group variables, the components of this vector are exactly the net displacements of the system in the x and y directions. It is not unique to this problem that we may simplify the relevant geometry with a gait-related assumption; the following model for legged locomotion hinges upon restriction to a particular family of gaits. 5.2. Tripod Gait in a Six-legged Robot Consider the six-legged robot shown in Figure 7. Assume that the robot walks with a tripod gait, alternating movement of legs 1-4-5 with movement of legs 2-3-6. We will suppose that where (x , y, 0) = planar position and orientation of the center of mass $i = angle of legs, & E [0, E ] hi = height of legs, l z i E [0, 11 The variable hl denotes the height of legs 1-4-5 and the variable h2 the height of legs 2-3-6. A height of 1 indicates that a leg is in contact with the ground, while a height of 0 indicates that the leg is raised free. The angles 4, describe the horizontal positions of the legs"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure16.16-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure16.16-1.png",
        "caption": "Figure 16.16 (a) Bridge modelled as a hinged beam with uniformly distributed movable load; (b) influence line for the shear force at E; (c) the load that causes the maximum positive shear force; (d) the load that causes the maximum negative shear force; (e) for railway bridges, trains are an uninterrupted load.",
        "texts": [
            "15, a change in sign occurs in (q(AC) \u2212 q(CB)). Figure 16.15h therefore gives the most unfavourable position of the set of loads in relation to the bending moment at C. 16 Influence Lines 761 The maximum bending moment is MC = ( 8 3 + 4 44 12 + 40 12 ) (m) \u00d7 (60 kN) = 820 kNm. The first term between brackets includes the influence values to be found from the influence line at the position of each of the point loads. Example 2 \u2013 Most unfavourable placement of a uniformly distributed load The hinged beam in Figure 16.16a is a model of a bridge. The traffic load on the bridge, consisting of a large number of vehicles in a line, is modelled as a uniformly distributed moving load of 90 kN/m. Figure 16.16b shows the influence line for the shear force at E. We will determine the maximum shear force at E in an absolute sense. Solution: Since gaps may appear in traffic jams, the uniformly distributed load is sometimes interrupted. The maximum positive shear force occurs when the uniformly distributed load is present in all fields where the influence line is positive as indicated in Figure 16.16c. This gives VE = { 1 2 \u00d7 (30 m) \u00d7 ( + 1 3 ) + 1 2 \u00d7 (20 m) \u00d7 ( + 2 3 )} \u00d7 (90 kN/m) = 1050 kN. The maximum negative shear force is found for the load in Figure 16.16d: VE = { 1 2 \u00d7 (10 m) \u00d7 ( \u2212 1 3 ) + 1 2 \u00d7 (30 m) \u00d7 ( \u2212 1 3 )} \u00d7 (90 kN/m) = \u2212600 kN. The maximum shear force in an absolute sense is therefore 1050 kN and occurs with the load shown in Figure 16.16c. 762 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM The fact that the positive shear force is predominant is clear from the influence line: the positive area under the influence line is larger than the negative area. In contrast to bridges for standard traffic, loads for trains on railway bridges are uninterrupted loads, that may consist partly of empty carriages, for which one then assumes a lesser load. For a railway bridge, the maximum shear force at E in an absolute sense occurs for the load given in Figure 16.16e. Assume the uniformly distributed load is again 90 kN/m, but now with a minimum of 15 kN/m for the empty carriages. The maximum shear force is then VE = { 1 2 \u00d7 (30 m) \u00d7 ( + 1 3 ) + 1 2 \u00d7 (20 m) \u00d7 ( + 2 3 )} \u00d7 (90 kN/m) + + 1 2 \u00d7 (10 m) \u00d7 ( \u2212 1 3 ) \u00d7 (15 kN/m) = 1025 kN. It is up to the reader to check that neither a distributed load over the entire length AD nor a distributed load over BD are predominant. 16 Influence Lines 763 General comment: If the influence line of a quantity X is requested, the positive direction of this quantity must be stated beforehand"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure10.5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure10.5-1.png",
        "caption": "Figure 10.5 Linked up with the modelling as line element, the section forces (interaction forces) are said to act at the normal centre NC, the intersection of the member axis with the cross-sectional plane. Here there are three different section forces: a normal force of 40 kN, a shear force of 30 kN and a bending moment of 120 kNm.",
        "texts": [
            " 2 2 390 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM the forces in a section as acting in the member axis, or in other words, at the normal centre NC. The vertical component of the stress resultant R at section C is 30 kN and can be shifted directly along its line of action to the member axis. The horizontal component of R is 40 kN and has to be shifted 3 m in section C parallel to its line of action. This gives a moment of (40 kN)(3 m) = 120 kNm. The section forces in section C, acting at the member axis, are shown in Figure 10.5. The section forces on the left- and right-hand sides of the section are equal and opposite. Section forces are interaction forces and always occur in pairs. You should always keep this in mind, even if you are drawing only one of the member segments to the right or left of the section. In the case shown in Figure 10.5, we can distinguish between the following three section forces: \u2022 A normal force: this is the pair of forces of 40 kN with their lines of action along the member axis; a normal force acts normal to the crosssectional plane. \u2022 A shear force: this is the pair of forces of 30 kN in the cross-sectional plane; a shear force acts transverse to the member axis. \u2022 A bending moment: this is the pair of couples of 120 kNm in a plane normal to the cross-sectional plane. For normal force, shear force and bending moment1 we use the symbols N , V and M respectively"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure11.15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure11.15-1.png",
        "caption": "Figure 11.15 (a) For the cantilever beam with a uniformly distributed load along the entire length we have to distinguish two fields. (b) The joining condition, M(1) = M(2), at support B is found from the moment equilibrium of a small beam segment with length x \u2192 0.",
        "texts": [
            " The support reactions can be found from the V diagram as the shear forces on the beam ends: x = 0 : V = + 1 6\u03c1gbh2, x = h : V = \u2212 1 3\u03c1gbh2. These shear forces on the boundaries of the beam are shown in Figure 11.14. As a check, one can be examine whether the beam as a whole is in equilibrium. The resultant R = 1 2\u03c1gbh2 of the distributed load acts at x = 2 3h. This gives \u2211 Fz = R \u2212 1 6\u03c1gbh2 \u2212 1 3\u03c1gbh2 = 0,\u2211 Ty |B = R \u00b7 1 3h \u2212 1 6\u03c1gbh2 \u00b7 h = 0. Force and moment equilibrium therefore are satisfied. 11 Mathematical Description of the Relationship between Section Forces and Loading 451 Example 4 Cantilever beam ABC in Figure 11.15a is simply supported at A and B, and has an overhang BC at B. The beam is carrying a uniformly distributed load of 40 kN/m over its entire length. The dimensions are given in the figure. Question: Using the differential equations for equilibrium, determine the variation of the shear force and the bending moment. Solution: The as yet unknown support reaction at B gives a discontinuity in the distributed load on the isolated member. At this point, the differential equations for the equilibrium are not valid (see Section 11",
            " For field (2) with (5 m)<x<(7 m): qz = 40 kN/m, d2M(2) dx2 = \u2212qz = \u221240 kN/m, V (2) = dM(2) dx = \u2212 \u222b 40 dx = (\u221240x + C (2) 1 ) kN, M(2) = \u222b V (2) dx = \u222b (\u221240x + C (2) 1 ) dx = (\u221220x2 + C (2) 1 x + C (2) 2 ) kNm. There are four boundary conditions available for solving the total of four unknown integration constants C (1) 1 , C (1) 2 , C (2) 1 and C (2) 2 : 1. end condition at A: x = 0; M(1) = 0. 2. joining condition at B: x = 5; M(1) = M(2). 3. end condition at C: x = 7; V (2) = 0. 4. end condition at C: x = 7; M(2) = 0. For the joining condition at B, we will show below how this can be derived from the equilibrium of a member segment with length x ( x \u2192 0) at the joining of the two fields. Figure 11.15b shows the small member segment with the four section forces acting on it and the unknown support reaction at B. If B is not located 11 Mathematical Description of the Relationship between Section Forces and Loading 453 in the middle of the element, but at a distance \u03be x from the left-hand section plane, respectively (1 \u2212 \u03be) x from the right-hand section plane (0 < \u03be < 1), then the equation for the moment equilibrium about B is\u2211 Ty |B = \u2212M(1) + M(2) \u2212 V (1)\u03be x \u2212 V (2)(1 \u2212 \u03be) x + qz x ( \u03be \u2212 1 2 ) x = 0"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003524_0005-1098(94)90008-6-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003524_0005-1098(94)90008-6-Figure7-1.png",
        "caption": "FIG. 7. Simplified model of the robot effector acting upon a surface.",
        "texts": [
            " Assume for simplicity that the whole dynamics of the system are determined by the robot effector and the surface. A further simplification is the restriction to one degree of freedom, which means that we only have one coordinate in our model. Furthermore, let the effector be modeled as a mass-spring-damper system and the surface as a spring. The corresponding differential equation is m \"y + c \u2022 0 ~ - ) ? o ) - m .g + K1 \" (y - Y o - / ) ) + K 2 \u2022 (y -- y,. + r + A) = 0, (35) Equation (36) is a force balance equation (see Fig. 7): N t , + F D - F c ; + F t + F o = 0 (36) with FT the inertial force, F o the damping force, F~ the weighting force, FF the spring force and Fo the reaction force within the surface. Let the sensor force be represented by the spring force FF. Then having that FF= K , ( y - y o - D) ; PF= K , ( ~ - po); je.. = g~(y - Yo), (37) we obtain the equation for the spring (sensor) force FF: F t : - m C . ~ . + K ~ . g - m . K \" F o - K , . y o. (38) In this equation u = - K 1 \u2022 Yo (39) is the control variable which has to be produced by the controller"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure2.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure2.2-1.png",
        "caption": "FIGURE 2.2. Measuring the force under the front wheels.",
        "texts": [
            "34m, are: Fz1 = 2000N (2.17) Fz2 = 1800N (2.18) Therefore, the longitudinal position of the car\u2019s mass center is at a1 = 2l mg Fz2 = 2 2.34 2 (2000 + 1800) \u00d7 1800 = 1.1084m (2.19) a2 = 2l mg Fz1 = 2 2.34 2 (2000 + 1800) \u00d7 2000 = 1.2316m. (2.20) Example 41 Longitudinal mass center determination. The position of mass center C can be determined experimentally. To determine the longitudinal position of C, we should measure the total weight of the car as well as the force under the front or the rear wheels. Figure 2.2 illustrates a situation in which we measure the force under the front wheels. Assuming the force under the front wheels is 2Fz1 , the position of the mass center is calculated by static equilibrium conditionsX Fz = 0 (2.21)X My = 0. (2.22) Applying the equilibrium equations 2Fz1 + 2Fz2 \u2212mg = 0 (2.23) \u22122Fz1a1 + 2Fz2a2 = 0 (2.24) 42 2. Forward Vehicle Dynamics provide the longitudinal position of C and the reaction forces under the rear wheels. a1 = 2l mg Fz2 = 2l mg (mg \u2212 2Fz1) (2.25) Fz2 = 1 2 (mg \u2212 2Fz1) (2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure5.53-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure5.53-1.png",
        "caption": "Figure 5.53 The forces in the bars (a) and (c) follow from the force equilibrium of joint C\u2032: in bar (a) there is a tensile force and in bar (c) there is a compressive force.",
        "texts": [
            " Suppose there is a tensile force in bar (e) of N(e). The magnitude of N(e) follows from the moment equilibrium about S of the left-hand or right-hand part. The simpler equation is obtained with the unloaded right-hand part: \u2211 T (SB) z |S = (20 kN)(6 m) \u2212 N(e) \u00d7 (3 m) = 0 \u21d2 N(e) = 40 kN. Bar (e) is therefore a tension member. The moment equilibrium of the left-hand part about S can be used to check the solution. The forces in the bars (a) and (c) follow from the force equilibrium of joint C\u2032. In Figure 5.53 these forces have been determined using a force polygon. In bar (a) there is a tensile force, while there is a compressive force in bar (c): N(a) = 40 \u221a 2 kN and N(c) = \u221240 kN. In the same way, the force equilibrium of joint D\u2032 gives N(b) = 40 \u221a 2 kN and N(d) = \u221240 kN. c. Figure 5.54 shows all the forces acting on the beam segments AS and SB. The components of the hinge force S follow from the force equilibrium of the part to the left or to the right of S. 186 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM d"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000402_s00170-011-3566-1-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000402_s00170-011-3566-1-Figure3-1.png",
        "caption": "Fig. 3 An example of balling induced SLM forming defect",
        "texts": [
            "2 Negative factor of balling Although the existence of small-sized balling is universal in SLM, it will not block the powder paving process; thus, the negative factor of small-sized balling can be ignored, while the big-sized balling can not only induce a coarse surface but also tends to block the powder paving process, leading to severe forming defects, which can be listed as follows: (1) The metal part can be fractured by severe friction between paving roller and balling; (2) the forming process could not be continued when high friction resistance occurred. To better illustrating the negative factor of balling, an example of balling caused defect, as shown in Fig. 3, is presented here. According to our SLM experiment, the blocking effect induced by balling is depicted in Fig. 4, and the detailed processes are as follows: 1. First layer paving (Fig. 4a): A powder layer with the thickness of H is paved on the forming substrate. 2. Laser scanning of the first layer (Fig. 4b): Laser scanning induces the big-sized balling, which height exceeds the layer thickness H. 3. Second layer paving (Fig. 4c): Although the paving roller can constrainedly pass the top surface of balling, the roller fluctuates up and down during the powder paving process and yields an uneven powder layer which thickness exceeds 2H"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001498_tie.2010.2045322-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001498_tie.2010.2045322-Figure2-1.png",
        "caption": "Fig. 2. n-phase PM motor.",
        "texts": [
            " It is found that by applying suitable constraints of equal fundamental current components and relaxation of the sixth-harmonic pulsating torque component, higher output torque with low ripple can be achieved. Solutions of faulttolerant currents under single-phase open, nonadjacent doublephase open, and adjacent double-phase open conditions are presented. The presented solutions can achieve higher torque than the reported solutions in [20]. Detailed experimental results of a five-phase PM machine under various fault conditions are presented for the validation of the proposed fault-tolerant control techniques. Consider an n-phase two-pole PM machine, as shown in Fig. 2, where phase 1 is under fault condition. The polarities of the back EMFs, induced in the stator windings, and the direction of the excitation currents are also shown in this figure. The rotor angular speed is \u03c9. The generalized expression for the back EMF induced in any gth phase can be written as eg = \u221e\u2211 m=1 Em \u00b7 cos {m(\u03c9t \u2212 2\u03c0g/n)} . (1) Consider, under this fault condition, that the fault-tolerant current in the gth phase can be defined as ig = \u221e\u2211 m=1 Im \u00b7 cos(m\u03c9t \u2212 \u03b8m). (2) Consider phase k, which is symmetrically located to phase g, with respect to the fault axis (phase 1)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure6.23-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure6.23-1.png",
        "caption": "Fig. 6.23 (a)World frame representation of a force acting on a point (b)World frame representation of a moment acting on a point",
        "texts": [
            " 206 6 Dynamic Simulation In this section, all equations of motion are represented in the world frame, thus the force and the moment must be written with respect to the world frame. For example, even for the gravity acting on the CoM of an object, we must consider its moment around the world frame origin. Although this seems complicated, we can correctly calculate the interaction between multi rigid bodies. The basic rules to convert the force and moment into the world frame are as follows. Rule 1: A force f acting on a point p is converted to a force f and a torque p\u00d7 f with respect to the world frame (Fig. 6.23(a)). Rule 2: A moment \u03c4 acting on a point p is the same \u03c4 with respect to the world frame (Fig. 6.23(b)). In general, when a force fp and a moment \u03c4 p are simultaneously acting on a point p, they are converted a force and moment fo, \u03c4 o around the origin given by [ fo \u03c4 o ] = [ fp p\u00d7 fp + \u03c4 p ] . (6.50) One might think them obvious, we tend to make a curious mistake for this sort of calculation. The quiz shown in Fig. 6.24 might help to understand this problem. 6.6 Appendix 207 6.6.2 Subroutines 210 6 Dynamic Simulation 1. Aldebaran robotics, http://www.aldebaran-robotics.com 2. Carnegie mellon university TARTAN RESCUE, http://www"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure2.11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure2.11-1.png",
        "caption": "Figure 2.11 A trestle is loaded by a (tensile) force of 30 kN that has to be resolved into components along the lines a and b.",
        "texts": [
            " VOLUME 1: EQUILIBRIUM 2 Statics of a Particle 29 The fact that Fa and Fb are both positive means that they act in the directions we chose in Figure 2.9b. If we had chosen the directions of Fa and Fb in the opposite sense, we would have found that Fa and Fb were negative. A force F can be resolved graphically into two components Fa and Fb, with given lines of action a en b using the parallelogram rule in Figure 2.10a or the force polygon in Figure 2.10b. The graphical approach has the advantage that you can at once see in which directions the components Fa and Fb are working. Example A force F = 30 kN acts on the trestle in Figure 2.11a. This has to be resolved into the components Fa and Fb with lines of action a and b. Solution: In Figure 2.12, the force has been resolved using a force polygon. The force scale is 1 cm =\u0302 5 kN. By measuring, you find (in the force polygon) Fa and Fb have lengths 5.7 cm and 6.3 cm respectively, so that Fa \u2248 5.7 \u00d7 5 kN = 28.5 kN, Fb \u2248 6.3 \u00d7 5 kN = 31.5 kN. In Figure 2.11b forces Fa and Fb are shown as they act on the trestle. It is clear that force Fa is pulling on the beam while force Fb is pushing against it. Together, they exert the same load as the single force F . The forces Fa and Fb are statically equivalent (equal in an equilibrium consideration) to their resultant F . 30 Check: The magnitude of Fa and Fb can also be calculated from the force triangle by using the sine rule (see Figure 2.12): Fa sin 55\u25e6 = Fb sin 65\u25e6 = F sin 60\u25e6 so that: Fa = F \u00b7 sin 55\u25e6 sin 60\u25e6 = 28"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure14.58-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure14.58-1.png",
        "caption": "Figure 14.58 (a) The cable replaced by a bar structure. The bar structure is kinematically indeterminate, but the equilibrium is stable (reliable). (b) If the bar structure is folded over, the signs change in all the bars. The equilibrium is now unreliable (unstable).",
        "texts": [
            " VOLUME 1: EQUILIBRIUM structural shapes in which the bending moments are as small as possible. Example 1 The beam in Figure 14.56a is subject to bending by the two forces F . The M diagram is shown in Figure 14.56b. In Figure 14.57, the same load is carried by a cable with compression bar. The cable and compression bar transfer normal forces only. The cable has the same shape as the M diagram in Figure 14.56b. With a cable sag the scale factor is H = F = F. H is the (compressive) force in the bar that is equal to the horizontal component of the (tensile) force in the cable. In Figure 14.58a the straight cable parts have been replaced by bars. Plus and minus signs indicate whether the bar forces are tensile or compressive. The structure can be considered a kind of arch under tension that is kept together by a compression bar. If the bar structure in Figure 14.58a is \u201cturned over\u201d with equal loads as shown in Figure 14.58b, all the signs in the bars change. The structure has now changed into an arch under compression with tension bar (tie rod). In the position shown in Figure 14.58b, the bar structure is in equilibrium. However, the equilibrium is unstable (unreliable): a small change in position will cause the equilibrium to fail and the bar structure will collapse.1 The bar structure is kinematically indeterminate. This collapse can be pre- 1 To prove this we have to investigate the equilibrium of the structure in its deformed state. However, this topic is beyond the scope of this book. Here we assume that the reader is acquainted with this phenomenon of instbility on the basis of some practical experience. 14 Cables, Lines of Force and Structural Shapes 681 vented by making the structure kinematically determinate, for example by introducing bracing members, and changing the bar structure into a truss (see Figures 14.59 and 14.60b). If we calculate the member forces for the given load, we find that all the interior members are zero members. The cable in Figure 14.57 and the bar structure in Figure 14.58a are also kinematically indeterminate. The equilibrium is stable (reliable) in this case as the load makes the structure go back into the original equilibrium position after a disruption. In Figure 14.60a, the bar structure in Figure 14.58a has been changed into a kinematically determinate truss. All the interior members are zero. As in contrast to the cable, the truss has the benefit that the shape does not change when the load changes. In Figure 14.61, the forces on the truss are shifted to the horizontal plane through the supports. The verticals are no longer zero members. These 682 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM trusses, in which all the members are subject to extension (or are zero members), can be an alternative for the beam subject to bending in Figure 14.56. Figure 14.62a shows the bar structure from Figure 14.58b, but now without tension member. This kinematically indeterminate structure can be made kinematically determinate not only by changing it into a truss, but also by replacing the hinged joints between the bars by rigid joints. The structure then becomes a bent member, recognisable in Figure 14.62b as a two-hinged frame. One problem is that the frame is statically indeterminate to the first degree. As such, it is not possible to determine the horizontal support reaction H directly from the equilibrium"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure8.13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure8.13-1.png",
        "caption": "Figure 8.13 The maximum shear stress \u03c4g;max is reached when the soil slides. One can distinguish two cases: (a) the soil element slides (moves downwards) and the shear stress in the slide plane PQ works upwards, and (b) the soil element is upset (moves upwards) and the shear stress in the slide plane PQ is directed downwards. The condition after sliding or upsetting is shown with a dashed line.",
        "texts": [
            " Suppose grain stresses \u03c3g and \u03c4g are acting on the oblique section PQ, and horizontal grain pressure \u03c3g;h is acting in the vertical section OP. Vertical grain pressure \u03c3g;v is acting on the horizontal section OQ. From the moment equilibrium in the plane of the drawing, about the middle of PQ, it follows that no shear stresses can be acting in the horizontal section OQ. Check it! If shear stress \u03c4g on the oblique section PQ has reached its maximum \u03c4g;max, the soil mass will slide. Here one can distinguish two situations, depending on the direction in which the soil slides, and therefore the direction of the shear stress \u03c4g;max. Figure 8.13 shows both cases; the situation after sliding is shown by a dashed line. The soil element in Figure 8.13a is sliding (moving downwards); the shear stress in the slide plane PQ is acting upwards. In Figure 8.13b, the soil element is upset (moving upwards) and the shear stress in the slide plane PQ is directed downwards. For both cases with given \u03c3g;v we can look for the angle \u03b1 for which the horizontal grain pressure \u03c3g;h is extreme. It is conventional to express the horizontal grain pressure in the vertical grain pressure using a coefficient K: \u03c3g;h = K\u03c3g;v. 294 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM The slide in Figure 8.13a leads to the so-called active earth pressure: \u03c3g;h = Ka\u03c3g;v. The active earth pressure is the largest horizontal pressure for which the soil yields sideways (with smaller pressures the soil will certainly yield). The upset in Figure 8.13b leads to the so-called passive earth pressure: \u03c3g;h = Kp\u03c3g;v. The passive earth pressure is the smallest horizontal pressures for which the soil is upset (with larger pressure, the soil will certainly be upset). The distinction between active and passive is derived from the way in which the soil mass acts on its surroundings: active when (part of) a structure yields under the influence of earth pressure, and passive when the soil offers resistance to the displacement of (part of) a structure. The active earth pressure on a wall has the same direction as the one in which the wall yields; the passive earth pressure acts opposite to the direction in which the wall moves"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000013_j.actamat.2016.07.019-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000013_j.actamat.2016.07.019-Figure1-1.png",
        "caption": "Fig. 1. Schematics of an LBM machine (left), and of the three process steps iterated during the build (right).",
        "texts": [
            " For this reason, a description emphasizing major characteristics and highlighting themost significant differences of these processes is put first. This is to support the understanding of formation of microstructure and development of mechanical properties of parts resulting from the LBM, EBM and LMD processes. LBM is a powder bed-based process in which metal powder is spread in thin layers with a typical layer thickness DS of 20 mme100 mm [19] across a work area with dimensions reaching from 50 mm 50 mm up to 800 mm 400 mm at present [14]. A sketch of a typical LBM system is presented in Fig. 1. The metal powder is either fed by a hopper or provided by a reservoir next to the work area [20]. For uniform distribution of the powder a levelling system or a recoater blade is used (Fig. 1, Step I. Powder Deposition). With the help of a galvanometer scanner a laser beamwith a power PL in-between 20 W [21] and 1 kW [14] is directed with a scan speed vs of up to 15 m/s [12] across the deposited powder layer (Fig. 1, Step II. Laser Exposure). Laser beam sources that are predominantly used in LBM are single mode fibre lasers in continuous wave mode that emit radiation with a wavelength of 1060 nme1080 nm in the near infrared. Typical spot sizes of the laser beam in the focal plane are between 50 mm [22] and 180 mm [23], depending on the manufacturing system used. According to the cross section of the part, the metal powder is selectively exposed to the laser beam in the x-y-plane. By exceeding the melting temperature of the material the powder is completely melted along the parts contour and filling (Fig. 1, B-B). Usually, the sequence of the individual melt tracks follows a pattern, the socalled scan strategy, whereby the melt tracks overlap with a certain hatch distance hs. The volume energy EV \u00bc PL=\u00f0vs*hs*DS\u00de (1) supplied to the powder layer causes not only the exposed material to melt but also reaches areas adjacent to the melt pool due to heat conduction. During solidification of the melt the individual melt tracks and the already solidified layer below are fused [24]. After selective exposure of the powder bed to the laser beam, the build plate is lowered (Fig. 1, Step III. Lower Build Plate), another powder layer is applied and the process of melting the newly deposited powder layer is repeated. These three steps are iterated until the part is completed. After completion unmelted metal powder can be sieved and reintroduced into a subsequent LBM process. The part itself is fixed to a build plate, quite often connected by so-called support structures. Support structures are lattice-like structures which are necessary for heat dissipation and fixation of the part in the powder bed, especially for supporting its horizontally oriented and overhanging surfaces, cf. Fig. 1, Detail A, and also cf. Fig. 20 (left). That way deformation of the part is prevented. The support structures need to be removed later on to finish the part. In addition to support structures, pre-heating of the build plate can be used to avoid distortion of parts by lowering thermal gradients, resulting in a reduction of residual stresses which evolve during the LBM process [27]. For LBM fabrication of Ti-6Al-4V parts, typical pre-heating temperatures are 200 C [25] up to 500 C [26]. The LBM process is carried out in a closed process chamber in which an inert gas atmosphere is continuously maintained so that the residual oxygen content is less than 0",
            " the energy density, the thickness of the layer and the pre-heating temperature, if applied [96]. In addition, the temperature gradients are also affected by the surrounding material, e.g. different heat conduction of powder and solidified material adjacent to the melt pool in powder bed-based processes. Thus, microstructures of components fabricated by AM are also influenced by the geometry of the part, and may vary within a manufactured part, especially between bulk material and surface areas [35]. As heat conduction in build direction (z in Fig. 1) is typically higher than in the other spatial directions (x, y) as a result of the solidified material from lower, previously built layers, anisotropy in both microstructure and properties has been observed in selected cases [2,35,107,113]. Under certain conditions and for several materials, an epitaxial grain growth in build direction has been reported for e.g. LBM of Ti-6Al4V [25] and Ta [113], EBM [36] and LMD [2] (both Ti-6Al-4V), yielding grain sizes in build direction that exceed layer thickness"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000637_978-3-642-83006-8-Figure3.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000637_978-3-642-83006-8-Figure3.2-1.png",
        "caption": "Fig. 3.2. Spatial biped model with massless legs (after [20])",
        "texts": [
            " The results of this analysis definitely prove that for large-scale active spatial mechanisms, a li nearized model is rather far from the actual system. Thus the control synthesis based only on the linearized model equations, i.e., neglec ting the nonlinear effects, cannot satisfy the requirements of speed and efficiency of stabilization of the system. All the above references are related only to planar models. Gubina[20] considered a spatially placed massive body supported on two massless legs of variable length (Fig. 3.2). The body of massmt , has main iner tia moments J x ' J y ' Jz' The variable length of legs replaces the knee function, and the body positing in space is achieved by leg extension (it is supposed that the \"knee\" can generate force F). The body orien- 189 tation is maintained by three torques P1' P2 and P3 applied at the mechanism's hip which rotate the body in the frontal and sagittal pla ne, as well as around the vertical body axis (self-rotation). Only single-support phase is considered. A spherical coordinate frame has been chosen because it is closest to that used by humans"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-FigureC.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-FigureC.3-1.png",
        "caption": "FIGURE C.3 Partial self-inductance for round wire section.",
        "texts": [
            "13) Note that e represents the end of the object and s is used for the start. Equation (C.8) can also be used as an approximation to the round wire partial inductance especially if the wire diameter is small compared to the spacing. Of course, other cross section can be approximated efficiently with several filaments and numerical integration in the cross-section direction. C.1.2 Lp11 for Round Wire We give two formulas for the partial self-inductance for round shapes. The popular equation for the partial self-inductance for the section of round wire shown in Fig. C.3 is PARTIAL INDUCTANCE FORMULAS FOR ORTHOGONAL GEOMETRIES 387 Lp11 = 1 2 1 \ud835\udf070 4\ud835\udf0b \u222b xe1 xs1 \u222b1 \u222b x\u2032 e1 x\u2032 s1 \u222b\u2032 1 1 R1,1\u2032 d\u2032 1 dx\u20321 d1 dx1, (C.14) where R1,1\u2032 is the usual distance between the source and observation points. 1 and \u2032 1 are the cross sections. The first formula is a simple approximation of a wire partial self- inductance used in Refs [1, 2], which is based on the assumption that the above filament formula can be used to represent a round wire by placing one of the filaments at the center of the wire, and the second one on the surface of the wire at distance a = d\u22152 away. The approximate result for this case is Lp11 \u2243 \ud835\udf070 2\ud835\udf0b \ud835\udcc1 \u23a1\u23a2\u23a2\u23a3ln \u239b\u239c\u239c\u239d\ud835\udcc1a + \u221a( \ud835\udcc1 a )2 + 1) \u239e\u239f\u239f\u23a0 \u2212 \u221a 1 + ( a \ud835\udcc1 )2 + a \ud835\udcc1 \u23a4\u23a5\u23a5\u23a6 , (C.15) where \ud835\udcc1 = (xe \u2212 xs) is the length of the segment. We could view this approximation technique as the most simple case of numerical integration based on the analytical result for the longitudinal direction. A second analytical formula is for the partial inductance of a zero thickness cylindrical tube with a radius a = d\u22152 shown in Fig. C.3. The integrals to evaluate are Lp11 = \ud835\udf070 4\ud835\udf0b2 \u222b \ud835\udf0b \ud835\udf19=0 \u222b \ud835\udcc1 x=0 \u222b \ud835\udcc1 x\u2032=0 dx\u2032 dx d\ud835\udf19\u221a 4a2sin2(\ud835\udf19\u22152) + (x \u2212 x\u2032)2 . (C.16) by recognizing that symmetry can be used to reduce the fourfold integral to the threefold integral. We were able to analytically solve the integrals for the case of interest where the length \ud835\udcc1 of the wire is longer than the diameter d = 2 a [8]. The result for the tube conductor is given by 388 COMPUTATION OF PARTIAL INDUCTANCES Lp11 = \ud835\udf070\ud835\udcc1 4 [( k2 480 + k4 1280 + 1 3600 ) \ud835\udf0b 3 + ( 1 18 \u2212 k2 24 ) \ud835\udf0b + ( 2 \u2212 2 log(\ud835\udcc1) + 6 log(2) + 2 log(a) \u2212 4 log(k\ud835\udf0b) ) 1 \ud835\udf0b + 8 a \ud835\udcc1 \ud835\udf0b2 ] , (C"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.22-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.22-1.png",
        "caption": "FIGURE 7.22. A two-axle vehicle with a trailer is steered according to the Ackerman condition.",
        "texts": [
            "65) \u03b4 = cot\u22121 \u00b5 cot \u03b4o + cot \u03b4i 2 \u00b6 = 0.19684 rad \u2248 11.278 deg (7.66) R = q a22 + l2 cot2 \u03b4 = 520.46 in \u2248 13.219m (7.67) \u03b8 = 2 tan\u22121 \u2219 1 b1 \u2212 b2 \u00b5 Rt \u00b1 q R2t \u2212 b21 + b22 \u00b6\u00b8 = \u00bd \u22123.0132 rad \u2248 \u2212172.64 deg 0.22121 rad \u2248 12.674 deg (7.68) Example 271 F Space requirement. The kinematic steering condition can be used to calculate the space requirement of a vehicle with a trailer during a turn. Consider that the front wheels of a two-axle vehicle with a trailer are steered according to the Ackerman geometry, as shown in Figure 7.22. The outer point of the front of the vehicle will run on the maximum radius RMax, whereas a point on the inner side of the wheel at the trailer\u2019s rear 402 7. Steering Dynamics axle will run on the minimum radius Rmin. The maximum radius RMax is RMax = r\u00b3 R1 + wv 2 \u00b42 + (l + g)2 (7.69) where R1 = q (Rmin + wt) 2 + b22 \u2212 b21. (7.70) and the width of the vehicle is shown by wv. The required space for turning the vehicle and trailer is a ring with a width 4R, which is a function of the vehicle and trailer geometry"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000801_j.finel.2014.04.003-Figure12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000801_j.finel.2014.04.003-Figure12-1.png",
        "caption": "Fig. 12. Wall build using quiet elements (1C, case wq2).",
        "texts": [
            " Then, layer by layer elements are switched to quiet, and then switched individually to active based on the heat input similarly to the quiet element method implemented in cases wq1 and wq2. In this approach, equation numbering and solver initialization is repeated only when each layer is activated resulting into slightly faster computer run times of about 976 s wall (6284 s using 1 core) per case. In cases wq2, wi2, wiq2, and wiq3, the temperature of quiet or inactive elements before activation n 1T is set to the initial temperature 0T , while in cases wq1, wi1, wiq1, it is not. The temperature result during the deposition of the 61st layer for case wq2 is illustrated in Fig. 12. Fig. 13 illustrates the temperature result for case wq2 at the same time instance with all quiet element removed. The temperature result during the deposition of the 61st layer for case wi2 is illustrated in Fig. 14. Temperature results along line AA are extracted and plotted in Fig. 15. Results from cases wq2 and wi3 are also plotted in the same figure. Using case wiq2 as reference, the maximum % error along the line is listed in Table 5. As seen in Table 5, neglecting to reset n 1T upon element activation leads to errors (3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure6.8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure6.8-1.png",
        "caption": "Fig. 6.8. In-wheel motor, PM ring and stator winding of a coreless AFPM brushless motor for solar-powered vehicles. Courtesy of CSIRO , Lindfield, NSW, Australia.",
        "texts": [
            " CSIRO Telecommunications and Industrial Physics, Lindfield, NSW, Australia manufactures in-wheel coreless AFPM brushless motors for solar powered vehicles [225]. Solar-powered vehicles compete in the World Solar Challenge solar car race organized every two years in Australia between Darwin and Adelaide which has become a well-known international event. CSIRO provides in-wheel AFPM brushless motors both with surface PMs glued to steel discs and with PMs arranged in Halbach array. The motor structure, PM disc and stator winding are shown in Fig. 6.8 [225]. Fig. 6.9 shows a coreless brushless motor with film coil stator winding and twin PM rotor manufactured by EmBest , Soeul, South Korea. The coreles stator has a foil winding at both sides. The 8-pole PM rotor is designed as a twin external rotor. A low speed, three-phase, Y-connected AFPM brushless motor rated at 10 kW, 750 rpm, 28.5 A has been designed and investigated. The rotor does not have any ferromagnetic core and consists of trapezoidal coils embedded in a high mechanical integrity resin"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001514_j.engfailanal.2015.06.004-Figure26-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001514_j.engfailanal.2015.06.004-Figure26-1.png",
        "caption": "Fig. 26. (a) 12 DOF model [70], (b) 12 DOF model [71].",
        "texts": [
            " [28] adopted a 6 DOF model of a gear system to analyze the effects of crack on the system vibration response [67] (see Fig. 25a). To study the influence of tooth friction, Bartelmus [67] presented an 8 DOF model (see Fig. 25b) where two lateral degrees of freedom (DOFs) are added into the 6 DOF model. The same models can also be found in Refs. [68,69]. Considering the effects of local tooth damage, changing load and speed, Chaari et al. [70] presented a two stage gearbox model with 12 DOFs (see Fig. 26a). Mohammed et al. [71] developed a 12 DOF gear dynamic model including the gyroscopic and friction effects (see Fig. 26b). In order to investigate dynamic characteristics of the system when the crack in the pinion grows, Zhou et al. [46] adopted a 16 DOF mathematic model, which is developed by Howard et al. [50] for a one-stage gear system. In the model, the gear engagement is described by a TVMS function k(t) and a viscous damping c(t) (see Fig. 27a). Using a lumped parameter technique, Endo et al. [59] established a 16 DOF LMM (Fig. 27b) of a gear test rig to reproduce fault signals. Jia and Howard [51] presented a 26 DOF mathematic model including two pairs of meshing gears (see Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001620_j.addma.2017.11.001-Figure11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001620_j.addma.2017.11.001-Figure11-1.png",
        "caption": "Fig. 11. Rotated",
        "texts": [
            " This is necessary to reduce the temperature gradient and is achieved by first reducing the scan line length and then the difference in length between the two adjacent filling lines. In the splitting mode, the multi-outline section is divided into a smaller number of separated sub-inner areas perpendicular to the filling lines. Each sub-inner area is then filled with scanning lines, but the length of the lines should be as short as possible. In order to obtain short scanning lines, the rotated scanning mode is applied. Fig. 11 (rectangular section) demonstrates the application of the splitting mode for a 12 M. Galati, L. Iuliano / Additive Manufacturing 19 (2018) 1\u201320 scann r s s t c i T t r A e o l t a r T t t u w l s s w t ectangular section [89]. For = 90\u25e6, the first scan line is the longer ide of the rectangle, and for = 0\u25e6 the first scan line is the shorter ide of the rectangle. A rotation angle, , between 0\u25e6 and 90\u25e6, leads o a shorter first scan line than those for = 0 or = 90 and, as a onsequence, the probability of first line scan balling or instabilty is decreased while the quality of the first scan line is increased"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002366_j.jmatprotec.2016.08.006-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002366_j.jmatprotec.2016.08.006-Figure2-1.png",
        "caption": "Fig. 2. STL images for the support structures. (a) A 16-unit \u2018IY\u2019 support structure supporting a thin plate. (b) A 16-unit \u2018Y\u2019-support structure supporting a thin plate. This model illustrates the unequal distribution of overhang-support contact points with l4 > l3. (c) An array of Pin supports (red) supporting a cuboid where d is the separation space between the pins. (For interpretation of the references to colour in this figure legend, the reader is referred to the web version of this article.)",
        "texts": [
            " In the first group, the support structures were built to support a cuboid, while the support structures in the second group were built to support a thin plate of similar cross-section. Figs. 2(a) and (b) show the STereoLithography (STL) illustrations of a 16-unit \u2018IY\u2019 and \u2018Y\u2019 support structure model \u2013 each supporting a thin square plate respectively. The number of support structures were varied at 16, 25, and 36 in each group. The \u2018IY\u2019 support structures are placed such that the vertical struts are distributed equally within the supported geometry\u2019s area (Fig. 2(a)). As for the \u2018Y\u2019 design, the support structures are distributed within the supported geometry\u2019s area, in addition to the equally spaced vertical struts in contact with the substrate plate. Referring to Fig. 2(b), having four inclined struts branching out from each vertical strut, a single \u2018Y\u2019 support structure has four times more overhang-support contact points than a single \u2018IY\u2019 support structure. In addition, due to the arrangement described above, the inclined struts of the \u2018Y\u2019 support structures intersects, leading to unequal overhangsupport contact spacing (l3 > l4). Furthermore, the centre row from point indicated y has twice the contact area as compared to other points. On the contrary, the equally spaced vertical struts produce equal overhang-support contact spacing, measured centre to centre (l3 = l4), for the \u2018IY\u2019 support structure. Fig. 2(c) illustrates the Pin support structure, generated on overhangs of 5 mm width, supporting a cuboid. The centre to centre separation d of the pin support structure generated, as indicated in Fig. 2(c), was varied at 0.2, 0.4 and 0.6 mm (hereafter referred to as d0.2, d0.4, and d0.6 respectively). The height of the pins ph and overhangs oh are 4 mm and 6 mm respectively. The thickness ot and downskin length dl of an overhang is 1 mm and 4 mm respectively. In addition, the supported area was segmented into nine equal areas to facilitate support structure removal. Furthermore, maintaining minimal contacts between the overhangs and the build plate allow users to simply chisel built parts off eliminating the need for time-consuming wire cutting process"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003690_1.1802311-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003690_1.1802311-Figure1-1.png",
        "caption": "Fig. 1 Spindle-bearing system",
        "texts": [
            " The stiffness of the bearings, contact forces on bearing balls, natural frequencies, time history response, and frequency response functions can be obtained by applying the cutting forces to the spindle for a given preload, including the rotating effects of the spindle on both bearing stiffness and the dynamics of the spindle system. The model is experimentally verified with Frequency Response Test measurements conducted on an instrumented research spindle. 004 by ASME NOVEMBER 2004, Vol. 126 \u00d5 1089 13 Terms of Use: http://asme.org/terms Downloaded F 2 Finite Element Model of Spindle-Bearing Systems Figure 1 shows the spindle designed and built by a spindle manufacturer. The spindle has a standard CAT 40 toolholder interface, and is designed to operate at up to 15,000 rev/min with a 15 kW motor connected to the shaft with a pulley-belt system. The finite element model of the spindle-bearing system is shown in Fig. 2. The Timoshenko beam is used to model the spindle shaft and housing. The black dots represent nodes, where each node has three translational and two rotational degrees of freedom. The pulley is modeled as a rigid disk",
            " The axial displacements relative to the spindle housing at the spindle nose are measured by the axial displacement sensor integrated to the spindle and laser displacement sensor. Deformations of the Spindle Nose. As mentioned in Sec. 2, the whole spindle is self-balanced in the axial direction under the preload. The displacement in the axial direction is tested by simply fixing the spindle housing on a table so that it cannot move under a preload. The preload force is applied by hydraulic unit as shown in Fig. 1. The spindle already has an initial preload applied during assembly, and it is estimated as follows. First, the axial deformations of the spindle nose are measured under different preloads. Then, the different hydraulic preloads plus the same estimated initial preload are applied to the FE model to obtain the simulated displacements. By trying to match the two curves from both measurement and simulation, the initial preload is predicted as 300 N. Figure 9 shows the axial deformations of the spindle nose under different preloads"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure14.60-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure14.60-1.png",
        "caption": "Figure 14.60 The bar structures in Figure 14.58 changed into trusses.",
        "texts": [
            " 14 Cables, Lines of Force and Structural Shapes 681 vented by making the structure kinematically determinate, for example by introducing bracing members, and changing the bar structure into a truss (see Figures 14.59 and 14.60b). If we calculate the member forces for the given load, we find that all the interior members are zero members. The cable in Figure 14.57 and the bar structure in Figure 14.58a are also kinematically indeterminate. The equilibrium is stable (reliable) in this case as the load makes the structure go back into the original equilibrium position after a disruption. In Figure 14.60a, the bar structure in Figure 14.58a has been changed into a kinematically determinate truss. All the interior members are zero. As in contrast to the cable, the truss has the benefit that the shape does not change when the load changes. In Figure 14.61, the forces on the truss are shifted to the horizontal plane through the supports. The verticals are no longer zero members. These 682 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM trusses, in which all the members are subject to extension (or are zero members), can be an alternative for the beam subject to bending in Figure 14"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure9.46-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure9.46-1.png",
        "caption": "Figure 9.46 Section for determining the forces in members 9 and 12. They are found from the moment equilibrium about respectively A and B.",
        "texts": [
            " Four unknown member forces are acting in the section. Since the lines of action of the forces N5, N6 and N7 intersect one another at point A, only one force is unknown in the equation for the moment equilibrium about A, which can be determined directly. This gives \u2211 Tz|A = \u2212(60 kN)(2 m) \u2212 N9 \u00d7 (3 m) = 0 \u2192 N9 = \u221240 kN. 9 Trusses 347 Member 9 is a compression member. The same equation is found from the moment equilibrium about A of the part to the left of the section over members 7, 8, 9 and 12 (see Figure 9.46). This section offers the advantage that the forces in the members of both the top chord and the bottom chord can be found. In this way, force N12 in member 12 follows directly from the moment equilibrium about B: \u2211 Tz|B = \u2212(240 kN)(3 m) \u2212 (60 kN)(2 m) + N12 \u00d7 (3 m) = 0 \u21d2 N12 = +280 kN. A tensile force is acting in member 12. The section in Figure 9.46 has the additional benefit that the values found for N9 and N12 can be checked using the horizontal force equilibrium of the isolated part, without having to know the forces in the diagonal members or verticals: \u2211 Fx = \u2212(240 kN) + N9 + N12 = \u2212 (240 kN) + (\u221240 kN) + (280 kN) = 0. The isolated section in Figure 9.46 therefore meets the conditions for horizontal force equilibrium. With the section in Figure 9.46, we can quickly determine the forces in the top chord member 9 and bottom chord member 12, but not the forces in the verticals 7 and 8. The forces in these verticals are found from the equilibrium of joints A and B, but we do not have enough information to do so yet. 348 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM The forces in the diagonal members 10 and 11 are found using the section in Figure 9.47 over members 9, 10, 11 and 12. Since we already know N9, we can find N10 from the moment equilibrium about C"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002683_j.jmatprotec.2020.117023-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002683_j.jmatprotec.2020.117023-Figure6-1.png",
        "caption": "Fig. 6. Schematic of the formation mechanism during the hybrid DED and thermal milling process of 316 L SS components: (a) DED process and (b) milling process.",
        "texts": [
            " However, the densification level began to decrease when v continued to increase; when v reached 12 mm/s, the densification level (\u03c1rel = 93.11 %) decreased remarkably, and many pores were observed in the microstructure compared with the number of pores observed for a v of 8 mm/s. This was due to the discontinuous scan tracks and the resulting interlaminar pores at a high v. The experimental results show that the large and irregular pores were transformed into very small pores, thereby increasing the density. When the laser beam was focused onto the previous layer, the distribution of the laser heat source was Gaussian (seen in Fig. 6(a)). Thus, a large temperature gradient was produced. Moreover, the interaction between the laser beam and metal powder determined the wetting of the liquid phase, which had a substantial influence on the bonding ability between the layers (Gu et al., 2019). A lower v caused a high laser energy input and could lead to the high temperature of the liquid phase, which led to a high thermo-capillary (Marangoni) flow, which can be calculated as follows (Yadollahi et al., 2015): Ma = \u2202\u03c3 \u2202T r2 m[T \u2212 Tm] \u03bc\u03b1 (1) where rm represents the radius of the melt pool, \u2202\u03c3/ \u2202T is the surface tension coefficient, and \u03bc and \u03b1 are the dynamic viscosity and thermal diffusivity, respectively",
            " The stronger liquid metal flow caused the energy exchange rate between the centre and edge of the molten pool to accelerate. Therefore, a higher densification degree was obtained. With increasing v, the flow behaviour in the molten pool also changed. When v was 12 mm/s, the circulating current that was formed in the molten pool was weak, thus giving rise to a weak reflux at the centre of the edge. Therefore, the pool depth away from the centre became shallow, resulting in defects, including pores, discontinuities, and unmelted powder particles, which reduced the density. Fig. 6(b) illustrates the impact of the subtractive thermal milling process on the layer morphology and defect distribution. When the DED process was completed, the unfused powders adhered to the surface, and well-defined scanning paths emerged on the surface due to the wettability of the molten pool, as illustrated by Gu et al. (2012). Additionally, there were still pores in the sample because the gas contained in the argon and powder was intercepted during the solidification process, as indicated by Hu et al"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure4.14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure4.14-1.png",
        "caption": "Fig. 4.14: Basic movements for gripper fingers: a) swinging motion; b) translation parallel motion.",
        "texts": [],
        "surrounding_texts": [
            "Two-finger grippers are widely used because two-finger grasp can be used in most cases. Most of the objects that are used in human or industrial manipulations have a size that is comparable with the size of the human hand or gripping devices and they have a socalled \u2018regular geometry\u2019, which describes the shape of basic easily graspable figures. Figure 4.15 summarizes feasible mechanical designs for two-finger grippers. In general, fingers can be designed as articulated fingers to mimic the human hand, as shown in Fig. 4.15. b) and c), or as coupler links to use planar linkages as gripping mechanisms, as shown in Fig. 4.15. a). A further peculiarity can be recognized in flexible tips for the contacts, which are installed on fingertips and finger phalanges with a suitable curvature in order to be easily adapted to the shape of specific objects, as schematically shown in Fig. 4.15 by small grey tips. These tips can be of flexible material to help the shape adaptation and mainly to enlarge the contact surface so that the contact pressure can be limited to avoid surface Fundamentals of the Mechanics of Robotic Manipulation 255 damage of the grasped object. Indeed, the need for a suitable low value of contact pressure is a requirement to adopt two, three, four, and even more contacts, although static equilibrium for grasping can be ensured with two contacts only in planar cases. Since the main purpose of a two-finger gripper can be recognized as performing a planar grasp, some fundamental design considerations may be deduced by approaching specifically a two-finger grasp and its mechanics. The fundamental characteristics of a two-finger grasp can be modeled as in Fig. 4.16, which has been obtained to describe all the phases of a grasping action as: in Fig. 4.16a) one finger impacts the object and starts the grasping while the fingers are moving to close with a suitable approaching motion; in Fig. 4.16b) the finger pushes the object against the other finger while the motion of the fingers continues the closing; in Fig. 4.16c) the second final impact of the fingers to the object concludes the approaching and pushing motion; in Fig. 4.16d) the object is grasped and a static equilibrium of the object between the fingers is ensured; in Fig. 4.16e) dynamic force may change the equilibrium but still the object is statically grasped by the fingers; in Fig. 4.16a, f) an external disturbance, such as an impact with another object in the Chapter 4 Fundamentals of the Mechanics of Grasp256 environment, may change the equilibrium and the object may move to another static equilibrium configuration between the fingers. The above-mentioned analysis can be outlined similarly for any other case of grasping devices, even with multiple contacts, such as those in Fig. 4.15. The analysis has been focused on the interactions among object and fingers through a purely mechanical viewpoint, but the feasibility of finger configurations strongly depends on the mechatronic gripper design and operation that guide the finger motion and action. The analysis of Fig. 4.16 refers to planar grasp but it can also be extended to threedimensional cases by looking similarly at the phases in planes of a Cartesian frame reference for a three-dimensional grasp."
        ]
    },
    {
        "image_filename": "designv10_0_0001447_j.isatra.2016.09.019-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001447_j.isatra.2016.09.019-Figure1-1.png",
        "caption": "Fig. 1. Quadrotor UAV.",
        "texts": [
            " In addition, it is noted that the coefficients of the defined sliding mode manifolds (more detail seen from the literature [9]) were calculated based on the condition of Hurwitz stability. Therefore, the finite-time stabilization of the quadrotor can be ensured through global fast dynamic TSMC. The organization of this work is arranged as follows. In Section 2, the dynamical model of a small quadrotor UAV is presented. The flight control method is detailedly introduced in Section 3. In Section 4, the simulation results and analysis are presented. The conclusion is given in Section 5. 2. Quadrotor model The dynamical model of the given quadrotor UAV is presented in Fig. 1, the more detail are seen from the literature [8], where the state vector [x, y, z]' denotes the position of the center of the gravity of the quadrotor and the vector [u, v, w]' denotes its linear velocity in the body-frame, the vector [p, q, r]' denotes its angle velocity in the body-frame. m denotes the total mass. g represents the accelerant of gravity. l denotes the distance from the center of each rotor the center of gravity. The dynamical model of the quadrotor UAV can be described by the following equations [8,9] \u20acx\u00bc 1 m\u00f0 cos \u03d5 sin \u03b8 cos \u03c8\u00fe sin \u03d5 sin \u03c8 \u00deu1 K1 _x m \u20acy\u00bc 1 m\u00f0 cos \u03d5 sin \u03b8 sin \u03c8 sin \u03d5 cos \u03c8 \u00deu1 K2 _y m \u20acz\u00bc 1 m\u00f0 cos \u03d5 cos \u03b8\u00deu1 g K3 _z m \u20ac\u03d5\u00bc _\u03b8 _\u03c8 Iy Iz Ix \u00fe Jr Ix _\u03b8\u03a9r\u00fe l Ix u2 K4l Ix _\u03d5 \u20ac\u03b8\u00bc _\u03c8 _\u03d5Iz Ix Iy Jr Iy _\u03d5\u03a9r\u00fe l Iy u3 K5l Iy _\u03b8 \u20ac\u03c8 \u00bc _\u03d5 _\u03b8Ix Iy Iz \u00fe 1 Iz u4 K6 Iz _\u03c8 8>>>>>>< >>>>>>: \u00f01\u00de where the three Euler angles [\u03d5; \u03b8; \u03c8 ]' denotes the roll, the pitch and the yaw, respectively, and \u03d5A\u00f0 \u03c0=2;\u03c0=2\u00de, \u03b8A\u00f0 \u03c0=2;\u03c0=2\u00de, \u03c8A \u00f0 \u03c0=2;\u03c0=2\u00de, those mean that the different acrobatic flying is Please cite this article as: Xiong J-J, Zhang G-B"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001514_j.engfailanal.2015.06.004-Figure20-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001514_j.engfailanal.2015.06.004-Figure20-1.png",
        "caption": "Fig. 20. (a) Crack schematic, (b) FE model of cracked tooth in Ref. [62].",
        "texts": [
            " [56] calculated the TVMS of a spur gear pair with a tooth root crack by applying a constant load along the tooth width, and analyzed the influence of the crack sizes and crack propagation directions along the tooth root on the TVMS. Based on the static analysis of MSC.MARC software and Loaded Tooth Contact Analysis (LTCA) [57], Endo et al. [58\u201360] established the FE models of gears with spalling and tooth crack in which the elastic deformations of the engaging teeth and the gear body as well as the Hertzian deformation are considered. Combining the global deformation caused by the tooth bending with the local contact deformation, Del Rincon et al. [61,62] presented a model to calculate meshing forces (see Fig. 20), and deduced some quantities such as the mesh stiffness, loaded transmission error and load sharing ratio. Their method, which avoids establishing a FE model with contact elements, is more computationally efficient. Moreover, their method can obtain more accurate meshing forces and has a higher calculation accuracy. Based on the conventional photoelasticity technique (see Fig. 21), Pandya and Parey [63] measured the mesh stiffness of a gear pair, and calculated the mesh stiffness changes of a pinion under different crack depths"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000801_j.finel.2014.04.003-Figure13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000801_j.finel.2014.04.003-Figure13-1.png",
        "caption": "Fig. 13. Wall build not showing quiet elements (1C, case wq2).",
        "texts": [
            " In this approach, equation numbering and solver initialization is repeated only when each layer is activated resulting into slightly faster computer run times of about 976 s wall (6284 s using 1 core) per case. In cases wq2, wi2, wiq2, and wiq3, the temperature of quiet or inactive elements before activation n 1T is set to the initial temperature 0T , while in cases wq1, wi1, wiq1, it is not. The temperature result during the deposition of the 61st layer for case wq2 is illustrated in Fig. 12. Fig. 13 illustrates the temperature result for case wq2 at the same time instance with all quiet element removed. The temperature result during the deposition of the 61st layer for case wi2 is illustrated in Fig. 14. Temperature results along line AA are extracted and plotted in Fig. 15. Results from cases wq2 and wi3 are also plotted in the same figure. Using case wiq2 as reference, the maximum % error along the line is listed in Table 5. As seen in Table 5, neglecting to reset n 1T upon element activation leads to errors (3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002278_tfuzz.2017.2774185-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002278_tfuzz.2017.2774185-Figure1-1.png",
        "caption": "Fig. 1. Electromechanical system.",
        "texts": [
            " However, if the nonlinear functions system (1) is of strict-feedback structure, the result of [41] can be obtained as a particular case. IV. SIMULATION RESULTS The effectiveness of the developed technique is demonstrated in the following technical example. The simulation results confirm the necessity of considering the prediction error between a fuzzy observer model and a serial-parallel estimation model, which improves the accuracy of estimation error between nonlinear functions and FLSs. Consider the following electromechanical system [51] shown in Fig. 1: \u0393q\u0308 + \u039eq\u0307 + \u03a0 sin(q) = I, F I\u0307 = V\u03b5 \u2212RI \u2212Kq\u0307. Let x1 = q, x2 = q\u0307, x3 = I, u = V\u03b5. Then the above equations can be rewritten as x\u03071 = x2, x\u03072 = x3/\u0393\u2212\u03a0 sin(x1)/\u0393\u2212 \u039ex2/\u0393, x\u03073 = (g1u1 + g2u2)/F \u2212 x2K/F \u2212 x3R/F, y = x1, where \u0393 = J Kt + mF 2 0 3Kt + \u03930F 2 0 Kt + 2\u03930R 2 0 5Kt , \u039e = \u039e0 Kt , \u03a0 = mF0G 2Kt + \u03930F0G Kt , m stands for the mass of the link, R0 is the load radius, \u03930 is the mass of the load, J represents the rotor inertia, F 0 means the length of the link, G represents the coefficient of the gravity, I(t) is the motor armature current, Kt is the coefficient of the electromechanical conversion between armature current and torque, R corresponds to the armature resistance, q(t) represents the position of the angular motor, V\u03b5 means the input control voltage, \u039e0 stands for the viscous friction coefficient of the joint, F stands for the armature inductance and K represents the back-emf coefficient"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003403_978-1-4615-5633-6-Figure7.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003403_978-1-4615-5633-6-Figure7.2-1.png",
        "caption": "Figure 7.2. Voltage and Current waveforms in TCR",
        "texts": [
            " Thyristor Controlled Reactor [Miller (1982), Gyugyi(1988)) The current in a TCR can be continuously varied from zero( corresponding to zero conduction angle) to maximum(corresponding to conduction angle of 1800 )by phase control in which the firing angle a( with respect to the zero crossing of the voltage) is varied from 1800 to 90 0 \u2022 The instantaneous current iTCR over half a cycle is given by =1;(cosa-coswt), a<wt<a+(7} = 0, a + (7 < wt < a + 7r iTCR (7.1 ) where V is the rms voltage applied, XL is the fundamental frequency reactance ofthe reactor. The current waveform in a single phase TCR is shown in Fig.7.2. The current is non-sinusoidal and contains odd harmonics which are func tions of the conduction angle (7. INTERACTIONS WITH SHUNT COMPENSATORS 173 The fundamental component of the TCR current h, is given by (7.2) where B (T - sm (T TCR = rrXL (7.3) n = 6k \u00b1 1 (7.4) where k is an integer. The phase and line current waveforms in a delta connected TCR are shown in Fig. 7.3 for 3 different values of a. It is assumed that the TCR currents are identical in waveform and magnitude in the three phases and only phase shifted from each other by 1200 \u2022 This is possible only if a is the same in all three phases"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure4.16-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure4.16-1.png",
        "caption": "Figure 4.16 (a) The roller support has (b) two degrees of freedom (a rotation and a movement along the rolling surface) and generates (c) one support reaction (a force perpendicular to the roller track).",
        "texts": [
            "14 the degrees of freedom and support reactions are shown. Sometimes, motions are depicted by means of open arrows, while forces are depicted by closed arrows. Figure 4.15 is a schematic representation of a roller support. For a roller support at A, the body can move parallel to the so-called roller track, and can also rotate freely about A. Only motion of A perpendicular to the roller track is prevented; this is the direction in which the interaction force is exerted. For the roller support in Figure 4.16, with the roller track parallel to the x axis, the following applies for the motion at A: ux;A = unknown (free motion), uy;A = 0 (prescribed motion), \u03d5z;A = unknown (free motion). ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM 4 Structures 123 The following applies for the forces in A in the coordinate system given: Fx;A = 0 (prescribed force), Fy;A = unknown (free force), Tz;A = 0 (prescribed force). The roller support therefore has two degrees of freedom and generates one support reaction. Note the parallel with a bar support"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000206_978-0-387-74315-8-Figure6-1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000206_978-0-387-74315-8-Figure6-1-1.png",
        "caption": "Figure 6-1:",
        "texts": [
            " The calculation of articulated-body inertias is described in thenext section, so here we shall concentrate on the forward dynamics problem. We could use one of the 92 mcthods of Chapter 5 to solvc this problem, but the one-joint case IS so simple that it is easily solved directly. One-joint robot with moving base Let a one-joint robot have a base moving with velocity vb and . . . acceleration a b' a link with inertia 1, a joint with axis s, and let the joint velocity, acceleration and force be q, q and Q respectively (see Figure 6-1). acceleration of the link are given by (6.2) and al = a b + vb X s q + s q , (6.3) and the equation of motion of the link is ... ... ......... f=Ial+'l XI'l' (6.4) where fis the net force applied to the link (ignoring gravity). fis applied via the joint, so its component on the joint axis is Q, \u00ce.e. 93 (6.5) Substituting Equations (6.3 and 6.4) into Equation (6.5), we obtain an expression for q in terms of Q: \u00b78 .\u2022 \u2022 .\u2022\u2022 Q = s (1 a, + 'l Xl 'l ) so q- (6.6) Equation (6.6) is the equation of motion for a one-joint robot with a moving base"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure2.11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure2.11-1.png",
        "caption": "FIGURE 2.11. A car moving on an inclined road and pulling a trailer.",
        "texts": [
            "144) The effect of changing the position of the mass center on the maximum achievable braking deceleration is shown in Figure 2.10 for a sample car with \u03bcx = 1 h = 0.56m (2.145) l = 2.6m. 2. Forward Vehicle Dynamics 59 Passenger cars are usually in the range 0.4 < (a1/l) < 0.6. In this range, (afwb/g) < (arwb/g) and therefore, front-wheel-brake cars can reach better forward deceleration than rear-wheel-brake cars. Hence, front brakes are much more important than the rear brakes. Example 60 F A car with a trailer. Figure 2.11 depicts a car moving on an inclined road and pulling a trailer. To analyze the car-trailer motion, we need to separate the car and trailer to see the forces at the hinge, as shown in Figure 2.12. We assume the mass center of the trailer Ct is at distance b3 in front of the only axle of the trailer. If Ct is behind the trailer axle, then b3 should be negative in the following equations. For an ideal hinge between a car and a trailer moving in a straight path, there must be a horizontal force Fxt and a vertical force Fzt ",
            "163) Fzt = h1 \u2212 h2 b2 \u2212 b3 mt (a+ g sin\u03c6) + b3 b2 \u2212 b3 mt g cos\u03c6 (2.164) 62 2. Forward Vehicle Dynamics Fz1 = b3 2l \u00b5 2a2 \u2212 b1 b2 \u2212 b3 mt + a2 b3 m \u00b6 g cos\u03c6 (2.165) + 1 2l \u2219 2a2 \u2212 b1 b2 \u2212 b3 (h1 \u2212 h2)mt \u2212 h1mt \u2212 hm \u00b8 (a+ g sin\u03c6) Fz2 = b3 2l \u00b5 a1 \u2212 a2 + b1 b2 \u2212 b3 mt + a1 b3 m \u00b6 g cos\u03c6 (2.166) + 1 2l \u2219 a1 \u2212 a2 + b1 b2 \u2212 b3 (h1 \u2212 h2)mt + h1mt + hm \u00b8 (a+ g sin\u03c6) Fz3 = 1 2 mt b2 \u2212 b3 (b2 g cos\u03c6+ (h1 \u2212 h2) (a+ g sin\u03c6)) (2.167) l = a1 + a2. Example 61 F Maximum inclination angle for a car with a trailer. For a car and trailer as shown in Figure 2.11, the maximum inclination angle \u03c6M is the angle at which the car cannot accelerate the vehicle. Substituting a = 0 and \u03c6 = \u03c6M in Equation (2.155) shows that sin\u03c6M = 2 (m+mt) g (Fx1 + Fx2) . (2.168) The value of maximum inclination angle \u03c6M increases by decreasing the total weight of the vehicle and trailer (m+mt) g or increasing the traction force Fx1 + Fx2 . The traction force is limited by the maximum torque on the drive wheel and the friction under the drive tire. Let\u2019s assume the vehicle is four-wheeldrive and friction coefficients at the front and rear tires are equal"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure4.4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure4.4-1.png",
        "caption": "Fig. 4.4: Examples of end-effectors with suction-cups: a) a scheme for a multi suctioncup end-effector; b) a palm end-effector with several suction-cups.",
        "texts": [
            " 4.2b) an application is reported in which a multi-fingered end-effector grasps an electronic card with several expanded fingers. The vacuum systems are widely used with cups that can be properly shaped and sized as a function of the grasped objects that must have a suitable smooth surface. In Fig. 4.3 examples of suction-cups are shown as those that are used in robot pneumatic end-effectors. Examples of pneumatic vacuum end-effectors are shown for the case of a multi suction-cup end-effector in Fig. 4.4a) and for a palm end-effector with several suction-cups in Fig. 4.4b). Chapter 4 Fundamentals of the Mechanics of Grasp244 The electrostatic and magnetic systems can be used for objects with ferromagnetic properties so that they can be grasped through the action of electromagnetic forces. In Fig. 4.5 a scheme for a typical electromagnetic end-effector is shown together with a slider-crank mechanism, which is useful for smooth releasing. However, all grasping devices should fulfil the following requirements, mainly for installation on robots: low-cost, robust design and operation, simplicity in the mechanical design and operation, and easy integration"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002151_j.engfailanal.2014.11.015-Figure15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002151_j.engfailanal.2014.11.015-Figure15-1.png",
        "caption": "Fig. 15. 3.9 mm manually made tooth crack in the sun gear.",
        "texts": [
            " Thus, the rotation speed of the sun gear of the second stage planetary gearbox is 46.66 r/min which is the same as the simulated sun gear speed in Section 2.3. To simulate the crack effect on the vibration signals, tooth crack was manually created in the sun gear tooth using the electro discharge machining. When the crack length is small, fault symptoms may submerge in the noise and hard to distinguish. To amplify the fault symptoms, 3.9 mm tooth crack was created in the sun gear tooth. The crack goes through the whole tooth width as shown in Fig. 15. Fig. 16 shows the experimental acceleration signals of the second stage planetary gearbox in the perfect condition. The symbol ay denotes the vertical direction acceleration of the gearbox. Amplitude modulation presents in the perfect condition of the gearbox and signal envelope fluctuates four times in one revolution of the carrier. Fig. 17 describes the experimental acceleration signals of the second stage planetary gearbox both in the perfect condition and in the cracked tooth condition. In the cracked tooth condition, some even bigger spikes show up"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001256_j.ijmachtools.2021.103729-Figure69-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001256_j.ijmachtools.2021.103729-Figure69-1.png",
        "caption": "Fig. 69. Schematic showing tests on two non-standard specimens. (a) Small Punch test. (b) Two Bar Specimen test. This illustrates the potential of small and non-conventional specimen testing in Powder Bed Fusion.",
        "texts": [
            " The former can normally provide information on creep fracture. For example, Brodin et al. observed the creep fracture surface of LPBF Hastelloy X tested at 815 \u25e6C with a tensile load [384]. Compressive creep tests are performed to analyse specimens creep rate. The tests were usually interrupted when a predetermined plastic deformation was achieved. Some non-standard creep specimens, such as 2 bar specimens are also used [75]. Small punch creep specimens are another commonly used specimen geometry, these tests are shown schematically in Fig. 69. Wang et al. used a small punch creep test to compare the creep performance of forged, cast and LPBF material [411]. The results (Fig. 70) indicated that the AB LPBF specimens possessed similar creep lifetime to the forged ones, but far lower than the cast specimen. Laves phase at the grain boundaries may have led to poor creep performance in LPBF specimens. Moreover, the local tensile stress induced by the experimental loading in the fracture region was believed to be another reason for the reduced time to rupture"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure3.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure3.1-1.png",
        "caption": "Fig. 3.1 Definition of Zero-Moment Point (ZMP) [90]",
        "texts": [
            " 1 Definition of ZMP In 1972, Vukobratovic\u0301 and Stepanenko defined the Zero-Moment Point (ZMP) at the beginning of the paper on control of humanoid robots1. Everything of the argument regarding the ZMP starts from here. 1 We can find the same definition in the book [81]. Later, some delicate aspects of the ZMP definition were discussed by Vukobratovic\u0301 and Borovac [88]. S. Kajita et al., Introduction to Humanoid Robotics, 69 Springer Tracts in Advanced Robotics 101, DOI: 10.1007/978-3-642-54536-8_3, c\u00a9 Springer-Verlag Berlin Heidelberg 2014 70 3 ZMP and Dynamics In Fig. 3.1 an example of force distribution across the foot is given. As the load has the same sign all over the surface, it can be reduced to the resultant force R, the point of attack of which will be in the boundaries of the foot. Let the point on the surface of the foot, where the resultant R passed, be denoted as the zero-moment point, or ZMP in short. 2 ZMP and Support Polygon We explain the support polygon which is another important concept related to the ZMP. As shown in Fig. 3.2, let us consider the region formed by enclosing all the contact points between the robot and the ground by using an elastic cord braid",
            "3(a), when a human stands on the ground, the ZMP coincides with the ground projection of CoM. In such a case, a human can keep balance if the ground projection of CoM is included strictly inside of the support polygon. On the other hand, when a human moves dynamically as shown in Fig. 3.3(b), the ground projection of CoM may exist outside the support polygon. However, the ZMP never exists outside the support polygon. In the following, we will explain the reason why the ZMP is always included in the support polygon. 1 ZMP in 2D In Fig. 3.1, although only the vertical component of the ground reaction force is shown, the horizontal component of it also exists due to the friction between the ground and the soles of the feet. In Fig. 3.4(a) and (b), we separately show the vertical component\u03c1(\u03be) and the horizontal component \u03c3(\u03be) of the ground reaction force per unit length of the sole. These forces simultaneously act on a humanoid robot. 72 3 ZMP and Dynamics Let us replace the forces distributed over the sole by a equivalent force and moment at a certain point in the sole"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure14.13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure14.13-1.png",
        "caption": "FIGURE 14.13. A double A-arm suspension mechanism at (a)- equilibrium, (b)upper limit, and (c)- lower limit.",
        "texts": [
            " On the contrary, decreasing \u03be or fn decreases the value of \u03c92nSX\u0308/SZ , which is equivalent to making the suspension softer. Softening of a suspension decreases the body acceleration, however it requires a large room for relative displacement. Due to physical constraints the wheel travel is limited and hence, we must design the suspension to use the available suspension travel as much as possible, and decrease the body acceleration as low as possible. Mathematically it is equivalent to (14.79) and (14.80). Example 505 Wheel travel calculation. Figure 14.13(a) illustrates a double A-arm suspension mechanism at its equilibrium position. To limit the motion of the wheel with respect to the body, two stoppers must be employed. There are many possible options for the type and position of stoppers. Most stoppers are made of stiff rubber balls and mounted somewhere on the body or suspension mechanism or both. It is also possible that the damper acts as a stopper. Figure 14.13(a) shows an example. The gap sizes \u03b4u and \u03b4l indicate the upper and lower distances that a mechanism can move. However, the maximum motion of the wheel must be calculated at the center of the wheel. So, we transfer \u03b4u and \u03b4l to the center of the wheel and show them by du and dl. du \u2248 bu au \u03b4u (14.108) dl \u2248 bl al \u03b4l (14.109) Figure 14.13(b) and (c) show the mechanism at the upper and lower limits respectively. The distance du is called the upper wheel travel, and dl is called the lower wheel travel. The upper wheel travel is important in ride comfort and the lower wheel travel is important for safety. To have better ride comfort, the upper wheel travel should be as high as possible to make the suspension as soft as possible. Although the upper and lower wheel travels may be different, for practical purposes, we may assume dl = du and design the suspension based on a unique wheel travel"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.4-1.png",
        "caption": "FIGURE 8.4. (a) Adding an anti-tramp bar to guide a solid axle. (b) Equivalent kinematic model.",
        "texts": [
            " These forces tend to push the springs into an S-shaped profile, as shown in Figure 8.3. The driving and braking flexibility of leaf springs, generates a negative caster and increases instability. Long springs provide better ride. However, long sprigs exaggerate their bending and twisting under different load conditions. Example 301 Leaf spring suspension and flexibility solution. To reduce the effect of a horizontal force and S-shaped profile appearance in a solid axle with leaf springs, the axle may be attached to the chassis by a longitudinal bar as Figure 8.4(a) shows. Such a bar is called an anti-tramp 8. Suspension Mechanisms 457 bar, and the suspension is the simplest cure for longitudinal problems of a Hotchkiss drive. A solid axle with an anti-tramp bar may be kinematically approximated by a four-bar linkage, as shown in Figure 8.4(b). Although an anti-tramp bar may control the shape of the leaf spring, it introduces a twisting angle problem when the axle is moving up and down, as shown in Figure 8.5. Twisting the axle and the wheel about the axle is called caster. The solid axle is frequently used to help keeping the wheels perpendicular to the road. Example 302 Leaf spring location problem. The front wheels need room to steer left and right. Therefore, leaf springs cannot be attached close to the wheel hubs, and must be placed closer to the middle of the axle"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure13.40-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure13.40-1.png",
        "caption": "Figure 13.40 The components of the snow load qsn on a member segment with length a measured horizontally.",
        "texts": [
            " Superposing the N diagrams in Figures 13.39a and 13.39b gives the N diagram in Figure 13.39c. This is the requested N diagram due to the dead weight qdw = 5 kN/m. Figure 13.39 (a) The N diagram due to the component normal to the roof plane, superposed on (b) the N diagram due to the component parallel to the roof plane gives (c) the requested N diagram due to the dead weight. 13 Calculating M, V and N Diagrams 581 c. Snow load Over a length a measured horizontally, the resultant of the snow load is aqsn (see Figure 13.40). The components of this force normal to and parallel to the axis are respectively aqsn cos \u03b1 and aqsn sin \u03b1. They act on a member segment with length a/cos \u03b1. For the components of the distributed load normal to and parallel to the beam axis we now find qsn;tr = aqsn cos \u03b1 a/ cos\u03b1 = qsn cos2 \u03b1, = (5 kN/m) cos2 21.8\u25e6 = 4.310 kN/m, qsn;pa = aqsn sin \u03b1 a/ cos\u03b1 = qsn sin \u03b1 cos \u03b1, = (5 kN/m) sin 21.8\u25e6 cos 21.8\u25e6 = 1.724 kN/m. The distributed load qsn = 5 kN/m due to the snow (Figure 13.41a) has components of 4"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002351_s00170-017-0426-7-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002351_s00170-017-0426-7-Figure7-1.png",
        "caption": "Fig. 7 Thermocouple locations: 1\u20134 upper inner group (blue), 5\u2013 10 lower outer group (green), and 11\u201316 upper outer group (red). Note: thermocouples 1\u20134 are not present in the conventional insert due to tool access constraints during subtractive manufacturing",
        "texts": [
            " The insert base surface was initially milled to form a datum, after which the insert was removed from the SLM build plate with electro discharge machining (EDM). Subsequently, the insert was assembled with the moving mould half plate and aligned with dowels prior to machining of the outer insert surface (Fig. 6(iii)). Both manufactured inserts were also incorporated with thermocouple channels for temperature measurement; however, due to tool access restrictions, the conventional insert included fewer channels (Fig. 7). An experimental and numerical study was conducted for both the conventionally and conformally cooled mould design, in order to assess the prediction capabilities of Autodesk Moldflow Insight 2016 commercial IM modelling software, in estimating the transient mould and part temperatures, as well as the expected part warpage after ejection. The experimental study was conducted over a range of moulding materials, mould temperatures, and coolant parameters (Table 8). An Arburg Allrounder 520S injection moulding machine was used to mould all parts",
            " Higher temperature variation in the part indicates cooling non-uniformity which can cause defects such as part warping. The results show that due to the more uniform distribution of cooling channels, the conformal cooling insert (Fig. 8(i)) exhibits higher uniformity in temperature compared to the standard insert (Fig. 8(ii)) with a difference in peak temperatures of 10 \u00b0C. The results highlight the performance benefits of conformal cooling. Similarly, Fig. 9 shows numerically predicted and experimentally measured temperatures at the mould thermocouple locations (Fig. 7) throughout the cycle for case studies C1.1 (conformal insert) and B1.1 (conventional insert). The results show several characteristics: & Inner thermocouples (1\u20134) showed the lowest increase in temperatures, attributable to close proximity of surrounding cooling channels and the associated high rate of cooling. & Outer thermocouples (11\u201316) showed higher overall temperatures due to: relatively close proximity to the high temperature injection point (top centre of part face (Fig. 3(vii)), and less direct conductive cooling pathways due to being located in the insert corner where the mould cavity side cooling line proximity is large (Fig",
            " 10(i)) and conformally cooled mould inserts (test case C1.1, Fig. 10(ii)). For the conventionally cooled insert, the numerical maximum temperature predictions were closely approximated, with deviations between the experimental values generally below 2 \u00b0C. For the conformal insert, the maximum temperatures showed a greater deviation that increased with the polymer flow path distance, that is the deviation was within approximately 1 \u00b0C for the top thermocouples closest to the injection point (1\u20134, Fig. 7), while the upper (11\u201316) and lower (5\u201310) peripheral sensor locations were under-predicted by \u223c3 and \u223c6 \u00b0C, respectively. The lower level of prediction accuracy for the conformal insert, as well as the increasing under prediction with polymer flow path distance, may be attributable to the following: & potential differences in the thermal conductivity of the conventional and SLM-manufactured H13 material thermal properties assumed in the simulation and & wall boundary heat transfer coefficient differences introduced by the high surface roughness (Rz \u2248 80 \u2212 100 \u03bcm, Section 3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.6-1.png",
        "caption": "FIGURE 8.6. A solid axle suspension with a triangulated linkage.",
        "texts": [
            " The wheels remain upright and hence, do not roll on a side. However, a solid axle shifts laterally from its static plane and its center does not remain on the vehicle\u2019s longitudinal axis under a lateral force. 458 8. Suspension Mechanisms A solid axle produces bump-camber when single-wheel bump occurs. If the right wheel goes over a bump, the axle is raised at its right end, and that tilts the left wheel hub, putting the left wheel at a camber angle for the duration of deflection. Example 303 Triangular linkage. A triangulated linkage, as shown in Figure 8.6, may be attached to a solid axle to provide lateral and twist resistance during acceleration and braking. Example 304 Panhard arm. High spring rate is a problem of leaf springs. Reducing their stiffness by narrowing them and using fewer leaves, reduces the lateral stiffness and increases the directional stability of the suspension significantly. A Panhard arm is a bar that attaches a solid axle suspension to the chassis laterally. Figure 8.7 illustrates a solid axle and a Panhard arm to guide the axle"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002546_j.isatra.2018.12.046-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002546_j.isatra.2018.12.046-Figure2-1.png",
        "caption": "Fig. 2. Configuration of a two-link robotic manipulator.",
        "texts": [
            " It yields that v\u0302i is increasing, which induces larger oscillation. On the other hand, if \u03b5 is too large, \u2225\u03c3\u2225 is evolving around \u03b5, and it follows that controller accuracy is not as good as possible. Considering all this, \u03b5 should rather be larger than smaller. In this section, the simulations are conducted for demonstrating the effectiveness of the proposed control scheme and compared with the existing controllers in [22,23]. The designed adaptive SOFNTSM controller is applied for the trajectory tracking of a two-link rigid robotic manipulator shown in Fig. 2. The dynamic of robotic manipulator in (1) and (2) by Lagrangian equation is represented as M(q) = \u23a1\u23a3 (m1 + m2)l21 + m2l22 +2m2l1l2 cos(q2) + J1 m2l22 + m2l1l2 cos(q2) m2l22 + m2l1l2 cos(q2) m2l22 + J2 \u23a4\u23a6 , C(q, q\u0307) = [ \u2212m2l1l2 sin(q2)q\u03072 \u2212m2l1l2 sin(q2)(q\u03071 + q\u03072) m2l1l2 sin(q2)q\u03071 0 ] , G(q) = [ (m1 + m2)l1g cos(q1) + m2l2g cos(q1 + q2) m2l2g cos(q1 + q2) ] , (22) where q1(t) and q2(t) are the angular positions of revolute joints 1 and 2, \u03c4 = [\u03c41, \u03c42]T is the input torque. Moreover, the two-link robotic manipulator has four inner states x1(t) = q1(t), x2(t) = q\u03071(t), x3(t) = q2(t), x4(t) = q\u03072(t), two outputs y1(t) = q1(t), y2(t) = q2(t) and two inputs u1(t) = \u03c41(t), u2(t) = \u03c42(t)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure11.4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure11.4-1.png",
        "caption": "FIGURE 11.4. Top view of a car and the forces system acting at the tireprints.",
        "texts": [
            "100) C\u03b2i = d\u03b2i dp (11.101) 11. F Vehicle Roll Dynamics 677 Therefore, \u03b2f = tan\u22121 \u00b5 vy + a1 r \u2212 C\u03b2f p vx \u00b6 (11.102) \u03b2r = tan\u22121 \u00b5 vy \u2212 a2 r \u2212 C\u03b2rp vx \u00b6 . (11.103) Assuming small angles for slip angles \u03b2f , \u03b2, and \u03b2r, the tire sideslip angles for the front and rear wheels, \u03b1f and \u03b1r, may be approximated as \u03b1f = 1 vx (vy + a1r \u2212 zfp)\u2212 \u03b4 \u2212 \u03b4\u03d5f = \u03b2 + a1 r vx \u2212 C\u03b2f p vx \u2212 \u03b4 \u2212 C\u03b4\u03d5f\u03d5 (11.104) \u03b1r = 1 vx (vy \u2212 a2r \u2212 zrp)\u2212 \u03b4\u03d5r = \u03b2 \u2212 a2 r vx \u2212 C\u03b2r p vx \u2212 C\u03b4\u03d5r\u03d5. (11.105) 11.3.3 F Body Force Components on a Two-wheel Model Figure 11.4 illustrates a top view of a car and the force systems acting at the tireprints of a front-wheel-steering four-wheel vehicle. When we consider the roll motion of the vehicle, the xy-plane does not remain parallel to the road\u2019s XY -plane, however, we may still use a two-wheel model for the vehicle. Figure 11.5 illustrates the force system and Figure 11.6 illustrates the kinematics of a two-wheel model for a vehicle with roll and yaw rotations. The rolling two-wheel model is also called the bicycle model"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure9.12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure9.12-1.png",
        "caption": "Figure 9.12 From the expected deformation due to a full load, we can deduce that (a) falling diagonals will extend and be subject to tension, and that (b) rising diagonals will shorten and be subject to compression.",
        "texts": [
            " If this is not the case, as in truss (e), it is called a halfparabolic truss beam. Truss (g) is found in large spans. By creating an auxiliary truss within the main truss, additional points of support are created for the bridge deck, allowing the structure to be lighter. Trusses (a) and (d) to (g) have falling diagonals, truss (b) has rising diagonals, and in truss (c) the diagonals alternate between falling and rising. Due to the dead load, falling diagonals are loaded by tensile forces, and rising diagonals are loaded by compressive forces. This is shown in Figure 9.12 in a general sketch of the expected deformation of the truss beam subject to full loading. The falling diagonals in case (a) extend and are loaded by tensile forces. The rising diagonals in case (b) shorten and are loaded by compressive forces. 9 Trusses 327 In steel trusses, falling diagonals (tension diagonals) are used most frequently, as (usually slender) steel members subject to compression run the risk of buckling. Preferably apply them as tension members. In contrast, rising diagonals (compression diagonals) are most often used in wooden trusses, as in general a wooden joint is more suitable to transfer compressive forces rather than tensile forces"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure4.19-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure4.19-1.png",
        "caption": "Figure 4.19 (a) The hinged support has (b) one degree of freedom (a rotation) and generates (c) two support reactions (the two components of a force).",
        "texts": [
            " Generally speaking, the weight of a structure, such as a bridge, is sufficiently large to ensure that the roller support is continuously loaded by compressive forces. If a roller support also has to be able to transfer tensile forces, special structural provisions have to be made. It is assumed here that a roller support can transfer both tensile and compressive forces. 124 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM A hinged support is a hinge between the structure and its immovable environment (see Section 4.2.1). A hinged support is modelled in Figure 4.19a (the open circle is often omitted). In a hinged support at A, the displacement of the body at A is prevented and the body can only rotate about A. The support cannot transfer a couple, but can transfer a force. The interaction force is unknown with respect to both magnitude and direction. For the hinged support in Figure 4.19 with the coordinate system shown, the following applies for motion at A: ux;A = 0 (prescribed motion), uy;A = 0 (prescribed motion), \u03d5z;A = unknown (free motion). and for the forces at A: Fx;A = unknown (free force), Fy;A = unknown (free force), Tz;A = 0 (prescribed force). A hinged support therefore has one degree of freedom (a rotation) and generates two support reactions (the two components of a force). 4 Structures 125 Figure 4.20 is a good example of a hinged support. Figure 4.21 shows how a steel hinged support can be used in small bridges"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003619_bf02665211-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003619_bf02665211-Figure4-1.png",
        "caption": "Fig. 4 - - T h e attenuation of the laser power owing to the powder particles.",
        "texts": [
            " (1) Before the laser beam reaches the workpiece, it passes through the gas-powder jet which \"attenuates\" part of the laser power. Denoting Pa, as the attenuated power, the amount of the laser-beam power directly absorbed by the workpiece is The t e r m eat/Pl corresponds to the ratio of the projected area of the powder particles in the laser beam to the area of the beam. This ratio may be determined by simple geometrical considerations, assuming that the laser beam and the powder stream are two intersecting cylinders and that shadowing of one powder particle by another is negligible (Figure 4). This later assumption is reasonable as the volume fraction of powder in the incoming jet is usually smaller than 1 pct. This attenuation term is given by Pat__ = mp if rjet < rz [3(a)] P, 2prtrpvp cos (0jet) rhp if rj~ >>- rt [3(b)] 20rjo,rpvp cos (0je,) (2) The powder particles absorb part of the attenuated power and surrender it to the workpiece only if they METALLURGICAL AND MATERIALS TRANSACTIONS B VOLUME 25B, APRIL 1994--283 strike and enter into the molten pool. Let/3p be the absorption of the gas-powder jet and r/p the powder efficiency"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure6.30-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure6.30-1.png",
        "caption": "FIGURE 6.30. A slender bar AB sliding at points A and C.",
        "texts": [
            " Point I is called the roll center of the wheel and body. 6. Applied Mechanisms 351 Example 245 The instant centers of rotation may not be stationary. When a mechanism moves, the instant centers of rotation may move, if they are not at a fixed joint with the ground. Figure 6.29 illustrates a four-bar linkage at a few different positions and shows the instant centers of rotation for the coupler with respect to the ground I13. Point I13 will move when the linkage moves, and traces a path shown in the figure. Example 246 Sliding a slender on the wall. Figure 6.30 illustrates a slender bar AB sliding at points A and C. We have the velocity axis of two points A and C, and therefore, we can find the instant center of rotation I. The coordinates of point I are a function of the parameter \u03b8 as follows: xI = h cot \u03b8 (6.219) yI = h+ xI cot \u03b8 = h \u00a1 1 + cot2 \u03b8 \u00a2 . (6.220) Eliminating \u03b8 between x and y, generates the path of motion for I. yI = h \u00b5 1 + x2I h2 \u00b6 (6.221) Example 247 F Plane motion of a rigid body. The plane motion of a rigid body is such that all points of the body move only in parallel planes"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure6.22-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure6.22-1.png",
        "caption": "FIGURE 6.22. Two configurations of an inverted slider crank mechanism for \u03b82 = 45deg.",
        "texts": [
            " For example, consider an inverted slider-crank mechanism at \u03b82 = \u03c0/4 rad = 45 deg with the lengths a = 1 e = 0.5 d = 3. (6.192) The parameters of Equation (6.172) are equal to G = d\u2212 e\u2212 a cos \u03b82 = 1.792 9 H = 2a sin \u03b82 = 1.414 2 I = a cos \u03b82 \u2212 d\u2212 e = \u22122.792 9 (6.193) Now, Equation (6.183) gives two real values for \u03b84 \u03b84 \u2248 \u00bd 1.48 rad \u2248 84.8 deg \u22122.08 rad \u2248 \u2212120 deg . (6.194) 6. Applied Mechanisms 343 Using the known values and for \u03b84 = 1.48 rad, Equations (6.188) will provide b \u2248 2.33. (6.195) When \u03b84 = \u22121.732 rad we get b = \u22122.28. (6.196) Figure 6.22 depicts the two configurations of the mechanism for \u03b82 = 45deg. Example 239 Velocity analysis of an inverted slider-crank mechanism. The velocity analysis of a slider-crank mechanism can be found by taking a time derivative of Equations (6.178) and (6.179), d dt (a sin \u03b82 + b cos \u03b84 \u2212 e sin \u03b84) = a\u03c92 cos \u03b82 \u2212 b \u03c94 sin \u03b84 + b\u0307 cos \u03b84 \u2212 e\u03c94 cos \u03b84 = 0 (6.197) d dt (a cos \u03b82 + b cos \u03b84 \u2212 e cos \u03b84 \u2212 d) = \u2212a\u03c92 sin \u03b82 \u2212 b \u03c94 sin \u03b84 + b\u0307 cos \u03b84 + e\u03c94 sin \u03b84 = 0 (6.198) where \u03c92 = \u03b8\u03072 \u03c94 = \u03c93 = \u03b8\u03074. (6"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure3.35-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure3.35-1.png",
        "caption": "FIGURE 3.35. Front view of a laterally deflected tire.",
        "texts": [
            " C\u03b1 = lim \u03b1\u21920 \u2202 (\u2212Fy) \u2202\u03b1 = \u00af\u0304\u0304\u0304 lim \u03b1\u21920 \u2202Fy \u2202\u03b1 \u00af\u0304\u0304\u0304 (3.132) The lateral force Fy is at a distance ax\u03b1 behind the centerline of the tireprint and makes a moment Mz called aligning moment. Mz = Mz k\u0302 (3.133) Mz = Fy ax\u03b1 (3.134) For small \u03b1, the aligning moment Mz tends to turn the tire about the z-axis and make the x-axis align with the velocity vector v. The aligning moment always tends to reduce \u03b1. Proof. When a wheel is under a constant load Fz and then a lateral force is applied on the rim, the tire will deflect laterally as shown in Figure 3.35. The tire acts as a linear spring under small lateral forces Fy = ky\u2206y (3.135) with a lateral stiffness ky. The wheel will start sliding laterally when the lateral force reaches a maximum value FyM . At this point, the lateral force approximately remains constant and is proportional to the vertical load FyM = \u03bcy Fz (3.136) 3. Tire Dynamics 137 where, \u03bcy is the tire friction coefficient in the y-direction. A bottom view of the tireprint of a laterally deflected tire is shown in Figure 3.36. If the laterally deflected tire is turning forward on the road, the tireprint will also flex longitudinally"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000242_j.jmatprotec.2017.05.042-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000242_j.jmatprotec.2017.05.042-Figure1-1.png",
        "caption": "Figure 1: Chessboard SLM scanning strategy",
        "texts": [
            " The powder grains were spherical in shape with sizes in the range 15-45 \u00b5m. Samples were produced by SLM using a SISMA MYSINT100 system with a fiber laser of wavelength 1030 nm, power up to 150 W and spot diameter of 50 \u03bcm. All samples were built in an Ar environment with a residual oxygen content of 0.5 %. Component orientation, process parameters and scanning strategy were designed by the MARCAM AutoFab software. A chessboard hatching style was utilized for the bulk volume, with each layer divided into 3x3 mm2 square blocks (Fig.1a), each of which was labeled either black or white as per a real chessboard. Blocks labeled black were melted first, followed by those labeled white, after which the laser beam contoured the layer along its perimeter twice. Each block was scanned with parallel tracks (Fig. 1a). In order to obtain a low temperature gradient in the bulk volume, each layer (n+1) was rotated 30\u00b0 with respect to the previous layer (n) as depicted in Fig.1b. A preliminary study was performed to investigate the process parameter window that enabled fabrication of components with a relative density of no less than 98 %. Samples for microstructural characterization, density measurements and mechanical tests were produced based on these results, 3 for each process parameter given in Table 2. Fatigue samples were fabricated using the process parameters given in Table 3, which had produced the best trade-off between ultimate tensile strength (UTS) and elongation to failure (E%)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000029_j.actamat.2010.02.004-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000029_j.actamat.2010.02.004-Figure1-1.png",
        "caption": "Fig. 1. A typical SLM machine layout [2].",
        "texts": [
            " Published by Elsevier Ltd. All doi:10.1016/j.actamat.2010.02.004 * Corresponding author. Tel.: +32 16 321286; fax: +32 16 321270. E-mail address: lore.thijs@mtm.kuleuven.be (L. Thijs). and can be reused. To counteract curling of the material due to the build-up of thermal stresses during the SLM process, the part is built on a solid substrate. Due to the high reactivity of Ti alloys, the process needs to be conducted under an inert argon atmosphere. A typical SLM machine layout is presented in Fig. 1 [2]. Compared to conventional manufacturing techniques, SLM offers a wide range of advantages, namely a lower time-to-market, a near-net-shape production without the need of expensive moulds, a high material utilization rate, direct production based on a CAD model, and a high level of flexibility (e.g. products with different geometry can be produced in the same batch). Moreover, due to the additive and layer-wise production, the SLM process is capable of producing complex geometrical features that cannot be obtained using conventional production routes"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure9.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure9.2-1.png",
        "caption": "FIGURE 9.2 Section of ground plane with layers for current flow.",
        "texts": [
            " We can use a 1D EXP model for current penetration or skin depth \ud835\udeff given by Ramo et al. [22] \ud835\udeff = 1\u221a \ud835\udf0bf\ud835\udf07\ud835\udf0e , (9.1) where f is the frequency, \ud835\udf0e is the conductivity of the conductor and \ud835\udf07 is the permeability, which is usually \ud835\udf070. It is evident from this that at high frequencies, the skin depth restricts the current to a well-defined path close to the surface. We assume that a conductor or ground plane is, in this case, thicker than the skin depth \ud835\udeff. Hence, we can assume that the two sides of the plane are isolated from each other. As shown in Fig. 9.2, we assume that we cut out a surface cell of a size \u0394x and \u0394y where at a given angular frequency \ud835\udf14, the equivalent current penetration thickness is \ud835\udeff. This results in an equivalent impedance of the cell of Zs = Rs + j\ud835\udf14Ls, with Rs = \u0394x \ud835\udf0e\ud835\udeff\u0394y , Ls = \u0394x \ud835\udf14\ud835\udf0e\ud835\udeff\u0394y . (9.2) ONE DIMENSIONAL CURRENT FLOW TECHNIQUES 217 Further, due to the small diffusion distance \ud835\udeff into the conductor, we also can assume that the lateral loss coupling between two neighboring conductor surface cells is small. As a consequence, all the skin-effect loss couplings to other surface cells can be ignored",
            " At least for thin conductors, the approach can be less costly than the VFI model, since many couplings are ignored. The solution of the current diffusion problem inside the conductor is fundamentally different from the external PEEC propagation model. For the thin conductor problem, the skin-effect model must represent the EXP current penetration on opposite surfaces. Hence, the mesh cells on both side surfaces need to be lined up and can be coupled. This corresponds to top and bottom cells in Fig. 9.2. Due to the assumed small conductor thickness, the coupling to other surface cells including the neighboring cells is ignored. This leads to a very sparse internal impedance coupling matrix. For broadband applications, the skin-depth changes widely over the frequencies of interest. Hence, the impact of the skin effect on the impedance is large for thin conductor areas. The next task is to provide an EM and finally a circuit model for the diffusion of current inside the thin conductor. Especially for the time domain, we want to keep the order or the number of poles or circuit elements of this skin-effect model as low as possible",
            " The 218 SKIN EFFECT MODELING fundamental idea is based on the derivation of an appropriate equivalent circuit for the EXP function based on [1]. This skin-effect model was designed for one-dimensional current dependence for a single surface. However, for PEEC we want a more general model that includes both sides of the thin conductor. Fortunately, the approach can be applied in the context of partial inductances [32], while the original model [1] was designed for a 2D TL model with a 1D current flow. We approximate the model by breaking the thickness into layers as shown in Fig. 9.2. As we consider below, we do not use uniform layers. This leads to an equivalent circuit model in the z-direction as is indicated by the dashed lines. This is the dominant skin-effect direction for the thin layer structures. The Maxwell equations for the interior of the conductor are \u2207 \u00d7 E = \u2212\ud835\udf07\ud835\udf15H \ud835\udf15t (9.3) and \u2207 \u00d7 H = J = \ud835\udf0eE. (9.4) We observe that for the conductor, a simplified solution is that the following fields are nonzero: E = x\u0302Ex and H = y\u0302Hy. Therefore, this reduces (9.3) to \ud835\udf15Ex \ud835\udf15z = \u2212\ud835\udf07 \ud835\udf15Hy \ud835\udf15t . (9.5) Further, (9.4) will also reduce to a simple form \ud835\udf15Hy \ud835\udf15z = \u2212\ud835\udf0eEx, (9.6) since in the derivation we discretize the plane thickness in the z-direction only as shown in Fig. 9.2. Of course, we are also using the usual subdivision of the conductor into PEEC cells for the currents in the x- and y-directions shown in Fig. 6.2. We start with z1 = 0 and we label the thickness of the layers as k = 1, \u2026 ,Nk where Nk is the number of layers. Hence, the top layer ends at zNk = d. Also, the thickness of the layers is given by \u0394z = zk+1 \u2212 zk. To derive an equivalent circuit, we start with (9.5) and we multiply it by the cell length \u0394x. The voltage drop along the layer k in the x-direction is given by \u0394xEx,k = Vx,k, which leads to Vx,k+1 \u2212 Vx,k = \u2212\ud835\udf07\u0394x\u0394z \ud835\udf15Hy,k \ud835\udf15t ",
            " It was shown in Ref. [45] that large inductance variations can occur for nonstraight conductors. In fact, in this example low-frequency inductance was shown to be almost 10 times larger than the high-frequency inductance. Nonuniform meshing is a key factor in the efficiency of the VFI model. The large impact is evident from Sections 9.2.3 and 9.2.6. For true 3D models, it is important to use the nonuniform meshes in all three space dimensions such that all outside layers are thin. In fact, the same rules as in Fig. 9.2 can be applied for the choice of the outside layer thickness being a fraction of the skin depth \ud835\udeff at the highest frequency of interest. The layer thickness should be increased by a factor of 1.1 to 3 depending on the specifics as suggested in Section 8.4.1. With such an increase, it is clear that only a few layers are needed with such a substantial thickness increase for most geometries. This results in large cells in the interior of the conductor. Without any speedup technique, only medium-sized problems can be solved on a computer with moderate memory",
            " [11], that for some geometries, a set of template current cell can be precomputed so that some of the computational work is done before the actual computation of the current. We give an example of the meshing and equivalent circuit for a 3D VFI model for a metal body with a finite thickness. To give details on the meshing, we show the conventional horizontal x\u2013y PEEC cells for a particular level in Fig. 9.21. For most conventional cases, the cells at all levels must be vertically aligned as is shown in Fig. 9.2. The vertical layer-to-layer connection and a top view of this connection is shown in Fig. 9.22. Hence, all three directions of currents internal to the conductors are properly represented in this conventional PEEC mesh. The PEEC equivalent circuit corresponding to a conductor node is shown in Fig. 9.23. Different approximations have been made to include the partial inductance couplings [46\u201348]. 3D VOLUME FILAMENT (VFI) SKIN-EFFECT MODEL 235 In Ref. [27], we attempted to construct a full three-dimensional model using the internal\u2013external inductance approximations for general 3D conductors"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001282_0006-2952(74)90626-1-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001282_0006-2952(74)90626-1-Figure5-1.png",
        "caption": "FIG. 5. L i n e w e a v e r - B u r k p lo ts showing inhib i t ion of ty ramine ox ida t ion (I) , by 2-phenylp iper id ine , 10-4 M t\u00ae), 3-phenylpiper id ine , 10 -4 M (O) and 4-phenylp iper id ine , 10- 3 M ( x ). pH 7.4, 30 \u00b0.",
        "texts": [],
        "surrounding_texts": [
            "TABLI: 2. K~,, VALUES FOR ,COME SUBSTRA'IES o r M A O AT 30 '> OBTAINED BY RECALCULATION O1- g , , VALUES, ASSUMING THAT I]IE ACIIVE SPECIES OF AMINE IS THE NEUTRAL MOI,ECUI,E. Till! QUOTED p g , VALI,IES WERE MEASURED AT THE SAME \"I'EMPERAFURE Substrate pKo K ' , (M) Benzylamine 9.22 + 0.02 9-3 x 10 6 . fl-Phcnylethylamine 9.64 + 0.02 6.3 x 10 s f 7 - P h e n y l p r o p y l a m i n e 10.03 -4- 0.02 1.3 x 10-~* 6-Phenyl-n-butylamine 10.18 __- 0.03 1.8 \u00d7 10 8? p-Methoxyphenylethylamine 9.90 + 0.04 4.8 x 10 ~t p-Fluorobenzylaminc 9.20 + 0.02 3.2 \u00d7 1 0 - s * o-Fluorophenylcthylamine 9.65 + 0.02 1.1 x 10 ~ t * Calculated from experimental data at various pH values, t Derived from K . value at p H 7.4 only. Figure3 showstheeffects o fpH on the rate of oxidation for four amines. All measurements were made at 1 mM concentrations of substrates. This examination was extended by measuring oxidation rates at varying concentrations of four different substrates over a pH range of about 7-9. Figure 4 is a series of Lineweaver-Burk plots obtained from such measurements for bcnzylamine and indicates that little change in Vma x Occurs over a pH range in which there is a large fall in K , values. Similar results were obtained for p-fluorobenzylamine and phenylpropylamine. At pH values higher than those indicated in Fig. 4, effects upon Vma x were apparent, but oxygen uptake became non-linear and in some cases terminated quickly so that reliable data were not readily obtained. These pH effects were found to be similar to those described by McEwen and his co-workers for M A O from human liver. 2t ,22 These workers showed that K,, values Monoamine oxidase--I 621 3OO I 7.4 > I / / \" P a.3 i ,oo - , I I I B 24 I Es-T m M-' FIc. 4. Lineweaver Burk plots for oxidation of I~nzylamine at indicated pH values and 30': in phosphate IP) of borate (B) buffer. for substrates of MAO were related to their pK~ values in a modified form of the Henderson-Hasselbach equation. K,, = K~, [1 + antilog (pKa - pH)] where K~, is the Michaelis constant expressed in terms of the concentration of unionized substrate, This value was found to be constant for a given substrate at all pH values, indicating that the unprotonated amine is the species which interacts with the enzyme. Table 2 gives such values calculated from the present data and it can be seen that K',, values show little change with pH. In inhibition studies, initial velocities were determined for a range of substrate concentrations in the presence of a fixed concentration of inhibitor. For studies at pH 7.4 tyramine hydrochloride was used as substrate. Data obtained were plotted by the Lineweaver-Burk method 18 to yield apparent Michaelis constants Kp which were used to calculate K i values thus: i K i - Kp/K, , - 1 where i is the concentration of inhibitor. Table 3 lists Ki values obtained at 30 \u00b0. Figures 5 and 6 show Lineweaver-Burk plots in presence of inhibitors at pH 7-4. The effects of pH on inhibition of MAO were studied in a similar way; phenylpropylamine replacing tyramine as substrate to avoid possible influences of the second ionizable grouping in the latter compound. In keeping with the finding for substrates, inhibitor constants were dependent upon pH in a way which suggested that the unprotonated amine interacts with the enzyme. Figure 7 shows the effects of pH on the inhibition of phenylpropylamine oxidation by 2-phenylpiperidine. Similar results were obtained using 4-phenylpiperidine and amphetamine as inhibitors. Ki values calculated from the graphical data appear in Table 3. 622 C . H . WILLIAMS Monoamine oxidase--I 623 I - - 2 0 0 V 7-7' /7\"7/ / 8 \" 0 / / / / / / , / / / \" , \" / \" 8.o / / / / ,,o e\" / o \" _ / o ,.x.::--' -I0 0 _2_1 [S] mM-2 FiG. 7. Lineweaver-Burk plots for phenylpropylamine oxidation (@ O), showing inhibition by 2- phenylpiperidine, 10 -4 M, (O---O)at indicated pH values and 30 ~."
        ]
    },
    {
        "image_filename": "designv10_0_0000040_41.184818-Figure9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000040_41.184818-Figure9-1.png",
        "caption": "Fig. 9. Boundary layer regularization.",
        "texts": [
            " In particular, taking into account delay or hysteresis of a switching element, small time constants neglected in an ideal model, replacing a discontinuous function by continuous approximation, are the examples of regularization since discontinuity points (if they exist) are isolated. In our opinion, the universal approach to regularization consists of introducing a boundary layer llsll < A around manifold s = 0 where an ideal discontinuous control is replaced by a real one such that the state trajectories of system (1) are not confined to this manifold but run arbitrarily inside the layer (Fig. 9). The only assumption for this motion is that the solution exists in ordinary sense. If, with the width of the boundary layer tending to zero, the limit of this solution exists, it is taken as a 26 IEEE TRANSACTIONS ON INDUSTRIAL ELECTRONICS, VOL. 40, NO. 1 , FEBRUARY 1993 solution to the system with ideal sliding modes. Otherwise we have to recognize that the equations beyond discontinuity surfaces do not derive unambiguously the motion equation in their intersection. Boundary-layer regularization enables substantiation of the so-called equicalent control method [ 141 intended for deriving a sliding mode equation in the systems depending on linear control: X = f ( x , t ) + B ( x , t ) u (4) where B ( x , t ) is an n x m matrix"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure13.5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure13.5-1.png",
        "caption": "FIGURE 13.5. A double pendulum.",
        "texts": [
            "58) When there is no slip, there is a constraint between \u03b8 and \u03d5 R\u03b8 = r\u03d5 (13.59) 834 13. Vehicle Vibrations which can be used to eliminate \u03d5 from K. K = 3 4 m (R\u2212 r) 2 \u03b8\u0307 2 (13.60) Based on the following partial derivatives: d dt \u00b5 \u2202L \u2202\u03b8\u0307 \u00b6 = 3 2 m (R\u2212 r)2 \u03b8\u0308 (13.61) \u2202L \u2202\u03b8 = \u2212mg(R\u2212 r) sin \u03b8 (13.62) we find the equation of motion for the oscillating disc. 3 2 (R\u2212 r) \u03b8\u0308 + g sin \u03b8 = 0 (13.63) When \u03b8 is very small, this equation is equivalent to a mass-spring system with meq = 3 (R\u2212 r) and keq = 2g. Example 477 F A double pendulum. Figure 13.5 illustrates a double pendulum made by a series of two pendulums. There are two massless rods with lengths l1 and l2, and two point masses m1 and m2. The variables \u03b81 and \u03b82 can be used as the generalized coordinates to express the system configuration. To calculate the Lagrangean of the system and find the equations of motion, we start by defining the global position of the masses. 13. Vehicle Vibrations 835 x1 = l1 sin \u03b81 (13.64) y1 = \u2212l1 cos \u03b81 (13.65) x2 = l1 sin \u03b81 + l2 sin \u03b82 (13.66) y2 = \u2212l1 cos \u03b81 \u2212 l2 cos \u03b82 (13"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure3.11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure3.11-1.png",
        "caption": "Fig. 3.11. Simple solids: (a) rectangular prism, (b) cylinder, (c) half-cylinder, (d) one-quarter of a cylinder, (e) half-ring, (f) one-quarter of a ring, (g) one-quarter of a sphere, (h) one-eighth of a sphere, (i) one-quarter of a shell, (j) one-eighth of a shell.",
        "texts": [
            " Stabilization means the PM is demagnetized up to a value which is slightly higher than the most dangerous demagnetization field during the operation of a system where the PM is installed. In magnetic 102 3 Materials and Fabrication circuits with stabilized PMs the operating point describing the state of the PM is located on the recoil line. More details about how to find the operating point of a PM graphically and analytically can be found in [106]. Permeances of air gaps and permeances for leakage fluxes can be found analytically by dividing the magnetic field into simple solids. Permeances of simple solids shown in Fig. 3.11 can be found using the following formulae: (a) Rectangular prism (Fig. 3.11a) G = \u00b50 wM lM g H (3.22) (b) Cylinder (Fig. 3.11b) G = \u00b50 \u03c0d2 M 4g (3.23) (c) Half-cylinder (Fig. 3.11c) G = 0.26\u00b50lM (3.24) where the average air gap gav = 1.22g and the average surface Sav = 0.322glM [15] (d) One-quarter of a cylinder (Fig. 3.11d) G = 0.52\u00b50lM (3.25) (e) Half-ring (Fig. 3.11e) G = \u00b50 2lM \u03c0(g/wM + 1) (3.26) For g < 3wM , G = \u00b50 lM \u03c0 ln ( 1 + 2wM g ) (3.27) (f) One-quarter of a ring (Fig. 3.11f) G = \u00b50 2lM \u03c0(g/c+ 0.5) (3.28) For g < 3c, G = \u00b50 2lM \u03c0 ln ( 1 + c g ) (3.29) 3.2 Rotor Magnetic Circuits 103 104 3 Materials and Fabrication (g) One-quarter of a sphere (Fig. 3.11g) G = 0.077\u00b50g (3.30) (h) One-eighth of a sphere (Fig. 3.11h) G = 0.308\u00b50g (3.31) (i) One-quarter of a shell (Fig. 3.11i) G = \u00b50 c 4 (3.32) (j) One-eighth of a shell (Fig. 3.11j) G = \u00b50 c 2 (3.33) Fig. 3.12 shows a model of a flat electrical machine with smooth armature core (without slots) and surface PM excitation system. The armature is of steel laminations. The PMs are fixed to the mild steel yoke. The pole pitch is \u03c4 , the width of each PM is wM , and its length is lM . In an AFPM machine lM = 0.5(Dout \u2212Din) (3.34) The space between the pole face and the armature core is divided into a prism (1), four quarters of cylinders (2 and 4), four quarters of rings (3 and 5), four pieces of 1/8 of a sphere (6) and four pieces of 1/8 of a shell (7)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003681_j.optlastec.2006.01.008-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003681_j.optlastec.2006.01.008-Figure1-1.png",
        "caption": "Fig. 1. The schematic arrangement of LRM setup.",
        "texts": [
            " The higher-the-better criterion is chosen, as the quality characteristics is deposition rate. This criterion is given by S N \u00bc 10 log10 1 n Xn i\u00bc1 1 yi \" # , where n represents the total number of tests in an experimental trail (n \u00bc 3) and yi represents the deposition rate of the specimen corresponding to the ith test (i \u00bc 1, 2 or 3). The result of S/N ratio calculations are tabulated in Table 3. The LRM setup consists of a high power laser system integrated with the beam delivery system, powder-feeding system and job/beam manipulation system. Fig. 1 presents the schematic arrangement of the LRM machine. In the present experiment, laser beam was generated using a 3.5 kW continuous wave (CW) CO2 laser system [14] and was transferred to the fabrication point at 3-axis laser workstation [15] using a couple of water-cooled goldcoated plane mirrors and a concave mirror of 600mm radius of curvature. At fabrication point, a defocused laser beam spot of 2mm was used for powder deposition. Inconel-625 powder (size range: 45\u2013106 mm) was fed into the molten pool using a volumetric-controlled powder feeder [16] through a co-axial powder-feeding nozzle [17]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure9.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure9.6-1.png",
        "caption": "Fig. 9.6. The stator and rotor layout of the 2 kW 500 rpm air-cored AFPM wind generator with concentrated coil topology developed by Durham University: 1 \u2014 bobbin wound air-cored stator, 2 \u2014 rotor disc. Courtesy of Durham University , UK.",
        "texts": [
            " Figure 9.4 shows a medium-power AFPM generator with twin external rotor and inner coreless stator for a 5-m blade diameter wind turbine. The stator system consists of 15 non-overlapping concentrated coils, one coil per phase. The thickness of coil is 12 mm and diameter of wire 0.6 mm. Each coil is individually rectified to d.c., to reduce the commutation torque ripple effect and provide better control over the voltage. The rotor has 2p = 16 poles. Performance characteristics are shown in Fig. 9.5. Figure 9.6 shows the construction detail of a 2 kW 500 rpm AFPM aircored wind turbine-driven generator with concentrated coil topology developed by Durham University, UK [27]. The stator is located between two rotor discs and contains 12 circular bobbin armature coils while each rotor disc has 16 circular NdFeB magnets. The rotor discs are mechanically joined by an outer location ring and through bolts. The magnets are secured and positioned 9.1 Power Generation 285 286 9 Applications in a retainer ring. The stator is supported from the inside using a double-shaft arrangement"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003971_s00332-004-0650-9-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003971_s00332-004-0650-9-Figure4-1.png",
        "caption": "Fig. 4. A three-link mechanism submerged in a perfect fluid.",
        "texts": [
            "6) which, in view of \u03be3 = g\u22121 3 g\u03073, yields, using the terminology of Bloch, Krishnaprasad, Marsden, and Murray [1996], the following reconstruction equation: g\u03073 = \u2212g3A(x)x\u0307 . (6.7) It is convenient for studying locomotion to assume that the relative motion x evolve according to prescribed functions of time.6 The net locomotion g3 is then fully determined by solving (6.7) and can be thought of as a geometric phase, or holonomy, over closed curves in X traced by the shape variables (x1, x2); see Figure 3. Consider the example analyzed in Radford [2003] of three articulated, submerged bodies that are constrained to undergo only planar motions, as shown in Figure 4. A Planar Three-Link Fish. For concreteness, assume the three bodies are identical and made of a homogeneous material and have an ellipsoidal geometry. Let a denote the length of the major semi-axis of the ellipses and b denote the length of the minor semiaxis. Let the joints be placed a distance c away from the tips of the ellipses along their major axes, as shown in Figure 4. 6 One can imagine that the joints are equipped with motors and controllers to enforce the prescribed motions. Of course, biological fish use their muscles to control their shape changes. Locomotion of Articulated Bodies in a Perfect Fluid 273 Consider a fixed inertial frame {ek} and three body-fixed frames attached at the center of mass of each body, assumed to coincide with its geometric center. Let (\u03b2, x, y) denote the rigid motion of the middle ellipse relative to {ek}, and let \u03b81 and \u03b82 denote, respectively, the orientation of B1 and B2 relative to B3. One can readily verify that the motions x\u03b1 of B1 relative to B3 and B2 relative to B3 can be represented by x\u03b1 = cos \u03b8\u03b1 \u2212 sin \u03b8\u03b1 \u00b1l(1 + cos \u03b8\u03b1) sin \u03b8\u03b1 cos \u03b8\u03b1 \u00b1l sin \u03b8\u03b1 0 0 1  , (7.1) where the upper sign, i.e., the (+), in \u00b1 is associated with \u03b81, and l = a+c is the distance between the mass center and the joint along the major semi-axis of the corresponding ellipse (see Figure 4). It should be clear from (7.1) that in this example the base spaceX actually reduces to S1 \u00d7S1, with coordinates \u03b8 = (\u03b81, \u03b82). Further, one may also verify that \u03b6 T \u03b1 = (\u03b8\u0307\u03b1, 0,\u00b1l \u03b8\u0307\u03b1). (7.2) Two Approaches. We present two solutions to the planar three-link fish problem. In Section 7.1, we assume that the submerged bodies are hydrodynamically decoupled, as done in Radford [2003]. This assumption means that the added masses associated with a given body are not affected by the presence of the other bodies, which is clearly not true for a fish model",
            " Specifically, one can also define a coadjoint action Ad\u2217 of G on g\u2217, the dual of the Lie algebra, as well as a coadjoint map ad\u2217: g\u2217 \u2192 g\u2217, such that \u3008\u03b4, ad\u03be \u03b7\u3009 = \u3008ad\u2217 \u03be \u03b4, \u03b7\u3009, where \u03b4 \u2208 g\u2217 and \u3008, \u3009 represents the pairing between g and g\u2217. To conclude this review, we discuss SE(2), the subgroup of SE(3) that corresponds to planar rigid body motion, and its Lie algebra se(2), see, e.g., Chapter 2 of Kelly (1998). In this case, one may represent elements of SE(2) by (\u03b2, x, y), where \u03b2 is the orientation of the body relative to an inertial frame and (x, y) are the coordinates of a given material point of the body (this is illustrated in Figure 4 forB3). Further, elements of se(2)may be represented as \u03b7 = ( \u03b2, vx , vy), where, clearly, \u03b2 is the angular velocity and (vx , vy) is the translational velocity. The conjugation map as well as the adjoint actions are defined in a similar way as above. In this appendix, we derive formally some of the results presented in Sections 4 and 5. 286 E. Kanso, J. E. Marsden, C. W. Rowley, and J. Melli-Huber The Bundle Structure. We first show thatR is a principal bundle with structure group G and diagonal action of G onR, G \u00d7R\u2192 R, given by h(g1, g2, g3) = (hg1, hg2, hg3); we do this following Cendra and Marsden (2004), which encountered a similar structure for the case of asteroid pairs"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure9.27-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure9.27-1.png",
        "caption": "Fig. 9.27. Two-phase, four-coil, single-sided 15-mm AFPM brushless motor. Photo courtesy of Moving Magnet Technologies, SA, Besancon, France.",
        "texts": [
            " A 400-\u00b5m eight pole PM and a three-strand, 110-\u00b5m disc shaped, lithographically produced stator winding have been used [154]. Plastic bound NdFeB magnets are a cost effective solution. However, the maximum torque is achieved with sintered NdFeB magnets. A miniature ball bearing has a diameter of 3 mm. Pennymotors find applications in miniaturized hard disc drives, cellular phones as vibration motors, mobile scanners and consumer electronics. Moving Magnet Technologies (MMT) two or three phase, miniature AFPM brushless motors (Fig. 9.27) are well adapted to high volume low cost production. In order to obtain rigidity the stator coils are overmoulded. The stator is mounted on a single-sided printed circuit board. Each part can be manufactured using standard moulding, stamping techniques and automatic coil winding with a specific design for simple and efficient assembly. Three position 9.8 Miniature AFPM Brushless Motors 309 sensing methods for the closed-loop control are used: (a) digital Hall probes located in the stator, (b) EMF signal (sensorless mode) and (c) absolute analogue position sensor (position sensor mode)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure10.40-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure10.40-1.png",
        "caption": "Figure 10.40 (a) A simply supported beam, loaded by a single point load. (b) and (c) When using deformation symbols, we are free to choose at which side of the member axis the shear force is plotted. (d) The bending moment is plotted at the side where the bending moment causes tension, so at the convex side of the member axis. The open side of the bending symbol therefore always faces towards the member axis. The gradient of the M diagram ( M/ x), shown as a \u201cstep\u201d, now corresponds directly to the shear symbol for the shear force. (e) If however the M diagram is plotted at the wrong side of the member axis, the relationship with the deformation symbol for the shear force (the \u201cstep\u201d in the M diagram) is lost.",
        "texts": [
            " \u2022 Shear symbol (deformation symbol for shear forces) Figure 10.39a again shows a small member segment, but now with shear forces. When subject to shear forces, one sectional plane will try to shift with respect to the other. This effect can be visualised by applying an (imaginary) slide joint within the segment, so that both parts can move with respect to one another (see Figure 10.39b). The step change formed by the moved member axes is used as the deformation symbol for shear forces (see Figure 10.39c). 10 Section Forces 419 Figure 10.40 shows the V diagram and the M diagram with the deformation symbols for a simply supported beam, loaded by a single point load. Since there is no coordinate system, we are in principle free to choose on which side of the member axis we plot the bending moment and the shear force, as long as we use the correct deformation symbols. See the shear force diagrams in Figures 10.40b and 10.40c, which are both correct. We make an exception for the bending moment. It is agreed that the M values are plotted at the side where the bending moment causes tension, so at the convex side of the member axis (see Figure 10.40d). The open side of the deformation symbol is therefore always faced towards the member axis.1 The advantage of this is that the gradient of the M diagram ( M/ x), shown in Figure 10.40d as a \u201cstep\u201d, now corresponds directly with the shear symbol for the shear force. This allows us to easily and directly check the relationship between the moment diagram and the shear force diagram. In Figure 10.40e, the M diagram has been plotted at the wrong side of the member axis. Although you will come across this often in books, we strongly recommend that you do not draw the moment diagram in this way, as the relationship with the deformation symbol for the shear force (the \u201cstep\u201d in the M diagram) is lost. 1 Thanks to this agreement, the deformation symbol in the M diagram is actually unnecessary. The deformation symbol is nevertheless always shown for clarity. 420 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM When drawing the N , V and M diagrams, you can use: \u2022 plus and minus signs, related to a (local) coordinate system with the x axis along the member axis; \u2022 deformation symbols (only for the V diagram and the M diagram)",
            " 1 The upper index refers to the segment in which shear force V is acting. 466 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM The bending moment MB at B1 is found from the moment equilibrium (about B) of the isolated part to the left or right of B. This bending moment is 60 kNm; the calculation is left to the reader. As this moment causes tension at the lower side of the beam, this value in the M diagram has to be plotted at the underside of the member axis. See the conventions discussed earlier in Sections 10.2.4 and 10.2.5 and see also Figure 10.40. In the same way, the bending moment found at C is 60 kNm, also with tension at the lower side of the beam. In the M diagram in Figure 12.1d, the two known and the two calculated values are shown by means of dots. The M diagram is completed by drawing straight lines between these values. The M diagram shows that in field BC, where the shear force is zero, the bending moment is indeed constant (according to rule 1). Example 2 In Figure 12.2a, the simply supported beam AD is subject to a uniformly distributed load of 12"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure13.24-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure13.24-1.png",
        "caption": "Figure 13.24 (a) Isolated beam ABC with (b) shear force diagram and (c) bending moment diagram.",
        "texts": [
            " To draw the M and V diagrams, the distributed loads in fields AB and BC have been replaced by their resultants RAB and RBC. For the triangular load on BC RBC = 1 2 \u00d7 (6 kN/m)(2 \u221a 2 m) = 6 \u221a 2 kN. The trapezoidal load on AB is divided into triangular loads (1) and (2); their resultants can be calculated easier: R(1) = 1 2 \u00d7 (18 kN/m)(4 \u221a 2 m) = 36 \u221a 2 kN, R(2) = 1 2 \u00d7 (6 kN/m)(4 \u221a 2 m) = 12 \u221a 2 kN, RAB = R(1) + R(2) = 48 \u221a 2 kN. The location of the line of action of RAB is found from the moment about A: aRAB = 1 3 \u00d7 4 \u221a 2 \u00d7 R(1) + 2 3 \u00d7 4 \u221a 2 \u00d7 R(2) = 160 kNm so that a = 160 kNm 48 \u221a 2 kN = 5 3 \u221a 2 m. c. Figure 13.24a shows all the forces acting normal to the beam axis. They generate shear forces and bending moments in the beam. The distributed loads on AB and BC have been replaced by their load resultants. In Figures 13.24b and 13.24c, the V and M diagrams due to the load resultants are shown by means of dashed lines. The solid lines are the actual V and M diagrams. 13 Calculating M, V and N Diagrams 567 The actual V diagram has a parabolic variation with a step change at B. At C, the distributed load is zero, and the V diagram has a \u201chorizontal\u201d tangent"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure3.53-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure3.53-1.png",
        "caption": "Figure 3.53 (a) A flat slab of 6\u00d75 m2 in the horizontal xz plane is loaded by five vertical forces. The grid lines are 1 m apart. (b) There is no resultant force, but there is a resultant couple of which the moment vector is in the xz plane. The resultant couple acts in a plane perpendicular to the moment vector.",
        "texts": [
            "51b). By shifting the resultant force \u2211 F parallel to itself one can provide that the resultant force vector and moment vector have the same direction (see Figure 3.51c). The combination of a force and a moment of which the vectors have the same direction is called a screw.1 The following represents three examples that relate to the determination of the resultant of a number of forces and/or couples. Example 1 A flat slab of 6 \u00d7 5 m2 in the horizontal xy plane is loaded by six vertical forces (see Figure 3.53a). The grid lines are 1 m apart. Question: Determine the resultant force R as to magnitude and direction and the location at which it acts on the slab. Solution: The units used are kN and m. The units are omitted in the interim calculations. The x and y components of all the forces given are zero, as are their moments about the z axis, therefore \u2211 Fx = 0,\u2211 Fy = 0,\u2211 Tz = 0. 1 Reducing a system of forces and couples into a screw is an interesting academic problem, but is of little practical use and therefore not covered in further detail. 90 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM In addition, \u2211 Fz = +15 + 20 \u2212 40 + 30 \u2212 30 + 35 = +30 kN,\u2211 Tx = \u221240 \u00d7 1 + 30 \u00d7 1 + 35 \u00d7 2 + 20 \u00d7 3 + 15 \u00d7 4 \u2212 30 \u00d7 4 = +60 kNm,\u2211 Ty = 15 \u00d7 0 \u2212 20 \u00d7 1 + 40 \u00d7 2 \u2212 30 \u00d7 4 + 30 \u00d7 4 \u2212 35 \u00d7 6 = \u2212150 kNm. So the system of forces can be replaced by a downward force R = 30 kN at O, together with two couples of 150 kNm and 60 kNm of which the moment vectors are along the x and y axis respectively (see Figure 3.53b). Since the moment vectors are perpendicular to force R, they can be eliminated by shifting R to another point of application. Imagine (x, y) is the new point of application (see Figure 3.53c). We can find (x, y) from the condition that R = 30 kN has to generate the same moment about the x and y axis as all the forces together, so that \u2211 Tx = Ry = 60 kNm \u21d2 y = 60 kNm R = 60 kNm 30 kN = 2 m, \u2211 Ty = \u2212Rx = \u2212150 kNm \u21d2 x = 150 kNm R = 150 kNm 30 kN = 5 m. 3 Statics of a Rigid Body 91 Example 2 In Figure 3.53a, a flat slab of 6 \u00d7 5 m2 in the horizontal xy plane is loaded by five vertical forces. The distance between the grid lines is 1 m. Question: Determine the resultant of this set of forces. Solution: The units used are kN and m. The units are omitted in the interim calculations. When determining the resultant of this system of parallel forces, only the force sum in the z direction and the moment sum about the x and y axis are relevant: \u2211 Fy = +25 \u2212 15 + 20 \u2212 45 + 15 = 0 kN,\u2211 Tx = 25 \u00d7 0 + 15 \u00d7 1 \u2212 20 \u00d7 3 \u2212 15 \u00d7 4 + 15 \u00d7 4 + 45 \u00d7 4 = +75 kN,\u2211 Tz = +25 \u00d7 6 \u2212 15 \u00d7 2 + 20 \u00d7 5 \u2212 45 \u00d7 3 + 15 \u00d7 1 = +100 kN. There is no resultant force, but there is a resultant couple T of which the moment vector is in the xy plane (see Figure 3.53b). Its magnitude is T = \u221a 752 + 1002 = 125 kNm. The resultant couple acts in a plane perpendicular to the moment vector. 92 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Example 3 Figure 3.54 shows a junction of three coplanar tubes that are rigidly connected at equal angles of 120\u25e6. The tubes are loaded (by torsion) by the couples TA, TB and TC. The resultant couple on the junction is zero. Question: How large are the couples TA and TB if TC = 75 Nm? Solution: In Figure 3.55a, the couples are represented by their moment vectors"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001035_tac.2007.908319-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001035_tac.2007.908319-Figure7-1.png",
        "caption": "Fig. 7. DF analysis.",
        "texts": [
            " Then, we can exploit the given relationships between the hysteresis parameters and the oscillation parameters in order to obtain the following expression for the DF of the G-SO algorithm: (59) A periodic solution can be found from the harmonic balance equation [2], where the negative reciprocal of the DF (59) is as follows: (60) As usual, the periodic solutions correspond to the points of intersection of the and loci, the latter being a straight line backing out of the origin with a slope that depends only on parameter , as depicted in Fig. 7. Therefore, a periodic motion may occur if at some frequency the phase characteristic of the actuator-plant transfer function is equal to . If such a requirement is fulfilled, so that intersection between the two plots occurs, then the frequency and the amplitude of the periodic solution can be derived from the \u201ccrossover\u201d frequency and from the magnitude of vector in Fig. 7, respectively. An intersection point will certainly exist if the overall relative degree of the combined actuator-plant degree is three or higher. The DF analysis given previously provides a simple and systematic, but approximate, evaluation of the magnitude and frequency of the periodic motions in linear systems driven by the G-SO algorithm (12). An exact solution, which does not require the actuator to be fast, can be obtained via application of the Tsypkin\u2019s method [43]. The Tsypkin locus approach involves computing the following complex function , called the Tsypkin locus: (61) where is the magnitude of the relay output"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure13.80-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure13.80-1.png",
        "caption": "Figure 13.80 Support reactions.",
        "texts": [
            "79b shows the closed force polygon for the force equilibrium of the joint (scale: 1 square = 2.5 kN). The force polygon gives NBC = +7.5 \u221a 2 kN, NCS = +5 kN. c. With NCS = +5 kN the vertical equilibrium of CSD gives Dv = 5 kN (\u2191). 13 Calculating M, V and N Diagrams 607 With V CS = 10 kN the horizontal equilibrium of ABC gives Ah = 10 kN (\u2190). From the force equilibrium of the structure as a whole follows Dh = 10 kN (\u2190), Av = 15 kN (\u2191). Finally, the fixed-end moment reactions at A and D follow from the moment equilibrium about S of DS and ABCS respectively. In Figure 13.80, the support reactions are shown as they act in reality. d. In Figures 13.81a to 13.81c the M , V and N diagrams are shown. At B and C, the bending moment \u201cgoes round the corner\u201d. The slopes of the M diagram are in line with the magnitudes and the deformations symbols of the shear forces. General comment: When asked to draw an M , V or N diagram, please draw the diagrams including the (deformation) symbols (or plus and minus signs) and the values at relevant points. Self-contained structures (Section 13"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000824_j.jtbi.2005.04.004-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000824_j.jtbi.2005.04.004-Figure5-1.png",
        "caption": "Fig. 5. Various collisional models for the step-to-step transition in walking. The hodographs show candidate trajectories for the tip of the velocity vector, during collision, as it moves from v to v\u00fe. All are for walking models with collisional impulses from two legs. Several cases are shown. (i) In passive-dynamic downhill walking, there is only impulse along the new stance leg. This case is the same for walking as for passive running with a dead leg. At every instant on path AC dW \u00bc mv dvo0 and the collision is exclusively absorbing (the length of v is decreasing throughout). The work required to make-up the length of v\u00fe (path CB) is supplied by gravity. An energetically equivalent model has power supplied by the trailing old stance leg after motion is constrained to the new stance arc. (ii) In the toe-off then heel-strike model of walking there is a purely generative phase with a force along the old stance leg, followed by a purely absorbing heel-strike phase along the new stance leg. (iii) Using only impulses from the two legs the vertical path requires overlapping impulses. (iv) The least energetically favorable collision using two compressional impulses along the legs, in which the heel-strike impulse is complete before push-off starts. (v) In the new pseudo-elastic model proposed here there are two branches. One along the virtual leg from hip to toe (f=4 forward of the rear leg) and one along the virtual leg from hip to heel (f=4 behind the front leg). This path is the minimum-cost two-leg simultaneous collision for given f. (vi) A trajectory of constant energy. This trajectory has precisely zero \u2018external work\u2019 yet requires leg work. Path (vi) can be traversed at no energy cost using a (slightly) rounded foot whose length is the step length.",
        "texts": [
            " Assuming a given speed v and stride length d the triplet collisions use three times less energy as would a pronk gait with one collision at each stride. This factor of 3 improvement holds for plastic, pseudo-elastic or with any other values of eg and r, so long as the same values are used for the triplet as for the pronk. Here we consider the step-to-step transition in human walking from the point-mass glancing-collision perspective. The relevant issues are conveniently displayed (e.g. Kuo, 2001) by a nineteenth century visualization technique called a \u2018\u2018hodograph\u2019\u2019 (Fig. 5). If the velocity vector is animated as it changes during the step-to-step transition, always putting its tail at the origin, the tip traces a curve. This velocity-trajectory curve is a hodograph. Fig. 5 shows the velocity before collision (v with tip at A) and just after collision (v\u00fe with tip at B). In this model v and v\u00fe are assumed to be orthogonal to the old and new stance legs, respectively. During a collision the tip of the velocity vector traces a path on the hodograph. Shown are several such velocitytip trajectories, each with a different collisional meaning and each with a different energetic cost. Even though we track the velocity trajectory in detail, the collisional model assumes that the path is traversed ARTICLE IN PRESS A",
            " All but one of these has the same net collisional impulse and the same impulses at each leg as the other cases. Yet, these different contact strategies have quite variable energetics. A special simple case for the model here is the famous rimless wheel (Bekker, 1956; Margaria, 1976) with mass only in the hub (see Fig. 6a) analysed in detail in McGeer (1990a). The collision analysis is the same as for the Simplest Walker (Garcia et al., 1998). In this case the only collisional impulse is from the new stance leg. The trajectory is labeled (i) in Fig. 5. The deficit in the magnitude of v\u00fe must be made up by gravity (walking down an incline) if the system is strictly passive. Using the formalism here, the cost of this collision is, from Eq. (15), using r \u00bc 0; eg \u00bc 1, Em \u00bc bf2v2m=2. (26) This corresponds with path (i) on the hodograph of Fig. 5, the most costly model for walking. As mentioned in Tucker (1975) and again in McGeer (1993), the collisional cost of walking can be reduced by preceding the absorbing heel-strike collision with a pushoff (plantar extension) which we model as a generative collision. At a given speed and step length the collisional cost can be reduced by a factor of 4 this way (Kuo, 2001). Using the formalism here we can look at a sequence of two collisions with f1 \u00bc f2 \u00bc f=2. The first is a purely generative push-off collision with eg1 \u00bc 1 and the second a purely absorbing collision with eg2 \u00bc 1. Using Eqs. (15)\u2013(17) and ri \u00bc 0, Em \u00bc bf2v2m=8 (27) ARTICLE IN PRESS A. Ruina et al. / Journal of Theoretical Biology 237 (2005) 170\u2013192 183 giving, comparing with Eq. (26), a factor of 4 reduction in cost as compared to passive walking. Pushoff before heelstrike is shown as path (ii) on Fig. 5. For given f this two leg walking collision is energetically identical to a single-leg pseudo-elastic collision (see Eq. (24)). Both have generating and absorbing phases with deflection angle f=2. For walking, however, the \u2018restitution\u2019 occurs before the absorption. There are a few ways to understand the efficiency gain of using pushoff before heel-strike as compared to the purely passive, rimless-wheel collision. By pushing off before heel-strike the collision loss at heel-strike is reduced. Or, the angle at the absorbing part of the collision is cut in half, reducing the absorption Ea by 1 4 ",
            " So kinetic energy fluctuations are less than the muscle work that causes them. Assume f and v are given, the two impulses are assumed to be along the legs, and the net collision conserves energy \u00f0DE \u00bc 0\u00de. Balance of momentum determines the two impulses and balance of energy determines that their total work is zero. But the collisional cost is not determined. The collision cost depends on the relative timing of the two impulses. For example, consider truly simultaneous collisions. Assume the two forces at all times proportional, giving path (iii) in Fig. 5. The path integral from the appendix (Eq. (42)) gives Em \u00bc 1 2 mv2f2=2. (28) Comparing with Eq. (27) we see that truly simultaneous push-off and heel-strike has twice the cost of push-off entirely before heel-strike. For heel-strike entirely before push-off, path (iv) in Fig. 5, we get Em \u00bc 3 4 mv2f2=2. (29) This is a peculiar limiting case. The lead foot first applies an impulse, then waits for the trailing leg to apply an impulse, and then (non-impulsively) begins single stance. Considering only cases with compressional impulses applied between stance phases this cost, three times more than that in Eq. (27), is the most costly down-to-up redirection. A case mimicking passive downhill walking, path (i) in Fig. 5 is also possible on level ground. First a heelstrike collision occurs (with the full f and with eg \u00bc 1) following this both legs apply forces to bring the magnitude of v up to its pre-collision value along path CB in Fig. 5. This requires a physically unrealistic nonworking tension force from the lead leg as the rear leg pushes off. Should this be implemented by a person or robot, the time of collision could be spread and the tension could be replaced by gravity, with identical net energetics (as implemented in Collins et al., 2005). Energetically this push-off-during-next-stance case is identical to that for passive walking (e.g. Eq. (26)). Another illustrative case is given by path (vi) in Fig. 5. At every instant in time the net force (the sum of the two leg forces) is orthogonal to the path. The kinetic energy is constant throughout the collision. The \u2018external work\u2019 is precisely zero. But each of the legs does some mixture of positive and negative work, a little more than in the sequential collision model (Fig. 5 path ii). As presented in the appendix, for this zero-external-work case we have Em \u00bc bmv2f2=6. (30) This cost is 4 3 the cost of the push off entirely before heel-strike (Eq. (27)). This zero-\u2018external-work\u2019 example suggests that modeling double stance or a horse gallop as a sequence of independent collisions may be closer to the appropriate work accounting (e.g. the \u2018individual-limbs\u2019 method of Donelan et al., 2002b) than just associating a cost with the center-of-mass energy fluctuations (e.g",
            " The variety of possible collision strategies raises the following question. Given v and v\u00fe, if one assumes that the collision is mediated by forces along two virtual legs (not necessarily orthogonal to v and v\u00fe), how should these legs be oriented and how should the separate force histories vary so as to minimize the net collisional cost? ARTICLE IN PRESS A. Ruina et al. / Journal of Theoretical Biology 237 (2005) 170\u2013192184 A slightly involved argument shows that the solution is the path (v) of Fig. 5; a sequence of two pseudo-elastic collisions oriented at the quarter and three-quarter point of the splayed double-stance legs. All other trajectories based on forces along two legs, no matter what the leg orientations or how the forces are distributed in time, have greater cost. The energetic cost of one such two-collision episode (n \u00bc 2, egi \u00bc 0, fi \u00bc f=2) is one-eighth the cost of passive-dynamic walking and one half of toe-off before heel-strike. This situation is essentially identical to the model for a horse canter, but with two beats here instead of the horse\u2019s three",
            " The convex polygon rolls well and the concave polygon loses a considerable fraction of its energy at each collision. Thus, the regular polygon is (in math language) a distinguished limit with different behavior depending on whether it is considered an extreme case of concavity or of convexity. A rolling, slightly convex polygon has motions that alternate between inverted pendulum motion (hinged at a vertex) and a collisional phase. The collisional phase is not dissipative but rather tracks path (vi) in Fig. 5, all in an instant. This is equivalent to having a continuum of legs. From the collisional perspective we can treat rounded feet, even if close to flat, as a means to track path (vi) in Fig. 5. If the foot length is the step length d then there is no dissipation. If the foot length df is less than d then the relevant collisional length is d df . This should replace d in any of our formulas involving collisional loss but not in formulas relating step length to speed. Thus, at a given speed and with a given collisional mechanism (be it passive walking, or push-off before heel-strike), use of a rounded foot multiplies the collisional cost per step by \u00f0d df \u00de 2=d2; a rounded foot with length of half the step-length quarters the collisional cost"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000952_j.jmatprotec.2014.07.030-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000952_j.jmatprotec.2014.07.030-Figure2-1.png",
        "caption": "Fig. 2. Experimental setup used for each case.",
        "texts": [
            " Table 1 provides a summary of the changes in dwell time used for each material as well as the time required to deposit a single track and the total cooling time between the layers for each case. In this work, the dwell time is defined as the pause programmed into the build between the completion of one layer and the beginning of the next, not including laser travel time. Dwell times of 0, 20, and 40 s are used for each material. 2.2. In situ distortion measurement In situ measurements of distortion in the substrate are made using the experimental set up shown in Fig. 2. In this experimental set up, the substrate material is mounted and clamped on one end, allowing the opposite end to be free to distort during the deposition. In situ distortion measurements are taken with a Keyence LK-031 E.R. Denlinger et al. / Journal of Materials Processing Technology 215 (2015) 123\u2013131 125 Table 1 Description of build for all cases. Case number 1 2 3 4 5 6 Material Ti\u20136Al\u20134V Inconel\u00ae 625 Powder feed rate (g/min) 8 8 8 16 16 16 Dwell time (s) 0 20 40 0 20 40 Deposition time per track (s) 9"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure4.9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure4.9-1.png",
        "caption": "Fig. 4.9: An example of a specific two-finger gripper for long objects.",
        "texts": [
            ") for grasping not only solid objects, and it can be dated to several centuries B.C. Therefore this chapter is focused mainly on two-finger grippers that can be designed and operated in industrial and non-industrial applications with the aforementioned characteristics for high reliability and robustness. Fundamentals of the Mechanics of Robotic Manipulation 247 In addition, two-finger grippers are also useful for an easy application for more complicated tasks. Figures 4.9 and 4.10 show examples of such typical applications: Fig. 4.9 shows how two fingers can be properly designed and located to grasp long objects by their extremities only; Fig. 4.10 shows alternative grasps by using two or more two-finger grippers to grasp long objects and even large 3D shaped objects. Chapter 4 Fundamentals of the Mechanics of Grasp248 In Fig. 4.11 a general scheme is shown for an industrial gripper with two fingers composed of: - an actuation system, that usually is of a pneumatic type with suitable pipelines and electrovalve; - a transmission system, connecting the actuator to a driving mechanism; - a driving mechanism that transmits movement and force to the fingers; - fingers and fingertips, that can be specifically designed and shaped for contacting specific objects"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001282_0006-2952(74)90626-1-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001282_0006-2952(74)90626-1-Figure3-1.png",
        "caption": "FIG. 3. p H vs ve loc i ty curves for oxidation of benzylamine (A)~ tyramine (B), p-fluorophenylethylamine (C) and p-methoxyphenylethylamine [D). 30L 1 m M concentration. Buffers: 0.1 M a c c t a t c ( & ) , 0.1 M borate (OI. 0.05 M p h o s p h a t e (O1.",
        "texts": [
            " QUOTED p g , VALI,IES WERE MEASURED AT THE SAME \"I'EMPERAFURE Substrate pKo K ' , (M) Benzylamine 9.22 + 0.02 9-3 x 10 6 . fl-Phcnylethylamine 9.64 + 0.02 6.3 x 10 s f 7 - P h e n y l p r o p y l a m i n e 10.03 -4- 0.02 1.3 x 10-~* 6-Phenyl-n-butylamine 10.18 __- 0.03 1.8 \u00d7 10 8? p-Methoxyphenylethylamine 9.90 + 0.04 4.8 x 10 ~t p-Fluorobenzylaminc 9.20 + 0.02 3.2 \u00d7 1 0 - s * o-Fluorophenylcthylamine 9.65 + 0.02 1.1 x 10 ~ t * Calculated from experimental data at various pH values, t Derived from K . value at p H 7.4 only. Figure3 showstheeffects o fpH on the rate of oxidation for four amines. All measurements were made at 1 mM concentrations of substrates. This examination was extended by measuring oxidation rates at varying concentrations of four different substrates over a pH range of about 7-9. Figure 4 is a series of Lineweaver-Burk plots obtained from such measurements for bcnzylamine and indicates that little change in Vma x Occurs over a pH range in which there is a large fall in K , values. Similar results were obtained for p-fluorobenzylamine and phenylpropylamine"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000366_tcst.2004.825052-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000366_tcst.2004.825052-Figure1-1.png",
        "caption": "Fig. 1. The four-rotor rotorcraft.",
        "texts": [
            " The rotorcraft is one of the most complex flying machines. Its complexity is due to the versatility and maneuverability to perform many types of tasks [1]. The classical helicopter is conventionally equipped with a main rotor and a tail rotor. However, other types of helicopters exist, including the twin rotor or tandem helicopter and the coaxial rotor helicopter. In this paper, we are particularly interested in controlling a mini rotorcraft having four rotors. Four-rotor rotorcrafts, like the one shown in Fig. 1, have some advantages over conventional helicopters. Given that the front and rear motors rotate counterclockwise while the other two rotate clockwise, gyroscopic effects and aerodynamic torques tend to cancel in trimmed flight. This four-rotor rotorcraft does not have a swashplate. In fact it does not need any blade pitch control. The collective input (or throttle input) is the sum of the thrusts of each motor. Pitch movement is obtained by increasing (reducing) the speed of the rear motor while reducing (increasing) the speed of the front Manuscript received July 10, 2002; revised July 4, 2003",
            " Dzul is with the Divisi\u00f3n de Estudios de Posgrado e Investigaci\u00f3n, Instituto Tecnol\u00f3gico de la Laguna, 27000 Torre\u00f3n, Coahuila, Mexico (e-mail: dzul@hds.utc.fr). Digital Object Identifier 10.1109/TCST.2004.825052 motor. The roll movement is obtained similarly using the lateral motors. The yaw movement is obtained by increasing (decreasing) the speed of the front and rear motors while decreasing (increasing) the speed of the lateral motors. This should be done while keeping the total thrust constant. In view of its configuration, the four-rotor rotorcraft in Fig. 1 has some similarities with planar vertical take off and landing (PVTOL) aircraft problem [2], [6], [7]. The PVTOL is a mathematical model of a flying object that evolves in a vertical plane. It has three degrees of freedom ( , , and ) corresponding to its position and orientation in the plane. The PVTOL is composed of two independent thrusters that produce a force and a moment on the flying machine, see Fig. 2. The PVTOL is an underactuated system since it has three degrees of freedom and only two inputs"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002467_tsmc.2021.3050616-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002467_tsmc.2021.3050616-Figure6-1.png",
        "caption": "Fig. 6. Performance of co-simulation for tracking S path.",
        "texts": [
            " The corresponding demonstrations are set as below in the following. 1) In order to illustrate the control accuracy and flexibility of the proposed SMPF algorithm, we set up different tracking trajectories, such as straight lines, arcs, and obstacles for verification. 2) Likewise, an individual experiment of path tracking is studied to evaluate the strategy (SMPF) in practical engineering applications. 3) In contrast, the path tracking experiment using PID and SMPF at high speed is first presented, and SMC and SMPF at low speed are considered. Fig. 6 displays the s-shaped trajectory tracking result. In order to evaluate the control precision of the proposed algorithm, we first set up three narrow aisles, and the width is only 5 mm wider than the robot body. At the same time, a 7-m long straight trajectory and a arc with 5-m radius are designed. Besides, we set e(\u03b2) \u2208 [\u22123,+3] and e(\u03c9r) \u2208 [\u22120.5,+0.5]. Moreover, we load a 200-kg weight on the robot. Figs. 7\u20139 are the simulation results of the proposed SMPF algorithm. The tracking performance of mobile robot is excellent not only in the straight line section but also in the curve section"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure2.24-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure2.24-1.png",
        "caption": "Fig. 2.24: A model for the mechanical design of the loading station.",
        "texts": [
            " Although the robot is equipped with a proper suction cup end-effector for a stable grasp of kinescopes, a safe bed has been located in the working area in order to mainly avoid the broken pieces of rarely lost kinescopes to be split everywhere in the work-cell. The suction cup end-effector is provided with a depressor sensor so that the approaching motion to a kinescope can be easily controlled by the signal of this sensor. Chapter 2: Analysis of manipulations 67 Fundamentals of the Mechanics of Robotic Manipulation68 The loading station is composed of mechanisms, which work to free a loaded scales conveyor from the transmission chain, to stop it for a loading operation by the robot, and to let the conveyor go back to the transmission chain. Figure 2.24 gives a simplified model of those mechanisms. The blocking system is an electropneumatically-actuated crank-sliding mechanism whose crank is fork shaped in order to fix the scales conveyor. Although this simple efficient system, the scales conveyor oscillates since oscillations can be produced previously during the path on the conveyor line. After the robot manipulation a kinescope is placed on a suitable plane at the unloading station, Fig. 2.25. This plane is determined by four fingers of a lifting system that, after a centering operation, gives the kinescope to another eight-finger gripper"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001538_j.jmatprotec.2011.09.020-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001538_j.jmatprotec.2011.09.020-Figure3-1.png",
        "caption": "Fig. 3. SLM chamber model",
        "texts": [
            " Gas flow and manifold efficiency were characterised using a modelling technique. Dimensional measurements of the chamber were recorded and a 3D model was constructed in Pro/Engineer Wildfire B. Ferrar et al. / Journal of Materials Processing Technology 212 (2012) 355\u2013 364 357 Table 1 Builds completed for study. Build No. of builds Machine status for flow condition Reference eliver rmedi l gas fl 3 s G ( l d t i AF1-AF5 3 As d BF1-BF3 3 Inte CF1-CF3 3 Fina (PTC, USA). Particular attention was paid to the inlet rail dimenion. A CAD model (Fig. 3) was then used to generate a 3D mesh in AMBIT V.2.4.6 (Fluent Inc., USA). The mesh was exported to the Computational Fluid Dynamics CFD) software suite Fluent (Fluent Inc., USA), where the inlet, outet, walls and gas properties (argon, density = 1.449 kg/m3) were efined allowing an iterative numerical solution to the flow within he chamber to be obtained. This analysis was run until continuty and directional mass flow residuals (X, Y and Z directions) had ed Initial ate stage gas rail modification Intermediate ow rail design Completed converged to ensure stability of the results"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure2.12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure2.12-1.png",
        "caption": "Figure 2.12 The forces drawn to scale in a force polygon.",
        "texts": [
            " If we had chosen the directions of Fa and Fb in the opposite sense, we would have found that Fa and Fb were negative. A force F can be resolved graphically into two components Fa and Fb, with given lines of action a en b using the parallelogram rule in Figure 2.10a or the force polygon in Figure 2.10b. The graphical approach has the advantage that you can at once see in which directions the components Fa and Fb are working. Example A force F = 30 kN acts on the trestle in Figure 2.11a. This has to be resolved into the components Fa and Fb with lines of action a and b. Solution: In Figure 2.12, the force has been resolved using a force polygon. The force scale is 1 cm =\u0302 5 kN. By measuring, you find (in the force polygon) Fa and Fb have lengths 5.7 cm and 6.3 cm respectively, so that Fa \u2248 5.7 \u00d7 5 kN = 28.5 kN, Fb \u2248 6.3 \u00d7 5 kN = 31.5 kN. In Figure 2.11b forces Fa and Fb are shown as they act on the trestle. It is clear that force Fa is pulling on the beam while force Fb is pushing against it. Together, they exert the same load as the single force F . The forces Fa and Fb are statically equivalent (equal in an equilibrium consideration) to their resultant F . 30 Check: The magnitude of Fa and Fb can also be calculated from the force triangle by using the sine rule (see Figure 2.12): Fa sin 55\u25e6 = Fb sin 65\u25e6 = F sin 60\u25e6 so that: Fa = F \u00b7 sin 55\u25e6 sin 60\u25e6 = 28.4 kN, Fb = F \u00b7 sin 65\u25e6 sin 60\u25e6 = 31.4 kN. The values measured in Figure 2.12 correspond closely to the calculated values. If not all the forces are in the same plane, the analytical approach is generally simpler than the graphical approach. For example, for the components Rx , Ry and Rz of the resultant R: Rx = \u2211 Fx, Ry = \u2211 Fy, Rz = \u2211 Fz. The summation symbol means that all the forces exerted on the particle have to be added together. ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM 2 Statics of a Particle 31 For the magnitude R of R therefore R = \u221a R2 x + R2 y + R2 z . The direction of R is determined by the angles \u03b1x , \u03b1y and \u03b1z which R makes with respectively the x, y and z axis, see Figure 2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure14.25-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure14.25-1.png",
        "caption": "FIGURE 14.25. A three-dimensional view of SX\u0308 for different SZ and fn, to show the optimal curve.",
        "texts": [
            " If a suspension is optimized for one passenger, it is still near optimal when the number of passengers, and hence the body mass, is changed. Example 512 Application of the optimal chart. Select a desired value for the relative displacement as a traveling space (or a desired value for the maximum absolute acceleration), and find the associated values for \u03c9n and \u03be at the intersection of the associated vertical (or horizontal) line with the optimal curve. 14. Suspension Optimization 909 910 14. Suspension Optimization Example 513 F Three-dimensional view of the optimal curve. Figure 14.25 illustrates a 3D view of SX\u0308 for different SZ and fn, to show the optimal curve in 3D. Theoretically, we may show the surface by SX\u0308 = SX\u0308 (SZ , fn) (14.146) and therefore, the optimal curve can be shown by the condition OSX\u0308 \u00b7 e\u0302SZ = 0 (14.147) where e\u0302SZ is the unit vector along the SZ-axis and OSX\u0308 is the gradient of the surface SX\u0308 . Example 514 F Suspension trade-off and trivial optimization. Reduction of the absolute acceleration is the main goal in the optimization of suspensions, because it represents the transmitted force to the body"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001195_j.msea.2013.06.055-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001195_j.msea.2013.06.055-Figure4-1.png",
        "caption": "Fig. 4. 3D view of the tensile test specimens; A vertical build-up; B horizontal build-up.",
        "texts": [
            " A shielding gas with an outlet pressure of 15 mbar was fed into the SLM chamber, keeping the residual oxygen level below 0.5%. A series of cubic samples with dimensions of 5 5 5 mmwere first produced (Fig. 3) to determine the influence that the main experimental SLM parameters might have on the density and structure of the solidified material. Subsequently, tensile test specimens were fabricated with vertical (model A) and horizontal (model B) positions on the fabrication platform to assess their mechanical properties, as shown in Fig. 4. Samples were cut perpendicular to the incident direction of the laser beam and then polished following classical procedures. To study the microstructural features, samples were etched in a HF solution (8 ml HF and 5 ml HCl per 100 ml CH3OH) for metallographic examination. In addition, a weak 9% Nital reagent (9 ml HNO3 per 100 ml CH3OH) was used to determine the thermal influence zone. Microstructures were observed using a scanning electron microscope (JEOL JSM-5800LV, Japan) and an optical microscope (Nikon, Japan)",
            " This phenomenon not only influences the dimensional accuracy of specimens but it may also obstruct the deposition of the next powder layer for processing. It is known that a low scanning velocity can lead to a high cooling rate [22] and that high cooling rates can cause larger thermal residual stresses. From previous simulation results [23], it appears that thermal residual stresses concentrate on both edges of the scanning track. In the case of fixed laser parameters, increasing the powder bed temperature can be considered an effective method for reducing deformations during the SLM process. Vertical placement (model A, Fig. 4) is the most commonly applied model in industrial layer processing technology because it allows for the placement of a large number of parts on the base plate, which is convenient for producing large quantities. Comparatively, horizontal placement (model B) encompasses fewer layers, and a shorter production time can be obtained for smaller amounts of parts. Model A and model B samples with preheating temperatures of 50 1C and 150 1C were tested to determine the potential differences in preheating with respect to the distance to the base plate"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001215_tsmc.2017.2697447-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001215_tsmc.2017.2697447-Figure1-1.png",
        "caption": "Fig. 1. Reference frames of AUV in the vertical plane.",
        "texts": [
            " To simplify the control law design, the motion of AUVs is usually considered as two independent parts, i.e., the motion in a vertical plane [58], [59] and the motion in a horizontal plane [58], [59]. According to [58] and [59], the motion of under-actuated AUVs in the vertical plane is described with kinematics \u23a7 \u23a8 \u23a9 x\u0307 = u cos \u03b8 + w sin \u03b8 z\u0307 = \u2212u sin \u03b8 + w cos \u03b8 \u03b8\u0307 = q (1) and kinetics \u23a7 \u23a8 \u23a9 muu\u0307 = fu(t, u, w, q) + \u03c4u + \u03c4du mww\u0307 = fw(t, u, w, q) + 0 + \u03c4dw mqq\u0307 = fq(t, u, w, q) + \u03c4q + \u03c4dq (2) where x and z denote the position and depth in XE \u2212 ZE coordinate as shown in Fig. 1; u \u2208 u and w \u2208 w denote the surge and heave velocities in XB \u2212 ZB coordinate where u = {u|umin \u2264 u \u2264 umax} and w = {w|wmin \u2264 w \u2264 wmax} with umax, umin, wmax, and wmin being constants; \u03b8 and q \u2208 r are pitch angle and pitch velocity where q = {q|qmin \u2264 q \u2264 qmax} with qmin and qmax being constants; \u03c4du, \u03c4dw, and \u03c4dq denote environmental forces; \u03c4u and \u03c4q are control inputs; fu(t, u, w, q), fw(t, u, w, q), and fq(t, u, w, q) represent hydrodynamic damping forces which are unknown; mu, mw, and mq are mass including hydrodynamic added mass"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure2.17-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure2.17-1.png",
        "caption": "Fig. 2.17: A scheme for conventional manual manufacturing of 3D filament winding of an ACM plait of rovings and a mechanical design for the corresponding winding support.",
        "texts": [
            " The human manufacturing process has been analyzed in detail through visual inspection and video recording in order to understand the basic manipulation requirements. The video means can also help to optimize the actual manual manufacturing as well as to individuate the operations to be or not to be mimicked by a robot for a suitable automation. The material used in this manufacturing is an advanced composite material (ACM), consisting of carbon fibers impregnated into a thermosetting resin matrix. The conventional manufacturing is performed by a manual winding of a plait PS of roving on a suitable winding support WD as shown in Fig. 2.17. The labor of a human operator is repetitive with very poor technological and decisional requirements and contributions. Recurrent human operations can be recognized as a variable grasp from a firm to a slipping mode, and arm movements from pulling to wrapping a roving plait on the winding support. Specifically, basic operations can be modeled to be mimicked by a suitable robotic manipulator and they can be summarized as shown in Fig. 2.18 as: - slipping of the hand along the plait direction to perform a feeding of the plait during the deposition; - pulling, which consists of straining a plait to ensure straight-line and compact deposition; - slipping and pulling to provide a feeding and a straining of the plait after the wrapping on curved surfaces of the winding support; - rolling the plait about points A and B to wrap the plait on the curved surfaces of the winding support"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure3.13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure3.13-1.png",
        "caption": "FIGURE 3.13. A model of normal stress \u03c3z(x, y) in the tireprint area for a stationary tire.",
        "texts": [
            "20)Z AP \u03c4x(x, y) dA = 0 (3.21)Z AP \u03c4y(x, y) dA = 0 (3.22) 3.3.1 Static Tire, Normal Stress Figure 3.11 illustrates a stationary tire under a normal load Fz along with the generated normal stress \u03c3z applied on the ground. The applied loads on the tire are illustrated in the side view shown in Figure 3.12. For a stationary tire, the shape of normal stress \u03c3z(x, y) over the tireprint area depends on tire and load conditions, however its distribution over the tireprint is generally in the shape shown in Figure 3.13. 3. Tire Dynamics 105 106 3. Tire Dynamics The normal stress \u03c3z(x, y) may be approximated by the function \u03c3z(x, y) = \u03c3zM \u00b5 1\u2212 x6 a6 \u2212 y6 b6 \u00b6 (3.23) where a and b indicate the dimensions of the tireprint, as shown in Figure 3.14. The tireprints may approximately be modeled by a mathematical function x2n a2n + y2n b2n = 1 n \u2208 N. (3.24) For radial tires, n = 3 or n = 2 may be used, x6 a6 + y6 b6 = 1 (3.25) while for non-radial tires n = 1 is a better approximation. x2 a2 = y2 b2 = 1. (3.26) Example 79 Normal stress in tireprint"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003549_robot.1996.509185-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003549_robot.1996.509185-Figure2-1.png",
        "caption": "Fig. 2: Planar Force-Angle stability measure.",
        "texts": [
            " height is always desirable from a stability point-of-view, heaviness on the other hand is stabilizing at low velocities and destabilizing at high velocities. In this work we are concerned with low velocity systems possibly exerting large forces on the environment, hence, heaviness will be considered a stabilizing influence. 3 Force-Angle Stability Measure To help frame the discussion of the general form of the Force-Angle stability measure we first present a planar cxample which highlights its simple graphical nature. 3.1 Planar Example Shown in Figure 2 is a two contact point. planar system whose center-of-mass (c.m.) is subject to a net force f, which is the sum of all forces acting on the vehicle body except the supporting reaction forces (which do not contribute to a tipover motion instability). This force vector subtends two angles, B1 and 0 2 , with the two tip-over axis normals 11 and 12. The Force-Angle stability measure, cy, is given by the minimum of the two angles, weighted by the magnitude of the force vector (as denoted by lifril) for heaviness sensitivity: (1) Q = 81 ' llfrll Critical tipover stability occurs when 6' goes to zero (and therefore f, coincides with 11 or 12) or, when the magnitude off, goes to zero and even the smallest disturbance may topple the vehicle"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000946_j.addma.2014.10.003-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000946_j.addma.2014.10.003-Figure7-1.png",
        "caption": "Fig. 7. The simulated temperature distribution within each deposition when it is h T",
        "texts": [],
        "surrounding_texts": [
            "J.C. Heigel et al. / Additive Manufacturing 5 (2015) 9\u201319 13\nc i s a 1 s m\nc p\n%\nw b t t\n5\nc u m s i d o h c T u\nm T ( q d h s i r w t 2 t p\nr l i a T b ( v f e m h\nxture\u2019s aluminum clamp.\nlamp used to hold the substrate in the measurement fixture is ncluded in the mesh to account for the heat transfer from the subtrate into it. Each deposition layer is 1 element tall (0.203 mm) nd 2 elements wide, equating to each element being equal to /4 of the laser diameter. The model is mechanically constrained uch that it is cantilevered and allowed to deform in the same anner as the experiments. Errors between measurement and simulation results are calulated by comparing instances in time, or by calculating the ercent error over the deposition time:\nError = 100 \u2211n\ni=1|[(Tmeas)i \u2212 (Tnode)i]/(Tmeas)i| n\n(9)\nhere n is the total number of simulated time increments etween the beginning and end of the deposition, i is the current ime increment, Tnode is the simulated temperature, and Tmeas is he measured temperature.\n. Simulation cases\nt b s e v 2 f h t\nCase 1) generates the highest temperature, since the heat is input uickly and the mass of the deposition is relatively small. The eposition of a second wall with no dwell (Case 2) experiences igh temperatures in the wall, but lower temperatures in the ubstrate compared to the first wall in Case 1. This is due to the ncreased wall height that adds thermal mass to the part and also esists conduction into the substrate. Additionally, the increased all height provides a greater surface area for the convection o extract heat from the part. The single wall deposition with a 0 s dwell between each layer (Case 3) experiences the lowest emperatures because the dwell allows for more cooling of the art.\nFig. 8 presents the thermal measurements and simulation esults of the single wall deposition using no dwell between ayers (Case 1). The measurements made using TC 1, which s located on the center of the bottom surface of the substrate, re presented along with the corresponding simulation results. he measurement-based forced convection model results in the est thermal simulation results during the deposition process 0\u2013287 s), with a percent error of 2.4%, whereas the free conection model results in a percent error of 15.4%. It is clear rom these results that the free convection model does not allow nough heat transfer and thus the temperatures are too high. The easurement-based forced convection model results in greater eat transfer that produce more accurate thermal results.\nFig. 9(a) presents the thermal measurements and simula-\nion results of the deposition of the second wall using no dwell etween layers (Case 2). As the surface area increases from the ingle wall (Case 1) to the double wall (Case 2), the error from ach convection model increases. The error from the free conection model increases from 15.4% (Case 1) to 22.2% (Case ), indicating that as the part size increases, the assumption of ree convection leads to greater simulation error. On the other and, the error from the forced convection model also increases, hough by a lesser extent, from 2.1% (Case 1) to 4.1% (Case 2).",
            "14 J.C. Heigel et al. / Additive Manufacturing 5 (2015) 9\u201319\nT a\nm\n0 50 100 150 200 250 300 350 0\n100\n200\n300\n400\n500\n600\nProcess time (s)\nT em\npe ra\ntu re\n( \u00ba C\n)\nTC 1 measurement Forced conv. model Free conv. model\nF d c\ns t C m i a i w T t m u\nt d m\nalf complete resulting from the measurement-based forced convection model. he temperature scale is in \u25e6C.\nhis indicates that although the forced convection model is more ccurate than the free convection model, it can be improved.\nFig. 9(b) compares the simulation results to the measureents made on the wall using TC 3. It was not possible to w 4 f\nig. 9. The temperature history during the deposition of the second wall with no well between layers (Case 2). The dashed vertical line indicates the deposition onclusion.\nhield the thermocouple from the effect of convection because he thermocouple was welded on the wall close to the deposition. onsequently, the convection cools the thermocouple junction, aking it experience lower temperatures while the jets are flowng. The temperature measurement rises quickly when the jets re shut off at approximately 293 s. Despite this difference durng the depositions comparisons can be made at a time of 297 s, hich is after the point that the argon gas supply was shut off. he forced convection produces a percent error of 2.2% at this ime, while the percent error resulting from the free convection odel is 16.9%. This further demonstrates the improved results\nsing the measurement-based forced convection model. The greatest errors between the measurements and simula-\nions results occur during the single wall deposition with a 20 s well between each layer (Case 3), as shown in Fig. 10. The easurement-based convection model results in a 10.4% error,\nhile the free convection model produces a percent error of\n3.8%. The increased processing time during this case allows\nor a greater amount of heat to be evacuated through convection.",
            "J.C. Heigel et al. / Additive Manufacturing 5 (2015) 9\u201319 15\n0 500 1000 1500 0\n100\n200\n300\n400\n500\n600\nProcess time (s)\nT em\npe ra\ntu re\n(\u00b0 C\n) TC 2 meassurement\nForced conv. model Free conv. model\nFig. 10. The temperature history of the single wall deposition with a 20 s dwell between layers (Case 3). Measurements are made using TC 2, which is located o d\nb t t d t m t a d T t c t p\n6\nw t i s b\nt d m c t s\nF n T d\nn A t a l d c o s t\nc g g o a m t\nn the top surface of the substrate, and compared to the simulation results. The ashed vertical line indicates the deposition conclusion.\nIt has been demonstrated in each case that the measurementased forced convection produces superior results compared to he assumption of free convection; however, the forced convecion model can be improved. This is understandable since it is eveloped from the measurements made at three locations on wo different surfaces. One approach to develop an improved\nodel is to perform more detailed measurements to investigate he effect of specific part geometries on convection. Another pproach would be to use convection measurements to valiate CFD of the gas flow over the evolving deposition surface. he CFD approach would enable a broader range of deposiion geometries to be modeled without necessitating further onvection measurements. Whichever approach is used, convecion measurements of the specific deposition equipment must be erformed.\n.2. Deflection history\nFig. 11 shows the final simulated deformation of the single all deposition with no dwell between layers (Case 1). The disortion has been scaled up by a factor of 5 so that the deformation s evident. The deposition process bends the part upwards and hrinks the wall from its nominal size. Each case exhibits this ehavior.\nig. 11. The simulated final deformation of the single wall deposition with o dwell between layers (Case 1) resulting from the forced convection model. he deformation has been scaled by a factor of 5 to emphasize the substrate eformation.\nc\nt 2 fi d v fi t d B f t i t c\nine indicates the deposition conclusion.\numbered layers and the beginning of the odd numbered layers. s the laser moves away from the the clamp, that material con-\nracts as it cools, causing the substrate to deflect up, producing consistent net increase in deflection during the deposition of ayers 1\u201314 (0\u201362 s). The net deflection decreases during the eposition of layers 15\u201335 (62\u2013155 s). The remaining layers ause a net increase in deflection, though the amplitude of the scillation decreases as the wall height increases. Once the depoition concludes, indicated by the vertical dashed line at 287 s, he deflection increases rapidly until it reaches a steady state.\nThe measurement-based convection model produces the best orrelation with the experimental measurements in the sinle wall deposition (Case 1). The lower predicted deflection enerated from the free convection model is a consequence f the greater amount of instantaneous annealing and creep t the higher temperatures. These results demonstrate that a easurement-based convection model is required to produce hermal and deflection results when instantaneous annealing and reep occur.\nFig. 13 presents the measured and simulated deflection for he double wall deposition with no dwell between layers (Case ). The residual deflection in the part from the deposition of the rst wall in Case 1 has been subtracted from the results. The eflection measured during the deposition of the double wall is ery different from that measured during the deposition of the rst in Case 1. The measured deflection decreases rapidly as he wall experiences thermal expansion during the first several eposition layers, then increases after the sixth track (at 32 s). oth of the simulation cases capture this trend very well. The ree convection model produces more accurate results compared o the measurement-based forced convection model despite the nferior thermal results. This indicates that the material properies are not fully captured in the mechanical model. One possible ause for this is that the instantaneous annealing and creep are"
        ]
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure12.48-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure12.48-1.png",
        "caption": "FIGURE 12.48. A car with mass m on an oscillating platform hung from point A.",
        "texts": [
            "470) where \u03c9n is the natural frequency of the oscillation. \u03c9n can be assumed as the frequency of small oscillation about the point A when the platform is set free after a small deviation from equilibrium position. The natural period of oscillation Tn = 2\u03c0/\u03c9n is what we can measure, and therefore, the moment of inertia I0 is equal to I0 = 1 4\u03c02 Mgh1T 2 n . (12.471) The natural period Tn may be measured by an average period of a few cycles, or more accurately, by an accelerometer. Now consider the swing shown in Figure 12.48. A car with mass m at C is on the platform such that C is exactly above the mass center of the platform. Because the location of the mass center C is known, the distance between C and the fulcrum A is also known as h2. 12. Applied Vibrations 803 To find the car\u2019s pitch mass moment of inertia Iy about C, we apply the Euler equation about point A, when the oscillator is deviated from the equilibrium condition. X My = IA \u03b8\u0308 (12.472) \u2212Mgh1 sin \u03b8 \u2212mgh2 sin \u03b8 = I0 + Iy +mh22 (12.473) Assuming very small oscillation, we may use sin \u03b8 \u2248 \u03b8 and then Equation (12"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001609_j.ijmecsci.2010.05.005-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001609_j.ijmecsci.2010.05.005-Figure1-1.png",
        "caption": "Fig. 1. Rolling elements replaced by spring and dash-pot.",
        "texts": [
            " To study the rolling element bearing structural vibration characteristics, the rolling element\u2013raceway contact can be considered as a spring mass system, in which the outer race is fixed in a rigid support and the inner race is fixed rigidly with the motor shaft. Elastic deformation between raceways and rolling elements produces a non-linear phenomenon between force and deformation, which is obtained by the Hertzian theory. The rolling element bearing is considered as non-linear contact spring as shown in Fig. 1. In the model, the outer race of the bearing is fixed in a rigid support and inner race is held rigidly on the shaft. A constant radial load acts on the bearing. The contact force is calculated using the Hertzian contact deformation theory. According to the Hertzian contact deformation theory, the non-linear relation load\u2013deformation is given by [10] F \u00bc Kdn r \u00f01\u00de where K is the load\u2013deflection factor or constant for Hertzian contact elastic deformation, dr the radial deflection or contact deformation and n the load\u2013deflection exponent; n\u00bc3/2 for ball bearing and 10/9 for roller bearing"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure13.68-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure13.68-1.png",
        "caption": "Figure 13.68 Statically indeterminate portal structure with the support reactions at A and C.",
        "texts": [
            " At A and B, there are acting support reactions and (components of the) compressive forces exerted by members AD and BD. Check: By reducing the support reactions (pointed upwards) at A and B by the vertical component (pointed downwards) of these member forces we 600 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM find the same shear forces as from the V diagram. The horizontal component of the compressive member forces at A and B results in a tensile force of 80/3 kN in ACB, see the N diagram in Figure 13.66b. With the load given, the support reactions at A and C for the structure are given in Figure 13.68. Questions: a. Determine the degree of static indeterminacy of the structure. b. Determine the support reactions at B. c. Isolate ADEBH, and draw all the forces acting on it, with the additional information of a compressive force of 6 kN acting in EG. d. For ADE and BH, draw the M , V and N diagrams with the deformation symbols. At D and E, draw the tangents to the M diagram. Solution (units kN and m): a. The structure consists of two singly-cohesive sub-structures (see Figure 13.69). There are v = 2 + 2 = 4 unknown joining forces acting in the hinged joints between both sub-structures"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure10.13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure10.13-1.png",
        "caption": "FIGURE 10.13. A two-wheel vehicle and its force system, including the steer moment reactions.",
        "texts": [
            " However, most of the rear steering construction vehicles work at a big steer angle. Therefore, these equations cannot predict the behavior of the construction vehicles very well. Example 387 F Better model for two-wheel vehicles. Because of the steer angle, a reaction moment appears at the tireprints of the front and rear wheels, which act as external moments M1 and M2 on the wheels. So the total steering reaction moment on the front and rear wheels are M1 \u2248 2D\u03b4fMz (10.227) M2 \u2248 2D\u03b4rMz (10.228) where D\u03b4f = dMz d\u03b4f (10.229) D\u03b4r = dMz d\u03b4r . (10.230) Figure 10.13 illustrates a two-wheel vehicle model. The force system on the vehicle is Fx \u2248 Fxf + Fxr (10.231) Fy \u2248 Fyf + Fyr (10.232) Mz \u2248 a1Fyf \u2212 a2Fyr +M1 +M2. (10.233) 618 10. Vehicle Planar Dynamics Example 388 F The race care 180 deg quick turn from reverse. You have seen that race car drivers can turn a 180 deg quickly when the car is moving backward. Here is how they do this. The driver moves backward when the car is in reverse gear. To make a fast 180 deg turn without stopping, the driver may follow these steps: 1\u2212 The driver should push the gas pedal to gain enough speed, 2\u2212 free the gas pedal and put the gear in neutral, 3\u2212 cut the steering wheel sharply around 90 deg, 4\u2212 change the gear to drive, and 5\u2212 push the gas pedal and return the steering wheel to 0 deg after the car has completed the 180 deg turn"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure9.8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure9.8-1.png",
        "caption": "Fig. 9.8. Hybrid electric gasoline car: 1 - gasoline combustion engine, 2 - integrated motor-generator, 3 - cranking clutch, 4 - gearbox, 5 - inverter, 6 - battery.",
        "texts": [
            " HEVs are now at the forefront of transportation technology development. HEVs combine the internal combustion engine of a conventional vehicle with the electric motor of an EV, resulting in twice the fuel economy of conventional vehicles. The electric motor/generator is usually located between the combustion engine and gear box. One end of the rotor shaft of the electric motor/generator is bolted to the combustion engine crankshaft, while the opposite end can be bolted to the flywheel or gearbox via the clutch (Fig. 9.8). The electric motor serves a number of functions, i.e.: \u2022 assisting in vehicle propulsion when needed, allowing the use of a smaller internal combustion engine; 288 9 Applications \u2022 operating as a generator, allowing excess energy (during braking) to be used to recharge the battery; \u2022 replacing the conventional alternator, providing energy that ultimately feeds the conventional low voltage, e.g. 12 V electrical system; \u2022 starting the internal combustion engine very quickly and quietly. This allows the internal combustion engine to be turned off when not needed, without any delay in restarting on demand; \u2022 damping crankshaft speed variations, leading to smoother idle. In a hybrid electric gasoline car the electric motor assists the gasoline engine in the low speed range by utilizing the high torque of the electric motor, as shown in Fig. 9.9. Currently manufactured hybrid electric gasoline cars 9.2 Electric Vehicles 289 (Fig. 9.8) are equipped either with cage induction motors or PM brushless motors. A PM brushless motor can increase the overall torque by more than 50%. In most applications, the rated power of electric motors is from 10 to 75 kW. Because of limited space between the combustion engine and gear box as well as the need to increase the flywheel effect, electric motors for HEVs are short and have large diameters. AFPM brushless machines are pancaketype high torque density motors and fit perfectly the HEV requirements"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000047_5.4400-Figure12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000047_5.4400-Figure12-1.png",
        "caption": "Fig. 12. Typical closed-loop system trajectories for the control (a) (73, and (b) (7.18).",
        "texts": [
            "20) Substituting (7.18) into (7.20) and manipulating produces i / = a J - + + J - f f g J - f - a J - aa aa aa aa at ax ax at (7.21) verifying the negative definiteness of V. This establishes attractiveness to the switching surface. A Comparison of the VSC and Deterministic Controllers The similarity of the structure of (7.7) and (7.18) i s clear in view of the uN term. The controllers, however, generate different responses with respect to different switching surfaces. This is best seen by viewing Fig. 12. Fig. 12(a) illustrates a typical trajectory of the closed loop system (7.2) driven by the controller (7.7). Notice that the control switches with respect to the surface BJV,V = 0 where V i s the Lyapunov function of the nominally stable system. In some cases for sufficiently large p(t, x), the response will behave as a VSC response in a vicinity of the origin. In fact, if p(t, x) i s sufficiently large the controller exhibits the standard VSC response once the trajectory intercepts the surface B 'V,V = 0 [50]. In Fig. 12(b)we have the usual VSC responsewith respect to the user-chosen switching surface a(t, x) = 0. In addition to stabilization of the closed loop system one may obtain tracking properties as implicitly built into the user-designed switching surface. Chattering The VSC controllers developed in Section VI and the uncertain system controllers of this section assure the desired behavior of the closed loop system. These controllers, however, requirean infinitely(in the ideal case) fast switching mechanism"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure1.12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure1.12-1.png",
        "caption": "FIGURE 1.12. A sample of tire tread to show lugs and voids.",
        "texts": [
            " The lugs are the sections of rubber that make contact with the road and voids are the spaces that are located between the lugs. Lugs are also called slots or blocks, and voids are also called grooves. The tire tread pattern of block-groove configurations affect the tire\u2019s traction and noise level. Wide and straight grooves running circumferentially have a lower noise level and high lateral friction. More lateral grooves running from side to side increase traction and noise levels. A sample of a tire tread is shown in Figure 1.12. Tires need both circumferential and lateral grooves. The water on the road is compressed into the grooves by the vehicle\u2019s weight and is evacuated from the tireprint region, providing better traction at the tireprint contact. Without such grooves, the water would not be able to escape out to the sides of the wheel. This would causes a thin layer of water to remain between the road and the tire, which causes a loss of friction with the road surface. Therefore, the grooves in the tread provide an escape path for water"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure12.23-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure12.23-1.png",
        "caption": "Figure 12.23 (a) Three-hinged portal frame with load and support reactions. (b) Normal force diagram. (c) Shear force diagram. (d) Bending moment diagram.",
        "texts": [
            " This also fixes the value at E. The V diagram can subsequently be calculated from the M diagram. We can change the order: first draw the V diagram and calculate the values at C, E and D from the areas of the V diagram. Note: The shear forces at A and B are not equal to the support reactions at A and B! Why not? 1 In Figure 12.19b, the \u201csteps\u201d in the M diagrams are no longer shown. 502 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Example 5 The support reactions for the three-hinged portal frame in Figure 12.23a were calculated in Section 5.3, Example 1. As there are no distributed loads, the normal forces and shear forces in each field are constant, and the bending moment varies linearly (rule 1). N and V diagrams To draw the N and V diagrams, we have to investigate the force equilibrium of the separate parts. The necessary calculations are left to the reader. The result is shown in Figures 12.23b and 12.23c. Due to the concentrated load at E, a step change of 60 kN occurs in the V diagram (rule 14). To draw the M diagram, we have to know only the bending moments at C and D",
            " The M diagram for AC and BD varies linearly, from 0 to 40 and 80 kNm respectively. The M diagram for CSD consists of two straight lines that have a bend at the concentrated load at E. In addition, the M diagram passes through hinge S where the bending moment is zero. We therefore have to draw a straight line from the value of 40 kNm at C, through S, up to 20 kN on the opposite side at E. From there, we continue with a straight line to the value of 80 kNm at D. The M diagram is shown for the entire three-hinged portal frame in Figure 12.23d. Check 1: We can read from the M diagram that the bending moment at the position of the point load is 20 kNm, with tension at the underside of the beam. This can be checked using the moment equilibrium of the isolated part BDE. Check 2: Note that the magnitude of the shear forces and the deformation symbols agree with the slopes of the M diagram. This relationship between the M and V diagram represents a simple and fast way of checking their correctness. Example 6 All the forces on the isolated members ACE and BDE for the structure in Figure 12"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000416_s0022-460x(88)80365-1-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000416_s0022-460x(88)80365-1-Figure4-1.png",
        "caption": "Figure 4. Typical models for gear dynamics: (a) torsional model; (b) torsional and translational model. I .. I. =mass moment of inertias of prime mover and load; 12\u2022 IJ =mass moment of inertias of gears; k\" = torsional stillness of shaft i; k\", =tooth mesh stiffness; k 2\u2022 kJ =stillnesses representing lateral nexibility due to shafts and bearings.",
        "texts": [
            " HOUSER two mutually perpendicular directions are considered. However, considering two coupled lateral vibrations of a gear shaft system makes the problem a rotor dynamics problem. Such models will be discussed in a subsequent section. Still, some of the models for studying the lateral vibrations in two directions will be mentioned in this group because of the difficulty in drawing absolute divisions between mathematical model t}'pes. Typical models used for torsional, and torsional and lateral vibrations of gears are shown in Figure 4. In an earlier work Johnson [83] used a receptance coupling technique to calculate the natural frequencies from the receptance equation obtained by first separately finding the receptances at the meshing point of each of a pair of geared shafts. In the model, the varying mesh stiffness was replaced by a constant stiffness equal to the mean value of the varying stiffness and thus, a linear system was obtained. His work was one of the first attempts at using a mesh stiffness in coupling the vibration of gear shafts"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure11.10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure11.10-1.png",
        "caption": "Figure 11.10 The same beam supported in three different ways, with the associated V and M diagrams: (a) simply supported at A and B, (b) fixed at A and (c) fixed at B.",
        "texts": [
            " The bending moment M is negative everywhere and is linear. The bending moment (in the absolute sense) has its maximum at the fixed end: |M|max = F . Example 2 In Figure 11.9, a uniformly distributed load q is acting normal to the beam axis over the entire length of beam AB. The method of support in A and/or B is given below. 11 Mathematical Description of the Relationship between Section Forces and Loading 445 Question: For the following three cases, determine the V and M diagrams using the differential equations for equilibrium (see Figure 11.10): (a) The beam is simply supported at A and B; (b) The beam has a fixed support at A and a roller support at B; (c) The beam has a fixed support at B and a roller support at A. Solution: In all three cases, with qz = q the following applies: d2M dx2 = \u2212q, dM dx = V = \u2212 \u222b q dx = \u2212qx + C1, M = \u222b V dx = \u222b (\u2212qx + C1) dx = \u2212 1 2qx2 + C1x + C2. Due to a uniformly distributed load, the shear force is linear, and the bending moment is quadratic (parabolic). The constants C1 and C2 are determined by the boundary conditions (associated with the type of support) at A and/or B",
            " VOLUME 1: EQUILIBRIUM \u2022 Elaboration of the boundary conditions in case (a): x = 0 : M = C2 = 0 \u21d2 C2 = 0, x = : M = \u2212 1 2q 2 + C1 + C2 = 0 \u21d2 C1 = 1 2q from which it follows that V = \u2212qx + 1 2q = 1 2q( \u2212 2x), M = \u2212 1 2qx2 + 1 2q x = 1 2qx( \u2212 x). \u2022 Elaboration of the boundary conditions in case (b): x = : V = \u2212q + C1 = 0 \u21d2 C1 = q , x = : M = \u2212 1 2q 2 + C1 + C2 = 0 \u21d2 C2 = \u2212 1 2q 2 V = \u2212qx + q = q( \u2212 x), M = \u2212 1 2qx2 + q x \u2212 1 2q 2 = \u2212 1 2q( \u2212 x)2. \u2022 Elaboration of the boundary conditions in case (c): x = 0 : V = C1 = 0 \u21d2 C1 = 0, x = 0 : M = C2 = 0 \u21d2 C2 = 0 from which it follows that V = \u2212qx, M = \u2212 1 2qx2. Figure 11.10 shows the V and M diagrams for all three cases. The tangents to the M diagram are also shown at A and B. These intersect in x = 1 2 , at mid-span. In the figure, an important variable p is shown: p = 1 8q 2. We will make use of p in Chapter 12. from which it follows that 11 Mathematical Description of the Relationship between Section Forces and Loading 447 Below we again show how, for two cases, the boundary conditions (end conditions) can be derived from the equilibrium of a small element with length x ( x \u2192 0) at the beam ends"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003933_robot.1989.99999-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003933_robot.1989.99999-Figure5-1.png",
        "caption": "Figure 5 : Geometry of a 4R Elbow-Like Manipulator",
        "texts": [
            " M; can be parametrized by a set of P independent parameters, $ = { $ I , . . . ,gr}, which can be thought of as a natural coordinate system for the self-motions. For given x, the parametrization is generally unique (up to isomorphism), but the parametrization can vary as x varies in the workspace. The self-motion manifolds are best illustrated using two examples: a planar 3R manipulator (Figure 4) which is redundant with respect to the position of its end-effector, and a 4R regional manipulator which is similar to an LLelbown manipulator (Figure 5) . The self-motion manifolds of the planar manipulator can be computed as follows. Let g, which is the orientation of the third link relative to a fixed reference system, be the parameter describing the internal motion of the manipulator. There are two possible sets of joint angles, Pa = {ela,e2a,e3a} and = (816,82b,&b}, which place the end-effector a t (Xee , yec ) with given (I. These solutions can be determined by evaluating the following equations: where 11,12,13 are the lengths of links 1, 2, and 3",
            " The two distinct self-motion manifolds in the preimage of point 1 physically corresponds to \u201cup elbow\u201d and \u201cdown elbow\u201d self-motions which are an analogous generalization of the \u201cup elbow\u201d and \u201cdown elbow\u201d configurations of a non-redundant two-link manipulator. Self-motions can be thought of as a natural generalization of the nonredundant manipulator concept of \u201cpose\u201d to redundant manipulators. In both cases, the self-motion manifolds are diffeomorphic2 to a circle. However, the preimage of point 1 contains a 2 x joint rotation while the other preimage does not. Now consider the 4R manipulator in Figure 5 . The kinematic parameters of this arm (following the conventions in [9]) are: a4 = I ; dl = dz = d3 = d4 = 0. Let the self-motion parameter, 4, be the angle between the plane containing the third A smooth map f : X -+ Y (where X and Y are manifolds) is a diffeomorphism ifit is one-to-one and onto, and f-\u2019: Y - X is smooth. X and Y are diffeomorphic if such an f exists CUO = 0; Q1 = 0 2 = X / 2 ; a 3 = - X / 2 ; a0 = a1 = U2 = 0; U j = 266 - ,Point 1 Point 2 - pc Preimage Figure 4: Self-Motions of Planar 3R Manipulator Point 2 \u2018i and fourth links and the vertical plane passing through joint axis 1",
            " This phenomenon can be demonstrated by consider- A fibre bundle is a four-tuple, ( = (2, X , F,p) , consisting o b total space z! a base space X, a fibre F, and a bundle projection p : X -+ X such that there exists an open covering {U} of X , and for each U 6 {U}, a homeomorphism 4: U x F - p-\u2019(U) such that the composition of maps, U x F -4 p - \u2018 ( U ) +P U , is a projection to the first factor, U. 270 ing a 7R arm which is formed by adding a 3 R wrist (with finite hand length) to the 4R manipulator of Figure 5 . In most parts of the workspace, the self-motion consists of a rotation of the plane containing links 3 and 4 about the line between the shoulder and the wrist. However, when axis 7 aligns with link 4, axes 1, 2, 3, and 4 are locked, and the selfmotion manifests itself as a self-motion of the wrist joints. This self-motion is physically very different from the nominal self-motion, and is separated in C from the nominal selfmotions by a coregular surface. By analyzing the singularities (a method is given in [a] for synthesizing the singularities of arbitrary n R ) , it is possible to predict where the physical nature of the self-motion may drastically change"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002201_j.addma.2014.09.004-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002201_j.addma.2014.09.004-Figure4-1.png",
        "caption": "Fig. 4. Fini",
        "texts": [
            " Material was added ahead of the heat source at each tep at specified time increments to simulate the travel velocty as shown in Fig. 3, and for all simulations in this study a alue of \u03c6 = 0.77 is used. Because the deposition takes place in vacuum, the 3-D models do not include convective heat transer on their vertical and top surfaces. The nodes have an initial emperature of T0 = 373 K and a constant temperature of 373 K pecified at the base of the model, which is similar to the temeratures in the EBF3 process. Eight-node, linear brick elements ere used throughout the model. As shown in Fig. 4, the mesh s biased toward the top surface and was refined in the region here data were extracted, in order to reduce computation time hile providing sufficient element density to resolve solidifica- ion cooling rates and thermal gradients. Studies were conducted o insure convergence of the thermal gradients and cooling rates xtracted from the simulations as a function of element density n and around the melt pool. Element sizes in and around the elt pool were scaled with melt pool size to maintain a suffi- ient number of elements across all simulations"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003542_iros.2006.282250-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003542_iros.2006.282250-Figure1-1.png",
        "caption": "Fig. 1. Stereo camera mounted at the top of the DLR Light-Weight Robot 3.",
        "texts": [
            ". INTRODUCTION This work concerns the use of visual sensors at the endeffector of robot manipulators/arms. In particular, hand-eye calibration consists of identifying the unknown position and orientation (pose) of the camera frame SC with respect to (w.r.t.) the robot end-effector frame (also known as hand or Tool Center Point TCP frame) ST , when the camera is rigidly mounted on the robot hand \u2013 see Fig. 1. There is a strong need for an accurate hand-eye calibration. The reasons are twofold: i) to map sensor-centered measurements into the robot/world frame and ii) to allow for an accurate prediction of the pose of the sensor on the basis of the arm motion \u2013 in fact these are often complementary aspects of the same problem. When performing hand-eye calibration on the basis of both the pose of ST w.r.t. the robot base frame SB bT t, and the pose of SC w.r.t. the world/object frame S0 0T c, there are two main approaches in order to estimate the hand-eye transformation: 1) Move the hand and observe/perceive the movement of the eye: or AX = XB, where A is the robot TCP motion t1T t2, B the induced camera motion c1T c2, and X is the hand-eye transformation tT c to be determined"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure9.21-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure9.21-1.png",
        "caption": "Fig. 9.21. Novel oil beam pump with AFPM brushless motor: 1 - horse head, 2 - walking beam, 3 - counterweights, 4 - Pitmans arm, 5 - disc type motor 6 - Samson beams, 7 - casing, 8- flow line, 9 - polishing rod.",
        "texts": [
            " An electric motor drives one end of a heavy beam via pair of cranks with counterweights and transmission. The other end of the beam is equipped with the so called horse head with attached vertical cable. The cable is connected to a polished rod of a pump at the bottom of the well. The cranks move the beam up and down and 9.6 Oil Beam Pumps 303 actuates the pump. Typically, a beam pump delivers 5 to 40 liters of a crude oil-water mixture per each stroke. 304 9 Applications 9.7 Elevators 305 The novel beam oil pump (Fig. 9.21) uses a low speed AFPM brushless motor, which replaces an induction motor, sheaves, belts and transmission [130]. There is a counterweight on the motor side of the beam that has an approximate weight as the horse head. Applications of an AFPM brushless motors simplifies the construction and maintenance of a beam pump, reduces its cost and improves the overall efficiency of the machine. Longitudinal section of the prototype of AFPM brushless motor is shown in Fig. 9.22 and specifications are given in Table 9"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.9-1.png",
        "caption": "FIGURE 8.9. Double triangle suspension mechanism.",
        "texts": [
            " High spring rate is a problem of leaf springs. Reducing their stiffness by narrowing them and using fewer leaves, reduces the lateral stiffness and increases the directional stability of the suspension significantly. A Panhard arm is a bar that attaches a solid axle suspension to the chassis laterally. Figure 8.7 illustrates a solid axle and a Panhard arm to guide the axle. Figure 8.8 shows a triangular linkage and a Panhard arm combination for guiding a solid axle. A double triangle mechanism, as shown in Figure 8.9, is an alternative design to guide the axle and support it laterally. Example 305 Straight line linkages. There are many mechanisms that can provide a straight line motion. The simplest mechanisms are four-bar linkages with a coupler point moving straight. Some of the most applied and famous linkages are shown in Figure 8.10. By having proper lengths, the Watt, Robert, Chebyshev, and Evance linkages can make the coupler point C move on a straight line vertically. Such a mechanism and straight motion may be used to guide a solid axle"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure9.73-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure9.73-1.png",
        "caption": "Figure 9.73 (a) If two members meet in an unloaded joint and the line of action of the load is in a direct line with one of the members, the other member is a zero-force member. (b) The forces acting on isolated joint C.",
        "texts": [
            " VOLUME 1: EQUILIBRIUM There are three situations of frequent occurrence in which zero-force members can be easily recognised: 1. If only two members meet in an unloaded joint, both are zero-force members (see Figure 9.71). 2. If three members meet in an unloaded joint of which two are in a direct line with one another, then the third is a zero-force member (see Figure 9.72). 3. If two members meet in an unloaded joint and the line of action of the load coincides with one of the members, the other member is a zero-force member (see Figure 9.73). These three rules are the direct consequence of the joint equilibrium, as shown below for each of the cases. Rule 1. Two members meet in unloaded joint A in Figure 9.71. The force in one of the members has a component normal to the direction of the other member. If we write down the equilibrium of joint A in the given (local) xy coordinate system, we find \u2211 Fx = N1 + N2 cos \u03b1 = 0,\u2211 Fy = N2 sin \u03b1 = 0 with the solution (because sin \u03b1 = 0):1 N1 = N2 = 0. Equilibrium is possible only if both member forces are zero",
            " The force in member 2 has a component normal to members 1 and 3. There can be equilibrium only if this component is zero, or in other words, if N2 = 0. If we write down the equilibrium of joint B in the given (local) xy coordinate system, we find \u2211 Fx = N1 + N2 cos \u03b1 \u2212 N3 = 0,\u2211 Fy = N2 sin \u03b1 = 0 so that N2 = 0 and N1 = N3. In addition to the fact that member 2 is a zero-force member, the normal forces in the continuous members 1 and 3, which are in a direct line with one another, are equal. Rule 3. The situation in Figure 9.73 is clearly similar to that in Figure 9.71. The equations for the equilibrium of joint C are \u2211 Fx = N1 + N2 cos \u03b1 \u2212 F = 0,\u2211 Fy = N2 sin \u03b1 = 0 so that N2 = 0 and N2 = F. By using rules 1 to 3 to determine the zero-force members first, you can often shorten the required calculation. A fourth rule with which we can shorten the calculation relates to an unloaded joint, in which four members meet and in pairs are in a direct line with one another. This situation is 366 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM shown in Figure 9"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000248_j.jmps.2005.03.007-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000248_j.jmps.2005.03.007-Figure4-1.png",
        "caption": "Fig. 4. Sketches of the total energy U as a function of the wrinkle amplitude A. The function has the form U \u00bc U0 \u00fe a\u00f0hE\u0304f jN0 11j\u00deA 2 \u00fe bA4, where a and b are positive and independent of A. (a) For a small membrane force, jN0 11johE\u0304f , the energy minimizes at the state A \u00bc 0, and the film is flat. (b) For a large membrane force jN0 11j4hE\u0304f , the flat film corresponds to a local energy maximum, and the film wrinkles to an equilibrium amplitude Aeq.",
        "texts": [
            " The thicker the substrate, the smaller the normal stress needed to deform the top surface by a certain amount. We will comment on the effect of Poisson\u2019s ratio ns later.The energy in the substrate is U s \u00bc k 2p Z 2p=k 0 1 2 T3wdx1 \u00bc E\u0304s 4 gkA2. (11) The sum of the above three contributions gives the total energy of the film/ substrate system: U\u00f0A; k\u00de \u00bc U0 \u00fe 1 4 \u00f0hE\u0304f jN0 11j\u00dek 2A2 \u00fe hE\u0304 32 k4A4, (12) with f \u00bc \u00f0kh\u00de2 12 \u00fe gE\u0304s \u00f0kh\u00deE\u0304 . (13) Similar to a debonded film (Audoly, 2000), the energy for the bonded film and substrate is a fourth order polynomial of A. Fig. 4 sketches this energy as a function of the amplitude of the wrinkles. When jN0 11johE\u0304f , the energy minimizes at the state A \u00bc 0, and the film is flat. When jN0 11j4hE\u0304f , the flat film corresponds to a local energy maximum, and the energy minimizes when the film wrinkles with the amplitude A \u00bc 2 k ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi jN0 11j E\u0304h f s . (14) Inserting (14) into (12), we express the energy as a function of the wave number U\u00f0k\u00de \u00bc U0 hE\u0304 2 jN0 11j hE\u0304 f 2 . (15) The equilibrium wrinkles select the wave number keq that minimizes the function U\u00f0k\u00de or, equivalently, minimizes the function f \u00f0kh\u00de"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001394_tro.2010.2043757-Figure11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001394_tro.2010.2043757-Figure11-1.png",
        "caption": "Fig. 11. Virtual-linkage model in bipedal stance. (a) Top view of the robot. (b) Virtual-linkage model involving two spherical joints and one prismatic joint. The point P is the ZMP and the points C are the contact CoP\u2019s. As discussed earlier, the ZMP lies on the line connecting the two CoP\u2019s. The black and white circles correspond to the CoMs of the feet links. To simplify the control of CoP\u2019s, we choose them to lie in a line parallel to the segment connecting the feet CoM\u2019s. (c) Six internal forces and moments that are independently controllable given ZMP trajectories. This corresponds to one tension force, two normal moments, and three out of four coordinates of the CoP\u2019s.",
        "texts": [
            " Therefore, a unified force/position operational task is created to control the right hand. Postural behavior is controlled by optimizing a criterion that minimizes the distance with respect to a human prerecorded posture using a method similar to the one described in [7]. During the multicontact phases (i.e., three or more supporting contacts) virtual-linkage models similar to that shown in Fig. 4 are implemented. During the biped phase, the special case of virtual-linkage model, as discussed in Fig. 11 in the Appendix, is used. Details on the algebraic expressions of the matrices involved in the virtual-linkage models are also given in the Appendix. The first two rows of the accompanying data correspond to the phase with three supporting contacts, where the right hand is controlled to clean the panel. In this phase, the horizontal position of the CoM is commanded to remain fixed. We display data on the tension forces occurring on the virtual-linkage model assigned between the left hand and the right foot, the moments about the CoP on the left hand, the normal force on the panel, and the sagittal and vertical trajectories of the CoM",
            " Equation (13) defines the relationship between CoM and internal forces and moments with respect to reaction forces. The six CoM forces and moments determine the ZMP\u2019s behavior and, therefore, impose the dependencies with respect to the CoP\u2019s, thus limiting the number of internal moments that can be controlled. Therefore, we define a different virtual-linkage model for bipedal coplanar contacts involving the tension between the two contacts, the two normal moments of the contacts, and only three out of four tangential moments (see Fig. 11). The reduced set of internal forces and moments describing the virtual-linkage model for bipedal contact is, therefore, equal to Fint = ( fint mint ) =  ft mcop(rf )(2d) mcop(lf )(1d) mn(rf ) mn(lf )  \u03b5R6 . (82) The behavior of the CoM with respect to contact reaction forces, as described in (12), remains the same. The case of noncoplanar bipedal contact stance needs also a special virtual-linkage model, since there exist also six independent internal-force and moment quantities. However, in noncoplanar bipedal contact stance, the four coordinates describing the two CoP\u2019s can now lie outside of the line connecting the ZMP (as analyzed in Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002216_j.ijfatigue.2021.106317-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002216_j.ijfatigue.2021.106317-Figure7-1.png",
        "caption": "Fig. 7. An equivalent ellipsoid model of the larger \u2018pancake\u2019 defects: (a) the coordinate system for the ellipsoid model: z, corresponds to the axial loading direction of the sample; a, b and c are the three principal semi-axes of the ellipsoid (representing the defect) respectively; the angle, \u03b8, characterising axis c of the ellipsoid compared to the loading direction of a given specimen; (b) spatial configuration of defect at different orientation angles.",
        "texts": [
            " 3 that the larger (LOF) defects are essentially pancake shaped and mostly lie in the plane of the build. As a result, they project very different profiles in the plane perpendicular to the tensile axis for the V-HCF and H-HCF test-pieces (see Fig. 6 ). Z. Wu et al. International Journal of Fatigue 151 (2021) 106317 In order to quantify their orientation, each defect is modelled as an equivalent ellipsoid having principal semi axes a, b, c [24,31]. The coordinate system for the model is shown in Fig. 7(a). The angle, \u03b8, characterising the angle of the minimum semi-axis of the ellipsoid, c, with respect to the loading direction, is used to define the orientation angle of the defects. Fig. 7(b) shows the projected area of the same defect at different orientation angles. By definition, the projected area of the defect is the largest when \u03b8 = 0\u25e6 and decreases with increasing \u03b8. The orientation angle \u03b8 as a function of defect size is plotted in Fig. 8 accordingly. A strong dependence of defect orientation angle \u03b8 on build Z. Wu et al. International Journal of Fatigue 151 (2021) 106317 direction can be observed. Defects of large size (e.g., >160 \u03bcm) clearly show preferred orientation tending towards \u03b8 =0\u25e6 for the V-HCF and \u03b8 =90\u25e6 for the H-HCF test-pieces",
            " The extreme defect size \u221aareaeff in the i-th sub-volume (i = 1, 2, 3, \u2026, n) can be expressed as \u221aareaeffi and \u221aareaeff1 \u2264 \u221aareaeff2 \u2264 \u22ef \u2264 \u221aareaeffn. The cumulative probability of the maximum defect \u221aareaeffi in the i-th sub-volume can be estimated using: G ( \u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305 area \u221a effi ) = i/(n + 1) = exp[ \u2212 exp ( \u2212 ( \u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305 area \u221a effi \u2212 \u03bb )/ \u03b1 ] (4) \u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305 area \u221a effi = \u03b1\u22c5 [ \u2212 ln ( \u2212 ln ( G ( \u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305\u0305 area \u221a effi ))] + \u03bb (5) where \u221aareaeff is the square root of the area of the equivalent ellipsoid perpendicular to the stress direction obtained from X-ray CT data (see Fig. 7). Here the largest defects \u221aareaeffi (i = 1, 2, 3, \u2026, 10) from ten sub-volumes (30 mm3) was extracted from X-ray tomograph of a V-HCF specimen and a H-HCF specimen respectively. The location and scale parameters, \u03bb and \u03b1 in Eq. (4), were determined using the least square method based on the linear relationship between (\u2212 ln(\u2212 ln G(\u221aareaeff))) and \u221aareaeff, as shown in Fig. 16(a). It is thus found that the maximum \u221aareaeff within the ten sub-volumes both from H-HCF and V-HCF specimens can be well described by the Gumbel distribution"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure10.7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure10.7-1.png",
        "caption": "FIGURE 10.7. A vehicle at time t = 0 and t = dt moving with a lateral velocity vy, a yaw rate r, and a forward velocity vx at a heading angle \u03c8.",
        "texts": [
            " When we find the translational and rotational velocities of a rigid vehicle, vx, vy, r, we may find the path of motion for the vehicle by integration. \u03c8 = \u03c80 + Z r dt (10.59) x = Z (vx cos\u03c8 \u2212 vy sin\u03c8) dt (10.60) y = Z (vx sin\u03c8 + vy cos\u03c8) dt (10.61) Example 380 F Equations of motion using principal method. The equations of motion for a rigid vehicle in a planar motion may also be found by principle of differential calculus. Consider a vehicle at time t = 0 that has a lateral velocity vy, a yaw rate r, and a forward velocity vx. The longitudinal x-axis makes angle \u03c8 with a fixed X-axis as shown in Figure 10.7. Point P (x, y) indicates a general point of the vehicle. The velocity components of point P are vPx = vx \u2212 y r (10.62) vPy = vy + x r (10.63) 10. Vehicle Planar Dynamics 595 because BvP = BvC + B G\u03c9B \u00d7 BrP = \u23a1\u23a3 vx vy 0 \u23a4\u23a6+ \u23a1\u23a3 0 0 r \u23a4\u23a6\u00d7 \u23a1\u23a3 x y 0 \u23a4\u23a6 . (10.64) After an increment of time, at t = dt, the vehicle has moved to a new position. The velocity components of point P at the second position are v0Px = (vx + dvx)\u2212 y (r + dr) (10.65) v0Py = (vy + dvy) + x (r + dr) . (10.66) However, vPx + dvPx = v0Px cos d\u03c8 \u2212 v0Py sin d\u03c8 (10"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000637_978-3-642-83006-8-Figure3.5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000637_978-3-642-83006-8-Figure3.5-1.png",
        "caption": "Fig. 3.5. Biped mechanism a) general spatial case, b) planar case (after [25])",
        "texts": [
            " Our further discussion will be con cerned with the results on the control of locomotion of anthropomor phic systems. We shall review first some of the results of Beletskii [25], whose sys tematic studies encompassed the control of planar models of locomotion mechanisms. The models were mostly simplified up to the level allowing analytic solutions, which led to detailed analyses and discussions of the results. 196 supported on massless legs. Such a system in three-dimensional space is shown in Fig. 3.5.a). Point C is the mass centre of the body (and, in this case, of thewhole system), while 0 is the point at which legs are connected to the torso. ~ ~ + The r c ' r o ' and rj are the radius vectors from the external coordinate frame origin to points C and 0, and to the supporting point of thej-th ~ leg (j=1,2) on the ground, respectively; p is distance vector from the ~ pOint 0 to C; Q is the body weight; and Rj (j=1,2) is the reaction for+ 2 ~ ce of the supporting leg. In general, R L R. corresponds to the sup~ j=1 J port phase; only one R"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001457_j.phpro.2010.08.083-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001457_j.phpro.2010.08.083-Figure8-1.png",
        "caption": "Fig. 8. 3D models and functional prototypes with a complex geometry fabricated by SLM: (a) light-weight model with complex inner structure and two spiral cooling channels from SS grade 316L powder; (b) mixing element Kenics with diameter 3 mm and height 15 mm and (c) sample with developed structure (pin diameter is 400 \u03bcm, height is 1 mm) from SS grade 904L powder; (d) part of combustion engine from Inconel 625 powder",
        "texts": [
            " The \u00abhorizontal\u00bb ones broke at the plane perpendicular to the laser scanning direction [30]. The analysis of the mechanical properties of the samples manufactured by SLM technique from Inconel 625, SS grade 316L and Co212-F powder has shown their excellent mechanical strength equivalent to the wrought materials. Based on the obtained results, several 3D models and functional prototypes with a complex geometry were fabricated from metal powders for different industrial applications: from aerospace and automotive to biomedical and chemical (Fig. 8). The analysis of the formation of single tracks from metal powders by SLM showed that the process has a threshold character: there are \u201cstability zones\u201d where the laser melted track is continuous and \u201cinstability zones\u201d 558 I. Yadroitsev, I. Smurov / Physics Procedia 5 (2010) 551\u2013560 Author name / Physics Procedia 00 (2010) 000\u2013000 where the tracks are not continuous. Instabilities appear at low scanning speed in the form of distortions and irregularities, and, on the contrary, excessively high speed gives rise to the balling effect"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure10.12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure10.12-1.png",
        "caption": "Figure 10.12 The stresses in the cross-section reflect the interaction through the area A ( A \u2192 0), of the right-hand part on the left-hand part, and vice versa. The normal stress \u03c3xx acts normal to the cross-sectional plane; the shear stresses \u03c3xy and \u03c3xz act in cross-sectional plane.",
        "texts": [
            " The stress vector is defined in a particular point and for a particular sectional plane. If we want to indicate the force transfer (interaction) at a point of the cross-section, the stress vector p alone is not enough, as we also have to indicate the status of the sectional plane that is considered. This is done by means of the unit normal vector n on that plane. To describe the action of the forces that the matter to the right of the section exerts on the matter to the left, and vice versa, we introduce the following quantities, which are known as cross-sectional stresses (see Figure 10.12): \u03c3xx = lim A\u21920 Fx A \u00b7 nx , \u03c3xy = lim A\u21920 Fy A \u00b7 nx , \u03c3xz = lim A\u21920 Fz A \u00b7 nx . Here, nx is the x component of the unit normal vector n on the sectional plane that is considered. 394 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM The kernel symbol \u03c3 for stress has two sub-indices. The first index relates to the normal of the plane on which the stress is acting; the second index relates to the direction of the stress (that is, the direction of the corresponding force component on that plane). If we look at two corresponding equal areas A to the right and to the left of the section, they are subject to two equal and opposite forces F ",
            "13 shows the positive stresses acting on the sides of an (infinitesimally) small rectangular block. The block is bounded by six planes, of which three are positive and three are negative. \u03c3ij \u2022 on a small area with the unit normal vector parallel to the i axis (1st index), \u2022 due to a force component parallel to the j axis (2nd index). The stress \u03c3ij is a normal stress when the indices are the same (i = j ) and a shear stress when the indices are different (i = j ). In a member cross-section, there are only normal stresses \u03c3xx and shear stresses \u03c3xy and \u03c3xz (see Figure 10.12). These stresses are as yet unknown functions of y and z, so that \u03c3xx = \u03c3xx(y, z), \u03c3xy = \u03c3xy(y, z) and \u03c3xz = \u03c3xz(y, z). 1 The stresses \u03c3ij (i, j = x, y, z) are the components of a quantity (the so-called stress tensor) that in a certain point for each arbitrary plane links the components of the stress vector p and the components of the unit normal vector n. is the stress 396 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM The resultant of the normal stresses on a small area A around a point P is a small force N : N = \u03c3xx A"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001766_j.cirp.2019.05.004-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001766_j.cirp.2019.05.004-Figure4-1.png",
        "caption": "Fig. 4. Early wire fed weld deposition for creation of pressure rollers [300].",
        "texts": [
            " An example of a part built up by a \u201ccut and stack\u201d method is shown in Fig. 2. Nakagawa at Tokyo University advanced layered production of tooling to include blanking, press forming, and injection moulding tools [153,194,195]. The use of a moving weld head to build up objects was initially patented by Baker in 1925 [20]. Fig. 3 illustrates a sketch from the patent. Starting in the 1960s, a number of approaches arose for producing tooling using a moving weld head, often to build parts on a moving axis or rotating mandrel [39,40,84,300]. Fig. 4 shows an embodiment from 1964 for creation of pressure rollers using a wire-fed welding head. For precursor processes, the earliest is the 1972 disclosure by Ciraud in France [56]. As shown in Fig. 5, the process contains many elements of directed energy deposition, including metal powder, an energy source (laser, electron beam or plasma beam), a part built up from a base plate and even powder recycling. The Housholder patent, issued in 1981, describes laser sintering of powdered material [111], eight years before Deckard\u2019s patent [61]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure2.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure2.1-1.png",
        "caption": "Fig. 2.1 1. Kinematic scheme of the manipulator with three degrees of freedom",
        "texts": [
            " Since kinematic modelling is an inevitable step in modern robot control, in this chapter we will consider the main principles of manipulator kinematic model generation. In this section we will introduce some basic notations and definitions relevant to manipulator kinematics formulation. We will be concerned with the manipu lator structure, link, kinematic pair, kinematic chain, joint coordinates, world coordinates, direct and inverse kinematic problems and redundancy. Let us consider the manipulator shown in Fig. 2.1. It consists of n rigid bodies interconnected by joints. The joints are either revolute or prismatic 20 2 Manipulator Kinematic Model (sliding). Revolute joints enable rotational motion of one link with respect to the other (Figure 2.2). In prismatic joints the motion is translational (Figure 2.3). The mechanical structure of the mechanism depends on the type of the joints that are included in the given robot and the disposition of the joints. For example, the robot shown in Fig. 2.1 has six revolute degrees of freedom, with the second and the third always being in parallel (anthropomor phic manipulator). 2.2 Definitions 21 Different schematic representations have been introduced in order to de scribe manipulator configurations simply. Figure 2.4{aHd) shows several types of robots. Here, revolute joints are denoted by cylinders, while prismatic joints are represented by parallelepipeds. Manipulator link is a rigid body described by its kinematic and dynamic parameters. Figure 2",
            " The second advantage is that the analytical inverse kinematics gives all the solutions (for six degrees-of-freedom manipulators there are eight solutions). The solutions are exact and no numeric errors are accumulated. When deriving the inverse kinematic expressions, the manipulator singularities are easily detected, so they can be treated separately. Let us illustrate solving inverse kinematics analytically by a simple example. We will consider the three degrees-of-freedom manipulator shown in Fig.2.1l. Its direct kinematics has already been obtained, i.e. the elements of the transformation 'matrix \u00b0Tn = \u00b0T3 are expressed as functions of q1' q2 and q3 (Eqs. (2.12)-{2.19)). Since we are dealing with a three degrees-of-freedom manipulator in this example, we cannot control its orientation, but only position x, y, z in the reference frame. Equations (2.19) and (2.21) yield x= C1(a3C23 +a2C2) y=S1(a3 C23+ a2C2) z=a3 S23 +a2 S2 (2.44) (2.45) (2.46) By multiplying Eq. (2.44) by S 1 and Eq. (2.45) by C 1 and subtracting these two equations, we obtain or q1 =arctg ~+k1t x (2",
            " Then, knowing the orts of the hand coordinate frame in the reference system\" it is feasible to determine the position of the end of the third link in the reference frame. This yields the first three angles q b q2 and q3. Finally, knowing angles ql' q2' q3 the orientation of the third link is determined and at last angles q4' qs and q6\u00b7 It should be mentioned that the majority of industrial robots produced in the world have such a mechanical structure that the analytical inverse kinematics is obtainable (most of them are of the type shown in Fig. 2.1). However, some non-redundant robots as well as redundant manipulators require the ap plication of purely numerical methods in evaluating joint coordinates, given the manipulator hand position and orientation [7-9]. As we have seen, solving the inverse kinematics is equivalent to seeking for the roots of the set of non-linear equations (2.42), within the work space and the joint coordinates space. Therefore, any of the standard techniques known from numerical analysis may be applied. These procedures are general and do not depend on mechanism configuration",
            ") represents the essential par ameters of an algorithm (numerical complexity). Numerical efficiency of robot kinematics and dynamics depends on the number of joints and relative dis position of the joint axes, etc. We pointed out in Sect. 5.2 that the floating-point arithmetics, including some trigonometric calculations and square roots, are carried out within specialized mathematical coprocessors. In order to have a clear insight into the numerical complexity of robot kinematics, let us consider two typical industrial robots: The first, shown in Fig. 2.1 (p. 19), has three degrees of freedom, while the second is a six degree-of freedom robot (Fig. 5.9). Both of them have kinematic chains for which the inverse kinematic solution in analytical form can be found. The number of numerical operations for both robots are given in Table 5.1 (see p. 173). We see 172 5 Microprocessor Implementation of Control Algorithms that the complexity is practically the same for the direct and inverse kinematic problem. Either the direct or inverse kinematic problem for this six-degree-of freedom robot can be solved within 5 to 10 ms by the use of an up-to-date 16-bit microcomputer"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000386_j.measurement.2013.11.012-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000386_j.measurement.2013.11.012-Figure3-1.png",
        "caption": "Fig. 3. (a) A planetary gearbox test rig and (b) its schematic model.",
        "texts": [
            " [19,20] provided the fundamentals by developing the characteristic frequencies of all three kinds of planetary gearboxes, which have influence on understanding the problems connected with condition monitoring and diagnosis of planetary gearboxes. 2.3. Illustrations using a planetary gearbox test rig After introducing the behaviors and evaluating the characteristic frequencies of planetary gearboxes, we next use the vibration signals acquired from a test rig to illustrate the challenging issues in fault diagnosis of planetary gearboxes [21]. Fig. 3 displays the test rig and its schematic model. The test rig includes two gearboxes, a 3-hp motor for driving the gearboxes, and a magnetic brake for loading. The motor rotating speed is controlled by a speed controller. The load is provided by the magnetic brake and can be adjusted by a brake controller. As shown in Fig. 3b, there are two gearboxes in the test rig: a two-stage planetary gearbox and a two-stage fixed-axis gearbox. The two-stage planetary gearbox is the focus of our illustrations. In each stage of the planetary gearbox, an inner sun gear is surrounded by three or four rotating planet gears and a standstill outer ring gear. Torque is transmitted through the sun gear to the planet gears, which ride on a planetary carrier. The planetary carrier, in turn, transmits torque to the output shaft. A vibration signal measured by an accelerometer with a sampling frequency of 5120 Hz under the healthy condition of the test rig is illustrated in Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure3.26-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure3.26-1.png",
        "caption": "Figure 3.26 Resolving the force F graphically along three given lines of action; (a) line of action figure and (b) force polygon.",
        "texts": [
            " The line of action of ( F \u2212 Fa) passes through the intersection A of the lines of action of F and Fa. The line of action of ( Fb + Fc) passes through the intersection Sab of the lines of action b and c. Therefore, ASbc is the line of action of both ( F \u2212 Fa) and ( Fb + Fc). F at A can now be resolved into Fa with line of action a and ( Fb + Fc) with line of action ASbc. Subsequently ( Fb + Fc) at Sbc can be resolved into Fb and Fc. This is shown graphically in Figure 3.25b in a single force polygon. The order in which F is resolved is irrelevant. In Figure 3.26 F is first resolved at B into Fb and ( Fa + Fc) and subsequently ( Fa + Fc) at Sac, the intersection of the lines of action a and c, is resolved into Fa and Fc. The force polygon now has a different shape, as the forces were resolved in a different order, but the result is the same. The name Culmann1 is associated with this graphical method in the literature. Resolving F into three forces Fa, Fb and Fc along given lines of action a, b, and c, can of course also be done analytically. Of the many possible methods, the method below is based on Varignon\u2019s first moment theorem: 1 Karl Culmann (1821\u20131881), a German engineer, was involved in the design and construction of important railway bridges and was especially known for his graphical methods for calculating structures"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002683_j.jmatprotec.2020.117023-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002683_j.jmatprotec.2020.117023-Figure3-1.png",
        "caption": "Fig. 3. Schematic diagram: (a) the chessboard scanning strategy was used for the DED process, (b) the scanning strategy was used for tensile specimens in the horizontal building direction, and. (c) the final dimensions of the tensile test specimen.",
        "texts": [
            " Gas-atomized 316 L SS powder was applied as the starting material, and its chemical composition (wt. %) was 0.02C-0.5Si-1.2Mn-2.5Mo-13.0Ni-17.0Cr-Fe. Fig. 2(a) shows the 316 L SS powder and high-magnification electron micrographs of the powder morphology. The high-magnification image in the inset shows that the powder grains were spherical. Fig. 2(b) shows a schematic of the particle size distribution. The particle size was below 130 \u03bcm, and the average particle diameter was approximately 60 \u03bcm. Fig. 3(a) shows the chessboard scanning strategy with dimensions of 7 \u00d7 7 mm2 and a scan direction rotation of 90\u25e6 between two adjacent layers that was used for the DED process to balance the residual stress in the samples. During the deposition process, there were four layers in a cycle period. The red dotted arrow represents the beginning direction of each of these layers. Then, the blue region was scanned. Finally, the black regions were scanned until the layer was finished. Because the deposition layer appeared in the form of multi-layer and multi-pass lapping, the surface of the deposition layer was not smooth due to the existence of gas trapped in voids, an irregular undulant surface and unmelted powder particles",
            " The relative density of the 316 L SS specimens was determined by the Archimedes method, and further details of this test method were reported by Yang et al. (2018). The hardness was measured by a 430SVD Vickers microhardness tester. The final hardness value was determined from 20 indentations, and the varying distance between the two successive microhardness indentations was 0.15 mm. At least three measurements were taken for each position to ensure statistical robustness. Tensile tests were performed using subsize Y. Yang et al. Journal of Materials Processing Tech. 291 (2021) 117023 samples with a gauge size of 16 \u00d7 3 \u00d7 2 mm3. Fig. 3(b) presents a schematic for the scanning strategy for samples sectioned from the 316 L SS specimens in the horizontal building direction. Fig. 3(c) displays the final sizes of the tensile specimens based on the ISO 6892\u2212 1 standard. Notably, all samples were manufactured using the horizontal build direction, and that direction lay on the same horizontal line as the tensile loading axis. Moreover, the specimens were extracted from the middle layer of the part that underwent the hybrid DED and thermal milling process. Three samples were fabricated and tested under each test case to reduce the effect of measurement error. The fracture surfaces were investigated by SEM"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure3.27-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure3.27-1.png",
        "caption": "Figure 3.27 Analytically resolving force F along three given lines of action.",
        "texts": [
            " Resolving F into three forces Fa, Fb and Fc along given lines of action a, b, and c, can of course also be done analytically. Of the many possible methods, the method below is based on Varignon\u2019s first moment theorem: 1 Karl Culmann (1821\u20131881), a German engineer, was involved in the design and construction of important railway bridges and was especially known for his graphical methods for calculating structures. so that 70 the moment of F about an arbitrary point is equal to the sum of the moments of Fa, Fb and Fc about that same point. In Figure 3.27, the directions of the as yet unknown forces Fa, Fb, and Fc have been assumed. In addition, a coordinate system has been assumed in order to be able to indicate the sign of the moments (the direction of rotation). If the moment theorem is applied with respect to Sbc, the intersection of the lines of action of Fb and Fc, then these forces do not contribute to the sum of the moments, and one can determine Fa directly: \u2211 Ty |Sbc = \u2212F \u00b7p = \u2212Fa\u00b7ha \u21d2 Fa = p ha F. Note: The signs are related to the xz coordinate system shown. By applying the moment theorem in the same way with respect to Sac and Sab respectively, we also find Fa and Fc directly. Since the direction of rotation of F about Sab is opposite to that of Fc about Sab the value of Fc will be negative. This means that the force Fc works opposite to the direction assumed in Figure 3.27. The analytical approach can also be used for resolving a couple into three forces along given lines of action. Example The block in Figure 3.28a is subject to the three forces Fa, Fb and Fc, along given lines of action a, b and c. The resultant is the couple T with the direction shown in the figure. Question: Determine the three forces if T = 80 kNm. ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM 3 Statics of a Rigid Body 71 Solution: In Figure 3.28b an assumption was made with respect to the directions of the forces"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure2.14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure2.14-1.png",
        "caption": "Figure 2.14 A secured mast.",
        "texts": [
            " In the latter case, the third direction cosine is given by the other two as cos2 \u03b1x + cos2 \u03b1y + cos2 \u03b1z = 1. Working with forces in space is illustrated using two examples. The first example relates to resolving a force into its components (Section 2.2.1). The second example relates to compounding forces (Section 2.2.2). In order to be able to resolve a force into its x, y and z component, we first have to calculate the direction cosines. This is illustrated by means of an example. 32 Example In Figure 2.14, a mast is being kept upright by three ropes. Rope AC is subject to a tensile force of 35 kN. Question: Find the x, y and z component of force F that the rope exerts on the foundation block in C (see Figure 2.15). Note that here we use the formal vector notation. Solution: The force F that is working on the foundation block has the same direction as the vector CA (directed from C to A). This vector, which indicates the direction of F is hereby referred to as d ,1 see Figure 2.15. F and d have the same direction cosines, so that cos \u03b1x = Fx F = dx d , cos \u03b1y = Fy F = dy d , cos \u03b1z = Fz F = dz d "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001514_j.engfailanal.2015.06.004-Figure11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001514_j.engfailanal.2015.06.004-Figure11-1.png",
        "caption": "Fig. 11. Proposed crack propagation path approximation [7].",
        "texts": [
            " Their results show that the improved method calculates the TVMS more accurately than the traditional method does. On the foundation of Ref. [35], Ma et al. [43] proposed an improved mesh stiffness model for the cracked spur gear pair where two parabolic curves are used to simulate the crack propagation path and limiting line, respectively [42] (see Fig. 10). Based on the crack propagation paths determined by FRANC, Pandya and Parey [7] presented an improved TVMS calculation method, which adopts straight lines with different slopes rather than only one straight to simulate the predicted crack path (see Fig. 11), and compared the difference of the TVMS under two crack propagation paths: a straight line and multiple straight lines. Based on the a curved crack propagation path obtained using FRANC (see Fig. 12), Zhao et al. [44] proposed a potential energy method to calculate the TVMS of a cracked gear pair. Considering the combined effects of tooth crack and the plastic inclination, Shao and Chen [45] developed a tooth plastic inclination model for a spur gear pair by considering the cracked tooth as a variable cross-section cantilever beam (see Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure5.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure5.1-1.png",
        "caption": "Fig. 5.1. AFPM brushless e-TORQTM motor with coreless stator windings: (a) general view; (b) motor integrated with wheel of a solar powered car. Photo courtesy of Bodine Electric Company , Chicago, IL, U.S.A.",
        "texts": [
            " On the other hand, owing to the increased nonmagnetic air gap, such a machine uses more PM material than an equivalent machine with a ferromagnetic stator core. Typical coil shapes used in the winding of a coreless stator are shown in Figs 3.16 and 3.17. In this chapter AFPM brushless machines with coreless stator and conventional PM excitation, i.e. PM fixed to backing steel discs, will be discussed. Bodine Electric Company , Chicago, IL, U.S.A. manufactures 178-mm (7-inch) and 356-mm (14-inch) diameter e-TORQTM AFPM brushless motors with coreless stator windings and twin external PM rotors with steel backing discs (Fig. 5.1a). The coreless stator design eliminates the so called cogging (detent) torque, improves low speed control, yields linear torque-current characteristics due to the absence of magnetic saturation and provides peak torque up to ten 154 5 AFPM Machines Without Stator Cores times the rated torque. Motors can run smoothly at extremely low speeds, even when powered by a standard solid state converter. In addition, the high peak torque capability can allow, in certain applications, the elimination of costly gearboxes and reduce the risk of lubricant leaks. The 356-mm diameter e-TORQTM motors have been used successfully by students of North Dakota State University for direct propulsion of a solar powered car participating in 2003 American Solar Challenge (Fig. 5.1b). A well designed solar-powered vehicle needs a very efficient and very light electric motor to convert the maximum amount of solar energy into mechanical energy at minimum rolling resistance. Coreless AFPM brushless motors satisfy these requirements. Small ironless motors may have printed circuit stator windings or film coil windings. The film coil stator winding has many coil layers while the printed circuit winding has one or two coil layers. Fig. 5.2 shows an ironless brushless motor with film coil stator winding manufactured by EmBest , Soeul, South Korea"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.25-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.25-1.png",
        "caption": "FIGURE 8.25. A semi-trailing arm suspension.",
        "texts": [
            " The variation in camber angle for a swing arm suspension is maximum, compared to the other suspension mechanisms. Example 312 Trailing arm suspension. Figure 8.24 illustrates a trailing arm suspension that is a longitudinal arm with a lateral axis of rotation. The camber angle of the wheel, supported by a trailing arm, will not change during the up and down motion. Trailing arm suspension has been successfully using in a variety of frontwheel-drive vehicles, to suspend their rear wheels. Example 313 Semi-trailing arm Semi-trailing arm suspension, as shown in Figure 8.25, is a compromise between the swing arm and trailing arm suspensions. The joint axis may have any angle, however an angle not too far from 45 deg is more applied. Such suspensions have acceptable camber angle change, while they can handle both, the lateral and longitudinal forces. Semi-trailing design has successfully applied to a series of rear-wheel-drive cars for several decades. Example 314 Antiroll bar and roll stiffness. Coil springs are used in vehicles because they are less stiff with better 470 8"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-FigureD.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-FigureD.3-1.png",
        "caption": "FIGURE D.3 Partial potential coefficient for round tube.",
        "texts": [
            "14c) PARTIAL POTENTIAL COEFFICIENTS FOR ORTHOGONAL GEOMETRIES 413 D.1.4 Pp11 for Round Tube Cell Shape We give a formula for the partial self-potential coefficient for a round tube cell, which is of importance for several models such as thin wire-type structures. Here, a new formula is used to evaluate the coefficient. However, the approach is similar to the one in Ref. [2]. The partial coefficient of potential for a zero thickness cylindrical tube with a radius a = d\u22152 is similar to the partial inductance in Section C.1.2 for Lp (Fig. D.3). The integral that is evaluated in this case is Pp11 = 1 4\ud835\udf0b\ud835\udf16 2\u222b\u222b\u2032 1 R1,1\u2032 d \u2032 d , (D.15) where  represents the surface of the tube and where R1,1\u2032 = \u221a (x1 \u2212 x \u2032 1)2 + (y1 \u2212 y \u2032 1)2 + (z1 \u2212 z \u2032 1)2. (D.16) Cylindrical coordinates and symmetry are used to reduce the fourfold integral to a threefold integral as was done for the partial inductance. The solution of the integrals is based on the assumption that the length \ud835\udcc1 of the wire is longer than the diameter d = 2a. The result for the tube conductor is Pp11 = 1 4 \ud835\udf160\ud835\udcc1 [( k2 480 + k4 1280 + 1 3600 ) \ud835\udf0b 3 + ( 1 18 \u2212 k2 24 ) \ud835\udf0b + ( 2 \u2212 2 log(\ud835\udcc1) + 6 log(2) + 2 log(a) \u2212 4 log(k\ud835\udf0b) ) 1 \ud835\udf0b + 8a \ud835\udcc1\ud835\udf0b2 ] , (D"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure13.34-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure13.34-1.png",
        "caption": "Figure 13.34 Isolated beam ACD with the support reactions in A and C due to a uniformly distributed load parallel to the roof plane.",
        "texts": [
            " The M diagram has two extreme values: the (in an absolute sense) smallest moment Mmin is the bending moment at C and the (in an absolute sense) largest moment Mmax is the bending moment in field AC: Mmin = 2.320 kNm ( ). The maximum moment in field AC is found 2.262 m from A. The magnitude can be determined from the area of the V diagram: Mmax = 1 2 (2.262 m)(2.262 kN) = 2.558 kNm ( ). Figure 13.33 (c) Shear force diagram and (d) bending moment diagram due to a uniformly distributed load of 1 kN/m normal to the roof plane. 13 Calculating M, V and N Diagrams 577 \u2022 N , V and M diagrams due to qpa = 1 kN/m (Figure 13.31b) In Figure 13.34, ACD has been isolated and the joining forces acting on ACD are shown. The distributed load qpa = 1 kN/m has been replaced by its resultant Rpa: Rpa = qpa ACD = (1 kN/m)(7.539 m) = 7.539 kN. The moment equilibrium about A gives Ctr = 0 and so Cpa = 0. The force equilibrium of ACD gives Atr = 0 and Apa = Rpa = 7.539 kN. In Figure 13.35a, ACD is shown again with the forces determined. In Figure 13.35b, the associated N diagram is shown: due to a uniformly distributed load the normal force is linear"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure9.8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure9.8-1.png",
        "caption": "Figure 9.8 Trusses with (a) rising diagonals (rd), (b) falling diagonals (fd) and (c) alternating falling and rising diagonals.",
        "texts": [
            " The members along the chord or perimeter of the truss are called chord members (ch), the others are referred to as bracing members (br). Chord members can be divided into top chord members (t) and bottom chord members (b). For bracing members, we distinguish between verticals (v) and diagonals (d), depending on whether the members are positioned vertically or obliquely. Vertical chord members are also referred to as verticals. In certain cases, one distinguishes between rising diagonals (rd) and falling diagonals (fd), depending on their position, seen from the perspective of the nearest support, towards the centre (see Figure 9.8). 9 Trusses 325 In the following you will find a number of types of trusses. Several trusses have been named after their designer or after the region where they were developed. We will not discuss this nomenclature further, which differs per language area. We will also not address the benefits and disadvantage of the various trusses. We will briefly discuss only the motive for choosing rising or falling diagonals. Figure 9.9 shows a number of trusses that are commonly used in roofs. In a Belgian truss (a) the bracing consists of members at right angles to the top chord, and diagonals"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure3.25-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure3.25-1.png",
        "caption": "Figure 3.25 Resolving the force F graphically along three given lines of action; (a) line of action figure and (b) force polygon.",
        "texts": [
            "24 shows the resultant R with its line of action. Note: If one performs the calculation using a picture, all the unknown quantities that are related to the coordinate system in that picture have to be shown positively. In Figure 3.23a that would be x, y, Rx and Ry , in Figure 3.23b this only relates to x and y. A force F , with given magnitude, direction, and line of action, can be resolved along three given lines of action a, b and c, which do not intersect in one point, into the forces Fa, Fb and Fc (see Figure 3.25a). Here F = Fa + Fb + Fc, ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM 3 Statics of a Rigid Body 69 ( F \u2212 Fa) = ( Fb + Fc). ( F \u2212 Fa) and ( Fb + Fc) are equal and therefore have the same line of action. The line of action of ( F \u2212 Fa) passes through the intersection A of the lines of action of F and Fa. The line of action of ( Fb + Fc) passes through the intersection Sab of the lines of action b and c. Therefore, ASbc is the line of action of both ( F \u2212 Fa) and ( Fb + Fc). F at A can now be resolved into Fa with line of action a and ( Fb + Fc) with line of action ASbc. Subsequently ( Fb + Fc) at Sbc can be resolved into Fb and Fc. This is shown graphically in Figure 3.25b in a single force polygon. The order in which F is resolved is irrelevant. In Figure 3.26 F is first resolved at B into Fb and ( Fa + Fc) and subsequently ( Fa + Fc) at Sac, the intersection of the lines of action a and c, is resolved into Fa and Fc. The force polygon now has a different shape, as the forces were resolved in a different order, but the result is the same. The name Culmann1 is associated with this graphical method in the literature. Resolving F into three forces Fa, Fb and Fc along given lines of action a, b, and c, can of course also be done analytically"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure9.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure9.3-1.png",
        "caption": "FIGURE 9.3. A body point mass moving with velocity GvP and acted on by force df .",
        "texts": [
            "33) which shows that the terminal speed of the mass is v2 = r 2K(tf ) m \u2248 81.7m/ s. (9.34) Example 348 Direct dynamics. When the applied force is time varying and is a known function, then, F(t) = m r\u0308. (9.35) The general solution for the equation of motion can be found by integration. r\u0307(t) = r\u0307(t0) + 1 m Z t t0 F(t)dt (9.36) r(t) = r(t0) + r\u0307(t0)(t\u2212 t0) + 1 m Z t t0 Z t t0 F(t)dt dt (9.37) This kind of problem is called direct or forward dynamics. 528 9. Applied Dynamics 9.2 Rigid Body Translational Dynamics Figure 9.3 depicts a moving body B in a global coordinate frame G. Assume that the body frame is attached at the mass center of the body. Point P indicates an infinitesimal sphere of the body, which has a very small mass dm. The point mass dm is acted on by an infinitesimal force df and has a global velocity GvP . According to Newton\u2019s law of motion we have df = GaP dm. (9.38) However, the equation of motion for the whole body in a global coordinate frame is GF = mGaB (9.39) which can be expressed in the body coordinate frame as BF = mB GaB +m B G\u03c9B \u00d7 BvB (9"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002049_j.ymssp.2021.107945-Figure10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002049_j.ymssp.2021.107945-Figure10-1.png",
        "caption": "Fig. 10. Honeycomb: (a) regular hexagons, and (b) inverted hexagons or re-entrant honeycomb.",
        "texts": [
            " Auxetic honeycomb structures are attractive for various engineering applications due to their volume change control capacities and excellent impact energy absorption performance. Yang et al. [143] established an analytical model of a 3D re-entrant honeycomb structure based on a large deflection beam model. Analytical solutions for the modulus, Poisson\u2019s ratio and yield strength of the cellular structure in principal directions were obtained and compared with the experimental results and finite element analysis. Fig. 10 shows a regular honeycomb with positive Poisson\u2019s ratio and a re-entrant honeycomb with negative Poisson\u2019s ratio. Li et al. [144] built a series of 2D random cellular solid models with re-entrant structures in light of a modified Voronoi tessellation technique. Poisson\u2019s ratio and energy absorption capacity of cellular solid models with different initial relative densities were studied and the minimum Poisson\u2019s ratio of 0:38 was obtained. They showed that the cellular structures with minimum negative Poisson\u2019s ratio could have the highest energy absorption capacity in this series of designs"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003443_0278364903022001005-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003443_0278364903022001005-Figure4-1.png",
        "caption": "Fig. 4. 3R2Txy 5-DoF 3 \u2212 xPzRzR(iRjR)N PM.",
        "texts": [
            " Note that cylindrical pairs and universal joints are kinematically equal to a specific combination of revolute pairs and prismatic pairs. By setting the first revolute pair axis of the 3R or 2R spherical subchain perpendicular to its anterior revolute pair axis and assuming the intersection of the two axes, we can obtain a universal joint. Similarly, by setting the first revolute pair axis of the 3R or 2R spherical subchain parallel to its anterior prismatic pair, we can obtain a cylindrical pair. A 3\u2212xPzRzR(iRjR)N PM is shown in Figure 4. Obviously, the mechanism is mechanism-symmetrical. Following the steps in Section 2.4, we can obtain the appropriate input selection. Any two prismatic pairs and one revolute pairs in each 2R subchain can be chosen as an actuated pair. Figure 5 shows another 5-DoF 3 \u2212 xPzR(iRjRkR)N PM. The two mechanisms are not instantaneous. With 2R Spherical Subchain With 3R Spherical Subchain p \u2212 zRzRzR(iRjR)N p \u2212 zRzRxP(iRjR)N p \u2212 zRzR(iRjRkR)N p \u2212 zRxPzR(iRjR)N p \u2212 xPyP(iRjRkR)N p \u2212 xPzRzR(iRjR)N p \u2212 zRxP(iRjRkR)N p \u2212 xPyPzR(iRjR)N p \u2212 xPzR(iRjRkR)N p \u2212 zRxPyP(iRjR)N p \u2212 xPzRyP(iRjR)N The 5-DoF PM with three translational DoFs and two rotational DoFs is denoted by 2R3T"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure3.24-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure3.24-1.png",
        "caption": "FIGURE 3.24. Spring structure of a tire.",
        "texts": [],
        "surrounding_texts": [
            "The tire is the main component interacting with the road. The performance of a vehicle is mainly influenced by the characteristics of its tires. Tires affect a vehicle\u2019s handling, traction, ride comfort, and fuel consumption. To understand its importance, it is enough to remember that a vehicle can maneuver only by longitudinal, vertical, and lateral force systems generated under the tires. Figure 3.1 illustrates a model of a vertically loaded stationary tire. To model the tire-road interactions, we determine the tireprint and describe the forces distributed on the tireprint. 3.1 Tire Coordinate Frame and Tire Force System To describe the tire-road interaction and force system, we attach a Cartesian coordinate frame at the center of the tireprint, as shown in Figure 3.2, assuming a flat and horizontal ground. The x-axis is along the intersection line of the tire-plane and the ground. Tire plane is the plane made by narrowing the tire to a flat disk. The z-axis is perpendicular to the ground, opposite to the gravitational acceleration g, and the y-axis makes the coordinate system a right-hand triad. To show the tire orientation, we use two angles: camber angle \u03b3 and sideslip angle \u03b1. The camber angle is the angle between the tire-plane and 96 3. Tire Dynamics the vertical plane measured about the x-axis. The camber angle can be recognized better in a front view as shown in Figure 3.3. The sideslip angle \u03b1, or simply sideslip, is the angle between the velocity vector v and the x-axis measured about the z-axis. The sideslip can be recognized better in a top view, as shown in Figure 3.4. The force system that a tire receives from the ground is assumed to be located at the center of the tireprint and can be decomposed along x, y, and z axes. Therefore, the interaction of a tire with the road generates a 3D force system including three forces and three moments, as shown in Figure 3.2. 1. Longitudinal force Fx. It is a force acting along the x-axis. The resultant longitudinal force Fx > 0 if the car is accelerating, and Fx < 0 if the car is braking. Longitudinal force is also called forward force. 2. Normal force Fz. It is a vertical force, normal to the ground plane. The resultant normal force Fz > 0 if it is upward. Normal force is also called vertical force or wheel load. 3. Lateral force Fy. It is a force, tangent to the ground and orthogonal to both Fx and Fz. The resultant lateral force Fy > 0 if it is in the y-direction. 4. Roll moment Mx. It is a longitudinal moment about the x-axis. The resultant roll moment Mx > 0 if it tends to turn the tire about the x-axis. The roll moment is also called the bank moment, tilting torque, or overturning moment. 3. Tire Dynamics 97 98 3. Tire Dynamics 5. Pitch moment My. It is a lateral moment about the y-axis. The resultant pitch momentMy > 0 if it tends to turn the tire about the y-axis and move forward. The pitch moment is also called rolling resistance torque. 6. Yaw moment Mz. It is an upward moment about the z-axis. The resultant yaw moment Mz > 0 if it tends to turn the tire about the z-axis. The yaw moment is also called the aligning moment, self aligning moment, or bore torque. The moment applied to the tire from the vehicle about the tire axis is called wheel torque T . Example 75 Origin of tire coordinate frame. For a cambered tire, it is not always possible to find or define a center point for the tireprint to be used as the origin of the tire coordinate frame. It is more practical to set the origin of the tire coordinate frame at the center of the intersection line between the tire-plane and the ground. So, the origin of the tire coordinate frame is at the center of the tireprint when the tire is standing upright and stationary on a flat road. Example 76 SAE tire coordinate system. The tire coordinate system adopted by the Society of Automotive Engineers (SAE) is shown in Figure 3.5. The origin of the coordinate system is at the center of the tireprint when the tire is standing stationary. The x-axis is at the intersection of the tire-plane and the ground plane. The z-axis is downward and perpendicular to the tireprint. The y-axis is on the ground plane and goes to the right to make the coordinate frame a righthand frame. The sideslip angle \u03b1 is considered positive if the tire is slipping to the right, and the camber angle \u03b3 is positive when the tire leans to the right. The SAE coordinate system is as good as the coordinate system in Figure 3.2 and may be used alternatively. However, having the z-axis directed downward is sometimes inefficient and confusing. Furthermore, in SAE convention, the camber angle for the left and right tires of a vehicle have opposite signs. So, the camber angle of the left tire is positive when the tire leans to the right and the camber angle of the right tire is positive when the tire leans to the left. 3.2 Tire Stiffness As an applied approximation, the vertical tire force Fz can be calculated as a linear function of the normal tire deflection 4z measured at the tire center. Fz = kz4z (3.1) 3. Tire Dynamics 99 The coefficient kz is called tire stiffness in the z-direction. Similarly, the reaction of a tire to a lateral and a longitudinal force can be approximated by Fx = kx4x (3.2) Fy = ky4y (3.3) where the coefficient kx and ky are called tire stiffness in the x and y directions. Proof. The deformation behavior of tires to the applied forces in any three directions x, y, and z are the first important tire characteristics in tire dynamics. Calculating the tire stiffness is generally based on experiment and therefore, they are dependent on the tire\u2019s mechanical properties, as well as environmental characteristics. Consider a vertically loaded tire on a stiff and flat ground as shown in Figure 3.6. The tire will deflect under the load and generate a pressurized contact area to balance the vertical load. Figure 3.7 depicts a sample of experimental stiffness curve in the (Fz,4z) plane. The curve can be expressed by a mathematical function Fz = f (4z) (3.4) however, we may use a linear approximation for the range of the usual application. Fz = \u2202f \u2202 (4z) 4z (3.5) 100 3. Tire Dynamics F1 < F2 < F3 3. Tire Dynamics 101 The coefficient \u2202f \u2202(4z) is the slope of the experimental stiffness curve at zero and is shown by a stiffness coefficient kz kz = tan \u03b8 = lim 4z\u21920 \u2202f \u2202 (4z) . (3.6) Therefore, the normal tire deflection 4z remains proportional to the vertical tire force Fz. Fz = kz4z (3.7) The tire can apply only pressure forces to the road, so normal force is restricted to Fz > 0. The stiffness curve can be influenced by many parameters. The most effective one is the tire inflation pressure. Lateral and longitudinal force/deflection behavior is also determined experimentally by applying a force in the appropriate direction. The lateral and longitudinal forces are limited by the sliding force when the tire is vertically loaded. Figure 3.8 depicts a sample of longitudinal and lateral stiffness curves compared to a vertical stiffness curve. The practical part of a tire\u2019s longitudinal and lateral stiffness curves is the linear part and may be estimated by linear equations. Fx = kx4x (3.8) Fy = ky4y (3.9) The coefficients kx and ky are called the tire stiffness in the x and y directions. They are measured by the slope of the experimental stiffness curves 102 3. Tire Dynamics in the (Fx,4x) and (Fy,4y) planes. kx = lim 4x\u21920 \u2202f \u2202 (4x) (3.10) ky = lim 4y\u21920 \u2202f \u2202 (4y) (3.11) When the longitudinal and lateral forces increase, parts of the tireprint creep and slide on the ground until the whole tireprint starts sliding. At this point, the applied force saturates and reaches its maximum supportable value. Generally, a tire is most stiff in the longitudinal direction and least stiff in the lateral direction. kx > kz > ky (3.12) Figure 3.9 illustrates tire deformation under a lateral and a longitudinal force. Example 77 F Nonlinear tire stiffness. In a better modeling, the vertical tire force Fz is a function of the normal tire deflection 4z, and deflection velocity 4z\u0307. Fz = Fz (4z,4z\u0307) (3.13) = Fzs + Fzd (3.14) In a first approximation we may assume Fz is a combination of a static and a dynamic part. The static part is a nonlinear function of the vertical tire deflection and the dynamic part is proportional to the vertical speed of the tire. Fzs = k14z + k2 (4z)2 (3.15) Fzd = k3z\u0307 (3.16) 3. Tire Dynamics 103 The constants k1 and k2 are calculated from the first and second slopes of the experimental stiffness curve in the (Fz,4z) plane, and k3 is the first slope of the curve in the (Fz, z\u0307) plane, which indicates the tire damping. k1 = \u2202 Fz \u22024z \u00af\u0304\u0304\u0304 4z=0 (3.17) k2 = 1 2 \u22022 Fz \u2202 (4z) 2 \u00af\u0304\u0304\u0304 \u00af 4z=0 (3.18) k3 = \u2202 Fz \u2202 z\u0307 \u00af\u0304\u0304\u0304 z\u0307=0 (3.19) The value of k1 = 200000N/m is a good approximation for a 205/50R15 passenger car tire, and k1 = 1200000N/m is a good approximation for a X31580R22.5 truck tire. Tires with a larger number of plies have higher damping, because the plies\u2019 internal friction generates the damping. Tire damping decreases by increasing speeds. Example 78 F Hysteresis effect. Because tires are made from rubber, which is a viscoelastic material, the loading and unloading stiffness curves are not exactly the same. They are similar to those in Figure 3.10, which make a loop with the unloading curve below the loading. The area within the loop is the amount of dissipated energy during loading and unloading. As a tire rotates under the weight of a vehicle, it experiences repeated cycles of deformation and recovery, and it dissipates energy loss as heat. Such a behavior is a common property of hysteretic material and is called hysteresis. So, hysteresis is a characteristic of a deformable material such as rubber, that the energy of deformation is greater than the energy of recovery. The amount of dissipated energy depends on the mechanical characteristics of the tire. Hysteretic energy loss in rubber decreases as temperature increases. The hysteresis effect causes a loaded rubber not to rebound fully after load removal. Consider a high hysteresis race car tire turning over road irregularities. The deformed tire recovers slowly, and therefore, it cannot push the tireprint tail on the road as hard as the tireprint head. The difference in head and tail pressures causes a resistance force, which is called rolling resistance. Race cars have high hysteresis tires to increase friction and limit traction. Street cars have low hysteresis tires to reduce the rolling resistance and low operating temperature. Hysteresis level of tires inversely affect the stopping distance. A high hysteresis tire makes the stopping shorter, however, it wears rapidly and has a shorter life time. 104 3. Tire Dynamics 3.3 Tireprint Forces The force per unit area applied on a tire in a tireprint can be decomposed into a component normal to the ground and a tangential component on the ground. The normal component is the contact pressure \u03c3z, while the tangential component can be further decomposed in the x and y directions to make the longitudinal and lateral shear stresses \u03c4x and \u03c4y. For a stationary tire under normal load, the tireprint is symmetrical. Due to equilibrium conditions, the overall integral of the normal stress over the tireprint area AP must be equal to the normal load Fz, and the integral of shear stresses must be equal to zero. Z AP \u03c3z(x, y) dA = Fz (3.20)Z AP \u03c4x(x, y) dA = 0 (3.21)Z AP \u03c4y(x, y) dA = 0 (3.22) 3.3.1 Static Tire, Normal Stress Figure 3.11 illustrates a stationary tire under a normal load Fz along with the generated normal stress \u03c3z applied on the ground. The applied loads on the tire are illustrated in the side view shown in Figure 3.12. For a stationary tire, the shape of normal stress \u03c3z(x, y) over the tireprint area depends on tire and load conditions, however its distribution over the tireprint is generally in the shape shown in Figure 3.13. 3. Tire Dynamics 105 106 3. Tire Dynamics The normal stress \u03c3z(x, y) may be approximated by the function \u03c3z(x, y) = \u03c3zM \u00b5 1\u2212 x6 a6 \u2212 y6 b6 \u00b6 (3.23) where a and b indicate the dimensions of the tireprint, as shown in Figure 3.14. The tireprints may approximately be modeled by a mathematical function x2n a2n + y2n b2n = 1 n \u2208 N. (3.24) For radial tires, n = 3 or n = 2 may be used, x6 a6 + y6 b6 = 1 (3.25) while for non-radial tires n = 1 is a better approximation. x2 a2 = y2 b2 = 1. (3.26) Example 79 Normal stress in tireprint. A car weighs 800 kg. If the tireprint of each radial tire is AP = 4\u00d7a\u00d7b = 4 \u00d7 5 cm \u00d7 12 cm, then the normal stress distribution under each tire, \u03c3z, 3. Tire Dynamics 107 must satisfy the equilibrium equation. Fz = 1 4 800\u00d7 9.81 = Z AP \u03c3z(x, y) dA = Z 0.05 \u22120.05 Z 0.12 \u22120.12 \u03c3zM \u00b5 1\u2212 x6 0.056 \u2212 y6 0.126 \u00b6 dy dx = 1.7143\u00d7 10\u22122\u03c3zM (3.27) Therefore, the maximum normal stress is \u03c3zM = Fz 1.7143\u00d7 10\u22122 = 1.1445\u00d7 10 5 Pa (3.28) and the stress distribution over the tireprint is \u03c3z(x, y) = 1.1445\u00d7 105 \u00b5 1\u2212 x6 0.056 \u2212 y6 0.126 \u00b6 Pa. (3.29) Example 80 Normal stress in tireprint for n = 2. The maximum normal stress \u03c3zM for an 800 kg car having an AP = 4\u00d7 a\u00d7 b = 4\u00d7 5 cm\u00d7 12 cm, can be found for n = 2 as Fz = 1 4 800\u00d7 9.81 = Z AP \u03c3z(x, y) dA = Z 0.05 \u22120.05 Z 0.12 \u22120.12 \u03c3zM \u00b5 1\u2212 x4 0.054 \u2212 y4 0.124 \u00b6 dy dx = 1.44\u00d7 10\u22122\u03c3zM (3.30) 108 3. Tire Dynamics \u03c3zM = Fz 1.44\u00d7 10\u22122 = 1.3625\u00d7 10 5 Pa. (3.31) Comparing \u03c3zM = 1.3625 \u00d7 105 Pa for n = 2 to \u03c3zM = 1.1445 \u00d7 105 Pa for n = 3 shows that maximum stress for n = 2 is \u00a1 1\u2212 1.1445 1.3625 \u00a2 \u00d7100 = 16% more than n = 3. 3.3.2 Static Tire, Tangential Stresses Because of geometry changes to a circular tire in contact with the ground, a three-dimensional stress distribution will appear in the tireprint even for a stationary tire. The tangential stress \u03c4 on the tireprint can be decomposed in x and y directions. The tangential stress is also called shear stress or friction stress. The tangential stress on a tire is inward in x direction and outward in y direction. Hence, the tire tries to stretch the ground in the x-axis and compact the ground on the y-axis. Figure 3.15 depicts the shear stresses on a vertically loaded stationary tire. The force distribution on the tireprint is not constant and is influenced by tire structure, load, inflation pressure, and environmental conditions: The tangential stress \u03c4x in the x-direction may be modeled by the following equation. \u03c4x(x, y) = \u2212\u03c4xM \u00b5 x2n+1 a2n+1 \u00b6 sin2 \u00b3x a \u03c0 \u00b4 cos \u00b3 y 2b \u03c0 \u00b4 n \u2208 N (3.32) 3. Tire Dynamics 109 \u03c4x is negative for x > 0 and is positive for x < 0, showing an inward longitudinal stress. Figure 3.16 illustrates the absolute value of a \u03c4x distribution for n = 1. The y-direction tangential stress \u03c4y may be modeled by the equation \u03c4y(x, y) = \u2212\u03c4yM \u00b5 x2n a2n \u2212 1 \u00b6 sin \u00b3y b \u03c0 \u00b4 n \u2208 N (3.33) where \u03c4y is positive for y > 0 and negative for y < 0, showing an outward lateral stress. Figure 3.17 illustrates the absolute value of a \u03c4y distribution for n = 1. 3.4 Effective Radius Consider a vertically loaded wheel that is turning on a flat surface as shown in Figure 3.18. The effective radius of the wheel Rw, which is also called a rolling radius, is defined by Rw = vx \u03c9w (3.34) where, vx is the forward velocity, and \u03c9w is the angular velocity of the wheel. The effective radius Rw is approximately equal to Rw \u2248 Rg \u2212 Rg \u2212Rh 3 (3.35) and is a number between the unloaded or geometric radius Rg and the loaded height Rh. Rh < Rw < Rg (3.36) 110 3. Tire Dynamics Proof. An effective radius Rw = vx/\u03c9w is defined by measuring a wheel\u2019s angular velocity \u03c9w and forward velocity vx. As the tire turns forward, each part of the circumference is flattened as it passes through the contact area. A practical estimate of the effective radius can be made by substituting the arc with the straight length of tireprint. The tire vertical deflection is Rg \u2212Rh = Rg (1\u2212 cos\u03d5) (3.37) and therefore Rh = Rg cos\u03d5 (3.38) a = Rg sin\u03d5. (3.39) If the motion of the tire is compared to the rolling of a rigid disk with radius Rw, then the tire must move a distance a = Rw\u03d5 for an angular rotation \u03d5. a = Rg sin\u03d5 = Rw\u03d5 (3.40) Hence, Rw = Rg sin\u03d5 \u03d5 . (3.41) Expanding sin\u03d5 \u03d5 in a Taylor series show that Rw = Rg \u00b5 1\u2212 1 6 \u03d52 +O \u00a1 \u03d54 \u00a2\u00b6 . (3.42) 3. Tire Dynamics 111 Using Equation (3.37) we may approximate cos\u03d5 \u2248 1\u2212 1 2 \u03d52 (3.43) \u03d52 \u2248 2 (1\u2212 cos\u03d5) \u2248 2 \u00b5 1\u2212 Rh Rg \u00b6 (3.44) and therefore, Rw \u2248 Rg \u00b5 1\u2212 1 3 \u00b5 1\u2212 Rh Rg \u00b6\u00b6 = 2 3 Rg + 1 3 Rh. (3.45) Because Rh is a function of tire load Fz, Rh = Rh (Fz) = Rg \u2212 Fz kz (3.46) the effective radius Rw is also a function of the tire load. The angle \u03d5 is called tireprint angle or tire contact angle. The vertical stiffness of radial tires is less than non-radial tires under the same conditions. So, the loaded height of radial tires, Rh, is less than the non-radials\u2019. However, the effective radius of radial tires Rw, is closer to their unloaded radius Rg. As a good estimate, for a non-radial tire, Rw \u2248 0.96Rg, and Rh \u2248 0.94Rg, while for a radial tire, Rw \u2248 0.98Rg, and Rh \u2248 0.92Rg. 112 3. Tire Dynamics Generally speaking, the effective radius Rw depends on the type of tire, stiffness, load conditions, inflation pressure, and wheel\u2019s forward velocity. Example 81 Compression and expansion of tires in the tireprint zone. Because of longitudinal deformation, the peripheral velocity of any point of the tread varies periodically. When it gets close to the starting point of the tireprint, it slows down and a circumferential compression results. The tire treads are compressed in the first half of the tireprint and gradually expanded in the second half. The treads in the tireprint contact zone almost stick to the ground, and therefore their circumferential velocity is close to the forward velocity of the tire center vx. The treads regain their initial circumferential velocity Rg\u03c9w after expanding and leaving the contact zone. Example 82 Tire rotation. The geometric radius of a tire P235/75R15 is Rg = 366.9mm, because hT = 235\u00d7 75% = 176.25mm \u2248 6.94 in (3.47) and therefore, Rg = 2hT + 15 2 = 2\u00d7 6.94 + 15 2 = 14.44 in \u2248 366.9mm. (3.48) Consider a vehicle with such a tire is traveling at a high speed such as v = 50m/ s = 180 km/h \u2248 111.8mi/h. The tire is radial, and therefore the effective tire radius Rw is approximately equal to Rw \u2248 0.98Rg \u2248 359.6mm. (3.49) After moving a distance d = 100 km, this tire must have been turned n1 = 44259 times because n1 = d \u03c0D = 100\u00d7 103 2\u03c0 \u00d7 359.6\u00d7 10\u22123 = 44259. (3.50) Now assume the vehicle travels the same distance d = 100 km at a low inflation pressure, such that the effective radius of the tire remained close to at the loaded radius Rw \u2248 Rh \u2248 0.92Rg = 330.8mm. (3.51) 3. Tire Dynamics 113 This tire must turn n2 = 48112 times to travel d = 100km, because, n2 = d \u03c0D = 100\u00d7 103 2\u03c0 \u00d7 330.8\u00d7 10\u22123 = 48112. (3.52) Example 83 F Radial motion of tire\u2019s peripheral points in the tireprint. The radial displacement of a tire\u2019s peripheral points during road contact may be modeled by a function d = d (x, \u03b8) . (3.53) We assume that a peripheral point of the tire moves along only the radial direction during contact with the ground, as shown in Figure 3.19. Let\u2019s show a radius at an angle \u03b8, by r = r (x, \u03b8). Knowing that cos \u03b8 = Rh r (3.54) cos\u03c6 = Rh Rg (3.55) we can find r = Rg cos\u03c6 cos \u03b8 . (3.56) 114 3. Tire Dynamics Thus, the displacement function is d = Rg \u2212 r (x, \u03b8) = Rg \u00b5 1\u2212 Rh Rg cos \u03b8 \u00b6 \u2212 \u03c6 < \u03b8 < \u03c6. (3.57) Example 84 Tread travel. Let\u2019s follow a piece of tire tread in its travel around the spin axis when the vehicle moves forward at a constant speed. Although the wheel is turning at constant angular velocity \u03c9w, the tread does not travel at constant speed. At the top of the tire, the radius is equal to the unloaded radius Rg and the speed of the tread is Rg\u03c9w relative to the wheel center. As the tire turns, the tread approaches the leading edge of tireprint, and slows down. The tread is compacted radially, and gets squeezed in the heading part of the tireprint area. Then, it is stretched out and unpacked in the tail part of the tireprint as it moves to the tail edge. In the middle of the tireprint, the tread speed is Rh\u03c9w relative to the wheel center. The variable radius of a tire during the motion through the tireprint is r = Rg cos\u03c6 cos \u03b8 \u2212 \u03c6 < \u03b8 < \u03c6 (3.58) where \u03c6 is the half of the contact angle, and \u03b8 is the angular rotation of the tire, as shown in Figure 3.19. The angular velocity of the tire is \u03c9w = \u03b8\u0307 and is assumed to be constant. Then, the radial velocity r\u0307 and acceleration r\u0308 of the tread with respect to the wheel center are r\u0307 = Rg\u03c9w cos\u03c6 sin \u03b8 cos2 \u03b8 (3.59) r\u0308 = 1 2 Rg\u03c9 2 w cos\u03c6 cos3 \u03b8 (3\u2212 cos 2\u03b8) . (3.60) Figure 3.20 depicts r, r\u0307, and r\u0308 for a sample car with the following data: Rg = 0.5m (3.61) \u03c6 = 15deg (3.62) \u03c9w = 60 rad/ s (3.63) 3.5 Rolling Resistance A turning tire on the ground generates a longitudinal force called rolling resistance. The force is opposite to the direction of motion and is proportional to the normal force on the tireprint. Fr = \u2212Fr \u0131\u0302 (3.64) Fr = \u03bcr Fz (3.65) 3. Tire Dynamics 115 The parameter \u03bcr is called the rolling friction coefficient. \u03bcr is not constant and mainly depends on tire speed, inflation pressure, sideslip and camber angles. It also depends on mechanical properties, speed, wear, temperature, load, size, driving and braking forces, and road condition. Proof.When a tire is turning on the road, that portion of the tire\u2019s circumference that passes over the pavement undergoes a deflection. Part of the energy that is spent in deformation will not be restored in the following relaxation. Hence, a change in the distribution of the contact pressure makes normal stress \u03c3z in the heading part of the tireprint be higher than the tailing part. The dissipated energy and stress distortion cause the rolling resistance. Figures 3.21 and 3.22 illustrate a model of normal stress distribution across the tireprint and their resultant force Fz for a turning tire. Because of higher normal stress in the front part of the tireprint, the resultant normal force moves forward. Forward shift of the normal force makes a resistance moment in the \u2212y direction, opposing the forward rotation. Mr = \u2212Mr j\u0302 (3.66) Mr = Fz\u2206x (3.67) The rolling resistance momentMr can be substituted by a rolling resistance 116 3. Tire Dynamics 3. Tire Dynamics 117 force Fr parallel to the x-axis. Fr = \u2212Fr \u0131\u0302 (3.68) Fr = 1 Rh Mr = \u2206x Rh Fz (3.69) Practically the rolling resistance force can be defined using a rolling friction coefficient \u03bcr. Fr = \u03bcr Fz (3.70) Example 85 A model for normal stress of a turning tire. We may assume that the normal stress of a turning tire is expressed by \u03c3z = \u03c3zm \u00b5 1\u2212 x2n a2n \u2212 y2n b2n + x 4a \u00b6 (3.71) where n = 3 or n = 2 for radial tires and n = 1 for non-radial tires. We may determine the stress mean value \u03c3zm by knowing the total load on the tire. As an example, using n = 3 for an 800 kg car with a tireprint AP = 4\u00d7 a\u00d7 b = 4\u00d7 5 cm\u00d7 12 cm, we have Fz = 1 4 800\u00d7 9.81 = Z AP \u03c3z(x, y) dA = Z 0.05 \u22120.05 Z 0.12 \u22120.12 \u03c3zm \u00b5 1\u2212 x6 0.056 \u2212 y6 0.126 + x 4\u00d7 0.05 \u00b6 dy dx = 1.7143\u00d7 10\u22122\u03c3zm (3.72) and therefore, \u03c3zm = Fz 1.7143\u00d7 10\u22122 = 1.1445\u00d7 10 5 Pa (3.73) Example 86 Deformation and rolling resistance. The distortion of stress distribution is proportional to the tire-road deformation that is the reason for shifting the resultant force forward. Hence, the rolling resistance increases with increasing deformation. A high pressure tire on concrete has lower rolling resistance than a low pressure tire on soil. To model the mechanism of dissipation energy for a turning tire, we assume there are many small dampers and springs in the tire structure. Pairs of parallel dampers and springs are installed radially and circumstantially. Figures 3.23 and 3.24 illustrate the damping and spring structure of a tire. 118 3. Tire Dynamics 3. Tire Dynamics 119 3.5.1 F Effect of Speed on the Rolling Friction Coefficient The rolling friction coefficient \u03bcr increases with a second degree of speed. It is possible to express \u03bcr = \u03bcr(vx) by the function \u03bcr = \u03bc0 + \u03bc1 v 2 x. (3.74) Proof. The rolling friction coefficient increases by increasing speed experimentally. We can use a polynomial function \u03bcr = nX i=0 \u03bci v i x (3.75) to fit the experimental data. Practically, two or three terms of the polynomial would be enough. The function \u03bcr = \u03bc0 + \u03bc1 v 2 x (3.76) is simple and good enough for representing experimental data and analytic calculation. The values of \u03bc0 = 0.015 (3.77) \u03bc1 = 7\u00d7 10\u22126 s2/m2 (3.78) are reasonable values for most passenger car tires. However, \u03bc0 and \u03bc1 should be determined experimentally for any individual tire. Figure 3.25 depicts a comparison between Equation (3.74) and experimental data for a radial tire. Generally speaking, the rolling friction coefficient of radial tires show to be less than non-radials. Figure 3.26 illustrates a sample comparison. Equation (3.74) is applied when the speed is below the tire\u2019s critical speed. Critical speed is the speed at which standing circumferential waves appear and the rolling friction increases rapidly. The wavelength of the standing waves are close to the length of the tireprint. Above the critical speed, overheating happens and tire fails very soon. Figure 3.27 illustrates the circumferential waves in a rolling tire at its critical speed. Example 87 Rolling resistance force and vehicle velocity. For computer simulation purposes, a fourth degree equation is presented to evaluate the rolling resistance force Fr Fr = C0 + C1 vx + C2 v 4 x. (3.79) The coefficients Ci are dependent on the tire characteristics, however, the following values can be used for a typical raided passenger car tire: C0 = 9.91\u00d7 10\u22123 C1 = 1.95\u00d7 10\u22125 (3.80) C2 = 1.76\u00d7 10\u22129 120 3. Tire Dynamics Example 88 Road pavement and rolling resistance. The effect of the pavement and road condition is introduced by assigning a value for \u03bc0 in equation \u03bcr = \u03bc0 + \u03bc1 v 2 x. Table 3.1 is a good reference. 122 3. Tire Dynamics Example 89 Tire information tips. A new front tire with a worn rear tire can cause instability. Tires stored in direct sunlight for long periods of time will harden and age more quickly than those kept in a dark area. Prolonged contact with oil or gasoline causes contamination of the rubber compound, making the tire life short. Example 90 F Wave occurrence justification. The normal stress will move forward when the tire is turning on a road. By increasing the speed, the normal stress will shift more and concentrate in the first half of the tireprint, causing low stress in the second half of the tireprint. High stress in the first half along with no stress in the second half is similar to hammering the tire repeatedly. Example 91 F Race car tires. Racecars have very smooth tires, known as slicks. Smooth tires reduce the rolling friction and maximize straight line speed. The slick racing tires are also pumped up to high pressure. High pressure reduces the tireprint area. Hence, the normal stress shift reduces and the rolling resistance decreases. Example 92 F Effect of tire structure, size, wear, and temperature on the rolling friction coefficient. The tire material and the arrangement of tire plies affect the rolling friction coefficient and the critical speed. Radial tires have around 20% lower \u03bcr, and 20% higher critical speed. Tire radius Rg and aspect ratio hT /wT are the two size parameters that affect the rolling resistance coefficient. A tire with larger Rg and smaller hT /wT has lower rolling resistance and higher critical speed. Generally speaking, the rolling friction coefficient decreases with wear in both radial and non-radial tires, and increases by increasing temperature. 3.5.2 F Effect of Inflation Pressure and Load on the Rolling Friction Coefficient The rolling friction coefficient \u03bcr decreases by increasing the inflation pressure p. The effect of increasing pressure is equivalent to decreasing normal load Fz. The following empirical equation has been suggested to show the effects of both pressure p and load Fz on the rolling friction coefficient. \u03bcr = K 1000 \u00b5 5.1 + 5.5\u00d7 105 + 90Fz p + 1100 + 0.0388Fz p v2x \u00b6 (3.81) The parameter K is equal to 0.8 for radial tires, and is equal to 1.0 for nonradial tires. The value of Fz, p, and vx must be in [ N], [ Pa], and [m/ s] respectively. 3. Tire Dynamics 123 Example 93 Motorcycle rolling friction coefficient. The following equations are suggested for calculating rolling friction coefficient \u03bcr applicable to motorcycles. They can be only used as a rough lower estimate for passenger cars. The equations consider the inflation pressure and forward velocity of the motorcycle. \u03bcr = \u23a7\u23aa\u23aa\u23a8\u23aa\u23aa\u23a9 0.0085 + 1800 p + 2.0606 p v2x vx \u2264 46m/ s (\u2248 165 km/h) 1800 p + 3.7714 p v2x vx > 46m/ s (\u2248 165 km/h) (3.82) The speed vx must be expressed in m/ s and the pressure p must be in Pa. Figure 3.28 illustrates this equation for vx \u2264 46m/ s (\u2248 165 km/h). Increasing the inflation pressure p decreases the rolling friction coefficient \u03bcr. Example 94 Dissipated power because of rolling friction. Rolling friction reduces the vehicle\u2019s power. The dissipated power because of rolling friction is equal to the rolling friction force Fr times the forward velocity vx. Using Equation (3.81), the rolling resistance power is P = Fr vx = \u2212\u03bcr vx Fz = \u2212K vx 1000 \u00b5 5.1 + 5.5\u00d7 105 + 90Fz p + 1100 + 0.0388Fz p v2x \u00b6 Fz. (3.83) 124 3. Tire Dynamics The resultant power P is in [W] when the normal force Fz is expressed in [ N], velocity vx in [m/ s], and pressure p in [ Pa]. The rolling resistance dissipated power for motorcycles can be found based on Equation (3.82). P = \u23a7\u23aa\u23aa\u23a8\u23aa\u23aa\u23aa\u23a9 \u00b5 0.0085 + 1800 p + 2.0606 p v2x \u00b6 vxFz vx \u2264 46m/ s (\u2248 165 km/h)\u00b5 1800 p + 3.7714 p v2x \u00b6 vxFz vx > 46m/ s (\u2248 165 km/h) (3.84) Example 95 Rolling resistance dissipated power. If a vehicle is moving at 100 km/h \u2248 27.78m/ s \u2248 62mi/h and each radial tire of the vehicle is pressurized up to 220 kPa \u2248 32 psi and loaded by 220 kg, then the dissipated power, because of rolling resistance, is P = 4\u00d7 K vx 1000 \u00b5 5.1 + 5.5\u00d7 105 + 90Fz p + 1100 + 0.0388Fz p v2x \u00b6 Fz = 2424.1W \u2248 2.4 kW. (3.85) To compare the given equations, assume the vehicle has motorcycle tires with power loss given by Equation (3.84). P = \u00b5 0.0085 + 1800 p + 2.0606 p v2x \u00b6 vxFz = 5734.1W \u2248 5.7 kW. (3.86) It shows that if the vehicle uses motorcycle tires, it dissipates more power. Example 96 Effects of improper inflation pressure. High inflation pressure increases stiffness, which reduces ride comfort and generates vibration. Tireprint and traction are reduced when tires are over inflated. Over-inflation causes the tire to transmit shock loads to the suspension, and reduces the tire\u2019s ability to support the required load for cornerability, braking, and acceleration. Under-inflation results in cracking and tire component separation. It also increases sidewall flexing and rolling resistance that causes heat and mechanical failure. A tire\u2019s load capacity is largely determined by its inflation pressure. Therefore, under-inflation results in an overloaded tire that operates at high deflection with a low fuel economy, and low handling. Figure 3.29 illustrates the effect of over and under inflation on tire-road contact compared to a proper inflated tire. Proper inflation pressure is necessary for optimum tire performance, safety, and fuel economy. Correct inflation is especially significant to the endurance and performance of radial tires because it may not be possible to find a 5 psi \u2248 35 kPa under-inflation in a radial tire just by looking. 3. Tire Dynamics 125 However, under-inflation of 5 psi \u2248 35 kPa can reduce up to 25% of the tire performance and life. A tire may lose 1 to 2 psi (\u2248 7 to 14 kPa) every month. The inflation pressure can also change by 1 psi\u2248 7 kPa for every 10 \u25e6F \u2248 5 \u25e6C of temperature change. As an example, if a tire is inflated to 35 psi \u2248 240 kPa on an 80 \u25e6F \u2248 26 \u25e6C summer day, it could have an inflation pressure of 23 psi \u2248 160 kPa on a 20 \u25e6F \u2248 \u22126 \u25e6C day in winter. This represents a normal loss of 6 psi \u2248 40 kPa over the six months and an additional loss of 6 psi \u2248 40 kPa due to the 60 \u25e6F \u2248 30 \u25e6C change. At 23 psi \u2248 160 kPa, this tire is functioning under-inflated. Example 97 Small / large and soft / hard tires. If the driving tires are small, the vehicle becomes twitchy with low traction and low top speed. However, when the driving tires are big, then the vehicle has slow steering response and high tire distortion in turns, decreasing the stability. Softer front tires show more steerability, less stability, and more wear while hard front tires show the opposite. Soft rear tires have more rear traction, but they make the vehicle less steerable, more bouncy, and less stable. Hard rear tires have less rear traction, but they make the vehicle more steerable, less bouncy, and more stable. 3.5.3 F Effect of Sideslip Angle on Rolling Resistance When a tire is turning on the road with a sideslip angle \u03b1, a significant increase in rolling resistance occurs. The rolling resistance force Fr would 126 3. Tire Dynamics then be Fr = Fx cos\u03b1+ Fy sin\u03b1 (3.87) \u2248 Fx \u2212 C\u03b1\u03b1 2 (3.88) where, Fx is the longitudinal force opposing the motion, and Fy is the lateral force. Proof. Figure 3.30 illustrates the top view of a turning tire on the ground under a sideslip angle \u03b1. The rolling resistance force is defined as the force opposite to the velocity vector of the tire, which has angle \u03b1 with the xaxis. Assume a longitudinal force Fx in \u2212x-direction is applied on the tire. Sideslip \u03b1 increases Fx and generates a lateral force Fy. The sum of the components of the longitudinal force Fx and the lateral force Fy makes the rolling resistance force Fr. Fr = Fx cos\u03b1+ Fy sin\u03b1 (3.89) For small values of the sideslip \u03b1, the lateral force is proportional to \u2212\u03b1 and therefore, Fr \u2248 Fx \u2212 C\u03b1\u03b1 2. (3.90) 3. Tire Dynamics 127 3.5.4 F Effect of Camber Angle on Rolling Resistance When a tire travels with a camber angle \u03b3, the component of rolling moment Mr on rolling resistance Fr will be reduced, however, a component of aligning moment Mz on rolling resistance will appear. Fr = \u2212Fr \u0131\u0302 (3.91) Fr = 1 Rh Mr cos \u03b3 + 1 Rh Mz sin \u03b3 (3.92) Proof. Rolling moment Mr appears when the normal force Fz shifts forward. However, only the component Mr cos \u03b3 is perpendicular to the tireplane and prevents the tire\u2019s spin. Furthermore, when a moment in zdirection is applied on the tire, only the component Mz sin \u03b3 will prevent the tire\u2019s spin. Therefore, the camber angle \u03b3 will affect the rolling resistance according to Fr = \u2212Fr \u0131\u0302 Fr = 1 h Mr cos \u03b3 1 h Mz sin \u03b3 (3.93) where Mr may be substituted by Equation (3.66) to show the effect of normal force Fz. Fr = \u2206x h Fz cos \u03b3 1 h Mz sin \u03b3 (3.94) 3.6 Longitudinal Force The longitudinal slip ratio of a tire is s = Rg\u03c9w vx \u2212 1 (3.95) where, Rg is the tire\u2019s geometric and unloaded radius, \u03c9w is the tire\u2019s angular velocity, and vx is the tire\u2019s forward velocity. Slip ratio is positive for driving and is negative for braking. To accelerate or brake a vehicle, longitudinal forces must develop between the tire and the ground. When a moment is applied to the spin axis of the tire, slip ratio occurs and a longitudinal force Fx is generated at the tireprint. The force Fx is proportional to the normal force, Fx = Fx \u0131\u0302 (3.96) Fx = \u03bcx (s) Fz (3.97) where the coefficient \u03bcx (s) is called the longitudinal friction coefficient and is a function of slip ratio s as shown in Figure 3.31. The friction coefficient 128 3. Tire Dynamics reaches a driving peak value \u03bcdp at s \u2248 0.1, before dropping to an almost steady-state value \u03bcds. The friction coefficient \u03bcx (s) may be assumed proportional to s when s is very small \u03bcx (s) = Cs s s << 1 (3.98) where Cs is called the longitudinal slip coefficient. The tire will spin when s & 0.1 and the friction coefficient remains almost constant. The same phenomena happens in braking at the values \u03bcbp and \u03bcbs. Proof. Slip ratio, or simply slip, is defined as the difference between the actual speed of the tire vx and the equivalent tire speeds Rw\u03c9w. Figure 3.32 illustrates a turning tire on the ground. The ideal distance that the tire would freely travel with no slip is denoted by dF , while the actual distance the tire travels is denoted by dA. Thus, for a slipping tire, dA > dF , and for a spinning tire, dA < dF . The difference dF \u2212 dA is the tire slip and therefore, the slip ratio of the tire is s = dF \u2212 dA dA . (3.99) To have the instant value of s, we must measure the travel distances in an infinitesimal time length, and therefore, s \u2261 d\u0307F \u2212 d\u0307A d\u0307A . (3.100) 3. Tire Dynamics 129 If the angular velocity of the tire is \u03c9w then, d\u0307F = Rg\u03c9w and d\u0307A = Rw\u03c9w where, Rg is the geometric tire radius and Rw is the effective radius. Therefore, the slip ratio s can be defined based on the actual speed vx = Rw\u03c9w, and the free speed Rg\u03c9w s = Rg\u03c9w \u2212Rw\u03c9w Rw\u03c9w = Rg\u03c9w vx \u2212 1 (3.101) A tire can exert longitudinal force only if a longitudinal slip is present. Longitudinal slip is also called circumferential or tangential slip. During acceleration, the actual velocity vx is less than the free velocity Rg\u03c9w, and therefore, s > 0. However, during braking, the actual velocity vx is higher than the free velocity Rg\u03c9w and therefore, s < 0. The frictional force Fx between a tire and the road surface is a function of normal load Fz, vehicle speed vx, and wheel angular speed \u03c9w. In addition to these variables there are a number of parameters that affect Fx, such as tire pressure, tread design, wear, and road surface. It has been determined empirically that a contact friction of the form Fx = \u03bcx(\u03c9w, vx)Fz can model experimental measurements obtained with constant vx, \u03c9w. Example 98 Slip ratio based on equivalent angular velocity \u03c9eq. It is possible to define an effective angular velocity \u03c9eq as an equivalent angular velocity for a tire with radius Rg to proceed with the actual speed vx = Rg\u03c9eq. Using \u03c9eq we have vx = Rg\u03c9eq = Rw\u03c9w (3.102) 130 3. Tire Dynamics and therefore, s = Rg\u03c9w \u2212Rg\u03c9eq Rg\u03c9eq = \u03c9w \u03c9eq \u2212 1. (3.103) Example 99 Slip ratio is \u22121 < s < 0 in braking. When we brake, a braking moment is applied to the wheel axis. The tread of the tire will be stretched circumstantially in the tireprint zone. Hence, the tire is moving faster than a free tire Rw\u03c9w > Rg\u03c9w (3.104) and therefore, s < 0. The equivalent radius for a braking tire is more than the free radius Rw > Rg. (3.105) Equivalently, we may express the condition using the equivalent angular velocity \u03c9e and deduce that a braking tire turns slower than a free tire Rg\u03c9eq > Rg\u03c9w. (3.106) The brake moment can be high enough to lock the tire. In this case \u03c9w = 0 and therefore, s = \u22121. It shows that the longitudinal slip would be between \u22121 < s < 0 when braking. \u22121 < s < 0 for a < 0 (3.107) Example 100 Slip ratio is 0 < s <\u221e in driving. When we drive, a driving moment is applied to the tire axis. The tread of the tire will be compressed circumstantially in the tireprint zone. Hence, the tire is moving slower than a free tire Rw\u03c9w < Rg\u03c9w (3.108) and therefore s > 0. The equivalent radius for a driving tire is less than the free radius Rw < Rg. (3.109) Equivalently, we may express the condition using the equivalent angular velocity \u03c9e and deduce that a driving tire turns faster than a free tire Rg\u03c9eq < Rg\u03c9w. (3.110) The driving moment can be high enough to overcome the friction and turn the tire on pavement while the car is not moving. In this case vx = 0 3. Tire Dynamics 131 and therefore, s = \u221e. It shows that the longitudinal slip would be between 0 < s <\u221e when accelerating. 0 < s <\u221e for a > 0 (3.111) The tire speed Rw\u03c9w equals vehicle speed vx only if acceleration is zero. In this case, the normal force acting on the tire and the size of the tireprint are constant in time. No element of the tireprint is slipping on the road. Example 101 Power and maximum velocity. Consider a moving car with power P = 100 kW \u2248 134 hp can attain 279 km/h \u2248 77.5m/ s \u2248 173.3mi/h. The total driving force must be Fx = P vx = 100\u00d7 103 77.5 = 1290.3N. (3.112) If we assume that the car is rear-wheel-drive and the rear wheels are driving at the maximum traction under the load 1600N, then the longitudinal friction coefficient \u03bcx is \u03bcx = Fx Fz = 1290.3 1600 \u2248 0.806. (3.113) Example 102 Slip of hard tire on hard road. A tire with no slip cannot create any tangential force. Assume a toy car equipped with steel tires is moving on a glass table. Such a car cannot accelerate or steer easily. If the car can steer at very low speeds, it is because there is sufficient microscopic slip to generate forces to steer or drive. The glass table and the small contact area of the small metallic tires deform and stretch each other, although such a deformation is very small. If there is any friction between the tire and the surface, there must be slip to maneuver. Example 103 Samples for longitudinal friction coefficients \u03bcdp and \u03bcds. Table 3.2 shows the average values of longitudinal friction coefficients \u03bcdp and \u03bcds for a passenger car tire 215/65R15. It is practical to assume \u03bcdp = \u03bcbp, and \u03bcds = \u03bcbs. 132 3. Tire Dynamics Example 104 Friction mechanisms. Rubber tires generate friction in three mechanisms: 1\u2212 adhesion, 2\u2212 deformation, and 3\u2212 wear. Fx = Fad + Fde + Fwe. (3.114) Adhesion friction is equivalent to sticking. The rubber resists sliding on the road because adhesion causes it stick to the road surface. Adhesion occurs as a result of molecular binding between the rubber and surfaces. Because the real contact area is much less than the observed contact area, high local pressure make molecular binding, as shown in Figure 3.33. Bound occurs at the points of contact and welds the surfaces together. The adhesion friction is equal to the required force to break these molecular bounds and separate the surfaces. The adhesion is also called cold welding and is attributed to pressure rather than heat. Higher load increases the contact area, makes more bounds, and increases the friction force. So the adhesion friction confirms the friction equation Fx = \u03bcx (s) Fz. (3.115) The main contribution to tire traction force on a dry road is the adhesion friction. The adhesion friction decreases considerably on a road covered by water, ice, dust, or lubricant. Water on a wet road prevents direct contact between the tire and road and reduces the formation of adhesion friction. The main contribution to tire friction when it slides on a road surface is the viscoelastic energy dissipation in the tireprint area. This dissipative energy is velocity and is time-history dependent. Deformation friction is the result of deforming rubber and filling the microscopic irregularities on the road surface. The surface of the road has many peaks and valleys called asperities. Movement of a tire on a rough 3. Tire Dynamics 133 surface results in the deformation of the rubber by peaks and high points on the surface. A load on the tire causes the peaks of irregularities to penetrate the tire and the tire drapes over the peaks. The deformation friction force, needed to move the irregularities in the rubber, comes from the local high pressure across the irregularities. Higher load increases the penetration of the irregularities in the tire and therefore increases the friction force. So the deformation friction confirms the friction equation (3.115). The main contribution to the tire traction force on a wet road is the deformation friction. The adhesion friction decreases considerably on a road covered by water, ice, dust, or lubricant. Deformation friction exists in relative movement between any contacted surfaces. No matter how much care is taken to form a smooth surface, the surfaces are irregular with microscopic peaks and valleys. Opposite peaks interact with each other and cause damage to both surfaces. Wear friction is the result of excessive local stress over the tensile strength of the rubber. High local stresses deform the structure of the tire surface past the elastic point. The polymer bonds break, and the tire surface tears in microscopic scale. This tearing makes the wear friction mechanism. Wear results in separation of material. Higher load eases the tire wear and therefore increases the wear friction force. So the wear friction confirms the friction equation (3.115). Example 105 Empirical slip models. Based on experimental data and curve fitting methods, some mathematical equations are presented to simulate the longitudinal tire force as a function of longitudinal slip s. Most of these models are too complicated to be useful in vehicle dynamics. However, a few of them are simple and accurate enough to be applied. The Pacejka model, which was presented in 1991, has the form Fx (s) = c1 sin \u00a1 c2 tan \u22121 \u00a1c3s\u2212 c4 \u00a1 c3s\u2212 tan\u22121 (c3s) \u00a2\u00a2\u00a2 (3.116) where c1, c2, and c3 are three constants based on the tire experimental data. The 1987 Burckhardt model is a simpler equation that needs three numbers. Fx (s) = c1 \u00a1 1\u2212 e\u2212c2s \u00a2 \u2212 c3s (3.117) There is another Burckhardt model that includes the velocity dependency. Fx (s) = \u00a1 c1 \u00a1 1\u2212 e\u2212c2s \u00a2 \u2212 c3s \u00a2 e\u2212c4v (3.118) This model needs four numbers to be measured from experiment. By expanding and approximating the 1987 Burckhardt model, the simpler model by Kiencke and Daviss was suggested in 1994. This model is Fx (s) = ks s 1 + c1s+ c2s2 (3.119) 134 3. Tire Dynamics where ks is the slope of Fx (s) versus s at s = 0 ks = lim s\u21920 4Fs 4s (3.120) and c1, c2 are two experimental numbers. Another simple model is the 2002 De-Wit model Fx (s) = c1 \u221a s\u2212 c2s (3.121) that is based on two numbers c1, c2. In either case, we need at least one experimental curve such as shown in Figure 3.31 to find the constant numbers ci. The constants ci are the numbers that best fit the associated equation with the experimental curve. The 1997 Burckhardt model (3.118) needs at least two similar tests at two different speeds. Example 106 F Alternative slip ratio. An alternative method for defining the slip ratio is s = \u23a7\u23aa\u23aa\u23a8\u23aa\u23a9 1\u2212 vx Rg\u03c9w Rg\u03c9w > vx driving Rg\u03c9w vx \u2212 1 Rg\u03c9w < vx braking (3.122) where vx is the speed of the wheel center, \u03c9w is the angular velocity of the wheel, and Rg is the tire radius. In another alternative definition, the following equation is used for longitudinal slip: s = 1\u2212 \u00b5 Rg\u03c9w vx \u00b6n where n = \u00bd +1 Rg\u03c9w \u2264 vx \u22121 Rg\u03c9w > vx (3.123) s \u2208 [0, 1] In this definition s is always between zero and one. When s = 1, then the tire is either locked while the car is sliding, or the tire is spinning while the car is not moving. Example 107 F Tire on soft sand. Figure 3.34 illustrates a tire turning on sand. The sand will be packed when the tire passes. The applied stresses from the sand on the tire are developed during the angle \u03b81 < \u03b8 < \u03b82 measured counterclockwise from vertical direction. It is possible to define a relationship between the normal stress \u03c3 and tangential stress \u03c4 under the tire \u03c4 = (c+ \u03c3 tan \u03b8) \u00b3 1\u2212 e r k [\u03b81\u2212\u03b8+(1\u2212s)(sin \u03b8)\u2212sin \u03b81] \u00b4 (3.124) 3. Tire Dynamics 135 where s is the slip ratio defined in Equation (3.122), and \u03c4M = c+ \u03c3 tan \u03b8 (3.125) is the maximum shear stress in the sand applied on the tire. In this equation, c is the cohesion stress of the sand, and k is a constant. Example 108 F Lateral slip ratio. Analytical expressions can be established for the force contributions in x and y directions using adhesive and sliding concept by defining longitudinal and lateral slip ratios sx and sy sx = Rg\u03c9w vx \u2212 1 (3.126) sy = Rg\u03c9w vy (3.127) where vx is the longitudinal speed of the wheel and vy is the lateral speed of the wheel. The unloaded geometric radius of the tire is denoted by Rg and \u03c9w is the rotation velocity of the wheel. At very low slips, the resulting tire forces are proportional to the slip Fx = Csx sx (3.128) Fy = Csy sy (3.129) where Csx is the longitudinal slip coefficient and Csy is the lateral slip coefficient. 3.7 Lateral Force When a turning tire is under a vertical force Fz and a lateral force Fy, its path of motion makes an angle \u03b1 with respect to the tire-plane. The angle 136 3. Tire Dynamics is called sideslip angle and is proportional to the lateral force Fy = Fy j\u0302 (3.130) Fy = \u2212C\u03b1 \u03b1 (3.131) where C\u03b1 is called the cornering stiffness of the tire. C\u03b1 = lim \u03b1\u21920 \u2202 (\u2212Fy) \u2202\u03b1 = \u00af\u0304\u0304\u0304 lim \u03b1\u21920 \u2202Fy \u2202\u03b1 \u00af\u0304\u0304\u0304 (3.132) The lateral force Fy is at a distance ax\u03b1 behind the centerline of the tireprint and makes a moment Mz called aligning moment. Mz = Mz k\u0302 (3.133) Mz = Fy ax\u03b1 (3.134) For small \u03b1, the aligning moment Mz tends to turn the tire about the z-axis and make the x-axis align with the velocity vector v. The aligning moment always tends to reduce \u03b1. Proof. When a wheel is under a constant load Fz and then a lateral force is applied on the rim, the tire will deflect laterally as shown in Figure 3.35. The tire acts as a linear spring under small lateral forces Fy = ky\u2206y (3.135) with a lateral stiffness ky. The wheel will start sliding laterally when the lateral force reaches a maximum value FyM . At this point, the lateral force approximately remains constant and is proportional to the vertical load FyM = \u03bcy Fz (3.136) 3. Tire Dynamics 137 where, \u03bcy is the tire friction coefficient in the y-direction. A bottom view of the tireprint of a laterally deflected tire is shown in Figure 3.36. If the laterally deflected tire is turning forward on the road, the tireprint will also flex longitudinally. A bottom view of the tireprint for such a laterally deflected and turning tire is shown in Figure 3.37. Although the tire-plane remains perpendicular to the road, the path of the wheel makes an angle \u03b1 with tire-plane. As the wheel turns forward, undeflected treads enter the tireprint region and deflect laterally as well as longitudinally. When a tread moves toward the end of the tireprint, its lateral deflection increases until it approaches the tailing edge of the tireprint. The normal load decreases at the tail of the tireprint, so the friction force is lessened and the tread can slide back to its original position when leaving the tireprint region. The point where the laterally deflected tread slides back is called sliding line. A turning tire under lateral force and the associated sideslip angle \u03b1 are shown in Figure 3.38. Lateral distortion of the tire treads is a result of a tangential stress distribution \u03c4y over the tireprint. Assuming that the tangential stress \u03c4y is proportional to the distortion, the resultant lateral force Fy Fy = Z AP \u03c4y dAp (3.137) is at a distance ax\u03b1 behind the center line. ax\u03b1 = 1 Fy Z AP x \u03c4y dAp (3.138) The distance ax\u03b1 is called the pneumatic trail, and the resultant moment 138 3. Tire Dynamics Mz is called the aligning moment. Mz = Mz k\u0302 (3.139) Mz = Fy ax\u03b1 (3.140) The aligning moment tends to turn the tire about the z-axis and make it align with the direction of tire velocity vector v. A stress distribution \u03c4y, the resultant lateral force Fy, and the pneumatic trail ax\u03b1 are illustrated in Figure 3.38. There is also a lateral shift in the tire vertical force Fz because of slip angle \u03b1, which generates a slip moment Mx about the forward x-axis. Mx = \u2212Mx \u0131\u0302 (3.141) Mx = Fz ay\u03b1 (3.142) The slip angle \u03b1 always increases by increasing the lateral force Fy. However, the sliding line moves toward the tail at first and then moves forward by increasing the lateral force Fy. Slip angle \u03b1 and lateral force Fy work as action and reaction. A lateral force generates a slip angle, and a slip angle generates a lateral force. Hence, we can steer the tires of a car to make a slip angle and produce a lateral force to turn the car. Steering causes a slip angle in the tires and creates a lateral force. The slip angle \u03b1 > 0 if the tire should be turned about the z-axis to be aligned with the velocity vector v. A positive slip angle \u03b1 generates a negative lateral force Fy. Hence, steering to the right about the \u2212z-axis makes a positive slip angle and produces a negative lateral force to move the tire to the right. A sample of measured lateral force Fy as a function of slip angle \u03b1 for a constant vertical load is plotted in Figure 3.39. The lateral force Fy is linear for small slip angles, however the rate of increasing Fy decreases 3. Tire Dynamics 139 for higher \u03b1. The lateral force remains constant or drops slightly when \u03b1 reaches a critical value at which the tire slides on the road. Therefore, we may assume the lateral force Fy is proportional to the slip angle \u03b1 for low values of \u03b1. Fy = \u2212C\u03b1 \u03b1 (3.143) C\u03b1 = lim \u03b1\u21920 \u2202 (\u2212Fy) \u2202\u03b1 (3.144) The cornering stiffness C\u03b1 of radial tires are higher than C\u03b1 for non-radial tires. This is because radial tires need a smaller slip angle \u03b1 to produce the same amount of lateral force Fy. Examples of aligning moments for radial and non-radial tires are illustrated in Figure 3.40. The pneumatic trail ax\u03b1 increases for small slip angles up to a maximum value, and decreases to zero and even negative values for high slip angles. Therefore, the behavior of aligning momentMz is similar to what is shown in Figure 3.40. The lateral force Fy = \u2212C\u03b1 \u03b1 can be decomposed to Fy cos\u03b1, parallel to the path of motion v, and Fy sin\u03b1, perpendicular to v as shown in Figure 3.41. The component Fy cos\u03b1, normal to the path of motion, is called cornering force, and the component Fy sin\u03b1, along the path of motion, is called drag force. Lateral force Fy is also called side force or grip. We may combine the lateral forces of all a vehicle\u2019s tires and have them acting at the car\u2019s mass center C. 140 3. Tire Dynamics 3. Tire Dynamics 141 Example 109 Effect of tire load on lateral force curve. When the wheel load Fz increases, the tire treads can stick to the road better. Hence, the lateral force increases at a constant slip angle \u03b1, and the slippage occurs at the higher slip angles. Figure 3.42 illustrates the lateral force behavior of a sample tire for different normal loads. Increasing the load not only increases the maximum attainable lateral force, it also pushes the maximum of the lateral force to higher slip angles. Sometimes the effect of load on lateral force is presented in a dimensionless variable to make it more practical. Figure 3.43 depicts a sample. Example 110 Gough diagram. The slip angle \u03b1 is the main affective parameter on the lateral force Fy and aligning moment Mz = Fyax\u03b1. However, Fz and Mz depend on many other parameters such as speed v, pressure p, temperature, humidity, and road conditions. A better method to show Fz and Mz is to plot them versus each other for a set of parameters. Such a graph is called a Gough diagram. Figure 3.44 depicts a sample Gough diagram for a radial passenger car tire. Every tire has its own Gough diagram, although we may use an average diagram for radial or non-radial tires. Example 111 Effect of velocity. The curve of lateral force as a function of the slip angle Fy (\u03b1) decreases as velocity increases. Hence, we need to increase the sideslip angle at higher velocities to generate the same lateral force. Sideslip angle increases by 142 3. Tire Dynamics 3. Tire Dynamics 143 increasing the steer angle. Figure 3.45 illustrates the effect of velocity on Fy for a radial passenger tire. Because of this behavior, a fixed steer angle, the curvature of a one-wheel-car trajectory, increases by increasing the driving speed. Example 112 F A model for lateral force. When the sideslip angle is not small, the linear approximation (3.131) cannot model the tire behavior. Based on a parabolic normal stress distribution on the tireprint, the following third-degree function was presented in the 1950s to calculate the lateral force at high sideslips Fy = \u2212C\u03b1 \u03b1 \u00c3 1\u2212 1 3 \u00af\u0304\u0304\u0304 C\u03b1 \u03b1 FyM \u00af\u0304\u0304\u0304 + 1 27 \u00b5 C\u03b1 \u03b1 FyM \u00b62! (3.145) where FyM is the maximum lateral force that the tire can support. FyM is set by the tire load and the lateral friction coefficient \u03bcy. Let\u2019s show the sideslip angle at which the lateral force Fy reaches its maximum value FyM by \u03b1M . Equation (3.145) shows that \u03b1M = 3FyM C\u03b1 (3.146) and therefore, Fy = \u2212C\u03b1 \u03b1 \u00c3 1\u2212 \u03b1 \u03b1M + 1 3 \u00b5 \u03b1 \u03b1M \u00b62! (3.147) Fy FyM = 3\u03b1 \u03b1M \u00c3 1\u2212 \u03b1 \u03b1M + 1 3 \u00b5 \u03b1 \u03b1M \u00b62! . (3.148) 144 3. Tire Dynamics Figure 3.46 shows the cubic curve model for lateral force as a function of the sideslip angle. The Equation is applicable only for 0 \u2264 \u03b1 \u2264 \u03b1M . Example 113 F A model for lateral stress. Consider a tire turning on a dry road at a low sideslip angle \u03b1. Assume the developed lateral stress on tireprint can be expressed by the following equation: \u03c4y(x, y) = c\u03c4yM \u00b3 1\u2212 x a \u00b4\u00b5 1\u2212 x3 a3 \u00b6 cos2 \u00b3 y 2b \u03c0 \u00b4 (3.149) The coefficient c is proportional to the tire load Fz sideslip \u03b1, and longitudinal slip s. If the tireprint AP = 4 \u00d7 a \u00d7 b = 4 \u00d7 5 cm \u00d7 12 cm, then the lateral force under the tire, Fy, for c = 1 is Fy = Z AP \u03c4y(x, y) dA = Z 0.05 \u22120.05 Z 0.12 \u22120.12 \u03c4yM \u00b3 1\u2212 x 0.05 \u00b4\u00b5 1\u2212 x3 0.053 \u00b6 cos2 \u00b3 y\u03c0 0.24 \u00b4 dy dx = 0.0144\u03c4yM . (3.150) If we calculate the lateral force Fy = 1000N by measuring the lateral acceleration, then the maximum lateral stress is \u03c4yM = Fz 0.014 4 = 69444Pa (3.151) and the lateral stress distribution over the tireprint is \u03c4y(x, y) = 69444 \u00b3 1\u2212 x 0.05 \u00b4\u00b5 1\u2212 x3 0.053 \u00b6 cos2 \u00b3 y\u03c0 0.24 \u00b4 Pa. (3.152) 3. Tire Dynamics 145 3.8 Camber Force Camber angle \u03b3 is the tilting angle of tire about the longitudinal x-axis. Camber angle generates a lateral force Fy called camber trust or camber force. Figure 3.47 illustrates a front view of a cambered tire and the generated camber force Fy. Camber angle is assumed positive \u03b3 > 0, when it is in the positive direction of the x-axis, measured from the z-axis to the tire. A positive camber angle generates a camber force along the \u2212y-axis. The camber force is proportional to \u03b3 at low camber angles, and depends directly on the wheel load Fz. Therefore, Fy = Fy j\u0302 (3.153) Fy = \u2212C\u03b3 \u03b3 (3.154) where C\u03b3 is called the camber stiffness of tire. C\u03b3 = lim \u03b3\u21920 \u2202 (\u2212Fy) \u2202\u03b3 (3.155) In presence of both, camber \u03b3 and sideslip \u03b1, the overall lateral force Fy on a tire is a superposition of the corner force and camber trust. Fy = \u2212C\u03b3 \u03b3 \u2212 C\u03b1 \u03b1 (3.156) Proof. When a wheel is under a constant load and then a camber angle is applied on the rim, the tire will deflect laterally such that it is longer in 146 3. Tire Dynamics the cambered side and shorter in the other side. Figure 3.48 compares the tireprint of a straight and a cambered tire, turning slowly on a flat road. As the wheel turns forward, undeflected treads enter the tireprint region and deflect laterally as well as longitudinally. However, because of the shape of the tireprint, the treads entering the tireprint closer to the cambered side, have more time to be stretched laterally. Because the developed lateral stress is proportional to the lateral stretch, the nonuniform tread stretching generates an asymmetric stress distribution and more lateral stress will be developed on the cambered side. The result of the nonuniform lateral stress distribution over the tireprint of a cambered tire produces the camber trust Fy in the cambered direction. Fy = Fy j\u0302 (3.157) Fy = Z AP \u03c4y dA (3.158) The camber trust is proportional to the camber angle for small angles. Fy = \u2212C\u03b3 \u03b3 (3.159) The camber trust Fy shifts a distance ax\u03b3 forward when the cambered tire turns on the road. The resultant moment Mz = Mz k\u0302 (3.160) Mz = Fy ax\u03b3 (3.161) 3. Tire Dynamics 147 is called camber torque, and the distance ax\u03b3 is called camber trail. Camber trail is usually very small and hence, the camber torque can be ignored in linear analysis of vehicle dynamics. Because the tireprint of a cambered tire deforms to be longer in the cambered side, the resultant vertical force Fz Fz = Z AP \u03c3z dA (3.162) that supports the wheel load, shifts laterally to a distance ay\u03b3 from the center of the tireprint. ay\u03b3 = 1 Fz Z AP y \u03c3z dAp (3.163) The distance ay\u03b3 is called the camber arm, and the resultant moment Mx 148 3. Tire Dynamics is called the camber moment. Mx = Mx k\u0302 (3.164) Mx = \u2212Fz ay\u03b3 (3.165) The camber moment tends to turn the tire about the x-axis and make the tire-plane align with the z-axis. The camber arm ay\u03b3 is proportional to the camber angle \u03b3 for small angles. ay\u03b3 = Cy\u03b3 \u03b3 (3.166) Figure 3.49 shows the camber force Fy for different camber angle \u03b3 at a constant tire load Fz = 4500N. Radial tires generate lower camber force due to their higher flexibility. It is better to illustrate the effect of Fz graphically to visualize the camber force. Figure 3.50 depicts the variation of camber force Fy as a function of normal load Fz at different camber angles for a sample radial tire. If we apply a slip angle \u03b1 to a turning cambered tire, the tireprint will distort similar to the shape in Figure 3.51 and the path of treads become more complicated. The resultant lateral force would be at a distance ax\u03b3 and ay\u03b3 from the center of the tireprint. Both distances ax\u03b3 and ay\u03b3 are functions of angles \u03b1 and \u03b3. Camber force due to \u03b3, along with the corner force due to \u03b1, give the total lateral force applied on a tire. Therefore, the lateral force can be calculated as Fy = \u2212C\u03b1 \u03b1\u2212 C\u03b3 \u03b3 (3.167) 3. Tire Dynamics 149 150 3. Tire Dynamics that is acceptable for \u03b3 . 10 deg and \u03b1 . 5 deg. Presence of both camber angle \u03b3 and slip angle \u03b1 makes the situation interesting because the total lateral force can be positive or negative. Figure 3.52 illustrates an example of lateral force as a function of \u03b3 and \u03b1 at a constant load Fz = 4000N. Similar to lateral force, the aligning moment Mz can be approximated as a combination of the slip and camber angle effects Mz = CM\u03b1 \u03b1+ CM\u03b3 \u03b3. (3.168) For a radial tire, CM\u03b1 \u2248 0.013Nm/deg and CM\u03b3 \u2248 0.0003Nm/deg, while for a non-radial tire, CM\u03b1 \u2248 0.01Nm/deg and CM\u03b3 \u2248 0.001Nm/deg. Example 114 Banked road. Consider a vehicle moving on a road with a transversal slope \u03b2, while its tires remain vertical. There is a downhill component of weight, F1 = mg sin\u03b2, that pulls the vehicle down. There is also an uphill camber force due to camber \u03b3 \u2248 \u03b2 of tires with respect to the road F2 = C\u03b3 \u03b3. The resultant lateral force Fy = C\u03b3 \u03b3 \u2212 mg sin\u03b2 depends on camber stiffness C\u03b3 and determines if the vehicle goes uphill or downhill. Since the camber stiffness C\u03b3 is higher for non-radial tires, it is more possible for a radial tire to go downhill and a non-radial uphill. The effects of cambering are particularly important for motorcycles that produce a large part of the cornering force by cambering. For cars and trucks, the cambering angles are much smaller and in many applications their effect can be negligible. However, some suspensions are designed to make the wheels camber when the axle load varies. 3. Tire Dynamics 151 Example 115 Camber importance and tireprint model. Cambering of a tire creates a lateral force, even though there is no sideslip. The effects of cambering are particularly important for motorcycles that produce a large part of the lateral force by camber. The following equations are presented to model the lateral deviation of a cambered tireprint from the straight tireprint, and expressing the lateral stress \u03c4y due to camber y = \u2212 sin \u03b3 \u00b3q R2g \u2212 x2 \u2212 q R2g \u2212 a2 \u00b4 (3.169) \u03c4y = \u2212\u03b3k \u00a1 a2 \u2212 x2 \u00a2 (3.170) where k is chosen such that the average camber defection is correct in the tireprint Z a \u2212a \u03c4y dx = Z a \u2212a y dx. (3.171) Therefore, k = 3 sin \u03b3 4a3\u03b3 \u00b5 \u2212a q R2g \u2212 a2 +R2g sin \u22121 a Rg \u00b6 (3.172) \u2248 3 4 Rg q R2g \u2212 a2 a2 (3.173) and \u03c4y = \u2212 3 4 \u03b3 Rg q R2g \u2212 a2 a2 \u00a1 a2 \u2212 x2 \u00a2 . (3.174) 3.9 Tire Force Tires may be considered as a force generator with two major outputs: forward force Fx, lateral force Fy, and three minor outputs: aligning moment Mz, roll moment Mx, and pitch moment My. The input of the force generator is the tire load Fz, sideslip \u03b1, longitudinal slip s, and the camber angle \u03b3. Fx = Fx (Fz, \u03b1, s, \u03b3) (3.175) Fy = Fy (Fz, \u03b1, s, \u03b3) (3.176) Mx = Mx (Fz, \u03b1, s, \u03b3) (3.177) My = My (Fz, \u03b1, s, \u03b3) (3.178) Mz = Mz (Fz, \u03b1, s, \u03b3) (3.179) Ignoring the rolling resistance and aerodynamic force, and when the tire is under a load Fz plus only one more of the inputs \u03b1, s, or \u03b3, the major 152 3. Tire Dynamics output forces can be approximated by a set of linear equations Fx = \u03bcx (s) Fz (3.180) \u03bcx (s) = Cs s Fy = \u2212C\u03b1 \u03b1 (3.181) Fy = \u2212C\u03b3 \u03b3 (3.182) where, Cs is the longitudinal slip coefficient, C\u03b1 is the lateral stiffness, and C\u03b3 is the camber stiffness. When the tire has a combination of tire inputs, the tire forces are called tire combined force. The most important tire combined force is the shear force because of longitudinal and sideslips. However, as long as the angles and slips are within the linear range of tire behavior, a superposition can be utilized to estimate the output forces. Driving and braking forces change the lateral force Fy generated at any sideslip angle \u03b1. This is because the longitudinal force pulls the tireprint in the direction of the driving or braking force and hence, the length of lateral displacement of the tireprint will also change. Figure 3.53 illustrates how a sideslip \u03b1 affects the longitudinal force ratio Fx/Fz as a function of slip ratio s. Figure 3.54 illustrates the effect of sideslip \u03b1 on the lateral force ratio Fy/Fz as a function of slip ratio s. Figure 3.55 and 3.56 illustrate the same force ratios as Figures 3.53 and 3.54 when the slip ratio s is a parameter. Proof. Consider a turning tire under a sideslip angle \u03b1. The tire develops a lateral force Fy = \u2212C\u03b1 \u03b1. Applying a driving or braking force on this tire will reduce the lateral force while developing a longitudinal force Fx = \u03bcx (s) Fz. Experimental data shows that the reduction in lateral force in presence of a slip ratio s is similar to Figure 3.54. Now assume the sideslip \u03b1 is reduced to zero. Reduction \u03b1 will increase the longitudinal force while decreasing the lateral force. Increasing the longitudinal force is experimentally similar to Figure 3.55. A turning tire under a slip ratio s develops a longitudinal force Fx = \u03bcx (s) Fz. Applying a sideslip angle \u03b1 will reduce the longitudinal force while developing a lateral force. Experimental data shows that the reduction in longitudinal force in presence of a sideslip \u03b1 is similar to Figure 3.53. Now assume the slip ratio s and hence, the driving or breaking force is reduced to zero. Reduction s will increase the lateral force while decreasing the longitudinal force. Increasing the lateral force is similar to Figure 3.54. Example 116 Pacejka model. An approximate equation is presented to describe force Equations (3.175) 3. Tire Dynamics 153 154 3. Tire Dynamics 3. Tire Dynamics 155 or (3.176). This equation is called the Pacejka model. F = A sin \u00a9 B tan\u22121 \u00a3 Cx\u2212D \u00a1 Cx\u2212 tan\u22121 (Cx) \u00a2\u00a4\u00aa (3.183) A = \u03bcFz (3.184) C = C\u03b1 AB (3.185) B,D = shape factors (3.186) The Pacejka model is substantially empirical. However, when the parameters A, B, C, D, C1, and C2 are determined for a tire, the equation expresses the tire behavior well enough. Figure 3.57 illustrates how the parameters can be determined from a test force-slip experimental result. Example 117 Friction ellipse. When the tire is under both longitudinal and sideslips, the tire is under combined slip. The shear force on the tireprint of a tire under a combined slip can approximately be found using a friction ellipse model.\u00b5 Fy FyM \u00b62 + \u00b5 Fx FxM \u00b62 = 1 (3.187) A friction ellipse is shown in Figure 3.58. Proof. The shear force Fshear, applied on the tire at tireprint, parallel to the ground surface, has two components: the longitudinal force Fx and the lateral force Fy. Fshear = Fx \u0131\u0302+ Fy j\u0302 (3.188) Fx = Cs sFz (3.189) Fy = \u2212C\u03b1 \u03b1 (3.190) 156 3. Tire Dynamics These forces cannot exceed their maximum values FyM and FxM . FyM = \u03bcy Fz FxM = \u03bcx Fz The tire shown in Figure 3.58 is moving along the velocity vector v at a sideslip angle \u03b1. The x-axis indicates the tire-plane. When there is no sideslip, the maximum longitudinal force is FxM = \u03bcx Fz = \u2212\u2192 OA. Now, if a sideslip angle \u03b1 is applied, a lateral force Fy = \u2212\u2212\u2192 OE is generated, and the longitudinal force reduces to Fx = \u2212\u2212\u2192 OB. The maximum lateral force would be FyM = \u03bcy Fz = \u2212\u2212\u2192 OD when there is no longitudinal slip. In presence of the longitudinal and lateral forces, we may assume that the tip point of the maximum shear force vector is on the following friction ellipse: \u00b5 Fy FyM \u00b62 + \u00b5 Fx FxM \u00b62 = 1 (3.191) When \u03bcx = \u03bcy = \u03bc, the friction ellipse would be a circle and Fshear = \u03bcFz. (3.192) 3. Tire Dynamics 157 Example 118 Wide tires. A wide tire has a shorter tireprint than a narrow tire. Assuming the same vehicle and same tire pressure, the area of tireprint would be equal in both tires. The shorter tireprint at the same sideslip has more of its length stuck to the road than longer tireprint. So, a wider tireprint generates more lateral force than a narrower tireprint for the same tire load and sideslip. Generally speaking, tire performance and maximum force capability decrease with increasing speed in both wide and narrow tires. Example 119 sin tire forces model. A few decades ago, a series of applied sine functions were developed based on experimental data to model tire forces. The sine functions, which are explained below, may be used to model tire forces, especially for computer purposes, effectively. The lateral force of a tire is Fy = A sin \u00a9 B tan\u22121 (C\u03a6) \u00aa (3.193) \u03a6 = (1\u2212E) (\u03b1+ \u03b4)\u03bcFz (3.194) C = C\u03b1 AB (3.195) C\u03b1 = C1 sin \u00b5 2 tan\u22121 Fz C2 \u00b6 (3.196) A,B = Shape factors (3.197) C1 = Maximum cornering stiffness (3.198) C2 = Tire load at maximum cornerin stiffness (3.199) 3.10 Summary We attach a coordinate frame (oxyz) to the tire at the center of the tireprint, called the tire frame. The x-axis is along the intersection line of the tire-plane and the ground. The z-axis is perpendicular to the ground, and the y-axis makes the coordinate system right-hand. We show the tire orientation using two angles: camber angle \u03b3 and sideslip angle \u03b1. The camber angle is the angle between the tire-plane and the vertical plane measured about the x-axis, and the sideslip angle \u03b1 is the angle between the velocity vector v and the x-axis measured about the z-axis. A vertically loaded wheel turning on a flat surface has an effective radius Rw, called rolling radius Rw = vx \u03c9w (3.200) where vx is the forward velocity, and \u03c9w is the angular velocity of the 158 3. Tire Dynamics wheel. The effective radius Rw is approximately equal to Rw \u2248 Rg \u2212 Rg \u2212Rh 3 (3.201) and is a number between the unloaded or geometric radius Rg and the loaded height Rh. Rh < Rw < Rg (3.202) A turning tire on the ground generates a longitudinal force called rolling resistance. The force is opposite to the direction of motion and is proportional to the normal force on the tireprint. Fr = \u03bcr Fz (3.203) The parameter \u03bcr is called the rolling friction coefficient and is a function of tire mechanical properties, speed, wear, temperature, load, size, driving and braking forces, and road condition. The tire force in the x-direction is a combination of the longitudinal force Fx and the roll resistance Fr. The longitudinal force is Fx = \u03bcx (s) Fz (3.204) where s is the longitudinal slip ratio of the tire s = Rg\u03c9w vx \u2212 1 (3.205) \u03bcx (s) = Cs s s << 1 (3.206) The wheel force in the tire y-direction, Fy, is a combination of the lateral force and the tire roll resistance Fr. The lateral force is Fy = \u2212C\u03b3 \u03b3 \u2212 C\u03b1 \u03b1 (3.207) where \u2212C\u03b3\u03b3 is called the camber trust and C\u03b1\u03b1 is called the sideslip force. 3. Tire Dynamics 159 3.11 Key Symbols a \u2261 x\u0308 acceleration a, b semiaxes of AP ax\u03b1 ax\u03b3 camber trail ay\u03b3 camber arm AP tireprint area c1, c2, c3, c4 coefficients of the function Fx = Fx (s) C0, C1, C2 coefficients of the polynomial function Fr = Fr (vx) Cs longitudinal slip coefficient Csx , Csy longitudinal and lateral slip coefficients C\u03b1 sideslip coefficient C\u03b3 camber stiffness d distance of tire travel dF no slip tire travel dA actual tire travel D tire diameter E Young modulus f function fk spring force Fr Fr rolling resistance force Fx longitudinal force, forward force Fy lateral force FyM pneumatic trail Fz normal force, vertical force, wheel load g g gravitational acceleration k stiffness k1, k2, k3 nonlinear tire stiffness coefficients keq equivalent stiffness ks slope of Fx (s) versus s at s = 0 kx tire stiffness in the x-direction ky tire stiffness in the y-direction kz tire stiffness in the z-direction K radial and non-radial tires parameter in \u03bcr = \u03bcr (p, vx) m mass Mr Mr rolling resistance moment Mx, Mx roll moment, bank moment, tilting torque, My pitch moment, rolling resistance torque Mz yaw moment, aligning moment, self aligning moment, bore torque n exponent for shape and stress distribution of AP n1 number of tire rotations p tire inflation pressure P rolling resistance power 160 3. Tire Dynamics r radial position of tire periphery r = \u03c9/\u03c9n frequency ratio Rg geometric radius Rh loaded height Rw rolling radius s longitudinal slip sy lateral slip T wheel torque v \u2261 x\u0307, v velocity x, y, z, x displacement x, y, z coordinate axes 4x tire deflection in the x-direction, rolling resistance arm 4y tire deflection in the y-direction 4z tire deflection in the z-direction z\u0307 tire deflection rate in the z-direction \u03b1 sideslip angle \u03b1M maximum sideslip angle \u03b2 transversal slope \u03b3 camber angle \u03b4 deflection 4x tire deflection in the x-direction, rolling resistance arm 4y tire deflection in the y-direction 4z tire deflection in the z-direction \u03b8 tire angular rotation \u03bc0, \u03bc1 nonlinear rolling friction coefficient \u03bcr rolling friction coefficient \u03bcx (s) longitudinal friction coefficient \u03bcdp friction coefficient driving peak value \u03bcds friction coefficient steady-state value \u03c3zM maximum normal stress \u03c3z(x, y) normal stress over the tireprint \u03c3zm normal stress mean value \u03c4x(x, y), \u03c4y(x, y) shear stresses over the tireprint \u03c4xM , \u03c4yM maximum shear stresses \u03d5 contact angle, angular length of AP \u03c9eq equivalent tire angular velocity \u03c9w angular velocity of a wheel \u03c9w actual tire angular velocity 3. Tire Dynamics 161"
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        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure13.38-1.png",
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        "caption": "Figure 13.38 (a) The dead weight of beam ACD of 5 kN/m, resolved into components (b) normal to the roof plane and (c) parallel to the roof plane.",
        "texts": [
            " Dead weight In Figure 13.37 we take a closer look at a member segment with length a. The dead weight of this member segment is aqdw with components aqdw cos \u03b1 and aqdw sin \u03b1 respectively normal to and parallel to the beam axis. For the components of the distributed load normal to and parallel to the beam axis we find qdw;tr = aqdw cos \u03b1 a = qdw cos \u03b1 = (5 kN/m) cos 21.8\u25e6 = 4.642 kN/m, qdw;pa = aqdw sin \u03b1 a = qdw sin \u03b1 = (5 kN/m) sin 21.8\u25e6 = 1.857 kN/m. The distributed load qdw = 5 kN/m due to the dead weight (Figure 13.38a) has components of 4.642 kN/m normal to the beam axis (Figure 13.38b) and 1.857 kN/m parallel to the beam axis (Figure 13.38c). The bending moment in ACD is caused by the load of 4.642 kN/m normal to the beam axis. The M diagram is equal to that in Figure 13.33d, but 4.462 as large, so that the extreme values of the bending moments are Mdw;max = (2.558 kNm) \u00d7 4.642 = 11.87 kNm, Mdw;min = (2.320 kNm) \u00d7 4.642 = 10.77 kNm. 13 Calculating M, V and N Diagrams 579 580 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM The N diagram in Figure 13.39a due to the 4.642 kN/m load normal to the beam axis is equal to the N diagram in Figure 13"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure5.15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure5.15-1.png",
        "caption": "Fig. 5.15. Laboratory set for measuring eddy current losses in the stator winding: (a) specially designed stator; (b) experimental machine; (c) schematic of experimental set-up.",
        "texts": [
            " The resistance limited eddy loss in the stator of an AFPM machine may be experimentally determined by measuring the difference in input shaft powers of the AFPM machine at the same speed, first with the stator in, and then by replacing it with a dummy stator (no conductors). The dummy stator has the same dimensions and surface finish as the real stator and is meant to keep the windage losses the same. A schematic representation of the experimental test setup for measuring eddy current losses in stator conductors is shown in Fig. 5.15. The shaft of the tested prototype and the shaft of the prime mover (driving machine) are coupled together via a torque meter. The stator is positioned in the middle of the two rotor discs with the outer end ring (see Fig. 5.15a) mounted on the outside supporting frame. Temperature sensors are also attached to the conductors in order to take into account temperature factor in the measurement. Initially, the prototype (with the coreless stator placed in it) was driven by a variable speed motor for a number of different speeds. The corresponding torque measurements were taken. Replacing the real stator with a dummy one, the tests were repeated for the same speeds. Both the torque and the temperature values were recorded. The difference of the torques multiplied by the speed gives the eddy current loss"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002769_j.cirpj.2021.04.010-Figure62-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002769_j.cirpj.2021.04.010-Figure62-1.png",
        "caption": "Fig. 62. Hollow pipe with multiple branches (a) CAD model, (b) fabricated part [44].",
        "texts": [
            " It was reported that grooves\u2019 inclination angle plays a vital role in complete filling of these grooves. Fabrication error was less than 0.3%, with no porosity, crack, and inclusions or infusions. The buy-to-fly ratio was about 92.12%, which indicates high material utilization. Fabrication of multi-directional parts through GMAW-AW GMAW based additive manufacturing was also employed to fabricate multi-directional parts and lattice structures by tilting the GMAW torch at different angles. A hollow pipe with multiple branches (Fig. 62) was manufactured using YHJ507M wire [44]. Here dimensional and angle error was within 1 mm and 0.5 . A lattice structure was fabricated in Ref. [67] using ER2319 aluminium alloy wire (Fig. 63). Geometrical accuracy of lattice structure directly depends upon strut diameter and angle, as the interconnection struts formed it. Strut diameter was depends upon ig. 56. Relationship between volume ratio of YS308L and chemical composition at he interface [49]. 427 S. Pattanayak and S.K. Sahoo CIRP Journal of Manufacturing Science and Technology 33 (2021) 398\u2013442 droplet diameter, number of droplets transferred onto the substrate, welding current, voltage, and pulse on time"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000655_ac801854j-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000655_ac801854j-Figure2-1.png",
        "caption": "Figure 2. Cyclic voltammograms of (a) CPE in 0.1 M PBS (pH 7.0), (b) CPE in 0.5 mM EP, (c) 2PHCMCNPE in 0.1 M PBS, (d) CNPE in 0.5 mM EP, (e) 2PHCMCPE in 0.5 mM EP, and (f) 2PHCMCNPE in 0.5 mM EP. In all cases the scan rate was 30 mV s-1.",
        "texts": [
            " The peak separation potential, \u2206E p () Epa - Epc), is greater than the (59/ n) mV expected for a reversible system, which indicates a quasi reversible behavior for the mediator in an aqueous medium. In addition, the effect of the scan rate of the potential on the electrochemical properties of the 2PHCMCNPE was studied in PBS by cyclic voltammetry. The plots of the anodic and cathodic peak currents were linearly dependent on the sweep rate (v) (Figure 1C). This behavior indicates that the nature of the redox process is diffusionless controlled. Electrocatalytic Oxidation of EP. Figure 2 depicts the cyclic voltammetric responses from the electrochemical oxidation of 0.5 mM EP at 2PHCMCNPE (curve f), 2PHC modified CPE (2PHCMCPE) (curve e), CNPE (curve d), and bare CPE (curve b). As can be seen, the anodic peak potential for the oxidation of EP at 2PHCMCNPE (curve f) and 2PHCMCPE (curve e) is about 210 mV, while at the CNPE (curve d) and bare CPE(curve b), peak potentials are about 410 and 450 mV, respectively. From these results it is concluded that the best electrocatalytic effect for EP oxidation is observed at 2PHCMCNPE (curve f)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure5.9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure5.9-1.png",
        "caption": "Fig. 5.9. Layout of different coreless windings with (a) normal overlap winding, (b) non-overlap winding, and (c) phase-group non-overlap winding.",
        "texts": [
            " Furthermore, non-overlap coil PM machines with ferromagnetic stator core are known for their additional losses in the current conducting PMs and rotor ferromagnetic yoke due to higher harmonic flux pulsations. In coreless non-overlap winding AFPM machines this is almost completely absent due to the low armature reaction effect. There are, thus, no disadvantages in using non-overlap coils with concentrated parameters in coreless stator AFPM machines, except for the lower torque performance in the case of some winding and PM configurations. In Fig. 5.9 the layouts of overlap and non-overlap windings are shown. Congruent with these layouts, three air-cored AFPM stators (1 kW) with normal overlap, non-overlap and phase-group windings are shown in Fig. 5.10(a)-(c). These stators have been designed and built for the PM rotor discs of the AFPM machine under test shown in Fig. 5.10(d). For the tests an a.c. electromechanical drive has been used as prime mover and a torque transducer to measure the shaft torque as shown in Fig. 5.10(d). For the numeric modeling of the AFPM machine the 2D FEM analysis has been used"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001730_tie.2017.2739700-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001730_tie.2017.2739700-Figure1-1.png",
        "caption": "Fig. 1. Body-fixed frame and earth-fixed frame for the quadrotor",
        "texts": [
            ", kn > 0 be such that the polynomial sn + kns n\u22121 + ...+ k2s+ k1 is Hurwitz, and consider multivariable integrator system x\u03071 = x2, ..., x\u0307n\u22121 = xn, x\u0307n = u (2) with xi,u \u2208 Rm. There exists \u03f5 \u2208 (0, 1) such that, for every \u03b1 \u2208 (1 \u2212 \u03f5, 1), the origin is a globally finite time stable equilibrium for the system under the feedback control law u = \u2212k1\u2308x1\u230b\u03b11 \u2212 ...\u2212 kn\u2308xn\u230b\u03b1n (3) where \u03b11, ..., \u03b1n satisfy \u03b1i\u22121 = \u03b1i\u03b1i+1 2\u03b1i+1 \u2212 \u03b1i , i = 2, ..., n (4) with \u03b1n = \u03b1, \u03b1n+1 = 1. Proof: See Appendix A. The configuration of the quadrotor UAV is given in Fig.1. It is composed by a rigid cross frame and four rotors. The position (x, y and z direction) and attitude (roll, pitch and yaw) motion can be achieved through an appropriate combination of the rotors 1 to 4 [10]. The 3DOF rotational motions of a quadrotor UAV is described by \u0398\u0307 = W\u2126 (5) I\u2126\u0307 = \u2212\u2126\u00d7 I\u2126+ \u03c4 +\u2206(t) (6) where \u0398 = [\u03d5, \u03b8, \u03c8] is the Euler angle expressed in inertial frame, \u2126 = [p, q, r]T denotes the attitude angular velocity expressed in the body frame, \u2206(t) represents the bounded external disturbances, \u03c4 = [\u03c41, \u03c42, \u03c43] T \u2208 R3 is the control torque vector, and I = diag[Ix, Iy, Iz] is a symmetric positive definite constant matrix"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure9.21-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure9.21-1.png",
        "caption": "FIGURE 9.21. A compound pendulum attached with a linear spring at the tip point.",
        "texts": [],
        "surrounding_texts": [
            "Dynamics of a rigid vehicle may be considered as the motion of a rigid body with respect to a fixed global coordinate frame. The principles of Newton and Euler equations of motion that describe the translational and rotational motion of the rigid body are reviewed in this chapter. 9.1 Force and Moment In Newtonian dynamics, the forces acting on a system of connected rigid bodied can be divided into internal and external forces. Internal forces are acting between connected bodies, and external forces are acting from outside of the system. An external force can be a contact force, such as the traction force at the tireprint of a driving wheel, or a body force, such as the gravitational force on the vehicle\u2019s body. External forces and moments are called load, and a set of forces and moments acting on a rigid body, such as forces and moments on the vehicle shown in Figure 9.1, is called a force system. The resultant or total force F is the sum of all the external forces acting on a body, and the resultant or 522 9. Applied Dynamics total moment M is the sum of all the moments of the external forces. F = X i Fi (9.1) M = X i Mi (9.2) Consider a force F acting on a point P at rP . The moment of the force about a directional line l passing through the origin is Ml = lu\u0302 \u00b7 (rP \u00d7F) (9.3) where u\u0302 is a unit vector on l. The moment of the force F, about a point Q at rQ is MQ = (rP \u2212 rQ)\u00d7F (9.4) so, the moment of F about the origin is M = rP \u00d7F. (9.5) The moment of a force may also be called torque or moment. The effect of a force system is equivalent to the effect of the resultant force and resultant moment of the force system. Any two force systems are equivalent if their resultant forces and resultant moments are equal. If the resultant force of a force system is zero, the resultant moment of the force system is independent of the origin of the coordinate frame. Such a resultant moment is called couple. When a force system is reduced to a resultant FP andMP with respect to a reference point P , we may change the reference point to another point Q and find the new resultants as FQ = FP (9.6) MQ = MP + (rP \u2212 rQ)\u00d7FP = MP + QrP \u00d7FP . (9.7) The momentum of a moving rigid body is a vector quantity equal to the total mass of the body times the translational velocity of the mass center of the body. p = mv (9.8) The momentum p is also called translational momentum or linear momentum. Consider a rigid body with momentum p. The moment of momentum, L, about a directional line l passing through the origin is Ll = lu\u0302 \u00b7 (rC \u00d7 p) (9.9) 9. Applied Dynamics 523 where u\u0302 is a unit vector indicating the direction of the line, and rC is the position vector of the mass center C. The moment of momentum about the origin is L = rC \u00d7 p. (9.10) The moment of momentum L is also called angular momentum. A bounded vector is a vector fixed at a point in space. A sliding or line vector is a vector free to slide on its line of action. A free vector is a vector that may move to any point as long as it keeps its direction. Force is a sliding vector and couple is a free vector. However, the moment of a force is dependent on the distance between the origin of the coordinate frame and the line of action. The application of a force system is emphasized by Newton\u2019s second and third laws of motion. The second law of motion, also called the Newton\u2019s equation of motion, states that the global rate of change of linear momentum is proportional to the global applied force. GF = Gd dt Gp = Gd dt \u00a1 mGv \u00a2 (9.11) The third law of motion states that the action and reaction forces acting between two bodies are equal and opposite. The second law of motion can be expanded to include rotational motions. Hence, the second law of motion also states that the global rate of change of angular momentum is proportional to the global applied moment. GM = Gd dt GL (9.12) Proof. Differentiating from angular momentum (9.10) shows that Gd dt GL = Gd dt (rC \u00d7 p) = \u00b5 GdrC dt \u00d7 p+ rC \u00d7 Gdp dt \u00b6 = GrC \u00d7 Gdp dt = GrC \u00d7 GF = GM. (9.13) Kinetic energy K of a moving body point P with mass m at a position GrP , and having a velocity GvP , is K = 1 2 mGv2P = 1 2 m \u00b3 Gd\u0307B + BvP + B G\u03c9B \u00d7 BrP \u00b42 . (9.14) 524 9. Applied Dynamics The work done by the applied force GF on m in moving from point 1 to point 2 on a path, indicated by a vector Gr, is 1W2 = Z 2 1 GF \u00b7 dGr. (9.15) However, Z 2 1 GF \u00b7 dGr = m Z 2 1 Gd dt Gv \u00b7 Gvdt = 1 2 m Z 2 1 d dt v2dt = 1 2 m \u00a1 v22 \u2212 v21 \u00a2 = K2 \u2212K1 (9.16) that shows 1W2 is equal to the difference of the kinetic energy between terminal and initial points. 1W2 = K2 \u2212K1 (9.17) Equation (9.17) is called principle of work and energy. Example 342 Position of center of mass. The position of the mass center of a rigid body in a coordinate frame is indicated by BrC and is usually measured in the body coordinate frame. BrC = 1 m Z B Br dm (9.18) \u23a1\u23a3 xC yC zC \u23a4\u23a6 = \u23a1\u23a2\u23a2\u23a3 1 m R B x dm 1 m R B y dm 1 m R B z dm \u23a4\u23a5\u23a5\u23a6 (9.19) Applying the mass center integral on the symmetric and uniform L-section rigid body with \u03c1 = 1 shown in Figure 9.2 provides the position of mass center C of the section. The x position of C is xC = 1 m Z B xdm = 1 A Z B x dA = \u2212b 2 + ab\u2212 a2 4ab+ 2a2 (9.20) and because of symmetry, we have yC = \u2212xC = b2 + ab\u2212 a2 4ab+ 2a2 . (9.21) 9. Applied Dynamics 525 When a = b, the position of C reduces to yC = \u2212xC = 1 2 b. (9.22) Example 343 F Every force system is equivalent to a wrench. The Poinsot theorem states: Every force system is equivalent to a single force, plus a moment parallel to the force. Let F andM be the resultant force and moment of a force system. We decompose the moment into parallel and perpendicular components,Mk andM\u22a5, to the force axis. The force F and the perpendicular moment M\u22a5 can be replaced by a single force F0 parallel to F. Therefore, the force system is reduced to a force F0 and a moment Mk parallel to each other. A force and a moment about the force axis is called a wrench. The Poinsot theorem is similar to the Chasles theorem that states: Every rigid body motion is equivalent to a screw, which is a translation plus a rotation about the axis of translation. Example 344 F Motion of a moving point in a moving body frame. The velocity and acceleration of a moving point P as shown in Figure 5.12 are found in Example 200. GvP = Gd\u0307B + GRB \u00a1 BvP + B G\u03c9B \u00d7 BrP \u00a2 (9.23) GaP = Gd\u0308B + GRB \u00a1 BaP + 2 B G\u03c9B \u00d7 BvP + B G\u03c9\u0307B \u00d7 BrP \u00a2 +GRB \u00a1 B G\u03c9B \u00d7 \u00a1 B G\u03c9B \u00d7 BrP \u00a2\u00a2 (9.24) Therefore, the equation of motion for the point mass P is GF = mGaP = m \u00b3 Gd\u0308B + GRB \u00a1 BaP + 2 B G\u03c9B \u00d7 BvP + B G\u03c9\u0307B \u00d7 BrP \u00a2\u00b4 +m GRB \u00a1 B G\u03c9B \u00d7 \u00a1 B G\u03c9B \u00d7 BrP \u00a2\u00a2 . (9.25) 526 9. Applied Dynamics Example 345 Newton\u2019s equation in a rotating frame. Consider a spherical rigid body, such as Earth, with a fixed point that is rotating with a constant angular velocity. The equation of motion for a moving point vehicle P on the rigid body is found by setting Gd\u0308B = B G\u03c9\u0307B = 0 in the equation of motion of a moving point in a moving body frame (9.25) BF = mBaP +mB G\u03c9B \u00d7 \u00a1 B G\u03c9B \u00d7 BrP \u00a2 + 2mB G\u03c9B \u00d7 B r\u0307P (9.26) 6= mBaP which shows that the Newton\u2019s equation of motion F = ma must be modified for rotating frames. Example 346 Coriolis force. The equation of motion of a moving vehicle point on the surface of the Earth is BF = mBaP +mB G\u03c9B \u00d7 \u00a1 B G\u03c9B \u00d7 BrP \u00a2 + 2mB G\u03c9B \u00d7 BvP (9.27) which can be rearranged to BF\u2212mB G\u03c9B \u00d7 \u00a1 B G\u03c9B \u00d7 BrP \u00a2 \u2212 2mB G\u03c9B \u00d7 BvP = mBaP . (9.28) Equation (9.28) is the equation of motion for an observer in the rotating frame, which in this case is an observer on the Earth. The left-hand side of this equation is called the effective force Feff , Feff = BF\u2212mB G\u03c9B \u00d7 \u00a1 B G\u03c9B \u00d7 BrP \u00a2 \u2212 2mB G\u03c9B \u00d7 BvP (9.29) because it seems that the particle is moving under the influence of this force. The second term is negative of the centrifugal force and pointing outward. The maximum value of this force on the Earth is on the equator r\u03c92 = 6378.388\u00d7 103 \u00d7 \u00b5 2\u03c0 24\u00d7 3600 366.25 365.25 \u00b62 = 3.3917\u00d7 10\u22122m/ s2 (9.30) which is about 0.3% of the acceleration of gravity. If we add the variation of the gravitational acceleration because of a change of radius from R = 6356912m at the pole to R = 6378388m on the equator, then the variation of the acceleration of gravity becomes 0.53%. So, generally speaking, a sportsman such as a pole-vaulter who has practiced in the north pole can show a better record in a competition held on the equator. The third term is called the Coriolis force or Coriolis effect, FC, which is perpendicular to both \u03c9 and BvP . For a mass m moving on the north hemisphere at a latitude \u03b8 towards the equator, we should provide a 9. Applied Dynamics 527 lateral eastward force equal to the Coriolis effect to force the mass, keeping its direction relative to the ground. FC = 2mB G\u03c9B \u00d7 Bvm = 1.4584\u00d7 10\u22124 Bpm cos \u03b8 kgm/ s2 (9.31) The Coriolis effect is the reason why the west side of railways, roads, and rivers wears. The lack of providing the Coriolis force is the reason for turning the direction of winds, projectiles, flood, and falling objects westward. Example 347 Work, force, and kinetic energy in a unidirectional motion. A vehicle with mass m = 1200 kg has an initial kinetic energy K = 6000 J. The mass is under a constant force F = F I\u0302 = 4000I\u0302 and moves from X(0) = 0 to X(tf ) = 1000m at a terminal time tf . The work done by the force during this motion is W = Z r(tf ) r(0) F \u00b7 dr = Z 1000 0 4000 dX = 4\u00d7 106Nm = 4MJ (9.32) The kinetic energy at the terminal time is K(tf ) =W +K(0) = 4006000 J (9.33) which shows that the terminal speed of the mass is v2 = r 2K(tf ) m \u2248 81.7m/ s. (9.34) Example 348 Direct dynamics. When the applied force is time varying and is a known function, then, F(t) = m r\u0308. (9.35) The general solution for the equation of motion can be found by integration. r\u0307(t) = r\u0307(t0) + 1 m Z t t0 F(t)dt (9.36) r(t) = r(t0) + r\u0307(t0)(t\u2212 t0) + 1 m Z t t0 Z t t0 F(t)dt dt (9.37) This kind of problem is called direct or forward dynamics. 528 9. Applied Dynamics 9.2 Rigid Body Translational Dynamics Figure 9.3 depicts a moving body B in a global coordinate frame G. Assume that the body frame is attached at the mass center of the body. Point P indicates an infinitesimal sphere of the body, which has a very small mass dm. The point mass dm is acted on by an infinitesimal force df and has a global velocity GvP . According to Newton\u2019s law of motion we have df = GaP dm. (9.38) However, the equation of motion for the whole body in a global coordinate frame is GF = mGaB (9.39) which can be expressed in the body coordinate frame as BF = mB GaB +m B G\u03c9B \u00d7 BvB (9.40)\u23a1\u23a3 Fx Fy Fz \u23a4\u23a6 = \u23a1\u23a3 max +m (\u03c9yvz \u2212 \u03c9zvy) may \u2212m (\u03c9xvz \u2212 \u03c9zvx) maz +m (\u03c9xvy \u2212 \u03c9yvx) \u23a4\u23a6 . (9.41) In these equations, GaB is the acceleration vector of the body mass center C in the global frame,m is the total mass of the body, and F is the resultant of the external forces acted on the body at C. 9. Applied Dynamics 529 Proof. A body coordinate frame at the mass center is called a central frame. If frame B is a central frame, then the center of mass, C, is defined such that Z B Brdm dm = 0. (9.42) The global position vector of dm is related to its local position vector by Grdm = GdB + GRB Brdm (9.43) where GdB is the global position vector of the central body frame, and therefore, Z B Grdm dm = Z B GdB dm+ GRB Z m Brdm dm = Z B GdB dm = GdB Z B dm = mGdB. (9.44) A time derivative of both sides shows that mGd\u0307B = mGvB = Z B Gr\u0307dm dm = Z B Gvdm dm (9.45) and another derivative is mGv\u0307B = mGaB = Z B Gv\u0307dm dm. (9.46) However, we have df = Gv\u0307P dm and therefore, mGaB = Z B df . (9.47) The integral on the right-hand side accounts for all the forces acting on the body. The internal forces cancel one another out, so the net result is the vector sum of all the externally applied forces, F, and therefore, GF = m GaB = m Gv\u0307B . (9.48) In the body coordinate frame we have BF = BRG GF = m BRG GaB = m B GaB = m BaB +m B G\u03c9B \u00d7 BvB. (9.49) 530 9. Applied Dynamics The expanded form of the Newton\u2019s equation in the body coordinate frame is then equal to BF = m BaB +m B G\u03c9B \u00d7 BvB\u23a1\u23a3 Fx Fy Fz \u23a4\u23a6 = m \u23a1\u23a3 ax ay az \u23a4\u23a6+m \u23a1\u23a3 \u03c9x \u03c9y \u03c9z \u23a4\u23a6\u00d7 \u23a1\u23a3 vx vy vz \u23a4\u23a6 = \u23a1\u23a3 max +m (\u03c9yvz \u2212 \u03c9zvy) may \u2212m (\u03c9xvz \u2212 \u03c9zvx) maz +m (\u03c9xvy \u2212 \u03c9yvx) \u23a4\u23a6 . (9.50) 9.3 Rigid Body Rotational Dynamics The rigid body rotational equation of motion is the Euler equation BM = Gd dt BL = BL\u0307+B G\u03c9B \u00d7 BL = BI B G\u03c9\u0307B + B G\u03c9B \u00d7 \u00a1 BI B G\u03c9B \u00a2 (9.51) where L is the angular momentum BL = BI B G\u03c9B (9.52) and I is the moment of inertia of the rigid body. I = \u23a1\u23a3 Ixx Ixy Ixz Iyx Iyy Iyz Izx Izy Izz \u23a4\u23a6 (9.53) The elements of I are functions of the mass distribution of the rigid body and may be defined by Iij = Z B \u00a1 r2i \u03b4mn \u2212 ximxjn \u00a2 dm , i, j = 1, 2, 3 (9.54) where \u03b4ij is Kronecker\u2019s delta. \u03b4mn = \u00bd 1 if m = n 0 if m 6= n (9.55) The expanded form of the Euler equation (9.51) is Mx = Ixx\u03c9\u0307x + Ixy\u03c9\u0307y + Ixz\u03c9\u0307z \u2212 (Iyy \u2212 Izz)\u03c9y\u03c9z \u2212Iyz \u00a1 \u03c92z \u2212 \u03c92y \u00a2 \u2212 \u03c9x (\u03c9zIxy \u2212 \u03c9yIxz) (9.56) 9. Applied Dynamics 531 My = Iyx\u03c9\u0307x + Iyy\u03c9\u0307y + Iyz\u03c9\u0307z \u2212 (Izz \u2212 Ixx)\u03c9z\u03c9x \u2212Ixz \u00a1 \u03c92x \u2212 \u03c92z \u00a2 \u2212 \u03c9y (\u03c9xIyz \u2212 \u03c9zIxy) (9.57) Mz = Izx\u03c9\u0307x + Izy\u03c9\u0307y + Izz\u03c9\u0307z \u2212 (Ixx \u2212 Iyy)\u03c9x\u03c9y \u2212Ixy \u00a1 \u03c92y \u2212 \u03c92x \u00a2 \u2212 \u03c9z (\u03c9yIxz \u2212 \u03c9xIyz) . (9.58) which can be reduced to M1 = I1\u03c9\u03071 \u2212 (I2 \u2212 I2)\u03c92\u03c93 M2 = I2\u03c9\u03072 \u2212 (I3 \u2212 I1)\u03c93\u03c91 (9.59) M3 = I3\u03c9\u03073 \u2212 (I1 \u2212 I2)\u03c91\u03c92 in a special Cartesian coordinate frame called the principal coordinate frame. The principal coordinate frame is denoted by numbers 123 to indicate the first, second, and third principal axes. The parameters Iij , i 6= j are zero in the principal frame. The body and principal coordinate frame sit at the mass center C. Kinetic energy of a rotating rigid body is K = 1 2 \u00a1 Ixx\u03c9 2 x + Iyy\u03c9 2 y + Izz\u03c9 2 z \u00a2 \u2212Ixy\u03c9x\u03c9y \u2212 Iyz\u03c9y\u03c9z \u2212 Izx\u03c9z\u03c9x (9.60) = 1 2 \u03c9 \u00b7 L (9.61) = 1 2 \u03c9T I \u03c9 (9.62) that in the principal coordinate frame reduces to K = 1 2 \u00a1 I1\u03c9 2 1 + I2\u03c9 2 2 + I3\u03c9 2 3 \u00a2 . (9.63) Proof. Let mi be the mass of the ith particle of a rigid body B, which is made of n particles and let ri = Bri = \u00a3 xi yi zi \u00a4T (9.64) be the Cartesian position vector of mi in a central body fixed coordinate frame Oxyz. Assume that \u03c9 = B G\u03c9B = \u00a3 \u03c9x \u03c9y \u03c9z \u00a4T (9.65) is the angular velocity of the rigid body with respect to the ground, expressed in the body coordinate frame. 532 9. Applied Dynamics The angular momentum of mi is Li = ri \u00d7mir\u0307i = mi [ri \u00d7 (\u03c9 \u00d7 ri)] = mi [(ri \u00b7 ri)\u03c9 \u2212 (ri \u00b7 \u03c9) ri] = mir 2 i\u03c9 \u2212mi (ri \u00b7 \u03c9) ri. (9.66) Hence, the angular momentum of the rigid body would be L = \u03c9 nX i=1 mir 2 i \u2212 nX i=1 mi (ri \u00b7 \u03c9) ri. (9.67) Substitution for ri and \u03c9 gives us L = \u00b3 \u03c9x \u0131\u0302+ \u03c9y j\u0302+ \u03c9z k\u0302 \u00b4 nX i=1 mi \u00a1 x2i + y2i + z2i \u00a2 \u2212 nX i=1 mi (xi\u03c9x + yi\u03c9y + zi\u03c9z) \u00b7 \u00b3 xi\u0131\u0302+ yij\u0302+ zik\u0302 \u00b4 (9.68) and therefore, L = nX i=1 mi \u00a1 x2i + y2i + z2i \u00a2 \u03c9x \u0131\u0302 + nX i=1 mi \u00a1 x2i + y2i + z2i \u00a2 \u03c9y j\u0302 + nX i=1 mi \u00a1 x2i + y2i + z2i \u00a2 \u03c9z k\u0302 \u2212 nX i=1 mi (xi\u03c9x + yi\u03c9y + zi\u03c9z)xi\u0131\u0302 \u2212 nX i=1 mi (xi\u03c9x + yi\u03c9y + zi\u03c9z) yij\u0302 \u2212 nX i=1 mi (xi\u03c9x + yi\u03c9y + zi\u03c9z) zik\u0302 (9.69) or L = nX i=1 mi \u00a3\u00a1 x2i + y2i + z2i \u00a2 \u03c9x \u2212 (xi\u03c9x + yi\u03c9y + zi\u03c9z)xi \u00a4 \u0131\u0302 + nX i=1 mi \u00a3\u00a1 x2i + y2i + z2i \u00a2 \u03c9y \u2212 (xi\u03c9x + yi\u03c9y + zi\u03c9z) yi \u00a4 j\u0302 + nX i=1 mi \u00a3\u00a1 x2i + y2i + z2i \u00a2 \u03c9z \u2212 (xi\u03c9x + yi\u03c9y + zi\u03c9z) zi \u00a4 k\u0302 (9.70) 9. Applied Dynamics 533 which can be rearranged as L = nX i=1 \u00a3 mi \u00a1 y2i + z2i \u00a2\u00a4 \u03c9x\u0131\u0302 + nX i=1 \u00a3 mi \u00a1 z2i + x2i \u00a2\u00a4 \u03c9y j\u0302 + nX i=1 \u00a3 mi \u00a1 x2i + y2i \u00a2\u00a4 \u03c9zk\u0302 \u2212 \u00c3 nX i=1 (mixiyi)\u03c9y + nX i=1 (mixizi)\u03c9z ! \u0131\u0302 \u2212 \u00c3 nX i=1 (miyizi)\u03c9z + nX i=1 (miyixi)\u03c9x ! j\u0302 \u2212 \u00c3 nX i=1 (mizixi)\u03c9x + nX i=1 (miziyi)\u03c9y ! k\u0302. (9.71) By introducing the moment of inertia matrix I with the following elements, Ixx = nX i=1 \u00a3 mi \u00a1 y2i + z2i \u00a2\u00a4 (9.72) Iyy = nX i=1 \u00a3 mi \u00a1 z2i + x2i \u00a2\u00a4 (9.73) Izz = nX i=1 \u00a3 mi \u00a1 x2i + y2i \u00a2\u00a4 (9.74) Ixy = Iyx = \u2212 nX i=1 (mixiyi) (9.75) Iyz = Izy = \u2212 nX i=1 (miyizi) (9.76) Izx = Ixz = \u2212 nX i=1 (mizixi) . (9.77) we may write the angular momentum L in a concise form Lx = Ixx\u03c9x + Ixy\u03c9y + Ixz\u03c9z (9.78) Ly = Iyx\u03c9x + Iyy\u03c9y + Iyz\u03c9z (9.79) Lz = Izx\u03c9x + Izy\u03c9y + Izz\u03c9z (9.80) 534 9. Applied Dynamics or in a matrix form L = I \u00b7 \u03c9 (9.81)\u23a1\u23a3 Lx Ly Lz \u23a4\u23a6 = \u23a1\u23a3 Ixx Ixy Ixz Iyx Iyy Iyz Izx Izy Izz \u23a4\u23a6\u23a1\u23a3 \u03c9x \u03c9y \u03c9z \u23a4\u23a6 . (9.82) For a rigid body that is a continuous solid, the summations must be replaced by integrations over the volume of the body as in Equation (9.54). The Euler equation of motion for a rigid body is BM = Gd dt BL (9.83) where BM is the resultant of the external moments applied on the rigid body. The angular momentum BL is a vector defined in the body coordinate frame. Hence, its time derivative in the global coordinate frame is GdBL dt = BL\u0307+B G\u03c9B \u00d7 BL. (9.84) Therefore, BM = dL dt = L\u0307+ \u03c9 \u00d7 L = I\u03c9\u0307 + \u03c9\u00d7 (I\u03c9) (9.85) or in expanded form BM = (Ixx\u03c9\u0307x + Ixy\u03c9\u0307y + Ixz\u03c9\u0307z) \u0131\u0302 +(Iyx\u03c9\u0307x + Iyy\u03c9\u0307y + Iyz\u03c9\u0307z) j\u0302 +(Izx\u03c9\u0307x + Izy\u03c9\u0307y + Izz\u03c9\u0307z) k\u0302 +\u03c9y (Ixz\u03c9x + Iyz\u03c9y + Izz\u03c9z) \u0131\u0302 \u2212\u03c9z (Ixy\u03c9x + Iyy\u03c9y + Iyz\u03c9z) \u0131\u0302 +\u03c9z (Ixx\u03c9x + Ixy\u03c9y + Ixz\u03c9z) j\u0302 \u2212\u03c9x (Ixz\u03c9x + Iyz\u03c9y + Izz\u03c9z) j\u0302 +\u03c9x (Ixy\u03c9x + Iyy\u03c9y + Iyz\u03c9z) k\u0302 \u2212\u03c9y (Ixx\u03c9x + Ixy\u03c9y + Ixz\u03c9z) k\u0302 (9.86) and therefore, the most general form of the Euler equations of motion for a rigid body in a body frame attached to C are Mx = Ixx\u03c9\u0307x + Ixy\u03c9\u0307y + Ixz\u03c9\u0307z \u2212 (Iyy \u2212 Izz)\u03c9y\u03c9z \u2212Iyz \u00a1 \u03c92z \u2212 \u03c92y \u00a2 \u2212 \u03c9x (\u03c9zIxy \u2212 \u03c9yIxz) (9.87) My = Iyx\u03c9\u0307x + Iyy\u03c9\u0307y + Iyz\u03c9\u0307z \u2212 (Izz \u2212 Ixx)\u03c9z\u03c9x \u2212Ixz \u00a1 \u03c92x \u2212 \u03c92z \u00a2 \u2212 \u03c9y (\u03c9xIyz \u2212 \u03c9zIxy) (9.88) Mz = Izx\u03c9\u0307x + Izy\u03c9\u0307y + Izz\u03c9\u0307z \u2212 (Ixx \u2212 Iyy)\u03c9x\u03c9y \u2212Ixy \u00a1 \u03c92y \u2212 \u03c92x \u00a2 \u2212 \u03c9z (\u03c9yIxz \u2212 \u03c9xIyz) . (9.89) 9. Applied Dynamics 535 Assume that we are able to rotate the body frame about its origin to find an orientation that makes Iij = 0, for i 6= j. In such a coordinate frame, which is called a principal frame, the Euler equations reduce to M1 = I1\u03c9\u03071 \u2212 (I2 \u2212 I2)\u03c92\u03c93 (9.90) M2 = I2\u03c9\u03072 \u2212 (I3 \u2212 I1)\u03c93\u03c91 (9.91) M3 = I3\u03c9\u03073 \u2212 (I1 \u2212 I2)\u03c91\u03c92. (9.92) The kinetic energy of a rigid body may be found by the integral of the kinetic energy of a mass element dm, over the whole body. K = 1 2 Z B v\u03072dm = 1 2 Z B (\u03c9 \u00d7 r) \u00b7 (\u03c9 \u00d7 r) dm = \u03c92x 2 Z B \u00a1 y2 + z2 \u00a2 dm+ \u03c92y 2 Z B \u00a1 z2 + x2 \u00a2 dm+ \u03c92z 2 Z B \u00a1 x2 + y2 \u00a2 dm \u2212\u03c9x\u03c9y Z B xy dm\u2212 \u03c9y\u03c9z Z B yz dm\u2212 \u03c9z\u03c9x Z B zx dm = 1 2 \u00a1 Ixx\u03c9 2 x + Iyy\u03c9 2 y + Izz\u03c9 2 z \u00a2 \u2212Ixy\u03c9x\u03c9y \u2212 Iyz\u03c9y\u03c9z \u2212 Izx\u03c9z\u03c9x (9.93) The kinetic energy can be rearranged to a matrix multiplication form K = 1 2 \u03c9T I \u03c9 (9.94) = 1 2 \u03c9 \u00b7 L. (9.95) When the body frame is principal, the kinetic energy will simplify to K = 1 2 \u00a1 I1\u03c9 2 1 + I2\u03c9 2 2 + I3\u03c9 2 3 \u00a2 . (9.96) Example 349 A tilted disc on a massless shaft. Figure 9.4 illustrates a disc with mass m and radius r, mounted on a massless shaft. The shaft is turning with a constant angular speed \u03c9. The disc is attached to the shaft at an angle \u03b8. Because of \u03b8, the bearings at A and B must support a rotating force. We attach a principal body coordinate frame at the disc center as shown in the figure. The angular velocity vector in the body frame is B G\u03c9B = \u03c9 cos \u03b8 \u0131\u0302+ \u03c9 sin \u03b8 j\u0302 (9.97) 536 9. Applied Dynamics and the mass moment of inertia matrix is BI = \u23a1\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3 mr2 2 0 0 0 mr2 4 0 0 0 mr2 4 \u23a4\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6 . (9.98) Substituting (9.97) and (9.98) in (9.90)-(9.92), with 1 \u2261 x, 2 \u2261 y, 3 \u2261 z, provides that Mx = 0 (9.99) My = 0 (9.100) Mz = mr2 4 \u03c9 cos \u03b8 sin \u03b8. (9.101) Therefore, the bearing reaction forces FA and FB are FA = \u2212FB = \u2212Mz l = \u2212mr2 4l \u03c9 cos \u03b8 sin \u03b8. (9.102) Example 350 Steady rotation of a freely rotating rigid body. The Newton-Euler equations of motion for a rigid body are GF = mGv\u0307 (9.103) BM = I B G\u03c9\u0307B + B G\u03c9B \u00d7 BL. (9.104) 9. Applied Dynamics 537 Consider a situation in which the resultant applied force and moment on the body are zero. GF = BF = 0 (9.105) GM = BM = 0 (9.106) Based on the Newton\u2019s equation, the velocity of the mass center will be constant in the global coordinate frame. However, the Euler equation reduces to \u03c9\u03071 = I2 \u2212 I3 I1 \u03c92\u03c93 (9.107) \u03c9\u03072 = I3 \u2212 I1 I22 \u03c93\u03c91 (9.108) \u03c9\u03073 = I1 \u2212 I2 I3 \u03c91\u03c92 (9.109) that show the angular velocity can be constant if I1 = I2 = I3 (9.110) or if two principal moments of inertia, say I1 and I2, are zero and the third angular velocity, in this case \u03c93, is initially zero, or if the angular velocity vector is initially parallel to a principal axis. Example 351 Angular momentum of a two-link manipulator. A two-link manipulator is shown in Figure 9.5. Link A rotates with angular velocity \u03d5\u0307 about the z-axis of its local coordinate frame. Link B is attached to link A and has angular velocity \u03c8\u0307 with respect to A about the xA-axis. We assume that A and G were coincident at \u03d5 = 0, therefore, the rotation matrix between A and G is GRA = \u23a1\u23a3 cos\u03d5(t) \u2212 sin\u03d5(t) 0 sin\u03d5(t) cos\u03d5(t) 0 0 0 1 \u23a4\u23a6 . (9.111) Frame B is related to frame A by Euler angles \u03d5 = 90deg, \u03b8 = 90deg, and \u03c8, hence, ARB = \u23a1\u23a3 c\u03c0c\u03c8 \u2212 c\u03c0s\u03c0s\u03c8 \u2212c\u03c0s\u03c8 \u2212 c\u03c0c\u03c8s\u03c0 s\u03c0s\u03c0 c\u03c8s\u03c0 + c\u03c0c\u03c0s\u03c8 \u2212s\u03c0s\u03c8 + c\u03c0c\u03c0c\u03c8 \u2212c\u03c0s\u03c0 s\u03c0s\u03c8 s\u03c0c\u03c8 c\u03c0 \u23a4\u23a6 \u23a1\u23a3 \u2212 cos\u03c8 sin\u03c8 0 sin\u03c8 cos\u03c8 0 0 0 \u22121 \u23a4\u23a6 (9.112) 538 9. Applied Dynamics and therefore, GRB = GRA ARB (9.113) = \u23a1\u23a3 \u2212 cos\u03d5 cos\u03c8 \u2212 sin\u03d5 sin\u03c8 cos\u03d5 sin\u03c8 \u2212 cos\u03c8 sin\u03d5 0 cos\u03d5 sin\u03c8 \u2212 cos\u03c8 sin\u03d5 cos\u03d5 cos\u03c8 + sin\u03d5 sin\u03c8 0 0 0 \u22121 \u23a4\u23a6 . The angular velocity of A in G, and B in A are G\u03c9A = \u03d5\u0307K\u0302 (9.114) A\u03c9B = \u03c8\u0307\u0131\u0302A. (9.115) Moment of inertia matrices for the arms A and B can be defined as AIA = \u23a1\u23a3 IA1 0 0 0 IA2 0 0 0 IA3 \u23a4\u23a6 (9.116) BIB = \u23a1\u23a3 IB1 0 0 0 IB2 0 0 0 IB3 \u23a4\u23a6 . (9.117) These moments of inertia must be transformed to the global frame GIA = GRB AIA GRT A (9.118) GIB = GRB BIB GRT B . (9.119) 9. Applied Dynamics 539 The total angular momentum of the manipulator is GL = GLA + GLB (9.120) where GLA = GIA G\u03c9A (9.121) GLB = GIB G\u03c9B = GIB \u00a1 G A\u03c9B + G\u03c9A \u00a2 . (9.122) Example 352 Poinsot\u2019s construction. Consider a freely rotating rigid body with an attached principal coordinate frame. HavingM = 0 provides a motion under constant angular momentum and constant kinetic energy L = I \u03c9 = cte (9.123) K = 1 2 \u03c9T I \u03c9 = cte. (9.124) Because the length of the angular momentum L is constant, the equation L2 = L \u00b7 L = L2x + L2y + L2z = I21\u03c9 2 1 + I22\u03c9 2 2 + I23\u03c9 2 3 (9.125) introduces an ellipsoid in the (\u03c91, \u03c92, \u03c93) coordinate frame, called the momentum ellipsoid. The tip of all possible angular velocity vectors must lie on the surface of the momentum ellipsoid. The kinetic energy also defines an energy ellipsoid in the same coordinate frame so that the tip of the angular velocity vectors must also lie on its surface. K = 1 2 \u00a1 I1\u03c9 2 1 + I2\u03c9 2 2 + I3\u03c9 2 3 \u00a2 (9.126) In other words, the dynamics of moment-free motion of a rigid body requires that the corresponding angular velocity \u03c9(t) satisfy both Equations (9.125) and (9.126) and therefore lie on the intersection of the momentum and energy ellipsoids. For clarity, we may define the ellipsoids in the (Lx, Ly, Lz) coordinate system as L2x + L2y + L2z = L2 (9.127) L2x 2I1K + L2y 2I2K + L2z 2I3K = 1. (9.128) Equation (9.127) is a sphere and Equation (9.128) defines an ellipsoid with\u221a 2IiK as semi-axes. To have a meaningful motion, these two shapes must intersect. The intersection may form a trajectory, as shown in Figure 9.6. 540 9. Applied Dynamics It can be deduced that for a certain value of angular momentum there are maximum and minimum limit values for acceptable kinetic energy. Assuming I1 > I3 > I3 (9.129) the limits of possible kinetic energy are Kmin = L2 2I1 (9.130) Kmax = L2 2I3 (9.131) and the corresponding motions are turning about the axes I1 and I3 respectively. Example 353 F Alternative derivation of Euler equations of motion. Assume that the moment of the small force df is shown by dm and a mass element is shown by dm, then, dm = Grdm \u00d7 df = Grdm \u00d7 Gv\u0307dm dm. (9.132) The global angular momentum dl of dm is equal to dl = Grdm \u00d7 Gvdm dm (9.133) and according to (9.12) we have dm = Gd dt dl (9.134) Grdm \u00d7 df = Gd dt \u00a1 Grdm \u00d7 Gvdm dm \u00a2 . (9.135) 9. Applied Dynamics 541 Integrating over the body results inZ B Grdm \u00d7 df = Z B Gd dt \u00a1 Grdm \u00d7 Gvdm dm \u00a2 = Gd dt Z B \u00a1 Grdm \u00d7 Gvdm dm \u00a2 . (9.136) However, utilizing Grdm = GdB + GRB Brdm (9.137) where GdB is the global position vector of the central body frame, can simplify the left-hand side of the integral toZ B Grdm \u00d7 df = Z B \u00a1 GdB + GRB Brdm \u00a2 \u00d7 df = Z B GdB \u00d7 df + Z B G Brdm \u00d7 df = GdB \u00d7 GF+ GMC (9.138) where MC is the resultant external moment about the body mass center C. The right-hand side of Equation (9.136) is Gd dt Z B \u00a1 Grdm \u00d7 Gvdm dm \u00a2 = Gd dt Z B \u00a1\u00a1 GdB + GRB Brdm \u00a2 \u00d7 Gvdm dm \u00a2 = Gd dt Z B \u00a1 GdB \u00d7 Gvdm \u00a2 dm+ Gd dt Z B \u00a1 G Brdm \u00d7 Gvdm \u00a2 dm = Gd dt \u00b5 GdB \u00d7 Z B Gvdmdm \u00b6 + Gd dt LC = Gd\u0307B \u00d7 Z B Gvdmdm+ GdB \u00d7 Z B Gv\u0307dmdm+ d dt LC . (9.139) We use LC for angular momentum about the body mass center. Because the body frame is at the mass center, we haveZ B Grdm dm = mGdB = mGrC (9.140)Z B Gvdmdm = mGd\u0307B = mGvC (9.141)Z B Gv\u0307dmdm = mGd\u0308B = mGaC (9.142) and therefore, Gd dt Z B \u00a1 Grdm \u00d7 Gvdm dm \u00a2 = GdB \u00d7 GF+ Gd dt GLC . (9.143) 542 9. Applied Dynamics Substituting (9.138) and (9.143) in (9.136) provides the Euler equation of motion in the global frame, indicating that the resultant of externally applied moments about C is equal to the global derivative of angular momentum about C. GMC = Gd dt GLC . (9.144) The Euler equation in the body coordinate can be found by transforming (9.144) BMC = GRT B GMC = GRT B Gd dt LC = Gd dt GRT B LC = Gd dt BLC = BL\u0307C + B G\u03c9B \u00d7 BLC . (9.145) 9.4 Mass Moment of Inertia Matrix In analyzing the motion of rigid bodies, two types of integrals arise that belong to the geometry of the body. The first type defines the center of mass and is important when the translation motion of the body is considered. The second is the moment of inertia that appears when the rotational motion of the body is considered. The moment of inertia is also called centrifugal moments, or deviation moments. Every rigid body has a 3 \u00d7 3 moment of inertia matrix I, which is denoted by I = \u23a1\u23a3 Ixx Ixy Ixz Iyx Iyy Iyz Izx Izy Izz \u23a4\u23a6 . (9.146) The diagonal elements Iij , i = j are called polar moments of inertia Ixx = Ix = Z B \u00a1 y2 + z2 \u00a2 dm (9.147) Iyy = Iy = Z B \u00a1 z2 + x2 \u00a2 dm (9.148) Izz = Iz = Z B \u00a1 x2 + y2 \u00a2 dm (9.149) and the off-diagonal elements Iij , i 6= j are called products of inertia Ixy = Iyx = \u2212 Z B xy dm (9.150) 9. Applied Dynamics 543 Iyz = Izy = \u2212 Z B yz dm (9.151) Izx = Ixz = \u2212 Z B zx dm. (9.152) The elements of I for a rigid body, made of discrete point masses, are defined in Equation (9.54). The elements of I are calculated about a body coordinate frame attached to the mass center C of the body. Therefore, I is a frame-dependent quantity and must be written like BI to show the frame it is computed in. BI = Z B \u23a1\u23a3 y2 + z2 \u2212xy \u2212zx \u2212xy z2 + x2 \u2212yz \u2212zx \u2212yz x2 + y2 \u23a4\u23a6 dm (9.153) = Z B \u00a1 r2I\u2212 r rT \u00a2 dm (9.154) = Z B \u2212r\u0303 r\u0303 dm. (9.155) Moments of inertia can be transformed from a coordinate frame B1 to another coordinate frame B2, both installed at the mass center of the body, according to the rule of the rotated-axes theorem B2I = B2RB1 B1I B2RT B1 . (9.156) Transformation of the moment of inertia from a central frame B1 located at B2rC to another frame B2, which is parallel to B1, is, according to the rule of parallel-axes theorem, B2I = B1I +mr\u0303C r\u0303TC . (9.157) If the local coordinate frame Oxyz is located such that the products of inertia vanish, the local coordinate frame is called the principal coordinate frame and the associated moments of inertia are called principal moments of inertia. Principal axes and principal moments of inertia can be found by solving the following equation for I:\u00af\u0304\u0304\u0304 \u00af\u0304 Ixx \u2212 I Ixy Ixz Iyx Iyy \u2212 I Iyz Izx Izy Izz \u2212 I \u00af\u0304\u0304\u0304 \u00af\u0304 = 0 (9.158) det ([Iij ]\u2212 I [\u03b4ij ]) = 0. (9.159) Since Equation (9.159) is a cubic equation in I, we obtain three eigenvalues I1 = Ix I2 = Iy I3 = Iz (9.160) 544 9. Applied Dynamics that are the principal moments of inertia. Proof. Two coordinate frames with a common origin at the mass center of a rigid body are shown in Figure 9.7. The angular velocity and angular momentum of a rigid body transform from the frame B1 to the frame B2 by vector transformation rule B2\u03c9 = B2RB1 B1\u03c9 (9.161) B2L = B2RB1 B1L. (9.162) However, L and \u03c9 are related according to Equation (9.52) B1L = B1I B1\u03c9 (9.163) and therefore, B2L = B2RB1 B1I B2RT B1 B2\u03c9 = B2I B2\u03c9 (9.164) which shows how to transfer the moment of inertia from the coordinate frame B1 to a rotated frame B2 B2I = B2RB1 B1I B2RT B1 . (9.165) Now consider a central frame B1, shown in Figure 9.8, at B2rC , which rotates about the origin of a fixed frame B2 such that their axes remain parallel. The angular velocity and angular momentum of the rigid body transform from frame B1 to frame B2 by B2\u03c9 = B1\u03c9 (9.166) B2L = B1L+ (rC \u00d7mvC) . (9.167) 9. Applied Dynamics 545 Therefore, B2L = B1L+mB2rC \u00d7 \u00a1 B2\u03c9\u00d7B2rC \u00a2 = B1L+ \u00a1 m B2 r\u0303C B2 r\u0303TC \u00a2 B2\u03c9 = \u00a1 B1I +m B2 r\u0303C B2 r\u0303TC \u00a2 B2\u03c9 (9.168) which shows how to transfer the moment of inertia from frame B1 to a parallel frame B2 B2I = B1I +mr\u0303C r\u0303TC . (9.169) The parallel-axes theorem is also called the Huygens-Steiner theorem. Referring to Equation (9.165) for transformation of the moment of inertia to a rotated frame, we can always find a frame in which B2I is diagonal. In such a frame, we have B2RB1 B1I = B2I B2RB1 (9.170) or \u23a1\u23a3 r11 r12 r13 r21 r22 r23 r31 r32 r33 \u23a4\u23a6\u23a1\u23a3 Ixx Ixy Ixz Iyx Iyy Iyz Izx Izy Izz \u23a4\u23a6 = \u23a1\u23a3 I1 0 0 0 I2 0 0 0 I3 \u23a4\u23a6\u23a1\u23a3 r11 r12 r13 r21 r22 r23 r31 r32 r33 \u23a4\u23a6 (9.171) which shows that I1, I2, and I3 are eigenvalues of B1I. These eigenvalues can be found by solving the following equation for \u03bb:\u00af\u0304\u0304\u0304 \u00af\u0304 Ixx \u2212 \u03bb Ixy Ixz Iyx Iyy \u2212 \u03bb Iyz Izx Izy Izz \u2212 \u03bb \u00af\u0304\u0304\u0304 \u00af\u0304 = 0. (9.172) 546 9. Applied Dynamics The eigenvalues I1, I2, and I3 are principal moments of inertia, and their associated eigenvectors are called principal directions. The coordinate frame made by the eigenvectors is the principal body coordinate frame. In the principal coordinate frame, the rigid body angular momentum is\u23a1\u23a3 L1 L2 L3 \u23a4\u23a6 = \u23a1\u23a3 I1 0 0 0 I2 0 0 0 I3 \u23a4\u23a6\u23a1\u23a3 \u03c91 \u03c92 \u03c93 \u23a4\u23a6 . (9.173) Example 354 Principal moments of inertia. Consider the inertia matrix I I = \u23a1\u23a3 20 \u22122 0 \u22122 30 0 0 0 40 \u23a4\u23a6 . (9.174) We set up the determinant (9.159)\u00af\u0304\u0304\u0304 \u00af\u0304 20\u2212 \u03bb \u22122 0 \u22122 30\u2212 \u03bb 0 0 0 40\u2212 \u03bb \u00af\u0304\u0304\u0304 \u00af\u0304 = 0 (9.175) which leads to the following characteristic equation. (20\u2212 \u03bb) (30\u2212 \u03bb) (40\u2212 \u03bb)\u2212 4 (40\u2212 \u03bb) = 0 (9.176) Three roots of Equation (9.176) are I1 = 30.385, I2 = 19.615, I3 = 40 (9.177) and therefore, the principal moment of inertia matrix is I = \u23a1\u23a3 30.385 0 0 0 19.615 0 0 0 40 \u23a4\u23a6 . (9.178) Example 355 Principal coordinate frame. Consider the inertia matrix I I = \u23a1\u23a3 20 \u22122 0 \u22122 30 0 0 0 40 \u23a4\u23a6 (9.179) the direction of a principal axis xi is established by solving\u23a1\u23a3 Ixx \u2212 Ii Ixy Ixz Iyx Iyy \u2212 Ii Iyz Izx Izy Izz \u2212 Ii \u23a4\u23a6\u23a1\u23a3 cos\u03b1i cos\u03b2i cos \u03b3i \u23a4\u23a6 = \u23a1\u23a3 0 0 0 \u23a4\u23a6 (9.180) 9. Applied Dynamics 547 for direction cosines, which must also satisfy cos2 \u03b1i + cos 2 \u03b2i + cos 2 \u03b3i = 1. (9.181) For the first principal moment of inertia I1 = 30.385 we have\u23a1\u23a3 20\u2212 30.385 \u22122 0 \u22122 30\u2212 30.385 0 0 0 40\u2212 30.385 \u23a4\u23a6\u23a1\u23a3 cos\u03b11 cos\u03b21 cos \u03b31 \u23a4\u23a6 = \u23a1\u23a3 0 0 0 \u23a4\u23a6 (9.182) or \u221210.385 cos\u03b11 \u2212 2 cos\u03b21 + 0 = 0 (9.183) \u22122 cos\u03b11 \u2212 0.385 cos\u03b21 + 0 = 0 (9.184) 0 + 0 + 9.615 cos \u03b31 = 0 (9.185) and we obtain \u03b11 = 79.1 deg (9.186) \u03b21 = 169.1 deg (9.187) \u03b31 = 90.0 deg . (9.188) Using I2 = 19.615 for the second principal axis\u23a1\u23a3 20\u2212 19.62 \u22122 0 \u22122 30\u2212 19.62 0 0 0 40\u2212 19.62 \u23a4\u23a6\u23a1\u23a3 cos\u03b12 cos\u03b22 cos \u03b32 \u23a4\u23a6 = \u23a1\u23a3 0 0 0 \u23a4\u23a6 (9.189) we obtain \u03b12 = 10.9 deg (9.190) \u03b22 = 79.1 deg (9.191) \u03b32 = 90.0 deg . (9.192) The third principal axis is for I3 = 40\u23a1\u23a3 20\u2212 40 \u22122 0 \u22122 30\u2212 40 0 0 0 40\u2212 40 \u23a4\u23a6\u23a1\u23a3 cos\u03b13 cos\u03b23 cos \u03b33 \u23a4\u23a6 = \u23a1\u23a3 0 0 0 \u23a4\u23a6 (9.193) which leads to \u03b13 = 90.0 deg (9.194) \u03b23 = 90.0 deg (9.195) \u03b33 = 0.0 deg . (9.196) 548 9. Applied Dynamics Example 356 Moment of inertia of a rigid rectangular bar. Consider a homogeneous rectangular link with mass m, length l, width w, and height h, as shown in Figure 9.9. The local central coordinate frame is attached to the link at its mass center. The moments of inertia matrix of the link can be found by the integral method. We begin with calculating Ixx Ixx = Z B \u00a1 y2 + z2 \u00a2 dm = Z v \u00a1 y2 + z2 \u00a2 \u03c1dv = m lwh Z v \u00a1 y2 + z2 \u00a2 dv = m lwh Z h/2 \u2212h/2 Z w/2 \u2212w/2 Z l/2 \u2212l/2 \u00a1 y2 + z2 \u00a2 dx dy dz = m 12 \u00a1 w2 + h2 \u00a2 (9.197) which shows Iyy and Izz can be calculated similarly Iyy = m 12 \u00a1 h2 + l2 \u00a2 (9.198) Izz = m 12 \u00a1 l2 + w2 \u00a2 . (9.199) Since the coordinate frame is central, the products of inertia must be zero. To show this, we examine Ixy. Ixy = Iyx = \u2212 Z B xy dm = Z v xy\u03c1dv = m lwh Z h/2 \u2212h/2 Z w/2 \u2212w/2 Z l/2 \u2212l/2 xy dxdy dz = 0 (9.200) 9. Applied Dynamics 549 Therefore, the moment of inertia matrix for the rigid rectangular bar in its central frame is I = \u23a1\u23a3 m 12 \u00a1 w2 + h2 \u00a2 0 0 0 m 12 \u00a1 h2 + l2 \u00a2 0 0 0 m 12 \u00a1 l2 + w2 \u00a2 \u23a4\u23a6 . (9.201) Example 357 Translation of the inertia matrix. The moment of inertia matrix of the rigid body shown in Figure 9.10, in the principal frame B(oxyz) is given in Equation (9.201). The moment of inertia matrix in the non-principal frame B0(ox0y0z0) can be found by applying the parallel-axes transformation formula (9.169). B0 I = BI +m B0 r\u0303C B0 r\u0303TC (9.202) The mass center is at B0 rC = 1 2 \u23a1\u23a3 l w h \u23a4\u23a6 (9.203) and therefore, B0 r\u0303C = 1 2 \u23a1\u23a3 0 \u2212h w h 0 \u2212l \u2212w l 0 \u23a4\u23a6 (9.204) that provides B0 I = \u23a1\u23a3 1 3h 2m+ 1 3mw2 \u221214 lmw \u221214hlm \u221214 lmw 1 3h 2m+ 1 3 l 2m \u221214hmw \u221214hlm \u2212 14hmw 1 3 l 2m+ 1 3mw2 \u23a4\u23a6 . (9.205) 550 9. Applied Dynamics Example 358 Principal rotation matrix. Consider a body inertia matrix as I = \u23a1\u23a3 2/3 \u22121/2 \u22121/2 \u22121/2 5/3 \u22121/4 \u22121/2 \u22121/4 5/3 \u23a4\u23a6 . (9.206) The eigenvalues and eigenvectors of I are I1 = 0.2413 , \u23a1\u23a3 2.351 1 1 \u23a4\u23a6 (9.207) I2 = 1.8421 , \u23a1\u23a3 \u22120.8511 1 \u23a4\u23a6 (9.208) I3 = 1.9167 , \u23a1\u23a3 0 \u22121 1 \u23a4\u23a6 . (9.209) The normalized eigenvector matrix W is equal to the transpose of the required transformation matrix to make the inertia matrix diagonal W = \u23a1\u23a3 | | | w1 w 2 w 3 | | | \u23a4\u23a6 = 2RT 1 = \u23a1\u23a3 0.856 9 \u22120.515 6 0.0 0.364 48 0.605 88 \u22120.707 11 0.364 48 0.605 88 0.707 11 \u23a4\u23a6 . (9.210) We may verify that 2I \u2248 2R1 1I 2RT 1 =WT 1I W = \u23a1\u23a3 0.2413 \u22121\u00d7 10\u22124 0.0 \u22121\u00d7 10\u22124 1.842 1 \u22121\u00d7 10\u221219 0.0 0.0 1.916 7 \u23a4\u23a6 . (9.211) Example 359 F Relative diagonal moments of inertia. Using the definitions for moments of inertia (9.147), (9.148), and (9.149) it is seen that the inertia matrix is symmetric, andZ B \u00a1 x2 + y2 + z2 \u00a2 dm = 1 2 (Ixx + Iyy + Izz) (9.212) and also Ixx + Iyy \u2265 Izz (9.213) Iyy + Izz \u2265 Ixx (9.214) Izz + Ixx \u2265 Iyy. (9.215) 9. Applied Dynamics 551 Noting that (y \u2212 z) 2 \u2265 0 it is evident that \u00a1 y2 + z2 \u00a2 \u2265 2yz and therefore Ixx \u2265 2Iyz (9.216) and similarly Iyy \u2265 2Izx (9.217) Izz \u2265 2Ixy. (9.218) Example 360 F Coefficients of the characteristic equation. The determinant (9.172)\u00af\u0304\u0304\u0304 \u00af\u0304 Ixx \u2212 \u03bb Ixy Ixz Iyx Iyy \u2212 \u03bb Iyz Izx Izy Izz \u2212 \u03bb \u00af\u0304\u0304\u0304 \u00af\u0304 = 0 (9.219) for calculating the principal moments of inertia, leads to a third-degree equation for \u03bb, called the characteristic equation. \u03bb3 \u2212 a1\u03bb 2 + a2\u03bb\u2212 a3 = 0 (9.220) The coefficients of the characteristic equation are called the principal invariants of [I]. The coefficients of the characteristic equation can directly be found from the following equations: a1 = Ixx + Iyy + Izz = tr [I] (9.221) a2 = IxxIyy + IyyIzz + IzzIxx \u2212 I2xy \u2212 I2yz \u2212 I2zx = \u00af\u0304\u0304\u0304 Ixx Ixy Iyx Iyy \u00af\u0304\u0304\u0304 + \u00af\u0304\u0304\u0304 Iyy Iyz Izy Izz \u00af\u0304\u0304\u0304 + \u00af\u0304\u0304\u0304 Ixx Ixz Izx Izz \u00af\u0304\u0304\u0304 = 1 2 \u00a1 a21 \u2212 tr \u00a3 I2 \u00a4\u00a2 (9.222) a3 = IxxIyyIzz + IxyIyzIzx + IzyIyxIxz \u2212 (IxxIyzIzy + IyyIzxIxz + IzzIxyIyx) = IxxIyyIzz + 2IxyIyzIzx \u2212 \u00a1 IxxI 2 yz + IyyI 2 zx + IzzI 2 xy \u00a2 = det [I] (9.223) 552 9. Applied Dynamics Example 361 F The principal moments of inertia are coordinate invariants. The roots of the inertia characteristic equation are the principal moments of inertia. They are all real but not necessarily different. The principal moments of inertia are extreme. That is, the principal moments of inertia determine the smallest and the largest values of Iii. Since the smallest and largest values of Iii do not depend on the choice of the body coordinate frame, the solution of the characteristic equation is not dependent of the coordinate frame. In other words, if I1, I2, and I3 are the principal moments of inertia for B1I, the principal moments of inertia for B2I are also I1, I2, and I3 when B2I = B2RB1 B1I B2RT B1 . We conclude that I1, I2, and I3 are coordinate invariants of the matrix [I], and therefore any quantity that depends on I1, I2, and I3 is also coordinate invariant. The matrix [I] has only three independent invariants and every other invariant can be expressed in terms of I1, I2, and I3. Since I1, I2, and I3 are the solutions of the characteristic equation of [I] given in (9.220), we may write the determinant (9.172) in the form (\u03bb\u2212 I1) (\u03bb\u2212 I2) (\u03bb\u2212 I3) = 0. (9.224) The expanded form of this equation is \u03bb3 \u2212 (I1 + I2 + I3)\u03bb 2 + (I1I2 + I2I3 + I3I1) a2\u03bb\u2212 I1I2I3 = 0. (9.225) By comparing (9.225) and (9.220) we conclude that a1 = Ixx + Iyy + Izz = I1 + I2 + I3 (9.226) a2 = IxxIyy + IyyIzz + IzzIxx \u2212 I2xy \u2212 I2yz \u2212 I2zx = I1I2 + I2I3 + I3I1 (9.227) a3 = IxxIyyIzz + 2IxyIyzIzx \u2212 \u00a1 IxxI 2 yz + IyyI 2 zx + IzzI 2 xy \u00a2 = I1I2I3. (9.228) Being able to express the coefficients a1, a2, and a3 as functions of I1, I2, and I3 determines that the coefficients of the characteristic equation are coordinate-invariant. Example 362 F Short notation for the elements of inertia matrix. Taking advantage of the Kronecker\u2019s delta (5.138) we may write the el- 9. Applied Dynamics 553 ements of the moment of inertia matrix Iij in short notation forms Iij = Z B \u00a1\u00a1 x21 + x22 + x23 \u00a2 \u03b4ij \u2212 xixj \u00a2 dm (9.229) Iij = Z B \u00a1 r2\u03b4ij \u2212 xixj \u00a2 dm (9.230) Iij = Z B \u00c3 3X k=1 xkxk\u03b4ij \u2212 xixj ! dm (9.231) where we utilized the following notations: x1 = x x2 = y x3 = z. (9.232) Example 363 F Moment of inertia with respect to a plane, a line, and a point. The moment of inertia of a system of particles may be defined with respect to a plane, a line, or a point as the sum of the products of the mass of the particles into the square of the perpendicular distance from the particle to the plane, line, or point. For a continuous body, the sum would be definite integral over the volume of the body. The moments of inertia with respect to the xy, yz, and zx-plane are Iz2 = Z B z2dm (9.233) Iy2 = Z B y2dm (9.234) Ix2 = Z B x2dm. (9.235) The moments of inertia with respect to the x, y, and z axes are Ix = Z B \u00a1 y2 + z2 \u00a2 dm (9.236) Iy = Z B \u00a1 z2 + x2 \u00a2 dm (9.237) Iz = Z B \u00a1 x2 + y2 \u00a2 dm (9.238) and therefore, Ix = Iy2 + Iz2 (9.239) Iy = Iz2 + Ix2 (9.240) Iz = Ix2 + Iy2 . (9.241) 554 9. Applied Dynamics The moment of inertia with respect to the origin is Io = Z B \u00a1 x2 + y2 + z2 \u00a2 dm = Ix2 + Iy2 + Iz2 = 1 2 (Ix + Iy + Iz) . (9.242) Because the choice of the coordinate frame is arbitrary, we can say that the moment of inertia with respect to a line is the sum of the moments of inertia with respect to any two mutually orthogonal planes that pass through the line. The moment of inertia with respect to a point has similar meaning for three mutually orthogonal planes intersecting at the point. 9.5 Lagrange\u2019s Form of Newton\u2019s Equations of Motion Newton\u2019s equation of motion can be transformed to d dt \u00b5 \u2202K \u2202q\u0307r \u00b6 \u2212 \u2202K \u2202qr = Fr r = 1, 2, \u00b7 \u00b7 \u00b7n (9.243) where Fr = nX i=1 \u00b5 Fix \u2202fi \u2202q1 + Fiy \u2202gi \u2202q2 + Fiz \u2202hi \u2202qn \u00b6 . (9.244) Equation (9.243) is called the Lagrange equation of motion, where K is the kinetic energy of the n degree-of-freedom (DOF ) system, qr, r = 1, 2, \u00b7 \u00b7 \u00b7 , n are the generalized coordinates of the system, F = \u00a3 Fix Fiy Fiz \u00a4T is the external force acting on the ith particle of the system, and Fr is the generalized force associated to qr. Proof. Let mi be the mass of one of the particles of a system and let (xi, yi, zi) be its Cartesian coordinates in a globally fixed coordinate frame. Assume that the coordinates of every individual particle are functions of another set of coordinates q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, and possibly time t. xi = fi(q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, t) (9.245) yi = gi(q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, t) (9.246) zi = hi(q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, t) (9.247) If Fxi, Fyi, Fzi are components of the total force acting on the particle mi, then the Newton equations of motion for the particle would be Fxi = mix\u0308i (9.248) Fyi = miy\u0308i (9.249) Fzi = miz\u0308i. (9.250) 9. Applied Dynamics 555 We multiply both sides of these equations by \u2202fi \u2202qr \u2202gi \u2202qr \u2202hi \u2202qr respectively, and add them up for all the particles to have nX i=1 mi \u00b5 x\u0308i \u2202fi \u2202qr + y\u0308i \u2202gi \u2202qr + z\u0308i \u2202hi \u2202qr \u00b6 = nX i=1 \u00b5 Fxi \u2202fi \u2202qr + Fyi \u2202gi \u2202qr + Fzi \u2202hi \u2202qr \u00b6 (9.251) where n is the total number of particles. Taking a time derivative of Equation (9.245), x\u0307i = \u2202fi \u2202q1 q\u03071 + \u2202fi \u2202q2 q\u03072 + \u2202fi \u2202q3 q\u03073 + \u00b7 \u00b7 \u00b7+ \u2202fi \u2202qn q\u0307n + \u2202fi \u2202t (9.252) we find \u2202x\u0307i \u2202q\u0307r = \u2202 \u2202q\u0307r \u00b5 \u2202fi \u2202q1 q\u03071 + \u2202fi \u2202q2 q\u03072 + \u00b7 \u00b7 \u00b7+ \u2202fi \u2202qn q\u0307n + \u2202fi \u2202t \u00b6 = \u2202fi \u2202qr . (9.253) and therefore, x\u0308i \u2202fi \u2202qr = x\u0308i \u2202x\u0307i \u2202q\u0307r = d dt \u00b5 x\u0307i \u2202x\u0307i \u2202q\u0307r \u00b6 \u2212 x\u0307i d dt \u00b5 \u2202x\u0307i \u2202q\u0307r \u00b6 . (9.254) However, x\u0307i d dt \u00b5 \u2202x\u0307i \u2202q\u0307r \u00b6 = x\u0307i d dt \u00b5 \u2202fi \u2202qr \u00b6 = x\u0307i \u00b5 \u22022fi \u2202q1\u2202qr q\u03071 + \u00b7 \u00b7 \u00b7+ \u22022fi \u2202qn\u2202qr q\u0307n + \u22022fi \u2202t\u2202qr \u00b6 = x\u0307i \u2202 \u2202qr \u00b5 \u2202fi \u2202q1 q\u03071 + \u2202fi \u2202q2 q\u03072 + \u00b7 \u00b7 \u00b7+ \u2202fi \u2202qn q\u0307n + \u2202fi \u2202t \u00b6 = x\u0307i \u2202x\u0307i \u2202qr (9.255) and we have x\u0308i \u2202x\u0307i \u2202q\u0307r = d dt \u00b5 x\u0307i \u2202x\u0307i \u2202q\u0307r \u00b6 \u2212 x\u0307i \u2202x\u0307i \u2202qr (9.256) 556 9. Applied Dynamics which is equal to x\u0308i x\u0307i q\u0307r = d dt \u2219 \u2202 \u2202q\u0307r \u00b5 1 2 x\u03072i \u00b6\u00b8 \u2212 \u2202 \u2202qr \u00b5 1 2 x\u03072i \u00b6 . (9.257) Now substituting (9.254) and (9.257) in the left-hand side of (9.251) leads to nX i=1 mi \u00b5 x\u0308i \u2202fi \u2202qr + y\u0308i \u2202gi \u2202qr + z\u0308i \u2202hi \u2202qr \u00b6 = nX i=1 mi d dt \u2219 \u2202 \u2202q\u0307r \u00b5 1 2 x\u03072i + 1 2 y\u03072i + 1 2 z\u03072i \u00b6\u00b8 \u2212 nX i=1 mi \u2202 \u2202qr \u00b5 1 2 x\u03072i + 1 2 y\u03072i + 1 2 z\u03072i \u00b6 = 1 2 nX i=1 mi d dt \u2219 \u2202 \u2202q\u0307r \u00a1 x\u03072i + y\u03072i + z\u03072i \u00a2\u00b8 \u22121 2 nX i=1 mi \u2202 \u2202qr \u00a1 x\u03072i + y\u03072i + z\u03072i \u00a2 . (9.258) where 1 2 nX i=1 mi \u00a1 x\u03072i + y\u03072i + z\u03072i \u00a2 = K (9.259) is the kinetic energy of the system. Therefore, the Newton equations of motion (9.248), (9.249), and (9.250) are converted to d dt \u00b5 \u2202K \u2202q\u0307r \u00b6 \u2212 \u2202K \u2202qr = nX i=1 \u00b5 Fxi \u2202fi \u2202qr + Fyi \u2202gi \u2202qr + Fzi \u2202hi \u2202qr \u00b6 . (9.260) Because of (9.245), (9.246), and (9.247), the kinetic energy is a function of q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn and time t. The left-hand side of Equation (9.260) includes the kinetic energy of the whole system and the right-hand side is a generalized force and shows the effect of changing coordinates from xi to qj on the external forces. Let us assume that the coordinate qr alters to qr+ \u03b4qr while the other coordinates q1, q2, q3, \u00b7 \u00b7 \u00b7 , qr\u22121, qr+1, \u00b7 \u00b7 \u00b7 , qn and time t are unaltered. So, the coordinates of mi are changed to xi + \u2202fi \u2202qr \u03b4qr (9.261) yi + \u2202gi \u2202qr \u03b4qr (9.262) zi + \u2202hi \u2202qr \u03b4qr (9.263) 9. Applied Dynamics 557 Such a displacement is called virtual displacement. The work done in this virtual displacement by all forces acting on the particles of the system is \u03b4W = nX i=1 \u00b5 Fxi \u2202fi \u2202qr + Fyi \u2202gi \u2202qr + Fzi \u2202hi \u2202qr \u00b6 \u03b4qr. (9.264) Because the work done by internal forces appears in opposite pairs, only the work done by external forces remains in Equation (9.264). Let\u2019s denote the virtual work by \u03b4W = Fr (q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, t) \u03b4qr. (9.265) Then we have d dt \u00b5 \u2202K \u2202q\u0307r \u00b6 \u2212 \u2202K \u2202qr = Fr (9.266) where Fr = nX i=1 \u00b5 Fxi \u2202fi \u2202qr + Fyi \u2202gi \u2202qr + Fzi \u2202hi \u2202qr \u00b6 . (9.267) Equation (9.266) is the Lagrange form of equations of motion. This equation is true for all values of r from 1 to n. We thus have n second-order ordinary differential equations in which q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn are the dependent variables and t is the independent variable. The coordinates q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn are called generalized coordinates and can be any measurable parameters to provide the configuration of the system. The number of equations and the number of dependent variables are equal, therefore, the equations are theoretically sufficient to determine the motion of all mi. Example 364 Equation of motion for a simple pendulum. A pendulum is shown in Figure 9.11. Using x and y for the Cartesian position of m, and using \u03b8 = q as the generalized coordinate, we have x = f(\u03b8) = l sin \u03b8 (9.268) y = g(\u03b8) = l cos \u03b8 (9.269) K = 1 2 m \u00a1 x\u03072 + y\u03072 \u00a2 = 1 2 ml2\u03b8\u0307 2 (9.270) and therefore, d dt \u00b5 \u2202K \u2202\u03b8\u0307 \u00b6 \u2212 \u2202K \u2202\u03b8 = d dt (ml2\u03b8\u0307) = ml2\u03b8\u0308. (9.271) The external force components, acting on m, are Fx = 0 (9.272) Fy = mg (9.273) 558 9. Applied Dynamics and therefore, F\u03b8 = Fx \u2202f \u2202\u03b8 + Fy \u2202g \u2202\u03b8 = \u2212mgl sin \u03b8. (9.274) Hence, the equation of motion for the pendulum is ml2\u03b8\u0308 = \u2212mgl sin \u03b8. (9.275) Example 365 A pendulum attached to an oscillating mass. Figure 9.12 illustrates a vibrating mass with a hanging pendulum. The pendulum can act as a vibration absorber if designed properly. Starting with coordinate relationships xM = fM = x (9.276) yM = gM = 0 (9.277) xm = fm = x+ l sin \u03b8 (9.278) ym = gm = l cos \u03b8 (9.279) we may find the kinetic energy in terms of the generalized coordinates x and \u03b8. K = 1 2 M \u00a1 x\u03072M + y\u03072M \u00a2 + 1 2 m \u00a1 x\u03072m + y\u03072m \u00a2 = 1 2 Mx\u03072 + 1 2 m \u00b3 x\u03072 + l2\u03b8\u0307 2 + 2lx\u0307\u03b8\u0307 cos \u03b8 \u00b4 (9.280) Then, the left-hand side of the Lagrange equations are d dt \u00b5 \u2202K \u2202x\u0307 \u00b6 \u2212 \u2202K \u2202x = (M +m)x\u0308+ml\u03b8\u0308 cos \u03b8 \u2212ml\u03b8\u0307 2 sin \u03b8 (9.281) d dt \u00b5 \u2202K \u2202\u03b8\u0307 \u00b6 \u2212 \u2202K \u2202\u03b8 = ml2\u03b8\u0308 +mlx\u0308 cos \u03b8. (9.282) 9. Applied Dynamics 559 The external forces acting on M and m are FxM = \u2212kx (9.283) FyM = 0 (9.284) Fxm = 0 (9.285) Fym = mg. (9.286) Therefore, the generalized forces are Fx = FxM \u2202fM \u2202x + FyM \u2202gM \u2202x + Fxm \u2202fm \u2202x + Fym \u2202gm \u2202x = \u2212kx (9.287) F\u03b8 = FxM \u2202fM \u2202\u03b8 + FyM \u2202gM \u2202\u03b8 + Fxm \u2202fm \u2202\u03b8 + Fym \u2202gm \u2202\u03b8 = \u2212mgl sin \u03b8 (9.288) and finally the Lagrange equations of motion are (M +m)x\u0308+ml\u03b8\u0308 cos \u03b8 \u2212ml\u03b8\u0307 2 sin \u03b8 = \u2212kx (9.289) ml2\u03b8\u0308 +mlx\u0308 cos \u03b8 = \u2212mgl sin \u03b8. (9.290) Example 366 Kinetic energy of the Earth. Earth is approximately a rotating rigid body about a fixed axis. The two motions of the Earth are called revolution about the sun, and rotation about an axis approximately fixed in the Earth. The kinetic energy of the Earth due to its rotation is K1 = 1 2 I\u03c921 = 1 2 2 5 \u00a1 5.9742\u00d7 1024 \u00a2\u00b56356912 + 6378388 2 \u00b62\u00b5 2\u03c0 24\u00d7 3600 366.25 365.25 \u00b62 = 2.5762\u00d7 1029 J 560 9. Applied Dynamics and the kinetic energy of the Earth due to its revolution is K2 = 1 2 Mr2\u03c922 = 1 2 \u00a1 5.9742\u00d7 1024 \u00a2 \u00a1 1.49475\u00d7 1011 \u00a22\u00b5 2\u03c0 24\u00d7 3600 1 365.25 \u00b62 = 2.6457\u00d7 1033 J where r is the distance from the sun and \u03c92 is the angular speed about the sun. The total kinetic energy of the Earth is K = K1 +K2. However, the ratio of the revolutionary to rotational kinetic energies is K2 K1 = 2.6457\u00d7 1033 2.5762\u00d7 1029 \u2248 10000. Example 367 F Explicit form of Lagrange equations. Assume the coordinates of every particle are functions of the coordinates q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn but not the time t. The kinetic energy of the system made of n massive particles can be written as K = 1 2 nX i=1 mi \u00a1 x\u03072i + y\u03072i + z\u03072i \u00a2 = 1 2 nX j=1 nX k=1 ajkq\u0307j q\u0307k (9.291) where the coefficients ajk are functions of q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn and ajk = akj . (9.292) The Lagrange equations of motion d dt \u00b5 \u2202K \u2202q\u0307r \u00b6 \u2212 \u2202K \u2202qr = Fr r = 1, 2, \u00b7 \u00b7 \u00b7n (9.293) are then equal to d dt nX m=1 amrq\u0307m \u2212 1 2 nX j=1 nX k=1 ajk \u2202qr q\u0307j q\u0307k = Fr (9.294) or nX m=1 amrq\u0308m + nX k=1 nX n=1 \u0393rk,nq\u0307kq\u0307n = Fr (9.295) where \u0393ij,k is called the Christoffel operator \u0393ij,k = 1 2 \u00b5 \u2202aij \u2202qk + \u2202aik \u2202qj \u2212 \u2202akj \u2202qi \u00b6 . (9.296) 9. Applied Dynamics 561 9.6 Lagrangian Mechanics Assume for some forces F = \u00a3 Fix Fiy Fiz \u00a4T there is a function V , called potential energy, such that the force is derivable from V F = \u2212\u2207V. (9.297) Such a force is called potential or conservative force. Then, the Lagrange equation of motion can be written as d dt \u00b5 \u2202L \u2202q\u0307r \u00b6 \u2212 \u2202L \u2202qr = Qr r = 1, 2, \u00b7 \u00b7 \u00b7n (9.298) where L = K \u2212 V (9.299) is the Lagrangean of the system and Qr is the nonpotential generalized force. Proof. Assume the external forces F = \u00a3 Fxi Fyi Fzi \u00a4T acting on the system are conservative. F = \u2212\u2207V (9.300) The work done by these forces in an arbitrary virtual displacement \u03b4q1, \u03b4q2, \u03b4q3, \u00b7 \u00b7 \u00b7 , \u03b4qn is \u2202W = \u2212\u2202V \u2202q1 \u03b4q1 \u2212 \u2202V \u2202q2 \u03b4q2 \u2212 \u00b7 \u00b7 \u00b7 \u2202V \u2202qn \u03b4qn (9.301) then the Lagrange equation becomes d dt \u00b5 \u2202K \u2202q\u0307r \u00b6 \u2212 \u2202K \u2202qr = \u2212 \u2202V \u2202q1 r = 1, 2, \u00b7 \u00b7 \u00b7n. (9.302) Introducing the Lagrangean function L = K \u2212 V converts the Lagrange equation to d dt \u00b5 \u2202L \u2202q\u0307r \u00b6 \u2212 \u2202L \u2202qr = 0 r = 1, 2, \u00b7 \u00b7 \u00b7n (9.303) for a conservative system. The Lagrangean is also called kinetic potential. If a force is not conservative, then the virtual work done by the force is \u03b4W = nX i=1 \u00b5 Fxi \u2202fi \u2202qr + Fyi \u2202gi \u2202qr + Fzi \u2202hi \u2202qr \u00b6 \u03b4qr = Qr \u03b4qr (9.304) and the equation of motion would be d dt \u00b5 \u2202L \u2202q\u0307r \u00b6 \u2212 \u2202L \u2202qr = Qr r = 1, 2, \u00b7 \u00b7 \u00b7n (9.305) where Qr is the nonpotential generalized force doing work in a virtual displacement of the rth generalized coordinate qr. 562 9. Applied Dynamics Example 368 Spherical pendulum. A pendulum analogy is utilized in modeling of many dynamical problems. Figure 9.13 illustrates a spherical pendulum with mass m and length l. The angles \u03d5 and \u03b8 may be used as describing coordinates of the system. The Cartesian coordinates of the mass as a function of the generalized coordinates are \u23a1\u23a3 X Y Z \u23a4\u23a6 = \u23a1\u23a3 r cos\u03d5 sin \u03b8 r sin \u03b8 sin\u03d5 \u2212r cos \u03b8 \u23a4\u23a6 (9.306) and therefore, the kinetic and potential energies of the pendulum are K = 1 2 m \u00b3 l2\u03b8\u0307 2 + l2\u03d5\u03072 sin2 \u03b8 \u00b4 (9.307) V = \u2212mgl cos \u03b8. (9.308) The kinetic potential function of this system is then equal to L = 1 2 m \u00b3 l2\u03b8\u0307 2 + l2\u03d5\u03072 sin2 \u03b8 \u00b4 +mgl cos \u03b8 (9.309) which leads to the following equations of motion: \u03b8\u0308 \u2212 \u03d5\u03072 sin \u03b8 cos \u03b8 + g l sin \u03b8 = 0 (9.310) \u03d5\u0308 sin2 \u03b8 + 2\u03d5\u0307\u03b8\u0307 sin \u03b8 cos \u03b8 = 0. (9.311) Example 369 Controlled compound pendulum. A massive arm is attached to a ceiling at a pin joint O as illustrated in Figure 9.14. Assume that there is viscous friction in the joint where an ideal motor can apply a torque Q to move the arm. The rotor of an ideal motor has no moment of inertia by assumption. 9. Applied Dynamics 563 The kinetic and potential energies of the manipulator are K = 1 2 I\u03b8\u0307 2 = 1 2 \u00a1 IC +ml2 \u00a2 \u03b8\u0307 2 (9.312) V = \u2212mg cos \u03b8 (9.313) where m is the mass and I is the moment of inertia of the pendulum about O. The Lagrangean of the manipulator is L = K \u2212 V = 1 2 I\u03b8\u0307 2 +mg cos \u03b8 (9.314) and therefore, the equation of motion of the pendulum is M = d dt \u00b5 \u2202L \u2202\u03b8\u0307 \u00b6 \u2212 \u2202L \u2202\u03b8 = I \u03b8\u0308 +mgl sin \u03b8. (9.315) The generalized force M is the contribution of the motor torque Q and the viscous friction torque \u2212c\u03b8\u0307. Hence, the equation of motion of the manipulator is Q = I \u03b8\u0308 + c\u03b8\u0307 +mgl sin \u03b8. (9.316) Example 370 An ideal 2R planar manipulator dynamics. An ideal model of a 2R planar manipulator is illustrated in Figure 9.15. It is called ideal because we assume the links are massless and there is no friction. The masses m1 and m2 are the mass of the second motor to run 564 9. Applied Dynamics the second link and the load at the endpoint. We take the absolute angle \u03b81 and the relative angle \u03b82 as the generalized coordinates to express the configuration of the manipulator. The global positions of m1 and m2 are\u2219 X1 Y2 \u00b8 = \u2219 l1 cos \u03b81 l1 sin \u03b81 \u00b8 (9.317)\u2219 X2 Y2 \u00b8 = \u2219 l1 cos \u03b81 + l2 cos (\u03b81 + \u03b82) l1 sin \u03b81 + l2 sin (\u03b81 + \u03b82) \u00b8 (9.318) and therefore, the global velocity of the masses are\u2219 X\u03071 Y\u03071 \u00b8 = \u2219 \u2212l1\u03b8\u03071 sin \u03b81 l1\u03b8\u03071 cos \u03b81 \u00b8 (9.319) \u2219 X\u03072 Y\u03072 \u00b8 = \u23a1\u23a3 \u2212l1\u03b8\u03071 sin \u03b81 \u2212 l2 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 sin (\u03b81 + \u03b82) l1\u03b8\u03071 cos \u03b81 + l2 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 cos (\u03b81 + \u03b82) \u23a4\u23a6 . (9.320) The kinetic energy of this manipulator is made of kinetic energy of the masses and is equal to K = K1 +K2 = 1 2 m1 \u00b3 X\u03072 1 + Y\u0307 2 1 \u00b4 + 1 2 m2 \u00b3 X\u03072 2 + Y\u0307 2 2 \u00b4 = 1 2 m1l 2 1\u03b8\u0307 2 1 + 1 2 m2 \u00b5 l21\u03b8\u0307 2 1 + l22 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b42 + 2l1l2\u03b8\u03071 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 cos \u03b82 \u00b6 . (9.321) 9. Applied Dynamics 565 The potential energy of the manipulator is V = V1 + V2 = m1gY1 +m2gY2 = m1gl1 sin \u03b81 +m2g (l1 sin \u03b81 + l2 sin (\u03b81 + \u03b82)) . (9.322) The Lagrangean is then obtained from Equations (9.321) and (9.322) L = K \u2212 V (9.323) = 1 2 m1l 2 1\u03b8\u0307 2 1 + 1 2 m2 \u00b5 l21\u03b8\u0307 2 1 + l22 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b42 + 2l1l2\u03b8\u03071 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 cos \u03b82 \u00b6 \u2212 (m1gl1 sin \u03b81 +m2g (l1 sin \u03b81 + l2 sin (\u03b81 + \u03b82))) . which provides the required partial derivatives as follows: \u2202L \u2202\u03b81 = \u2212 (m1 +m2) gl1 cos \u03b81 \u2212m2gl2 cos (\u03b81 + \u03b82) (9.324) \u2202L \u2202\u03b8\u03071 = (m1 +m2) l 2 1\u03b8\u03071 +m2l 2 2 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 +m2l1l2 \u00b3 2\u03b8\u03071 + \u03b8\u03072 \u00b4 cos \u03b82 (9.325) d dt \u00b5 \u2202L \u2202\u03b8\u03071 \u00b6 = (m1 +m2) l 2 1\u03b8\u03081 +m2l 2 2 \u00b3 \u03b8\u03081 + \u03b8\u03082 \u00b4 +m2l1l2 \u00b3 2 \u03b8\u03081 + \u03b8\u03082 \u00b4 cos \u03b82 \u2212m2l1l2\u03b8\u03072 \u00b3 2\u03b8\u03071 + \u03b8\u03072 \u00b4 sin \u03b82 (9.326) \u2202L \u2202\u03b82 = \u2212m2l1l2\u03b8\u03071 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 sin \u03b82 \u2212m2gl2 cos (\u03b81 + \u03b82) (9.327) \u2202L \u2202\u03b8\u03072 = m2l 2 2 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 +m2l1l2\u03b8\u03071 cos \u03b82 (9.328) d dt \u00b5 \u2202L \u2202\u03b8\u03072 \u00b6 = m2l 2 2 \u00b3 \u03b8\u03081 + \u03b8\u03082 \u00b4 +m2l1l2\u03b8\u03081 cos \u03b82 \u2212m2l1l2\u03b8\u03071\u03b8\u03072 sin \u03b82 (9.329) Therefore, the equations of motion for the 2R manipulator are Q1 = d dt \u00b5 \u2202L \u2202\u03b8\u03071 \u00b6 \u2212 \u2202L \u2202\u03b81 = (m1 +m2) l 2 1\u03b8\u03081 +m2l 2 2 \u00b3 \u03b8\u03081 + \u03b8\u03082 \u00b4 +m2l1l2 \u00b3 2 \u03b8\u03081 + \u03b8\u03082 \u00b4 cos \u03b82 \u2212m2l1l2\u03b8\u03072 \u00b3 2\u03b8\u03071 + \u03b8\u03072 \u00b4 sin \u03b82 +(m1 +m2) gl1 cos \u03b81 +m2gl2 cos (\u03b81 + \u03b82) (9.330) 566 9. Applied Dynamics Q2 = d dt \u00b5 \u2202L \u2202\u03b8\u03072 \u00b6 \u2212 \u2202L \u2202\u03b82 = m2l 2 2 \u00b3 \u03b8\u03081 + \u03b8\u03082 \u00b4 +m2l1l2\u03b8\u03081 cos \u03b82 \u2212m2l1l2\u03b8\u03071\u03b8\u03072 sin \u03b82 +m2l1l2\u03b8\u03071 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 sin \u03b82 +m2gl2 cos (\u03b81 + \u03b82) . (9.331) The generalized forces Q1 and Q2 are the required forces to drive the generalized coordinates. In this case, Q1 is the torque at the base motor and Q2 is the torque of the motor at m1. The equations of motion can be rearranged to have a more systematic form Q1 = \u00a1 (m1 +m2) l 2 1 +m2l2 (l2 + 2l1 cos \u03b82) \u00a2 \u03b8\u03081 +m2l2 (l2 + l1 cos \u03b82) \u03b8\u03082 \u22122m2l1l2 sin \u03b82 \u03b8\u03071\u03b8\u03072 \u2212m2l1l2 sin \u03b82 \u03b8\u0307 2 2 +(m1 +m2) gl1 cos \u03b81 +m2gl2 cos (\u03b81 + \u03b82) (9.332) Q2 = m2l2 (l2 + l1 cos \u03b82) \u03b8\u03081 +m2l 2 2\u03b8\u03082 +m2l1l2 sin \u03b82 \u03b8\u0307 2 1 +m2gl2 cos (\u03b81 + \u03b82) . (9.333) Example 371 Mechanical energy. If a system of masses mi are moving in a potential force field Fmi = \u2212\u2207iV (9.334) their Newton equations of motion will be mir\u0308i = \u2212\u2207iV i = 1, 2, \u00b7 \u00b7 \u00b7n. (9.335) The inner product of equations of motion with r\u0307i and adding the equations nX i=1 mir\u0307i \u00b7 r\u0308i = \u2212 nX i=1 r\u0307i \u00b7\u2207iV (9.336) and then, integrating over time 1 2 nX i=1 mir\u0307i \u00b7 r\u0307i = \u2212 Z nX i=1 ri \u00b7\u2207iV (9.337) shows that K = \u2212 Z nX i=1 \u00b5 \u2202V \u2202xi xi + \u2202V \u2202yi yi + \u2202V \u2202zi zi \u00b6 = \u2212V +E (9.338) 9. Applied Dynamics 567 where E is the constant of integration. E is called the mechanical energy of the system and is equal to kinetic plus potential energies. E = K + V (9.339) Example 372 Falling wheel. Figure 9.16 illustrates a wheel turning, without slip, over a cylindrical hill. We may use the conservation of mechanical energy to find the angle at which the wheel leaves the hill. At the initial instant of time, the wheel is at point A. We assume the initial kinetic and potential, and hence, the mechanical energies are zero. When the wheel is turning over the hill, its angular velocity, \u03c9, is \u03c9 = v r (9.340) where v is the speed at the center of the wheel. At any other point B, the wheel achieves some kinetic energy and loses some potential energy. At a certain angle, where the normal component of the weight cannot provide more centripetal force, mg cos \u03b8 = mv2 R+ r . (9.341) the wheel separates from the surface. Employing the conservation of energy, we have EA = EB (9.342) KA + VA = KB + VB. (9.343) The kinetic and potential energy at the separation point B are KB = 1 2 mv2 + 1 2 IC\u03c9 2 (9.344) VB = \u2212mg (R+ r) (1\u2212 cos \u03b8) (9.345) 568 9. Applied Dynamics where IC is the mass moment of inertia for the wheel about its center. Therefore, 1 2 mv2 + 1 2 IC\u03c9 2 = mg (R+ r) (1\u2212 cos \u03b8) (9.346) and substituting (9.340) and (9.341) provides\u00b5 1 + IC mr2 \u00b6 (R+ r) g cos \u03b8 = 2g (R+ r) (1\u2212 cos \u03b8) (9.347) and therefore, the separation angle is \u03b8 = cos\u22121 2mr2 IC + 3mr2 . (9.348) Let\u2019s examine the equation for a disc wheel with IC = 1 2 mr2. (9.349) and find the separation angle. \u03b8 = cos\u22121 4 7 (9.350) \u2248 0.96 rad \u2248 55.15 deg Example 373 Turning wheel over a step. Figure 9.17 illustrates a wheel of radius R turning with speed v to go over a step with height H < R. We may use the principle of energy conservation and find the speed of the wheel after getting across the step. Employing the conservation of energy, 9. Applied Dynamics 569 we have EA = EB (9.351) KA + VA = KB + VB (9.352) 1 2 mv21 + 1 2 IC\u03c9 2 1 + 0 = 1 2 mv22 + 1 2 IC\u03c9 2 2 +mgH (9.353)\u00b5 m+ IC R2 \u00b6 v21 = \u00b5 m+ IC R2 \u00b6 v22 + 2mgH (9.354) and therefore, v2 = vuutv21 \u2212 2gH 1 + IC mR2 . (9.355) The condition for having a real v2 is v1 > vuut 2gH 1 + IC mR2 . (9.356) The second speed (9.355) and the condition (9.356) for a solid disc are v2 = r v21 \u2212 4 3 Hg (9.357) v1 > r 4 3 Hg (9.358) because we assumed that IC = 1 2 mR2. (9.359) Example 374 Trebuchet. A trebuchet, shown schematically in Figure 9.18, is a shooting weapon of war powered by a falling massive counterweight m1. A beam AB is pivoted to the chassis with two unequal sections a and b. The figure shows a trebuchet at its initial configuration. The origin of a global coordinate frame is set at the pivot point. The counterweight m1 is at (x1, y1) and is hinged at the shorter arm of the beam at a distance c from the end B. The mass of the projectile is m2 and it is at the end of a massless sling with a length l attached to the end of the longer arm of the beam. The three independent variable angles \u03b1, \u03b8, and \u03b3 describe the motion of the device. We consider the parameters a, b, c, d, l, m1, and m2 constant, and determine the equations of motion by the Lagrange method. Figure 9.19 illustrates the trebuchet when it is in motion. The position coordinates of masses m1 and m2 are x1 = b sin \u03b8 \u2212 c sin (\u03b8 + \u03b3) (9.360) y1 = \u2212b cos \u03b8 + c cos (\u03b8 + \u03b3) (9.361) 570 9. Applied Dynamics and x2 = \u2212a sin \u03b8 \u2212 l sin (\u2212\u03b8 + \u03b1) (9.362) y2 = \u2212a cos \u03b8 \u2212 l cos (\u2212\u03b8 + \u03b1) . (9.363) Taking a time derivative provides the velocity components x\u03071 = b\u03b8\u0307 cos \u03b8 \u2212 c \u00b3 \u03b8\u0307 + \u03b3\u0307 \u00b4 cos (\u03b8 + \u03b3) (9.364) y\u03071 = b\u03b8\u0307 sin \u03b8 \u2212 c \u00b3 \u03b8\u0307 + \u03b3\u0307 \u00b4 sin (\u03b8 + \u03b3) (9.365) x\u03072 = l (c\u2212 \u03b1\u0307) cos (\u03b1\u2212 \u03b8)\u2212 a\u03b8\u0307 cos (\u03b8) (9.366) y\u03072 = a\u03b8\u0307 sin \u03b8 \u2212 l \u00b3 \u03b8\u0307 \u2212 \u03b1\u0307 \u00b4 sin (\u03b1\u2212 \u03b8) . (9.367) 9. Applied Dynamics 571 which shows that the kinetic energy of the system is K = 1 2 m1v 2 1 + 1 2 m2v 2 2 = 1 2 m1 \u00a1 x\u030721 + y\u030721 \u00a2 + 1 2 m2 \u00a1 x\u030722 + y\u030722 \u00a2 = 1 2 m1 \u00b3\u00a1 b2 + c2 \u00a2 \u03b8\u0307 2 + c2\u03b3\u03072 + 2c2\u03b8\u0307\u03b3\u0307 \u00b4 \u2212m1bc\u03b8\u0307 \u00b3 \u03b8\u0307 + \u03b3\u0307 \u00b4 cos \u03b3 + 1 2 m2 \u00b3\u00a1 a2 + l2 \u00a2 \u03b8\u0307 2 + l2\u03b1\u03072 \u2212 2l2\u03b8\u0307\u03b1\u0307 \u00b4 \u2212m2al\u03b8\u0307 \u00b3 \u03b8\u0307 \u2212 \u03b1\u0307 \u00b4 cos (2\u03b8 \u2212 \u03b1) . (9.368) The potential energy of the system can be calculated by y-position of the masses. V = m1gy1 +m2gy2 = m1g (\u2212b cos \u03b8 + c cos (\u03b8 + \u03b3)) +m2g (\u2212a cos \u03b8 \u2212 l cos (\u2212\u03b8 + \u03b1)) (9.369) Having the energies K and V , we can set up the Lagrangean L. L = K \u2212 V (9.370) Using the Lagrangean, we are able to find the three equations of motion. d dt \u00b5 \u2202L \u2202\u03b8\u0307 \u00b6 \u2212 \u2202L \u2202\u03b8 = 0 (9.371) d dt \u00b5 \u2202L \u2202\u03b1\u0307 \u00b6 \u2212 \u2202L \u2202\u03b1 = 0 (9.372) d dt \u00b5 \u2202L \u2202\u03b3\u0307 \u00b6 \u2212 \u2202L \u2202\u03b3 = 0. (9.373) The trebuchet appeared in 500 to 400 B.C. China and was developed by Persian armies around 300 B.C. It was used by the Arabs against the Romans during 600 to 1200 A.D. The trebuchet is also called the manjaniq, catapults, or onager. The \"Manjaniq\" is the root of the words \"machine\" and \"mechanic\". 9.7 Summary The translational and rotational equations of motion for a rigid body, expressed in the global coordinate frame, are GF = Gd dt Gp (9.374) GM = Gd dt GL (9.375) 572 9. Applied Dynamics where GF and GM indicate the resultant of the external forces and moments applied on the rigid body, measured at the mass center C. The vector Gp is the momentum and GL is the moment of momentum for the rigid body at C p = mv (9.376) L = rC \u00d7 p. (9.377) The expression of the equations of motion in the body coordinate frame are BF = Gp\u0307+ B G\u03c9B \u00d7 Bp = m BaB +m B G\u03c9B \u00d7 BvB (9.378) BM = BL\u0307+B G\u03c9B \u00d7 BL = BI B G\u03c9\u0307B + B G\u03c9B \u00d7 \u00a1 BI B G\u03c9B \u00a2 (9.379) where I is the moment of inertia for the rigid body. I = \u23a1\u23a3 Ixx Ixy Ixz Iyx Iyy Iyz Izx Izy Izz \u23a4\u23a6 . (9.380) 9. Applied Dynamics 573 The elements of I are functions of the mass distribution of the rigid body and are defined by Iij = Z B \u00a1 r2i \u03b4mn \u2212 ximxjn \u00a2 dm , i, j = 1, 2, 3 (9.381) where \u03b4ij is Kronecker\u2019s delta. Every rigid body has a principal body coordinate frame in which the moment of inertia is in the form BI = \u23a1\u23a3 I1 0 0 0 I2 0 0 0 I3 \u23a4\u23a6 . (9.382) The rotational equation of motion in the principal coordinate frame simplifies to M1 = I1\u03c9\u03071 \u2212 (I2 \u2212 I2)\u03c92\u03c93 M2 = I2\u03c9\u03072 \u2212 (I3 \u2212 I1)\u03c93\u03c91 (9.383) M3 = I3\u03c9\u03073 \u2212 (I1 \u2212 I2)\u03c91\u03c92. The equations of motion for a mechanical system having n DOF can also be found by the Lagrange equation d dt \u00b5 \u2202L \u2202q\u0307r \u00b6 \u2212 \u2202L \u2202qr = Qr r = 1, 2, \u00b7 \u00b7 \u00b7n (9.384) L = K \u2212 V (9.385) where L is the Lagrangean of the system, K is the kinetic energy, V is the potential energy, and Qr is the nonpotential generalized force. Qr = nX i=1 \u00b5 Qix \u2202fi \u2202q1 +Qiy \u2202gi \u2202q2 +Qiz \u2202hi \u2202qn \u00b6 (9.386) The parameters qr, r = 1, 2, \u00b7 \u00b7 \u00b7 , n are the generalized coordinates of the system, Q = \u00a3 Qix Qiy Qiz \u00a4T is the external force acting on the ith particle of the system, and Qr is the generalized force associated to qr. When (xi, yi, zi) are Cartesian coordinates in a globally fixed coordinate frame for the particle mi, then its coordinates may be functions of another set of generalized coordinates q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn and possibly time t. xi = fi(q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, t) (9.387) yi = gi(q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, t) (9.388) zi = hi(q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, t) (9.389) 574 9. Applied Dynamics 9.8 Key Symbols a, b, w, h length a acceleration C mass center d position vector of the body coordinate frame df infinitesimal force dm infinitesimal mass dm infinitesimal moment E mechanical energy F force FC Coriolis force g gravitational acceleration H height I moment of inertia matrix I1, I2, I3 principal moment of inertia K kinetic energy l directional line L moment of momentum L = K \u2212 V Lagrangean m mass M moment p momentum P,Q points in rigid body r radius of disc r position vector R radius R rotation matrix t time u\u0302 unit vector to show the directional line v \u2261 x\u0307, v velocity V potential energy w eigenvector W work W eigenvector matrix x, y, z, x displacement \u03b4ij Kronecker\u2019s delta \u0393ij,k Christoffel operator \u03bb eigenvalue \u03d5, \u03b8, \u03c8 Euler angles \u03c9,\u03c9 angular velocity k parallel \u22a5 orthogonal 9. Applied Dynamics 575 Exercises 1. Kinetic energy of a rigid link. Consider a straight and uniform bar as a rigid bar. The bar has a mass m. Show that the kinetic energy of the bar can be expressed as K = 1 6 m (v1 \u00b7 v1 + v1 \u00b7 v2 + v2 \u00b7 v2) where v1 and v2 are the velocity vectors of the endpoints of the bar. 2. Discrete particles. There are three particles m1 = 10kg, m2 = 20 kg, m3 = 30 kg, at r1 = \u23a1\u23a3 1 \u22121 1 \u23a4\u23a6 r1 = \u23a1\u23a3 \u22121\u22123 2 \u23a4\u23a6 r1 = \u23a1\u23a3 2 \u22121 \u22123 \u23a4\u23a6 . Their velocities are v1 = \u23a1\u23a3 2 1 1 \u23a4\u23a6 v1 = \u23a1\u23a3 \u221210 2 \u23a4\u23a6 v1 = \u23a1\u23a3 3 \u22122 \u22121 \u23a4\u23a6 . Find the position and velocity of the system at C. Calculate the system\u2019s momentum and moment of momentum. Calculate the system\u2019s kinetic energy and determine the rotational and translational parts of the kinetic energy. 3. Newton\u2019s equation of motion in the body frame. Show that Newton\u2019s equation of motion in the body frame is\u23a1\u23a3 Fx Fy Fz \u23a4\u23a6 = m \u23a1\u23a3 ax ay az \u23a4\u23a6+ \u23a1\u23a3 0 \u2212\u03c9z \u03c9y \u03c9z 0 \u2212\u03c9x \u2212\u03c9y \u03c9x 0 \u23a4\u23a6\u23a1\u23a3 vx vy vz \u23a4\u23a6 . 4. Work on a curved path. A particle of mass m is moving on a circular path given by GrP = cos \u03b8 I\u0302 + sin \u03b8 J\u0302 + 4 K\u0302. Calculate the work done by a force GF when the particle moves from \u03b8 = 0 to \u03b8 = \u03c0 2 . (a) GF = z2 \u2212 y2 (x+ y) 2 I\u0302 + y2 \u2212 x2 (x+ y) 2 J\u0302 + x2 \u2212 y2 (x+ z) 2 K\u0302 576 9. Applied Dynamics (b) GF = z2 \u2212 y2 (x+ y) 2 I\u0302 + 2y x+ y J\u0302 + x2 \u2212 y2 (x+ z) 2 K\u0302 5. Principal moments of inertia. Find the principal moments of inertia and directions for the following inertia matrices: (a) [I] = \u23a1\u23a3 3 2 2 2 2 0 2 0 4 \u23a4\u23a6 (b) [I] = \u23a1\u23a3 3 2 4 2 0 2 4 2 3 \u23a4\u23a6 (c) [I] = \u23a1\u23a3 100 20 \u221a 3 0 20 \u221a 3 60 0 0 0 10 \u23a4\u23a6 6. Rotated moment of inertia matrix. A principal moment of inertia matrix B2I is given as [I] = \u23a1\u23a3 3 0 0 0 5 0 0 0 4 \u23a4\u23a6 . The principal frame was achieved by rotating the initial body coordinate frame 30 deg about the x-axis, followed by 45 deg about the z-axis. Find the initial moment of inertia matrix B1I. 7. Rotation of moment of inertia matrix. Find the required rotation matrix that transforms the moment of inertia matrix [I] to an diagonal matrix. [I] = \u23a1\u23a3 3 2 2 2 2 0.1 2 0.1 4 \u23a4\u23a6 8. F Cubic equations. The solution of a cubic equation ax3 + bx2 + cx+ d = 0 9. Applied Dynamics 577 where a 6= 0, can be found in a systematic way. Transform the equation to a new form with discriminant 4p3 + q2, y3 + 3py + q = 0 using the transformation x = y \u2212 b 3a , where, p = 3ac\u2212 b2 9a2 q = 2b3 \u2212 9abc+ 27a2d 27a3 . The solutions are then y1 = 3 \u221a \u03b1\u2212 3 p \u03b2 y2 = e 2\u03c0i 3 3 \u221a \u03b1\u2212 e 4\u03c0i 3 3 p \u03b2 y3 = e 4\u03c0i 3 3 \u221a \u03b1\u2212 e 2\u03c0i 3 3 p \u03b2 where, \u03b1 = \u2212q + p q2 + 4p3 2 \u03b2 = \u2212q + p q2 + 4p3 2 . For real values of p and q, if the discriminant is positive, then one root is real, and two roots are complex conjugates. If the discriminant is zero, then there are three real roots, of which at least two are equal. If the discriminant is negative, then there are three unequal real roots. Apply this theory for the characteristic equation of the matrix [I] and show that the principal moments of inertia are real. 9. Kinematics of a moving car on the Earth. The location of a vehicle on the Earth is described by its longitude \u03d5 from a fixed meridian, say, the Greenwich meridian, and its latitude \u03b8 from the equator, as shown in Figure 9.20. We attach a coordinate frame B at the center of the Earth with the x-axis on the equator\u2019s plane and the y-axis pointing to the vehicle. There are also two coordinate frames E and G where E is attached to the Earth and G is the global coordinate frame. Show that the angular velocity of B and the velocity of the vehicle are B G\u03c9B = \u03b8\u0307 \u0131\u0302B + (\u03c9E + \u03d5\u0307) sin \u03b8 j\u0302B + (\u03c9E + \u03d5\u0307) cos \u03b8 k\u0302 B GvP = \u2212r (\u03c9E + \u03d5\u0307) cos \u03b8 \u0131\u0302B + r\u03b8\u0307 k\u0302. Calculate the acceleration of the vehicle. 578 9. Applied Dynamics \u03b8 X Y Z x y z r \u03c9E Z (a) (b) G B Y z x y E \u03d5 \u03b8 y z \u03c9E P P 9. Applied Dynamics 579 (c) the pivot O has a uniform motion on a circle rO = R cos\u03c9t I\u0302 +R sin\u03c9t J\u0302. 13. Equations of motion from Lagrangean. Consider a physical system with a Lagrangean as L = 1 2 m (ax\u0307+ by\u0307) 2 \u2212 1 2 k (ax+ by) 2 . and find the equations of motion. The coefficients m, k, a, and b are constant. 14. Lagrangean from equation of motion. Find the Lagrangean associated to the following equations of motions: (a) mr2\u03b8\u0308 + k1l1\u03b8 + k2l2\u03b8 +mgl = 0 580 9. Applied Dynamics (b) r\u0308 \u2212 r \u03b8\u0307 2 = 0 r2 \u03b8\u0308 + 2r r\u0307 \u03b8\u0307 = 0 15. Trebuchet. Derive the equations of motion for the trebuchet shown in Figure 9.18. 16. Simplified trebuchet. Three simplified models of a trebuchet are shown in Figures 9.23 to 9.25. Derive and compare their equations of motion. 9. Applied Dynamics 581 10"
        ]
    },
    {
        "image_filename": "designv10_0_0000209_tsmc.1980.4308393-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000209_tsmc.1980.4308393-Figure1-1.png",
        "caption": "Fig. 1. Standard coordinate axes definitions for connected links and relationships between neighboring coordinate axes.",
        "texts": [
            " Part 3) was delineated in [12] for the Newton- Euler method, and a similar preference for 3 x 3 versus 4x 4 matrices was expressed for reasons of computational efficiency. UICKER/KAHN LAGRANGIAN FORMULATION The standard formulation for manipulator dynamics is derived from Uicker [1], who used 4 x 4 matrices to set up the Lagrangian-based dynamics for a somewhat more general linkage problem. This formulation was particularized to open-loop kinematic chains by Kahn [2]. Borrowing Kahn's notation (Fig. 1), the links of a manipulator are numbered consecutively from 1 to n starting from the base to the tip. By convention the reference frame is numbered link 0. The joints are the points of articulation between links and are numbered so that joint i connects links i- 1 and i. An orthogonal coordinate system is fixed in each link as follows: zi xi Yi is directed along the axis of joint i+ 1, lies along the common normal from zi1 to z1, and completes the right coordinate. The relative position of two adjacent links is completely described by the following four parameters: ai distance between the origins of coordinate systems i- I and i measured along xi, si distance between xi_ l and x, measured along zi-1, ai angle between the zi- 1 and zi axes measured in a right-hand sense about xi, and 9i angle between the xi_, and the xi axes measured in the right-hand sense about zi- I If the joint is rotational the joint variable will be 9,; if translational the joint variable will be si"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003694_nature01232-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003694_nature01232-Figure1-1.png",
        "caption": "Figure 1 The layout of the experiment. A glass fibre (F) is inserted into a flowing soap-film tunnel (partly shown). The fibre is supported by a thin stainless-steel rod (S), which is",
        "texts": [
            " The fibre undergoes a bending transition, producing shapes that are self-similar; for such configurations, the drag scales with the length of self-similarity, rather than the fibre profile width. These predictions are supported by our experimental data. Experiments that cleanly reveal the nature of interactions between deformable bodies and flows are difficult to perform. Complications include controlling three-dimensional flow effects and visualizing the flows. Soap film is a convenient experimental system described by two-dimensional hydrodynamics in many aspects6\u201310. Our soap-film flow tunnel is illustrated in Fig. 1. Driven by gravity, soapy water (1.5% Dawn dish detergent; density r \u00bc 1 gm cm23) leaves an elevated reservoir and spreads into a vertical soap film (thickness f \u00bc 1\u20133 mm) descending between two straight nylon lines (tunnel width, 9.0 cm). Adjusting reservoir efflux rate adjusts flow velocity U through the range 0.5\u2013 3.0 m s21; breakage occurs at velocities outside this range. Half-way down the tunnel, a thin, flexible glass fibre (length L \u00bc 1\u20135 cm; diameter, 34 mm; rigidity E \u00bc 2.8 erg cm), glued at its midpoint to a thin rod, is inserted transverse to the flow"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure16.4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure16.4-1.png",
        "caption": "Figure 16.4 (a) Hinged beam with influence lines for (b) the vertical support reaction at B, (c) the bending moment at E, (d) the bending moment at G and (e) the shear force at G; (f) positive directions for bending moment and shear force.",
        "texts": [
            " 16 Influence Lines 747 If the load is to the right of C ( 1 2 <x \u2264 ) then the shear force at C is equal to the support reaction Av at A: VC = Av = F \u2212 x . This determines the influence line for the shear force at C. The influence line is shown in Figure 16.3c. If the point load is to the left of C the influence line is negative. The shear force acts in the direction opposite to the one assumed in Figure 16.3b. The shear force at C is a maximum if the point load F is directly to the left or right of C: VC = (\u00b10.5) \u00d7 F = \u00b1 1 2F. A change of sign occurs at C. Figure 16.4 shows the influence lines for a hinged beam. The positive direction of the support reaction Bv at B is shown in Figure 16.4a. For the bending moment and the shear force, the positive directions according to the xz coordinate system are given in Figure 16.4f. We do not address the calculation of each of these influence lines in detail here. They can be found by, for various positions of the point load, using the equilibrium equations to calculate the magnitude of the various quantities. In this case, that leads to a fair amount of work as there are so many positions of the load to be considered. It is preferable to investigate a number of characteristic positions of the point load (e.g. just above the support loads, or at the hinges) and to remember that the influence line between certain points has to be linear",
            " Using the influence line, determine the value of X due to the set of loads when F2 is at S2. f. Using the influence line, determine the value of X when the uniformly distributed load q acts over the entire length of the beam. g. Where must the uniformly distributed load q be placed so that X is a maximum? Determine this maximum value. h. Where must the uniformly distributed load q be placed so that X is a minimum? Determine this minimum value. 16.4: 1\u201310 A point load F , a set of loads F1; F2; F3 and a uniformly distributed load q can move across the hinged beam in Figure 16.4. Use F = F1 = F2 = F3 = 200 kN and q = 24 kN/m. The same questions are asked for 10 different quantities X: 1. X = vertical support reaction at A. 2. X = vertical support reaction at B. 3. X = bending moment at B. 4. X = bending moment at E. 5. X = bending moment at G. 6. X = shear force at S1. 7. X = shear force at E. 8. X = shear force at G. 9. X = shear force directly to the left of C. 10. X = shear force directly to the right of C. Questions: a. Draw the influence line for X. b. Where must the load F be placed so that X is a maximum"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.33-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.33-1.png",
        "caption": "FIGURE 7.33. Sign convention for steer angles.",
        "texts": [
            "73) where, wf and wr are the front and rear tracks, \u03b4if and \u03b4of are the steer angles of the front inner and outer wheels, \u03b4ir and \u03b4or are the steer angles of the rear inner and outer wheels, and l is the wheelbase of the vehicle. 410 7. Steering Dynamics We may also use the following more general equation for the kinematic condition between the steer angles of a 4WS vehicle cot \u03b4fr \u2212 cot \u03b4fl = wf l \u2212 wr l cot \u03b4fr \u2212 cot \u03b4fl cot \u03b4rr \u2212 cot \u03b4rl (7.74) where, \u03b4fl and \u03b4fr are the steer angles of the front left and front right wheels, and \u03b4rl and \u03b4rr are the steer angles of the rear left and rear right wheels. If we define the steer angles according to the sign convention shown in Figure 7.33 then, Equation (7.73) expresses the kinematic condition for both, positive and negative 4WS systems. Employing the wheel coordinate frame (xw, yw, zw), we define the steer angle as the angle between the vehicle x-axis and the wheel xw-axis, measured about the z-axis. Therefore, a steer angle is positive when the wheel is turned to the left, and it is negative when the wheel is turned to the right. Proof. The slip-free condition for wheels of a 4WS in a turn requires that the normal lines to the center of each tire-plane intersect at a common point",
            "93) Similarly, we may eliminate R1 R1 = 1 2 wr + c2 tan \u03b4ir (7.94) = \u22121 2 wr + c2 tan \u03b4or (7.95) between (7.89) and (7.90) to provide the kinematic condition between the rear steering angles \u03b4ir and \u03b4or. cot \u03b4or \u2212 cot \u03b4ir = wr c2 (7.96) Using the following constraint c1 \u2212 c2 = l (7.97) we may combine Equations (7.93) and (7.96) wf cot \u03b4of \u2212 cot \u03b4if \u2212 wr cot \u03b4or \u2212 cot \u03b4ir = l (7.98) to find the kinematic condition (7.73) between the steer angles of the front and rear wheels for a negative 4WS vehicle. Using the sign convention shown in Figure 7.33, we may re-examine Figures 7.35 and 7.34. When the steer angle of the front wheels are positive then, the steer angle of the rear wheels are negative in a negative 4WS system, and are positive in a positive 4WS system. Therefore, Equation (7.74) cot \u03b4fr \u2212 cot \u03b4fl = wf l \u2212 wr l cot \u03b4fr \u2212 cot \u03b4fl cot \u03b4rr \u2212 cot \u03b4rl (7.99) 7. Steering Dynamics 415 can express the kinematic condition for both, positive and negative 4WS systems. Similarly, the following equations can uniquely determine c1 and c2 regardless of the positive or negative 4WS system"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001674_j.mechmachtheory.2012.10.008-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001674_j.mechmachtheory.2012.10.008-Figure4-1.png",
        "caption": "Fig. 4. Geometric gap for Involute\u2013Tip rounding contacts.",
        "texts": [
            " The where working distance dT and the pressure angle \u03c6T after gear centre displacements with respect to the reference position are obtained from Eqs. (3) and (4) \u03c8 \u00bc \u2212 arctan x1\u2212x2 d0 \u00fe y1\u2212y2 \u00f02\u00de dT \u00bc ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x1\u2212x2\u00f0 \u00de2 \u00fe d0 \u00fe y1\u2212y2\u00f0 \u00de2 q \u00f03\u00de \u03c6T \u00bc arccos \u03c11 \u00fe \u03c12 dT \u00f04\u00de When considering the contact Involute\u2013Tip rounding (in gear 2), the normal to the profiles is the tangent to the base circle of the involute profile passing through the centre of the tip rounding arc (Cr) (see Fig. 4). On account of this, a new tangent circle in gear 2 is defined (referred to as equivalent base circle) whose origin is at the centre of the gear 2 and whose radius \u03c12r2 is obtained from \u03c12 r2 \u00bc C2Q2 r2 \u00bc dT cos \u03c6T r2 \u2212\u03c11 \u00f05\u00de Thus the Involute\u2013Tip rounding may be represented in a similar way to Involute\u2013Involute contact, i.e. by a line tangential to the two gear base circles. The parameters that define this contact are expressed by incorporating the superscript r2 to facilitate their distinction from the Involute\u2013Involute contact, leading to the following formulation \u03b4Ev\u2212r2 \u00bc \u03c11 tan\u03c60 \u00fe \u03c6T r2\u2212\u03c60 \u00fe \u03c8\u2212\u03b81\u00f0 \u00de \u00fe RExt2\u2212rTip2 sin \u03c6T r2\u2212\u03bb2 \u00fe rTip2\u2212dT sin \u03c6T r2 \u00f06\u00de When the rounding is on gear 1, the last equation remains valid, but exchanging the subscripts 1 and 2",
            " 5, the inverse contact profile is rotated with respect to the reference position at an angle \u03b8* whose value is derived from the ratio between the thickness of the tooth in the initial pitch circle (e10) and the corresponding radius using the well known expression for calculating the thickness of the involute tooth profile, so that \u03b8 \u00bc e10 R10 \u00bc m \u03c0 2 \u00fe x tan\u03c6 \u00fe 2 Ev \u03c6\u00f0 \u00de\u2212Ev \u03c60\u00f0 \u00de\u00bd \u00f07\u00de Then, the corresponding distances for the Involute \u2013 Involute and Involute \u2013 Tip rounding for the reverse contacts (designated by the subscript I) can be expressed as \u03b4EvI EvI \u00bc \u03c11 \u00fe \u03c12\u00f0 \u00de tan\u03c60 \u00fe \u03c6T\u2212\u03c60\u2212 \u03c8\u2212\u03b8 \u00f0 \u00de\u00f0 \u00de \u00fe \u03c11\u03b81 \u00fe \u03c12\u03b82\u00f0 \u00de\u2212dT sin \u03c6T\u00f0 \u00de \u00f08\u00de \u03b4EvI\u2212r2I \u00bc \u03c11 tan\u03c60 \u00fe \u03c6TI r2\u2212\u03c60\u2212 \u03c8\u2212 \u03b81 \u00fe \u03b8 \u00f0 \u00de\u00f0 \u00de \u00fe RExt2\u2212rpunta2 sin \u03c6TI r2\u2212\u03bb2 \u00fe rpunta2\u2212dT sin \u03c6TI r2 \u00f09\u00de According to the above expressions, the tooth profiles have three possible types of contact: Involute\u2013Involute, Involute (gear 1)\u2013Tip rounding (gear 2) and Involute (gear 2)\u2013Tip rounding in (gear 1). Therefore, a procedure should be established to discernwhich of them is applicable. Noting Fig. 4, contact at the tip rounding begins at points A or B depending on the direction of rotation. When contact occurs at point A, the Involute\u2013Involute pressure angle and contact base radii are identical to those of Involute\u2013Tip rounding (in gear 2). From that point until point B, Involute\u2013Tip rounding pressure angle increases its value compared to the Involute\u2013Involute one. The same approach is valid in the case of analysing the Involute\u2013Tip rounding (gear 1). Therefore, the active contact will be the one that presents the maximum pressure angle"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001466_j.mechmachtheory.2014.02.001-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001466_j.mechmachtheory.2014.02.001-Figure7-1.png",
        "caption": "Fig. 7. Structure of a planetary gear set with the ring gear fixed.",
        "texts": [
            " It can be seen that even the radius difference of the base circle and the root circle is only 1 mm for the planet gear; the mesh stiffness decreased about 17% in the double-tooth-pair meshing duration and about 23% in the single-tooth-pair meshing duration. This percentage change is smaller than the external\u2013external gear case because the same stiffness equations are used for the internal gear in these two models. Though relatively small, this result indicates the importance of modeling the gear tooth of the external gear starting from the root circle as it is more realistic. Fig. 7 shows the structure of a planetary gear set which will be considered in this study. It is comprised of a sun gear (s), a ring gear (r) and four equally-spaced planets (p) which are held by a common carrier (c). In this structure, the ring gear is fixed. The parameters of this planetary gear set are listed in Table 1. A planetary gear set comprises several pairs of sun\u2013planet gears (external\u2013external gears) and ring\u2013planet gears (external\u2013internal gears) meshing simultaneously. While each of the sun\u2013planet meshes (or ring\u2013planet meshes) has the same shape of mesh stiffness variation, they are not in phase with each other [20]",
            " But, the expression of the tooth height, hx, changes as: hx \u00bc Rr sin\u03b3; if 0\u2264x\u2264d1 Rb \u03b12\u2212\u03b1\u00f0 \u00de cos\u03b1 \u00fe sin\u03b1\u00bd ; if d1bx\u2264d: \u00f05:17\u00de Applying the same procedure as in Case 1, the bending stiffness kb, and shear stiffness ks for the four conditions are derived and listed as follows: Condition 1: When ha \u2265 ho & \u03b11 N \u03b1a 1 kb \u00bc Z \u03b1r \u03b13 12 sin\u03b1 N cos\u03b10 N\u22122:5 \u2212 cos\u03b1 \u00fe cos\u03b14\u2212 cos\u03b1r\u2212 q1 Rr cos\u03c5 cos\u03b11 h i2 EL sin\u03b14 \u00fe sin\u03b1\u2212 q1 Rr sin\u03c5 3 d\u03b1 \u00fe Z \u03b15 \u2212\u03b1a 12 1\u00fe cos\u03b11 \u03b12\u2212\u03b1\u00f0 \u00de sin\u03b1\u2212 cos\u03b1\u00bd f g2 \u03b12\u2212\u03b1\u00f0 \u00de cos\u03b1 EL N\u22122:5 N cos\u03b10 sin\u03b14\u2212 q1 Rb sin\u03c5\u00fe sin\u03b1 \u00fe \u03b12\u2212\u03b1\u00f0 \u00de cos\u03b1 h i3 d\u03b1 \u00fe Z \u2212\u03b1a \u2212\u03b11 3 1\u00fe cos\u03b11 \u03b12\u2212\u03b1\u00f0 \u00de sin\u03b1\u2212 cos\u03b1\u00bd f g2 \u03b12\u2212\u03b1\u00f0 \u00de cos\u03b1 2EL sin\u03b1 \u00fe \u03b12\u2212\u03b1\u00f0 \u00de cos\u03b1\u00bd 3 d\u03b1 \u00f05:18\u00de 1 ks \u00bc Z \u03b1r \u03b13 2:4 1\u00fe \u03bd\u00f0 \u00de cos2\u03b11 sin\u03b1 EL sin\u03b14 \u00fe sin\u03b1\u2212 q1 Rr sin\u03c5 3d\u03b1 \u00fe Z \u2212\u03b1a \u2212\u03b11 1:2 1\u00fe \u03bd\u00f0 \u00de \u03b12\u2212\u03b1\u00f0 \u00de cos\u03b1 cos2\u03b11 EL sin\u03b1 \u00fe \u03b12\u2212\u03b1\u00f0 \u00de cos\u03b1\u00bd d\u03b1 \u00fe Z \u03b15 \u2212\u03b1a 2:4 1\u00fe \u03bd\u00f0 \u00de \u03b12\u2212\u03b1\u00f0 \u00de cos\u03b1 cos2\u03b11 EL N\u22122:5 N cos\u03b10 sin\u03b14\u2212 q1 Rb sin\u03c5\u00fe sin\u03b1 \u00fe \u03b12\u2212\u03b1\u00f0 \u00de cos\u03b1 d\u03b1: \u00f05:19\u00de Condition 2: When ha b ho or when ha \u2265 ho & \u03b11 \u2264 \u03b1a 1 kb \u00bc Z \u03b1r \u03b13 12 sin\u03b1 N cos\u03b10 N\u22122:5 \u2212 cos\u03b1 \u00fe cos\u03b14\u2212 cos\u03b1r\u2212 q1 Rr cos\u03c5 cos\u03b11 h i2 EL sin\u03b14 \u00fe sin\u03b1\u2212 q1 Rr sin\u03c5 3 d\u03b1 \u00fe Z \u03b15 \u2212\u03b11 12 1\u00fe cos\u03b11 \u03b12\u2212\u03b1\u00f0 \u00de sin\u03b1\u2212 cos\u03b1\u00bd f g2 \u03b12\u2212\u03b1\u00f0 \u00de cos\u03b1 EL N\u22122:5 N cos\u03b10 sin\u03b14\u2212 q1 Rb sin\u03c5\u00fe sin\u03b1 \u00fe \u03b12\u2212\u03b1\u00f0 \u00de cos\u03b1 h i3 d\u03b1 \u00f05:20\u00de 1 ks \u00bc Z \u03b1r \u03b13 2:4 1\u00fe \u03bd\u00f0 \u00de cos2\u03b11 sin\u03b1 EL sin\u03b14 \u00fe sin\u03b1\u2212 q1 Rr sin\u03c5 3d\u03b1 \u00fe Z \u03b15 \u2212\u03b11 2:4 1\u00fe \u03bd\u00f0 \u00de \u03b12\u2212\u03b1\u00f0 \u00de cos\u03b1 cos2\u03b11 EL N\u22122:5 N cos\u03b10 sin\u03b14\u2212 q1 Rb sin\u03c5\u00fe sin\u03b1 \u00fe \u03b12\u2212\u03b1\u00f0 \u00de cos\u03b1 d\u03b1: \u00f05:21\u00de Condition 3: When hc b ho or when hc \u2265 ho & \u03b11 \u2264 \u03b1c 1 kb \u00bc Z \u03b1r \u03b13 12 sin\u03b1 N cos\u03b10 N\u22122:5 \u2212 cos\u03b1 \u00fe cos\u03b14\u2212 cos\u03b1r\u2212 sin\u03b14 sin\u03c5 \u2212 q2 Rr cos\u03c5 cos\u03b11 h i2 EL sin\u03b1\u2212 q2 Rr sin\u03c5 3 d\u03b1 \u00fe Z \u03b15 \u2212\u03b11 12 1\u00fe cos\u03b11 \u03b12\u2212\u03b1\u00f0 \u00de sin\u03b1\u2212 cos\u03b1\u00bd f g2 \u03b12\u2212\u03b1\u00f0 \u00de cos\u03b1 EL sin\u03b1 \u00fe \u03b12\u2212\u03b1\u00f0 \u00de cos\u03b1\u2212 q2 Rb sin\u03c5 h i3 d\u03b1 \u00f05:22\u00de 1 ks \u00bc Z \u03b1r \u03b13 2:4 1\u00fe \u03bd\u00f0 \u00de cos2\u03b11 sin\u03b1 EL sin\u03b1\u2212 q2 Rr sin\u03c5 3 d\u03b1 \u00fe Z \u03b15 \u2212\u03b11 2:4 1\u00fe \u03bd\u00f0 \u00de \u03b12\u2212\u03b1\u00f0 \u00de cos\u03b1 cos2\u03b11 EL sin\u03b1 \u00fe \u03b12\u2212\u03b1\u00f0 \u00de cos\u03b1\u2212 q2 Rb sin\u03c5 d\u03b1: \u00f05:23\u00de Condition 4: When hc \u2265 ho & \u03b11 N \u03b1c 1 kb \u00bc Z \u03b1r \u03b13 12 sin\u03b1 N cos\u03b10 N\u22122:5 \u2212 cos\u03b1 \u00fe cos\u03b14\u2212 cos\u03b1r\u2212 sin\u03b14 sin\u03c5 \u2212 q2 Rr cos\u03c5 cos\u03b11 h i2 EL sin\u03b1\u2212 q2 Rr sin\u03c5 3 d\u03b1 \u00fe Z \u03b15 \u2212\u03b1c 12 1\u00fe cos\u03b11 \u03b12\u2212\u03b1\u00f0 \u00de sin\u03b1\u2212 cos\u03b1\u00bd f g2 \u03b12\u2212\u03b1\u00f0 \u00de cos\u03b1 EL sin\u03b1 \u00fe \u03b12\u2212\u03b1\u00f0 \u00de cos\u03b1\u2212 q2 Rb sin\u03c5 h i3 d\u03b1 \u00f05:24\u00de 1 ks \u00bc Z \u03b1r \u03b13 2:4 1\u00fe \u03bd\u00f0 \u00de cos2\u03b11 sin\u03b1 EL sin\u03b1\u2212 q2 Rr sin\u03c5 3 d\u03b1 \u00fe Z \u03b15 \u2212\u03b1c 2:4 1\u00fe \u03bd\u00f0 \u00de \u03b12\u2212\u03b1\u00f0 \u00de cos\u03b1 cos2\u03b11 EL sin\u03b1 \u00fe \u03b12\u2212\u03b1\u00f0 \u00de cos\u03b1\u2212 q2 Rb sin\u03c5 d\u03b1: \u00f05:25\u00de In this section, the crack effect on the mesh stiffness of a planetary gear set (see Fig. 7) is investigated when a crack takes place in the sun gear or the planet gear. Two cases existwhen a crack occurs in the planet gear (sun gear side or ring gear side). If it is in the sun gear side, the ring\u2013planetmesh stiffness is assumed not to be affected as the crack part can still bear the compressive stiffness as if no crack exists. Similarly, if it is in the ring gear side, the sun\u2013planetmesh stiffnesswill not be affected. Overall, three cases are considered in this study; crack in the sun gear, crack in the planet gear (sun side) and crack in the planet gear (ring side)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003858_tbme.2005.846734-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003858_tbme.2005.846734-Figure3-1.png",
        "caption": "Fig. 3. Motion of the needle tip [ _x _y _ ] with respect to the motion applied to the base [ _x _y _ ] determines needle \u201csteerability.\u201d Needle steerability depends on the tissue and needle configuration q .",
        "texts": [
            " There may be additional constraints due to tissue force limits and displacement constraints that avoid the puncture of obstacles within the tissue. If the needle configuration is specified using only the position and orientation of its tip, then the problem of determining a needle trajectory becomes that of guiding its tip along a feasible path. We say that a needle is \u201csteerable\u201d if base motion (motion applied at the base of the needle) can be used to control the position and orientation of the needle tip with respect to the target. This is illustrated in Fig. 3. A needle manipulation Jacobian embodies the concept of needle steerability, and is defined by the relationship between the needle tip and base velocities, and respectively. In the 2-D, three-degree-of-freedom case, is the 3 3 matrix that depends on the needle and tissue configuration (2) The velocity relationship can be written in terms of infinitesimal changes in position, as follows: (3) A closed-form expression for is not available due to the complexity of needle-tissue interaction. Indeed, the needle manipulation Jacobian depends upon both the needle and the tissue configurations at any given instant"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure3.12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure3.12-1.png",
        "caption": "Fig. 3.12: Design parameters and topology in the generation of so-called \u2018cylindroid ring\u2019 workspace for telescopic manipulator arms.",
        "texts": [
            "32), to determine the tori envelopes from the ring up to the N-R hyper-ring in the chain, by computing from the extremity to the base of the manipulator chain . The term hyper-ring has been used to stress revolution operations about more than three axes, which is the case usually referred to as the ring geometry. A similar approach can be used for manipulators with prismatic joints as an extension of the above-mentioned formulation. The procedure of Eqs (3.1.31) and (3.1.32) can also be conveniently adjusted for manipulators with prismatic joints, as Fig. 3.12 synthetically illustrates. In fact, from a descriptive geometry viewpoint, the workspace of a telescopic arm also with prismatic joints can be characterized by looking at the figures, which are generated by H because of successive full movements of the joints starting from the last up to first one. The first joint is fixed to the manipulator base. A telescopic arm is a RRP manipulator whose prismatic joint is the extremity joint. Thus, movements of the sliding joint will generate a straight-line segment",
            " Then, the second revolute pair of the chain perform a full rotation of the straight-line segment, which generates a circular cone. Indeed, depending on the orientation of the straight-line segment with respect to the revolute joint axis we may have a cylinder, a cone, or generally a hyperboloid. Finally, a full revolution of the first revolute joint will generate, by revolving the hyperboloid, a solid of revolution which can be generally depicted as a \u2019cylindroid ring\u2019, which contains a ring topology, as shown in Fig. 3.12. A \u2018cylindroid ring\u2019 shows a hollow bulk shape and its cross-section is characterized by straight-lines with possible cusps and two circular contours on the top and the bottom. Referring to Fig. 3.12, a cylindroid ring is generated by revolving a circular cone about Chapter 3: Fundamentals of the Mechanics of Robots92 axis Z1. Therefore, workspace volume W(H) of the cylindroid ring can be thought as the union of the points swept by revolving cone TR2P(H), due to the mobility in R2 and P joints, during the \u03b81 revolution about Z1 axis. This can be expressed in the form ( ) ( )\u03c0 =\u03d1 = 2 0 PR 1 2 HTHW . (3.1.33) Fundamentals of Mechanics of Robotic Manipulation 93 Alternatively, a cylindroid ring WRRP(H) can be considered as the union of the tori TR1R2(H), which are due to the mobility in R1 and R2 joints and are traced by all parallel circles which can be cut on the generating cylindrical cone TR2P(H) so that WRRP(H) can be expressed in the form ( ) ( )max min 21 d dd RRRRP HTHW = = ",
            " A numerical example of the workspace determination by using the aforementioned algebraic formulation has been reported in Fig. 3.16 in which the envelope contours of the generating workspace are shown together with the final results that are obtained by Chapter 3: Fundamentals of the Mechanics of Robots102 neglecting the internal branches not representing void boundary. A similar algebraic formulation can be deduced for the workspace of manipulators with prismatic joints. In particular, by referring to the case of a telescopic arm in Fig. 3.12, the procedure of Eq. (3.1.35) can be formulated as in the following. Figure 3.12 shows that the workspace boundary of the cylindroid ring is composed of Fundamentals of Mechanics of Robotic Manipulation 103 two different geometrical topologies: envelope segments and toroidal surfaces. The envelope segments are located in the lateral sides of the cross-section representation, and two toroidal surfaces are the top and bottom covers, respectively. This cross-section shows a shape that justifies once more the name cylindroid ring which we have given to workspace volume of telescopic arms, since it recalls the geometry both of a cylinder and a ring",
            "53) where the independent variable is the stroke parameter d. Equations (3.1.51) to (3.1.53) can be used to determine the workspace volume for a given telescopic arm by scanning the stroke interval from a minimum value dmin to a maximum value dmax to perform Eq. (3.1.34). Indeed, a torus is traced by using Eq. (3.1.51) to determine r by assuming z. This is obtained by scanning z within its range from minimum to maximum values, which can Chapter 3: Fundamentals of the Mechanics of Robots104 be calculated by using the geometry of the telescopic arm in Fig. 3.12 from ( ) 1122122 d+coscosd+h+sinsinr=z \u03b1\u03b1\u03b1\u03b8 (3.1.54) From this computation the toroidal segments can be determined as the top and bottom of the workspace boundary. The envelope segments can be determined by a formulation by using Eq. (3.1.35) to obtain a more efficient computational algorithm for workspace determination. In fact, enveloping all the tori traced by H during the scanning process generates the envelope segments. An analytical expression of the envelope can be obtained from Eq. (3.1"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000448_s11740-009-0192-y-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000448_s11740-009-0192-y-Figure6-1.png",
        "caption": "Fig. 6 Temperatures and stresses (von Mises) after the 8th layer",
        "texts": [
            " For the cooling cycle of the sample, radiation as well as convection (40 W/m2 K) is applied at the top of the solidified layer [29, 32]. Beyond that, the heat is basically transferred into the base plate due to thermal conductivity. A further solidification of single layers with 1.0 mm thickness leads to the effect, that the strains during the heating period are comparable to the first layers. Hence, the structure shows a bending in the negative z-direction, while the deformations are reversed within the cooling period. Comparable effects are observed for the 8th layer. In Fig. 6, the temperatures, deformations and equivalent stresses of the initial layer on the supports are displayed. Due to the higher thermal conductivity within the massive center of the cantilever compared to the supports, the heat flow to the base plate is considerably higher there. This effect leads to lower temperatures in the center (800 C), whereas regions positioned directly above the supports show a heat accumulation in correlation with higher temperatures of 1,013 C. Pertaining to the equivalent stresses and single stress peaks, noticeably larger values are observed after the cooling period of the 8th layer"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002794_j.rcim.2021.102165-Figure15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002794_j.rcim.2021.102165-Figure15-1.png",
        "caption": "Fig. 15. The comparison of the trajectories with no noise considered.",
        "texts": [
            " Robot kinematic calibration can reduce the deviation by compensating errors. Compensated trajectories are generated after calibration. G. Luo et al. Robotics and Computer-Integrated Manufacturing 71 (2021) 102165 The trajectory marked in green is generated after LM-DEH calibration. The medium olive and pale maroon ones are generated after L-M and DE calibration respectively. The comparison of the compensated trajectories, generated with no noise considered, with the actual trajectory and theoretical trajectory is shown in Fig. 15. After calibration with L-M and DE algorithms, the deviations have dropped to about 0.06 mm and 0.001 mm, the declining contents of deviations after LM-DEH calibration is almost the same as the decrease after DE calibration. The compensated trajectories shown in Fig. 16 are generated with measurement noise considered. In this case, the deviation dropped to 0.38 mm after L-M calibration, and both the deviations after DE and LM-DEH calibration have decreased by about 47.4% compared with that after L- M calibration"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure4.43-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure4.43-1.png",
        "caption": "Figure 4.43 The support is kinematically indeterminate as no horizontal force equilibrium is possible.",
        "texts": [
            " With this type of support, the three equilibrium conditions can be represented by \u23a1 \u23a2\u23a3 1 2 \u221a 2 0 0 1 2 \u221a 2 1 + 1 2 \u221a 2 0 0 0 \u23a4 \u23a5\u23a6 \u23a1 \u23a2\u23a3 F1 F2 F3 \u23a4 \u23a5\u23a6 = \u23a1 \u23a2\u23a3 Rx Ry R \u00b7 a \u23a4 \u23a5\u23a6 . The determinant of the coefficient matrix is now zero. In the last equilibrium equation, the moment equilibrium about D, the condition R \u00b7 a = 0 cannot be met. Neither can the support reactions for R \u00b7 a = 0 be determined. The method of support in Figure 4.42 allows a 4 Structures 135 rotation about D and is therefore kinematically indeterminate. This is in line with what we determined in the previous section for a support on three bars that pass through a single point. The support in Figure 4.43, on three parallel bars, is also kinematically indeterminate. No force equilibrium is possible in the direction normal to the bars and the block is able to move in that direction. With less than three support reactions, there are more equilibrium equations than unknowns. Here the support is also kinematically indeterminate: the conditions for all three equilibrium equations cannot be met for arbitrary loading. In Figure 4.44 in case (a) moment equilibrium is not possible and a rotation occurs about A"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002769_j.cirpj.2021.04.010-Figure14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002769_j.cirpj.2021.04.010-Figure14-1.png",
        "caption": "Fig. 14. (a) CAD model of part to be deposited, (b) tool path generated by MATLAB (c) final deposited component with129 layers [116].",
        "texts": [
            " This approach also facilitates an automated selection of process parameters for a defined bead geometry requirement (using artificial neural networks). This automated approach of path planning reduces the time required for analysing, slicing, and setting input process parameter values as per the CAD model. In another interesting study, Panchagnula et al. [116] had implemented MATLAB code for slicing the CAD model and generated tool path during fabrication 409 S. Pattanayak and S.K. Sahoo CIRP Journal of Manufacturing Science and Technology 33 (2021) 398\u2013442 of thin-walled inclined product (Fig. 14). An advanced path planning approach called Modular path planning (MPP) was developed by Michel et al. [117]. In this approach, each section\u2019s deposition pattern and deposition parameters of each zone/layers fabrication of A320 aft pylon bracket mount (Fig. 15(g\u2013h)) for Airbus. Fig. 15(g) and Fig. 15(h) represent the tool path and actual fabricated part, respectively. In another longitudinal study, Zhang et al. [118] had developed an advanced path planning approach for were different from one another"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure3.37-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure3.37-1.png",
        "caption": "Fig. 3.37: Recursive computation flows for the dynamics of a manipulator.",
        "texts": [
            "16) Fundamentals of Mechanics of Robotic Manipulation 157 The actions on the extremity link can be given from the manipulating task of the robot, including the case for a movement of unloaded manipulation. Equations (3.6.15) and (3.6.16), like the previous ones, can be computed after the kinematics has been solved to compute inertia actions. Thus, the dynamics of a manipulator can be computed in two steps, namely an outward computation for the kinematics and an inward computation for the dynamic equations, as synthetically summarized in Fig. 3.37. An iterative numerical algorithm can be outlined for serial manipulators with revolute and prismatic pairs by summarizing the kinematic analysis in 3.3 and force propagation of Eqs (3.6.15) and (3.6.16) in the following: - Outward kinematic iteration with i from 0 to (N-1), to compute: 1i 1i 1ii i i1i1i 1i R + + +++ + \u03b8+= Z 1i 1i 1ii i i1i1i 1i 1ii i i1i1i 1i xRR + + +++ + +++ + \u03b8+\u03b8+= ZZ ( ) 1i 1i 1i1i 1i 1i1i 1i i i 1iii i i i 1iii i i1i1i 1i ddx2xxxR + + ++ + ++ + ++++ + ++++= ZZvppv i i Giii i i i Giii i Gi i xxx vppv ++= (3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000556_bfb0036073-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000556_bfb0036073-Figure4-1.png",
        "caption": "Fig. 4?. Point stabilization with nonsmooth time-varying feedback (II): v2 (rad/sec) vs. time (sec)",
        "texts": [
            " The structure of change of coordinates (8) is interesting because it can be generalized to nonholonomic systems of higher dimension, such as the N-trailer robot [46]. In particular, the xl and xn coordinates can be always chosen as the x and y coordinates of the midpoint of the last trailer wheel axle. It is interesting to note that the (2, n) chained form can be also obtained starting from a different point of view. In particular, assume that the car-like robot must follow a given path which is parameterized by its arc length. With reference to Fig. 4, let d be the distance between the rear axle midpoint and the path, and s be the corresponding value of the path parameter. Denote by Ot the angle between the current tangent to the path and the x axis, and let Op = 0 - Or. The curvature along the path is defined as dot = Z 2 ' which implies Ot =c(s)a. (11) In the following, we assume that c(-) E C 1 and that the path satisfies some technical conditions (see [44] for details). It is easy to verify that --~ L~I COS Op q- 0 t d (12) d = Vl s i n 0 p "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001189_j.ijmachtools.2005.11.005-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001189_j.ijmachtools.2005.11.005-Figure6-1.png",
        "caption": "Fig. 6. Schematic illustration of in-process temperature measurement with a pyrometer.",
        "texts": [
            " Calibration of the pyrometer was done prior to its first use using a blackbody calibration source and has also been calibrated regularly after moderate usage. For heating-only tests, the pyrometer position is fixed along the circumferential plane close to the location of the cutting tool after the cutting tool was removed for these tests. It continuously measures the surface temperature from one spot while translating along with the laser. With a special fixture designed for holding the pyrometer probe, it could be moved 40\u2013601 circumferentially from the laser. Fig. 6 shows the pyrometer fixture design for heating-only tests. The heating tests have been conducted using this setup with the intention of measuring temperature that would be generated in the vicinity of cutting tool. A high performance ThermaCAM SC3000 infrared camera was later used to measure the surface temperature. Unlike the pyrometer, for measurements with the infrared camera, the workpiece is rotated in anticlockwise direction and the camera remains stationery while the laser Instrument Symbol Wavelength range Center wavelength Spectral emissivity Radiation pyrometer epyro 2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001514_j.engfailanal.2015.06.004-Figure31-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001514_j.engfailanal.2015.06.004-Figure31-1.png",
        "caption": "Fig. 31. Sliced teeth model [79].",
        "texts": [
            " There are some difficulties in dealing with 3D nature of helical gear tooth crack due to uneven distribution of loads and tooth complicated geometry, where numerous computational resources are needed. The efforts to the further explore of new methods for calculating TVMS of helical gear pair with tooth crack are expected. Some ideas that can be helpful for the calculation of TVMS of helical gear pair with tooth crack include: (1) approximating the helical gears by many sliced spur models in parallel [79] (see Fig. 31), the TVMS for each sliced spur with tooth crack can then be calculated using the method presented in Ref. [36]; (2) combined analytical-FE approach proposed in Refs. [61,62] can be further improved to fit for the helical gear pair with tooth crack. (4) Many researchers discussed the effects of the model of spur gear systems on the system vibration responses [68,71]. It is important to select an appropriate dynamic model for numerical simulation because a model with a large number of DOFs, such as FE model, will consume huge computer resources and a small DOF model may lead to a large discrepancy between the model and the real gear system"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure3.32-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure3.32-1.png",
        "caption": "Fig. 3.32: Kinematic model of a planar 2R manipulator.",
        "texts": [
            "3.16) An acceleration analysis of a manipulator can be carried out by using Eqs (3.3.12) and (3.3.15) from the base to the extremity, that is from 0 to N. Multiplying the dot vectors N\u03c9N and N vN by 0RN gives the acceleration of the extremity link in terms of the coordinates of the fixed base frame 0. The application of the outlined formulation for kinematic analysis of a manipulator is illustrated by referring to a planar 2R manipulator, since it is a basic architecture in robot structures. In Fig. 3.32 a planar 2R manipulator is modeled with its parameters. Reference frames are attached at the links and H\u2013D parameters are indicated. The transformation matrices can be computed as 1000 0100 00cs 00sc T 11 11 10 \u2212 = 1000 0100 00cs a0sc T 22 122 21 \u2212 = (3.3.17) Chapter 3: Fundamentals of the Mechanics of Robots142 1000 0100 0010 a001 T 2 32 = The last matrix can be thought as referring to a third link of a 3R manipulator that is fixed permanently at zero degrees. Thus, using the expressions for velocity propagation one can compute ( )t 11 00 \u03b8= ( )t 11 000=v ( )t 212 00 \u03b8+\u03b8= ( )t 1211211122 22 22 0casa 0 a 0 100 0cs 0sc \u03b8\u03b8=\u03b8\u2212=v (3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure6.22-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure6.22-1.png",
        "caption": "Figure 6.22 (a) A plate in the xy plane is loaded normal to its plane by an arbitrarily distributed load p(x, y). A small force R = p(x, y) A is acting on a small area A. (b) The resultant R of the distributed load and the location (xR, yR) where the line of action of R intersects the xy plane. The magnitude of R is equal to the volume of the load diagram. The line of action of R passes through the centroid of the load diagram.",
        "texts": [
            "21b; this resultant has to be replaced by the resultants of the distributed loads on the individual parts (see Figure 6.21c). With equations (a) and (b), an efficient way of obtaining results is to consider the moment equilibrium of SD: \u2211 T (SD) y |S = +Cv \u00d7 (2 m) \u2212 (12 kN)(1 m) = 0. (c) From (c) we find Cv = +6 kN which then gives the following from (b) and (a) Bv = +33 kN, Av = \u22123 kN. Figure 6.21d shows the support reactions in the directions in which they are really acting. Only the direction of the vertical support reaction at A was assumed falsely. In Figure 6.22a, a plate in the xy plane is loaded normal to its plane by an arbitrarily distributed load p(x, y). The resultant of the distributed load on a small area A is a small force R: R = p(x, y) A. 228 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM The magnitude of the resultant R of the distributed load is equal to the sum of all parallel forces R: R = \u2211 R = \u2211 p(x, y) A = \u222b A p(x, y) dA. Conclusion: The magnitude of R is equal to the volume of the load diagram. The location (xR, yR) of the line of action of R is found using Varignon\u2019s theorem1 (see Figure 6.22b): \u2211 Ty = \u2212xRR = \u2212\u2211 (x R) = \u2212 \u222b A xp(x, y) dA, \u2211 Tx = +yRR = +\u2211 (y R) = + \u222b A yp(x, y) dA so that xR = \u222b A xp(x, y) dA R = \u222b A xp(x, y) dA\u222b A p(x, y) dA , yR = \u222b A yp(x, y) dA R = \u222b A yp(x, y) dA\u222b A p(x, y) dA . By definition, xR and yR are the x and y coordinates of the centroid of the load diagram. 1 See also Examples 1 and 2 in Section 3.3.4. 6 Loads 229 Conclusion: The line of action of R passes through the centroid of the load diagram. Example In Figure 6.23a, a rectangular plate in the xy plane, with an area A = ab, is loaded by a distributed load normal to its plane: p(x, y) = p\u0302 x(b \u2212 y) ab "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure2.25-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure2.25-1.png",
        "caption": "Fig. 2.25 Calculating Inverse Kinematics of the Legs",
        "texts": [
            " Inverse Kinematics is necessary for such a case. There exist both an analytical method and a numerical method of solving Inverse Kinematics. First we will explain how to solve it analytically. Let\u2019s focus on the right leg of the model shown in Fig. 2.19. The position and attitude of the body and right leg will be (p1,R1) and (p7,R7) respectively. To simplify the equation we will define D, which is the distance between the Body origin and the hip joint. The upper leg length is A, and the lower leg length is B, as shown in Fig. 2.25(a). So therefore, the position of the hip would be 2.5 Kinematics of a Humanoid Robot 51 p2 = p1 +R1 \u23a1 \u23a3 0 D 0 \u23a4 \u23a6 . Next, we calculate the position of the crotch viewed from the ankle coordinate space r = RT 7 (p2 \u2212 p7) \u2261 [rx ry rz ] T . (2.61) From this we can calculate the distance between the ankle and the hip, which we will define as C = \u221a r2x + r2y + r2z . As shown in Fig. 2.25(b), if we consider the triangle ABC we get the angle of the knee q5. From the cosine rule we get, C2 = A2 +B2 \u2212 2AB cos(\u03c0 \u2212 q5). So the angle of the knees will be, q5 = \u2212 cos\u22121 ( A2 +B2 \u2212 C2 2AB ) + \u03c0. If we define the angle at the lower end of the triangle as \u03b1, from the sine rule we get, C sin(\u03c0 \u2212 q5) = A sin\u03b1 . 52 2 Kinematics So therefore, \u03b1 = sin\u22121 ( A sin(\u03c0 \u2212 q5) C ) . Next we will focus on the ankle local coordinates. As shown in Fig. 2.25(c), from vector r you can calculate the ankle roll and pitch angles. So, q7 = atan2(ry , rz) q6 = \u2212atan2 ( rx, sign(rz) \u221a r2y + r2z ) \u2212 \u03b1. The function atan2(y, x) calculates the angle between vector (x, y) and the x axis as it has already appeared in Section 2.2.7. It is built into Matlab and is also available as a built-in function in most programming languages. Also sign(x) is a function that returns +1 if x is a positive value and \u22121 if it is negative. What remains is the yaw, roll and pitch angles at the base of the leg"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure11.12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure11.12-1.png",
        "caption": "Figure 11.12 (a) The water pressure on a water-retaining slide (b) modelled as a line load on a line element.",
        "texts": [
            " For x \u2192 0 the terms with x disappear and we find the boundary condition at A: M = 0. Boundary conditions at the free member end B in case (b) In Figure 11.11b, the small \u201clast\u201d member segment at the free end B is shown, with all the forces acting on it. The element has a length x. The equations for the equilibrium are \u2211 Fz = \u2212V + q x = 0,\u2211 Ty |B = \u2212M \u2212 V x + q x \u00b7 1 2 x = 0. For x \u2192 0 the terms with x disappear in both equations and we find the boundary conditions at the free member end B: V = 0, M = 0. 448 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Example 3 Figure 11.12a shows a water-retaining slide of width b and height h, which is supported by a hinge at the top and supported against a sill below. The mass density of water is \u03c1. Question: Model the slide as a line element with a line load, and use the differential equations for equilibrium to find the variation of the shear force and the bending moment. Solution: The water pressure on the slide at a depth x is p = \u03c1gx, in which g is the gravitational field strength. The water pressure increases linearly with the depth. In Figure 11.12b the slide with width b is modelled as a line element (beam). The support at the base is considered a roller support. At a depth x the load on the slide is qz = pb = \u03c1gbx. It holds d2M dx2 = \u2212qz = \u2212\u03c1gbx, V = dM dx = \u2212 \u222b qz dx = \u2212 \u222b \u03c1gbx dx = \u2212 1 2\u03c1gbx2 + C1, M = \u222b V dx = \u222b ( \u2212 1 2\u03c1gbx2 + C1 ) dx = \u2212 1 6\u03c1gbx3 + C1x + C2. 11 Mathematical Description of the Relationship between Section Forces and Loading 449 Due to a linear distributed load, the shear force is a quadratic (parabolic) function in x, and the bending moment is a third degree (cubic) function in x"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003908_978-1-84882-730-1-Figure3.5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003908_978-1-84882-730-1-Figure3.5-1.png",
        "caption": "Figure 3.5 Motion variables for an ocean vessel moving in a horizontal plane",
        "texts": [
            "4 Standard Models for Ocean Vessels In this section, the main results of the previous section are simplified to give a set of standard models for surface ships and underwater vehicles. These standard models will be extensively used for the control design in the coming chapters. 3.4.1 Three Degrees of Freedom Horizontal Model 3.4.1.1 Standard Three Degrees of Freedom Horizontal Model The horizontal motion of a surface ship or an underwater vehicle moving in a horizontal plane is often described by the motion components in surge, sway, and yaw. Figure 3.5 illustrates the motion variables in this case. Therefore, we choose D \u0152x y T and v D \u0152u v r T . This model is obtained from the general model (3.31) under the following assumption. Assumption 3.1. 1. The motion in roll, pitch, and heave is ignored. This means that we ignore the dynamics associated with the motion in heave, roll, and pitch, i.e., z D 0, w D 0, D 0, p D 0, D 0, and q D 0. 2. The vessel has homogeneous mass distribution and xz-plane of symmetry so that Ixy D Iyz D 0: (3.40) 3. The center of gravity CG and the center of buoyancy, CB , are located vertically on the z-axis"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000321_j.actamat.2016.05.017-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000321_j.actamat.2016.05.017-Figure8-1.png",
        "caption": "Fig. 8. Schematic depicting the action of evaporated metal flux on the flow pattern of the surrounding Ar gas and displacement of particles in the powder bed. At high pressures the Knudsen number (Kn) is small as shown on the left, and the low pressure zone near the melt pool caused by high speed metal vapor flux induces Ar gas flow through the Bernoulli effect which in turn results in powder particle entrainment. As shown in the left diagram for Kn < 1, particles are either drawn into the melt pool, adding to melt pool material consolidation, or are ejected upward (and rearward) as shown in the high speed video. Conversely, at low pressures where the surrounding Ar gas transport is defined by a molecular flow, Kn is large and the metal vapor flux can expand laterally which will tend to push powder particles outward with minimal material consolidation into the melt pool. Particle collisions from glancing irradiation conditions as discussed in the text can also act to expel powder outward.",
        "texts": [
            " 7a) that the vapors from a fully developed melt pool are ejected partially back but also sideward due to the curvedmelt pool surface. Vapor flow from the curved melt pool surface can thus move particles outside the track in multiple directions. In addition, metal vapor flux emitted from the melt pool at low pressures may sufficiently expand into a wide plume as compared to a more narrow jet at higher pressures. At high ambient pressure the experimental results demonstrate that the inward particle motion results from a metal vapor jet and Bernoulli effect-driven gas flow, as depicted in Fig. 8; at low pressure this mechanism is arrested and denudation could be explained by outward gas expansion and the ejection of particles from the track region leading to particle pile up on the edge of denudation zone. Despite the unavailability of highspeed video data at low pressures, the pressure-dependent experiments help to clarify the relative roles of the competing effects that are at play during the LPBF process. Our experiments demonstrate the important effect of the surrounding gas on the powder particle movement for pressures above ~5 Torr",
            " The velocity of the incoming Ar gas flow over the adjacent powder is 10\u2019s of times smaller than the metal vapor jet velocity [21], about 10 m/s in our case, but can be high enough to entrain the particles and explain the observations for pressures greater than ~10 Torr. The entrainment of particles in the convective flow which was observed in our experiments can be expected from the arguments given above and will lead to particles far (several particle diameters) from the laser spot moving towards and becoming consolidated with the melt pool. As depicted in Fig. 8 for a simplified flow pattern, those particles which can attain significant vertical momentum from particle-particle collisions or are entrained by the vapor stream will tend to follow the vapor flow and are expelled, while those scattered downward or with negligible vertical momentum will collide and are available to merge with the melt pool. Thus, in terms of source material for the final weld bead produced by LPBF, the convective flow due to evaporation and subsequent particle entrainment are shown to be important factors"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001653_j.matdes.2017.11.042-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001653_j.matdes.2017.11.042-Figure7-1.png",
        "caption": "Fig. 7. Schematic illustration of SLM process integrated with a process map that relates laser power, scan speed, and energy density.",
        "texts": [
            " Proper fusionwas attained at the optimum energy density (71.43 J/mm3) accounting for the reduced balling and well spread scan tracks (Fig. 5(b)). Accordingly, relatively smooth surface with a roughness value of Ra = 35 \u03bcm was obtained under this optimum condition while poor surface qualities were attained at low and high values of energy density (Fig. 6). According to the above results, a process map with schematic illustration of the SLM process for maraging steel powders can be obtained as shown in Fig. 7. The intra-layer (dark ovals at the intersections between two adjacent scan tracks) and inter-layer (bright ovals at the intersections between neighboring layers) overlapping phenomena that cause powder fusion is also highlighted. Clearly, there exist several processing regions which correspond to different relative density values, microstructure, and hence mechanical properties of the SLMed products. Significantly low energy input is experienced in Region I due to low laser power and scan speed"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure3.50-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure3.50-1.png",
        "caption": "Fig. 3.50: The Eclipse robot developed at Seoul National University: a) a kinematic scheme; b) a first prototype, (Courtesy of Seoul National University).",
        "texts": [
            " Even the number of attempts may affect the numerical evaluation, since the distribution of the measure results can be more or less accurate, but other performances, such as heating and friction can modify the robot positioning over the time of the repeated movements. Thus, one can use short-term or long-term repeatability errors when the operation of a robot is evaluated for periods of time, e.g., hours or days, respectively. An interpretation of the relationship between accuracy and repeatability errors is illustrated in Fig. 3.50 referring to a general view from the Theory of Measures. The accuracy error can be interpreted as a measure of the difference between a desired position and the mean position of the statistical distribution of the measures. The repeatability error can be related to the dispersion size of the repeated measures and it can be represented with a sphere of radius erS inside which the measures are contained. Fundamentals of Mechanics of Robotic Manipulation 179 The smaller the sphere, the better repeatability the manipulator shows",
            " Its maximum velocity is 90 m/min and its acceleration is 2 g. The F-200iB robot in Fig. 3.49 is a 6-d.o.f.s servo-driven parallel manipulator commercialized by FANUC. It has been designed for manufacturing and automotive assembly. Fanuc F-200iB has a repeatability of \u00b1 0.1 mm. Currently it is used for several applications, primarily spot welding. Maximum declared velocities are 1,500 mm/sec on X and Y directions and 300 mm/sec on Z direction. Fundamentals of Mechanics of Robotic Manipulation 187 The Eclipse parallel manipulator in Fig. 3.50 has been conceived at Seoul National University. It has been designed to overcome the problem of the limited tilting angle of the Gough-\u2013Stewart Platform. It has a tilting angle of 90 degrees. The prototype has 6 d.o.f.s with the maximum speed of 1.2 m/min and the maximum acceleration of 0.1 g. It is used for 5-axis machining. A prototype of year 2000, shown in Figure 3.50, has been developed to test the Eclipse for rapid machining. The Eclipse parallel manipulator is an illustrative example of the variety of kinematic chain that can be conceived with the parallel architecture, as shown in the kinematic diagram of Fig. 3.50 a). The Eclipse consists of three vertical columns, each of which slides independently on the circular guides. The movement of each column along the circular guide is achieved by a servomotor and a pinion and ring gear transmission. Each column has a carriage, which moves vertically along the linear guide of the column. A fixed length rod is attached to each of the carriages through a pin joint. Each of the 2 R joints is actuated by a servomotor\u2013harmonic drive mechanism. The other end of the fixed length rod is attached to the tool spindle plate via a ball-socket joint"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure3.20-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure3.20-1.png",
        "caption": "Figure 3.20 A body loaded by couples Tz;i and forces Fi , with components Fx;i and Fy;i , at points i (i = 1, 2, 3, . . . ).",
        "texts": [
            " In fact, this means that all the forces are shifted to point A with addition of a couple (see Section 3.1.5). If we place the origin O of the coordinate system at A, and xi and yi are the 1 Pujol Varignon (1654\u20131722) was a French mathematician. ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM 3 Statics of a Rigid Body 65 Figure 3.21 (a) The resultant force R at O and the associated couple \u2211 Tz|O are statically equivalent to (b) a force R at a distance a = ( \u2211 Tz|O)/R from O. coordinates of the point of application of force Fi (or of another point on the line of action of Fi ), then (see Figure 3.20) Rx = \u2211 Fx;i = \u2211 Fi cos \u03b1i, Ry = \u2211 Fy;i = \u2211 Fi sin \u03b1i,\u2211 Tz|O = \u2211 {(Fy;ixi \u2212 Fx;iyi) + Tz;i}. The sum of the moments also includes the moments of the (concentrated) couples Tz;i that may be applied on the body. For the (instantaneous) movement or the equilibrium of a rigid body, one may replace the force system by a single resultant force R at O together with a couple \u2211 Tz|O (see Figure 3.21a). The resultant force R at O can be compounded with the couple \u2211 Tz|O into a single force R by shifting it parallel to itself to a line of action at a perpendicular distance a from O (see Figure 3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003403_978-1-4615-5633-6-Figure2.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003403_978-1-4615-5633-6-Figure2.3-1.png",
        "caption": "Figure 2.3. Synchronous machine with rotating armature windings",
        "texts": [
            " Out of these, '0' coil (in which zero-sequence current io flows) has no coupling with the rotor coils and may be neglected if io = O. Since the (transformed) mutual inductance terms between d,q coils and the rotor coils are constants, it can be interpreted that d and q coils rotate at the same speed as the rotor. Furthermore, as mutual inductances between the d-coil and the rotor coils on the q-axis are zero, it can be assumed that d-coil is aligned with the d-axis. Similarly, it MODELLING OF TURBINE GENERATOR 25 can be assumed that q coil is aligned along with the q-axis. This is shown in Fig.2.3 211\" 211\" cosO + cos(O - 3) + cos(O + 3) sinO + sin(O _ 2;) + sin(O + 2;) 2 2 211\" 211\" cos 0 + cos (0 - 3) + cos2 (0 + 3 ) sin20 + sin2(O _ 2311\") + sin2(O + 2;) o o 3 - 2 3 2 Although, the physical interpretation of Park's transformation is useful in gain ing an intuitive understanding of its implications, it must be understood that it is not essential in the mathematical analysis of the synchronous machine. This is true of any mathematical transformation whose main objective is to simplify the analysis"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure2.7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure2.7-1.png",
        "caption": "Fig. 2.7. World coordinates",
        "texts": [
            " Joint coordinate space Q eRn, is the set of all joint coordinate vectors when they vary between their physical constraints, Q={q:qimin::;;qi::;;qimax}' 2.2.6 External Coordinates External (operational, world) coordinates describe manipulator end-effector position and orientation with respect to some reference coordinate system. The reference system is chosen to suit a particular application. Most frequently, a fixed coordinate frame attached to manipulator base is considered as the reference system. The manipulator hand position is usually described by Cartesian coordinates x, y and z (Fig. 2.7), although sometimes cylindrical or spherical coordinates are used (if it is convenient to describe the manipulation task in these coordinates). Precise execution of manipulation tasks requires the control not only of the manipulator hand position, but also of hand orientation with the objects located in the manipulator work space. The orientation is usually specified by Euler angles between the coordinate frame attached to the last (n-th) link and the reference system (Fig. 2.7). We will consider the yaw angle 1/1, the pitch angle () 2.2 Definitions 23 and the roll angle cpo The yaw angle corresponds to a rotation t/I about the z axis, pitch corresponds to a rotation () about the new y axis, and roll corresponds to: a rotation cp about the new x axis. In the general case, the vector of external coordinates x. is an m-dimensional vector, where m is the number of external coordinates required to specify a certain class of manipulation tasks, m~n. However, most frequently m equals 6, i",
            " As we have already mentioned, the upper left 3 x 3 submatrix describes the rotation of the third coordinate frame 03X3Y3Z3 with respect to the reference frame Oxyz, while the fourth column contains the Cartesian coordinates x, y, z of the manipulator tip in the fixed frame I I X I I Y I I Z I -------1--- I o 0 0 I 1 I (2.21) Once the numerical values of the \u00b0A3 matrix are evaluated, the other three external coordinates describing the orientation (Euler angles) may also be determined. Then, the direct kinematic problem is completely solved. Manipulator hand orientation with respect to the reference frame is usually specified by three Euler angles (Fig. 2.7). One method for evaluation of Euler angles is the following. Let us consider two coordinate frames and describe rotation between them. The first system Qn is assigned to the last manipulator link, and its unit axes are denoted by qnl' qnz, qn3 (Fig. 2.12). The transformation matrix \u00b0An which maps vectors from system Qn to system \u00a2xyz is (2.22) 2.3 Direct Kinematic Problem 33 The rotation of system On Xn Yn Zn with respect to Oxyz can be described by the following sequence of rotations: a rotation by t/"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure8.24-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure8.24-1.png",
        "caption": "Figure 8.24 (a) The distribution of the vertical earth pressure and (b) the distribution of the passive and active (horizontal) earth pressure on the anchor plate.",
        "texts": [
            " Determine the influence of a vertical terrain load of 20 kN/m2 on the magnitude of the maximum anchor force. Solution: a. Due to the anchor force, the plate will tend to move to the left. An area of passive earth pressure develops in front of the plate, and an area of active earth pressure develops behind the plate. The coefficients Kp and 8 Earth Pressures 305 Ka, respectively for the active and passive earth pressure, are Kp = 1 + sin 24\u25e6 1 \u2212 sin 24\u25e6 = 2.37, Ka = 1 \u2212 sin 24\u25e6 1 + sin 24\u25e6 = 0.42. Figure 8.24a shows the distribution of the vertical earth pressure. Derived from the vertical earth pressure, Figure 8.24b shows the distribution of the (horizontal) passive and active earth pressure on the plate. Taking into account the width of 2 metres of the plate, the resultants Rp and Ra of respectively the passive and active earth pressure are Rp = (2 m) \u00d7 0.5 \u00d7 (85.32 kN/m2)(2 m) = 170.64 kN, Ra = (2 m) \u00d7 0.5 \u00d7 (15.12 kN/m2)(2 m) = 30.24 kN. The maximum anchor force F that the plate can provide is F = Rp \u2212 Ra = (170.64 kN) \u2212 (30.24 kN) = 140.4 kN. b. Any terrain load p increases the vertical grain pressure in the soil and therefore also increases the active earth pressure (resultant Ra) and the passive earth pressure (resultant Rp)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001467_j.mechmachtheory.2010.05.001-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001467_j.mechmachtheory.2010.05.001-Figure1-1.png",
        "caption": "Fig. 1. The twist screw and wrench screw of a rigid body.",
        "texts": [
            " In Section 5, the relationship between the new transmission index and the singularity of parallel manipulators is briefly discussed. Conclusions are given in the last section. Screw theory is undoubtedly an efficient mathematical tool for the study of spatial mechanisms. It has been used in the analysis of both kinematics and dynamics [21], and in the type synthesis [22] of parallel manipulators. In this contribution, screw theory will be employed as the mathematical resource for the analysis of motion/force transmission of parallel manipulators. As shown in Fig. 1, the instantaneous motion of a rigid body can be represented by a twist screw as $1 = w$1 = w \u03c9;v\u00f0 \u00de = w \u03c9; r1 \u00d7 + h1\u03c9\u00f0 \u00de = w L1;M1;N1; P1\u204e;Q1; \u204e R1 \u204e \u00f01\u00de w is the amplitude of the twist screw $1, $1 is the unit twist screw or the normalized twist screw, and \u03c9 represents the unit where rotational velocity of the rigid body. A wrench exerted on the rigid body can be expressed by a wrench screw as $2 = f $2 = f f ; \u03c4\u00f0 \u00de = f f ;r2 \u00d7 f + h2f\u00f0 \u00de = f L2;M2;N2; P2\u204e;Q2 \u204e;R2 \u204e \u00f02\u00de or where f is the amplitude of the wrench screw $2, $2 is the unit wrench screw or the normalized wrench screw, and f represents the where unit pure force vector"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure6.26-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure6.26-1.png",
        "caption": "FIGURE 6.26. The instant centers of rotation for a four-bar linkage, a slider-crank, and an inverted slider-crank mechanism.",
        "texts": [
            " So, there are three instant centers between three bodies. The number N of instant centers between n relatively moving bodies is N = n (n\u2212 1) 2 . (6.217) Thus, a four-bar linkage has six instant centers, I12, I13, I14, I23, I24, I34. The symbol Iij indicates the instant center of rotation between kinks i and j. Because two links have only one instant center, we have Iij = Iji. (6.218) The four instant centers of rotation for a four-bar linkage, a slider-crank, and an inverted slider-crank mechanisms are shown in Figure 6.26. The instant center of rotation for two links that slide on each other is at infinity, 350 6. Applied Mechanisms on a line normal to the common tangent. So, I14 in Figure 6.26(b) is on a line perpendicular to the ground, and I34 in Figure 6.26(c) is on perpendicular to the link 3. Figure 6.27 depicts the 15 instant centers for a six-link mechanism. Example 244 Application of instant center of rotation in vehicles. Figure 6.28 illustrates a double A-arm suspension and its equivalent kinematic model. The wheel will be fastened to the coupler link AB, witch connects the upper A-arm BN to the lower A-arm AN . The A-arms are connected to the body with two revolute joints at N and M . The body of the vehicle acts as the ground link for the suspension mechanism, which is a four-bar linkage"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure3.67-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure3.67-1.png",
        "caption": "Fig. 3.67: Closed-loop vector equation for the k-th leg.",
        "texts": [
            " Chapter 3: Fundamentals of the Mechanics of Robots216 They can be used to compute the velocity according to Eq. (3.9.40). A second approach for deriving the velocity equations consists of writing closed-loop velocity equations as function of linear and angular velocities of the links of the mechanism. This analysis leads to an invariant form of the Jacobian matrices. The velocity vector of an articulation point H k is formulated from two different loop-closure circuits. Each one consists of the FP, MP, and links of the leg, as shown in Fig. 3.67. The rates of unactuated joints in each leg are passive variables and performing a dot product of the velocity vector-loop equation with an appropriate vector, which is orthogonal to all vectors of unactuated joint rates, and can eliminate them. Thus, the resulting equations can be assembled in the Jacobian matrices. It is assumed that, unless otherwise indicated, all vectors and matrices are represented in the world frame. By considering Fig. 3.67, for each k-th leg a closed-loop equation can be written as kHHk 'R rpp += (3.9.51) r\u2019 k is the vector connecting H to H k and it is defined in the moving frame H-X P YP Z P and R is the rotation matrix which describes the orientation between the moving and fixed frame. By considering Fig. 3.67, Eq. (3.9.56) can also be written as kkkkHk hslup +++= (3.9.52) Differentiating Eqs (3.9.51) and (3.9.52) with respect to time yields kkkkH slrv +\u00d7=\u03a9+ (3.9.53) Fundamentals of Mechanics of Robotic Manipulation 217 where tRR=\u03a9 ; \u03c9 k is the angular velocity of the k-th leg with respect to the AP plane reference frame; s k is the linear velocity of the prismatic joint. In order to eliminate the passive joint rates in Eq. (3.9.52), we dot-multiply both sides by hk, which is a constant vector for all the three legs, to obtain kkkkkHk lhrhvh \u00d7\u22c5=\u03a9\u22c5+\u22c5 (3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure8.13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure8.13-1.png",
        "caption": "FIGURE 8.13 Example of \ud835\udefc-projection for layers under narrow conductor.",
        "texts": [
            " The cells on both sides need to be closely projected on each other. It becomes even more complicated for multiple close layers where the cell size should be different on both sides of layers. For this case, it is desirable to attempt a slow increasing or decreasing in the projection between the cell boundaries. We call the technique used for the cell size change \ud835\udefc-projection algorithm. The motivation is given by a permissible angle \ud835\udefc by which the cell size projection changes. We illustrate the \ud835\udefc projection issue in Fig. 8.13, where it is required that the cell width at the bottom surface should be larger than the small width W of the top conductor. This is accomplished by increasing the cell width for each layer in Fig. 8.13 to WL. For the layer that is at distance d from the narrow top conductor, the cell width is increased to WL = W + d tan(\ud835\udefc). (8.7) example in Fig. 8.13, rather than being the same for all layers. 200 GEOMETRICAL DESCRIPTION AND MESHING Another issue related to the example in Fig. 8.13 is that conductors also need to be subdivided along the cell width W. This represents the fact that the current also distributes toward the edge of the conductor. To accomplish this, further width subdivisions may be needed for the conductor with the overall width W. This is part of the VFI skin-effect model in Section 9.3. Hence, it would be very costly to also project the small width of the cells to the layers. Provided that we want to use nonorthogonal cells, we can use the node matching technique in Fig",
            " However, for these long-time solution the waveform details are less important For this case, it is desirable to decrease the meshing accuracy by switching to a coarser mesh. For a more coarse mesh, we found that it is easier to use the node positioning relaxation algorithm to let the contacts, etc., snap to locations in the new mesh. Usually, the simplified mesh leads to considerably reduced compute times. For the third application, we return to the \ud835\udefc-projection technique given in the previous section in Fig. 8.13. However, if the layers are approximated by rectangular type cells, the application of node relaxation can be used if the angle \ud835\udefc is small. In Fig. 8.16, we give an example where the tolerance spheres are overlapping such that the nodes are connected together. 202 GEOMETRICAL DESCRIPTION AND MESHING The mesh generation in this section consists of several examples. Hopefully, it provides the fundamental concepts for the generation of geometrical objects. We start out with a simple example. Today\u2019s PCBs have many layers"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.20-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.20-1.png",
        "caption": "FIGURE 7.20. A vehicle with a one-axle trailer.",
        "texts": [
            " Another option for multi-axle vehicles is to use self-steering wheels that can adjust themselves to minimize sideslip. Such wheels cannot provide lateral force, and hence, cannot help in maneuvers very much. Self-steering wheels may be installed on buggies and trailers. Such a self-steering axle mechanism for locomotive wagons is shown in Figure 7.19. 7.3 F Vehicle with Trailer If a four-wheel vehicle has a trailer with one axle, it is possible to derive a kinematic condition for slip-free steering. Figure 7.20 illustrates a vehicle with a one-axle trailer. The mass center of the vehicle is turning on a circle with radius R, while the trailer is turning on a circle with radius Rt. Rt = s\u00b5 l cot \u03b4i + 1 2 w \u00b62 + b21 \u2212 b22 (7.53) Rt = s\u00b5 l cot \u03b4o \u2212 1 2 w \u00b62 + b21 \u2212 b22 (7.54) At a steady-state condition, the angle between the trailer and the vehicle 7. Steering Dynamics 399 is \u03b8 = \u23a7\u23aa\u23aa\u23a8\u23aa\u23aa\u23a9 2 tan\u22121 \u2219 1 b1 \u2212 b2 \u00b3 Rt \u2212 p R2t \u2212 b21 + b22 \u00b4\u00b8 b1 \u2212 b2 6= 0 2 tan\u22121 1 2Rt (b1 + b2) b1 \u2212 b2 = 0 (7.55) Proof. Using the right triangle 4OAB in Figure 7.20, we may write the trailer\u2019s radius of rotation as Rt = q R21 + b21 \u2212 b22 (7.56) because the length OB is OB 2 = R2t + b22 = R21 + b21. (7.57) Substituting R1 from Equation (7.6) shows that the trailer\u2019s radius of ro- 400 7. Steering Dynamics tation is related to the vehicle\u2019s geometry by Rt = s\u00b5 l cot \u03b4i + 1 2 w \u00b62 + b21 \u2212 b22 (7.58) Rt = s\u00b5 l cot \u03b4o \u2212 1 2 w \u00b62 + b21 \u2212 b22 (7.59) Rt = q R2 \u2212 a22 + b21 \u2212 b22. (7.60) Using the equation Rt sin \u03b8 = b1 + b2 cos \u03b8 (7.61) and employing trigonometry, we may calculate the angle \u03b8 between the trailer and the vehicle as (7"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003824_1.575577-FigureI-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003824_1.575577-FigureI-1.png",
        "caption": "FIG. I. Surface displacements due to a point load Cal. and Cb) parameters used in tip-flat geometry calculations.",
        "texts": [
            " Therefore, since the local rigidity is in series with that of the macroscopic apparatus, there will always be a finite total rigidity irrespec tive of apparatus construction. For very fine tips, the local rigidity will be small enough that it controls the separation at which the surfaces jump together, and provides a limit to the proximity at which two bodies can be held apart. II. CONTINUUM ANALYSIS The response of an axisymmetric solid to a defined surface stress distribution may be found using Boussinesq's relation for the deformation due to a point load9 ; Dh = F /1rE*r with parameters as in Fig. I, and E * defined as l/E* = (1- '?)/E with E the Young's modulus and v Poisson's ratio. The total displacement at a given place due to a distributed load may then be obtained by integrating the contributions from all the loaded regions of the surface to the displacement at that place. The problem is particularly simple if, as here, we wish to know the surface displacement at the axis of symmetry. Then h = ('\" P(r)21Tr dr Jo 1TE *r and the total force acting is given by (1) 2490 J. Vac. Sci. Techno\\"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003785_j.jsv.2009.09.022-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003785_j.jsv.2009.09.022-Figure2-1.png",
        "caption": "Fig. 2. Dynamic models of (a) a sun-planet pi pair, (b) a ring-planet pi pair, and (c) a carrier-planet pi pair.",
        "texts": [
            " 1, only one of the N planet branches located at an arbitrary planet position angle ci is shown. In this figure, yjc (j=s,r,pi) are the nominal kinematic rotations and ej represents the initial position angle of the run-out (eccentricity) vector of each gear j. Here, es and er are both defined from the horizontal x-axis while epi is defined from the radial axis along the centers of the sun gear (O) and planet pi located at an angle ci. A gear pair (external or internal) sub-system and a pinion\u2013carrier pair sub-system, as shown in Fig. 2, will be formulated first before assembling the dynamic model of the entire gear set. Fig. 2(a) and (b) show dynamic models of external (sunplanet pi) and internal (ring-planet pi) gear pairs, respectively, with planet pi positioned at angle ci. Gears are modeled as rigid disks with masses ms, mr and mp, and mass moments of inertia Js, Jr and Jp. The gear mesh interfaces are represented by (i) periodically time-varying gear mesh stiffnesses kspi(t) and krpi(t) to account for the fluctuation of the stiffness with gear rotations [27], (ii) constant gear mesh damping elements csp and crp that are assumed to be the same for each s\u2013pi and r\u2013pi mesh, and (iii) periodic gear transmission error displacement excitations espi(t) and erpi(t) [27]",
            " These errors will be defined explicitly in the next section. Eqs. (2a)\u2013(2f) can be written in matrix form to obtain sub-matrices associated with the gear j-planet pi pair as Mjpi \u00bc mj 0 0 mpi \" # ; (3a) Cjpi \u00bc c\u00f011\u00de jpi c\u00f012\u00de jpi c\u00f012\u00de jpi c\u00f022\u00de jpi 2 4 3 5; (3b) Kjpi \u00bc k\u00f011\u00de jpi k\u00f012\u00de jpi k\u00f012\u00de jpi k\u00f022\u00de jpi 2 4 3 5 (3c) with displacement and force vectors Q jpi \u00bc qj qpi ( ) ; (3d) Fjpi \u00bc f j 0 ( ) ; (3e) Fjpi\u00f0t\u00de \u00bc f\u00f01\u00dejpi f\u00f02\u00dejpi 8< : 9= ;: (3f) Meanwhile, equations of motion for the carrier-planet pi sub-system, which is shown in Fig. 2(c) with uc=rcyc as the rotational carrier coordinate (rc is the radius of the circle passing through the planet centers), are defined as mc \u20acyc\u00f0t\u00de \u00fe cy\u00bd _yc\u00f0t\u00de _ypi\u00f0t\u00de cosci _uc\u00f0t\u00de \u00fe ky\u00bdyc\u00f0t\u00de ypi\u00f0t\u00de cosciuc\u00f0t\u00de \u00bc 0; (4a) mc \u20acxc\u00f0t\u00de \u00fe cx\u00bd _xc\u00f0t\u00de _xpi\u00f0t\u00de \u00fe sinci _uc\u00f0t\u00de \u00fe kx\u00bdxc\u00f0t\u00de xpi\u00f0t\u00de \u00fe sinciuc\u00f0t\u00de \u00bc 0; (4b) Jc r2 c \u20acuc\u00f0t\u00de \u00fe cx\u00bd _xc\u00f0t\u00de _xpi\u00f0t\u00de \u00fe sinci _uc\u00f0t\u00de sinci cy\u00bd _yc\u00f0t\u00de _ypi\u00f0t\u00de cosci _uc\u00f0t\u00de cosci \u00fe kx\u00bdxc\u00f0t\u00de xpi\u00f0t\u00de \u00fe sinciuc\u00f0t\u00de sinci ky\u00bdyc\u00f0t\u00de ypi\u00f0t\u00de cosciuc\u00f0t\u00de cosci \u00bc Tc Nrc ; (4c) mp \u20acypi\u00f0t\u00de cy\u00bd _yc\u00f0t\u00de _ypi\u00f0t\u00de cosci _uc\u00f0t\u00de ky\u00bdyc\u00f0t\u00de ypi\u00f0t\u00de cosciuc\u00f0t\u00de \u00bc 0; (4d) mp \u20acxpi\u00f0t\u00de cx\u00bd _xc\u00f0t\u00de _xpi\u00f0t\u00de \u00fe sinci _uc\u00f0t\u00de kx\u00bdxc\u00f0t\u00de xpi\u00f0t\u00de \u00fe sinciuc\u00f0t\u00de \u00bc 0: (4e) Here, Tc is the external torque applied to the carrier, and mc and Jc are the mass and the polar mass moment of inertia of the carrier, respectively. In this model, the planet pin-bearing assembly is represented by linear springs kx and ky, and dampers cx and cy as illustrated in Fig. 2(c). It is noted that above equations are not a function of planet rotation upi=rpypi since planet pi can rotate freely relative to the carrier. Eqs. (4a)\u2013(4e) are written in matrix form to define Mcpi \u00bc mc 0 0 mpi \" # ; (5a) Cpi \u00bc c\u00f011\u00de cpi c\u00f012\u00de cpi c\u00f012\u00de cpi c\u00f022\u00de cpi 2 4 3 5; (5b) Kcpi \u00bc k\u00f011\u00de cpi k\u00f012\u00de cpi k\u00f012\u00de cpi k\u00f022\u00de cpi 2 4 3 5; (5c) Q cpi \u00bc qc qpi ( ) ; (5d) Fcpi \u00bc f c 0 ( ) : (5e) With the matrices of every sub-system available, the overall equations of motion of an N-planet planetary gear set with planets spaced at ci (iA[1,N]) are constructed systematically as M \u20acQ \u00f0t\u00de \u00fe \u00bdC\u00fe Cb _Q \u00f0t\u00de \u00fe \u00bdK\u00f0t\u00de \u00fe Kb Q \u00f0t\u00de \u00bc F \u00fe F\u00f0t\u00de; (6a) where Q \u00bc qc qr qs qp1; ",
            " Similarly, Espi(t) and Erpi(t) are at frequencies om/Zs and om/Zr, respectively, representing the rotational velocities of the sun and ring gears relative to the carrier. Coefficients Es, Er and Epi are the amplitudes of these errors. The additional phase angles es, er and epi illustrated in Fig. 1 define the initial orientation of them, as previously discussed. In the absence of errors defined in Eqs. (9a)\u2013(9d), gear transmission error excitations defined as a part of the gear mesh interfaces shown in Fig. 2(a) and (b) are given in periodic form as [27] espi\u00f0t\u00de \u00bc XL \u2018\u00bc1 e\u00f0\u2018\u00desp sin\u00bd\u2018omt \u00fe \u2018Zsci \u00fe f\u00f0\u2018\u00desp ; (10a) erpi\u00f0t\u00de \u00bc XL \u2018\u00bc1 e\u00f0\u2018\u00derp sin\u00bd\u2018omt \u00fe \u2018Zrci \u00fe \u2018gsr \u00fe f\u00f0\u2018\u00derp (10b) where gsr is the phase angle between espi(t) and erpi(t). If the dynamic model proposed in the previous section were to be exercised with excitations as defined in Eqs. (9a)\u2013(10b) as it was done in Ref. [27], one would observe little or no modulation sidebands in dynamic gear mesh forces Fspi(t) and Frpi(t) as there is no apparent interaction between the two groups of excitations"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure3.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure3.1-1.png",
        "caption": "Fig. 3.1. Model of robot mechanism with n links and n joints",
        "texts": [
            " In this book, it was decided to use Newton-Euler's equations as the most suitable way of demonstrating the physical sense of the problem. These as- 52 3 Dynamics and Dynamic Analysis of Manipulation Robots sumptions were made: (a) links which are modelled by rigid bodies, ignoring the effects of elasticity; (b) the kinematic chain of robotic manipulator is neither branched nor closed; (c) there are no kinematically coupled degrees of freedom. Let us consider the model of the robot mechanism shown in Fig. 3.1. The model consists of n rigid bodies which represent the links of the mechanism chain. These links are interconnected by prismatic (sliding) or rotational joints. For a precise derivation of differential equations of the robot dynamics, a number of basic definitions are introduced. Configuration Configuration of the mechanism is represented by an arranged n-tuple (11' ... , In), such that for each iEN={l, ... , n}, IiE{R,T}, where R denotes a ro tational and T a prismatic joint. To illustrate this, the configuration RTT-RRR stands for a mechanism with six joints, where the second and third joints are prismatic, and the remaining ones are rotational"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000233_978-1-4614-0460-6-Figure2.23-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000233_978-1-4614-0460-6-Figure2.23-1.png",
        "caption": "Fig. 2.23 Damped free vibration response",
        "texts": [
            " For example, if z \u00bc 0.05 (5% viscous damping), then od \u00bc on ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 0:052 p \u00bc 0:999on. The roots in Eq. 2.66 can now be written in more compact notation: s1;2 \u00bc zon iod. Substitution in the assumed harmonic form enables us to rewrite Eq. 2.62 as shown in Eq. 2.68, where the first term in the product again describes the damping envelope that bounds the oscillation (at the damped natural frequency) defined by the second term in the product. See Fig. 2.23. x\u00f0t\u00de \u00bc e zont X1e iodt \u00fe X2e iod t (2.68) For the exponential term in Eq. 2.68, a higher damping ratio gives a more rapid decay rate. For the oscillatory part, if the period of vibration is t (expressed in seconds), the damped natural frequency (in Hz) is: fd \u00bc 1 t : (2.69) Also, the relationship between the damped natural frequency in rad/s,od, and Hz is od \u00bc 2pfd, as we saw for the undamped case. 50 2 Single Degree of Freedom Free Vibration Let\u2019s calculate the free vibration response for a spring\u2013mass\u2013damper system with the following parameters: \u2022 m \u00bc 1 kg \u2022 c \u00bc 20 N-s/m \u2022 k \u00bc 1 104 N/m \u2022 the initial displacement is x0 \u00bc 25 mm \u2022 the initial velocity is _x0 \u00bc 1;000 mm/s"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003774_j.1469-185x.1974.tb01299.x-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003774_j.1469-185x.1974.tb01299.x-Figure8-1.png",
        "caption": "Fig. 8. (a ) One point of application of force on a cilium; the position of P in relation to time ( t ) and distance along the ciliary axis (s) is used later. (b) The induced velocities in the fluid due to a force F. (c) Induced velocity at Q, is twice that at Q2, both being equidistant from the point force F at P. (d) Comparison of the force distribution between (i) longitudinal and (ii) transverse motion. ( e ) Velocity contours around an elongated body in motion. Note the \u2018capture zone\u2019 immediately surrounding the elongated body.",
        "texts": [
            " a flagellum), the influence of the cell wall in the case of a cilium, the concept of the theory of Gray & Hancock (1955)~ and the exterior flow around an organism. (a) Elongated inert body As an introduction to the fluid motion due to an elongated body at low Reynolds number, we will introduce the concept of the motion due to a force F at a point P in an infinite fluid. We could imagine this as being a point on the centre-line of a cilium or flagellum at an instant during its beating cycle (see Fig. 8a). In spherical polar coordinates (R, O), the radial u,- and azimuthal u,-induced velocities at Q due to this force are where R is the distance from P to Q and 8 is the angle PQ subtends to the direction of the force F (Fig. 8b). Thus, the velocity falls off inversely as the distance from the force while the &dependence indicates the preferred direction of fluid motion (i.e. in the direction of the force), which is equivalent both in front and behind the point of application of the force. From this definition we can note the following fundamental point for fluid flow at low Reynolds number (Fig. 8c): the induced velocity at the point Q1 ( R = d, 8 = 0) in Fig. 8(c) is twice that at Q2 (R = d, 0 = 90\u00b0), both points being equidistant from the point of application of the force on the fluid. For an elongated (rod-shaped) body we can imagine the velocity field in the fluid as being due to a \u2018line of force\u2019 distributed along its centre-line. I n a viscous fluid the layer of fluid adjacent to the surface of the elongated body moves at the same velocity as that part of the body (in fluid mechanical literature this is called the \u2018noslip \u2019 condition). Following the discussion and terminology of the previous paragraph, it is apparent that the force distribution needed to satisfy the \u2018no-slip\u2019 condition on an elongated body moving tangentially (Fig. 8d) will be half that for movement at the same velocity in the normal direction. This follows because the induced velocity at each point of a body due to a force at a different point on the elongated body is twice as large for tangential as it is for normal motion. This discussion forms the basis for the theory of Gray & Hancock (1955) of the distribution of forces along a flagellum. The previous discussion gives us some feeling for the force distribution along an elongated body, but little about the fluid motion around the body",
            " As the logarithmic function varies very slowly, we observe that the fluid velocity in the field close to the elongated body will be similar to that of the body itself, indicating a \u2018shielding\u2019 or \u2018capturing\u2019 effect. Further away from the elongated body, the velocity field reverts to the I/R field, because the force distribution can be regarded as effectively concentrated at one point. Mechanics of ciliary motion 99 T o give us some feeling for the \u2018 drop-off\u2019 rate of the velocity field away from the body, we have plotted in Fig. 8(e) a contour map of the ratio q of the fluid velocity z, against the velocity U of movement of a rigid elongated body; that is, q = v p (0 < q < I). (6) Elongated motile organelles (e.g. flagella) Most protozoologists would have little faith in the previous discussion in (a) , with regard to the fluid disturbance created by an elongated microscopic body. They observe the disturbance to be restricted to a thin region around the organism, which is in direct contrast to the large I/R field motion for an inert body, mentioned in the last section",
            " The bases of the cilia are distributed in a regular array over the infinite plane with spacing a in the xl-direction and b in the x2-direction. This is illustrated in Fig. 15. In protozoa the effective stroke of the cilium can be in any direction, but for the purposes of this model it is taken in the direction of increasing xl , although the beat need not be planar. We can represent the movement of a cilium in Cartesian coordinates by the vector function [(s, t), where s is the distance along the cilium from the base and t is time (Fig. 8a). Thus the mean velocity profile, in terms of components Ul and U2, can be obtained in terms of the following integral equation (Blake, 1972u), where and The bar over the integral sign indicates a time average, x3 is the vertical coordinate, &: the vertical coordinate of the cilium above the cell surface, Fl and F2 are the force I I 2 J. R. BLAKE AND M. A. SLEIGH components of each element of the cilium in the x, and x2 directions respectively, and c is the metachronal wave speed with the + or - in w(s, t), indicating that the wave is travelling to the left or right respectively"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001409_j.engfailanal.2014.05.018-Figure11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001409_j.engfailanal.2014.05.018-Figure11-1.png",
        "caption": "Fig. 11. Schematic of crack propagation in method 3: (a) q 6 q1max, (b) q > q1max.",
        "texts": [
            " 1) is a slight curve extending from the tooth root in the gear with a high backup ratio. Lewicki [7] noted that crack propagation paths are smooth, continuous, and in most cases, rather straight with only a slight curvature. Pandya and Parey [9] also pointed out the crack grew following a curved trajectory. In order to reduce the error generated by straight line assumption, for method 3 a parabolic curve is adopted to simulate the crack propagation instead of straight line and the limiting line is the same as that in method 2 (see Fig. 11). Assuming that the intersection angle t corresponding to the leftmost point of crack propagation is known, the parabolic equation can be determined by crack initial point Q and angle t, so effective area moment of inertia and area of the cross section at the position of y in method 3 can be calculated as follows. When q \u00bc q1 6 q1max (see Fig. 11a), the expressions of the effective area moment of inertia and area of the cross section at the position of y are the same as those acquired from method 2 (see Eqs. (23) and (24)). But different from method 2, q1 in method 3 stands for the arc length between crack initial point Q and crack tip P, not the straight-line distance between the two points and q1max in method 3 stands for the arc length between crack initial point Q and the leftmost point of crack propagation. When q = q1max + q2 > q1max (see Fig. 11b), I \u00bc 2 3 x3L; y < ymax 1 12 x ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi \u00f0y ymax\u00dex2 Q yQ ymax r\" #3 L; ymax 6 y 6 yP 1 12 xB\u00fexP \u00f0yB yP \u00de2 \u00f0y yP\u00de 2 xP \u00fe x h i3 L; y > yP 8>>>< >>>>: ; \u00f027\u00de A \u00bc 2xL; y < ymax x ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi \u00f0y ymax\u00dex2 Q yQ ymax r\" # L; ymax 6 y 6 yP xB\u00fexP \u00f0yB yP\u00de2 \u00f0y yP\u00de 2 xP \u00fe x h i L; y > yP 8>>>< >>>: : \u00f028\u00de The schematic of the calculation process in method 3 is basically the same with method 2 (see Fig. 9) except that B3, C3 are replaced with B3 \u00bc 1 12 x ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi \u00f0y ymax\u00dex2 Q yQ ymax r\" #3 L and C3 \u00bc x ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi \u00f0y ymax\u00dex2 Q yQ ymax r\" # L"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure2.17-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure2.17-1.png",
        "caption": "Figure 2.17 Schematisation of the situation surrounding a salvage operation. The wreckage of a crashed lorry is located on a slope at A, and is being salvaged with the cables AB and AC and winches",
        "texts": [
            " Using this information, we can calculate the components of F : Fx = F dx d = (35 kN) \u00d7 \u22128 m 28 m = \u221210 kN, Fy = F dy d = (35 kN) \u00d7 24 m 28 m = +30 kN, Fz = F dz d = (35 kN) \u00d7 12 m 28 m = +15 kN. Figure 2.16 shows the components of F as they are working on the foundation block. In order to determine the resultant of the forces on a particle in space, we first resolve all the forces into their x, y and z component, and then add all the associated components together. This is illustrated in an example. Example Figure 2.17 shows the schematised situation in a salvage operation. A shows the wreckage of a crashed lorry on a slope. People are trying to salvage the wreckage using cables AB and AC and winches in B and C. Cable AB is pulling on the wreckage with a force of magnitude F1 = 7.5 kN; cable AC is pulling on the wreckage with force of magnitude F2 = 10 kN. Question: Find the resultant force being exerted by the cables on the wreckage. 34 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Solution: First, we have to resolve the forces F1 and F2 into their components in the same way as described in the previous section: Fx = Fdx/d , etc"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure11.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure11.2-1.png",
        "caption": "FIGURE 11.2. A vehicle with roll, and yaw rotations.",
        "texts": [
            "5) The roll model vehicle dynamics can be expressed by four kinematic variables: the forward motion x, the lateral motion y, the roll angle \u03d5, and the roll angle \u03c8. In this model, we do not consider vertical movement z, and pitch motion \u03b8. 11.2 F Equations of Motion A rolling rigid vehicle has a motion with four degrees of freedom, which are translation in x and y directions, and rotation about the x and z axes. The Newton-Euler equations of motion for such a rolling rigid vehicle in the body coordinate frame B are: Fx = mv\u0307x \u2212mr vy (11.6) Fy = mv\u0307y +mr vx (11.7) Mz = Iz\u03c9\u0307z = Iz r\u0307 (11.8) Mx = Ix\u03c9\u0307x = Ixp\u0307. (11.9) Proof. Consider the vehicle shown in Figure 11.2. A global coordinate frame G is fixed on the ground, and a local coordinate frame B is attached to the vehicle at the mass center C. The orientation of the frame B can be expressed by the heading angle \u03c8 between the x and X axes, and the roll angle \u03d5 between the z and Z axes. The global position vector of the mass center is denoted by Gd. The rigid body equations of motion in the body coordinate frame are: BF = BRG GF = BRG \u00a1 mGaB \u00a2 = m B GaB = m Bv\u0307B +m B G\u03c9B \u00d7 BvB. (11.10) 11. F Vehicle Roll Dynamics 667 BM = Gd dt BL = B GL\u0307B = BL\u0307+ B G\u03c9B \u00d7 BL = BI B G\u03c9\u0307B + B G\u03c9B \u00d7 \u00a1 BI B G\u03c9B \u00a2 "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure4.24-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure4.24-1.png",
        "caption": "Fig. 4.24: A scheme for kinematic and accuracy analysis of a four-bar linkage gripping mechanism with coupler fingertip point M.",
        "texts": [
            " The method of closure equations can be used also to derive numerical procedures for the case of more complicated mechanisms with several loops when the kinematic circuits give a system of equations that are not suitable for hand computation or closed-form formulation. Several techniques are available from the field of mechanism analysis. Since in general gripping mechanisms are not very complicated, it can be convenient to deduce analysis formulation by using straightforwardly the closure loop equations as in the following example for a four-bar linkage mechanism. Figure 4.24 shows a general four-bar linkage in which a denotes the length of the frame, b and d are respectively the input and the output length links, and c is the length of the coupler whose fingertip point M is located by the length p and the angle \u0393. Referring to Fig. 4.24, if OXY is a fixed reference and assuming that \u03b1 is the input angle, positive counter-clockwise, \u03b2 the output angle, \u03b8 the angle between the generic position of c and X-axis, from the loop closure equations it follows C+B )C-B+(sin-sin tan2 1/2222 1- \u03b1\u03b1=\u03b2 (4.6.7) Fundamentals of the Mechanics of Robotic Manipulation 269 D+B )D-B+(sin-sin tan2 1/2222 1- \u03b1\u03b1=\u03b8 where b a cosB \u2212\u03b1= \u03b1\u2212+\u2212+= cos d a 2bd dcba C 2222 (4.6.8) 2bc dcba cos c a D 2222 \u2212++\u2212\u03b1= The coordinates of the fingertip point M can be computed as ( )\u03b8\u2212\u0393\u2212\u03b8+\u03b1= cospcosccosbx (4",
            "7) to (4.6.15) allow a kinematic analysis of a four-bar linkage gripping mechanism whose size is assigned. In addition, using the computations of the above-mentioned formulation one can carry out accuracy analysis. Accuracy of a gripping mechanism is usually referred to as the position error of a reference coupler fingertip point in reaching prescribed trajectory points during the motion, and it can be evaluated as the maximum of these deviation errors along a determined coupler segment. Referring to Fig. 4.24, let M be the fingertip point lying on the coupler curve, and R be the corresponding point on the circle of center G when the GR line intercepts M. In this case, the accuracy of a coupler segment, which is traced by M, can be investigated by means of the geometric ratio r MR=\u03b5 (4.6.16) in which r = GR is the radius of the osculating circle. Referring to the scheme of Fig. 4.24, the geometrical deviation MR can be computed as r-MG=MR (4.6.17) where ( ) ( )[ ]1/22 G 2 G y-y+ x-x=MG (4.6.18) and UU' UPo =r 2 (4.6.19) with Po the instantaneous center of velocity and U' the inflection point relating to U. MR is positive or negative when a coupler curve segment circumscribes or inscribes its osculating circular arc, respectively. Moreover, assuming \u03c4 as the angle between X-axis and GU, positive counter-clockwise, one can compute the positive clockwise angle between GM and GU as G G1- xx yy tan \u2212 \u2212 \u2212\u03c4=\u03d5 (4"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001205_s10846-016-0442-0-Figure15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001205_s10846-016-0442-0-Figure15-1.png",
        "caption": "Fig. 15 Parameters to determine tracking error",
        "texts": [
            " The location of control point can be located on the centre of front tire axle, centre of rear tire axle, and centre of gravity (COG) of the vehicle. Point on path may be defined by the lookahead distance or nearest point from control point. One of the most common tracking error formula is stated in Eq. 23 with lateral (ye), longitudinal (xe), and orientation errors (\u03b8e) relative to the local coordinates were formulated in terms of the distance between rear axle control point and a point of path within one lookahead away as shown in Fig. 15a. The same formula can be used with different location of control point. Stanley Method by Stanford University [26] used only lateral error in their formulation as shown in Fig. 11. Here, the control point situated on the front tire axle and the error was defined as the shortest perpendicular distance between the control point and the point on path. \u23a1 \u23a3 xe ye \u03b8e \u23a4 \u23a6 = \u23a1 \u23a3 cos \u03c8c sin \u03c8c 0 \u2212 sin \u03c8c cos \u03c8c 0 0 0 1 \u23a4 \u23a6 \u23a1 \u23a3 Xd \u2212 Xc Yd \u2212 Yc \u03c8d \u2212 \u03c8c \u23a4 \u23a6 (23) Basic tracking error in Eq. 23 has been used widely to measure the controller performance throughout the path tracking manoeuvre [23, 53, 66, 72, 86, 87, 94, 99, 114]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure3.12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure3.12-1.png",
        "caption": "Fig. 3.12 An Example of 6 Axis Force/Torque Sensor (courtesy of Nitta Corp.)",
        "texts": [
            " The rubber bushes and the dampers are positioned to prevent large impulse forces from being transmitted to the robot. Since the displacement of them is small, we do not consider the displacement when calculating the ZMP. A 6 axis force/torque sensor is coordinated to simultaneously measure the force f = [fx, fy, fz] and the moment \u03c4 = [\u03c4x \u03c4y \u03c4z] applied from outside the robot. This sensor is mainly used for measuring the force at the end effector of industrial robots. An example of 6 axis force/torque sensor is shown in Fig. 3.12. To measure the ZMP of a humanoid robot, the force/torque sensor must be light and must be strong enough to accept the large impulsive force applied to the sensor. To obtain the ZMP from the measured data of 6 axis force/torque sensor, we set N = 1 in (3.24) and (3.25). Let the position of the ZMP in the right and the left foot be pR and pL, respectively, as shown in Fig. 3.13. Especially when the center of measurement 80 3 ZMP and Dynamics of the sensor lies on the z axis of the reference coordinate system, the position of the ZMP of each foot can be obtained very simply"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000641_s11661-009-9949-3-Figure11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000641_s11661-009-9949-3-Figure11-1.png",
        "caption": "Fig. 11\u2014Tensile samples fabricated by laser powder deposition. (a) Thin-walled sample deposited on the wrought DA718 coupon for as-deposited and direct-age evaluations and (b) fully deposited block sample for STA and homogenization treatment evaluations.",
        "texts": [
            " Figure 10 shows the EDAX elemental analysis results for the major phases after homogenization heat treatment, i.e., NbC particles, the remaining d phase, and the matrix phase. Two types of deposition geometries were used for making the tensile specimens: (1) deposition of thinwalled geometry on the direct-aged wrought IN718 (DA718) coupon with the final size of 42 mm 9 81 mm 9 3 mm, as shown in Figures 11(a); and (2) fully deposited block with the size of 64 mm 9 15 mm 9 7 mm on stainless steel substrate (25 mm 9 25 mm 9 6 mm), as shown in Figure 11(b). The former sample is designed to represent the repair situations for compressor and blisk blades, where only direct-age heat treatment can be applied after repair. The strength of the interface between the deposited material and the wrought DA718 base material is evaluated in the tensile test. The base DA718 coupon was verified to have a microstructure consisting of fine equiaxed grains with a size of ASTM 11. The DA718 samples, as shown in Figure 11(a), were tested in two heat treatment conditions, i.e., as deposited and direct aged. No solution and homogenization heat treatments were applied to these samples due to the potential degradation risk to the 2414\u2014VOLUME 40A, OCTOBER 2009 METALLURGICAL AND MATERIALS TRANSACTIONS A base material as a result of recrystallization and grain size coarsening. The fully deposited samples (Figure 11(b)) were divided into two groups and heat treated by standard STA and the homogenization STA method, respectively. The goal was to evaluate the laser deposition as a potential viable rapid manufacturing technique for new make parts. After heat treatment, flat tensile bars 60.3 mm (L) 9 12.7 mm (W) 9 1.5 mm (T) that conformed to GE Aviation specifications were cut from the deposition samples by wire electrical discharge machining. For each heat treatment condition, two to four specimens were prepared and tested. All the tensile tests were conducted at room temperature. METALLURGICAL AND MATERIALS TRANSACTIONS A VOLUME 40A, OCTOBER 2009\u20142415 Figure 12 demonstrates the average ultimate stress, 0.2 pct yield stress, and plastic elongation of the tensile test results. These results were compared with different heat treatment conditions and the minimum properties from the AMS specifications for cast and wrought IN718, respectively. All the DA718 specimens (from Figure 11(a)) failed in the gage sections of the laserdeposited portions. The as-deposited material produces low yield stress (552 MPa) and ultimate stress (904 MPa) but relatively high plastic elongation (16.2 pct). After direct-age heat treatment, the ultimate stress increased by 47 pct to 1333 MPa, while the yield stress almost doubled and reached 1084 MPa. However, the plastic elongation of the direct-aged material compared with the as-deposited material dropped significantly from 16.2 to 8.4 pct. Although the direct-aged elongation value is still above the AMS cast property, it is much lower than the AMS wrought property, which is at 12 pct"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001383_025403-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001383_025403-Figure8-1.png",
        "caption": "Figure 8. Steady state calculation of the melt pool corresponding to the beginning of the building of the 3rd layer.",
        "texts": [
            " 293 K 473 K 673 K 1073 K 1893 K 1933 K 2273 K k (W m\u22121 K\u22121) 10 12 15 17 20 60 60 Cp (J Kg\u22121 K\u22121) 580 680 760 800 930 950 950 \u03c1 (kg m\u22123) 4420 4300 4200 3800 having incremental widths en and height per layer hn , until layers reach a constant width value. A step by step incremental approach allows us to describe the morphological evolution of manufactured walls in the (Oz) direction, and particularly en = f (n). In more detail, once the first layer bottom is calculated from the substrate melt pool, a first layer top width is chosen, artificially incremented by a d\u03b5 value, thus forming a non-rectangular first layer (figure 8). Then a thermal calculation is carried out, where laser irradiates the layer top surface. If the fusion zone coincides with e2 = e1 + d\u03b5 width, the new width e2 is validated. If the fusion zone width eF is smaller than e2 = e1+d\u03b5, d\u03b5 has to be reduced until both values fit exactly. A synoptic of the simulation-aided morphological calculation is presented in figure 9. 2.2.2. Thermal limitation to layer growth. For specific powder deposition conditions where for instance large mass feed rates and small melt pools are interacting, all the powder coming into the fusion zone will not necessarily contribute to the growth"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001283_s0022-0728(79)80227-2-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001283_s0022-0728(79)80227-2-Figure2-1.png",
        "caption": "Fig. 2. Structural configuration of Pt-ferrocene electrode surface.",
        "texts": [
            "1 V s -1 ) measurements were made with an Amel model 551/SU potentiostatgalvanostat connected to a Moseley 7030 AM X-Y recorder. For faster scan rates (0.1--1000 V s -1 ) a potential sweep generator, constructed in this laboratory, was employed in conjunction with a Tektronix type 564 storage oscilloscope. Positive feedback techniques were employed for residual resistance compensation. The potentials of current peaks could be determined from the oscilloscope trace with a precision of + 2--4 inV. Ideally, the chemical modification procedure described above should lead to a surface structure as shown in Fig.2. Molecular models indicate a distance of 6--8 A between the platinum substrate and the redox centre of the ferrocene grouping. To examine the electroactivity of the attached species, voltammetric measurements were carried out with the modified electrodes immersed in acetonitrile solutions containing supporting electrolyte alone. Fig.3 shows a typical cyclic voltammogram recorded at low potential scan rates. The close agreement between the potentials corresponding to the anodic and cathodic current peaks is characteristic of rapid surface charge transfer reactions, i"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure9.28-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure9.28-1.png",
        "caption": "Figure 9.28 Statically determinate trusses.",
        "texts": [
            " n > 0 The truss is statically indeterminate. There are more unknowns than equilibrium equations. One or more of the member forces and/or support reactions cannot be determined directly from the equilibrium. In principle, there is an infinite number of solutions that satisfy the equilibrium conditions (the solution is undetermined). The correct solution can be found by taking into account the deformation behaviour of the structure. The surplus of unknowns, n, is known as the degree of static indeterminacy. Figure 9.28 provides examples of statically determinate trusses. The shape-retaining truss ABCD in Figure 9.28a is immovable supported by a hinge at A and a bar at B. The hinged support provides two support reactions, and the bar support provides one, so that r = 3. With s = 25 and k = 14, n = 0. The truss is therefore statically determinate. If the bar support is considered as one of the truss members, B\u2032 has to be seen as a hinged support. In that case, r = 4, s = 26 and k = 15, and again n = 0. In the simple truss in Figure 9.28b, the diagonal members cross one another. The truss is immovable, supported on a roller and by a hinge. Here r = 3, s = 13 and k = 8, so that n = 0. The truss is therefore statically determinate. The compound truss in Figure 9.28c is also immovable supported. With r = 4, s = 14 and k = 9, n = 0. The truss is statically determinate. 336 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Figure 9.29 provides examples of statically indeterminate trusses. All the trusses are kinematically determinate. The degree of static indeterminacy can be determined with n = r + s \u2212 2k. The truss in Figure 9.29a, in which the diagonal members cross one another, has a hinged support and a bar support. With r = 3, s = 31, k = 14, one finds n = 6. The truss is six-fold statically indeterminate. If we compare the truss with the statically determinate structure in Figure 9.28a, we see that the truss has 6 redundant diagonal members. For the truss in Figure 9.29b, with crossing diagonals, r = 3, s = 26, k = 14 and so n = 1. The truss is therefore statically indeterminate to the first degree. For the compound truss in Figure 9.29c, r = 4, s = 16 and k = 9, so that n = 2. A member could be omitted in each of two self-contained parts (see also Figure 9.28c). The truss in Figure 9.29d is statically indeterminate to the first degree, with r = 4, s = 19, k = 11 and so n = 1. The structure can be made statically determinate by, for example, removing one of the roller supports. You could also remove an arbitrary top or bottom chord member. For the truss in Figure 9.29e, r = 6, s = 13 and k = 8, so that n = 3. The truss is statically indeterminate to the third degree. The simple truss has three redundant support reactions and/or members. In statically determinate trusses, all the force members and support reactions can be determined directly from the equilibrium"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure14.54-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure14.54-1.png",
        "caption": "Figure 14.54 The distribution of normal stresses in a cross-section due to (a) a bending moment M and (b) a normal force N .",
        "texts": [
            " All cross-sections between B and A have to transfer the resultant of the forces F at C and D. The line of force for AB coincides with the line of action of this resultant. The line of action figure shows that this line of action passes through D and is parallel to ABC. Since the normal force is a compressive force everywhere, all the lines of force are lines of pressure. 14 Cables, Lines of Force and Structural Shapes 679 The bending moment and the normal force are the resultants of the normal stresses in a cross-section. Figure 14.54 shows the stress distribution due to a bending moment M and a normal force N .1 A characteristic of stress distribution in bending is that the outermost fibres of the cross-section are most heavily loaded, while the fibres in the environment of the member axis are virtually unloaded. In contrast, the stresses due to a normal force are constant over the cross-section. In extension, all fibres are therefore loaded equally. If we compare the stress distributions due to bending and extension, the material in the cross-section is used far more efficiently in extension than in bending",
            " This is achieved by ensuring the member axis and line of force coincide as much as possible. Since a cable cannot transfer bending moments, it assumes a shape in which the line of force coincides with its axis everywhere. Taking the cable shape and line of force as basis, in the following four examples, we look for 1 In Volume 2, Stresses, Deformations, Displacements, we take a closer look at the exact development of the normal stresses in a cross-section and at the conditions under which the stress distribution in Figure 14.54 applies. 680 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM structural shapes in which the bending moments are as small as possible. Example 1 The beam in Figure 14.56a is subject to bending by the two forces F . The M diagram is shown in Figure 14.56b. In Figure 14.57, the same load is carried by a cable with compression bar. The cable and compression bar transfer normal forces only. The cable has the same shape as the M diagram in Figure 14.56b. With a cable sag the scale factor is H = F = F. H is the (compressive) force in the bar that is equal to the horizontal component of the (tensile) force in the cable"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.36-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.36-1.png",
        "caption": "FIGURE 8.36. The kinematic model for a double A-arm suspension mechanism.",
        "texts": [
            " A suspension roll center may be on, above, or below the road surface, as shown in Figure 8.35(a)-(c) for an external suspension roll center. When the suspension roll center is on the ground, above the ground, or below the ground, the vehicle roll center would be on the ground, below the ground, and above the ground, respectively. Example 318 F Camber variation of double A-arm suspension. When a wheel moves up and down with respect to the vehicle body, de- 474 8. Suspension Mechanisms 8. Suspension Mechanisms 475 pending on the suspension mechanism, the wheel may camber. Figure 8.36 illustrates the kinematic model for a double A-arm suspension mechanism. The mechanism is equivalent to a four-bar linkage with the ground link as the vehicle chasis. The wheel is always attached to a coupler point C of the mechanism. We set a local suspension coordinate frame (x, y) with the x-axis indicating the ground link MN . The x-axis makes a constant angle \u03b80 with the vertical direction. The suspension machanism has a length a for the upper A-arm, b for the coupler link, c for the lower A-arm, and d for the ground link",
            "14) where, D = J5 \u2212 J1 + (1 + J4) cos \u03b82 (8.15) E = \u22122 sin \u03b82 (8.16) F = J5 + J1 \u2212 (1\u2212 J4) cos \u03b82 (8.17) and J1 = d a (8.18) J2 = d c (8.19) J3 = a2 \u2212 b2 + c2 + d2 2ac (8.20) J4 = d b (8.21) J5 = c2 \u2212 d2 \u2212 a2 \u2212 b2 2ab . (8.22) Substituting (8.14) and (8.13), and then, eliminating \u03b82 between (8.13) and (8.8) provides the relationship between the vertical motion of the wheel, z, and the camber angle \u03b3. 478 8. Suspension Mechanisms Example 319 F Camber angle and wheel fluctuations. Consider the double A-arm suspension that is shown in Figure 8.36. The dimensions of the equivalent kinematic model are: a = 22.4 cm b = 22.1 cm c = 27.3 cm d = 17.4 cm \u03b80 = 24.3 deg (8.23) The coupler point C is at: e = 14.8 cm \u03b1 = 54.8 deg (8.24) If the angle \u03b82 at the rest position is at \u03b820 = 121.5 deg (8.25) then the initial angle of the other links are: \u03b830 = 18.36 deg \u03b840 = 107.32 deg (8.26) At the rest position, the coupler point is at: xC0 = \u221222.73 cm yC0 = 9.23 cm z0 = 16.92 cm (8.27) We may calculate h and \u03b3 by varying the parameter \u03b82. Figure 8.37 illustrates h as a function of the camber angle \u03b3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001288_j.actbio.2014.06.010-Figure9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001288_j.actbio.2014.06.010-Figure9-1.png",
        "caption": "Fig. 9. Schematic to depict the buckling and bending vectors of the load applied on (a) cubic, (b) G7 and (c) rhombic dodecahedron cell struts.",
        "texts": [
            " Based on these deformation modes, a simple model was established to estimate accurately mechanical properties such as elastic modulus and collapse strength using the properties of cell wall and the cell geometry as input [37,38]. The following investigations on the mechanical properties of cellular solids are based mainly on this model. The videos and in situ observations presented in Section 3.3 show that the cell struts are deformed mainly by buckling and/or bending in compression. Thus, different mechanical behavior of the meshes should be related to the interaction of two deformation mechanisms. To characterize their contributions, the load P applied to a strut in Fig. 9 was divided into two components P1 and P2 along the bending and buckling deformation directions, respectively. It is clear that the P2 component decreases in the order cubic, rhombic dodecahedron and G7, whereas P1 follows the reverse order. The above variation tendencies of P1 and P2 components with cell shapes are in agreement with those of the strength, modulus and deformation behavior described in Sections 3.2 and 3.3. That is, under identical density, higher P1 (lower P2) correspond to lower strength and modulus, and their stress\u2013strain curves are easily shown to have a ductile character. For example, Fig. 4 shows that strength and modulus decrease in turn for the cubic, rhombic dodecahedron and G7 cells, while their deformation behavior changes from brittle to ductile characteristics. The above empirical relation can be explained by the simple model presented in Fig. 9. As for the elastic modulus E of a mesh, the analyses in the Appendix (Eq. (A19)) shows that it is proportional to sin a= cos2 a, in which a is the angle between P1 and P. As a result, the larger a leads to a higher modulus. Therefore the meshes have higher and lower E if their structures tend to buckling (larger P2) and bending (larger P1), respectively. The analysis of strength also has a similar tendency. It should be noted that, for stochastic foams of ductile materials such as polymer and aluminum foams, previous studies have shown their deformations dominated mainly by the bending mechanism and neglected generally for the buckling mechanism. The study of reticulate meshes of Ti\u20136Al\u20134V alloy, however, found that their deformations were controlled by both mechanisms (Fig. 9). Since the increase in bending contribution tends to ductility of the meshes (Fig. 6), the following experiments were conducted to give a solid confirmation: for the rhombic dodecahedron meshes presented in Fig. 10, the a angles were designed as 36, 30 and 23 , corresponding to the increase in bending component P1. The compression tests (Fig. 10) found that the stress plateau gradually became smooth with the decrease in the angle, while the strength and modulus decreased correspondingly. The above analyses suggest that the brittle deformation behavior of the EBM meshes studied can be avoided by promoting the bending deformation mechanism through structure design"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure2.20-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure2.20-1.png",
        "caption": "Fig. 2.20 (a) Joint Axis Vector aj (b) Relative Position Vector bj and the Origin of Local Coordinates pj",
        "texts": [
            " In the same way, for each ankle, we assign the origins of two ankle frames on the same point where the two joint axes intersect. Then, all rotation matrices which describe attitude of links are set to match the world coordinates when the robot is in its initial state, standing upright with fully stretched knees. Therefore we set thirteen matrices as, R1 = R2 = . . . = R13 = E. We show the local coordinates defined at this stage in Fig. 2.19(b). Next we will define the joint axis vectors aj and the relative position vector bj as indicated in Fig. 2.20. The joint axis vector is a unit vector which defines the axis of the joint rotation in the parent link\u2019s local coordinates. The positive (+) joint rotation is defined as the way to tighten a right-hand screw placed in the same direction with the joint axis vector. Using the knee joints as an example, the joint axis vectors would be, a5,a11 = [0 1 0]T . When we rotate this link in the + direction, a straight knee will flex in the same direction as a human\u2019s. The relative position vector bj is the vector that indicates where the origin of a local coordinate lies in the parent link\u2019s 2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure9.18-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure9.18-1.png",
        "caption": "Fig. 9.18. AFPM synchronous machine for EMALS. 1 \u2014 PM rotor assembly, 2 \u2014 stator assembly, 3 \u2014 bearing, 4 \u2014 enclosure, 5 \u2014 mounting flange, 6 \u2014 brake, 7 \u2014 shaft encoder.",
        "texts": [
            " The EMALS technology employs linear induction or synchronous motors fed from disc-type alternators through cycloconverters. The electrical energy obtained from the carrier power plant is stored kinetically in the rotors of AFPM synchronous generators. The energy is then released as a fewsecond pulses to accelerate and launch an aircraft. The cycloconverter between the AFPM generator and linear motor raises the voltage and frequency. The EMALS operates in a \u201creal time\u201d closed loop control [79]. The requirements are given in Table 9.5. Specifications of the AFPM synchronous machine shown in Fig. 9.18 are given in Table 9.6 [79]. The EMALS uses four AFPM machines mounted in a torque frame and grouped in counter-rotating pairs to reduce the torque and gyroscopic effect. The rotor of the AFPM machine serves both as the kinetic energy storage when operating as a motor and a field excitation system when operating as a generator. The electric power from the on-board generators is fed to the AFPM machine via rectifier\u2013inverters. There are two separate stator windings for motoring and generating mode"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003605_027836499601500604-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003605_027836499601500604-Figure4-1.png",
        "caption": "Fig. 4. Telescopic ball bar. (With permission from Goswami et al. 1993b [&copy;IEEE]).",
        "texts": [
            " Wampler and Arai (1992) presented simulations for a two-loop planar mechanism comprised of three prismatic legs with passive rotary joints at their ends attached to a triangular stage (M = 3). Leg lengths plus attachment angles for one leg were measured (S = 5). 3.6. C = 1 Several articles have recently appeared that require only one component of pose to be measured (Sj = 1 and Dj = 0). Goswami et al. (1993a,b) employed a ball bar with linear extension measured by a Linear Variable Differential Transformer (Fig. 4). Driels and Swayze (1994) used a single wire potentiometer. Tang and Liu (1993) employed a single laser displacement meter; this method requires that the laser be positioned roughly perpendicular to the flat surfaces of a stack of blocks and that the nomi- nal parameter estimates are close to the correct estimates. Nahvi et al. (1994) used a multiple-loop formulation to calibrate a three-loop redundantly actuated shoulder joint (NI = 3 and S = 4). Boulet (1992) applied closedloop calibration to a mechanical two-loop system formed as a single joint (M = 1) with two antagonistic linear actuators (S = 2); this device was a precursor to the shoulder joint above"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003619_bf02665211-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003619_bf02665211-Figure6-1.png",
        "caption": "Fig. 6- -Workpiece absorption computation. Laser beam and melt pool projections onto the horizontal plane.",
        "texts": [
            " The Workpiece Absorption [3w One of the most difficult aspects of laser cladding is that the overall workpiece absorption depends on the processing conditions, tm,2~ Indeed, experiments have shown that the absorption of a laser beam onto an inclined plane depends on the angle of the normal to the plane with respect to the laser beam. Thus, as the powder-feed rate is increased, with all the other parameters remaining constant, the clad height will increase, leading to a greater angle between the normal to the liquid-air interface and the laser beam (Figure 5), thus increasing the absorption of the workpiece. A model has therefore been developed for computing the workpiece absorption and is presented subsequently. As described in Figure 6, the laser beam illuminates g,liq either the liquid surface o~ or the solid surface S~112 (either the substrate or the previous track). Assuming, as described in Figure 7, that the liquid-gas interface is planar and that the cross section of the previous track is triangular, the liquid surface illuminated by the beam makes an angle 0~q with the horizontal, given by 01iq = arctan [5] Xrnax - - Xmi n whereas the previous track illuminated by the beam makes an angle with the horizontal. Let fl~(0) be the workpiece absorption of a planar surface making an angle 0 with the horizontal (0 < 0 < 60 deg). As indicated previously (assumption A7 in Section II) flw(O) has the following form: /3w(0) = r (1 + a~0) [61 where a~ is a proportionality coefficient which has to be measured for each material 9 As the laser beam illuminates the substrate with absorption flw(0), the previously deposited track with absorption fl~(OsoO, and the liquid pool with absorption flw(O~iq) (Figure 6), the workpiece absorption [3w is given by l [ Slas~w(O) + Slas[~w( Osol) ] -}- Slasflw( Oliq) _ sol sol liq 2 fl~ = [7] a las c~iq = 1rr2. As shown in Figure 6, where S~as these areas are approximated by ellipses. Please note that the empirical law 6 can be roughly explained by the following argument: it is well known that the absorption of an inclined plane illuminated by a linearly polarized beam depends upon the angle of inclination of the plane (with respect to the horizontal) and upon the plane of polarization (Brewster effect), t2u As our laser beam is circularly polarized (assumption A8), it can be considered as the sum of an infinite number of linearly polarized beams about the Oz axis",
            "iq melt-pool surface reached by the powder particles, oj~,, and the projected area of the gas-powder jet on the horizontal plane, Sj~,: ~p = Sjet [15] As shown in Figure 8, these areas are approximated by ellipses. Finally, the mass conservation law between the injection nozzle and the cladded layer gives ~flhp = pv~h~s c [ 16] This model is now complete. Indeed, I0 equations are available, namely Eqs. [4], [7], and [9] through [16], to solve 10 unknowns, which are Pw, fl~, vj, ~lp, the, xc, Xm~n, X . . . . XO, and Ymax\" The ratio Pat/P~ in Eq. [14] is computed with Eq. [3], the temperature field T is com@iq in Eq. [7] puted using Eq. [8], and the areas ~~ and ~',a~ are evaluated as described in Figure 6. As explained at the beginning of Section II, the user of the program should prescribe the following data: Pt, r,, he, s~, Oj~,, rj~,, xj~,, rp, Vp, k, pc, ~w (0), a w (Eq. [6]), and tip. Then the following algorithm, based on our physical understanding of the process, is used to uncouple Eq. [4], [7], and [9] through [16]. Step 0 Initialization. The power reaching the workpiece Pw is set to flw(O)Pt. A first estimate is given for the initial laser velocity vt. Step 1 Equations [9] and [10] are solved simultaneously, leading to new values for vt and x~"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure10.37-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure10.37-1.png",
        "caption": "Figure 10.37 When you have to deal with bent members or structures consisting of several members, you have to introduce a local coordinate system along each straight member segment if you want to indicate the directions of M and V using the plus and minus signs. This soon becomes cumbersome and cluttered in manual calculations.",
        "texts": [
            " To interpret the signs in the M and V diagrams correctly, we therefore have to know the coordinate system. So far, in all the examples showing the M and V diagrams, the structure has consisted of a single straight member, and the coordinate system was always shown. When you have to deal with bent members or structures consisting of several members, you have to introduce a local coordinate system along each straight member segment if you want to indicate the directions of M and V using the plus and minus signs. For the simple structure in Figure 10.37, this already leads to three local coordinate systems: one for AB, one for BC and one for CD. This soon becomes cumbersome and cluttered. In manual calculations, we will therefore use deformation symbols: the bending symbol for bending moments, and the shear symbol for shear forces. The bending symbol and shear symbol symbolise the deformation of the member axis due to a bending moment and a shear force respectively. These deformation symbols can be used to set the direction of the section forces unequivocally, regardless of a coordinate system"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002121_amr.816-817.134-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002121_amr.816-817.134-Figure4-1.png",
        "caption": "Figure 4: Specimen geometries; (a) Tensile",
        "texts": [
            " Before shot-peening, abrasive material was blasted on the surface which resulted in Ra \u2248 2.9 \u00b5m. For the shot-peening process, Almen intensity was used as a measure of process standardization. According to the recommendations, the sample was shot-peened to achieve an Almen intensity (A) of 12 mm [8]. A saturation curve was established to determine the peening duration to achieve the required intensity as shown in Figure 2. Tensile tests were performed to obtain the quasi-static mechanical properties of the SLM generated parts. Figure 4a shows the specimen geometry for the tensile test. The test was performed at room temperature with a strain rate of 5.5\u221910 -6 1/s, and confirmed that the yield strength and the ultimate tensile strength of the SLM generated parts are almost consistent to those of wrought material. Ductility, however, is sufficiently reduced (percent elongation= 7.5) which is due to the high temperature gradients resulting in a very fine acicular microstructure. An exemplary result of the tensile test is plotted in Figure 3. test, (b) HCF test HCF tests were carried out according to ISO 12107:2003 (Figure 4b) at room temperature on a 100 KN servo-hydraulic systemMTS 318.10with an integrated load cell. Tests were performed with a stress ratio (R) of 0.1 at a frequency of 50 Hz. A fatigue life of 10 7 cycles was defined as the fatiguelimit. Specimens after failure were observed under scanning electron microscope (SEM) to determine the cause of failure. As-built sample. The fatigue limit for the as-built sample, averaged according to staircase method, turned out to be 210 MPa. The low value of fatigue limit is due to poorsurface quality (Ra \u224813 \u00b5m) obtained from the SLM process"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001415_978-3-319-74528-2-Figure7.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001415_978-3-319-74528-2-Figure7.1-1.png",
        "caption": "Fig. 7.1 Spatial temperature distribution for aggregation at the warmest spot. (a) Side view (b) Top view",
        "texts": [
            " By emulating some of the software-engineering speak, we say we go through a swarm robotics use case. Say, a customer orders software from you and specifies it in the following way. The software controls a robot swarm, which aggregates at a certain spot determined by sensor input and stays flexible to a dynamic environment. That sensory input could be anything but say it is temperature. Then we want aggregation at the warmest spot. Possible scenarios depend on the distribution of temperature, such as shown in Fig. 7.1. However, much more complex scenarios are possible. The temperature distribution could have local optima, many scattered spots instead of just one well-defined, smooth peak. There could also be systematic plateaus that give no information via the gradient (i.e., the spatial derivative is zero and we cannot guess in what direction we may find a peak). \u00a9 Springer International Publishing AG 2018 H. Hamann, Swarm Robotics: A Formal Approach, https://doi.org/10.1007/978-3-319-74528-2_7 163 164 7 Case Study: Adaptive Aggregation You may want to guess an efficient approach immediately but we first go through a few alternative solutions",
            " An alternative modeling approach could be the L\u00e9vy flight [248], which allows bigger jumps that could represent straight motion. Here, we stick to the Brownian-motion approach, which is also called 170 7 Case Study: Adaptive Aggregation Wiener process, which in turn is a special case of the L\u00e9vy flight. As before, B scales the intensity of the random behavior. In Fig. 7.3 you see example trajectories using Eq. (7.1) and the effect of varying parameter \u02db. The temperature distribution is still the same as in Fig. 7.1, that is, the desired warm spot is at the left border. For \u02db D 0:01 we see basically no bias towards the optimum. Then with increasing \u02db the robot approaches the optimum more and more directly and stays closer to it. Note that we have not yet defined a condition for stopping, hence, the robot keeps moving at all time. 7.4 Model 171 Next, we want to model the \u201cflat gradient information\u201d situation. The question is what the correct setting for parameter \u02db is. In the case of the young honeybee experiments, the situation is somewhat unclear [209, 372]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003728_1.3099008-Figure26-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003728_1.3099008-Figure26-1.png",
        "caption": "Fig 26. The dynamics of a mass-spring system with displacement dependent Coulomb friction is shown in a three-dimensional phase portrait. The friction force was modeled as a product \u2022between a \u2022 normal load and a friction coefficient. The spring-loaded normal force varied linearly with displacement of the mass, and the friction coefficient was taken as a constant. The phase portrait is plotted in cylindrical coordinates, with displacement as the radial axis, velocity as the longitudinal axis, and time as the circumferential coordinate (from Feeny, 1995).",
        "texts": [
            " In these systems, the motion collapses during stick-slip, leading to a one-dimensional map in the Poincar6 section. In the former case, the system was both belt driven and harmonically excited. The underlying map was a one-dimensional circle map. lntermittency was the predominant route to chaos. The strange attractor was on a folded toroidal structure (Fig 25). The latter study was on a forced oscillator with displacement-dependent dry damping. The underlying map was a one-dimensional single-humped map. The strange attractor was a stretching and folding branched manifold (Fig 26). The mechanism of dimensional collapse is as follows. The discontinuity of the friction law occurs on a surface D in the phase space. In regions R of D in which the vector field points towards D from both sides, trajectories will stick to this surface for an interval of time. Otherwise, they may pass through or depart from the surface. Suppose, for example in a three-dimensional phase space, a small enough three-dimensional blob of trajectories approaches R. Then at Some time, all of those trajectories will be in R in a two-dimensional blob"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure9.59-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure9.59-1.png",
        "caption": "Figure 9.59 (a) The forces in members 3 and 4 follow from the equilibrium of joint B. (b) The closed force polygon for the equilibrium of joint B. FB;1 is known. (c) Joint B with all the forces acting on it. From this figure we can see that N3 and N4 are compressive forces.",
        "texts": [
            " The order in which the joint equilibrium is determined, with no more than two unknowns per joint, is A \u21d2 B \u21d2 C \u21d2 D \u21d2 E \u21d2 G \u21d2 H. 356 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM For calculating the still unknown member forces, we now use the graphical method. After A, the next joint is B, where we can calculate the member forces. Joint B is subject to the forces FB;1, FB;3 and FB;4, of which FB;1 is known. Earlier, we found that the force in member 1 is a compressive force: N1 = \u22128F \u221a 2. Member 1 therefore exerts a compressive force on joint B of 8F \u221a 2, so that FB;1 = 8F \u221a 2 (see Figure 9.59a). The two unknowns FB;3 and FB;4 can be determined from the closed force polygon for the equilibrium of joint B (see Figure 9.59b): FB;3 = 4F, FB;4 = 4F \u221a 5. In Figure 9.59c, the forces from the force polygon are shown as they act on joint B in reality. Here we see that FB;3 and FB;4 are both compressive forces. Converted into the normal forces in the members 3 and 4, with the correct sign for tension and compression, we therefore get N3 = \u2212FB;3 = \u22124F, N4 = \u2212FB;4 = \u22124F \u221a 5. The following joint with only two unknowns is C. The forces that the members 2 and 3 exert on the joint are known (see Figure 9.60a): FC;2 = 4F \u221a 5, FC;3 = 4F. The unknown forces FC;5 and FC;6 follow from the force polygon in Figure 9"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure5.7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure5.7-1.png",
        "caption": "Fig. 5.7. Air-cooled single-stage synchronous AFPM machine: (a) rotor disc with surface mounted PM segments, (b) coreless stator with busbars, and (c) the assembled machine. Courtesy of the University of Stellenbosch, South Africa.",
        "texts": [
            "31) 164 5 AFPM Machines Without Stator Cores 5.6 Performance Characteristics 165 Specifications of e-TORQTM AFPM brushless motors manufactured by Bodine Electric Company , Chicago, IL, U.S.A. are given in Table 5.2. The EMF constant, torque constant, winding resistance and winding inductance are lineto-line quantities. Motors are designed for a maximum winding continuous temperature of 130oC. Steady-state performance characteristics of a 356-mm, 1-kW, 170-V e-TORQTM motor are shown in Fig. 5.6. Fig. 5.7 shows an air-cooled 160-kW AFPM brushless generator with a coreless stator built at the University of Stellenbosch, South Africa [260]. The stator winding consists of sixty single-layer trapezoidal-shape coils, which have the advantage of being easy to fabricate and have relatively short overhangs (Fig. 3.17). The winding coils are held together to form a disc-type stator by using composite material of epoxy resin and hardener. Sintered NdFeB PMs with Br \u2248 1.16 T and maximum allowable working temperature around 130oC have been used"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001415_978-3-319-74528-2-Figure1.8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001415_978-3-319-74528-2-Figure1.8-1.png",
        "caption": "Fig. 1.8 Centrifugal governor as example of negative feedback (public domain)",
        "texts": [
            " The effect of negative feedback is a force that reduces deviations, diminishes majorities, and damps overshoots. Examples are control systems such as the centrifugal governor or the baroreflex which keeps the blood pressure constant. An increased blood pressure effects a reduction of the heart frequency which then decreases the blood pressure. A too low blood pressure, in turn, inhibits the baroreflex which then increases the blood pressure. The centrifugal governor is a nice piece of engineering that can be used to regulate the rotational speed of a steam engine (see Fig. 1.8). The centrifugal force lifts up masses that in turn regulate the intake of steam via a valve. In swarm robotics we can make use of negative feedback, for example, to keep the swarm in a certain equilibrium state or to keep the swarm away from too extreme system states. 1.3 Self-Organization, Feedbacks, and Emergence 19 A dynamic example is Rayleigh-Benard convection. When a relatively flat layer of a fluid (e.g., oil or water) is heated homogeneously from below, flows of warm and cool fluid are generated"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001252_j.ymssp.2007.12.001-Figure9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001252_j.ymssp.2007.12.001-Figure9-1.png",
        "caption": "Fig. 9. The spur gear rig.",
        "texts": [
            " Randall / Mechanical Systems and Signal Processing 22 (2008) 1924\u20131951 1937 In order to compare the simulation results to those obtained experimentally, the different types of faults were tested under a 50-Nm load, while setting the output shaft speed to 10Hz (600 rpm). Vibration signals were collected using an accelerometer positioned on the top of the gearbox casing above the defective bearing. The 1.35-s (65,536 samples) signals were sampled at 48 kHz. A photo-reflective switch was placed near the output shaft to measure its speed by providing a once per rev tacho signal. The torque for each case was measured at the input shaft. A photograph of the rig showing the position of the accelerometer and the encoder at the output shaft is shown in Fig. 9. The bearings under test (Fig. 10) are double row self-aligning (Koyo 1205) with an angle of contact of 01, a ball diameter of 7.12mm and a pitch diameter of 38.5mm. The defect frequencies (frequencies at which the ball passes the defect on the outer race (BPFO) or the inner race (BPFI)) can be estimated for the outer race as 48.9Hz and for the inner race as 71.1Hz. The fundamental train frequency (FTF) is estimated at 4.1Hz, while the speed at which the balls spin (BSF) is 26.5Hz. ARTICLE IN PRESS N"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001599_iros.2010.5648837-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001599_iros.2010.5648837-Figure4-1.png",
        "caption": "Fig. 4. Posture control is achieved by a virtual torque on the torso. The original contact forces (dashed) are modified (solid) to compensate for this virtual torque.",
        "texts": [
            " In practice, small feedback torques, \u03c4fb, are also added to bias the joint angles and velocities to desired angles and velocities, \u03c4fb = Kp (qdes \u2212 q) +Kd (q\u0307des \u2212 q\u0307) (25) For a standing balance task where the robot is pushed from behind, the result of the DBFC controller is shown in Figure 3. This figure shows the total desired and measured FX and MY forces, as well as the state of the robot COM after the push. Maintaining torso posture can be a desireable goal during standing balance. Often this is handled by special consideration of the hip joint torque [12]. This example shows how torso posture control can be achieved via DBFC-VMC and implemented on a humanoid robot. The posture, or torso angle, can be corrected by a torque, Mtorso, to the torso, as shown in Figure 4. This torque can be written as a simple PD controller, Mtorso = Kp (\u03b8pos des \u2212 \u03b8pos)\u2212Kd\u03b8\u0307 pos (26) where \u03b8pos is the orientation of the torso. Using (21), F\u0302 can be determined by solving[ D1 D2 ] F = ( mC\u0308des + Fg Mtorso ) (27) and (22) can be used to determine the joint torques, G ( q\u0308 \u03c4dbfc ) = ( \u2212N(q, q\u0307) + JT F\u0302 + \u2211 i J T torsoMtorso \u2212J\u0307 q\u0307 ) (28) where Jtorso is the Jacobian associated with the torso body to which the virtual torque is applied. This controller was applied to the Sarcos Primus humanoid robot, which used an inertial measurement unit (IMU) attached to the hip to measure torso angle with respect to ground"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000840_j.electacta.2012.02.087-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000840_j.electacta.2012.02.087-Figure1-1.png",
        "caption": "Fig. 1. Schematic depiction of the processes in a membraneless enzymatic fuel cell, w r",
        "texts": [
            " Whilst some effort has been focused on assembling multiple enzymes at the anode of an EFC to provide for extraction of up to 24 electrons from glucose [37\u201343], most EFC research has focused to date on use of a single enzyme at the anode to oxidise glucose to gluconolactone, Eq. (2), providing only 2 electrons per mole of glucose and a maximum reversible cell voltage of 1.18 V. The maximum cell voltages for EFCs are usually determined by the difference between the formal redox potentials (E\u25e6\u2032) of the redox enzyme cofactors, in the active site, utilised for the anode and cathode, depicted in the scheme in Fig. 1. C6H12O6 + 6O2 \u2192 6CO2 + 6H2O (1) C6H12O6 + 1/2O2 \u2192 C6H10O6 + H2O (2) 2.1. Enzyme electron transfer Redox enzymes consist of an apoenzyme (the protein component of an enzyme) and cofactor(s): small nonproteinaceous electroactive species. The presence of the cofactor ensures electron transfer between enzyme and substrate/co-substrate. The cofactor can be tightly bound within the enzyme structure or released from the enzyme active site during the reaction. Common cofactors for glucose-oxidising enzymes include flavin adenine dinucleotide (FAD), nicotinamide adenine dinucleotide (NAD) and pyrroloquinoline quinone (PQQ) (Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure9.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure9.1-1.png",
        "caption": "Fig. 9.1. 100-kW, 8-disc AFPM synchronous generator. Courtesy of TurboGenset , London, U.K.",
        "texts": [
            " Although the minimization of windage losses of high speed generators requires rotors with small diameters, a multidisc design has the advantages of modular and compact construction, low synchronous reactance, good voltage regulation and very high efficiency in the case of coreless stators. A multidisc AFPM high speed generator is usually driven by a gas microturbine . The turbine rotor and PM rotors are mounted on the same shaft. Such a generator has compact construction, low mass and very high efficiency. According to TurboGenset Company , U.K., a 100-kW, 60 000 rpm multidisc generator has an outer diameter of 180 mm, length of 300 mm and weighs only 12 kg (Fig. 9.1)1. The high frequency output is rectified to d.c. 1 ULEV-TAP Newsletter, No. 2, 2000, www.ulev-tap.org 282 9 Applications and then inverted to 50, 60 or 400 Hz a.c. (Fig. 2.20). The generator is totally air cooled. Armed forces are interested in applications of microturbine driven PM synchronous generators to battery chargers for soldiers. The low energy density of portable bagpack batteries imposes a major constraint and challenge to future infantry operations. With recent advancement in PM brushless machine technologies, a lightweight miniature generator set can minimize the 9"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000840_j.electacta.2012.02.087-Figure10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000840_j.electacta.2012.02.087-Figure10-1.png",
        "caption": "Fig. 10. 3D drawing of the modular stack half-cell showing the central reaction chamber with reference electrode inlet (bottom right) and solution filling ports, the working electrode plate with inlet for the GC electrode (bottom left), and the counter",
        "texts": [
            " Other studies on assembled EFCs mostly focus on power genration under a defined condition (pH, temperature, biocatalyst, urface, etc.), usually with little consideration of the cell voltage, urrent or power benchmarks required for powering devices. Comarison of response from such assemblies is thus problematic. ntroduction of standardised half-cells [50], and normalisation of esponses to electrode volume and BET estimated surface areas has een suggested, to permit systematic comparison of electrodes for FCs [49]. The standardised half-cell [50], presented in Fig. 10, was sed to compare reproducibility of GC surface areas, of voltametry of deposited polymethylene green, and of response of the eposited film to NADH oxidation, with a view to use in EFCs fueled y NADH-dependent dehydrogenase reactions. Results from the se of such a standardised methodology should prove invaluable or comparison of surfaces, mediators and enzymes in half-cells nd EFCs. A noteworthy contribution focused on systematic optimisation f biocatalyst and either mediator or carbon support, to provide for lucose/oxygen EFCs [17,34\u201336,81,82,119,153,157,165]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure3.14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure3.14-1.png",
        "caption": "Fig. 3.14. Shapes of PM rotors of disc-type machines: (a) trapezoidal, (b) circular, (c) semicircular.",
        "texts": [
            "38) is for the demagnetizing armature flux and the lower sign is for the magnetizing armature flux. 3.2 Rotor Magnetic Circuits 107 The coefficient of the PM leakage flux (3.9) can also be expressed in terms of permeances, i.e. \u03c3lM = 1 + \u03a6lM \u03a6g = 1 + GlM Gg (3.41) Magnetic circuits of rotors of AFPM brushless machines provide the excitation flux and are designed as: \u2022 PMs glued to a ferromagnetic ring or disc which serves as a backing magnetic circuit (yoke); \u2022 PMs arranged into Halbach array without any ferromagnetic core. Shapes of PMs are usually trapezoidal, circular or semicircular (Fig. 3.14). The shape of PMs affects the distribution of the air gap magnetic field and contents of higher space harmonics. The output voltage quality (harmonics of the EMF) of AFPM generators depends on the PM geometry (circular, semicircular, trapezoidal) and distance between adjacent magnets [83]. Since the magnetic flux in the rotor magnetic circuit is stationary, mild steel (carbon steel) backing rings can be used. Rings can be cut from 4 to 6 mm mild steel sheets. Table 3.13 shows magnetization characteristics B\u2013H of a mild carbon steel and cast iron"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000252_j.automatica.2006.10.008-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000252_j.automatica.2006.10.008-Figure2-1.png",
        "caption": "Fig. 2. Contraction property of some 2-sliding homogeneous controllers.",
        "texts": [
            " Recall that a region is dilation-retractable iff, with each its point ( , \u0307), it contains all the points of the parabolic segment ( 2 , \u0307), 0 1. As follows from Corollary 1 a small violation of the conditions of Proposition 1 preserves the finite-time stability of controller (14), if Km \u2212 C < 2/2, but still Km > C. Proposition 2. Let C < Km C+ 2/2. Then with sufficiently small C+ 2/2\u2212 Km controller (14) provides for the twistinglike convergence to the finite-time-stable 2-sliding mode \u2261 0. Proof. The corresponding trajectories and dilation-retractable sets are shown in Fig. 2b. Define once more = \u0307 + | |1/2 sign . The boundaries of the regions are the trajectories of the differential equation \u0308 = \u2212C + Km with > 0 and \u0308 = C \u2212 Km with < 0. Consider the combined curve 1\u20132\u20133\u20134 (Fig. 2b). With sufficiently small C + 2/2 \u2212 Km point 4 is closer to zero than point 1. This requirement forms the convergence condition. Due to the homogeneity of the system, this geometric condition does not depend on the placement of point 1 on the axis \u0307. Direct calculation will replace it with an algebraic condition. The absolute value |\u0308| is separated from zero by Km \u2212 C. Therefore, the convergence time to the smaller set is estimated by 3\u0307M/(Km \u2212 C), where \u0307M is the maximal value of |\u0307| in the corresponding dilationretractable set. Hence, the finite-time stability is obtained due to Theorem 1. A new controller is obtained when ( 1, 1) = ( 2, 2) and all components are not zero, especially interesting are the cases when r1 = r2, 4r1Km \u2212 2C > max( 2 1/ 2 1, 2 2/ 2 2). In that case the trajectory is confined between two parabolic segments (Figs. 1c, 2a). The control vanishes in that region. The corresponding dilation-retractable sets are shown in Fig. 2a. The convergence time to the smaller set is estimated by 2\u0307M/(2Kmr1 \u2212 C), where \u0307M is the maximal value of |\u0307| in the corresponding dilation-retractable set. The idea of that controller is close to that of the sliding sector control (Furuta & Pan, 2000). The corresponding proposition as well as its proof are obvious and are omitted. An important class of HOSM controllers comprises recently proposed so-called quasi-continuous controllers, featuring control continuous everywhere except of the HOSM = \u0307=\u00b7 \u00b7 \u00b7= (r\u22121) = 0 itself"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000492_02783649922066376-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000492_02783649922066376-Figure5-1.png",
        "caption": "Fig. 5. An analysis of the CoP. In the foot/ground interface, we have the normal forces (left) and the frictional tangential forces (center). The CoP is the point (P ) where the resultant Rn of the normal forces acts. At the CoP, the tangential forces may be represented by a resultant force Rt and a moment M . The ground-reaction force is R = Rn + Rt .",
        "texts": [
            " The CoP and GCoM are used both in the robotics literature as well as in biomechanics, and are often a source of misconception and confusion. We will pay particular attention to the concept of the ZMP, and show that it is identical to the CoP. We show that the FRI point better reflects postural instability in a dynamic situation compared with the CoP and the GCoM. Although the concept of \u201cCoP\u201d most likely originated in the field of fluid mechanics, it is frequently used in the study of gait and postural balance. The CoP is defined as the point on the ground where the resultant of the ground-reaction force acts. As shown in Figure 5, two types of interaction forces act on the foot at the foot/ground interface. They are the normal forces f ni , always directed upward (Fig. 5, left) and the frictional tangential forces f t i (Fig. 5, center). The CoP may be defined as the point P where the resultant Rn = \u2211 f ni acts. With respect to a coordinate origin O, OP = \u2211 qifni\u2211 fni , where q i is the vector to the point of action of force f i and fi is the magnitude of f i . The unilaterality of the foot/ground constraint is a key feature of legged locomotion. This means that f ni \u2265 0, which translates to the fact that P must lie within the support polygon. The resultant of the tangential forces may be represented at P by a force Rt = \u2211 f t i and a moment M = \u2211 r i \u00d7 f t i , where r i is the vector from P to the point of application of\u2211 f t i . The complete picture is shown in Figure 5, right. The stance foot of the biped robot is subjected to a resultant ground- reaction force R = Rn + Rt and a ground-reaction moment M . An analysis with a continuous distribution of ground-reaction forces was performed earlier (Coussi and Bessonet 1995; Espiau 1998). We point out that contrary to what appeared in Shih\u2019s work (1996), R, and not Rn, is the total ground-reaction force. Please note that the CoP is identical to what has been termed the \u201ccenter of the actual ground-reaction force\u201d (C-ATGRF) in a recent paper (Hirai, et al"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003785_j.jsv.2009.09.022-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003785_j.jsv.2009.09.022-Figure1-1.png",
        "caption": "Fig. 1. Components and basic geometric parameters of a simple planetary gear set (only one planet branch is shown).",
        "texts": [
            " The model considers a simple planetary gear set with N planets (typically N=3\u20137 for most automotive and aerospace applications) that are positioned in any angular spacing allowed by the geometric constraints. As N and the spacing angles of planets ci, (iA[1,N] with c1=0) together with the number of teeth of gears were shown to define amplitude modulations due to carrier rotation [4], it is essential that the dynamic model should have the capability to model a gear set with any combinations of the same parameters. The dynamic model of the planetary gear set shown in Fig. 1 allows each gear and carrier body j to translate in x and y directions, denoted by xj and yj, and rotate about its axis that remains normal to the transverse plane of the gears. The fluctuations of the same member about its nominal rigid-body rotation are defined by yj. Each planet (pi) is in mesh with the sun gear (s) and the ring (internal) gear (r) while it is supported through its axis by a bearing held by the carrier (c). In Fig. 1, only one of the N planet branches located at an arbitrary planet position angle ci is shown. In this figure, yjc (j=s,r,pi) are the nominal kinematic rotations and ej represents the initial position angle of the run-out (eccentricity) vector of each gear j. Here, es and er are both defined from the horizontal x-axis while epi is defined from the radial axis along the centers of the sun gear (O) and planet pi located at an angle ci. A gear pair (external or internal) sub-system and a pinion\u2013carrier pair sub-system, as shown in Fig",
            " Defining uj=rjyj (j=s,r) and upi=rpiypi as the coordinates in place of yj and ypi where rj and rpi are the radii of the gears, and further defining the dynamic gear mesh forces as Fjpi\u00f0t\u00de \u00bc cjp _pjpi\u00f0t\u00de \u00fe kjpi\u00f0t\u00depjpi\u00f0t\u00de; (1a) where pjpi(t) is the relative gear mesh displacement given as pjpi\u00f0t\u00de \u00bc \u00bdyj\u00f0t\u00de ypi\u00f0t\u00de coscji \u00bdxj\u00f0t\u00de xpi\u00f0t\u00de sincji uj\u00f0t\u00de djupi\u00f0t\u00de ejpi\u00f0t\u00de \u00fe Ejpi\u00f0t\u00de Epij\u00f0t\u00de (1b) equations of motion of a gear pair sub-system are written as mj \u20acyj\u00f0t\u00de \u00fe hjpi coscjiFjpi\u00f0t\u00de \u00bc 0; (2a) mj \u20acxj\u00f0t\u00de hjpi sincjiFjpi\u00f0t\u00de \u00bc 0; (2b) Jj r2 j \u20acuj\u00f0t\u00de hjpiFjpi\u00f0t\u00de \u00bc Tj Nrj ; (2c) mp \u20acypi\u00f0t\u00de hjpi coscjiFjpi\u00f0t\u00de \u00bc 0; (2d) mp \u20acxpi\u00f0t\u00de \u00fe hjpi sincjiFjpi\u00f0t\u00de \u00bc 0; (2e) Jp r2 p \u20acupi\u00f0t\u00de djhjpiFjpi\u00f0t\u00de \u00bc 0: (2f) In these equations, dj=1 for j=s (external s\u2013pi gear pair) and dj= 1 for j=r (internal r\u2013pi gear pair) and cji=ci djaj, as shown in Fig. 1, where aj is the transverse pressure angle of the gears. The external torque Tj in Eq. (2c) represents the torque externally applied on the sun or the ring gear. An additional function hjpi is applied as a multiplier to Fjpi(t) to account for any nonlinear effects that might take place due to tooth separations. Here, hjpi=1 when the gear teeth are in contact (i.e. pjpi(t)Z0) and hjpi=0 when gears loose contact (i.e. pjpi(t)o0). Also included in Eqs. (1a) and (1b) are Ejpi(t) and Epij(t) to represent manufacturing errors of gear j relative the planet pi and manufacturing errors of planet pi relative to gear j, respectively",
            " Such errors include gear eccentricities, pitch-line run-out errors, tooth spacing and indexing errors, which are approximated here to harmonic forms for both the s\u2013pi and r\u2013pi meshes as Espi\u00f0t\u00de \u00bc Essin om Zs t \u00fe es csi ; (9a) Epis\u00f0t\u00de \u00bc Episin om Zp t \u00fe epi as ; (9b) Erpi\u00f0t\u00de \u00bc Ersin om Zr t \u00fe er \u00fe cri ; (9c) Epir\u00f0t\u00de \u00bc Episin om Zp t \u00fe epi \u00fe ar : (9d) Here, the gear mesh frequency is defined as om=Zs|os oc|=Zr|or oc|=Zp|op oc|, where oj and Zj (j=r,s,p) are absolute nominal angular velocity and number of teeth of gear j and oc is the absolute nominal angular velocity of the carrier. Accordingly, the excitations Epis(t) and Epir(t) of pinion pi are at a frequency om/Zp that is equal to the rotational velocity of pi relative to c. Similarly, Espi(t) and Erpi(t) are at frequencies om/Zs and om/Zr, respectively, representing the rotational velocities of the sun and ring gears relative to the carrier. Coefficients Es, Er and Epi are the amplitudes of these errors. The additional phase angles es, er and epi illustrated in Fig. 1 define the initial orientation of them, as previously discussed. In the absence of errors defined in Eqs. (9a)\u2013(9d), gear transmission error excitations defined as a part of the gear mesh interfaces shown in Fig. 2(a) and (b) are given in periodic form as [27] espi\u00f0t\u00de \u00bc XL \u2018\u00bc1 e\u00f0\u2018\u00desp sin\u00bd\u2018omt \u00fe \u2018Zsci \u00fe f\u00f0\u2018\u00desp ; (10a) erpi\u00f0t\u00de \u00bc XL \u2018\u00bc1 e\u00f0\u2018\u00derp sin\u00bd\u2018omt \u00fe \u2018Zrci \u00fe \u2018gsr \u00fe f\u00f0\u2018\u00derp (10b) where gsr is the phase angle between espi(t) and erpi(t). If the dynamic model proposed in the previous section were to be exercised with excitations as defined in Eqs"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure5.32-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure5.32-1.png",
        "caption": "FIGURE 5.32 Single-loop inductance example.",
        "texts": [
            " There, central distances are varied to check how accurate the complex inductance computation could be based on the fast multipole algorithm when it is compared with the direct computation results. Figures 5.30 and 5.31 show that excellent results are obtained. And the method could be used for close-coupling computation. The truncation number in this case is L = 6. PROBLEMS 5.1 Partial mutual inductance for two short wires Derive the formula for the partial mutual inductance for the two-wire geometry (C.2) in Appendix C. Make sure that your result agrees with the formula in (C.8). 126 INDUCTANCE COMPUTATIONS PROBLEMS 127 5.2 Single-loop inductance Compute the inductance of the loop in Fig. 5.32 in terms of the following partial inductance matrix. All the orthogonal partial inductances are zero as indicated. Compute the inductance L = VAB\u2215(sI(s)) of the loop by solving the circuit equations. Note that the VAB = V1 + V2 + V3 + V4. Use the sign rule given in Section 5.2. \u23a1\u23a2\u23a2\u23a2\u23a3 V1 V2 V3 V4 \u23a4\u23a5\u23a5\u23a5\u23a6 = \u23a1\u23a2\u23a2\u23a2\u23a3 Lp11 0 +Lp13 0 0 Lp22 0 +Lp24 +Lp31 0 Lp33 0 0 +Lp42 0 Lp44 \u23a4\u23a5\u23a5\u23a5\u23a6 \u23a1\u23a2\u23a2\u23a2\u23a3 sI1 sI2 sI3 sI4 \u23a4\u23a5\u23a5\u23a5\u23a6 . (5.78) 5.3 Inductance for loop with a gap For Problem 5.2, introduce a gap by reducing the length of branch 4 of the loop in the previous problem by half. Calculate the loop inductance for a 5 cm square loop with a square cross section of 1 mm. Compare the closed-loop inductance with the open-loop inductance with the 2.5-cm branch 4. For such an open geometry, different formulas can be used for the partial self- and mutual inductances. Round wires or rectangular ones can be used. 5.4 Long loop If the loop in Fig. 5.32 is oblong, or nonsquare, the opposite partial self-inductances are equal resulting in Lp11 = Lp33 and Lp22 = Lp44. Therefore, (5.23) is reduced to L = 2Lp11 \u2212 2Lp13 + 2Lp22 \u2212 2Lp24. (5.79) However, if the loop is very long in the \ud835\udcc1 direction and the wires are almost touching, then we can ignore the small values of 2Lp22 \u2212 2Lp24. Make comparisons for which aspect ratio of \ud835\udcc1\u2215w this contribution becomes small compared to the length inductances. 5.5 Two-loop transformer Compute the inductances L of the two-loop transformer using the following partial inductance matrix: [ VAB VCD ] = [ L11 L12 L21 L22 ] [ sI1 sI2 ] "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002467_tsmc.2021.3050616-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002467_tsmc.2021.3050616-Figure2-1.png",
        "caption": "Fig. 2. Kinematic model and tire model.",
        "texts": [
            " To aim at the lateral force of the wheel, we assume that lateral force is proportional to the angle, then F\u03beF = J\u03beF\u03c8\u03beF (6) F\u03beB = J\u03beB\u03c8\u03beB (7) \u03c8\u03beF = \u03b2 + H\u03c9c 2vx \u2212 \u03b4\u03beF (8) \u03c8\u03beB = \u03b2 + H\u03c9c 2vx \u2212 \u03b4\u03beB (9) where \u03be = L,R represents left and right, respectively. F\u03beF , F\u03beB, J\u03beF , J\u03beB, \u03c8\u03beF , and \u03c8\u03beB denote the lateral force of the inner and outer wheels, the cornering stiffness of the inner and outer wheels, and the slip angle, respectively. Consider the counterclockwise as the positive direction, and then the dynamic model can be addressed as follows: my\u0308 = FLF cos \u03b4LF + FRF cos \u03b4RF + FLB cos \u03b4LB + FRB cos \u03b4RB (10) Iz\u03d5\u0308 = H 2 (FLF cos \u03b4LF + FRF cos \u03b4RF \u2212 FLB cos \u03b4LB \u2212 FRB cos \u03b4RB). (11) Fig. 2 exhibits the kinematic model and wheel model \u23a1 \u23a3x\u0307c y\u0307c \u03d5\u0307c \u23a4 \u23a6 = \u23a1 \u23a2\u23a2\u23a3 cos(\u03d5c + \u03b2) sin(\u03d5c + \u03b2) 2 tan \u03b4 H(1 + tan \u03b4) \u23a4 \u23a5\u23a5\u23a6vc (12) \u03c9c = \u03d5\u0307c = 2 tan \u03b4 H(1 + tan \u03b4) vc (13) where (xc, yc) is the x-axis position and y-axis position of the robot, respectively. \u03d5c denotes the course angle and \u03b2 denotes Authorized licensed use limited to: Rutgers University. Downloaded on May 17,2021 at 03:04:21 UTC from IEEE Xplore. Restrictions apply. the slip angle. \u03c9c is the yaw rate. \u03b4 is the turning angle and vc is the line velocity. Further simplifying the model as \u03be\u0307kin = f (\u03bekin, ukin) (14) where the state variable \u03bekin = [xc, yc, \u03d5]T and the control variable ukin = [vc, \u03c9c]T . It can be seen that the control target of the robot position is determined by \u03c9c and \u03b2. On the other hand, it is necessary to manage the slip rate, so we describe the wheel model shown in Fig. 2 as follows: J\u03c9\u0307 = RFx \u2212 RF\u03c9 \u2212 Tb (15) Mv\u0307 = \u2212Fx \u2212 Fv (16) where R denotes the wheel radius. v denotes the robot leading speed, and \u03c9 denotes the tire angular velocity. M is the robot quality and J is the rotary inertia. Tb,Fx,Fv,F\u03c9,F\u03c9, and Fz represent the braking torque, road friction, air friction, rolling resistance, and ground reaction force, respectively. To obtain the wheel state model, we define the slip rate and road friction as S = v \u2212 \u03c9R v (17) Fx = \u03bcFz (18) where S is the slip rate and \u03bc denotes the adhesion coefficient",
            " For example, when the robot turns at a low adhesion condition, the driving torque of the wheel easily exceeds the limit value of the adhesion, dramatically causing the driving performance of the robot, which affects the stability and safety. In order to further implement soft constraints, we introduce a relaxation factor to dynamically adjust the constraint range to achieve better robustness and versatility ay,min \u2212 \u2202 \u2264 ay \u2264 ay,max + \u2202 (62) where \u2202 is the relaxation factor, and ay,min and ay,max are the limit of longitudinal acceleration. Thus, the ideal yaw rate can be presented as \u03c9\u2217 r = min { \u03c9d, \u2223\u2223\u2223\u2223\u03bcg vx \u2223\u2223\u2223\u2223 } \u2217 sign(\u03b4in) (63) where sign(\u2022) is the signum function. In Fig. 2, we can define the lateral acceleration and slip angle as y\u0308 = ay = v\u0307y + vx\u03c9r (64) \u03b2 = arctan vy vx \u2248 vy vx . (65) Considering vx as a constant, we transfer (65) with time derivative to v\u0307y = vx\u03b2\u0307 (66) where \u03b2 and \u03c9r can be derived from the dynamic model. The sideslip angle \u03b2 will directly affect the robot driving, so it is necessary to constrain it within a reasonable range. According to the vehicle stability study conducted by BOSCH [42], the limit of the sideslip angle is \u00b112\u25e6 on the road with good adhesion conditions, while the limit value is Authorized licensed use limited to: Rutgers University"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002797_j.jsv.2021.116029-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002797_j.jsv.2021.116029-Figure5-1.png",
        "caption": "Fig. 5. A friction-vibration dynamic model of ACBB.",
        "texts": [
            " The elastic hysteresis friction torque is represented as M h = 1 \u03c9 i N b \u2211 j=1 [ E hi + E ho ] (31) where, E hi and E ho are the energy loss produced by the elastic hysteresis for inner and outer raceways, respectively. The differential sliding friction torque is represented as M d = 1 2 \u03c0\u03c9 i N b \u2211 j=1 A j (32) where A j is the power from j th ball produced by the differential sliding. More details about the variables appearing in above equations are listed in Ref. [24] . The friction-vibration dynamic model of an ACBB is depicted in Fig. 5 . In this model, the translational motions of inner race in X, Y , and Z axes; the translational motions for balls in y j and z j axes; the rotations of balls around the x j , y j , and z j axes; the revolutions of balls and cage around the Z axis are considered. The external loads are located on the inner race in X, Y , and Z axes. The assumptions and considerations are: 1) The raceways are assumed as rigid ones; 2) The outer raceway is immobile; 3) The rotations of inner race around X and Y axes are ignored; 4) The ball-raceway contact follows Hertzian contact theory"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003628_91.705510-Figure10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003628_91.705510-Figure10-1.png",
        "caption": "Fig. 10. Structure of a double-inverted pendulum system.",
        "texts": [
            " Nevertheless, the duration of the action may not be long enough to reduce the speed of the cart to zero, as the cart passes through the origin. The value of must not be too large, otherwise the cart will be always oscillating around the origin. In the simulation, the following specifications are used: kg, kg, m, m/s , , , , , , , . and initial values are Figs. 7\u20139 are obtained with . It is possible to see that the pole and the cart are stabilized to the equilibrium. The structure of a double inverted pendulum system is illustrated in Fig. 10. Pole 1 is the pole connected to the cart and pole 2 is the one above pole 1. The system\u2019s dynamics is represented by (34) where angle of pole 1 with respect to the vertical axis; angular velocity of pole 1 with respect to the vertical axis; angle of pole 2 with respect to the vertical axis; angular velocity of pole 2 with respect to the vertical axis; position of the cart; velocity of the cart and the form of , , , , , are shown in Appendix B. The control objective is twofold: to keep the angle of both poles (1 and 2) close to zero in an upright position and to allow the cart to move freely without any boundary limits"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003964_tnn.2005.849824-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003964_tnn.2005.849824-Figure3-1.png",
        "caption": "Fig. 3. Trajectories of y(t) and y (t) (case 1) in Example 1.",
        "texts": [],
        "surrounding_texts": [
            "CONTROLLER VIA OUTPUT FEEDBACK In this section, our primary task is to design an observer that estimates the state vector in (10), to use the fuzzy-neural network to approximate the control and to the develop direct adaptive output-feedback update law to adjust the parameters of the fuzzy-neural network in order to achieve the control objective. First, we replace in (9) by the output of the fuzzy-neural network, in (12), i.e. (14) where . Next, consider the following observer that estimates the state vector in (10): (15) where is the observer gain vector, chosen such that the characteristic polynomial of is strictly Hurwitz because is observable. The control term is employed to compensate the modeling error. Although the state observer (15) includes the unknown function , a significant part of our design is that in Theorems 1 and 2, and so we can eliminate the unknown function from the state observer (15). Therefore, the proposed state observer (28) in Theorem 2 of this paper does not require the function , which is assumed as an unknown function. Define the observation errors as and . Subtracting (15) from (10), we have (16) where . Besides, the output error dynamics of (16) can be given as (17) where is the Laplace variable, and is the transfer function of (16). In order to derive the direct adaptive output feedback update law, the following assumption must be required. Assumption 1 [33]: Let and belong to compact sets and , respectively, where denotes the estimate of and and are the upper bounds of and , respectively. It is known that the optimal parameter vector lies in some convex region , where the radius is a design parameter. According to Assumption 1, (16) can be rewritten as (18) where is an approximation error. According to (14), (18) can be rewritten as (19) where . Since only the output in (19) is assumed to be measurable, we use the strictly-positive-real (SPR) Lyapunov design approach to analyze the stability of (19) and generate the direct adaptive output-feedback update law for . Equation (19) can be rewritten as (20) where is a known stable transfer function. In order to employ the SPR-Lyapunov design approach, (20) can be written as (21) where , , and . is chosen so that is a proper stable transfer function and is a proper SPR transfer function. Suppose that , where , such that is a proper SPR transfer function. The state\u2013space realization of (21) can be written as (22) where , and . For the purpose of stability analysis of the observer-based direct adaptive fuzzy-neural controller, the following assumptions and lemma must be required. Lemma 1 [14], [27]: Suppose that the adaptive laws are chosen as (23), shown at the bottom of the page, where the projection operator [14] is given as Then and . Assumption 2: The unknown function is bounded by (24) where and positive constants. Assumption 3: is assumed to satisfy (25) where is a positive constant. Remark 1: Due to (6) and (7), and the existence of in [32], the assumption of boundedness of in Assumption 1 is reasonable. Since denotes the estimation of , the assumption of boundedness of is also reasonable. The boundedness of follows that of (or ). The would be differentiable, if we choose differentiable functions, for example, exponential functions, to be the fuzzy membership functions. From Lemma 1, the vector is adjustable and differentiable. Therefore, is differentiable. Since is bounded, is bounded. Therefore, Assumption 3 is reasonable. The goal of this paper is to develop a controller , where is designed to approximate the controller in [32] and the control term is employed to compensate the modeling error, to guarantee that all signals involved are bounded and the output of the closed-loop system will asymptotically track the desired output trajectory. On the basis of the previous discussions, the following theorems can be obtained. Theorem 1: Consider the system (22) that satisfies Assumptions 1\u20133. Let be adjusted by the update law (23), and let be given as if if (26) where . Then converges to zero as . Proof: Given in the Appendix. Theorem 2: Consider the nonlinear system (1) that satisfies Assumptions 1\u20133. Suppose that the control law is (27) if or and if and (23) with the adaptive law (23). Let . The state observer (15) becomes (28) Then all signals in the closed-loop system are bounded, and converges to zero as . Proof: Given in the Appendix. According to the previous theorems, the design algorithm of the direct adaptive fuzzy-neural controller is described as the following. Design Algorithm Step 1) Select the feedback and observer gain vectors , such that the matrices and are Hurwitz matrices, respectively. Step 2) Choose appropriate values in (26), and in (23). In order to remedy the control chattering, (26) can be modified as if and , if and , if where is a positive constant. Step 3) Solve the state observer in (28). Step 4) Construct fuzzy sets for . From (13), compute the fuzzy basis vector . Step 5) Obtain the control law (27), and the update law (23). Remark 2: The initial values of should be determined before solving the adaptive laws in (23). The value of in (23) is obtained by trial and error according to the values of . In addition to compute the controller in (27), we need to decide . The chosen value of is obtained by trial and error such that and based on Assumptions 2\u20133, without using any adaptive tuning procedure in this paper. Larger results in larger control input according to (27). From (26), we see that the absolute value of the control term is the value of . The control term is employed to compensate for external disturbance and modeling error. Remark 3: Regarding Step 4) of the design algorithm, the number of fuzzy rules depends on the number of inputs and the number of fuzzy sets of each input. For example, we can generate fuzzy rules for inputs, in which each input has fuzzy sets. The input vector is . The membership function of each fuzzy set can be a bell-shaped form or others. Then, we can compute the values of the fuzzy bases from (13). To summarize, Fig. 2 shows the overall scheme of the observer-based direct adaptive fuzzy-neural control proposed in this paper. IV. ILLUSTRATIVE EXAMPLES This section presents the simulation results of the proposed direct adaptive fuzzy-neural controller to illustrate that the stability of the closed-loop system is guaranteed, and all signal involved are bounded. Example 1: Consider the nonlinear system (29) The control objective is to control the state of the system to track the reference trajectory (case 1) and (case 2). The design parameters are selected as , 5, and . The feedback and observer gain vectors are given as and , respectively. The filter is given as . The membership functions for , are given as The initial states are chosen to be (case 1), (case 2), (case 1), and (case 2). The computer simulation results are shown in Figs. 3\u20138. From Figs. 4 and 7, it is observed the state observer can generate the estimated state very fast and correct. Moreover, as shown in Figs. 3 and 6, it is observed that the tracking error is small, and the convergence of tracking error is fast and well. The control signals for two cases are shown in Figs. 5 and 8. The computer simulation results show that the observer-based direct adaptive fuzzy-neural controller can perform successful control and achieve desired performance for the nonaffine nonlinear systems. Example 2: Consider the nonlinear system [23] (30) The control objective is to control the state of the system to track the reference trajectory . The design parameters are selected as , 5, and . The feedback and observer gain vectors are given as and , respectively. The filter is given as . The membership functions for , are the same as those in Example 1. The initial states are chosen to be and . The computer simulation results are shown in Figs. 9\u201311. From the simulation results, it is observed that the state observer can generate the estimated state very fast and correct. Moreover, it is also observed that the tracking error is small, and the convergence of tracking error is fast and well. In comparison with the control input in [23], using the high gain observer and saturation method to overcome the peaking phenomenon in the transient behavior, the proposed control input shown in Fig. 11 is without saturation and smoother than that in [23], especially during the transient period."
        ]
    },
    {
        "image_filename": "designv10_0_0001662_j.jmapro.2020.07.025-Figure13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001662_j.jmapro.2020.07.025-Figure13-1.png",
        "caption": "Fig. 13. Finite element modelling of the DED process: (a) thermal analysis illustrating inactive or quiet elements from [220], (b) metallurgical analysis reproduced from [194], with the permission of the Laser Institute of America, (c) mechanical analysis from [219]. All images reproduced with permission.",
        "texts": [
            " Software such as ABAQUS/CAE offer coupled temperature-displacement analyses to model the residual stresses in one simulation but are often much more computationally expensive with little increase in precision. A study by Marion et al. [194] used this sequentially coupled approach to observe the residual stresses generated through AM deposition. A thermal analysis was conducted and the resulting temperature histories were used in a mechanical and metallurgical analysis. Phase volume fraction changes due to the temperature change from the metallurgical analysis were also input into the mechanical analysis for a more accurate representation of calculated stresses. Fig. 13 illustrates the thermal, mechanical and metallurgical models for the AM process and their coupling relationship. The solid arrows represent the included physical phenemona such as: the effect of temperature on phase changes (1) and mechanical deformation (5), and the phase volume fraction influence on the mechanical behavior (3). The dashed arrows represent small contributions to the model compared to the laser source which can be ignored such as: latent heat of fusion (2) and plasticity induced thermal dissipation (6)",
            " The quiet method employs a different approach in that the thermal conductivity and Young's modulus of inactive elements are scaled down to near-zero values so they have very little effect on the printed material. As elements are activated, these values are restored to their rightful values however, all elements (inactive and active) are included in the global stiffness matrix which increases computation times. Additionally, the scaling down of material properties can yield instabilities within the Jacobian matrix which can cause convergence issues during computation [234]. Fig. 13 illustrates thermal, metallurgical and mechanical analyses of the DED process which employ this type of element activation. Thermal modelling of the AM processes requires a transient analysis due to the non-homogeneous rapid heating and cooling that occurs after the laser comes in contact with the metal [223]. This process requires simulation of a moving heat source (laser or electron beam) through the use of user defined subroutines and scripts [224]. In many cases, heat transfer through mass transport is neglected [220] and some studies ignore heat loss due to radiation [219] since convective and conductive heat losses are much larger for laser based systems",
            " The accuracy of the thermal analysis is the most important component for the whole simulation since it determines the corresponding microstructure, mechanical properties and dimensional accuracy [223]. Moreover, the melt pool geometry has a significant effect on the resulting microstructure, and thus an accurate representation of the laser-material interaction is crucial in predicting realistic residual stresses throughout the build [224]. As expected, the highest temperature experienced in the melt pool is directly below the laser penetration depth and decreases and the distance increases away from the heat source [238]. The temperature profile of the melt pool can be seen in Fig. 13a. A sequentially coupled model is most common in the literature, where a thermal analysis is initially conducted and the nodal temperature history is used as a thermal load in a mechanical analysis to calculate the residual stresses throughout the build [219,239]. The mechanical analysis in FEM software assumes small strain and deformation and use elasto-plastic constitutive model [233]. The elastic behavior is modeled with the isotropic formulation of Hooke's law using temperature dependent values for both the Young's modulus and Poisson's ration [239]",
            " Von Mises yield criterion is often used and a timeindependent kinematic plasticity model is often used for plastic behavior to capture the cyclic non-linear work hardening called the Baushinger effect [239,233]. In addition to the Young's modulus and Poisson's ratio, the temperature strain-hardening rates (plastic tangent modulus) and yield strengths are required [239,233] for plasticity modelling. An example of mapped thermal stresses for an as-built AM processed component produced by Hajializadeh et al. is shown in Fig. 13c. The metallurgical model is highly dependent on both the thermal history and the state of the material [194\u2013196]. Due to the cyclic heating that occurs during the AM processes, there is a complex set of phase transformations that occur spatially throughout the build such as HAZ (similar to welding) which greatly affect the mechanical properties. Due to the rapid solidification of the melt pool during the AM process, kinetics have a strong influence on the resulting microstructure and thus equations for diffusion and diffusionless transformations similar to alloy specific time-temperature-transformation (TTT) curves can be used to map the grain morphology based on temperature and cooling rate",
            " This expression is required to be updated to consider the eventual amount of \u03b1\u2019 at the beginning of the transformation and the retained \u03b2 phase [194]: = \u2212 \u2212 \u2212\u2032z \u03b3 M T1 exp( ( ))\u03b1 s (3) For this equation, \u03b3 is a constant and Ms is the martensite start temperature. It can be noted that the K\u2013M equation yields a parabolic shape instead of a sigmoidal shape [241] as observed on TTT diagrams. This dissimilarity causes values to deviate from experimental data significantly for the first 10\u201320% of martensite formation [241]. Fig. 13 illustrates the computed fractions of martensite of a DED built wall. Various models have been proposed which apply temperature as a variable but are still an approximation and yield some degree of error. Thus a more accurate approach could be to conduct a thermo-mechanical FEM model and incorporate the metallurgical model using software which can model the grain growth and nucleation of the metal during solidification. Recent work conducted by Zhang et al. [242] used the temperature history from an FEM analysis for a metallurgical analysis of the AM process"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure15.18-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure15.18-1.png",
        "caption": "Figure 15.18 (a) The simply supported beam AC with a slide or shear force hinge at midspan B. (b) A kinematically admissible virtual displacement: the segments to the right and left of the slide remain parallel to one another. (c) The displacements in case the rotation \u03b4\u03d5 is small and the quadratic terms in \u03b4\u03d5 can be neglected.",
        "texts": [
            " VOLUME 1: EQUILIBRIUM demand is formulated by looking only at the first-order variation of the displacements in the virtual work equation (in other words, only the linear terms in the virtual displacements). This limitation with respect to the virtual displacements means that the \u03b4u\u03b4\u03d5 is neglected, and the normal force pair N performs no virtual work. In diagrams, the dimensions of the structure are greatly reduced and the displacements, even though they are infinitesimally small, are greatly enlarged. This can give rise to problems at first sight. As an example, consider the simply supported beam ABC modelled as a line element in Figure 15.18a, with a slide or shear force hinge at midspan B. The mechanism has one degree of freedom. For a kinematically admissible virtual displacement, the displacement must be consistent with the freedom of movement at the supports and the slide. The latter means that beam segments AB and BC must remain parallel to one another on both sides of the shear force hinge. Figures 15.18b and 15.18c show the virtual displacement for the mechanism in two different ways. Figure 15.18b would seem to be the correct one, but this is not so. One has to imagine that the virtual rotation \u03b4\u03d5 is very small. In that case the horizontal displacement of the beam ends with respect to one another at B, (\u03b4\u03d5)2, is a degree smaller than the vertical displacement \u03b4\u03d5. So the displaced beam ends at the shear force hinge B can be drawn directly above one another. Figure 15.18c gives therefore the correct representation of the virtual displacements in the mechanism. Another example is beam ABC in Figure 15.19a, that has been changed into a mechanism by the introduction of the hinge at S. Figure 15.19b shows a virtual displacement: a bend \u03b4\u03b8 occurs at hinge S. The small angle (rotation) \u03b4\u03b8 , that is drawn to a large scale, can be defined by the ratio \u03b4a/ . This value is known as the orthogonal value. The orthogonal value is not equal to the sine or tangent of the angle, nor to the value in radians"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003799_pnas.0703530104-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003799_pnas.0703530104-Figure3-1.png",
        "caption": "Fig. 3. Schematic representation of a swimming path in two dimensions in a radial concentration field of chemoattractant. The swimming path r(t) (black line) is a drifting circle whose center moves along the centerline R(t) (red line). In the limit of weak concentration gradients with the parameter small (see text) this centerline is a logarithmic spiral whose tangent encloses a constant angle of with the gradient direction at every point. The distance R R of the centerline to the source and the polar angle are used in the formula of the logarithmic spiral (see text).",
        "texts": [
            " 8 for the centerline can then be generalized as R\u0307 vdM c / c 2 [10] where the operator M( ) rotates the vector c in the plane by the angle in a counterclockwise sense. Eq. 10 represents a differential equation for the centerline R(t) if the gradient c of the concentration field is evaluated at any time at position R(t). This equation for the centerline can be illustrated for the case of a radial concentration field c(x) C( x ). In this case, according to Eq. 10, the centerline R(t) winds around the origin in a logarithmic spiral, R( ) R0 exp( cot( ) ), where is the polar angle and R R (see Fig. 3). It spirals inwards for /2 /2, which corresponds to a phase shift of the signaling system of 2 . Otherwise, it spirals outwards. We have compared solutions to the dynamic equations (Eqs. 1\u20134) for the swimming path to solutions of Eq. 10 for the center line in our perturbation calculation and found good quantitative agreement for 0.1. Furthermore, Eq. 10 could predict perfectly whether the path reached the egg. Motion in Three-Dimensional Space. Our numerical solutions for swimming paths in three dimensions shown in Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001236_j.euromechsol.2007.11.005-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001236_j.euromechsol.2007.11.005-Figure2-1.png",
        "caption": "Fig. 2. Geometrical parameters for the fillet-foundation deflection.",
        "texts": [
            ", 2004) which assumes linear and constant stress variations at root circle. This analytical expression is given by: \u03b4f = F cos2 \u03b1m WE { L\u2217 ( uf Sf )2 + M\u2217 ( uf Sf ) + P \u2217(1 + Q\u2217 tg2 \u03b1m )} (6) The coefficients L\u2217, M\u2217, P \u2217, Q\u2217 can be approached by polynomial functions (Sainsot et al., 2004): X\u2217 i (hf i, \u03b8f ) = Ai/\u03b8 2 f + Bih 2 f i + Cihf i/\u03b8f + Di/\u03b8f + Eihf i + Fi (7) the values of Ai , Bi , Ci , Di , Ei and Fi are given in Table 1. W is the tooth width. uf , Sf , hf i = rf /rint and \u03b8f are defined in Fig. 2. The corresponding fillet-foundation stiffness can be obtained by: kf = F \u03b4f (8) From the results derived by Yang and Sun (Yang and Sun, 1985), the stiffness of Hertzian contact of two meshing teeth (commonly nonlinear) is practically a constant along the entire line of action independent to both the position of contact and the depth of interpenetration. kh can be approximated by a constant value depending on the tooth width and the mechanical properties of the gear material: kh = \u03c0EW 4(1 \u2212 \u03bd2) (9) The local deformation is then expressed by: \u03b4h = F (10) kh For a one pair of teeth in contact the gearmesh stiffness Ke can be written as: Ke = 1 /( 1 kb1 + 1 kf 1 + 1 kb2 + 1 kf 2 + 1 kh ) (11) Subscripts 1 and 2 denote respectively the pinion and the gear"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000946_j.addma.2014.10.003-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000946_j.addma.2014.10.003-Figure1-1.png",
        "caption": "Fig. 1. An illustration of the LENS deposition head and powder nozzles.",
        "texts": [
            " Calibration and validation depositions Single track thin walls of Ti\u20136Al\u20134V are deposited using an ptomec\u00ae LENS MR-7 system with a 500 W IPG Photonics ber laser. The deposition occurs in a chamber with an argon tmosphere that has an oxygen content of less than 15 parts per illion. A 30 L/min argon jet is used to supply argon to the hamber, to protect the laser optics, and to shield the melt pool. he Ti\u20136Al\u20134V powder delivered to the melt pool is assisted y four argon jets that have a combined flow rate of 4 L/min. hese four jets exit nozzles positioned around the main nozzle nd aimed at the melt pool, as shown in Fig. 1. The powder has een sieved so that only particles with diameters between 44 and 49 m are delivered at a rate of 3.0 g/min. A LENS system is hosen because several researchers have investigated the effect f its processing parameters on the deposition geometry [42] nd the material properties [43,44] of Ti\u20136Al\u20134V. The model is validated using three different depositions, as hown in Fig. 2. Each case builds a wall that is designed to e 38.1 mm long, 12.7 mm tall, and 3 mm wide. These cases roduce different thermal and mechanical results that are used o validate the model: "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure12.5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure12.5-1.png",
        "caption": "FIGURE 12.5. A mass m hanging from a flexible frame.",
        "texts": [
            " keq = k1 + k2 + k3 (12.18) Parallel dampers have the same speed x\u0307, and a resultant force fc equal to the sum of individual forces. We may substitute parallel dampers with only one equivalent damping ceq that produces the same force fc under the same velocity. Consider three parallel dampers such as is shown in Figure 12.4. Their force balance and equivalent damping would be fc = \u2212c1x\u0307\u2212 c2x\u0307\u2212 c3x\u0307 (12.19) fc = \u2212ceqx\u0307 (12.20) ceq = c1 + c2 + c3. (12.21) 734 12. Applied Vibrations Example 425 Flexible frame. Figure 12.5 depicts a massm hanging from a frame. The frame is flexible, so it can be modeled by some springs attached to each other, as shown in Figure 12.6(a). If we assume that each beam is simply supported, then the equivalent stiffness for a lateral deflection of each beam at their midspan is k5 = 48E5I5 l35 (12.22) k4 = 48E4I4 l34 (12.23) k3 = 48E3I3 l33 . (12.24) When the mass is vibrating, the elongation of each spring would be similar to Figure 12.6(b). Assume we separate the mass and springs, and then apply a force f at the end of spring k1 as shown in Figure 12"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000730_s0022112006002631-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000730_s0022112006002631-Figure3-1.png",
        "caption": "Figure 3. Diagram of the geometry for two near-separated squirmers. The vector ez denotes the direction of squirmer 1 from squirmer 2, while e\u03c1 points radially out from that vector and e\u03c6 is azimuthal.",
        "texts": [
            " Let the two spheres, 1 and 2 say, have radii a and \u03b1a, orientation vectors e1 and e2 and squirming sets B(1) and B(2), respectively. Let the z-axis pass through the two sphere centres and let z = 0 be the plane containing the point on sphere 2 closest to sphere 1 (so that sphere 2 is in the region z 0 and sphere 1 in z > 0). Let the surfaces of the spheres be determined by z =h1 and z = h2 (for spheres 1 and 2 respectively). Furthermore let the minimum separation of the spheres be \u03b5a (with \u03b5 1) and assume that \u03b1 =O(1) (i.e. the spheres are comparable in size). This is illustrated in figure 3. To find the difference in velocity one can consider a frame in which sphere 2 is fixed and the centre of sphere 1 moves with velocity V \u2212 \u2126 \u2032 \u2227 r , where V is the separation velocity (i.e. the velocity of sphere 1 less the velocity of sphere 2), \u2126 \u2032 is the angular velocity of sphere 2 and r is the separation vector between the two sphere centres (r = r1 \u2212 r2). The problem is governed by the Stokes equations, \u00b5\u22072u = \u2207p, \u2207 \u00b7 u = 0, (3.1) where the velocity of the fluid is u, its viscosity is \u00b5 and p is the pressure",
            " The first-order solution Integrating the z-component of the first equation in (3.4) shows that p0 = p0(X, Y ), which is as yet an arbitrary function (assumed to have continuous second partial derivatives). The following solutions for u0 and v0 can be found by integrating the xand y-components: u0 = 1 2 \u2202p0 \u2202X (Z \u2212 H1)(Z \u2212 H2) + Z \u2212 H2 H u0 \u00b7 ex, (3.13) v0 = 1 2 \u2202p0 \u2202Y (Z \u2212 H1)(Z \u2212 H2) + Z \u2212 H2 H u0 \u00b7 ey. (3.14) Here H = H1 \u2212 H2 = 1 + \u03b1 + 1 2\u03b1 \u03c12 + O(\u03b5), (3.15) is the distance between the two spheres\u2019 surfaces at \u03c1 from their line of centres (see figure 3). Using the equation of continuity, these solutions can be combined to yield H 3 12 \u22072 \u22a5p0 + H 2 4 \u03b1 + 1 \u03b1 \u03c1 \u00b7 \u2207\u22a5p0 \u2212 \u03b1 \u2212 1 2\u03b1 \u2211 n BnWn(\u2212e \u00b7 ez)e \u00b7 \u03c1 = 0. (3.16) We seek a solution to (3.16) of the form p0(\u03c1, \u03c6) = q0(\u03c1) e \u00b7 e\u03c1 (3.17) and find that q0 satisfies a linear second-order ordinary differential equation in one variable, \u03c1: H 3 12\u03c12 \u2202 \u2202\u03c1 ( \u03c1 \u2202q0 \u2202\u03c1 ) \u2212 H 3 12\u03c13 q0 + H 2 4 \u03b1 + 1 \u03b1 \u2202q0 \u2202\u03c1 \u2212 \u2211 n \u03b1 \u2212 1 2\u03b1 BnWn(\u2212e \u00b7 ez) = 0. (3.18) This is the Reynolds equation for the problem (see Kim & Karrila 1992)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001584_elan.201000545-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001584_elan.201000545-Figure3-1.png",
        "caption": "Fig. 3. A schematic representation of the Brillouin zone in graphite (A), illustrating the long pockets of holes and electrons that run from H to K to H and from H\u2019 to K\u2019 to H\u2019 (after [35] and [37]). A schematic illustration of the upper half of a Fermi surface in graphite (B), from K to H only. (Reprinted from [35], Figure 7, with kind permission from Springer Science+Business Media.)",
        "texts": [
            " Of these four bands, one is empty, one is nearly full of electrons and the other two are degenerate. It is the shape of the degenerate bands near the Fermi energy that creates the pockets of electrons and holes responsi- ble for graphite s conductivity [35,36]. The resulting Fermi surfaces are long and narrow cigar-like shapes, running down the six edges of the hexagonal-prism Brillouin zone, from the top corners, H, to the midpoints of the edges, K, and down to the other H points at the bottom corners (Figure 3A). Along each edge, two pockets of holes are found, one at each H point; in between these lies the pocket of electrons, which contacts the hole pockets at each end through four narrow legs (Figure 3B). The strong directionality of graphite s bonding, structure and electronic properties is echoed in its electrochemistry. Heterogeneous electron transfer occurs far more rapidly at the edges of each graphite layer (a typical rate constant, k, is approximately 0.02 cm s 1) than at pristine basal planes (k<10 9 cm s 1) [38,39]. Consequently, the electrochemical activity of graphite-based electrodes, for most redox reactions, is effectively due to the edge planes only, despite the fact that the surface area of edges on some graphitic electrodes can be less than 0"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000637_978-3-642-83006-8-Figure3.34-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000637_978-3-642-83006-8-Figure3.34-1.png",
        "caption": "Fig. 3.34. Walk simulation, ZMP deviation compensated",
        "texts": [
            " It is interesting to note that the smallest torque deviations occur at the ankle joint which additionally controls the ZMP motion. The most interesting is Fig. 3.33. e), illustrating the ZMP behaviour. It can be noticed that ZMP excursions from the nominals are smallest upto now. Note that the scale for 6XZMP is enlarged approximately three ti mes comparing to previous cases, and it will be used for all subsequ entent examples. But, the behaviour of other joints, especially of the ankle, is not spoiled much by such very \"strong\" keeping ZMP position under control. Next example, Fig. 3.34, considers the same gait as in Fig. 3.33, but Gk for k ZMP = 1. After initial oscillations, the behaviour of ZMP is even better, and for 10 15 1 = 3 [rad/s 2 ] it almost coincides with the nomimaX nal value. For an easier feedback gain the same 0 15 maX comparison of ZMP behaviour with respect to the global value, k~~p, in Fig. 3.35. are given the diagrams for Gk Gk and k ZMP = 0.5 and k ZMP = 1. Gk As can be seen, the larger the k ZMP values are, the smaller are the 272 ankle joint, k~~p = 0.5, t:",
            " At the end of the consid,ered period of time, the mechanism is not ready for supporting foot switching. The gait has to be continued in single support phase till legs d,isposition is aPpropriate to start the d,ouble support phase. The next two stick diagrams (Figs. 3.40. and 3.41) illustrate the mec hanism behaviour fOr the initial disturbance assigned to the ankle joint. Fig. 3.40. shows the Case when the ZMP deviations are compensated by the ank.le joint, and. Fig. 3.41. shows cOmpensation by trunk joint. 278 Fig. 3.34. for In15 1=7 [rad/sl. ZMP displacement max 15 compensated by ankle, 6q (0) = 0.2 [radl Fig. 3.39. Stick diagram of the walk simulation presented at Fig. 3.32. for In 15 1=7 [rad/s 21. ZMP displacement max 15 compensated by hip, 6q (0) = 0.2 [radl Fig. 3.37. for In4 1=7 [rad/s2 ]. ZMP displacement max 4 \\ \\ \\ \\ \\ \\ 1 t II 1\\ 1 compensated by ankle, 6q (0)=0.1 [rad] , \\ \\ Fig. 3.41. Stick diagram of walk simulation presented at Fig. 3.30. for In4 1=7 [rad/s 2 ]. ZMP displacement max 4 compensated by trunk, ~q (0) = 0"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure6.14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure6.14-1.png",
        "caption": "Figure 6.14 (a) A distributed load, at an angle to a member, can be resolved into components (b) parallel, and (c) normal to the member axis.",
        "texts": [
            " In the special case that the distributed load is constant, we refer to a uniformly distributed load (see Figure 6.12c). The distributed load in Figure 6.12d is known as a linearly distributed load; it varies linearly from q(x1) = 3 kN/m to q(x2) = 5 kN/m. A distributed load can also act in the direction of the member axis. Figure 6.13, for example, shows the uniformly distributed load q on a column as a result of its dead weight. A distributed load q , acting at an angle to a member, can be resolved into directions parallel to and normal to the member axis (see Figure 6.14). In the xz coordinate system shown, qx = q cos \u03b1 and qz = q sin \u03b1 are called the components of q . 220 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Note: the distributed load has the dimension of force per length. So far, the length was always measured along the member axis. With inclined members, the distributed load is also sometimes expressed per length projected on the (horizontal) ground surface, for example in the case of a snow load. See Example 3 in this section. When considering the equilibrium of a system of forces on a structure, considered as a rigid body, we can replace the system of forces by its resultant"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure1.18-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure1.18-1.png",
        "caption": "FIGURE 1.18. Illustration of a wheel attched to the spindle axle.",
        "texts": [
            " Rim width, rim diameter, and offset are shown in Figure 1.15. Offset is 1. Tire and Rim Fundamentals 23 the distance between the inner plane and the center plane of the rim. A rim may be designed with a negative, zero, or positive offset. A rim has a positive offset if the spider is outward from the center plane. The flange shape code signifies the tire-side profile of the rim and can be B, C, D, E, F , G, J , JJ , JK, and K. Usually the profile code follows the nominal rim width but different arrangements are also used. Figure 1.18 illustrates how a wheel is attached to the spindle axle. Example 35 Wire spoke wheel. A rim that uses wires to connect the center part to the exterior flange is called a wire spoke wheel, or simply a wire wheel. The wires are called spokes. This type of wheel is usually used on classic vehicles. The highpower cars do not use wire wheels because of safety. Figure 1.19 depicts two examples of wire spoke wheels. Example 36 Light alloy rim material. Metal is the main material for manufacturing, rims, however, new composite materials are also used for rims occasionally"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure4.53-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure4.53-1.png",
        "caption": "Fig. 4.53: A scheme for an articulated finger at the grasp.",
        "texts": [
            " articulated finger that has been built by using the mechanism layout of Fig. 4.50 with a specific dimensional synthesis to obtain a compact design in which the linkages are hidden in the finger body. The mechanism solution for an articulated finger of Fig. 4.50 can also be considered as convenient because it permits the application of features and requirements for design and operation performances that can be deduced similarly to those for grippers. An alternative design that is widely used, is shown in Fig. 4.53 in which the architecture of a planar 3R manipulator can be easily recognized. In the scheme the problem of multiple grasping contacts is also indicated by means of the determination of the contact points P1, P2, P3 . The action of a finger grasping an object through the forces F1, F2, F3 at the contact points P1, P2, P3 can be evaluated in a straightforward way by applying the Principle of Virtual work, when friction forces are assumed as negligible, in the form 332211332211 vFvFvF ++=\u03b8\u03c4+\u03b8\u03c4+\u03b8\u03c4 (4"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000637_978-3-642-83006-8-Figure2.35-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000637_978-3-642-83006-8-Figure2.35-1.png",
        "caption": "Fig. 2.35. Compensating movements for the single-support gait up the stairs for T=1.5, S=0.6 and different ZIIP laws",
        "texts": [],
        "surrounding_texts": [
            "are shown the reaction forces and driving torques, and in Fig. 2.30. is shown the power at joints. [rad] 0.1 o 9 q 0.1 0.2 105 T=I, S=0.4 S=0.8 T=.5 S=0.6 10 q [rad] 0.3 106 t[s) Fig. 2.29.b. Ankle joint torques for the single-support gait upon level ground for T=1.5, 5=0.6 and different ZMP laws - supporting phase Fig. 2.29.c. Knee joint torques for the single-support gait upon level ground for T=1.5, 5=0.6 and different ZMP laws - supporting phase [Nm) P5 30 10 t[s) 4-------------------~~~~-=~~~~~~ O ~ .75 -50 -100 -150 .~-.-. I! /-...... ~~\"L I z ...... \u00b7-\u00b7\u00b7\u00b7~, .. ::::::.;;;o- 1 i \"/ . . . . I , i i i ; u Fig. 2.29.d. Hip joint torques for the single-support gait upon level ground for T=1.5, S=0.6 and different ZMP laws - supporting phase Fig. 2.29.e. Ankle joint torques for the single-support gait upon [Nm) level ground for T=1.5, S=0.6 and different ZMP laws P 7 - swing phase . 15 10 5 o t[s) 0.75 Fig. 2.29.f. Knee joint torques for the single-support gait upon level ground for T=1.5, S=0.6 and different ZMP laws - swing phase 107 108 [NmJ Ps 20 10 0 Fig. 2.29.g. [NmJ Pg 30 10 0 -50 -100 -150 t[sJ 0.75 Hip joint torques for the single-support gait upon level ground for T=1.5, S=0.6 and different ZMP laws - swing phase Fig. 2.29.h. Trunk joint torques in the sagittal plane for the single-support gait upon level ground for T=l .5, S=0.6 and different ZMP laws [NmJ 10 o -20 -40 -- t[sJ 0.75 .j Fig. 2.29.i. Trunk joint torques in the frontal plane for the single-support gait upon level ground for T=1.5, S=0.6 and different ZMP laws 109 [W) 10 t[s) \\ . ~ I 1.5 -10 l __ \\.j \\1 Fig. 2.30.a. Power at ankle joint for the single-support gait upon level ground for T=1.5, 5=0.6 and different ZMP laws n [W) 20 10 t[s) 0 1.5 -20 . ; -40 \\i Fig. 2.30.b. Power at knee joint for the single-support gait upon level ground for T=1.5, 5=0.6 and different ZMP laws [W) 60 40 Fig. 2.30.c. Power at hip joint for the single-support gait upon level ground for T=1.5, 5=0.6 and different ZMP laws 110 [Wj A H 50 30 10 0 v Fig. 2.30.d. Power at trunk joint in the sagittal plane for the single-support gait upon level ground for T=1.5, 8=0.6 and different Zl\u00b71P laws Fig. 2.30.e. Power at trunk joint in the frontal plane for the single-support gait upon level ground for T=1.5, 8=0.6 and different ZMP laws .5 111 Example 2. In this example we consider the anthropomorphic mechanism shown in Fig. 2.31, which consists of 17 links and 17 revolute joints of the 5-th class. The topplogic,al structure represented by the serief' of .'+\" joints is of the form: MS \u2022 [: 2 3 4 5 6 7 8 9 10 11 1 ~] 2 3 4 5 6 7 13 14 15 0 (2.8.3) 2 3 4 5 6 7 13 16 17 0 The kinematic and dynamic parameters of the mechanism are given in Table 2.2. Fig. 2.32. shows the q3 and q13 d.o.f. for compensation in the frontal (2.32a) and sagittal (Fig. 2.32.b.) plane, respectively, un der the condition link 7 (pelvis) is parallel to the ground and the projection of the legs are parallel in the frontal plane. This condi tion is fulfilled by putting q7+q7_q 3 andq8+q8+q3. As in Example 1, the mechanism's arms are fixed, Le., q14 = q16 = 1.862253 [rad] and q15 = q17 = 1.5707 [rad]. The other data are: N = 31 I1 = 1, I2 = 7, I3 = 81 K1 = 12, K23= 10, K3 = 101 NT=171 E;i = 0, i=1, .\u2022\u2022 ,171 E;~ = 1, E;~3: 11 values E;i is 1m for iE{3, 13} and 0 for other values of index i1 r o1 =(O, 0, -0.0001)~1 e1 = (1, 0, 0) T 1 the initial values of the compensating part of the. system (q3(O), q13(O), q3(O) and q13(O\u00bb1 the repeatibility conditions are: w -1 o o o o o o o o (2.8.4) o -1 o o o Figs. 2.33 - 2.38. show the compensating movements for the different parameters and gait types. Example 3. In this case we consider the realization of an artificial anthropomorphic gait by a mechanism with free arms (Fig. 2.39). The mechanism consists of 19 links and 19 revolute joints of the 5-th class. Its topological structure can be represented by a series of \"+\" joint in the form of the matrix 112 ~4 Mechanical scheme of the antropomorphic mechanism with fixed arms sagittal (b) pl~ne T a b le :. 2 . K in e m a ti c a n d d y n a m ic p a ra m e te rs o f th e m e c h a n is m !-l as s M om en t of i n er ti a [k gm 2 ] D is ta nc e of t he a xe s ce nt re s of j o in ts f ro m th e Li nk Jo in t u n it a xe s [k g] J x J y Jz 1i nk c en tr e [m ] 1 2 3 4 5 6 7 1 0. 0 0. 0 0. 0 0. 0 z T r 1 ,1 = ( 0, 0, 0. 00 01 ) ; 1 1 ,2 = ( 0, 0, -0 .G O G 1) T z T e 1 = ( 1, 0, 0) 2 1. 5 3 C .0 00 06 0. 00 05 5 0. 00 04 5 f 2 ,2 = ( 0, 0, 0. 03 0) T ; f 2 ,3 = ( 0, 0, -O .0 70 )T \"t T e 2 = (0 ,1 ,0 ) 3 0. 0 0. 0 0. 0 0. 0 z T r 3 ,3 = ( 0, 0, 0. O e0 1) ; z T r 3 ,4 = ( 0, 0, -0 .0 00 1) \"t T e 3 = ( 1, 0, 0) -- 4 3. 21 0. 00 39 3 0. 00 39 3 0. 00 03 8 7 T r 4 ,4 = ( 0, 0, 0. 21 0) ; 7 T r 4 ,5 = ( 0, 0, -0 .2 10 ) \"t T e 4 = ( 0, 1, 0) 5 8. 41 0. 01 12 0 0. 01 20 0 0. O C 30 0 z T r 5 ,5 = ( 0, 0, 0. 22 0) ; t 5 ,6 = ( 0, 0, -O .2 20 )T -zT e 5 = ( 0, 1, 0 ) 6 0. 0 0. 0 0. 0 0. 0 \"t T r 6 ,6 = ( 0, 0, 0. 00 01 ) ; -z- T r 6 ,7 = ( 0, 0, -0 .0 0G 1) -zT e 6 = ( 0 , 1, 0) 7 6. 96 0. 00 70 0 0. 00 56 5 0. 00 62 7 \"t T r l ,7 = ( 0, 0. 13 5, 0 .1 ) ; z T r 7 ,8 = ( 0, -0 .1 35 , 0. 1) ; -z- T e 7 = ( 1 , Q , 0) -zT r 7 ,1 3= (0 , C, -0 .0 5) ; 8 0. 0 0. 0 0. 0 0. 0 \"t T r 8 ,8 = ( 0 , 0, -0 .0 00 1) ; z T r 8 ,9 = ( 0, 0, 0. 00 01 ) 7 T e 8 = ( 1 , C, 0) 9 15 .41 0. 01 12 0 0. 01 20 0 0. 00 30 0 f 9 ,9 = ( 0, 0, -0 .2 20 )T ; f 9 ,1 0 = ( 0, 0, C .2 2C )T 7 T e 9 = ( 0 , -1 , 0) , 10 3. 21 0. 00 39 3 0. 00 39 3 0. 00 03 8 ~1 0, 10 = ( 0 , 0, -O .2 10 )T ; z T r 1 o ,1 1 = ( 0 , 0, 0. 21 0) -z' T e 1 0 = ( 0, - 1 , 0) 11 0. 0 0. 0 0. 0 0. 0 -zT \"t T \"t T r 1 1 ,1 1 = ( 0, 0, -0 .0 00 1) ; r 1 1 ,1 2 = ( 0 ,0 , O .O G 01 ) e 1 1 = (0 ,- 1 ,0 ) ~ c.> T a b le 2 .2 . C o n t. ~ ~ 1 2 3 4 5 6 7 12 1. 53 0. 00 00 6 0. 00 05 5 0. 00 04 5 ~ T r 1 2 ,1 2 = ( 0, 0, -0 .0 70 ) ~ T e 1 2 = (1 ,0 ,0 ) 13 30 .8 5 0. 15 14 0 0. 13 70 0 0. 02 83 0 7 T r 1 3 ,1 3 = ( 0, 0, 0. 34 ) ; f 1 3 ,1 4 = ( 0, 0. 2, _0 .6 )T ; 7 T e 13 = ( 0 ,1 , 0) ~ T r 1 3 ,1 6 = ( 0, -0 .2 , -0 .0 6) ~ T -+ ~ T 14 2. 07 0. 00 20 0 0. 00 20 0 0. O C 02 2 r 1 4 ,1 4 = ( 0, 0, -0 .1 54 ) ; f 1 4 ,1 5 = ( 0, 0, 0. 15 4) e 1 4 = (1 ,Q ,0 ) 15 1. 14 0. 00 25 0 0. 00 42 5 0. O O C1 4 = T r 1 5 ,1 5 = ( 0, 0, -0 .1 32 ) ~ T e 1 5 = ( 1 , 0, 0) 16 2. 07 0. 00 20 0 0. 00 20 0 0. 00 02 2 = T r 1 6 ,1 6 = ( 0, 0, -0 .1 54 ) ; = T r 1 6 ,1 7 = ( 0 , 0, 0. 15 4) ~ T e 1 6 = (- 1 ,0 ,0 ) 17 1. 1 4 0. 00 25 0 0. 00 42 5 0. 00 01 4 z T r 1 7 ,1 7 = ( 0, 0, -0 .1 32 ) ~ T e 1 7 = ( -1 , 0, 0) 115 parameters T and S 116 117 118 2 2 2 3 3 3 4 4 4 5 5 5 6 6 6 7 7 7 8 13 13 9 14 17 10 15 13 11 16 19 12 0 J o (2.8.5 ) The kinematic and dynamic parameters for this mechanism are given in Table"
        ]
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure6.8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure6.8-1.png",
        "caption": "FIGURE 6.8. A sample of double-arm parallel windshield wiper mechanism.",
        "texts": [
            " This equation can be utilized to design a four-bar linkage for three associated inputoutput angles. 322 6. Applied Mechanisms Figure 6.7 illustrates the four popular windshield wiper systems. Doublearm parallel method is the most popular wiping system that serves more than 90% of passenger cars. The double-arm opposing method has been used been using since last century, however, it was never very popular. The single-arm simple method is not very efficient, so the controlled single-arm is designed to maximize the wiped area. Wipers are used on windshields, and headlights. Figure 6.8 illustrates a sample of double-arm parallel windshield wiper mechanism. A four-bar linkage makes the main mechanism match the angular positions of the left and right wipers. A dyad or a two-link connects the driving motor to the main four-bar linkage and converts the rotational output of the motor into the back-and-forth motion of the wipers. The input and output links of the main four-bar linkage at three different positions are shown in Figure 6.9. We show the beginning and the end angles for the input link by \u03b821 and \u03b823, and for the output link by \u03b841 and \u03b843 respectively"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure2.18-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure2.18-1.png",
        "caption": "Fig. 2.18: Basic mechanical characteristics of a human manipulation of a composite material roving in a 3D winding manufacturing.",
        "texts": [
            " The conventional manufacturing is performed by a manual winding of a plait PS of roving on a suitable winding support WD as shown in Fig. 2.17. The labor of a human operator is repetitive with very poor technological and decisional requirements and contributions. Recurrent human operations can be recognized as a variable grasp from a firm to a slipping mode, and arm movements from pulling to wrapping a roving plait on the winding support. Specifically, basic operations can be modeled to be mimicked by a suitable robotic manipulator and they can be summarized as shown in Fig. 2.18 as: - slipping of the hand along the plait direction to perform a feeding of the plait during the deposition; - pulling, which consists of straining a plait to ensure straight-line and compact deposition; - slipping and pulling to provide a feeding and a straining of the plait after the wrapping on curved surfaces of the winding support; - rolling the plait about points A and B to wrap the plait on the curved surfaces of the winding support. During the manual winding the manipulative operations of the operator have been observed as being neither regular nor continuous",
            "4 execution times have been reported for each elementary action providing measured intervals during the operation of human operators in several occasions. Those times give measures of actions\u2019 significance and the possibility of achievable productivity improvements. In fact, Table 2.4 can be used to suggest speeding up some elementary actions more efficiently than others, as for example elementary actions n. 3, 7, 14, and 18, as well as to identify useless operations, as for example n. 4 for both hands. Chapter 2: Analysis of manipulations 61 In an analogous way, Table 2.4 and Fig. 2.18 can be useful for determining the elementary actions, which are needed for a better consideration both for the operator labor and manufacturing design. Thus, some elementary actions can be remarked to the operator as requiring particular attention for a manipulative path, because of fiber directions, or for straining action, because of material compactness. It is remarkable that the right hand is used much more than the left, which is mainly devoted to complementary elementary actions. Consequently, one robotic manipulator can be used to perform the basic elementary actions for the 3D winding, as for example those by right hand n"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure9.80-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure9.80-1.png",
        "caption": "Figure 9.80 A truss in which the diagonals cross one another.",
        "texts": [
            " When we talk about omitting zero-force members, this is done only to simplify the calculation. If the zero-force members are removed from the truss in reality, the truss becomes kinematically indeterminate. Zero-force members therefore have a genuine function in the truss. On the one hand they ensure the truss retains its shape, while on the other they can prevent buckling (in the plane of the structure) of (long) compressed members, such as the bottom chord in Figure 9.77, or the top chord in Figure 9.79. 9 Trusses 369 Example 3 You are given the truss in Figure 9.80. The diagonals are crossing members. Question: Determine all the zero-force members for the given load. Solution: In this truss, it is not possible to find a section across three members (that do not intersect in one point), nor is there a joint with less than two unknowns (member forces or support reactions). We therefore cannot determine the member forces with the method of sections, or with the method of joints, unless we first determine the support reactions. For determining the zero-force members in the truss, it is enough to know that the support reaction at the point of the roller is vertical, so that N2 = 0"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003721_s0924-0136(03)00283-8-Figure11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003721_s0924-0136(03)00283-8-Figure11-1.png",
        "caption": "Fig. 11. The two-body part shown in Fig. 6 (the gear mold and the base steel plate) was sintered at 1260 \u25e6C for 30 min under vacuum to increase the density to \u223c99% theoretical. The sound product shows that no shrinkage has occurred during sintering.",
        "texts": [
            " To close the remaining pores, homogenize the microstructure, and remove residual stresses it is recommended to post-sinter the laser processed part. Fig. 10b shows the effect of post-sintering on the residual porosity and dimensional change of the developed material. Technically full density parts (99% theoretical density) can be manufactured while the sintering shrinkage is close to zero. To verify the applicability of the method, the gear mold insert shown in Fig. 6 was sintered in a batch furnace at 1260 \u25e6C for 30 min under vacuum. It should be noted that the part was fixed on the base plate during sintering. Fig. 11 illustrates the corresponding part (no warpage or cracking is visible). It clearly depicts the dimensional stability of the laser-sintered artifacts during the post-sintering process, i.e. if any dimensional change occurs the shrinkage will lead to cracking. In addition, the post-sintering treatment improved the surface quality. The Rz and Ra parameters were reduced to around 30 and 8 m, respectively. The apparent hardness and the transverse rupture strength of the material were measured as 330 HV30 and 800 MPa, respectively"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure2.9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure2.9-1.png",
        "caption": "Fig. 2.9. Toroidal winding of a three-phase, six-pole, 18-coil AFPM machine with twin external rotor.",
        "texts": [
            "9) The winding factor for fundamental is the product of the distribution factor (eqn (2.8)) times pitch factor (eqn (2.9)), i.e. kw1 = kd1kp1 (2.10) The angle in electrical degrees between neighbouring slots is \u03b3 = 360o s1 p (2.11) Fig. 2.8 shows a single layer winding distributed in s1 = 36 slots of a threephase, 2p = 6 AFPM machine. Toroidal stator windings are used in twin-rotor double-sided AFPM machines (Fig. 2.4). The toroidal stator winding of a three-phase, six-pole AFPM machine with twin external rotor is shown in Fig. 2.9. Each phase of the winding has an equal number of coils connected in opposition so as to cancel the possible flux circulation in the stator core. Those coils are evenly distributed along the stator core diametrically opposing each other so the only possible number of poles are 2, 6, 10, . . . etc. The advantages of the toroidal stator winding [50, 245] are short end connection, simple stator core and easy design of any number of phases. Coreless stator windings are used in twin-rotor double-sided AFPM machines (Fig",
            " As with single-sided AFPM machines the stator ferromagnetic cores can be slotted or slotless, and the rotor magnets can be surface mounted, embedded or buried 124 4 AFPM Machines With Iron Cores (Fig. 2.6). Again, in the case of a slotless stator with a large air gap between the rotor and stator core the magnets are almost always surface mounted. The stator windings of double-sided AFPM machines can be flat wound (slotted or slotless) as shown in Fig. 2.8 or toroidally wound (normally slotless) as shown in Fig. 2.9. An example of a commercial double-sided AFPM servo motor with ferromagnetic core is shown in Figs. 4.1 and 4.2. External stators have slotted ring-shaped cores made of nonoriented electrotechnical steel ribbon. The inner rotor does not have any ferromagnetic material. PMs are mounted on a nonmagnetic rotating disc. Fig. 4.3 shows a double-sided AFPM synchronous generator with the stator core wound of amorphous alloy ribbon manufactured by LE Incorporated , Indianapolis, IN, U.S.A. The volume of LE AFPM generators is approximately 60% lower than that of classical synchronous generators of the same rating"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001483_0954405414567522-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001483_0954405414567522-Figure5-1.png",
        "caption": "Figure 5. Locally refined mesh.45",
        "texts": [
            " Certain regions, particularly those close to the heat source, exhibit steep temperature gradients and an accurate analysis would require a more dense mesh in those areas.58 For that reason, a local refinement technique is employed in order to update the mesh density in each load step, according to the needs stated above. In several studies, fine meshing is used for the region directly heated by the beam and a close surrounding area, while the mesh coarsens as moving away from the beam spot.17,34,35,44,45,49 An example of a mesh created using this technique is presented in Figure 5. Other studies utilize a uniform mesh for the whole model.36,50,52,59\u201362 In the two previous cases, the mesh thickness is usually equal to the layer thickness.37 The solid substrate can be meshed with larger element size than the overlying layers in order to reduce simulation time,40,43,51,63,64 as shown in Figure 6. The already solidified layers can also be meshed with a coarse mesh.48 The material in powder or liquid form does not contribute to the overall stiffness of the model. In order to reach accurate results when predicting the residual stresses and distortion, this fact has to be taken into account"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000091_s00170-015-7077-3-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000091_s00170-015-7077-3-Figure7-1.png",
        "caption": "Fig. 7 Schematic of electron beam freeform fabrication (EBF) system components [13]",
        "texts": [
            " The hardness of the fabricated sample is significantly influenced by the heat treatment rather than by the process parameters. The tensile properties of the additive manufactured blocks are correlated to the deposition direction due to the anisotropic property of the deposition. Electron beam freeform fabrication is a NASA-patented additive manufacturing process designed to build complex, near-net-shape parts requiring substantially less raw material and finish machining than traditional manufacturing methods. Figure 7 shows a schematic of the primary components in an electron beam freeform fabrication (EBF3) system [13]. The process introduces metal wire feedstock into a molten pool that is created and sustained using a focused electron beam in a high vacuum environment. The electron beam couples effectively with any electrically conductive material, including highly reflective alloys such as aluminium and copper. The EBF3 process is capable of bulk metal deposition at deposition rates over 2500 cm3/h as well as finer detailed deposition at lower deposition rates with the same piece of equipment, limited only by the positioning precision and wire feed capability"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure9.9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure9.9-1.png",
        "caption": "FIGURE 9.9 Cross section for round-wire skin-effect model.",
        "texts": [
            "3) where we used J = \ud835\udf0eE to get \u2207 \u00d7 J = \u2212\ud835\udf07\ud835\udf0e \ud835\udf15H \ud835\udf15t . (9.19) 224 SKIN EFFECT MODELING We have to take the cross-product in cylindrical coordinate for the conditions where J = Jzz\u0302. Also, the magnetic field intensity has only the component H = H \ud835\udf19 \u0302\ud835\udf53. This leads to \ud835\udf15Jz(\ud835\udf0c, t) \ud835\udf15\ud835\udf0c = \ud835\udf07\ud835\udf0e \ud835\udf15H \ud835\udf19 (\ud835\udf0c, t) \ud835\udf15t . (9.20) From this and (9.18), we finally get a first-order DE \ud835\udf15Jz(\ud835\udf0c, t) \ud835\udf15\ud835\udf0c = \ud835\udf07\ud835\udf0e 2\ud835\udf0b\ud835\udf0c \ud835\udf15Iz(\ud835\udf0c, t) \ud835\udf15t . (9.21) The goal is a circuit model for the internal part of the cylindrical conductor, which we subdivide into the \u0394\ud835\udf0c tube segments as is shown in Fig. 9.9. The current in each tube is of the form Jz(k, t) = Ik\u22121(t) \u2212 Ik(t) k (9.22) for tube k and k is its cross-sectional area and Ik(t) is the current in the tube segment. All currents will be in the z-direction. With this, we can approximate the derivative on the left-hand side of (9.21) such that the equation can be written for the diffusion case shown in Fig. 9.9 for the first two tubes as I0 \u2212 I1 \ud835\udf0e1 \u2212 I1 \u2212 I2 \ud835\udf0e2 = \ud835\udf07(\ud835\udf0cAV1 \u2212 \ud835\udf0cAV2) 2\ud835\udf0b\ud835\udf0c1 \ud835\udf15I1(\ud835\udf0c, t) \ud835\udf15t , (9.23) where the average radius of the tubes is \ud835\udf0cAV1 = (\ud835\udf0c0 + \ud835\udf0c1)\u22152, and so on. Since all currents represent the total z-directed current in each tube, we can rewrite (9.23) by multiplying by \u0394z = |ze \u2212 zs| in terms of circuit elements where L1 = \ud835\udf07|ze \u2212 zs|(\ud835\udf0cAV1 \u2212 \ud835\udf0cAV2) 2\ud835\udf0b\ud835\udf0c1 (9.24) ONE DIMENSIONAL CURRENT FLOW TECHNIQUES 225 as the differential diffusion inductance. Also, on the left-hand side of the equation, we recognize the resistors of the form R1 = |ze \u2212 zs|\u2215(\ud835\udf0e1). With this, we can recognize the circuit equation R1I1 + R2I1 + L1 \ud835\udf15I1 \ud835\udf15t \u2212 R2I2 = 0 (9.25) corresponds to an equivalent circuit. We should note in Fig. 9.9 that the current in tube 2 is I1 \u2212 I2. All other loops are of the same form since we are progressing toward the center of the round conductor. This leads to the equivalent circuit in Fig. 9.10. Note that the partial inductance of the outer zero thickness shell (9.28) has been added to the model. We emphasize that since this model is a self-term, it can be applied to any orientation for a cylindrical conductor in the global coordinate system. The overall solution approach is based on the approximate internal\u2013external inductance approximation for inductance [30]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002125_tfuzz.2016.2634162-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002125_tfuzz.2016.2634162-Figure2-1.png",
        "caption": "Fig. 2 Schematic of the electromechanical system.",
        "texts": [],
        "surrounding_texts": [
            "1063-6706 (c) 2016 IEEE. Personal use is permitted, but republication/redistribution requires IEEE permission. See http://www.ieee.org/publications_standards/publications/rights/index.html for more information.\nDifferentiating Vn will give\nV\u0307n \u2264 \u2212 n\u22121\u2211 i=1 ci\u03bd 2 i + n\u22121\u2211 i=1 1 2l2i \u03bd2i (\u2225Wi\u22252 \u2212 \u03b8\u0302)ST i Si\n+ 1\n2 n\u22121\u2211 i=1 (l2i + \u03b52i ) + gn\u22121(xn\u22121)\u03bdn\u22121\u03bdn\n+\u03bdn(fn(x) + gn(x)u\u2212 x\u0307n,c \u2212 \u03be\u0307n). (30)\nSimilarly, there exists WT n Sn(Zn) for any given \u03b5n > 0\nsuch that fn(Zn) = WT n Sn(Zn) + \u03b4n\nwith |\u03b4n| < \u03b5n and Zn = x. By the Young\u2019s inequality, one has\n\u03bdnfn(x) \u2264 1\n2l2n \u03bd2n \u2225Wn\u22252 ST\nn Sn + 1\n2 l2n +\n1 2 \u03bd2n + 1 2 \u03b52n. (31)\nSubstituting (31) into (30) yields\nV\u0307n \u2264 \u2212 n\u22121\u2211 i=1 ci\u03bd 2 i + n\u22121\u2211 i=1 1 2l2i \u03bd2i (\u2225Wi\u22252 \u2212 \u03b8\u0302)ST i Si\n+ 1\n2 n\u22121\u2211 i=1 (l2i + \u03b52i ) + \u03bdn(gn\u22121(xn\u22121)zn\u22121\n\u2212gn\u22121(xn\u22121)\u03ben\u22121 + 1\n2l2n \u03bdn \u2225Wn\u22252 ST n Sn\n+ 1\n2 \u03bdn + gn(x)u\u2212 x\u0307n,c \u2212 \u03be\u0307n) +\n1 2 l2n\n+ 1\n2 \u03b52n. (32)\nBy use of equation (15), (32) can be rewritten as\nV\u0307n \u2264 \u2212 n\u22121\u2211 i=1 ci\u03bd 2 i + n\u22121\u2211 i=1 1 2l2i \u03bd2i (\u2225Wi\u22252 \u2212 \u03b8\u0302)ST i Si\n+ 1\n2 n\u22121\u2211 i=1 (l2i + \u03b52i ) + \u03bdn(gn\u22121(xn\u22121)zn\u22121\n+ 1\n2l2n \u03bdn \u2225Wn\u22252 ST\nn Sn + 1\n2 \u03bdn + gn(x)u\n\u2212x\u0307n,c + cn\u03ben) + 1\n2 l2n +\n1 2 \u03b52n. (33)\nNow we design the true controller u as:\nu = \u03b1(1 + \u03c1(t))w(t) + \u03bb(t). (34)\nSubstituting (34) into (33) arrives at\nV\u0307n \u2264 \u2212 n\u22121\u2211 i=1 ci\u03bd 2 i + n\u22121\u2211 i=1 1 2l2i \u03bd2i (\u2225Wi\u22252 \u2212 \u03b8\u0302)ST i Si\n+ 1\n2 n\u22121\u2211 i=1 (l2i + \u03b52i ) + \u03bdn(gn\u22121(xn\u22121)zn\u22121\n+ 1\n2l2n \u03bdn \u2225Wn\u22252 ST\nn Sn + 1\n2 \u03bdn\n+gn(x)(\u03b1(1 + \u03c1(t))w(t) + \u03bb(t))\n\u2212x\u0307n,c + cn\u03ben) + 1\n2 l2n +\n1 2 \u03b52n. (35)\nBy use of (6), (8) and Assumption 1, one has\n\u03b1\u03c1(t)w(t) + \u03bb(t) \u2264 (\u03b1\u0304\u2212 \u03b1)Mn + \u03bb\u0304. (36)\nSubstituting (36) into (35), then we can obtain:\nV\u0307n \u2264 \u2212 n\u22121\u2211 i=1 ci\u03bd 2 i + n\u22121\u2211 i=1 1 2l2i \u03bd2i (\u2225Wi\u22252 \u2212 \u03b8\u0302)ST i Si\n+ 1\n2 n\u22121\u2211 i=1 (l2i + \u03b52i ) + \u03bdn(gn\u22121(xn\u22121)zn\u22121\n+ 1\n2l2n \u03bdn \u2225Wn\u22252 ST\nn Sn + 1\n2 \u03bdn\n+gn(x)(\u03b1w(t) + (\u03b1\u0304\u2212 \u03b1)Mn + \u03bb\u0304)\n\u2212x\u0307n,c + cn\u03ben) + 1\n2 l2n +\n1 2 \u03b52n. (37)\nThe control signal w(t) is constructed as\nw(t) = 1 gn(x)\u03b1 (\u2212cnzn \u2212 1 2 \u03bdn \u2212 gn\u22121(xn\u22121)zn\u22121\n+x\u0307n,c \u2212 1\n2l2n \u03bdn\u03b8\u0302S\nT n Sn)\u2212\n1 \u03b1 (\u03b1\u0304\u2212 \u03b1)Mn\n\u2212 1\n\u03b1 \u03bb\u0304.\nThen (37) can be recomputed by\nV\u0307n \u2264 \u2212 n\u2211\ni=1\nci\u03bd 2 i + n\u2211 i=1 1 2l2i \u03bd2i (\u2225Wi\u22252 \u2212 \u03b8\u0302)ST i Si\n+ 1\n2 n\u2211 i=1 (l2i + \u03b52i ). (38)\nWe define the error \u03b8\u0303 = \u03b8\u2212\u03b8\u0302. Then we design the Lyapunov function as\nV = Vn + 1\n2r \u03b8\u0303T \u03b8\u0303. (39)\nwith r being a positive parameter. Substituting (39) into (38), then differentiating V can pro-\nduce\nV\u0307 \u2264 \u2212 n\u2211\ni=1\nci\u03bd 2 i +\n1\n2 n\u2211 i=1 (l2i + \u03b52i )\n+ 1\nr \u03b8\u0303( n\u2211 i=1 1 2l2i r\u03bd2i S T i Si \u2212 \u02d9\u0302 \u03b8) (40)\nthen the adaptive law can be constructed as\n\u02d9\u0302 \u03b8 = n\u2211 i=1 1 2l2i r\u03bd2i S T i Si \u2212Mm\u03b8\u0302 (41)\nwith Mm > 0. Theorem 1. Consider system (1) with Assumptions 1-4\nand the given reference signal xd. The command filteringbased fuzzy controller (34), the virtual controllers (17) and the adaptive law (41) can guarantee that the tracking error of the closed-loop controlled system will converge into a sufficiently small neighborhood of the origin and all the closed-loop signals are bounded.",
            "1063-6706 (c) 2016 IEEE. Personal use is permitted, but republication/redistribution requires IEEE permission. See http://www.ieee.org/publications_standards/publications/rights/index.html for more information.\nProof. By using (41), (40) can be rewritten as\nV\u0307 \u2264 \u2212 n\u2211\ni=1\nci\u03bd 2 i +\n1\n2 n\u2211 i=1 (l2i + \u03b52i ) + Mm r \u03b8\u0303T \u03b8\u0302.\nAccording to Young\u2019s inequality, there exists \u03b8\u0303T \u03b8\u0302 \u2264 \u22121\n2 \u03b8\u0303 T \u03b8\u0303 + 1 2\u03b8 T \u03b8. Then we can get\nV\u0307 \u2264 \u2212 n\u2211\ni=1\nci\u03bd 2 i +\n1\n2 n\u2211 i=1 (l2i + \u03b52i )\u2212 Mm 2r \u03b8\u0303T \u03b8\u0303\n+ Mm\n2r \u03b8T \u03b8\n\u2264 \u2212aV + b (42)\nwhere a = min {2ci,Mm}, b = 1 2 n\u2211 i=1 (l2i + \u03b52i ) + Mm 2r \u03b8T \u03b8.\nFurthermore, (42) signifies\nV (t) \u2264 (V (t0)\u2212 b\na )e\u2212a(t\u2212t0) +\nb a \u2264 V (t0) + b a , \u2200t \u2265 t0.\n(43) It can be known that \u03bdi, \u03b8\u0303 are bounded for i = 1, 2, ..., n. \u03b8 is a constant, so \u03b8\u0302 is bounded in probability. \u2225\u03bei\u2225 is bounded, zi = \u03bdi + \u03bei, we can have the signal zi is bounded. So x(t) and all control signals are bounded over any time interval, by [14], the solution exists for t \u2208 [0,\u221e). So it can gain that\nlim t\u2192\u221e\n|z1| \u2264 \u221a 2b\na +\n\u00b5\u03c1 2c0 . (44)\nRemark 4: It can be seen from the definition of a and b that the larger r and the smaller li, \u03b5i lead to a smaller z1 after the parameters ci are selected.\nIV. ELECTROMECHANICAL SYSTEM APPLICATION\nThis section presents a comparison between our proposed approach and the DSC without the error compensation mechanism in [41] to certify the validity of command filtering fuzzy control method. The simulation is presented on an electromechanical dynamic system, which is displayed as the following figure.\nAnd its system model is given as [37]:{\nDq\u0308 +Bq\u0307 +N sin(q) = \u03c4 M\u03c4\u0307 +H\u03c4 = V0 \u2212KB q\u0307\n(45)\nin which the definitions and values of V0, R, L, K\u03c4 , KB , \u03c4 , J , B, m, L0, M0, R0, D, q, N , M , H can be seen in [37].\nIn order to simplify the system model, by introducing the notations x1 = q, x2 = q\u0307, x3 = \u03c4 , u = V0 and the system dynamics described by (45) can be rewritten as x\u03071 = x2 x\u03072 = f2(x1, x2) + 1 Dx3 x\u03073 = f3(x1, x2, x3) + 1 M u\ny = x1\n(46)\nwhere f2(x1, x2) = \u2212N D sin(x1)\u2212 B Dx2 and f3(x1, x2, x3) = \u2212KB\nM x2 \u2212 H M x3 are unknown functions. The reference signal xd(t) is chosen as xd(t) = 0.5 sin(4t)+0.3 sin(2t). The dead zone parameters are given as: or = 2.1, ol = 2, pr = 5 and pl = 5.\n(a). Now the command filtering fuzzy controller proposed in this paper is utilized to control the system (46). The definitions of errors are given as\nz1 = x1 \u2212 xd\nz2 = x2 \u2212 x2,c\nz3 = x3 \u2212 x3,c.\nThe virtual control signals are given as:\n\u03b11 = \u2212c1z1 + x\u0307d\n\u03b12 = D(\u2212c2z2 \u2212 1\n2 \u03bd2 \u2212 z1 + x\u03072,c \u2212\n\u03bd2\u03b8\u0302S T 2 S2\n2l22 ).\nThe compensating signals are designed as follows:\n\u03be\u03071 = \u2212c1\u03be1 + \u03be2 + (x2,c \u2212 \u03b11)\n\u03be\u03072 = \u2212c2\u03be2 \u2212 \u03be1 + 1\nD \u03be3 +\n1 D (x3,c \u2212 \u03b12)\n\u03be\u03073 = \u2212c3\u03be3 \u2212 1\nD \u03be2.\nThe compensated error signals are \u03bdi = zi\u2212\u03bei for i = 1, 2, 3 and the control signal w(t) is given as\nw(t) = M\n\u03b1 (\u2212c3z3 \u2212\n1 2 \u03bd3 \u2212 1 D z2 + x\u03073,c \u2212 1 2l23 \u03bd3\u03b8\u0302S T 3 S3)\n\u2212 1 \u03b1 (\u03b1\u0304\u2212 \u03b1)Mn \u2212 1 \u03b1 \u03bb\u0304.\nConstruct the adaptive law as\n\u02d9\u0302 \u03b8 = 3\u2211 i=2 r\u03bd2i S T i Si 2l2i \u2212Mm\u03b8\u0302\nfor i = 2, 3. Simulations for the electromechanical system are imple-\nmented under the zero initial condition. The system control parameters are chosen as c1 = 230, c2 = 115, c3 = 172.5, r = 0.005, l2 = 0.01, l3 = 0.01, Mm = 0.01, Mn = 1000, \u03b6 = 0.4, \u03c9n = 2000. 11 fuzzy sets are defined over interval [\u22125, 5] for all state variables, and the fuzzy membership functions are given as follows:\n\u00b5F 1 i (xi) = exp[\n\u2212(xi + 5)2\n2 ]\n\u00b5F 2 i (xi) = exp[\n\u2212(xi + 4)2\n2 ]",
            "1063-6706 (c) 2016 IEEE. Personal use is permitted, but republication/redistribution requires IEEE permission. See http://www.ieee.org/publications_standards/publications/rights/index.html for more information.\n\u00b5F 3 i (xi) = exp[\n\u2212(xi + 3)2\n2 ]\n\u00b5F 4 i (xi) = exp[\n\u2212(xi + 2)2\n2 ]\n\u00b5F 5 i (xi) = exp[\n\u2212(xi + 1)2\n2 ]\n\u00b5F 6 i (xi) = exp[\n\u2212(xi + 0)2\n2 ]\n\u00b5F 7 i (xi) = exp[\n\u2212(xi \u2212 1)2\n2 ]\n\u00b5F 8 i (xi) = exp[\n\u2212(xi \u2212 2)2\n2 ]\n\u00b5F 9 i (xi) = exp[\n\u2212(xi \u2212 3)2\n2 ]\n\u00b5F 10 i (xi) = exp[\n\u2212(xi \u2212 4)2\n2 ]\n\u00b5F 11 i (xi) = exp[\n\u2212(xi \u2212 5)2\n2 ]\nThen, according to Lemma 1, construct Si for i = 1, 2, 3.\n(b). The previous DSC approach proposed in [41] is also applied to control the electromechanical system (46). For comparison, the corresponding parameters of the controller are selected as c1 = 80, c2 = 40, c3 = 60, and all the other control parameters are the same as in Part (a).\nSimulation results can be seen in Figs. 3-7, respectively. Figs. 3(a)-5(a) and Figs. 6-7 are for our proposed control method. For a comparison, Figs. 3(b)-5(b) are shown for the DSC method. Fig. 3(a) and Fig. 3(b) display the tracking effect of the reference signal x1 and the output response xd. Fig. 4(a) and Fig. 4(b) give the curve of tracking error. From Figs. 3-4, it can be seen clearly that the system output signal x1 can track the given reference signal xd well. The control input signals are depicted in Fig. 5 where they are bounded. The curves of the \u03b1i (i = 1, 2) and xi,c (i = 2, 3) are displayed in Figs. 6-7 which bring out the good filtering effect of command filters, that is, the output of command filter xi,c can track the input signals \u03b1i well. Figs. 3-7 illustrate the effectiveness of our proposed command filtered adaptive fuzzy controllers.\nRemark 5: It can be observed from the simulation results that both of control approaches have satisfactory control effects, but our proposed command filtered fuzzy control approach has better control performance than DSC method in two aspects: 1) it gains better tracking performance and 2) it has smaller overshoots.\nRemark 6: The DSC method proposed in [41] does not consider the errors resulting from the filtering process. Thus, in order to deal with this drawback, our proposed control method introduces an error compensation mechanism to gain a better control effect in this paper. Simulation comparison results between Figs. 3(a)-5(a) and Figs. 3(b)-5(b) illustrate the validity of our proposed control method."
        ]
    },
    {
        "image_filename": "designv10_0_0002483_scirobotics.aau3038-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002483_scirobotics.aau3038-Figure2-1.png",
        "caption": "Fig. 2. Design modifications and evaluation. (A) An image of the HAMR-E including axes definitions. (B) Schematic representation of the electroadhesive pad and the three-DOF origami ankle with components labeled. Inset depicts a detailed view of the ankle\u2019s center of rotation. (C) Experimental (mean \u00b1 SD, n = 5) and theoretical normal adhesion pressure as a function of applied electrical field for the highest-performing pad designs (see fig. S2 for measurement details).",
        "texts": [
            " One key contribution described in this paper that enables robust locomotion at arbitrary inclines is the integration and characterization of low-voltage electroadhesive pads and passive origami ankles. Another innovation is the design and tuning of a parametric tripedal crawl gait. Last, we performed extensive experiments to evaluate these newly designed elements and the robot\u2019s locomotion performance. Our studies demonstrate that HAMR-E is one of the smallest legged robots capable of versatile locomotion on arbitrarily inclined and curved \u201creal-world\u201d conductive surfaces. We based our climbing microrobot (Fig. 2A) on the Harvard Ambulatory MicroRobot (HAMR-VI), a 1.43-g quadruped that is capable of high-speed running (25), climbing (26), and power and control autonomy (27). The robot has eight degrees of freedom (DOFs) actuated by high\u2013power density piezoelectric bending bimorph actuators (28). A spherical five-bar (SFB) transmission connects the two actuators to a single leg for independent control of swing (leg-x) and lift (leg-z) motions (fig. S1). The SFB transmission and actuators are sized as 1 of 12 by guest on D ecem ber 31, 2018 http://robotics",
            " The design and evaluation of the integrated leg structure are described in the next section, and the development of the parametric tripedal crawl gait is detailed in the \u201cGait design\u201d section. These advances were leveraged to demonstrate horizontal, vertical, and inverted locomotion on conductive surfaces (\u201cLocomotion characterization\u201d). Inverted and vertical locomotion requires normal and shear adhesion forces at least equal to the robot\u2019s body weight. To satisfy these requirements,wedeveloped a functional leg (electroadhesive footpad andpassive origami ankle; Fig. 2B) that provides robust adhesion for inverted and vertical locomotion while retaining horizontal running capabilities. We leveraged insights from previous work with electroadhesion in microrobots (26, 30) to develop a suitable footpad for HAMR. Using a standardmodel for electroadhesion (31), the adhesion force generated by an electrode on a conductive substrate is given by Fadh \u00bc Ae0ed V2 2d2 \u00bc Ae0ed E2 2 \u00f01\u00de where A is the electrode area, e0 is the permittivity of vacuum, ed is the dielectric constant of the dielectric material,V is the applied voltage differential, d is the dielectric thickness, and E = V/d is the applied electric field. Intuitively, the footpad and conductive substrate form a parallel plate capacitor, and Fadh is the attractive force between the plates. The Coulomb friction model states that the maximum shear force the footpad can support is Fshear \u00bc mFadh \u00f02\u00de where m is the coefficient of static friction between the footpad and the conductive substrate. de Rivaz et al., Sci. Robot. 3, eaau3038 (2018) 19 December 2018 For such footpads (Fig. 2B), the two major design considerations are the choices of electrode geometry and dielectricmaterial.Althoughelectrodedesign for engagement to arbitrary substrates is an active area of research (22, 31, 32), the solution ismore straightforward for conductive substrates. Adhesion force is governed by the area (linearly increases; Eq. 1) of the electrodes. To simplify their geometry, we considered only circular electrodes because they mitigate electric field concentrations near the edges and avoided preferential bending axes after repeated use",
            " The selection of dielectric material was dictated by trade-offs between minimizing the thickness and maximizing the dielectric constant, the coefficient of friction (Eq. 2), and the We evaluated themaximum shear force supported by footpads with varying geometries and dielectric materials (see Materials andMethods for details). The experimental measurements and theoretical predictions (Eq. 2) of maximum shear adhesion as a function of applied field are shown in fig. S2 for a variety of pad designs. Similar measurements for the best two footpads, with polyimide and silicone dielectrics, are shown in Fig. 2C.We found that polyimide dielectric (ed = 3.5) footpads generated normal adhesion comparable with silicone versions (ed = 2.7). Furthermore, footpads with a silicone dielectric (m = 0.75) generated larger shear adhesion than those with a polyimide dielectric (m = 0.25) at the same field strength. However, the silicone pads suffered from dielectric breakdown within the desired operating voltages (\u2264300V) because of the significantly lower dielectric strength of silicone (Emax \u2248 20 V mm\u22121) compared with polyimide (Emax \u2248 120 V mm\u22121)",
            "org/ D ow nloaded from with a pad; fig. S5B) is approximately 2.6 \u00b1 0.5 ms (n = 8 trials, two pads). Representative voltage traces during charge and discharge are shown in fig. S7 (A and B, respectively). The system time constant is likely dominated by that of the DC-DC amplifier and is too small to affect pad performance at typical operating frequencies (0.2 to 2 Hz). The footpad described in the previous section was attached to a passive three-DOF origami ankle joint located at the distal tip of the leg (Fig. 2B). These ankles (26) compensated for leg rotations induced by the transmission kinematics and body dynamics. In addition, the ankles also passively aligned to macroscale surface topology, increasing effective contact area. Each ankle joint is composed of 12 flexures embedded into themonolithic leg structure fabricated with the PC-MEMS processes. The laminate was then folded up into its final configuration in an origami-like fashion. The kinematics of the ankle allowed for leg rotations of \u00b190\u00b0 in roll and \u00b145\u00b0 in pitch and yaw",
            " The insulation (or backing) is a 12.5-mm-thick layer of acrylic adhesive (DuPont, Pyralux FR1500) laminated to the electrode with heat and pressure. Excess material for both footpad varieties was removed via laser machining (Oxford Lasers, E-Series), and both footpad varieties were bonded to the copper underside of the ankles (see the next section) with high-conductivity silver epoxy (MG Chemicals). de Rivaz et al., Sci. Robot. 3, eaau3038 (2018) 19 December 2018 Like the robot\u2019s chassis, the legs (Fig. 2B) were also manufactured with the PC-MEMS process. Each leg is a seven-layer composite laminate that consists of two rigid layers, a flexural layer, and a conductive layer that was bonded together with three adhesive layers (DuPont, Pyralux FR1500). The rigid layers were formed by curing five layers of woven fiberglass (TenCate, YLA FB9K387) at 0\u00b0-45\u00b0-0\u00b0-45\u00b0-0\u00b0 angles, and the cured fiberglass has a thickness of 100 mm. The flexural layer is a 7.5-mm-thick polyimide film (DuPont, Kapton), and the conductive layer for wiring electroadhesion signals to the footpads is a 5-mm conductive copper sheet"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000081_cr020719k-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000081_cr020719k-Figure2-1.png",
        "caption": "Figure 2. Alternative electron-transfer mechanisms. (a) Direct electron transfer (tunneling mechanism) from electrode surface to the active site of an enzyme. (b) Electron transfer via redox mediator.",
        "texts": [
            " Enzyme biocatalyst assemblies on electrode surfaces usually do not achieve significant electrontransfer communication between the redox center and the conductive support, mostly because of the electrical insulation of the biocatalytic site by the surrounding protein matrixes.70 During the past four decades, several methods have been proposed and investigated in the field of bioelectrochemical technology in an effort to establish efficient electrical communication between biocatalysts and electrodes.71-81 In general, electron transfer is classified by two different mechanisms (see Figure 2): mediated electron transfer (MET) and direct electron transfer (DET). In MET, a low-molecular-weight, redox-active species, referred to as a mediator, is introduced to shuttle electrons between the enzyme active site and the electrode.80 In this case, the enzyme catalyzes the oxidation or reduction of the redox mediator. The reverse transformation (regeneration) of the mediator occurs on the electrode surface. The major characteristics of mediator-assisted electron transfer are that (i) the mediator acts as a cosubstrate for the enzymatic reaction and (ii) the electrochemical transformation of the mediator on the electrode has to be reversible"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003678_mssp.2001.1416-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003678_mssp.2001.1416-Figure7-1.png",
        "caption": "Figure 7. The gear test rig.",
        "texts": [
            " This is because the fourth meshing harmonic is at the 108th shaft order and a narrowband around it can just catch the left half of the resonance centred at 121 shaft orders. The other two narrow bands are far away from the resonance centre, so the demodulated waveforms are dominated by modulating functions associated with normal gears, i.e. a m (t) and b m (t) in equation (1). The experiments of gear vibration were conducted earlier in [9] on a gear rig for the study of tooth crack propagation in spur gears. The test procedures were designed to simulate the natural development of gear tooth cracking. The test rig con\"guration is shown in Fig. 7. The test gearbox was driven by an electric motor through a belt drive that provides a full loading capacity of 45 kW. The output gear shaft was coupled via a belt drive with a hydraulic dynamometer where the torque load was generated. During the test, the torque load was controlled by a potentiometer, and the input speed to the gearbox was set to 2400 rpm (40 Hz) although the actual speed tended to drop slightly under heavy loads. The test gear was the input pinion, labelled G6, with a spark-eroded notch (length]width]depth\"2 mm]0"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.31-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.31-1.png",
        "caption": "FIGURE 7.31. A positive four-wheel steering vehicle.",
        "texts": [
            " Steering Dynamics 409 reduces dramatically to an acceptable value when the vehicle is moving. Furthermore, an offset design sometimes makes more room to attach the other devices, and simplifies manufacturing. So, it may be used for small off-road vehicles, such as a mini Baja, and toy vehicles. 7.5 F Four wheel steering. At very low speeds, the kinematic steering condition that the perpendicular lines to each tire meet at one point, must be applied. The intersection point is the turning center of the vehicle. Figure 7.31 illustrates a positive four-wheel steering vehicle, and Figure 7.32 illustrates a negative 4WS vehicle. In a positive 4WS situation the front and rear wheels steer in the same direction, and in a negative 4WS situation the front and rear wheels steer opposite to each other. The kinematic condition between the steer angles of a 4WS vehicle is cot \u03b4of \u2212 cot \u03b4if = wf l \u2212 wr l cot \u03b4of \u2212 cot \u03b4if cot \u03b4or \u2212 cot \u03b4ir (7.73) where, wf and wr are the front and rear tracks, \u03b4if and \u03b4of are the steer angles of the front inner and outer wheels, \u03b4ir and \u03b4or are the steer angles of the rear inner and outer wheels, and l is the wheelbase of the vehicle"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-FigureC.12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-FigureC.12-1.png",
        "caption": "FIGURE C.12 Two parallel zero thickness conductors.",
        "texts": [
            "43) Rmnij = \u221a (xvm \u2212 xvn)2 + (yvi \u2212 yvj)2 + (z1 \u2212 z2)2, where the subscripts are for m = 1 \u2236 vm = s1; and for m = 2 \u2236 vm = e1; (C.44) n = 1 \u2236 vn = s2; and for n = 2 \u2236 vn = e2; (C.45) PARTIAL INDUCTANCE FORMULAS FOR ORTHOGONAL GEOMETRIES 397 i = 1 \u2236 vi = s1; and for i = 2 \u2236 vn = e1; (C.46) j = 1 \u2236 vj = s2; and for j = 2 \u2236 vn = e2. (C.47) C.1.10 1\u2215R3 Kernel Integral for Orthogonal Rectangular Sheets A second important integral with a 1\u2215R2 Green\u2019s function is for two orthogonal rectangles shown in Fig. C.12. For this case, we want to give a solution for the following integral Ip12 = \u222b xe1 xs1 \u222b ye1 ys1 \u222b xe2 xs2 \u222b ze2 zs2 z1 \u2212 z2 R3 dx1 dy1 dx2 dz2, (C.48) where, R3 = [(x1 \u2212 x2)2 + (y1 \u2212 y2)2 + (z1 \u2212 z2)2]3\u22152. Then the final mutual high order coupling formulation Ip12 for two orthogonal current sheets is Ip12 = 2\u2211 m=1 2\u2211 n=1 2\u2211 i=1 2\u2211 j=1 (\u22121)m+n+i+j { \u2212 0.5(y2 \u2212 yvi)Rmnij + (y2 \u2212 yvi)(xvn \u2212 xvm) tanh\u22121 ( Rmnij xvn \u2212 xvm ) + 0.5[(xvn \u2212 xvm)2 \u2212 (zvj \u2212 z1)2] ln [(y2 \u2212 yvi) + Rmnij] + (xvn \u2212 xvm)(zvj \u2212 z1) tan\u22121 (zvj \u2212 z1) y2 \u2212 yvi \u2212 (xvn \u2212 xvm)(zvj \u2212 z1) tan\u22121 (zvj \u2212 z1)2 + (xvn \u2212 xvm)2 + Rmnij(xvn \u2212 xvm) (y2 \u2212 yvi)(zvj \u2212 z1) } (C"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure8.21-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure8.21-1.png",
        "caption": "Figure 8.21 In practice, one often uses a method developed by Coulomb based on flat slide planes. Here one assumes that the pressure on the wall is caused by a triangular piece of soil that slides along a flat slide plane. One can recognise the triangular soil element from Figure 8.15a on the edge of the slide plane.",
        "texts": [
            "5 m) + (24 kN)(1.5 m) + (60 kN)(1 m) = 117 kNm. These are the requested forces acting in A on the 1-metre wide vertical strip AB from the sheet piling. In practice, one often uses a method developed by Coulomb1 based on flat slide planes. Here, one assumes that the pressure on the wall is caused by a 1 Charles Auguste de Coulomb (1736\u20131806), French scientist, known for his experiments in friction and electrostatic forces. 8 Earth Pressures 301 triangular piece of soil that slides along a flat slide plane (see Figure 8.21). One can deduce that the most dangerous slide plane (in the active case of a yielding wall) makes an angle \u03b1 = \u03c0 4 \u2212 \u03d5 2 with the vertical. This turns out to be precisely the angle of the planes, as derived earlier, in which the shear stress is a maximum. The magnitude of the earth pressure according to Coulomb\u2019s method also agrees with the result above. The agreement can best be understood by means of the sliding wedge of soil in Figure 8.21, assuming that there are no shear stresses in the horizontal and vertical planes (there is therefore also no wall friction). For simplicity, we will again assume that the soil is entirely dry. One can recognise the triangular soil element we dealt with earlier on the edge of the slide plane, with stresses \u03c3g and \u03c4g;max on the oblique side. The weight of the vertical soil column determines the vertical grain pressure \u03c3g;v. The horizontal strip of soil transfers the horizontal grain pressure \u03c3g;h to the wall"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001125_316-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001125_316-Figure2-1.png",
        "caption": "Figure 2. Stick-slip according to the different friction models (for clarity of illustration, only the stick trace is shown): (a) fk and fs constant, (b ) fk constant, fs a function of time of stick, (C)fk a function of sliding velocity,fs a function of time of stick; the midpoint of the stick varies with the sliding velocity.",
        "texts": [
            " Having made these simplifying assumptions, we may now describe the nature of the oscillations at the three levels of sophistication considered in the literature, and show that a further complicating factor must be considered. 8 2. FIRST ASSUMPTION: fk CONSTANT,fs CONSTANT This is the simplest solution, in which the kinetic coefficient of friction fk 1s Independent of the sliding velocity ZJ, and the static coefficient fs is larger than fk but independent of the time of stationary contact t . Plotting the force In the spring as a function of time of sliding we then have the situation shown in figure 2(a), in which the force in the spring increases at a steady rate until it reaches a value f,L, where L is the applied load, and then during the slip stage the force drops to (2fk-fs)L7 as far below fkL as it started above it. If the frequency of the oscillation is changed, either by raising the sliding velocity o r by using a stiffer spring, the amplitude of the oscillations, which we will denote as .lf7 remains constant at 2( fs -fk). This presentation is essentially that due to Blok (1940), who realized that 111 actual practice the stick-slip amplitude becomes smaller as z, is raised, and attributed this to damping",
            " 6- 5- S 4 - 3- If we place our surfaces together and set the flat in motion at a velocity i , the spring force increases with time at a rate kat, and this may be plotted in figure 3 as a straight line through the origin with slope kr /L . =It point -1 slip occurs and, since the kinetic coefficient is assumed constant, we again have a situation where slip continues to point B, as far below the fk line as -1 was above it. At B the rider comes to rest, which we plot as point C, and the next stick-cycle takes us to D, E and F. If n e increase 'L' or k then our line becomes steeper and the stick-slip will have a sndlei zmplitude (figure 2 ( b ) ) . Finally, a steady state is reached. DATA OF DOKOS (1946) 5 T E E L ON STEEL UNLUBRICATED - 0 0 50 LBS IO0 0 150 A A 200 0 5 4. THIRD ASS~WIPTIOX : .fi, A FUSCTIOS OF z', f q A FLSCTIOS OF t In general the kinetic coefficient is known to be a function of the sliding velocit!, curves such as are shown in figure 5 being typical. The kinetic coefficients plotted as a function of sliding velocity are those for steel sliding on steel unlubricated, perfectly lubricated by a fatty acid derivative, and in intermediate stages of lubrication"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000456_tac.2005.858646-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000456_tac.2005.858646-Figure1-1.png",
        "caption": "Fig. 1. Kinematic car model.",
        "texts": [
            " The proof of the theorem is now finished by the similar to Lemma 2 proof that for any > 0 with sufficiently large the inequality j (r 1)+ r 1N 1=2 r 2;r r 2;rj N 1=2 r 2;r is established in finite time and kept afterwards. Consider a simple kinematic model of car control _x = v cos' _y = v sin' _' = v=l tan _ = u where x and y are Cartesian coordinates of the rear-axle middle point, ' is the orientation angle, v is the longitudinal velocity, l is the length between the two axles and is the steering angle (Fig. 1). The task is to steer the car from a given initial position to the trajectory y = g(x), where g(x) and y are assumed to be available in real time. Define = y g(x). Let v = const = 10 m/s, l = 5 m, g(x) = 10 sin(0:05x) + 5; x = y = ' = = 0 at t = 0. The relative degree of the system is 3 and the 3-sliding controller [15], [16] solves the problem, but its transient features some chattering (Fig. 2). Also 3-sliding controllerN 3 can be applied here. It was taken = 1; L = 400. The resulting output-feedback controller (8)\u2013(10) is u = [z2 + 2(jz1j+ jz0j 2=3) 1=2(z1 + jz0j 2=3signz0)]=[jz2j + 2(jz1j+ jz0j 2=3)1=2] _z0 = v0 v0 = 14:7361jz0 j2=3sign(z0 ) + z1 _z1 = v1 v1 = 30jz1 v0j 1=2sign(z1 v0) + z2 _z2 = 440 sign(z2 v1): The controller parameter is convenient to find by simulation [3], [14]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure3.7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure3.7-1.png",
        "caption": "Fig. 3.7. Electrical scheme of D.C. motor",
        "texts": [
            " Generally, actuator models can be presented in the following form where Ai = subsystem matrix, Xi = state vector of the i-th actuator, bi = input distribution vector, N(uJ = amplitude saturation nonlinearity, j-Uj for uj< -uj N(u;) = Ui for -Uj~Ui~Uj uj for ui>uj fi =load distribution vector, and Pi = vector of driving torque (force) of the i-th degree of freedom (3.25) The total system dynamic model S is obtained by uniting the model of the system mechanical part sm and the actuators model Si. In order to avoid matrix calculations and to reduce necessary computing time, we proceed as follows since the model of the total system is sufficiently complex. We start from the differential equations which describe the behaviour of electro-mechanical actu- 3.2 Mathematical Model of Manipulation Robot Dynamics 63 ator (D.C. motor electrical scheme is presented in Fig. 3.7) in the following way: Lrir + Rrir+ CEq = u -CMir+Jrq'+Bcq= -P where Lr[H] ir[mA] Rr[Q] CE[V/rad/s] q[rad] = rotor inductivity, = rotor current, = rotor resistance, = proportionality constant of the electromotor force, = rotational angle of the motor output shaft = voltage on the rotor of motor, =moment proportionality constant, (3.26) u[V] CM[Nm/mA] J r [kgm2] Be [Nm/rad/s] P[Nm] = rotor moment of inertia J M reduced to the output shaft = viscous friction coefficient, and = moment of the external motor load Here, it has to be noticed that CE=CeNy , CM=CmNM Bc=BcNvNM' Jr=JMNvNM where Ny and NM are the gear speed ratio and gear torque ratio, respectively; Ce, Cm, Be are particular catalogue values"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure3.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure3.3-1.png",
        "caption": "FIGURE 3.3. Front view of a tire and measurment of the camber angle.",
        "texts": [
            " Tire plane is the plane made by narrowing the tire to a flat disk. The z-axis is perpendicular to the ground, opposite to the gravitational acceleration g, and the y-axis makes the coordinate system a right-hand triad. To show the tire orientation, we use two angles: camber angle \u03b3 and sideslip angle \u03b1. The camber angle is the angle between the tire-plane and 96 3. Tire Dynamics the vertical plane measured about the x-axis. The camber angle can be recognized better in a front view as shown in Figure 3.3. The sideslip angle \u03b1, or simply sideslip, is the angle between the velocity vector v and the x-axis measured about the z-axis. The sideslip can be recognized better in a top view, as shown in Figure 3.4. The force system that a tire receives from the ground is assumed to be located at the center of the tireprint and can be decomposed along x, y, and z axes. Therefore, the interaction of a tire with the road generates a 3D force system including three forces and three moments, as shown in Figure 3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure8.12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure8.12-1.png",
        "caption": "Figure 8.12 The grain stresses on a triangular soil element.",
        "texts": [
            " 1 Cohesion is the resistance to sliding resulting from a certain bond between the soil particles because of sticking and tangling, the influence of capillary water, and/or the hooking of particles with an irregular shape (hook resistance). 8 Earth Pressures 293 To keep matters simple, we will consider only grainy matter, such as sand, for which cohesion c is practically zero, so that \u03c4g;max = \u03c3g tan \u03d5. In order to find the extreme values of the horizontal grain pressure, a triangular slice OPQ is isolated from the soil (see Figure 8.12). No shear stresses are acting on the vertical boundaries (section OP and the front and back of the slice). This was shown by means of symmetry in Section 8.2. Suppose grain stresses \u03c3g and \u03c4g are acting on the oblique section PQ, and horizontal grain pressure \u03c3g;h is acting in the vertical section OP. Vertical grain pressure \u03c3g;v is acting on the horizontal section OQ. From the moment equilibrium in the plane of the drawing, about the middle of PQ, it follows that no shear stresses can be acting in the horizontal section OQ"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure7.32-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure7.32-1.png",
        "caption": "Figure 7.32 The distribution of the water pressures on the tunnel element.",
        "texts": [
            " Dimensions: length = 60 m, width b = 9 m and height h = 6 m. Dead weight of the tunnel element: qdw = 524 kN/m. Weight of each of the temporary bulkheads: Fhead = 235 kN. Specific weight of water: \u03b3w = 10 kN/m3. Questions: a. Determine the water pressure at the base of the tunnel element. b. Determine the resultant of the horizontal water pressure on a bulk- head. 7 Gas Pressure and Hydrostatic Pressure 265 Solution: a. The total dead weight Rdw of the tunnel element is Rdw = qdw + 2Fhead = (524 kN/m)(60 m) + 2 \u00d7 (235 kN) = 31910 kN. Figure 7.32 shows the distribution of the water pressures on the tunnel element. With a specific weight \u03b3w, the water pressure pw at a depth d is pw = \u03b3wd. The vertical water pressure on the base of the tunnel element gives an upward force Rv;w: Rv;w = pwb = \u03b3wdb . The upward force is equal to the weight of the displaced water. The tunnel element will sink in the water until the upward force is in equilibrium with the total dead weight Rdw: Rdw = Rv;w = \u03b3wdb so that d = Rdw \u03b3wb = 31910 kN (10 kN/m3)(9 m)(60 m) = 5.91 m. The water pressure at the base of the tunnel element is therefore pw = \u03b3wd = (10 kN/m3)(5.91 m) = 59.1 kN/m2. 266 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM b. The resultant Rh;w of the horizontal water pressure on a bulkhead is equal to the volume of the load diagram (see Figure 7.32): Rh;w = 1 2pwbd = 1 2 \u00d7 (59.1 kN/m2)(9 m)(5.91 m) = 1572 kN. In the calculation, it was noted that the vertical water pressure on the tunnel element exerts an upward force that is equal in magnitude to the weight of the displaced water. This is not a coincidence, but applies in general, regardless of the shape of the body, and is known as Archimedes\u2019 Law.1 The general proof can be found below. Take a contained space of arbitrary shape within a fluid (see Figure 7.33). If there is an equilibrium, the vertical component of the hydrostatic pressures has to provide an upward force on the outside of the contained space that equals the weight of the fluid within the contained space"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure14.40-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure14.40-1.png",
        "caption": "Figure 14.40 If the anchor chain is taut (pk/ 1), the distributed load along the chain can be replaced by an equal distributed load along the chord. The anchor chain now has the shape of a parabola.",
        "texts": [
            "9 m) 3500 N ) = 100 m For the vertical force at B Bv = \u2212Vx= = H sinh q H = (3500 N) sinh ( 21 N/m)(100 m) 3500 N ) = 2228 N \u2248 2.23 kN. The maximum force in the anchor chain is Nmax = NB = \u221a H 2 + B2 v = \u221a (3.5 kN)2 + (2.23 kN)2 = 4.15 kN. Alternative solution: The load q is a vertical force measured per length along the cable. If the an- 14 Cables, Lines of Force and Structural Shapes 669 chor chain is taut, this load can be approximated by an equal vertical force per length measured along the chord (see Figure 14.40). The associated shape of the anchor chain is then a parabola. The resultant of the distributed load is R = q \u221a 2 + h2. The moment equilibrium of the isolated chain gives \u2211 T |B = Hh \u2212 R \u00b7 1 2 = Hh \u2212 1 2q \u221a 2 + h2 = 0 so that 2Hh q = \u221a 2 + h2. After squaring ( 2Hh q )2 = 2( 2 + h2) we find 4 + h2 2 \u2212 ( 2Hh q )2 = 0. With h = 30.9 m, H = 3500 N and q = 21 N/m this leads to 4 + (30.9 m)2 2 \u2212 ( 2 \u00d7 (3500 N)(30.9 m) 21 N/m )2 = 0 670 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM so that 4 + (9.54.81 m2) 2 \u2212 (106"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure3.58-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure3.58-1.png",
        "caption": "Figure 3.58 A cube with edge length a and weight G is kept in equilibrium by six forces F1 to F6. For the angle \u03b1 between the lines of action of the forces applies tan \u03b1 = 3/4.",
        "texts": [
            " VOLUME 1: EQUILIBRIUM 3 Statics of a Rigid Body 95 Next, we find Av directly from \u2211 Tx = 0: \u2211 Tx = \u2212Av \u00d7 (5 m) + (60 kN) \u00d7 (3 m) + (50 kNm) \u2212Cv \u00d7 (3 m) + (100 kN) \u00d7 (1 m) + (40 kN) \u00d7 (1 m) = 0 \u21d2 Av = +53 kN, after which Bv follows directly from \u2211 Fz = 0: \u2211 Fz = +{(100 + 40 + 60) kN} \u2212 Av \u2212 Bv \u2212 Cv = 0 \u21d2 Bv = +112 kN. Figure 3.57 shows the forces Av, Bv and Cv as they act on the structure in reality. To check, one could also have a look at the moment equilibrium at a point other than the origin, such as point A: \u2211 Tx |A = \u2212{(60 \u2212 35) kN} \u00d7 (2 m) + (50 kNm) \u2212 \u2212{(100 + 40) kN} \u00d7 (4 m) + (112 kN) \u00d7 (5 m) = 0. The moment equilibrium is also met about a line through A parallel to the x axis. Example 2 In Figure 3.58, a cube with edge length a and weight G is kept in equilibrium by the six forces F1 to F6. For the angle \u03b1 between the lines of action of the forces applies tan \u03b1 = 3/4. Question: Determine the six forces F1 to F6 if a = 1 m and G = 24 kN. 96 Solution: When writing down the equilibrium equations, it can sometimes be useful to project all the forces on the three coordinate planes (see Figure 3.59). In doing so, the forces F2, F4 and F6 are resolved into components according to the coordinate directions",
            " VOLUME 1: EQUILIBRIUM 3 Statics of a Rigid Body 97 Determining the forces F1, F2 and F3 from the three remaining equations (b), (d) and (f) demands some arithmetic. Sometimes one can reduce the amount of calculation by looking at the moment equilibrium about another point. Also here: \u2211 Tz|P = \u2212 ( F1 + 4 5F2 + 4 5F6 ) \u00b7 a = 0 (g) so that F1 = 0. Equation (d) now gives F3 = \u2212 1 6G = \u22124 kN. Finally, equation (f) gives F5 = \u2212 7 6G = \u221228 kN. Figure 3.60 depicts the forces (in kN) as they are acting on the cube in reality. The forces F3, F5 and F6 act in directions opposite to those shown in Figure 3.58. By using alternative equilibrium equation (g), equation (b) for the force equilibrium in y direction was not used, and can be used as a check. With the forces expressed in kN this gives \u2211 Fy = F1 + 4 5F2 + F3 + 4 5F4 + F5 + 4 5F6 = 0 + 4 5 \u00d7 20 \u2212 4 + 4 5 \u00d7 40 \u2212 28 \u2212 4 5 \u00d7 20 = 0. The conditions for force equilibrium in y direction are met. 98 Example 3 The cube that has been halved diagonally in Figure 3.61 is subject to a couple of 80 kNm in plane ABEF and a couple of 60 kNm in plane BCDE. The directions are shown in the figure"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure2.13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure2.13-1.png",
        "caption": "Fig. 2.13 The sum of angular velocity vectors: Object 1 rotates according to angular velocity vector \u03c91. Object 2 rotates relative to Object 1 at \u03c92. The angular velocity of Object 2 in world coordinates is \u03c91 + \u03c92.",
        "texts": [
            " If the velocity of object 1 is (v1,\u03c91) and the velocity of object 2 relative to object 1 is (vd,\u03c9d), the velocity of object 2 will be given as follows8 8 The linear and angular velocity of the object have simply been combined and called \u201cvelocity\u201d. 40 2 Kinematics v2 = v1 +R1vd + \u03c91 \u00d7 (p2 \u2212 p1) (2.52) \u03c92 = \u03c91 +R1\u03c9d. (2.53) Here R1vd and R1\u03c9d describe the relative velocity between the objects in world coordinates. If we replace these using Wvd, W\u03c9d we get, v2 = v1 + Wvd + \u03c91 \u00d7 (p2 \u2212 p1) (2.54) \u03c92 = \u03c91 + W\u03c9d. (2.55) Therefore barring the fact that we get an influence of the rotational velocity the third element of (2.54), the velocity of the object in world coordinates is simply a sum. In Fig. 2.13 we have tried to visualise the meaning of (2.55). A humanoid robot is a mechanism consisting of many links connect by joints. To start our analyze, we would like to divide it into smaller units. Among many possible ways of separation, we present two typical ways in Fig. 2.14. In Fig. 2.14(a) each joint gets included in the link that is further away from the trunk. In (b) each joint is defined as a part of the trunk or the link closer to the trunk. In terms of computer programming the former style is more convenient"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure2.19-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure2.19-1.png",
        "caption": "Fig. 2.19 (a) Structure of a 12 degree of freedom biped robot. Numbers in brackets refer to the ID number (b) Rotation matrix which describes attitude of each link.",
        "texts": [
            " If the ID number is not 0, then it will return the sum of the mass of the corresponding link, the total mass of the descendant links from the sister, and the total mass of the descendant links from the child (line 6). If you type TotalMass(1) on the command line, this function will visit all nodes in the tree and return the total mass of the links under the \u201cBODY\u201d link. Compared with the usual method which checks the existing links beforehand and sums them, I feel this is much smarter. What do you think? 2.5 Kinematics of a Humanoid Robot 45 To explain the kinematics of a humanoid robot, we will use a 12 DoF model shown in Fig. 2.19, which consists of two legs. The link names and their ID numbers are indicated in Fig. 2.19(a). Splitting this model by the manner of Fig. 2.14(a), each link will have just one joint to drive it except the BODY link. Thus we can identify a link by joint name or its ID number. For example, 46 2 Kinematics by the ID number 5, we refer the joint RLEG J3 as well as the link of the right lower leg. First, we must define the local coordinates for each link. An origin of each local coordinates can be set anywhere on its own joint axis. For each hip, however, it is reasonable to assign the origins of the three frames at the same point where the three joint axes intersect. In the same way, for each ankle, we assign the origins of two ankle frames on the same point where the two joint axes intersect. Then, all rotation matrices which describe attitude of links are set to match the world coordinates when the robot is in its initial state, standing upright with fully stretched knees. Therefore we set thirteen matrices as, R1 = R2 = . . . = R13 = E. We show the local coordinates defined at this stage in Fig. 2.19(b). Next we will define the joint axis vectors aj and the relative position vector bj as indicated in Fig. 2.20. The joint axis vector is a unit vector which defines the axis of the joint rotation in the parent link\u2019s local coordinates. The positive (+) joint rotation is defined as the way to tighten a right-hand screw placed in the same direction with the joint axis vector. Using the knee joints as an example, the joint axis vectors would be, a5,a11 = [0 1 0]T . When we rotate this link in the + direction, a straight knee will flex in the same direction as a human\u2019s",
            " For instance, suppose our robot is in the front of stairs and we want to place one of the foot on the first step whose height and depth are already known. We certainly need to determine the amount of 50 2 Kinematics joint rotation for the hip, knee and the ankle. Inverse Kinematics is necessary for such a case. There exist both an analytical method and a numerical method of solving Inverse Kinematics. First we will explain how to solve it analytically. Let\u2019s focus on the right leg of the model shown in Fig. 2.19. The position and attitude of the body and right leg will be (p1,R1) and (p7,R7) respectively. To simplify the equation we will define D, which is the distance between the Body origin and the hip joint. The upper leg length is A, and the lower leg length is B, as shown in Fig. 2.25(a). So therefore, the position of the hip would be 2.5 Kinematics of a Humanoid Robot 51 p2 = p1 +R1 \u23a1 \u23a3 0 D 0 \u23a4 \u23a6 . Next, we calculate the position of the crotch viewed from the ankle coordinate space r = RT 7 (p2 \u2212 p7) \u2261 [rx ry rz ] T ",
            " By expanding the left side of this equation and calculating the right hand side we get the following \u23a1 \u23a3 c2c4 \u2212 s2s3s4 \u2212s2c3 c2s4 + s2s3c4 s2c4 + c2s3s4 c2c3 s2s4 \u2212 c2s3c4 \u2212c3s4 s3 c3c4 \u23a4 \u23a6 = \u23a1 \u23a3 R11 R12 R13 R21 R22 R23 R31 R32 R33 \u23a4 \u23a6 where c2 \u2261 cos q2, and s2 \u2261 sin q2. By looking carefully at the left hand side of this equation we get, q2 = atan2(\u2212R12, R22) (2.62) q3 = atan2(R32,\u2212R12s2 +R22c2) (2.63) q4 = atan2(\u2212R31, R33). (2.64) An implementation of the above is shown in Fig. 2.2615. For the left leg, we could invert the sign on D and apply the same program. This implementation can only be applied to a robot which has the same layout as the one in Fig. 2.19. If the robot does not have three joint axes that intersect each other at one point, we need an entirely different algorithm. There are many different algorithms outlined in the robot textbooks 15 As a practical implementation, our program covers the target position exceeding the leg length and joint angle limits. 2.5 Kinematics of a Humanoid Robot 53 (for instance [96, 127]), but in general it requires a large amount of heavy calculation, so it is more common to use the numerical solution which we will go over in the next section"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure9.10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure9.10-1.png",
        "caption": "FIGURE 9.10. A rigid rectangular link in the principal and non principal frames.",
        "texts": [
            "199) Since the coordinate frame is central, the products of inertia must be zero. To show this, we examine Ixy. Ixy = Iyx = \u2212 Z B xy dm = Z v xy\u03c1dv = m lwh Z h/2 \u2212h/2 Z w/2 \u2212w/2 Z l/2 \u2212l/2 xy dxdy dz = 0 (9.200) 9. Applied Dynamics 549 Therefore, the moment of inertia matrix for the rigid rectangular bar in its central frame is I = \u23a1\u23a3 m 12 \u00a1 w2 + h2 \u00a2 0 0 0 m 12 \u00a1 h2 + l2 \u00a2 0 0 0 m 12 \u00a1 l2 + w2 \u00a2 \u23a4\u23a6 . (9.201) Example 357 Translation of the inertia matrix. The moment of inertia matrix of the rigid body shown in Figure 9.10, in the principal frame B(oxyz) is given in Equation (9.201). The moment of inertia matrix in the non-principal frame B0(ox0y0z0) can be found by applying the parallel-axes transformation formula (9.169). B0 I = BI +m B0 r\u0303C B0 r\u0303TC (9.202) The mass center is at B0 rC = 1 2 \u23a1\u23a3 l w h \u23a4\u23a6 (9.203) and therefore, B0 r\u0303C = 1 2 \u23a1\u23a3 0 \u2212h w h 0 \u2212l \u2212w l 0 \u23a4\u23a6 (9.204) that provides B0 I = \u23a1\u23a3 1 3h 2m+ 1 3mw2 \u221214 lmw \u221214hlm \u221214 lmw 1 3h 2m+ 1 3 l 2m \u221214hmw \u221214hlm \u2212 14hmw 1 3 l 2m+ 1 3mw2 \u23a4\u23a6 . (9.205) 550 9. Applied Dynamics Example 358 Principal rotation matrix"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000956_s00170-014-6214-8-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000956_s00170-014-6214-8-Figure1-1.png",
        "caption": "Fig. 1 Schematic overview of the SLM Process",
        "texts": [
            " In a first step, a thin layer of metal powder (20\u2013150 \u03bcm) is deposited by means of a powder deposition system. In the second step, a laser melts the powder according to a predefined scanning path [13]. This scanning path is a line pattern generated to fill the two-dimensional layer contours extracted from the three-dimensional CAD model. Once the layer is scanned by the laser, the part is lowered one layer thickness and the cycle can repeat itself. A schematic overview of the process is illustrated in Fig. 1. Due to the very local laser heat input, the material will be different in comparison to conventional production techniques for bulk materials, such as casting or extrusion. The localized heat input, induced by the laser, results in a high thermal gradient in space and time, creating thermal stresses. Thermal stresses can induce deformations which can even cause a job to fail [15, 17]. The high cooling rates result in a very fine microstructure, influencing the mechanical performances of the product [18]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure4.45-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure4.45-1.png",
        "caption": "Figure 4.45 If the support of a rigid body or self-contained structure has more than three support reactions, then the support is statically indeterminate.",
        "texts": [
            " No force equilibrium is possible in the direction normal to the bars and the block is able to move in that direction. With less than three support reactions, there are more equilibrium equations than unknowns. Here the support is also kinematically indeterminate: the conditions for all three equilibrium equations cannot be met for arbitrary loading. In Figure 4.44 in case (a) moment equilibrium is not possible and a rotation occurs about A. In case (b) horizontal force equilibrium is not possible, and the block will move horizontally. With more than three bar supports, such as in Figure 4.45, which do not all pass through a single point and are not all parallel, the support is immovable or kinematically determinate. The number of unknown support reactions is now larger than the available number of equilibrium equations and a unique solution is impossible. In fact, there is an infinity of solutions that satisfy the equilibrium equations. An immovable support of a rigid body or self-contained structure with more than three support reactions is therefore referred to as being statically indeterminate (or hyperstatic)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003908_978-1-84882-730-1-Figure11.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003908_978-1-84882-730-1-Figure11.3-1.png",
        "caption": "Figure 11.3 Ship parking problem",
        "texts": [
            "2 can be applied by setting u0 equal to zero. However, this will result in a yaw angle that may be very different from the desired parking one, at the desired parking position, since our proposed path-following controller is designed to drive ze to a small ball, not to zero for reasons of robustness. To resolve this problem, we first generate a regular curve, \u02ddp.xd ;yd /, which goes via the parking position and its tangent angle at the parking position is equal to the desired parking yaw angle, see Figure 11.3. For simplicity of calculation, the curve can be taken as a straight line in almost all cases of the vessel\u2019s initial conditions. Then the proposed path-following controller can be used to make the vessel follow \u02ddp.xd ;yd /. In this case, the velocity u0 should be chosen such that it is equal to zero or tends to zero when the virtual ship tends to the desired parking position, i.e., limzep!0u0 D 0 with zep Dp .xd xp/2C .yd yp/2. A simple choice can be taken as u0 D u 0.1 e 1zep /e 2ze ; (11.45) where i >0; i D 1;2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.48-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.48-1.png",
        "caption": "FIGURE 8.48. The tire, wheel, and wheel body frames of a steered wheel.",
        "texts": [
            " The associated rotation matrix is CRT = \u23a1\u23a3 cos \u03b4 \u2212 sin \u03b4 0 sin \u03b4 cos \u03b4 0 0 0 1 \u23a4\u23a6 (8.46) Therefore, the transformation between the tire and wheel-body frames can be expressed by Cr = CRT T r+ CdT (8.47) or equivalently, by a homogeneous transformation matrix CTT . CTT = \u2219 CRT CdT 0 1 \u00b8 = \u23a1\u23a2\u23a2\u23a3 cos \u03b4 \u2212 sin \u03b4 0 0 sin \u03b4 cos \u03b4 0 0 0 0 1 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 (8.48) As an example, the wheel-body coordinates of the point P on the tread of a 492 8. Suspension Mechanisms negatively steered tire at the position shown in Figure 8.48, are: Cr = CTT T rP = \u23a1\u23a2\u23a2\u23a3 cos\u2212\u03b4 \u2212 sin\u2212\u03b4 0 0 sin\u2212\u03b4 cos\u2212\u03b4 0 0 0 0 1 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 \u23a1\u23a2\u23a2\u23a3 Rw 0 Rw 1 \u23a4\u23a5\u23a5\u23a6 = \u23a1\u23a2\u23a2\u23a3 Rw cos \u03b4 \u2212Rw sin \u03b4 0 1 \u23a4\u23a5\u23a5\u23a6 (8.49) The homogeneous transformation matrix for tire to wheel-body frame TTC is: TTC = CT\u22121T = \u2219 CRT CdT 0 1 \u00b8\u22121 = \u2219 CRT T \u2212CRT T CdT 0 1 \u00b8 = \u2219 CRT T \u2212 T CdT 0 1 \u00b8 = \u23a1\u23a2\u23a2\u23a3 cos \u03b4 sin \u03b4 0 0 \u2212 sin \u03b4 cos \u03b4 0 0 0 0 1 Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 (8.50) Example 331 F Cycloid. Assume that the wheel in Figure 8.48 is turning with angular velocity \u03c9 and has no slip on the ground. If the point P is at the center of the tireprint 8. Suspension Mechanisms 493 when t = 0, MrP = \u23a1\u23a3 0 0 \u2212Rw \u23a4\u23a6 (8.51) then we can find its position in the wheel frame at a time t by employing another coordinate frame M . The frame M is called the rim frame and is stuck to the wheel at its center. Because of spin, the M frame turns about the yw-axis, and therefore, the rotation matrix to go from the rim frame to the wheel frame is: WRM = \u23a1\u23a3 cos\u03c9t 0 sin\u03c9t 0 1 0 \u2212 sin\u03c9t 0 cos\u03c9t \u23a4\u23a6 (8"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure2.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure2.1-1.png",
        "caption": "FIGURE 2.1. A parked car on level pavement.",
        "texts": [
            " We ignore air friction and examine the load variation under the tires to determine the vehicle\u2019s limits of acceleration, road grade, and kinematic capabilities. 2.1 Parked Car on a Level Road When a car is parked on level pavement, the normal force, Fz, under each of the front and rear wheels, Fz1 , Fz2 , are Fz1 = 1 2 mg a2 l (2.1) Fz2 = 1 2 mg a1 l (2.2) where, a1 is the distance of the car\u2019s mass center, C, from the front axle, a2 is the distance of C from the rear axle, and l is the wheel base. l = a1 + a2 (2.3) 40 2. Forward Vehicle Dynamics Proof. Consider a longitudinally symmetrical car as shown in Figure 2.1. It can be modeled as a two-axel vehicle. A symmetric two-axel vehicle is equivalent to a rigid beam having two supports. The vertical force under the front and rear wheels can be determined using planar static equilibrium equations. X Fz = 0 (2.4)X My = 0 (2.5) Applying the equilibrium equations 2Fz1 + 2Fz2 \u2212mg = 0 (2.6) \u22122Fz1a1 + 2Fz2a2 = 0 (2.7) provide the reaction forces under the front and rear tires. Fz1 = 1 2 mg a2 a1 + a2 = 1 2 mg a2 l (2.8) Fz2 = 1 2 mg a1 a1 + a2 = 1 2 mg a1 l (2.9) Example 39 Reaction forces under wheels"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003690_1.1802311-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003690_1.1802311-Figure5-1.png",
        "caption": "Fig. 5 Geometry of an angular contact ball bearing",
        "texts": [
            "$q%5$Fb% (23) where @M b# is the mass matrix, @M b#C is the mass matrix used for computing the centrifugal forces, @Gb# is the gyroscopic matrix which is skew-symmetric, @Kb# is the stiffness matrix, @Kb#P is the stiffness matrix due to the axial force, and $Fb% is the force vector that includes distributed and concentrated forces. The superscript b represents the beam. The details of the matrices can be found in Appendix A. The damping matrix is not included here and is estimated from experimentally identified modal damping. Contact Force and Deformation for Angular Contact Ball Bearings. The geometry of an angular contact ball bearing and coordinate system are shown in Fig. 5. The Hertzian contact forces between the inner ring and the balls, and the outer ring and the balls are expressed by @18#: Qik5Kid ik 3/2 (24) Qok5Kodok 3/2 where Ki and Ko can be obtained from the following formula @19,20#: Q5Kd3/2 K5 pkE8 3F A2ER F k51.0339S Ry Rx D 0.6360 (25) F51.527710.6023 lnS Ry Rx D E51.000310.5968S Rx Ry D and E852/(12ma 2)/Ea1(12mb 2)/Eb . For the inner ring contact, Rx and Ry are: Rx5 D 2 ~12g i!; Ry5D f i 2 f i21 (26) where g i5D cos uik /Dm . For the outer ring contact, Rx and Ry , are: 1092 \u00d5 Vol"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure6.12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure6.12-1.png",
        "caption": "Figure 6.12 Distributed loads normal to the member axis: (a) a distributed load q(x) as a function of x; (b) a distributed load that changes direction at A; (c) a uniformly distributed load; (d) a linearly distributed load.",
        "texts": [
            " For information concerning load The material factor accounts for insecurities in construction. As such, steel 6 Loads 219 factors and material factors, please refer to the regulations, building codes and relevant books. In this book, all the examples use only design values. When working with a distributed load, it can sometimes be useful to replace it (temporarily) by its resultant. This section addresses the calculation of the resultant of a distributed load. Most attention is devoted to line loads on a member. Figure 6.12a provides a schematic representation of a line load q on (a part of) a member. The direction of the distributed load is shown by means of arrows. The load in Figure 6.12a is acting normal to the member axis and is a function of x. Other examples of distributed loads acting normal to the member axis are shown in Figures 6.12b to 6.12d. In the special case that the distributed load is constant, we refer to a uniformly distributed load (see Figure 6.12c). The distributed load in Figure 6.12d is known as a linearly distributed load; it varies linearly from q(x1) = 3 kN/m to q(x2) = 5 kN/m. A distributed load can also act in the direction of the member axis. Figure 6.13, for example, shows the uniformly distributed load q on a column as a result of its dead weight. A distributed load q , acting at an angle to a member, can be resolved into directions parallel to and normal to the member axis (see Figure 6.14). In the xz coordinate system shown, qx = q cos \u03b1 and qz = q sin \u03b1 are called the components of q "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure9.30-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure9.30-1.png",
        "caption": "Fig. 9.30. Coin type AFPM vibration motor for mobile phones [62]: (a) two coil motor; (b) multicoil motor. 1 \u2014 phase coil, 2 \u2014 PM (mechanically unbalanced system), 3 \u2014 ferromagnetic yoke, 4 \u2014 cover, 5 \u2014 base plate, 6 \u2014 shaft, 7 \u2014 bearing, 8 \u2014 detent iron [62].",
        "texts": [
            " Since a mobile phone is now a gadget as necessary as a watch or wallet, small vibration motors with diameters from 6 to 14 mm are manufactured in unbelievable large quantities (Fig. 9.29). The trends in vibration motors for mobile phones include reduction of mass and size, minimization of energy consumption and guaranteed stable vibration alarming in any circumstances [62]. There are two types of brushless vibration motors for mobile phones: cylindrical or RFPM motor and coin type or AFPM motor (Fig. 9.30). The unbalanced exciting force generated by an eccentric rotor is F = m\u03b5\u2126 = 2\u03c0m\u03b5n2 (9.7) where m, \u03b5 and n denote the rotor mass, eccentricity and rotational speed respectively. The rotational speed is the most effective design parameter to increase the unbalanced exciting force [62]. 9.9 Vibration Motors 311 312 9 Applications Moving Magnet Technologies single-phase AFPM brushless vibration motors provide a strong vibration feeling and silent alert with a very simple design (Fig. 9.31). Fabrication of MMT vibration motors is cost effective due to contactless design, scalable size and slim shape"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure13.30-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure13.30-1.png",
        "caption": "FIGURE 13.30 Two tube or wire segments in series for coupling calculations.",
        "texts": [
            " It is clear that the large increase 356 STABILITY AND PASSIVITY OF PEEC MODELS in the resistive part shown in Fig. (13.28) is very desirable from a stability and passivity point of view. This issue is particularly important for closely located coupling elements due to their larger impact. In this section, we consider a specifice example based on the derivation in Section 5.8.1 for the partial inductances of coupling wire which are in line with retardation. The geometry for this case is again shown in Fig. 13.30. We are interested in an analytic solution for the SOLVER BASED STABILITY AND PASSIVITY ENHANCEMENT TECHNIQUES 357 long objects to save evaluation time. Both wires are less than \u03bb\u221520 in length for the highest frequency considered. Starting from Section 5.8.1, we start by adding and subtracting e\u2212j\ud835\udefdb2 Lp12 to get LpR 12 = e\u2212j\ud835\udefd\u2217b2 Lp12 \u2212 e\u2212j+\ud835\udefd\u2217b2 Lp12 + LpRM 12 . (13.32) where LpRM 12 is given in (5.67). It is based on inductance for the case \u2013 where the delay is taken into account only external to the integral \u2013 as opposed to the case where the retardation is included in the partial element",
            "28) as Re(LpM 12) = \ud835\udf07e1(e2 \u2212 b2)\ud835\udefd2 96\ud835\udf0bb3 2 [7b4 2 + 10e1b2 2 \u2212 17b3 2e2 + +13b2 2e2 2 + 6e1b2e2 2 + 12e1b3 2 + 3e3 1b2 + \u22126e2 1b2e2 \u2212 18e1b2 2e2 \u2212 3b2e3 2] (13.34) and the imaginary part is Im(LpM 12) = \ud835\udf07e1(e2 \u2212 b2) 96\ud835\udf0bb3 2 [\ud835\udefd(\u221212e1e2 2 + 6e3 2 \u2212 26b3 2 + +36e1e2b2 \u2212 6e3 1 \u2212 26e2 2b2 + 46e2b2 2 + \u221220e2 1b2 \u2212 36e1b2 2 + 12e2 1e2) + +\ud835\udefd3(2e1b4 2 + 2e2 1b3 2 \u2212 3b4 2e2 + e13b2 2 + \u2212b2 2e3 2 + 3b3 2e2 2 \u2212 2e2 1b2 2e2 + b5 2 + +2e1b2 2e2 2 \u2212 4e1b3 2e2)]. (13.35) where the physical parameters are specified in Fig. 13.30. Finally, the result for the entire coupling factor will be LpR 12 = e\u2212j\ud835\udefdb2 [Lp12 \u2212 Re(LpM 12) \u2212 jIm(LpM 12)] (13.36) which results in an additional contribution to the quasi-static partial inductance computation. Next, we give an example for the additional contribution of the magnitude of LpM 12 in comparison to the quasi-static impedance \ud835\udf14Lp12. In this comparison, we ignore the impact of the external retardation since it represents the same multiplier in both cases. In the example, we chose f = 10GHz, where each wire is \u03bb\u221520 = 1"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure3.38-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure3.38-1.png",
        "caption": "Figure 3.38 The right part of the three-hinged frame ABC as an example of a body subject to three forces at three points: (a) moment equilibrium exists because the lines of action pass through a single point and (b) there is force equilibrium because the forces form a closed force polygon.",
        "texts": [
            " The same closed force polygon (c) is applicable for both bodies (a) and (b) in Figure 3.37: there is therefore force equilibrium in both cases. In case (a) there is moment equilibrium. This is easily checked by determining the moment of the three forces about the intersection of the three lines of action: none of the forces contribute to the sum of the moments. There is no moment equilibrium in case (b). The system of forces forms a resultant couple. The magnitude of the couple is determined by deriving the sum of the moments about the intersection of two lines of action. Figure 3.38 shows the right-hand part BC of the three-hinged frame, mentioned earlier. This part of the frame is loaded by the vertical force F shown. In addition, the left frame part AB is exerting a force FB at B on BC and the foundation is exerting a force FC at C on BC. Moment equilibrium is only possible if the lines of action of the three forces F , FB and FC pass through a single point. The force equilibrium exists if the three forces form a closed force polygon. Example The block in Figure 3.39, loaded by two forces in C, is kept in equilibrium by the three forces Ah, Av and Bv"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.45-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.45-1.png",
        "caption": "FIGURE 8.45. A steerable wheel must have three DOF.",
        "texts": [],
        "surrounding_texts": [
            "The suspension is what links the wheels to the vehicle body and allows relative motion. This chapter covers the suspension mechanisms, and discusses the possible relative motions between the wheel and the vehicle body. The wheels, through the suspension linkage, must propel, steer, and stop the vehicle, and support the associated forces. 8.1 Solid Axle Suspension The simplest way to attach a pair of wheels to a vehicle is to mount them at opposite ends of a solid axle, such as the one that is shown in Figure 8.1. The solid axle must be attached to the body such that an up and down motion in the z-direction, as well as a roll rotation about the x-axis, is possible. So, no forward and lateral translation, and also no rotation about the axle and the z-axis, is allowed. There are many combinations of links and springs that can provide the kinematic and dynamic requirements. The simplest design is to clamp the axle to the middle of two leaf springs with their ends tied or shackled to the vehicle frame as shown schematically in Figure 8.1. A side view of a multi-leaf spring and solid axle is shown in Figure 8.2. A suspension with a solid connection between the left and right wheels is called dependent suspension. 456 8. Suspension Mechanisms The performance of a solid axle with leaf springs suspension can be improved by adding a linkage to guide the axle kinematically and provide dynamic support to carry the non z-direction forces. The solid axle with leaf spring combination came to vehicle industry from horse-drawn vehicles. Example 299 Hotchkiss drive. When a live solid axle is connected to the body with nothing but two leaf springs, it is called the Hotchkiss drive, which is the name of the car that used it first. The main problems of a Hotchkiss drive, which is shown in Figure 8.2, are locating the axle under lateral and longitudinal forces, and having a low mass ratio \u03b5 = ms/mu, where ms is the sprung mass and mu is the unsprung mass. Sprung mass refers to all masses that are supported by the spring, such as vehicle body. Unsprung mass refers to all masses that are attached to and not supported by the spring, such as wheel, axle, or brakes. Example 300 Leaf spring suspension and flexibility problem. The solid axle suspension systems with longitudinal leaf springs have many drawbacks. The main problem lies in the fact that springs themselves act as locating members. Springs are supposed to flex under load, but their flexibility is needed in only one direction. However, it is the nature of leaf springs to twist and bend laterally and hence, flex also in planes other than the tireplane. Leaf springs are not suited for taking up the driving and braking traction forces. These forces tend to push the springs into an S-shaped profile, as shown in Figure 8.3. The driving and braking flexibility of leaf springs, generates a negative caster and increases instability. Long springs provide better ride. However, long sprigs exaggerate their bending and twisting under different load conditions. Example 301 Leaf spring suspension and flexibility solution. To reduce the effect of a horizontal force and S-shaped profile appearance in a solid axle with leaf springs, the axle may be attached to the chassis by a longitudinal bar as Figure 8.4(a) shows. Such a bar is called an anti-tramp 8. Suspension Mechanisms 457 bar, and the suspension is the simplest cure for longitudinal problems of a Hotchkiss drive. A solid axle with an anti-tramp bar may be kinematically approximated by a four-bar linkage, as shown in Figure 8.4(b). Although an anti-tramp bar may control the shape of the leaf spring, it introduces a twisting angle problem when the axle is moving up and down, as shown in Figure 8.5. Twisting the axle and the wheel about the axle is called caster. The solid axle is frequently used to help keeping the wheels perpendicular to the road. Example 302 Leaf spring location problem. The front wheels need room to steer left and right. Therefore, leaf springs cannot be attached close to the wheel hubs, and must be placed closer to the middle of the axle. That gives a narrow spring-base, which means that a small side force can sway or tilt the body relative to the axle through a considerable roll angle due to weight transfer. This is uncomfortable for the vehicle passengers, and may also produce unwanted steering. The solid axle positively prevents the camber change by body roll. The wheels remain upright and hence, do not roll on a side. However, a solid axle shifts laterally from its static plane and its center does not remain on the vehicle\u2019s longitudinal axis under a lateral force. 458 8. Suspension Mechanisms A solid axle produces bump-camber when single-wheel bump occurs. If the right wheel goes over a bump, the axle is raised at its right end, and that tilts the left wheel hub, putting the left wheel at a camber angle for the duration of deflection. Example 303 Triangular linkage. A triangulated linkage, as shown in Figure 8.6, may be attached to a solid axle to provide lateral and twist resistance during acceleration and braking. Example 304 Panhard arm. High spring rate is a problem of leaf springs. Reducing their stiffness by narrowing them and using fewer leaves, reduces the lateral stiffness and increases the directional stability of the suspension significantly. A Panhard arm is a bar that attaches a solid axle suspension to the chassis laterally. Figure 8.7 illustrates a solid axle and a Panhard arm to guide the axle. Figure 8.8 shows a triangular linkage and a Panhard arm combination for guiding a solid axle. A double triangle mechanism, as shown in Figure 8.9, is an alternative design to guide the axle and support it laterally. Example 305 Straight line linkages. There are many mechanisms that can provide a straight line motion. The simplest mechanisms are four-bar linkages with a coupler point moving straight. Some of the most applied and famous linkages are shown in Figure 8.10. By having proper lengths, the Watt, Robert, Chebyshev, and Evance linkages can make the coupler point C move on a straight line vertically. Such a mechanism and straight motion may be used to guide a solid axle. Two Watt suspension mechanisms with a Panhard arm are shown in Figures 8.11 and 8.12. 460 8. Suspension Mechanisms 8. Suspension Mechanisms 461 Figures 8.13, 8.14, and 8.15 illustrate three combinations of Robert suspension linkages equipped with a Panhard arm. Example 306 Solid axle suspension and unsprung mass problem. A solid axle is counted as an unsprung member, and hence, the unsprung mass is increased where using solid axle suspension. A heavy unsprung mass ruins both, the ride and handling of a vehicle. Lightening the solid axle makes it weaker and increases the most dangerous problem in vehicles: axle breakage. The solid axle must be strong enough to make sure it will not break under any loading conditions at any age. As a rough estimate, 90% of the leaf spring mass may also be counted as unsprung mass, which makes the problem worse. The unsprung mass problem is worse in front, and it is the main reason that they are no longer used in street cars. However, front solid axles are still common on trucks and buses. These are heavy vehicles and solid axle 8. Suspension Mechanisms 463 464 8. Suspension Mechanisms suspension does not reduce the mass ratio \u03b5 = ms/mu very much. When a vehicle is rear-wheel-drive and a solid axle suspension is used in the back, the suspension is called live axle. A live axle is a casing that contains a differential, and two drive shafts. The drive shafts are connected to the wheel hubs. A live axle can be three to four times heavier than a dead I-beam axle. It is called live axle because of rotating gears and shafts inside the axle. Example 307 Solid axle and coil spring. To decrease the unsprung mass and increase vertical flexibility of solid axle suspensions, it is possible to equip them with coil springs. A sample of a solid axle suspension with coil spring is shown in Figure 8.16. The suspension mechanism is made of four longitudinal bars between the axle and chassis. The springs may have some lateral or longitudinal angle to introduce some lateral or longitudinal compliance. Example 308 De Dion axle. When a solid axle is a dead axle with no driving wheels, the connecting beam between the left and right wheels may have different shapes to do different jobs, usually to give the wheels independent flexibility. We may also modify the shape of a live axle to attach the differential to the chassis and reduce the unsprung mass. De Dion design is a modification of a beam axle that may be used as a dead axle or to attach the differential to the chassis and transfer the driving power to the drive wheels by employing universal joints and split shafts. Figure 8.17 illustrates a De Dion suspension. 8.2 Independent Suspension Independent suspensions is introduced to let a wheel to move up and down without affecting the opposite wheel. There are many forms and designs of independent suspensions. However, double A-arm and McPherson strut suspensions are the simplest and the most common designs. Figure 8.18 illustrates a sample of a double A-arm and Figure 8.19 shows a McPherson suspension. Kinematically, a double A-arm suspension mechanism is a four-bar linkage with the chassis as the ground link, and coupler as the wheel carrying link. A McPherson suspension is an inverted slider mechanism that has the chassis as the ground link and the coupler as the wheel carrying link. A double A-arm and a McPherson suspension mechanism on the left and right wheels are schematically shown in Figures 8.20 and 8.21 respectively. Double A-arm, is also called double wishbone, or short/long arm suspension. McPherson also may be written as MacPherson. Example 309 Double A-arm suspension and spring position. Consider a double A-arm suspension mechanism. The coil spring may be between the lower arm and the chassis, as shown in Figure 8.18. It is also possible to install the spring between the upper arm and the chassis, or between the upper and lower arms. In either case, the lower or the upper arm, which supports the spring, is made stronger and the other arm acts as a connecting arm. Example 310 Multi-link suspension mechanism. When the two side bars of an A-arm are attached to each other with a joint, as shown in Figure 8.22, then the double A-arm is called a multi-link mechanism. A multi-link mechanism is a six-bar mechanism that may have a better coupler motion than a double A-arm mechanism. However, multi- 466 8. Suspension Mechanisms 8. Suspension Mechanisms 467 link suspensions are more expensive, less reliable, and more complicated compare to a double A-arm four-bar linkage. There are vehicles with more than six-link suspension with possibly better kinematic performance. Example 311 Swing arm suspension. An independent suspension may be as simple as a triangle shown in Figure 8.23. The base of the triangle is jointed to the chassis and the wheel to the tip point. The base of the triangle is aligned with the longitudinal axis of the vehicle. Such a suspension mechanism is called a swing axle or swing arm. The variation in camber angle for a swing arm suspension is maximum, compared to the other suspension mechanisms. Example 312 Trailing arm suspension. Figure 8.24 illustrates a trailing arm suspension that is a longitudinal arm with a lateral axis of rotation. The camber angle of the wheel, supported by a trailing arm, will not change during the up and down motion. Trailing arm suspension has been successfully using in a variety of frontwheel-drive vehicles, to suspend their rear wheels. Example 313 Semi-trailing arm Semi-trailing arm suspension, as shown in Figure 8.25, is a compromise between the swing arm and trailing arm suspensions. The joint axis may have any angle, however an angle not too far from 45 deg is more applied. Such suspensions have acceptable camber angle change, while they can handle both, the lateral and longitudinal forces. Semi-trailing design has successfully applied to a series of rear-wheel-drive cars for several decades. Example 314 Antiroll bar and roll stiffness. Coil springs are used in vehicles because they are less stiff with better 470 8. Suspension Mechanisms ride comfort compared to leaf springs. Therefore, the roll stiffness of the vehicle with coil springs is usually less than in vehicles with leaf springs. To increase the roll stiffness of such suspensions, an antiroll bar must be used. Leaf springs with reduced layers, uni-leaf, trapezoidal, or nonuniform thickness may also need an antiroll bar to compensate for their reduced roll stiffness. The antiroll bar is also called a stabilizer. Figure 8.26 illustrates an anti-roll bare attached to a solid axle with coil springs. Example 315 Need for longitudinal compliance. A bump is an obstacle on the road that opposes the forward motion of a wheel. When a vehicle goes over a bump, the first action is a force that tends to push the wheel backward relative to the rest of the vehicle. So, the lifting force has a longitudinal component, which will be felt inside the vehicle unless the suspension system has horizontal compliance. There are situations in which the horizontal component of the force is even higher than the vertical component. Leaf springs can somewhat absorb this horizontal force by flattening out and stretching the distance from the forward spring anchor and the axle. Such a stretch is usually less than 1/2 in \u2248 1 cm. 8.3 Roll Center and Roll Axis The roll axis is the instantaneous line about which the body of a vehicle rolls. Roll axis is found by connecting the roll center of the front and rear suspensions of the vehicle. Assume we cut a vehicle laterally to disconnect the front and rear half of the vehicle. Then, the roll center of the front or rear suspension is the instantaneous center of rotation of the body with respect to the ground. 8. Suspension Mechanisms 471 472 8. Suspension Mechanisms Figure 8.27 illustrates a sample of the front suspensions of a car with a double A-arm mechanism. To find the roll center of the body with respect to the ground, we analyze the two-dimensional kinematically equivalent mechanism shown in Figure 8.28. The center of tireprint is the instant center of rotation of the wheel with respect to the ground, so the wheels are jointed links to the ground at their center of tireprints. The instant center I18 is the roll center of the body with respect to the ground. To find I18, we apply the Kennedy theorem and find the intersection of the line I12I28 and I13I38 as shown in Figure 8.29. The point I28 and I38 are the instant center of rotation for the wheels with respect to the body. The instant center of rotation of a wheel with respect to the body is called suspension roll center. So, to find the roll center of the front or rear half of a car, we should determine the suspension roll centers, and find the intersection of the lines connecting the suspension roll centers to the center of their associated tireprints. The Kennedy theorem states that the instant center of every three relatively moving objects are colinear. Example 316 McPherson suspension roll center. A McPherson suspension is an inverted slider crank mechanism. The instant centers of an example of inverted slider crank mechanism are shown in Figure 8.30. In this figure, the point I12 is the suspension roll center, which is the instant center of rotation for the wheel link number 2 with respect to the chassis link number 1. A car with a McPherson suspension system is shown in Figure 8.31. The kinematic equivalent mechanism is depicted in Figure 8.32. Suspension roll centers along with the body roll center are shown in Figure 8.33. To 8. Suspension Mechanisms 473 find the roll center of the front or rear half of a car, we determine each suspension roll center and then find the intersection of the lines connecting the suspension roll centers to the center of the associated tireprint. Example 317 Roll center of double A-arm suspension. The roll center of an independent suspension such as a double A-arm can be internal or external. The kinematic model of a double A-arm suspension for the front left wheel of a car is illustrated in Figure 8.34. The suspension roll center in Figure 8.34(a) is internal, and in Figure 8.34(b) is external. An internal suspension roll center is toward the vehicle body, while an external suspension roll center goes away from the vehicle body. A suspension roll center may be on, above, or below the road surface, as shown in Figure 8.35(a)-(c) for an external suspension roll center. When the suspension roll center is on the ground, above the ground, or below the ground, the vehicle roll center would be on the ground, below the ground, and above the ground, respectively. Example 318 F Camber variation of double A-arm suspension. When a wheel moves up and down with respect to the vehicle body, de- 474 8. Suspension Mechanisms 8. Suspension Mechanisms 475 pending on the suspension mechanism, the wheel may camber. Figure 8.36 illustrates the kinematic model for a double A-arm suspension mechanism. The mechanism is equivalent to a four-bar linkage with the ground link as the vehicle chasis. The wheel is always attached to a coupler point C of the mechanism. We set a local suspension coordinate frame (x, y) with the x-axis indicating the ground link MN . The x-axis makes a constant angle \u03b80 with the vertical direction. The suspension machanism has a length a for the upper A-arm, b for the coupler link, c for the lower A-arm, and d for the ground link. The configuration of the suspension is determined by the angles \u03b82, \u03b83, and \u03b84, all measured from the positive direction of the x-axis. When the suspension is at its equilibrium position, the links of the double A-arm suspension make initial angles \u03b820 \u03b830, and \u03b840 with the x-axis. The equilibrium position of a suspension is called the rest position. To determine the camber angle during the fluctuation of the wheel, we should determine the variation of the coupler angle \u03b83, as a function of vertical motion z of the coupler point C. Using \u03b82 as a parameter, we can find the coordinates (xC , yC) of the 476 8. Suspension Mechanisms coupler point C in the suspension coordinate frame (x, y) as xC = a cos \u03b82 + e cos (p\u2212 q + \u03b1) (8.1) yC = a sin \u03b82 + e sin (p\u2212 q + \u03b1) (8.2) where, q = tan\u22121 a sin \u03b82 d\u2212 a cos \u03b82 (8.3) p = tan\u22121 q 4b2f2 \u2212 (b2 + f2 \u2212 c2) 2 b2 + f2 \u2212 c2 (8.4) f = p a2 + d2 \u2212 2ad cos \u03b82. (8.5) The position vector of the coupler point is uC uC = xC \u0131\u0302+ yC j\u0302 (8.6) and the unit vector in the z-direction is u\u0302z = \u2212 cos \u03b80\u0131\u0302\u2212 sin \u03b80j\u0302. (8.7) Therefore, the displacement z in terms of xC and yC is: z = uC \u00b7 u\u0302z = \u2212xC cos \u03b80 \u2212 yC sin \u03b80 (8.8) 8. Suspension Mechanisms 477 The initial coordinates of the coupler point C and the initial value of z are: xC0 = a cos \u03b820 + e cos (p0 \u2212 q0 + \u03b1) (8.9) yC0 = a sin \u03b820 + e sin (p0 \u2212 q0 + \u03b1) (8.10) z0 = \u2212xC0 cos \u03b80 \u2212 yC0 sin \u03b80 (8.11) and hence, the vertical displacement of the wheel center can be calculated by h = z \u2212 z0 (8.12) The initial angle of the coupler link with the vertical direction is \u03b80 \u2212 \u03b830. Therefore, the camber angle of the wheel would be \u03b3 = (\u03b80 \u2212 \u03b83)\u2212 (\u03b80 \u2212 \u03b830) = \u03b830 \u2212 \u03b83 (8.13) The angle of the coupler link with the x-direction is equal to \u03b83 = 2 tan \u22121 \u00c3 \u2212E \u00b1 \u221a E2 \u2212 4DF 2D ! (8.14) where, D = J5 \u2212 J1 + (1 + J4) cos \u03b82 (8.15) E = \u22122 sin \u03b82 (8.16) F = J5 + J1 \u2212 (1\u2212 J4) cos \u03b82 (8.17) and J1 = d a (8.18) J2 = d c (8.19) J3 = a2 \u2212 b2 + c2 + d2 2ac (8.20) J4 = d b (8.21) J5 = c2 \u2212 d2 \u2212 a2 \u2212 b2 2ab . (8.22) Substituting (8.14) and (8.13), and then, eliminating \u03b82 between (8.13) and (8.8) provides the relationship between the vertical motion of the wheel, z, and the camber angle \u03b3. 478 8. Suspension Mechanisms Example 319 F Camber angle and wheel fluctuations. Consider the double A-arm suspension that is shown in Figure 8.36. The dimensions of the equivalent kinematic model are: a = 22.4 cm b = 22.1 cm c = 27.3 cm d = 17.4 cm \u03b80 = 24.3 deg (8.23) The coupler point C is at: e = 14.8 cm \u03b1 = 54.8 deg (8.24) If the angle \u03b82 at the rest position is at \u03b820 = 121.5 deg (8.25) then the initial angle of the other links are: \u03b830 = 18.36 deg \u03b840 = 107.32 deg (8.26) At the rest position, the coupler point is at: xC0 = \u221222.73 cm yC0 = 9.23 cm z0 = 16.92 cm (8.27) We may calculate h and \u03b3 by varying the parameter \u03b82. Figure 8.37 illustrates h as a function of the camber angle \u03b3. For this suspension mechanism, the wheel gains a positive camber when the wheel moves up, and gains a negative camber when the it moves down. The mechanism is shown in Figure 8.38, when the wheel is at the rest position and has a positive or a negative displacement. 8.4 F Car Tire Relative Angles There are four major wheel alignment parameters that affect vehicle dynamics: toe, camber, caster, and trust angle. 8. Suspension Mechanisms 479 8.4.1 F Toe When a pair of wheels is set so that their leading edges are pointed toward each other, the wheel pair is said to have toe-in. If the leading edges point away from each other, the pair is said to have toe-out. Toe-in and toe-out front wheel configurations of a car are illustrated in Figure 8.39. The amount of toe can be expressed in degrees of the angle to which the wheels are not parallel. However, it is more common to express the toein and toe-out as the difference between the track widths as measured at the leading and trailing edges of the tires. Toe settings affect three major performances: tire wear, straight-line stability, and corner entry handling. For minimum tire wear and power loss, the wheels on a given axle of a car should point directly ahead when the car is running in a straight line. Excessive toe-in causes accelerated wear at the outboard edges of the tires, while too much toe-out causes wear at the inboard edges. Toe-in increases the directional stability of the vehicle, and toe-out increases the steering response. Hence, a toe-in setting makes the steering function lazy, while a toe-out makes the vehicle unstable. With four wheel independent suspensions, the toe may also be set at the rear of the car. Toe settings at the rear have the same effect on wear, directional stability, and turn-in as they do on the front. However, we usually do not set up a rear-drive race car toed out in the rear, because of excessive instability. When driving torque is applied to the wheels, they pull themselves forward and try to create toe-in. Furthermore, when pushed down the road, a non-driven wheel or a braking wheel will tend to toe-out. Example 320 Toe-in and directional stability. Toe settings have an impact on directional stability. When the steering 480 8. Suspension Mechanisms 8. Suspension Mechanisms 481 wheel is centered, toe-in causes the wheels to tent to move along paths that intersect each other in front of the vehicle. However, the wheels are in balance and no turn results. Toe-in setup can increase the directional stability caused by little steering fluctuations and keep the car moving straight. Steering fluctuations may be a result of road disturbances. If a car is set up with toe-out, the front wheels are aligned so that slight disturbances cause the wheel pair to assume rolling directions that approach a turn. Therefore, toe-out encourages the initiation of a turn, while toe-in discourages it. Toe-out makes the steering quicker. So, it may be used in vehicles for a faster response. The toe setting on a particular car becomes a trade-off between the straight-line stability afforded by toe-in and the quick steering response by toe-out. Toe-out is not desirable for street cars, however, race car drivers are willing to drive a car with a little directional instability, for sharper turn-in to the corners. So street cars are generally set up with toe-in, while race cars are often set up with toe-out. Example 321 Toe-in and toe-out in the front and rear axles. Front toe-in: slower steering response, more straight-line stability, greater wear at the outboard edges of the tires. Front toe-zero: medium steering response, minimum power loss, minimum tire wear. Front toe-out: quicker steering response, less straight-line stability, greater wear at the inboard edges of the tires. Rear toe-in: straight-line stability, traction out of the corner, more steerability, higher top speed. 482 8. Suspension Mechanisms 8.4.2 F Caster Angle Caster is the angle to which the steering pivot axis is tilted forward or rearward from vertical, as viewed from the side. Assume the wheel is straight to have the body frame and the wheel frame coincident. If the steering axis is turned about the wheel yw-axis then the wheel has positive caster. If the steering axis is turned about the wheel \u2212yw-axis, then the wheel has negative caster. Positive and negative caster configurations on the front wheel of a car are shown in Figure 8.40. Negative caster aids in centering the steering wheel after a turn and makes the front tires straighten quicker. Most street cars are made with 4\u22126 deg negative caster. Negative caster tends to straighten the wheel when the vehicle is traveling forward, and thus is used to enhance straight-line stability. Example 322 Negative caster of shopping carts. The steering axis of a shopping cart wheel is set forward of where the wheel contacts the ground. As the cart is pushed forward, the steering axis pulls the wheel along, and because the wheel drags along the ground, it falls directly in line behind the steering axis. The force that causes the wheel to follow the steering axis is proportional to the distance between the steering axis and the wheel-to-ground contact point, if the caster is small. This distance is referred to as trail. The cars\u2019 steering axis intersects the ground at a point in front of the tireprint, and thus the same effect as seen in the shopping cart casters is achieved. While greater caster angles improves straight-line stability, they also cause an increase in steering effort. Example 323 Characteristics of caster in front axle. Zero castor provides: easy steering into the corner, low steering out of the corner, low straight-line stability. 8. Suspension Mechanisms 483 Negative caster provides: low steering into the corner, easy steering out of the corner, more straight-line stability, high tireprint area during turn, good turn-in response, good directional stability, good steering feel. When a castered wheel rotates about the steering axis, the wheel gains camber. This camber is generally favorable for cornering. 8.4.3 F Camber Camber is the angle of the wheel relative to vertical line to the road, as viewed from the front or the rear of the car. Figure 8.41 illustrates the wheel number 1 of a vehicle. If the wheel leans in toward the chassis, it is called negative camber and if it leans away from the car, it is called positive camber. The cornering force that a tire can develop is highly dependent on its angle relative to the road surface, and so wheel camber has a major effect on the road holding of a car. A tire develops its maximum lateral force at a small camber angle. This fact is due to the contribution of camber thrust, which is an additional lateral force generated by elastic deformation as the tread rubber pulls through the tire/road interface. To optimize a tire\u2019s performance in a turn, the suspension should provide a slight camber angle in the direction of rotation. As the body rolls in a turn, the suspension deflects vertically. The wheel is connected to the chassis by suspension mechanism, which must rotate to allow for the wheel deflection. Therefore, the wheel can be subject to large camber changes as the suspension moves up and down. So, the more the wheel must deflect from its static position, the more difficult it is to maintain an ideal camber angle. Thus, the relatively large wheel travel and soft roll stiffness needed to provide a smooth ride in passenger cars presents a difficult design challenge, while the small wheel travel and high roll stiffness inherent in racing cars reduces the problem. Example 324 Castor versus camber. Camber doesn\u2019t improve turn-in as the positive caster does. Camber is not generally good for tire wear. Camber in one wheel does not improve directional stability. Camber adversely affects braking and acceleration efforts. 8.4.4 F Trust Angle The trust angle \u03c5 is the angle between vehicle\u2019s centerline and perpendicular to the rear axle. It compares the direction that the rear axle is aimed with the centerline of the vehicle. A nonzero angle configuration is shown in Figure 8.42. Zero angle confirms that the rear axle is parallel to the front axle, and the wheelbase on both sides of the vehicle are the same. A reason for nonzero 484 8. Suspension Mechanisms 8. Suspension Mechanisms 485 trust angle would have unequal toe-in or toe-out on both sides of the axle. Example 325 Torque reaction. There are two kinds of torque reactions in rear-whel-drive: 1\u2212 the reaction of the axle housing to rotate in the opposite direction of the crown wheel rotation, and 2\u2212 the reaction of axle housing to spin about its own center, opposite to the direction of pinion\u2019s rotation. The first reaction leads to a lifting force in the differential causing a wind-up in springs. The second reaction leads to a lifting force on the right wheels. 8.5 Suspension Requirements and Coordinate Frames The suspension mechanism should allow a relative motion between the wheel and the vehicle body. The relative motions are needed to pass the road irregularities and steering. To function properly, a suspension mechanism should have some kinematic and dynamics requirements. 8.5.1 Kinematic Requirements To express the motions of a wheel, we attach a wheel coordinate system W (oxwywzw) to the center of the wheel. A wheel, as a rigid body, has six degrees-of-freedom with respect to the vehicle body: three translations and three rotations, as shown in Figure 8.43. 486 8. Suspension Mechanisms The axes xw, yw, and zw indicate the direction of forward, lateral, and vertical translations and rotations. In the position shown in the figure, the rotation about the xw-axis is the camber angle, about the yw-axis is the spin, and about the zw-axis is the steer angle. Consider a non-steerable wheel. Translation in zw-direction and spin about the yw-axis are the only two DOF allowed for such a wheel. So, we need to take four DOF. If the wheel is steerable, then translation in the zw-direction, spin about the yw-axis, and steer rotation about the zw-axis are the three DOF allowed. So, we must take three DOF of a steerable wheel. Kinematically, non-steerable and steerable wheels should be supported as shown in Figures 8.44 and 8.45 respectively. Providing the required freedom, as well as eliminating the taken DOF, are the kinematic requirements of a suspension mechanism. 8.5.2 Dynamic Requirements Wheels should be able to propel, steer, and stop the vehicle. So, the suspension system must transmit the driving traction and deceleration braking forces between the vehicle body and the ground. The suspension members must also resist lateral forces acting on the vehicle. Hence, the wheel suspension system must make the wheel rigid for the taken DOF. However, there must also be some compliance members to limit the untaken DOF. The most important compliant members are spring and dampers to provide returning and resistance forces in the z-direction. 8. Suspension Mechanisms 487 8.5.3 Wheel, wheel-body, and tire Coordinate Frames Three coordinate frames are employed to express the orientation of a tire and wheel with respect to the vehicle: the wheel frame W , wheel-body frame C, and tire frame T . A wheel coordinate frame W (xw, yw, zw) is attached to the center of a wheel. It follows every translation and rotation of the wheel except the spin. Hence, the xw and zw axes are always in the tire-plane, while the yw-axis is always along the spin axis. A wheel coordinate frame is shown in Figure 8.43. When the wheel is straight and the W frame is parallel to the vehicle coordinate frame, we attach a wheel-body coordinate frame C (xc, yc, zc) at the center of the wheel parallel to the vehicle coordinate axes. The wheelbody frame C is motionless with respect to the vehicle coordinate and does not follow any motion of the wheel. The tire coordinate frame T (xt, yt, zt) is set at the center of the tireprint. The zt-axis is always perpendicular to the ground. The xt-axis is along the intersection line of the tire-plane and the ground. The tire frame does not follows the spin and camber rotations of the tire however, it follows the steer angle rotation about the zc-axis. Figure 8.46 illustrates a tire and a wheel coordinate frames. Example 326 Visualization of the wheel, tire, and wheel-body frames. Figure 8.47 illustrates the relative configuration of a wheel-body frame C, a tire frame T , and a wheel frame W . If the steering axis is along the zc-axis then, the rotation of the wheel about the zc-axis is the steer angle \u03b4. Rotation about the xt-axis is the camber angle \u03b3. 488 8. Suspension Mechanisms Generally speaking, the steering axis may have any angle and may go through any point of the ground plane. Example 327 Wheel to tire coordinate frame transformation. If TdW indicates the T -expression of the position vector of the wheel frame origin relative to the tire frame origin, then having the coordinates of a point P in the wheel frame, we can find its coordinates in the tire frame using the following equation. T rP = TRW W rP + TdW (8.28) If W rP indicates the position vector of a point P in the wheel frame, W rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 (8.29) then the coordinates of the point P in the tire frame T rP are T rP = TRW W rP + Td = TRW W rP + TRW W T dW = \u23a1\u23a3 xP yP cos \u03b3 \u2212Rw sin \u03b3 \u2212 zP sin \u03b3 Rw cos \u03b3 + zP cos \u03b3 + yP sin \u03b3 \u23a4\u23a6 (8.30) 8. Suspension Mechanisms 489 where, WT dW is the W -expression of the position vector of the wheel frame in the tire frame, Rw is the radius of the tire, and TRW is the rotation matrix to go from the wheel frame W to the tire frame T . TRW = \u23a1\u23a3 1 0 0 0 cos \u03b3 \u2212 sin \u03b3 0 sin \u03b3 cos \u03b3 \u23a4\u23a6 (8.31) W T dW = \u23a1\u23a3 0 0 Rw \u23a4\u23a6 . (8.32) As an example, the center of the wheel W rP = W ro = 0 is the origin of the wheel frame W , that is at T ro = TdW = TRW W T dW = \u23a1\u23a3 0 \u2212Rw sin \u03b3 Rw cos \u03b3 \u23a4\u23a6 (8.33) in the tire coordinate frame T . 490 8. Suspension Mechanisms Example 328 F Tire to wheel coordinate frame transformation. If rP indicates the position vector of a point P in the tire coordinate frame, T rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 (8.34) then the position vector W rP of the point P in the wheel coordinate frame is W rP = WRT T rP \u2212 W T dW (8.35) = \u23a1\u23a3 xP yP cos \u03b3 + zP sin \u03b3 zP cos \u03b3 \u2212Rw \u2212 yP sin \u03b3 \u23a4\u23a6 because WRT = \u23a1\u23a3 1 0 0 0 cos \u03b3 sin \u03b3 0 \u2212 sin \u03b3 cos \u03b3 \u23a4\u23a6 (8.36) WdT = \u23a1\u23a3 0 0 Rw \u23a4\u23a6 (8.37) and we may multiply both sides of Equation (8.28) by TRT W to get TRT W T rP = W rP + TRT W TdW (8.38) = W rP + W T dW W rP = WRT T rP \u2212 W T dW . (8.39) As an example, the center of tireprint in the wheel frame is at W rP = \u23a1\u23a3 1 0 0 0 cos \u03b3 \u2212 sin \u03b3 0 sin \u03b3 cos \u03b3 \u23a4\u23a6T \u23a1\u23a3 0 0 0 \u23a4\u23a6\u2212 \u23a1\u23a3 0 0 Rw \u23a4\u23a6 = \u23a1\u23a3 0 0 \u2212Rw \u23a4\u23a6 . (8.40) Example 329 F Wheel to tire homogeneous transformation matrices. The transformation from the wheel to tire coordinate frame may also be expressed by a 4\u00d7 4 homogeneous transformation matrix TTW , T rP = TTW W rP = \u2219 TRW TdW 0 1 \u00b8 W rP (8.41) where TTW = \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 cos \u03b3 \u2212 sin \u03b3 \u2212Rw sin \u03b3 0 sin \u03b3 cos \u03b3 Rw cos \u03b3 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 . (8.42) 8. Suspension Mechanisms 491 The corresponding homogeneous transformation matrix WTT from the tire to wheel frame would be WTT = \u2219 WRT WdT 0 1 \u00b8 = \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 cos \u03b3 \u2212 sin \u03b3 0 0 sin \u03b3 cos \u03b3 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 . (8.43) It can be checked that WTT = TT\u22121W , using the inverse of a homogeneous transformation matrix rule. TT\u22121W = \u2219 TRW TdW 0 1 \u00b8\u22121 = \u2219 TRT W \u2212 TRT W TdW 0 1 \u00b8 = \u2219 WRT \u2212WRT TdW 0 1 \u00b8 (8.44) Example 330 F Tire to wheel-body frame transformation. The origin of the tire frame is at CdT in the wheel-body frame. CdT = \u23a1\u23a3 0 0 \u2212Rw \u23a4\u23a6 (8.45) The tire frame can steer about the zc-axis with respect to the wheel-body frame. The associated rotation matrix is CRT = \u23a1\u23a3 cos \u03b4 \u2212 sin \u03b4 0 sin \u03b4 cos \u03b4 0 0 0 1 \u23a4\u23a6 (8.46) Therefore, the transformation between the tire and wheel-body frames can be expressed by Cr = CRT T r+ CdT (8.47) or equivalently, by a homogeneous transformation matrix CTT . CTT = \u2219 CRT CdT 0 1 \u00b8 = \u23a1\u23a2\u23a2\u23a3 cos \u03b4 \u2212 sin \u03b4 0 0 sin \u03b4 cos \u03b4 0 0 0 0 1 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 (8.48) As an example, the wheel-body coordinates of the point P on the tread of a 492 8. Suspension Mechanisms negatively steered tire at the position shown in Figure 8.48, are: Cr = CTT T rP = \u23a1\u23a2\u23a2\u23a3 cos\u2212\u03b4 \u2212 sin\u2212\u03b4 0 0 sin\u2212\u03b4 cos\u2212\u03b4 0 0 0 0 1 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 \u23a1\u23a2\u23a2\u23a3 Rw 0 Rw 1 \u23a4\u23a5\u23a5\u23a6 = \u23a1\u23a2\u23a2\u23a3 Rw cos \u03b4 \u2212Rw sin \u03b4 0 1 \u23a4\u23a5\u23a5\u23a6 (8.49) The homogeneous transformation matrix for tire to wheel-body frame TTC is: TTC = CT\u22121T = \u2219 CRT CdT 0 1 \u00b8\u22121 = \u2219 CRT T \u2212CRT T CdT 0 1 \u00b8 = \u2219 CRT T \u2212 T CdT 0 1 \u00b8 = \u23a1\u23a2\u23a2\u23a3 cos \u03b4 sin \u03b4 0 0 \u2212 sin \u03b4 cos \u03b4 0 0 0 0 1 Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 (8.50) Example 331 F Cycloid. Assume that the wheel in Figure 8.48 is turning with angular velocity \u03c9 and has no slip on the ground. If the point P is at the center of the tireprint 8. Suspension Mechanisms 493 when t = 0, MrP = \u23a1\u23a3 0 0 \u2212Rw \u23a4\u23a6 (8.51) then we can find its position in the wheel frame at a time t by employing another coordinate frame M . The frame M is called the rim frame and is stuck to the wheel at its center. Because of spin, the M frame turns about the yw-axis, and therefore, the rotation matrix to go from the rim frame to the wheel frame is: WRM = \u23a1\u23a3 cos\u03c9t 0 sin\u03c9t 0 1 0 \u2212 sin\u03c9t 0 cos\u03c9t \u23a4\u23a6 (8.52) So the coordinates of P in the wheel frame are: W rP = WRM MrP = \u23a1\u23a3 \u2212Rw sin t\u03c9 0 \u2212Rw cos t\u03c9 \u23a4\u23a6 (8.53) The center of the wheel is moving with speed vx = Rw\u03c9 and it is at Gr =\u00a3 vxt 0 Rw \u00a4 in the global coordinate frame G on the ground. Hence, the coordinates of point P in the global frame G, would be GrP = W rP + \u23a1\u23a3 vxt 0 Rw \u23a4\u23a6 = \u23a1\u23a3 Rw (\u03c9t\u2212 sin t\u03c9) 0 Rw (1\u2212 cos t\u03c9) \u23a4\u23a6 . (8.54) The path of motion of point P in the (X,Z)-plane can be found by eliminating t between X and Z coordinates. However, it is easier to expressed the path by using \u03c9t as a parameter. Such a path is called cycloid. In general case, point P can be at any distance from the center of the rim frame. If the point is at a distance d 6= Rw, then its path of motion is called the trochoid. A trochoid is called a curtate cycloid when d < Rw and a prolate cycloid when d > Rw. Figure 8.49(a)-(c) illustrate a cycloid, curtate cycloid, and prolate cycloid respectively. Example 332 F Wheel to wheel-body frame transformation. The homogeneous transformation matrix CTW to go from the wheel frame 494 8. Suspension Mechanisms to the wheel-body frame can be found by combined transformation. CTW = CTT TTW (8.55) = \u23a1\u23a2\u23a2\u23a3 c\u03b4 \u2212s\u03b4 0 0 s\u03b4 c\u03b4 0 0 0 0 1 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 c\u03b3 \u2212s\u03b3 \u2212Rw sin \u03b3 0 s\u03b3 c\u03b3 Rw cos \u03b3 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 = \u23a1\u23a2\u23a2\u23a3 cos \u03b4 \u2212 cos \u03b3 sin \u03b4 sin \u03b3 sin \u03b4 Rw sin \u03b3 sin \u03b4 sin \u03b4 cos \u03b3 cos \u03b4 \u2212 cos \u03b4 sin \u03b3 \u2212Rw cos \u03b4 sin \u03b3 0 sin \u03b3 cos \u03b3 Rw cos \u03b3 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 If rP indicates the position vector of a point P in the wheel coordinate frame, W rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 (8.56) then the homogeneous position vector CrP of the point P in the wheel-body 8. Suspension Mechanisms 495 coordinate frame is: CrP = CTW W rP = \u23a1\u23a2\u23a2\u23a3 xP cos \u03b4 \u2212 yP cos \u03b3 sin \u03b4 + (Rw + zP ) sin \u03b3 sin \u03b4 xP sin \u03b4 + yP cos \u03b3 cos \u03b4 \u2212 (Rw + zP ) cos \u03b4 sin \u03b3 \u2212Rw + (Rw + zP ) cos \u03b3 + yP sin \u03b3 1 \u23a4\u23a5\u23a5\u23a6 (8.57) The position of the wheel center W r = 0, for a cambered and steered wheel is at Cr = CTW W r = \u23a1\u23a2\u23a2\u23a3 Rw sin \u03b3 sin \u03b4 \u2212Rw cos \u03b4 sin \u03b3 \u2212Rw(1\u2212 cos \u03b3) 1 \u23a4\u23a5\u23a5\u23a6 (8.58) The zc = Rw (cos \u03b3 \u2212 1) indicates how much the center of the wheel comes down when the wheel cambers. If the wheel is not steerable, then \u03b4 = 0 and the transformation matrix CTW reduces to CTW = \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 cos \u03b3 \u2212 sin \u03b3 \u2212Rw sin \u03b3 0 sin \u03b3 cos \u03b3 Rw (cos \u03b3 \u2212 1) 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 (8.59) that shows CrP = CTW W rP = \u23a1\u23a2\u23a2\u23a3 xP yP cos \u03b3 \u2212Rw sin \u03b3 \u2212 zP sin \u03b3 zP cos \u03b3 + yP sin \u03b3 +Rw (cos \u03b3 \u2212 1) 1 \u23a4\u23a5\u23a5\u23a6 (8.60) Example 333 F Tire to vehicle coordinate frame transformation. Figure 8.50 illustrates the first and fourth tires of a 4-wheel vehicle. There is a body coordinate frame B (x, y, z) attached to the mass center C of the vehicle. There are also two tire coordinate frames T1 (xt1 , yt1 , zt1) and T4 (xt4 , yt4 , zt4) attached to the tires 1 and 4 at the center of their tireprints. The origin of the tire coordinate frame T1 is at Bd1 BdT1 = \u23a1\u23a3 a1 \u2212b1 \u2212h \u23a4\u23a6 (8.61)"
        ]
    },
    {
        "image_filename": "designv10_0_0003946_robot.2005.1570348-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003946_robot.2005.1570348-Figure3-1.png",
        "caption": "Fig. 3. Slice of soft tissue in the yz plane. The bevel-tip needle is initially at the base coordinate p0 with orientation \u03b8. It is inserted a distance d, causing the surrounding soft tissue to deform.",
        "texts": [
            " We use explicit Euler time integration to integrate velocity and displacement for each free node for each time step. Time steps have duration h = 0.01 seconds. Without loss of generality, we set the coordinate axes of the world frame so that the default needle insertion axis is along the positive z-axis. The y-axis corresponds to the initial location degree of freedom. The needle tip is initially located at a base coordinate p0 = (y0, z0). The initial orientation of the needle is specified using \u03b8, as shown in Fig. 3. For simulation stability, we constrain \u03b8 between \u221245\u25e6 and 45\u25e6. The needle tip rotation is either bevel-right (0\u25e6) or bevel-left (180\u25e6). We assume the needle tip rotation is held constant during insertion due to planner efficiency and lack of experimental data for simulation, although we hope to relax this assumption in future work. We assume the flexible needle is supported so that it does not bend outside the tissue. Once the needle has entered the tissue, it will bend in the direction of the bevel-tip"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003721_s0924-0136(03)00283-8-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003721_s0924-0136(03)00283-8-Figure6-1.png",
        "caption": "Fig. 6. A gear mold insert made from the investigated low alloy steel powder mixture using DMLS process at building rates of 6.7 cm3/h in a nitrogen atmosphere. The sintered density is 95% theoretical.",
        "texts": [
            " The building process was performed under nitrogen atmosphere. The scan rate was in the range of 100\u2013275 mm/s. Fig. 5 presents the influence of the scan rate on the sintered density of the powder blend. It is evident that the sintered density strongly depends on the duration time of the laser beam on the surface of the powder particles. The densification response shows a plateau for duration over 4 ms. According to the results of several experiments the optimum building rate for manufacturing complex-shaped parts was found to be around 6.7 cm3/h. Fig. 6 depicts a gear mold in- sert made by direct laser sintering of the developed powder blend. For manufacturing of mold inserts the dimensional accuracy and surface quality are of special concern. Experiments showed that the shrinkage of the developed material during laser sintering is nearly zero. However, the heat-affected zone (HAZ) of the laser beam on the boundary of the part generates some dimensional offset. Therefore, pre-contouring and post-contouring techniques were used to decrease the HAZ and thus improving dimensional accuracy",
            " Moreover, rapid cooling of the material in DMLS as well as phase transformations may lead to accumulation of thermal stresses. To close the remaining pores, homogenize the microstructure, and remove residual stresses it is recommended to post-sinter the laser processed part. Fig. 10b shows the effect of post-sintering on the residual porosity and dimensional change of the developed material. Technically full density parts (99% theoretical density) can be manufactured while the sintering shrinkage is close to zero. To verify the applicability of the method, the gear mold insert shown in Fig. 6 was sintered in a batch furnace at 1260 \u25e6C for 30 min under vacuum. It should be noted that the part was fixed on the base plate during sintering. Fig. 11 illustrates the corresponding part (no warpage or cracking is visible). It clearly depicts the dimensional stability of the laser-sintered artifacts during the post-sintering process, i.e. if any dimensional change occurs the shrinkage will lead to cracking. In addition, the post-sintering treatment improved the surface quality. The Rz and Ra parameters were reduced to around 30 and 8 m, respectively"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001805_j.matdes.2019.108091-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001805_j.matdes.2019.108091-Figure4-1.png",
        "caption": "Figure 4. The micro-CT scanned volume is divided into various sections and variability in distributions of the defect characteristics in these sections are compared, (a) The scanned volume is divided into four equal control volumes sectioned along the build height, (b) The scanned volume is divided into four equal quadrants as control volumes, (c) The scanned volume is divided into four shells with equal thicknesses of ~ 0.3 mm as control volumes.",
        "texts": [
            " The specimens for which defect characteristics were analyzed and the number of specimens characterized for each condition are presented in Figure 3. The two AM machines had different sets of process parameters creating very different defect contents. All AM250 specimens considered here were vertically built and M290 specimens considered were vertically and diagonally built. Variabilities in the distributions of the defect characteristics gathered from 3D and 2D analysis results throughout the specimens were studied by statistical means. The micro-CT scanned model was divided into various sections as shown in Figure 4 and variability in distributions of the defect characteristics in these sections were compared. For 3D analysis, three different groups of distributions were considered. First, the scanned volume was divided into four equal control volumes sectioned along the build height, as shown in Figure 4(a). The objectives were to examine the defect characteristics distributions along the height and recognize if there is significant variation and if this is the case, are these variations following a specific trend for all the specimens. Based on the observations discussed later, the defects were found to be randomly distributed along the height, so for the rest of the comparisons, the full height of the scanned volume was considered. The scanned volume was then divided into equal quadrants around the specimen axis as control volumes, as shown in Figure 4(b). This was done to see if defects are randomly dispersed in the cross-section of the specimen around the axis of symmetry or if they are concentrated at specific locations around the specimen. Finally, the scanned volume was divided into four shells with equal thicknesses, as shown in Figure 4(c). This was performed by means of the known center of each defect, defined as the center position of the circumscribed sphere of the defect. The wall thickness consists of about 100 voxels which were divided into 4 smaller shells, each about 25 voxels thick. Each defect belonged to a shell if the center of that defect was located in that shell. Generally, defects are not randomly distributed through thickness and there might be a difference between density and size of the defects based on their distance from the surface",
            "06 Ti-6Al-4V M290 45\u02da Annealed Machined -0.13 0.004 -0.052 -0.031 Ti-6Al-4V M290 V HIPed Machined -0.106 -0.106 -0.125 -0.109 17-4 PH M290 V Heat Treated As-built 0.044 0.097 0.085 -0.015 4.2 Analysis of variability in defect characteristics based on location K-S test results based on the analysis of 3D characterization data are listed in Table 3. Distributions of diameter, volume, projected area, and sphericity of the defects based on their location in each specimen were considered. As shown in Figure 4 for each condition the scanned volume is divided into four control volumes. The distribution of each defect characteristic in each control volume is compared with the other three control volumes. This results in six comparisons in case of each characteristic for each condition per specimen. If equal to or more than half of the comparisons for each condition were showing significant differences, the overall result is highlighted with red color and the percentage of similarity is listed in the table",
            " If more than half of the comparisons are showing no significant difference, the overall result is highlighted with green color with the percentage of similarity listed in the table. The micro-CT scan results of two specimens for each condition are used for variability analysis. Defect were analyzed for the middle part of the gage section with uniform thickness considering each defect characteristic of interest. To inspect the variability of the defect characteristics distributions along the build direction as can be seen in Figure 4(a), the scanned Jo ur na l P re -p ro of volume was divided into four equal sub-volumes with 2.5 mm height and a 1 mm gap was considered in between every two sub-volumes. The exponential probability plots for the defect diameter, volume, projected area, and sphericity distributions were depicted to visualize the variability of these defect characteristics distributions along the height of each specimen. The exponential probability plots of defect diameter distributions for the four control volumes along the build direction for different conditions are shown in Figure 9",
            " The K-S test results as listed in Table 3, do not show any significant difference between the defect characteristics distributions gathered along the built height for any of the materials, machines, or surface conditions for the micro-CT scanned volume. This means that considering each of these sections would provide an acceptable representation of the defect distribution in the whole scanned volume of the gage section. These data can also be combined to a single population and this was done for the subsequent 3D analysis to study variability around the specimen axis and through the specimen thickness. Jo ur na l P re -p ro of axis of symmetry, as can be seen in Figure 4(b), the scanned volume was randomly divided into four equal quadrants parallel to specimen axis with equal heights of 10 mm each. Exponential probability plots of defect diameter distributions for the four quadrants as control volumes can be seen in Figure 10. Probability plots show distinct differences for the distribution of defect diameters in different quadrants around AM250 specimens for both as-built and machined surface conditions, with one or two quadrants containing a more defects with larger defect sizes",
            " This means that characterization of only one of these control volumes will not provide a comprehensive picture of the distribution of defects in the specimens and might cause missing critical information. K-S tests show no significant variability in quadrants for M290 annealed specimens and as expected no significant difference between the distributions after HIPing. To inspect the variability of the defect characteristics distributions through the wall thickness of the specimens, as shown in Figure 4(c), the scanned volume was divided into four shells with thicknesses of ~ 0.3 mm and heights of 10 mm. Exponential probability plots for the defect distribution in these four shells are shown in Figure 11. These plots show very distinct differences between the distributions of defect diameter in different shells for as-built surface specimens. These results were confirmed by the K-S test results from 3D analysis as listed in Table 3. The K-S test results in this table do not show significant differences between the shells of machined surface specimens and the distribution of defects through wall thickness is more uniform after removing the surface defects by machining"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000366_tcst.2004.825052-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000366_tcst.2004.825052-Figure3-1.png",
        "caption": "Fig. 3. Schema of the four-rotor rotorcraft.",
        "texts": [
            " The only potential energy which needs to be considered is the standard gravitational potential given by (3) The Lagrangian is (4) The model for the full rotorcraft dynamics is obtained from the Euler\u2013Lagrange equations with external generalized force where and is the generalized moments. is the translational force applied to the rotorcraft due to the control inputs. We ignore the small body forces because they are generally of a much smaller magnitude than the principal control inputs and then we write (5) where (see Fig. 3) and where is a constant and is the angular speed of motor \u201c \u201d ( , ), then (6) where is the transformation matrix representing the orientation of the rotorcraft. We use for and for . The generalized moments on the variables are (7) where where is the distance from the motors to the center of gravity and is the couple produced by motor . Since the Lagrangian contains no cross terms in the kinetic energy combining and (4), the Euler\u2013Lagrange equation can be partitioned into the dynamics for the coordinates and the dynamics",
            " The collective input \u201c \u201d in (13) is essentially used to make the altitude reach a desired value. The control input is used to set the yaw displacement to zero. is used to control the pitch and the horizontal movement in the axis. Similarly, is used to control the roll and horizontal displacement in the axis. In order to simplify let us propose a change of the input variables (15) where (16) are the new inputs. Then (17) Rewriting (13), (14) (18) (19) (20) (21) (22) (23) where and are the coordinates in the horizontal plane, and is the vertical position (see Fig. 3). is the yaw angle around the axis, is the pitch angle around the (new) axis, and is the roll angle around the (new) axis. The control inputs , , and are the total thrust or collective input (directed out the bottom of the aircraft) and the new angular moments (yawing moment, pitching moment and rolling moment). The control of the vertical position can be obtained by using the following control input: (24) where (25) where and are positive constants and is the desired altitude. The yaw angular position can be controlled by applying (26) Indeed, introducing (24)\u2013(26) into (18)\u2013(21) and provided that , we obtain (27) (28) (29) (30) The control parameters , , , and should be carefully chosen to ensure a stable well-damped response in the vertical and yaw axes"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000646_j.copbio.2007.03.007-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000646_j.copbio.2007.03.007-Figure4-1.png",
        "caption": "Figure 4",
        "texts": [
            " As such, when developing power curves, it is not sufficient to report the results per electrode surface area (cm2). Rather, it is important to report the geometry of the cell in terms of electrode placement, cell volume, etc. and, if possible, to maintain a fixed geometry across comparisons. This is particularly crucial when comparing the performance of various anodes and cathodes across laboratories. It is also most useful in modeling studies. One of the more promising geometries is the stack cell design and a typical example is presented in Figure 4. www.sciencedirect.com Fluorescence The entrapment of enzymes within polymer networks has found numerous applications in the design of bioelectrochemical devices [16,21], primarily owing to the simplicity and mild conditions of the procedure and its ability to preserve the catalytic activity of biomolecules. In the context of biosensor and biofuel cell development, electrochemistry has generally been the method of choice when characterizing the performance of polymer-immobilized enzymes. Other methods, however, such as scanning electron microscopy [22,23], infrared spectroscopy, X-ray diffraction [24\u201328], atomic force microscopy [18,24,29\u201331], and small angle neutron scattering [32\u201335] have also been employed to further characterize enzyme\u2013polymer systems"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002059_j.jmatprotec.2013.06.007-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002059_j.jmatprotec.2013.06.007-Figure1-1.png",
        "caption": "Fig. 1. Schematic representation of the laser deposition system.",
        "texts": [
            " Materials Plates of dimension 100 mm \u00d7 180 mm \u00d7 6 mm were machined from austenitic stainless steel AISI 304 and used as substrate material. These were grit blasted and cleaned with acetone before the deposition runs so as to improve substrate surface laser absorptivity and remove contaminants, respectively. The additive material is Inconel 625 wire of 1.2 mm diameter supplied by VBC group, Loughborough, UK. Table 1 gives the chemical compositions of Inconel 625 wire and AISI 304 stainless steel, as quoted by the manufacturers. 2.2. Laser processing A diagram of the laser deposition system used in this study is shown in Fig. 1. Deposition was performed using a 2 kW Ytterbium doped fibre laser (IPG Photonics) operating at 1070 nm wavelength. The beam was focused to a small round spot of approximately 3.1 mm at 20 mm away from focus giving a 212 mm working distance with a Gaussian energy distribution. Inconel 625 wire was \u201cfront fed\u201d at an angle of 42 \u00b1 1\u25e6 to the horizontal so as to aim the wire tip at the centre of the meltpool. A WF200DC wire feeder (Redman Controls and Electronic Ltd.) was used. Single tracks were deposited, at varying processing param- ) SEM o s l u n r w f s 1 t 2 T p ( g c s u T t Q i c t 2 e a t r i E T P f experimental runs required"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure4.5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure4.5-1.png",
        "caption": "Fig. 4.5. Robot UMS-2",
        "texts": [
            "19) Thus we obtain the gains which satisfy all necessary requirements to the maximum possible degree. It should be noted that, since the linear servosystems are applied not only in robotics but also for control in various other systems, it is possible to synthesize the feedback gains using various other methods developed in the automatic control theory. These methods can be easily found in the relevant references [2,4] (method in frequency domain, pole-placement, linear optimal regulator etc.). 130 4 Control of Robots Example For the first joint of the manipulator in Fig. 4.5 (the photograph of this robot is given in Fig. 1.6), synthesis of the servosystem gains should be performed. The joint is driven by a D.C. electromotor of the type IG2315-P20, the parameters of which are given in Table 4.1. The data on masses, moments of inertia, lengths and positions of centres of masses ofthe links are given in Table 4.2. It is easy to show that the moment of intertia of the mechanism around the axis of the first joint is given by (4.20) 4.2 Control of a Single Joint of the Robot 131 if the third link is fixed in the position qg = 0, the moment of inertia of the mechanism around the axis of the first joint is H jj =0",
            " The large reducers are specially inconvenient due to large backlash and dry friction that they might introduce in the system. In the last few years large efforts have been made towards introducing the so-called direct-drive actuators (i.e. motors without reducers). By this the problems with backlash and friction are reduced, but on the other hand variation of the moment of inertia of the mechanism might affect the servo system performance, and more complex con trollaw (with variable velocity gain, for example) has to be introduced. Example For the servosystem in the first joint of the robot in Fig. 4.5 in the previous example we have computed the gains if the third joint is in position qO = O. Considering the expression (4.20) for the moment of inertia of the mechanism around the axis of the first joint, it is obvious that if the third joint is set in the position qO > 0 the moment of inertia of the mechanism Hii will be greater and the damping factor will be less than 1. Using Eq. (4.26) the damping factor for the position of the third joint qg = 0.3 [m] can be calculated as: (1 = !J.~N~N~+Jz1 +Jz2+Jz3+m3l~ 02 =",
            "30), for the moment of inertia of the mechanism R jj , the least possible value should be taken (which this moment might have, depending on the position of the rest of the joints qJ The reason for this is to avoid overshoots. Thus, the procedure for choice of the moment of inertia of the mechanism for calculating the local nominal control is opposite to that for computation of the speed feedback gain. 142 4 Control of Robots Example Let us assume that at input of the servosystem for the first joint of the manipulator in Fig. 4.5 (the servo system has been synthesized in the example in Sect. 4.2.2), instead the position of the signal corresponding to joint trajectory q?(t) is fed. This trajectory is presented in Fig. 4.13 and can be described by: where a 1 = 1 [rad/s2 ] is the acceleration, and r = 2 [s] is the time duration of the movement. If the precompensator is not introduced, but we directly feed the desired trajectory to the input of the positioning servosystem, the real trajectory of the joint will be delayed to the nominal one, as can be seen in Fig",
            " However, if we have to ensure tracking of fast trajectories the influence of dynamic forces cannot be ignored. Since these forces act as external loads, if the servosystem feedback gains are high, the errors caused by these forces might be small, so that even in the case of fast trajectories we may accept the servosystems synthesized for isolated joints (Sect. 4.3.5). However, since high gains are limited, if we must ensure precise tracking of fast trajectories we cannot only apply local servosystems, but the dynamic forces must be compensated for. Example For the first joint of the robot in Figure 4.5 we have synthesized the servosystem as in the example in Sect. 4.2.3, assuming that all the remaining joints are fixed. The positioning of this joint is shown in Fig. 4.8. If simultaneously with the first joint the third joint is also moving, the Coriolis moment acts around the first joint given by m3(l3 + q3)2 4143 which acts as an external load upon the servo system. If we just consider positioning of the first joint (as in the Example in Sect. 4.2.3) but assuming that the third joint is moving with the velocity 4~ = 0",
            " Generally, this analysis requires application by the computer. Selec tion of adequate dynamic control of robots is therefore best performed by computer-aided control synthesis. It should be mentioned that various combinations of the control laws are possible. It is possible to apply a nominal programmed control (which com pensates the nominal dynamic moments), and the local servosystems and global control (which in this case has to compensate for the deviation ofthe real moments from the nominal ones, APi). Example For the first joint of the robot in Fig. 4.5 the nominal programmed control can be obtained in the following way, assuming that the nominal trajectories of the first and the third joint are given by Eq. (4.35) where for the first joint a l = 1 [rad/s2], and for the third joint a3=O.8 [m/s2], while .=2[s]. The nominal 4.3 Control of Simultaneous Motion of Several Robot Joints 153 driving torque is given by: P?=(JZl +Jz2 +Jz3 +m3(l3 +q~(t?)q?(t) + 2m3 (l3 + q~(t))q~(t)q?(t)) or (taking into account Eq. (4.35) and the parameters given in Table 4",
            "3 Program Modules 177 We see that in each sampling period 4 n multiplications and 5 n additions must be carried out. Here, n is the number of joints. It should be emphasized that in each period T actual joint coordinates and velocities must be sampled. The time interval T of course differs from the computational period of the kinematic module and is usually several times shorter. The value of the sampling period can influence the tracking quality of prescribed trajectories a great deal. In order to illustrate this, let us consider the experimental results obtained for the cylindrical robot shown in Fig. 4.5. The first experiment is carried out with the sampling time T=20 ms, while the second one with T=60 ms. In both cases, the PD regulator gains are computed for the same requirements on the tracking quality. The obtained feedback gains are given in Table 5.2. The deviations from the desired trajectories for the first and the third degree of freedom are shown in Fig. 5.11. The behaviour of the other degrees of freedom is practically the same as shown in Fig. 5.11. We see that the tracking quality is much better with T=20 ms than with T=60 ms"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001545_j.rcim.2015.01.003-Figure16-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001545_j.rcim.2015.01.003-Figure16-1.png",
        "caption": "Fig. 16. Geometry 5, thin-walled structure with varied characteristic thicknesses. (a) Geometry is represented by black lines, MAT represented by dotted red lines, and red solid lines stand for branches. (b) Generated trimmed path. (For interpretation of the references to color in this figure legend, the reader is referred to the web version of this article.)",
        "texts": [],
        "surrounding_texts": [
            "Five geometries sliced from CAD models are tested to validate the effectiveness and robustness of the developed algorithms. As shown in Figs. 12\u201316, various types of geometries are tested including solid structures with or without holes, and thin-walled structures. Table 2 provides basic information of the geometries. The MATs and the trimmed paths for all geometries are successfully generated using the developed algorithms. The material efficiency of AM, based on the proposed MATbased path at various step-over distances, is simulated for the different geometries as shown in Fig. 17. Material efficiency of the traditional CNC machining is also provided as a comparison. It is found that, in general, material efficiency decreases with the increasing of step-over distance. This is intuitive as step-over distance represents the resolution of the deposition system. To fill geometry with deposition material, the greater the step-over distance, the more excess material will need to be deposited. The results also show that AM technology has a much higher material efficiency compared to traditional subtractive manufacturing. The effects of the step-over distance on part building time are also discussed. The building time t is determined by the sum of the total path length L and the travel speed, V, of the deposition head. If the build time at the step-over distance of 1 mm, t1, is set to be the characteristic time, the non-dimensional building time T at the step- over distance of d is expressed as T t t L V L V L L/ ( / )/( / ) /d d d1 1 1= = = where Ld represents the total path length at the step-over distance of d; and the travel speed V is assumed to be the same at various stepover distances. The non-dimensional building time as the function of step-over distances are calculated as shown in Fig. 17. With the increasing of the step-over distance, the build time T declines exponentially. It reveals that it is always better to have a small stepover distance for highmaterial efficiency whereas that would be at the expense of productivity. For powder-based AM system, the step-over distance generally ranges from 0.01 mm to 2 mm. Material efficiency in this range decreases only slightly, indicating that the step-over distance does not have a significant effect on the material efficiency of powderbased AM. As inferred above, material efficiency is determined by the resolution of the deposition system in relation to the size of the geometry. Building time changes significantly in this range, therefore, build rate will be an important factor for powder-based AM system. On the other hand, for WAAM technology, the typical step-over distance varies from 2 mm to 12 mm for mild steel materials. Building time in this range as shown in Fig. 17 does not change y of the given geometry. Red lines represent the MAT. Green lines stand for the ath. (b) Trimmed paths. (For interpretation of the references to color in this figure significantly and therefore build rate for WAAM is not the major concern. Material efficiency in WAAM using the proposed MATbased paths is shown in Fig. 18. It is found that for solid structures (Geometry 1 and Geometry 2) material efficiency is relatively constant, with a slight decrease as step-over distance increases. However for thin-walled structures, such as Geometry 3, Geometry 4, and Geometry 5, the variation of material efficiency corresponding to the step-over distance are significant. Although the generally-descending trend of material efficiency can still be found, the variation cannot be predicted since the shapes of various geometries are very different. The optimal step-over distances for Geometry 1 and Geometry 2 are 4 mm, while for Geometry 3 is around 4 mm, for Geometry 4 is near 6 mm and for Geometry 5 is approximately 3 mm. Variations of material efficiency in WAAM for the five tested geometries are shown in Fig. 19. It is clear that while the variation of material efficiency for different step-over distances is minimal for solid structures, it is quite significant for thin-walled structures. For Geometry 5, by choosing the optimal step-over distance, material efficiency could be increased by 2.4 times from 38.63% to 94.15%. This indicates that step-over distance plays an important role on material efficiency when fabricating thin-walled structures using WAAM technology. Experimental results of the proposed MAT-based path generation strategy are conducted using a robotic welding system at the University of Wollongong as shown schematically in Fig. 20. The details of the system can be found in previous publications [24,27]. In this study, a section of Geometry 5 is fabricated using both the proposed MAT-based path patterns and the traditional contour path patterns. Comparisons between the proposed path patterns and the traditional contour path patterns are shown in Fig. 21. For the Fig. 20. Schematic diagram of the proposed path patterns (Fig. 21a), layers are generated by offsetting the deposition head from the medial axis to outside, and the deposited layers are slightly larger than real geometry. For the traditional path patterns (Fig. 21b), deposited layers are consistent to the real geometry boundaries; however, gaps are created since the thicknesses of the walls vary. After surface milling, it can be seen that, while the traditional contour path patterns (Fig. 21d) leaves gaps on the component, the proposed MAT-based path patterns are able to produce gap-free walls. Depending on the thickness of the walls, more material may need to be removed for the parts generated using the proposed MAT-based path patterns, in order to produce the desired geometry. However, for certain applications such as AM components subjected to high mechanical loading, the extra material removal is less of an issue than leaving gaps in the interior of the deposited parts using traditional contour path patterns."
        ]
    },
    {
        "image_filename": "designv10_0_0001182_b916831a-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001182_b916831a-Figure1-1.png",
        "caption": "Fig. 1 The polymer film is mounted vertically (start position shown with dashed lines) and oscillates around the horizontal plane of incidence of the laser beam with full oscillation angle 4.",
        "texts": [
            " These high speed, remotely accessed, oscillatory photodriven transducers have potential application in a variety of areas including microfluidics and remotely accessed optical elements. Notably, these fast and large amplitude oscillations are demonstrated for the first time using sunlight,31 the cheapest and most readily available actinic source. The phenomenon of light-induced oscillations is particularly efficient for monodomain azobenzene liquid crystal polymer network (azo-LCN) materials where oscillations of 28 Hz frequency and nearly 170 degrees amplitude were observed for a cantilever of geometry 5 mm (height) 1 mm (width) 50 mm (thickness).30 Shown in Fig. 1 is the alignment of the cantilever with respect to the incident Ar+ laser. If the cantilever base is exposed to laser intensity above a threshold power density, the cantilever tip has a sufficient moment of inertia to deflect through the focused (3 mm diameter) laser beam. The back surface of the cantilever, now exposed, reverses the direction of the surface 780 | Soft Matter, 2010, 6, 779\u2013783 contraction, starting a feedback loop between the upstroke and downstroke. Of critical importance to the effect is the generation of a photoelastic response (e",
            " surface contractile strain only generated in the presence of the actinic light). The frequency and amplitude of oscillations (the deflection angle of the cantilever tip) as a function of laser beam power density for an azo-LCN cantilever of 2.7 mm 0.8 mm 50 mm size are shown in the images of Fig. 2a and plotted in Fig. 2b, as well as in MOVIE1 (ESI\u2021). Increasing the laser power density increases the amplitude of the oscillations but does not increase the frequency. Since such behavior is typical to second order phase transitions, we expect the oscillation angle (see Fig. 1) to increase with power density as 4 (I Ith)1/2. Indeed, such a fit holds well for power density values exceeding the threshold by up to 50% (Fig. 2c). Thereafter, nonlinear processes are highly evident in the plot of azo-LCN cantilever deformation as a function of I1/2. The nonlinearity is related not only with the large amplitude of the oscillations, but also by the changing material parameters due to local temperature increases. Estimates and measures of the local heating confirm there is some rise depending on the power density although the increase is still below the glass transition point ( 65 C)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000637_978-3-642-83006-8-Figure2.42-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000637_978-3-642-83006-8-Figure2.42-1.png",
        "caption": "Fig. 2.42. Compensating movements for the single-support gait upon level ground with free arms for ZMP law I and different values of parameters T and 5",
        "texts": [],
        "surrounding_texts": [
            "move freely (due to the coupling from the powered joints). For unpowe red joints, the only moments considered are those arising from the for ces of viscous friction \u00b718 P 18 = kq , (k=-0.2) where k is the viscous friction coefficient. In this case, the compen sating dynamics comprises the trunk movements both in the sagittal and frontal plane and the arms movements in the sagittal plane only (Fig."
        ]
    },
    {
        "image_filename": "designv10_0_0000005_978-3-642-84379-2-Figure14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000005_978-3-642-84379-2-Figure14-1.png",
        "caption": "Fig. 14",
        "texts": [
            " Its derivative with respect to both arguments becomes discontinuous when any 3 Piecewise Smooth Lyapunov Functions 51 of these arguments vanishes. The value v obtained from the motion equations . ov. ov. . (. 2') V=-S1 +-s2=slgns1 -slgns1 + slgns2 OS1 OS2 + signs2( - 2 sign S1 - signs2) = - 2. is negative everywhere except for the discontinuity surfaces where this function is not defined. Hence, the functions v and v have different signs and, as it follows from consideration of the S1,S2 plane (Fig. 14), the system is stable indeed, i.e. a sliding mode does occur on the intersection of the discontinuity surfaces. To be able to correlate these two facts, let us give consideration to the second example dealing with a system with two-dimensional discontinuous control whose projection on the s 1, S2 plane may be described by the equations 51 = - 2 sign S1 - sign S2, 52 = - 2 sign S1 + sign S2' Let us take the positive definite function v = 41s11 + IS21. Its derivative OV OV . V=-5 1 +-52= -7-6slgns1s2 OS1 OS2 is negative everywhere except for the discontinuity surfaces"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000138_6.2007-6461-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000138_6.2007-6461-Figure8-1.png",
        "caption": "Figure 8. Diagram representing the effects of rotor blade flapping5 and the modeling of stiff rotor blades as hinged blades with an effective offset and blade stiffness k\u03b2.3 The rotor plane becomes tilted, resulting in a deflection of the thrust vector, and a moment is generated at the blade root.",
        "texts": [
            "5 This flapping of the blades tilts the rotor plane back away from the direction of motion, which has a variety of effects on the dynamics of the vehicle, in particular affecting stability in attitude.15 For this subsection, the effects on an individual rotor will be considered, so for readability, the rotor index, subscript i, is implied but not written. The backwards tilt of the rotor plane generates a longitudinal thrust, Tlon, Tb,lon = T sin a1s (18) where a1s is the angle by which the thrust vector T is deflected (see Figure 8). If the center of gravity of the vehicle is not aligned with the rotor plane, this longitudinal force will generate a moment about the c.g., Mb,lon = Tb,lonrcg, where rcg is the vertical distance from the rotor plane to the c.g. of the vehicle. For stiff rotors, as are used in most current quadrotor helicopters, the tilt of the blades also generates a moment at the rotor hub Mbs = k\u03b2a1s (19) where k\u03b2 is the stiffness of the rotor blade in Nm/rad. Coning (the upward flexure of the rotor blades from the lift force on each blade) also causes the impinging airflow to have unbalanced forcing of the blades which causes a lateral tilt of the rotor plane, the details of which are developed in the literature"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure10.26-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure10.26-1.png",
        "caption": "FIGURE 10.26. A vehicle in a steady state condition when it is turning in a circle.",
        "texts": [
            " The steering input is positive and therefore, the vehicle must turn left, in a positive direction of the y-axis. The yaw rate in Figure 10.21 is positive and correctly shows that the vehicle is turning about the z-axis. We can find the lateral velocity of the front and rear wheels, vy1 = vy + a1r (10.398) vy2 = vy \u2212 a2r (10.399) by having vy and r. The lateral speed of the front and rear wheels are shown in Figures 10.22 and 10.23. The sideslip angle \u03b2 = vy/vx and the radius of rotation R = vx/r are also shown in Figures 10.24 and 10.25. Figure 10.26 illustrates the vehicle at a steady-state condition when it is turning on a circle. The steer angle \u03b4 is shown, however, the angles \u03b1 and \u03b2 are too small to be shown. Example 403 F Time series and free response. The response of a vehicle to zero steer angle \u03b4 (t) = 0 (10.400) 10. Vehicle Planar Dynamics 639 640 10. Vehicle Planar Dynamics 10. Vehicle Planar Dynamics 641 642 10. Vehicle Planar Dynamics at constant speed is called free response. The equation of motion under free dynamics is q\u0307 = [A]q"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003473_j.aca.2004.10.007-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003473_j.aca.2004.10.007-Figure1-1.png",
        "caption": "Fig. 1. Cyclic voltammograms for 2.0 \u00d7 10\u22124 mol l\u22121 solutions of catechol (a) and caffeic acid (b), at: (1) Tyr-nAu-GCE; (2) Tyr-GCE; (3) Au-GCE; (4) GCE; v = 25 mV s\u22121. Supporting electrolyte: 0.05 mol l\u22121 phosphate buffer (pH 7.4).",
        "texts": [
            "5 ml of Folin\u2013Ciocalteau reagent phosphotungstic\u2013phosphomolybdic acid) were added to .5 ml of sample (in the case of red wines a 1:8 dilution was carried out). The mixture was stirred for about 1 min, and 1.0 ml of an 80% sodium carbonate solution and 4.2 ml of deionized water were added. The resulting solution was allowed to stand for 2 h at ambient temperature in darkness. The absorbance is then read at 730 nm. The same procedure was used to construct a calibration plot with standard solutions of caffeic acid. 3. Results and discussion Fig. 1 shows a comparison of cyclic voltammograms for catechol (Fig. 1a) and caffeic acid (Fig. 1b) obtained at a Tyr-nAu-GCE, at a biosensor in which Tyr was immobilized, under strictly identical conditions that those used for the TyrnAu-GCE, at a bare GCE (Tyr-GCE), and at a conventional naked GCE. As expected, no reduction signal was observed at the electrode with no enzyme. Moreover, although a slight reduction wave can be observed at the Tyr-GCE, the phenolic compound catalytic responses were remarkably higher at the Tyr-nAu-GCE, showing fairly well the advantages predicted for this type of biosensor configuration, commented in Section 1"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003935_s0022112079002482-Figure12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003935_s0022112079002482-Figure12-1.png",
        "caption": "FIGURE 12. (a) Trajectory of an organism during one cycle of the wave. (b ) Trajectory of ends of the flagellum during one cycle of the wave. Arrows indicate swimming diroction. Note the apparent increase in amplitude a t the trailing end.",
        "texts": [
            " In the overall picture, we see that the average flow in the wave direction and hence the swimming speed in the opposite direction are very small. Figure 11 ( b ) shows the pattern of streaklines one quarter-cycle later. At this instant, we see the same dominance of the normal forces, but the overall picture is somewhat different. The average flow in the wave direction is much larger, and the swimming speed is proportionately higher. The trajectory of the organism during one cycle is shown in figure 1 2 ( a ) . The positions of the cell body and flagellum are shown a t intervals of one-sixth of a cycle. Figure 12 ( b ) shows the trajectory of the end points of the flagellum during the cycle. This figure confirms the non-uniformity in the swimming speed over the cycle inferred from the flow patterns in figure 1 1 . We note that the actual distance travelled is much greater than the distance covered in the swimming direction. An interesting feature of the motion is that the constant amplitude wave appears to have increasing amplitude owing to the yawing motion during the cycle. We note that for any symmetric wave the organism has zero net rotation over a cycle, and its position a t the end of each cycle lies on a line which specifies the swimming direction"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001383_025403-Figure13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001383_025403-Figure13-1.png",
        "caption": "Figure 13. Transient thermal simulation of a 20 layers DMD (600 W, 0.006 m s\u22121) and position of 3 thermocouples.",
        "texts": [
            " \u03ba(T , t, x, z) = k(T ) \u00b7 (1 \u2212 flc2hs(x \u2212 x0 \u2212 V \u00b7 t, xscale)) \u00d7(1 \u2212 flc2hs(z \u2212 z0, zscale)) , (15) with flc2hs = Heaviside step function, xscale, zscale = transition intervals, x0, z0 = initial positions, V = scanning speed (m s\u22121). The only physical difference between our simulations and the real experimental DMD conditions came from the heat deposit zone: we apply the laser source in between plane layers, whereas the laser source is experimentally applied on a moving inclined plane. However, the global and local (near the melt pool) thermal fields obtained by simulations seem to be satisfactory (figure 13). Various parameters can be recorded during a LMP. Among these, one can mention: (1) the local temperature history T = f (t) by thermocouples or pyrometers measurements and (2) the melt-pool size by fast camera analysis. These parameters are necessary to validate thermal simulations of the DMD process carried out with different experimental conditions. Next, comparisons will be shown between experimental and simulated T = f (t) profiles and melt-pool sizes. First, 0.2 mm diameter type-K thermocouples were spotwelded at different locations in the substrate, as close as possible (0"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure5.42-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure5.42-1.png",
        "caption": "Figure 5.42 (a) A structure loaded by a vertical force of 40 kN on the left-hand rafter, with (b) its support reactions.",
        "texts": [
            " To demonstrate clearly how the shoring bars act on frame ASB, the frame and the shoring bars have been isolated from one another in Figure 5.40. 180 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM c. The force acting at S on SD is equal to the support reaction at A. The force that shoring bar (2) exerts on SD is also known. Still unknown are the components of the force exerted on SD at D. These are found via the force equilibrium of column BD (see Figure 5.41). Check: SD must be in equilibrium. Example 2 The structure in Figure 5.42a is loaded on rafter ACE by a vertical force F = 40 kN. Questions: a. Determine the support reactions. b. Determine the force in bar CD (with the correct sign for tension and compression). c. Determine the hinge force at E. Solution: a. The support reactions follow directly from the equilibrium of the structure as a whole. There are only vertical support reactions. They are shown in Figure 5.42b. b. Suppose the tensile force in CD is N(CD). In Figure 5.43, CD has been isolated from AEB. The magnitude of N(CD) follows from the moment equilibrium about E of one of the rafters AE or BE. The unloaded rafter BE is simpler with respect to the amount of arithmetic: \u2211 T (BE) z |E = \u2212N(CD) \u00d7 (4 m) + (15 kN(6 m) = 0 \u21d2 N(CD) = 22.5 kN CD is a tension member. c. The hinge force at E is subsequently found from the force equilibrium of one of the rafters AE or BE. Again, the unloaded right-hand rafter BE is preferable"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure6.11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure6.11-1.png",
        "caption": "FIGURE 6.11 Single loop impedance example.",
        "texts": [
            " Use the closed-form partial elements given in Appendices C and D to compute the value of all components. Hint: The model has two partial inductance and four capacitive cells. Each partial inductance consists of an inductive half cell where they are sharing a node at each end. 6.2 Assemble circuit matrix using element stamps Corresponding to the circuits in Fig. 6.8, assemble the circuit matrix (6.53) from the circuit equations using the MNA stamps in Appendix B. REFERENCES 155 6.3 Compute the impedance of a loop A wire loop is shown in Fig. 6.11. Make a (QS)PEEC model for the loop and compute the loop impedance between A and B. Then remove branch 4 from the loop and compute the impedance of the open loop and compare the impedance results for the two cases. Also, check the impedance for using 1, 2, and 4 partial inductances per side. Use zero thickness sheet conductors of 0.5 mm width for the model using equations (C.23) and (C.25). Assume that the square loop has a side length of 30 mm. Also compute the impedance of the loop arrangements if you also include the resistance of the conductors"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000857_j.matdes.2014.07.043-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000857_j.matdes.2014.07.043-Figure8-1.png",
        "caption": "Fig. 8. Dimensional geom",
        "texts": [
            " A non-assembly mechanism is one-step fabrication of multi-articulated systems without requiring assembly of their structural members and joints after fabrication [19,20]. Self-supporting angle is used to control the creation of supports on angled walls and surfaces. It represents the minimum angle of the part wall that will be built without supports. Facets that are closer to flat and having an inclination less than the self-supporting angle will be built on tops of supports. Four types of overhanging structures were investigated: downward sloping face, concave and convex radii, ledge (Fig. 8). Due to high thermal stresses typical of the SLM process, which can cause deformation of the part, downward sloping faces with angles smaller than 45 (angle between the xy-plane and the face) have to be supported. It is well known in literature [21\u201324] that for angles greater than 45 , they are self-supporting and the values of surface roughness decreases with increasing angle because the staircase effect is reduced. Oh the other hand, angles smaller than 30 lead to an increase in the staircase effect"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000613_978-94-015-9261-1-Figure1.8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000613_978-94-015-9261-1-Figure1.8-1.png",
        "caption": "Figure 1.8. Mobile robot and wheeled module",
        "texts": [],
        "surrounding_texts": [
            "where w is a given frequency. In general, the common base and energetic interaction of the pendulums provides their self-synchronization, The pro cess can be accelerated by relevant control actions associated with external synchronizing loops. The additional cont rol actions, introducing averaged loops of the sys tem , are needed to achieve the required frequency of oscillations, Then , in synchronous mode, (tl = (t2 = a(t) and the output variables (tl , (t 2 synchronously evolve in accordance with a periodic undamped solution of equation (1.7) (see Example 5.15). 0 A som ewhat more general mode of one-frequency synchronous oscil lat ions and ranking of their amplitudes can be given via arbitrary linear relations of the variables Yj and a more general model of the averaged oscillation y+2~wy +w2y = 0, (1.8) where ~ is the damping ratio, -1 < ~ < 1. The nonlinear relations (1.3) and the general definition of averaged motion y in the form (1.4) en able one to impose the requirement for nonlinear oscillations of arbitrary shapes that corr espond to so-called gen eralized problems 01 synchronization [173,24,89,179] (see example 5.16). The statement of the MIMO problem of coordinat ing control and synchronization, in particular , implies stabi lization about nontrivial spatial attractors and , besides , includes a simple single-input jsingle-output (SISO) task associated with the execution of the required generalized (averaged) motion. The rnost evident geometric nature is inherent in sunchrotiization via arranging orbital motion. Being stated as finding the control that provides attractivity of t.he closed orbit Sy arid rnaintaining a given longi tudinal rate y = Y\"'(t) determining the shap e of the oscillations [see Section 5.5 and Exarnple 5.17) , the problern refers to a dass of spatuil moiion conirol problems."
        ]
    },
    {
        "image_filename": "designv10_0_0000386_j.measurement.2013.11.012-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000386_j.measurement.2013.11.012-Figure2-1.png",
        "caption": "Fig. 2. (a) An elementary planetary gear set having 4 planet gears, (b) scheme of planetary gear set with the standstill ring gear, (c) scheme of planetary gear set with the standstill sun gear and (d) scheme of planetary gearbox set with the rotating sun gear and the rotating ring gear.",
        "texts": [
            " Condition monitoring and fault diagnosis of planetary gearboxes aims to prevent accidents and generate cost savings for users of planetary gearboxes. Condition monitoring and fault diagnosis of gearboxes has been attracting considerably increasing attention [4\u2013 6]. Most investigations, however, focus on fixed-axis gearboxes in which all gears are designed to rotate around their own fixed centers [7\u201310] (see Fig. 1). Fundamentally different from fixed-axis gearboxes, planetary gearboxes refer to those including the transmission structure of planetary gear sets. An elementary planetary gear set, shown in Fig. 2a, is a compound gear system. The gear system contains a ring gear, a sun gear that rotates around its own center, and several planet gears that not only rotate around their own centers but also revolve around the center of the sun gear. With such a complex gear transmission structure, planetary gearboxes exhibit some unique behaviors, which Transmission diagram of a fixed-axis gear system having two g pairs. invalidate fault diagnosis methods working well for fixedaxis gearboxes. Compared with fixed-axis gearboxes, studies on condition monitoring and fault diagnosis of planetary gearboxes are not that many",
            " Transmission structures and operation behaviors In this subsection, the transmission diagram of a fixedaxis gearbox containing two meshing pairs is given in Fig. 1 first before introducing that of a planetary gearbox. It is observed from Fig. 1 that all gears of the two meshing pairs rotate only around their own fixed centers. Gearboxes including only this kind of gear transmission structures are defined as fixed-axis ones accordingly. In contrast, a planetary gear set have several planet gears which rotate around unfixed centers. As shown in Fig. 2a, an example elementary planetary gear set has a ring gear, a sun gear that rotates around its own center, and four planet gears that not only rotate around their own unfixed centers but also revolve around the center of the sun gear. The planet gears are between the sun gear and the ring gear, and therefore mesh simultaneously with both of them. Generally, there are three elementary types of planetary gearboxes commonly used in modern industry. The three kinds of planetary gearboxes are given in Fig. 2b\u2013d, respectively. Fig. 2b\u2013d shows a planetary gearbox with the standstill ring gear, the standstill sun gear, and no standstill gear, respectively. Because of their transmission structure, planetary gearboxes generally have the following behaviors, which are not observed in fixed-axis gearboxes. (1) Multiple planet gears meshing simultaneously with the sun gear and the ring gear, and a large number of synchronous components (gears or bearings) in close proximity will excite similar vibrations in planetary gearboxes. These vibrations with different meshing phases couple with each other; as a result, some of the vibrations could be neutralized or cancelled [12]",
            " Evaluation of characteristic frequencies The characteristic frequencies, including the gear rotating frequency, the meshing frequency, etc., are critical to fault detection of gears. The identification of faults is related to the occurrence of the characteristic frequencies which is linked to the given fault. Hence, the characteristic frequencies of planetary gearboxes as well as fixed-axis gearboxes are provided in this subsection. The derivation of these characteristic frequencies is based on the gear transmission structures of the fixed-axis gearbox in Fig. 1 and the planetary gearbox with the standstill ring gear in Fig. 2b, respectively. 2.2.1. Fixed-axis gearboxes Define the following notation: Nj \u2013 is the number of teeth of gear j (j = 1,2,3,4). fj \u2013 is the rotating frequency of gear j (j = 1,2,3,4). The rotating frequency of gear 1. (f1) \u2013 is the input frequency of the whole gear transmission and is generally known beforehand. ik \u2013 is the transmission ratio of meshing pair k (k = 1,2), which is defined as the ratio between the rotating frequency of the driving gear and that of the driven gear in a meshing pair",
            " the rotating frequencies of each gear and the meshing frequencies of each meshing pair, can be expressed as a function of the input frequency (f1) and the number of teeth of gears as follows. The equations of the characteristic frequencies are also summarized in Table 1. f2 \u00bc f1 i1 \u00bc N1 N2 f1; \u00f01\u00de f3 \u00bc f2 \u00bc N1 N2 f1; \u00f02\u00de 2.2.2. Planetary gearboxes Based on the principles used in evaluating the characteristic frequencies of fixed-axis gearboxes, we can obtain the characteristic frequencies of the planetary gearbox in Fig. 2b. Make the following definitions first. NS, NP and NR \u2013 the number of teeth of the sun gear, a planet gear, and the ring gear, respectively. MP \u2013 the number of planet gears. fS, fP, fR and fC \u2013 the rotating frequency of the sun gear, a planet gear, the ring gear and the carrier, respectively. i \u2013 the transmission ratio of the planetary gearbox, which is defined as the ratio between the rotating frequency of the input shaft (the sun gear) and that of the output shaft (the carrier). fp\u2013p \u2013 the pass frequency of the planet gears. fm\u2013p \u2013 the meshing frequency of the planetary gearbox. Since the transmission ratio is the key to evaluation of the characteristic frequencies of planetary gearboxes, its calculation process will be illustrated as follows [18], which is relatively complicated compared to that of fixed-axis gearboxes. The ring gear of the planetary gear transmission shown in Fig. 2b is standstill. So, we have the following equation. fR \u00bc 0: \u00f06\u00de Transforming the planetary gear transmission into a fixed-axis transmission and utilizing the calculation principles of the fixed-axis one, the following equations are generated accordingly. fS fC fP fC \u00bc NP NS fP fC fR fC \u00bc NR NP 8< : : \u00f07\u00de Substituting Eq. (6) into Eq. (7), we obtain the transmission ratio i \u00bc fS fC \u00bc 1\u00fe NR NS : \u00f08\u00de Based on the calculated transmission ratio, the characteristic frequencies can be written as a function of the rotating frequency of the sun gear (fS) and the numbers of gear teeth by the following equations, which are summarized in Table 2 as well. fP \u00bc \u00f0NP NR\u00deNS \u00f0NR \u00fe NS\u00deNP fS; \u00f09\u00de fC \u00bc fS i \u00bc NS NR \u00fe NS fS; \u00f010\u00de fp p \u00bc MPfC \u00bc NSMP NR \u00fe NS fS; \u00f011\u00de and fm p \u00bc \u00f0fS fC\u00deNS \u00bc NRNS NR \u00fe NS fS: \u00f012\u00de For the other two types of planetary gearboxes in Fig. 2c and d, the characteristic frequencies can be calculated in the same way. Refs. [19,20] provided the fundamentals by developing the characteristic frequencies of all three kinds of planetary gearboxes, which have influence on understanding the problems connected with condition monitoring and diagnosis of planetary gearboxes. 2.3. Illustrations using a planetary gearbox test rig After introducing the behaviors and evaluating the characteristic frequencies of planetary gearboxes, we next use the vibration signals acquired from a test rig to illustrate the challenging issues in fault diagnosis of planetary gearboxes [21]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002501_j.jsv.2019.115144-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002501_j.jsv.2019.115144-Figure4-1.png",
        "caption": "Fig. 4. Changes in the contact zone between the roller and defective raceway.",
        "texts": [],
        "surrounding_texts": [
            "(a) (b) (c) (d) (e) To formulate the effects of the changes in the contact zone (\u2206\u03b8) on the TFE model, as depicted in Fig. 5, an improved TFE model based on the half-sine function is presented in this work. The proposed TFE model is described as ( )( )TFE TFE d d0 d sin mod ,2iF K H \u03c0 \u03b8 \u03c0 \u03b8 \u03b8   = \u2212    (14) where KTFE is the time-depended contact stiffness between the fault edge and roller, which can be obtain by the method in Ref. [41]; Hd is the maximum additional displacement, which is given by 22 dr r d 2 2 2 bd d H   = \u2212 \u2212        (15) where dr is the roller diameter. Furthermore, \u03b8d is the arc length of the TFE zone, which is formulated as b d d 2 d R \u03b8 \u03b8= + \u2206 (16) where Rd is the radius of the defective raceway; \u2206\u03b8 is the additional excitation zone caused by the destressing and restressing processes, which is given as c d2 b R \u03b8\u2206 = (17) where bc is the half contact width, which is formulated by [50] 1 2 2 2 r r in/ot c r in/ot 4 1 1 + Q v v b l E E\u03c0 \u03c1   \u2212 \u2212=   \u03a3    (18) where the subscript r and in denote the roller and inner raceway, respectively; Qr is the load applied on the roller; l is the roller length; \u03a3\u03c1 is the sum curvature of the mating bodies. In addition, \u03b8i is the ith roller angular position, which is defined as ( ) ( ) ( ) c in r c r 2 1 =1 = 2 1 =2 i i t q N i t q N \u03c0 \u03c9 \u03c9 \u03b8 \u03c0 \u03c9  \u2212 + \u2212    \u2212 +  (18) where Nr is the number of the rollers; \u03c9c and \u03c9in are the angular speed of the cage and inner raceway, respectively; and t is time; q denotes the fault location, q=1 for the fault on the inner raceway and q=2 for the fault on the outer raceway; and \u03b8d0 is the initial arc length between the first roller to the fault."
        ]
    },
    {
        "image_filename": "designv10_0_0000995_revmodphys.79.643-Figure12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000995_revmodphys.79.643-Figure12-1.png",
        "caption": "FIG. 12. The kite, a square of elastic material fastened in a hinged frame. Base line is shown as a dashed, horizontal line. The transverse midline goes from front to back. This line is the y axis of the material coordinates. The x axis is the ridge line. Ridge length X, assumed curvature radius Rc, and sag distance are indicated.",
        "texts": [
            " Via edge forces normal to the sheet, one then constrains its two long edges to lie in the folded surface. It is apparent from the figure that this smooth procedure creates a ridge region of concentrated bending between the two vertices. As discussed in Sec. V below, this ridge is amenable to systematic analysis that allows several aspects of its shape and energy to be calculated explicitly. In order to gain a concrete sense of how these ridges form, we consider a further variant called the kite shape. The kite shape is pictured in Fig. 12. It consists of a square sheet fastened to a frame at its edges. The frame is hinged so that it can fold along a diagonal, thus forc- Rev. Mod. Phys., Vol. 79, No. 2, April\u2013June 2007 ing the sheet to bend. If one makes such a kite out of, e.g., a piece of office paper stapled into a manilla envelope, one sees that the width of the bent region is significantly smaller than the width of the kite. One also sees directly that the sheet must be stretched. By sighting along the ridge line, one sees a clear sagging of the ridge line",
            " To proceed further, we must understand how the geometric constraints lead to a tradeoff of bending and stretching energy. We may show this tradeoff by a variational strategy in which we assume a particular shape with a free parameter and then optimize this parameter. To define this shape, we consider the transverse midline that proceeds from one transverse corner, across the ridge and then to the opposite corner. We define a material coordinate x running along the ridge and y running along the transverse midline, as shown in Fig. 12. For the moment, we suppose that this transverse midline retains the same length as in the unbent sheet: it does not stretch. We suppose this line consists of a circular arc of undetermined radius Rc crossing the ridge, connected to each corner by straight segments. It is useful to define the straight line in space between the two ends of the ridge. The unbent sheet passes through this line. We call it the baseline and denote its length by X. If Rc=0, the transverse midline also touches the baseline",
            " 24 We see that in the limit he /X\u21920, our approximation that Rc X is justified. Though we derived Eq. 23 under restrictive assumptions about the shape of the transverse midline, the result is more general. It requires only that a the deformed region is confined to a width of the order of the transverse radius of curvature at the ridge, b the dominant strain is a longitudinal strain confined to a similar region, and c there is relatively little energy outside this ridge region. For the kite shape of Fig. 12, we may justify assumption c and thus remove it. This assumption entered our argument through the requirement that the length of the transverse midline be unchanged as the surface was bent. We now allow the midline to stretch. In order to diminish the sag significantly, this line must increase its length by an amount of order Rc, incurring a strain of order Rc /X. This strain is larger than the ridge strain calculated above of order Rc /X 2 . Moreover, it extends over a region much larger than the ridge region"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure4.37-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure4.37-1.png",
        "caption": "Figure 4.37 Examples of trussed beams.",
        "texts": [
            " If the structure with hinged supports itself consists of two parts joined by a hinge, this is referred to as a three-hinged frame (see Figure 4.34). If the beam structure is not bent but arched, then the structure in Figure 4.35a is called a two-hinged arch, and the structure in Figure 4.35b a three-hinged arch. 130 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM It will be clear that a wide range of planar structures can be constructed using line elements. Two types of structure not mentioned in the earlier categories are shown here. The structure in Figure 4.36 is called a shored frame. The structures in Figure 4.37 go by the name of trussed beams. Although the structures in Figures 4.36 and 4.37 include hinges at all the connections, none of these structures are trusses. A characteristic of a truss is that all the ends of the members that merge in a connection are hinged together. This is not the case in the circled connections. Here the hinge is attached to the outside of a so-called continuous beam, and is not fitted internally in the beam. Structures are supported in such a way that all free movements are prevented"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001031_j.jsv.2008.03.038-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001031_j.jsv.2008.03.038-Figure4-1.png",
        "caption": "Fig. 4. The cracked tooth model for case 2.",
        "texts": [],
        "surrounding_texts": [
            "The gear mesh stiffness model described in this study was based on the work by Yang and Lin [3] in 1987. They used the potential energy method to analytically model the effective mesh stiffness. The total potential energy stored in the meshing gear system was assumed to include three components: Hertzian energy, bending energy, and axial compressive energy. This model was refined by Tian [4] in 2004 in which shear energy was taken into account as well. Thus, for the single-tooth-pair meshing duration, the total effective mesh stiffness can be expressed as [4] kt \u00bc 1 1=kh \u00fe 1=kb1 \u00fe 1=ks1 \u00fe 1=ka1 \u00fe 1=kb2 \u00fe 1=ks2 \u00fe 1=ka2 , (1) where kh, kb, ks, and ka represent the Hertzian, bending, shear, and axial compressive mesh stiffness, respectively. For the double-tooth-pair meshing duration, the total effective mesh stiffness is the sum of the two pairs\u2019 stiffnesses, which is shown as [4] kt \u00bc X2 i\u00bc1 1 1=kh;i \u00fe 1=kb1;i \u00fe 1=ks1;i \u00fe 1=ka1;i \u00fe 1=kb2;i \u00fe 1=ks2;i \u00fe 1=ka2;i , (2) where i \u00bc 1 represents the first pair of meshing teeth and i \u00bc 2 represents the second. The derivations of these two equations are given in Tian [4]. Calculation of each of the components in these two equations when there are no cracks in any gear is given in Refs. [3,4,15], where Ref. [15] is a shorter version of Ref. [4]. The expressions of these components when cracks are introduced will be provided later in this paper. For typical gear parameters given in Table 1, we wrote simple Matlab programs and obtained numerical values of the total effective mesh stiffness as a function of the gear rotation angle. This total effective mesh stiffness within one shaft period of the gear is plotted in Fig. 1. Fig. 1 represents the total meshing stiffness of the pair of gears when the gear teeth are perfect (that is, have no cracks). On crack development in a gear, Refs. [16\u201319] consider that a crack is developing at the root of a single tooth of the pinion. A tooth root crack typically starts at the point of the largest stress in the material. In Ref. [20], a computational model which applies the principles of linear elastic fracture mechanics is used to simulate gear tooth root crack propagation. Based on the computational results, the crack propagation path shows a slight curve extending from the tooth root as shown in the left side of Fig. 2 [20]. Lewicki [21] also indicates that crack propagation paths are smooth, continuous, and in most cases, rather straight with only a ARTICLE IN PRESS S. Wu et al. / Journal of Sound and Vibration 317 (2008) 608\u2013624610 slight curvature. He has also studied the effects of rim and web thickness on crack propagation path and showed that different paths exist. In this paper, based on the results shown in Ref. [20], we further simplify the crack model. We will consider the crack path to be a straight line as shown in the right side of Fig. 2. The crack starts at the root of the pinion and then proceeds as shown in Fig. 2. Further referring to Figs. 3\u20136, the intersection angle, u, between the crack and the central line of the tooth is set at a constant 45 . The crack length, q1, grows from zero with an increment size of Dq1 \u00bc 0:1 mm until the crack reaches the tooth\u2019s central line. At that point, q1, reaches its maximum value of 3.9mm. After that, the crack then changes direction to q2 (see Fig. 5), which is assumed to be exactly symmetric around the tooth\u2019s central line. Theoretically, the maximum length of q2 should be the ARTICLE IN PRESS S. Wu et al. / Journal of Sound and Vibration 317 (2008) 608\u2013624 611 same as q1, however, the tooth is expected to suffer sudden breakage before the crack runs through the whole tooth. Thus the maximum length of q2 is assumed to be 60% of q1max and the increment size, Dq2, is also 0.1mm. In later reference of the crack growth, we will use a relative length. The highest crack level will be 3:9\u00fe 60% 3:9 \u00bc 6:24mm, or 80% (\u00bc 6:24=7:8) of the theoretical total length of the full through crack. ARTICLE IN PRESS S. Wu et al. / Journal of Sound and Vibration 317 (2008) 608\u2013624612 With the crack introduced as described above, we need to calculate all components of the total mesh stiffness, that is, Hertzian stiffness, axial compressive stiffness, bending stiffness, and shear stiffness. Based on the work documented in Ref. [15], the Hertzian and axial compressive stiffnesses remain the same when a crack is introduced. However, the bending and shear stiffnesses will change due to the appearance of the crack, and their derivation are provided under each of the following four cases. Case 1 Tian [4]: When hc1Xhr & a14ag, where a1 \u00bc 90 \u00f0pressure angle\u00de. ARTICLE IN PRESS S. Wu et al. / Journal of Sound and Vibration 317 (2008) 608\u2013624 613 The potential energy stored in a meshing gear tooth can be calculated by Ub \u00bc Z d 0 M2 2EIx dx \u00bc Z d 0 \u00bdF b\u00f0d x\u00de Fah 2 2EIx dx, (3) Us \u00bc Z d 0 1:2F 2 b 2GAx dx \u00bc Z d 0 \u00bd1:2F cos a1 2 2GAx dx, (4) where Ix and Ax represent the area moment of inertia and area of the section where the distance from the tooth\u2019s root is x, and G represents the shear modulus. They can be obtained by Ix \u00bc 1 12 \u00f0hc1 \u00fe hx\u00de 3L if xpgc; 1 12 \u00f02hx\u00de 3L if x4gc; 8>< >: (5) Ax \u00bc \u00f0hc1 \u00fe hx\u00deL if xpgc; 2hxL if x4gc; ( (6) G \u00bc E 2\u00f01\u00fe n\u00de , (7) where hx represents the distance between the point on the tooth\u2019s curve and the tooth\u2019s central line where the horizontal distance from the tooth\u2019s root is x. The bending mesh stiffness of the cracked tooth is 1 kbcrack \u00bc Z a2 ag 12f1\u00fe cos a1\u00bd\u00f0a2 a\u00de sin a cos a g2\u00f0a2 a\u00de cos a EL\u00bdsin a2 \u00f0q1=Rb1\u00de sin u\u00fe sin a\u00fe \u00f0a2 a\u00de cos a 3 da \u00fe Z ag a1 3f1\u00fe cos a1\u00bd\u00f0a2 a\u00de sin a cos a g2\u00f0a2 a\u00de cos a 2EL\u00bdsin a\u00fe \u00f0a2 a\u00de cos a 3 da (8) and the shear mesh stiffness of the cracked tooth is 1 kscrack \u00bc Z a2 ag 2:4\u00f01\u00fe n\u00de\u00f0a2 a\u00de cos a\u00f0cos a1\u00de 2 EL\u00bdsin a2 \u00f0q1=Rb1\u00de sin u\u00fe sin a\u00fe \u00f0a2 a\u00de cos a da \u00fe Z ag a1 1:2\u00f01\u00fe n\u00de\u00f0a2 a\u00de cos a\u00f0cos a\u00de2 EL\u00bdsin a2 \u00f0q1=Rb1\u00de sin u\u00fe sin a\u00fe \u00f0a2 a\u00de cos a da: (9) Case 2 Tian [4]: When hc1ohr or when hc1Xhr & a1pag. The bending mesh stiffness of the cracked tooth is 1 kbcrack \u00bc Z a2 a1 12f1\u00fe cos a1\u00bd\u00f0a2 a\u00de sin a cos a g2\u00f0a2 a\u00de cos a EL\u00bdsin a2 \u00f0q1=Rb1\u00de sin u\u00fe sin a\u00fe \u00f0a2 a\u00de cos a 3 da (10) and the shear mesh stiffness of the cracked tooth is 1 kscrack \u00bc Z a2 a1 2:4\u00f01\u00fe n\u00de\u00f0a2 a\u00de cos a\u00f0cos a1\u00de 2 EL\u00bdsin a2 \u00f0q1=Rb1\u00de sin u\u00fe sin a\u00fe \u00f0a2 a\u00de cos a da. (11) Case 3: When hc1ohr or when hc1Xhr & a1pag. This case was not covered in Ref. [4]. The bending mesh stiffness of the cracked tooth is 1 kbcrack \u00bc Z a2 a1 12f1\u00fe cos a1\u00bd\u00f0a2 a\u00de sin a cos a g2\u00f0a2 a\u00de cos a EL\u00bdsin a \u00f0q2=Rb1\u00de sin u\u00fe \u00f0a2 a\u00de cos a 3 da (12) ARTICLE IN PRESS S. Wu et al. / Journal of Sound and Vibration 317 (2008) 608\u2013624614 and the shear mesh stiffness of the cracked tooth is 1 kscrack \u00bc Z a2 a1 2:4\u00f01\u00fe n\u00de\u00f0a2 a\u00de cos a\u00f0cos a1\u00de 2 EL\u00bdsin a \u00f0q2=Rb1\u00de sin u\u00fe \u00f0a2 a\u00de cos a da. (13) Case 4: When hc2Xhr & a14ag. This case was not covered in Ref. [4] either. We have found the bending mesh stiffness of the cracked tooth to be 1 kbcrack \u00bc Z a2 ag 12f1\u00fe cos a1\u00bd\u00f0a2 a\u00de sin a cos a g2\u00f0a2 a\u00de cos a EL\u00bdsin a \u00f0q2=Rb1\u00de sin u\u00fe \u00f0a2 a\u00de cos a 3 da (14) ARTICLE IN PRESS S. Wu et al. / Journal of Sound and Vibration 317 (2008) 608\u2013624 615 and the shear mesh stiffness of the cracked tooth is 1 kscrack \u00bc Z a2 ag 2:4\u00f01\u00fe n\u00de\u00f0a2 a\u00de cos a\u00f0cos a1\u00de 2 EL\u00bdsin a \u00f0q2=Rb1\u00de sin u\u00fe \u00f0a2 a\u00de cos a da. (15) With the expressions of the components of the total mesh stiffness provided above, we are able to find the total mesh stiffness value given each shaft rotation angle and each crack size. For a pair of standard steel involute spur teeth whose main parameters are given in Table 1, take four specific crack sizes given in Table 2 as an example. These selected crack sizes cover all the four cases classified in the above derivations. The total mesh stiffness under each of the four crack sizes has been calculated as a function of the shaft rotation angle and plotted in Fig. 7. From Fig. 7, it can be observed that as the size of the crack grows, the total mesh stiffness when the cracked tooth is in meshing becomes much lower. This is important information for fault detection and assessment."
        ]
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure8.7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure8.7-1.png",
        "caption": "Figure 8.7 In fully saturated soil, the vertical grain pressure \u03c3g;v is found from the vertical earth pressure \u03c3e;v by subtracting it by the water pressure \u03c3w. The vertical grain pressure is shown by means of a hatching.",
        "texts": [
            " (6) If the soil is entirely saturated with water up to the ground level, the vertical grain pressure is no longer equal to the vertical earth pressure: \u03c3g;v = \u03c3e;v. (7) In that case, the vertical grain pressure \u03c3g;v is deduced from the vertical earth pressure \u03c3e;v by reducing it by the water pressure \u03c3w: \u03c3g;v = \u03c3e;v \u2212 \u03c3w. If the water is stationary and all the pores are linked to one another, the water pressure increases hydrostatically from zero at ground level to \u03b3wz at a depth z: \u03c3w = \u03b3wz 290 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM so that \u03c3g;v = \u03c3e;v \u2212 \u03c3w = \u03b3ez \u2212 \u03b3wz = (\u03b3e \u2212 \u03b3w)z. (8) In Figure 8.7, the contribution of the vertical grain pressure \u03c3g;v in the vertical earth pressure is shown by means of a hatching. If, with fully saturated soil, the water level is raised to above the ground level, this influences the vertical earth pressures, but not the vertical grain pressures (see Figure 8.8): \u03c3e;v = \u03b3wd + \u03b3ez, \u03c3g;v = \u03c3e;v \u2212 \u03c3w = \u03b3wd + \u03b3ez \u2212 \u03b3w(d + z) = (\u03b3e \u2212 \u03b3w)z. (9) An extensive, uniformly distributed load p on the ground level increases both the vertical earth pressure and the vertical grain pressure by an amount p"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002049_j.ymssp.2021.107945-Figure15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002049_j.ymssp.2021.107945-Figure15-1.png",
        "caption": "Fig. 15. Examples of typical origami base structures and their geometrical representations corresponding to different crease patterns: (a) generic degree-4 vertex pattern (Miura-ori pattern), (b) stacked Miura-ori (SMO) pattern, (c) Tachi\u2013Miura polyhedron (TMP) pattern, and (d) waterbomb base.",
        "texts": [
            " There are many crease design patterns that can be applied to practical engineering, including generic degree-4 vertex pattern, Miura-ori pattern, the stacked Miura-ori (SMO) pattern, the Tachi\u2013Miura polyhedron (TMP) pattern, and waterbomb base pattern, cylindrical pattern and the other patterns. The origami architectures, generated from these patterns, will be referred to here as origami base structures. It should be mentioned that there are other phrases widely used in the literature, such as origami base, base element, and unit cell, in particular, unit cell is generally used in origami-based metamaterials. Sometimes, unit cells are alternatively referred to as base structures. Fig. 15 shows the examples of the typical origami base structures which are simply formed by common crease patterns. The base structure of the generic degree-4 vertex pattern is a structural element made of one central vertex and four parallelogram facets. It can possess one continuous degree of freedom for rigid folding [204]. Specifically, the well-known Miura-ori pattern,whichpossesses bothflat-foldability and collinear folds, is also amember of thedegree-4 vertex family. Fang et al. [205] studied the deformation mechanisms of the four types of generic degree-4 vertex base structures",
            " Poisson\u2019s ratios of the standard Miura-ori patterned metamaterial structures are equal but opposite for in-plane and out-of-plane deformations of the structure [75]. There are other types of folding patterns that have been developed on the basis of the Miura-ori pattern such as the zigzag pattern [207,208]. In a typical Miura-ori base structure, outer dimensions including four folding parameters can be described by its smallest facet size and the dihedral fold angle h. The facet is normally a parallelogram with side lengths of a b and an acute angle of c. By referring to Fig. 15(a), the folded pattern parameters can be determined by [75]: H \u00bc a sin h sin c S \u00bc b cos h tan cffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1\u00fecos2 h tan2 c p L \u00bc a ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 sin2 h sin2 c q V \u00bc b 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1\u00fecos2 h tan2 c p 8>>>< >>>>: \u00f025\u00de where geometric parameters H, S, L and V are shown in Fig. 15(a). Parameters n, w and u represent the angles between the facets and the reference planes. Their relationships are given by: tan n \u00bc cos h tan c sinw \u00bc sin h sin c cos c \u00bc cos n cosw sinu \u00bc sin n= sin c 8>>< >>: \u00f026\u00de The negative in-plane Poisson\u2019s ratio (tin plane) of the Miura-ori pattern can be characterized by measuring strains of the base structure during the folding motion. The out-of-plane Poisson\u2019s ratio (tout of plane) can be equal but opposite to that of the in-plane deformation. Constraints are normally required for the Miura-ori pattern without facet bending",
            " Different origami patterns with the successive layout could exploit the fruitful geometry of the unit cell. The self-locking function could be enabled through specific configurations [210], which could provide effective mitigation under impact by the self-assembly technique. Fang et al. [204] investigated the dualcomponent design of the SMO structure based on the generic degree-4 vertex pattern to achieve a programmable selflocking feature and special stiffness property. For the SMO-based structure combined by at least two layers of different Miura-ori bases (unit cells) shown in Fig. 15(b), two different unit cells share the same designing parameters at the connecting planes and include at least one independent parameter. The expansion coefficients are then different from those of the standard Miura-ori pattern and can be re-written as [211]: L \u00bc 2bI cos hI tan cI 1\u00fecos2 hI tan2 cI W \u00bc 2aI ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 sin2 hI sin 2 cI q H \u00bc aI sin cI ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi tan2 cII tan2 cI cos2 hI q sin hI 8>>>< >>: \u00f028\u00de where parameters L, W and H are indicated in Fig. 15(b). The joint edges of the top sheet (sheet II) and bottom sheet (sheet I) share the same facet side (bI \u00bc bII), and the other edges of each sheet are independent (aI; aII). The enclosed internal volume within a SMO base structure (unit cell) can be calculated according to its folding conditions as [212]: V \u00bc 2a2I bI sin 2 cI cos hI ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi tan2 cII tan2 cI cos2 hI s sin hI ! \u00f029\u00de Tachi-Miura polyhedron (TMP) pattern is a rigid folding origami base structure with the nonlinear stiffness developed on the basis of Miura-ori pattern [213]",
            " With lateral assembly of small Miura-ori base structures into one single integrated unit, a certain volume can be held within the TMP base structure. Since the structure only deforms along the crease lines and does not rely on the material elasticity of its facets, the facets are completely rigid during the folding process. Yasuha and Yang [213] analytically and experimentally studied the Poisson\u2019s ratio and bi-stability of the TMP-based structure. Nonlinear stiffness and dynamic response of the TMP structure were discussed in [214]. As illustrated in Fig. 15(c), a certain volume in the unit could be changed within a full folding motion cycle. The minimum volume can be zero. The outer dimensions of the TMP pattern are obtained from its flatted sheet parameters and folding angles as [213]: B \u00bc 2m sin hG \u00fe d cos hM W \u00bc 2l\u00fe d tana\u00fe 2m cos hG H \u00bc 2d sin hM 8>< >: \u00f030\u00de where the geometric parameters B, W , H, l and m are given in Fig. 15(c). It is interesting to note that in-plane Poisson\u2019s ratios of the TMP base structure with respect to either width B or H are different [213]: tHW \u00bc dW=W dH=H \u00bc 4m tana sin hG cos2 hG 2 2l\u00fe d tana\u00fe2m cos hG sin hM tan hM tHB \u00bc dB=B dH=H \u00bc 4m tana cos hG cos2 hG 2 \u00fed 2M sin hG\u00fed cos hM sin hM tan hM 8>< >: \u00f031\u00de From the above equations, it can be calculated that Poisson\u2019s ratio in the HW plane (the front plane as shown in Fig. 15(c)) would always be less than zero. Meanwhile in the HB plane (the side plane), Poisson\u2019s ratio could drop to zero and remains at the low level when the folding ratio increases. A waterbomb base is a single-vertex origami structure, which is normally folded from a square or circular sheet, as illustrated in Fig. 15(d). At least three groups of creases should be involved and each group includes one mountain and one valley. Characteristics of the waterbomb base include developable vertices, completely rigid facet, and rotationally symmetric geometry [215]. For a traditional n degree waterbomb base containing n pairs of mountain and valley folds, the geometric relationships for angles are given by: a \u00bc 180 =n b \u00bc 360 =n 0 h 180 a 8>< >: \u00f032\u00de where a is the facet sector angle, b denotes the angle between any two folds that are the same type, and h is the angle between the valley fold and the reference central axis"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001444_j.cej.2011.03.045-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001444_j.cej.2011.03.045-Figure3-1.png",
        "caption": "Fig. 3 shows the effects of temperature on the RBO3R adsorption ptake, qe (mg/g) for various initial dye concentrations. The RBO3R dsorption uptakes were found to increase with increase in solution emperature from 303 to 333 K for all initial concentrations. This ay be due to increase in the dye mobility to penetrate inside the ample pores at high temperature. Besides, it might also be due to he increase in chemical interaction between the adsorbate and sur-",
        "texts": [],
        "surrounding_texts": [
            "M.A. Ahmad, N.K. Rahman / Chemical Engin\nt c n T l t o a t t a t w 6\n2\n( c 6 D u\n2\no t R d s f R U i e\nq\nw R u\n2\nc i 2 fl\nubular reactor placed in a tube furnace. Carbonization step was arried out at 673 K with heating rate of 10 K/min under purified itrogen (99.99%) flown through at flowrate of 150 ml/min for 2 h. he char produced was mixed with sodium hydroxide (NaOH) pelets with equally weight ratio. Deionized water was then added o dissolve the NaOH pellets. The mixture was dehydrated in an ven at 378 K overnight to remove moisture. Then the sample was ctivated under the same condition as carbonization but at final emperature of 973 K. Once the final temperature was reached, he nitrogen gas flow was switched to carbon dioxide (CO2) and ctivation was held for 1 h. Then the sample was cooled to room emperature under nitrogen flow and washed with hot deionized ater and 0.1 M HCl until the pH of the washing solution reached\n.5\u20137.\n.2. Remazol Brilliant Orange 3R\nRemazol Brilliant Orange 3R (RBO3R) supplied by Sigma\u2013Aldrich M) Sdn Bhd, Malaysia was used as an adsorbate. RBO3R has a hemical formula of C20H17N3Na2O11S3 with molecular weight of 17.54 g/mol. The chemical structure of RBO3R is shown in Fig. 1. eionized water supplied by USF ELGA water treatment system was sed to prepare all the reagents and solutions.\n.3. Batch equilibrium studies\nBatch equilibrium tests were carried out for RBO3R adsorption n the CHAC prepared. The effects of initial RBO3R concentraion, contact time, solution temperature and solution pH on the BO3R uptake were investigated. Sample solutions were withrawn at equilibrium to determine the residual concentration. The olutions were filtered prior to analysis in order to minimize intererence of the carbon fine with the analysis. The concentration of BO3R solution before and after adsorption was determined using V\u2013vis spectrophotometer (UV-1800 Shimadzu, Japan) at it maxmum wavelength of 493 nm. The amount of RBO3R adsorbed at quilibrium, qe (mg/g) was calculated by:\ne = (Co \u2212 Ce)V W\n(1)\nhere Co and Ce (mg/l) are the liquid-phase concentrations of BO3R dye at initial and at equilibrium, respectively. V is the volme of the solution and W is the mass of dry adsorbent used.\n.3.1. Effect of contact time and RBO3R initial concentration\nIn order to study the effect of contact time and initial dye conentration on the RBO3R uptake, 100 ml of RBO3R solutions with nitial concentration of 25\u2013250 mg/l were prepared in a series of 50 ml Erlenmeyer flasks. 0.2 g of the CHAC was added into each ask covered with glass stopper and the flasks were then placed in\neering Journal 170 (2011) 154\u2013161 155\nan isothermal water bath shaker at temperature of 303 K with rotation speed of 120 rpm, until equilibrium point was reached. In this case, the solution pH was kept original without any pH adjustment.\n2.3.2. Effect of solution temperature on RBO3R adsorption The effect of solution temperature on the RBO3R adsorption process was examined by varying the adsorption temperature from 303, 318 and 333 K by adjusting the temperature controller of the water bath shaker, while other operating parameters were kept constant. The solution pH was kept constant without any adjustment.\n2.3.3. Effect of solution pH on RBO3R adsorption The effect of solution pH on the RBO3R adsorption process was studied by varying the initial pH of the solution from 2 to 12. The pH was adjusted using 0.1 M HCl and/or 0.1 M NaOH and was measured using pH meter (Model Delta 320, Mettler Toledo, China). The RBO3R initial concentration was fixed at 200 mg/l with adsorbent dosage of 0.2 g/100 ml. In this case, the solution temperature is at 303 K. The RBO3R percent removal was calculated by the following equation:\nRemoval (%) = (Co \u2212 Ce) Co \u00d7 100 (2)\n2.4. Batch kinetic studies\nFor the batch kinetic studies, the same procedure was followed, but the aqueous sample was taken at preset time interval. The concentration of RBO3R was similarly measured. The RBO3R uptake at any time, qt (mg/g) was calculated by:\nqt = (Co \u2212 Ct)V W\n(3)\nwhere Ct (mg/l) is the liquid-phase concentration of RBO3R at any time, t (h).\n2.5. Characterization of CHAC\nTextural characterization of the CHAC prepared was determined by nitrogen adsorption at 77 K using volumetric adsorption analyzer (Micromeritics ASAP 2020). The surface area of the sample was determined using Brunauer\u2013Emmett\u2013Teller (BET) equation. The total pore volume was estimated to be the liquid volume of nitrogen at a relative pressure of 0.98. The surface morphology of the sample was examined by using a scanning electron microscope (JEOL, JSM-6460 LV, Japan). The proximate analysis was carried out using thermogravimetric analyzer (Perkin Elmer TGA7, USA). The elemental analysis was performed using Elemental Analyzer (Perkin Elmer Series II 2400, USA). The functional group of the sample was estimated by Fourier transform infrared (FTIR) spectroscopy (FTIR-2000, Perkin Elmer).",
            "c\n[ 1 2 d b t a r\n5 o t r m h fi [ a p w t m\n3\nu a t m s t\nentrations.\n10]. The adsorption uptakes of RBO3R at equilibrium increase from 2.09 to 65.33 mg/g as the initial dye concentration increases from 5 to 250 mg/l. The initial dye concentration provides an important riving force to overcome the mass transfer resistance of the dye etween the aqueous and solid phases [11]. Therefore, at higher iniial dye concentration, the number of molecules competing for the vailable sites on the surface of activated carbon was high, hence, esulting in higher RBO3R adsorption capacity.\nFrom Fig. 2, it was also observed that an equilibrium time of h was needed for RBO3R dye solution with initial concentrations f 25\u201350 mg/l to reach equilibrium. However, for initial concenrations of 100\u2013250 mg/l, longer equilibrium times of 22\u201324 h are equired for the system to reach equilibrium. Initially, adsorbate olecules have to first encounter the boundary layer effect. Then it as to diffuse from boundary layer film onto adsorbent surface and nally, it has to diffuse into the porous structure of the adsorbent 12]. In fact, the ratio of the initial number of dye molecules to the vailable surface area is low at lower initial concentration comared to higher initial concentration. Therefore, RBO3R solution ith higher initial concentration would take relatively longer conact time to attain equilibrium due to the higher amount of RBO3R olecules.\n.2. Effect of solution temperature on RBO3R adsorption\nface functionalities of adsorbent [11]. The results indicate that the adsorption reaction of RBO3R adsorbed by CHAC is an endothermic process.\n3.3. Effect of initial solution pH on RBO3R adsorption\nThe effect of solution pH on the RBO3R removal was studied at initial pH values of 2\u201312. As shown in Fig. 4, the percentage removal of RBO3R was high at pH 2\u20135 before it decreases at higher pH. The high percentage removal of the RBO3R solution at acidic pH could be attributed to the electrostatic interactions between the positively charged adsorbent and the negatively charged RBO3R dye anions [11]. The surface charge of adsorbent has become generally positive charge in acidic medium and the negative charge in basic medium. Therefore, the removal of anionic dye by CHAC was high at lower pH. The adsorbent surface would attract the negatively charged functional groups located on the RBO3R dye and hence increase the adsorption capacity. Similar observation was obtained by Ada et al. [13] where the maximum adsorption capacity of reactive blue dye onto ZnO fine powder was obtained at pH 4.\n3.4. Adsorption isotherms\nEquilibrium isotherm equations are used to describe the experimental sorption data and the parameters obtained from the different models. The adsorption equilibrium data were analyzed using Langmuir, Freundlich and Temkin isotherms expression. Linear regression is commonly used to determine the best-fitting isotherm and the applicability of isotherm equations is compared by judging the correlation coefficients, R2.\nLangmuir model is based on the assumption that adsorption energy is constant and independent of surface coverage where the adsorption occurs on localized sites with no interaction between adsorbate molecules. The maximum adsorption occurs when the surface is covered by a monolayer of adsorbate [14]. The linear form of Langmuir isotherm equation is given as:\nCe qe = 1 QoKL + 1 Qo Ce (4)\nwhere Ce (mg/l) is the equilibrium concentration of the RBO3R, qe (mg/g) is the amount of RBO3R adsorbed per unit mass of adsorbent. Qo (mg/g) and KL (l/mg) are Langmuir constants related to adsorption capacity and rate of adsorption, respectively. A straight line with slope of 1/Qo and intercept of 1/QoKL is obtained when Ce/qe is plotted against Ce (Fig. 5). The essential characteristics of Langmuir",
            "F 3\ne f\nR\nw i ( s c v o i w\no a g\nl\nw c i w n o f i\nm a\nF 3\nig. 5. Langmuir isotherm plot for the adsorption of RBO3R dye on CHAC; 313 K (\u2666), 18 K ( ) and 333 K ( ).\nquation can be expressed in terms of dimensionless separation actor, RL defined by [15]:\nL = 1 (1 + KLCo)\n(5)\nhere Co is the highest initial solute concentration. The RL value mplies whether the adsorption is unfavorable (RL > 1), linear RL = 1), favorable (0 < RL < 1), or irreversible (RL = 0). Fig. 6 repreents the plot of the calculated RL values versus the initial dye oncentration at 303, 318 and 333 K. It was observed that all the RL alues obtained were between 0 and 1, showing that the adsorption f RBO3R on the CHAC was favorable. The RL values decrease with ncrease in initial dye concentration, indicating that the adsorption\nas more favorable at higher RBO3R concentration. Freundlich model is an empirical equation based on sorption\nn a heterogeneous surface or surface supporting sites of varied ffinities [16]. The well known logarithmic form of Freundlich is iven by the following equation:\nog qe = log Kf + 1 n log Ce (6)\nhere KF and n are Freundlich constants related to adsorption apacity and adsorption intensity, respectively. In general n > 1 llustrates that adsorbate is favorably adsorbed on the adsorbent\nhereas n < 1 demonstrates the adsorption process is chemical in ature. The plot of log qe versus log Ce gave a straight line with slope f 1/n and intercept of log KF (Fig. 7). In this study, the values found\nor n were between 3.61 and 4.32, which prove that the adsorption s favorable and process is physical in nature.\nTemkin isotherm assumes that the heat of adsorption of all the olecules in the layer would decrease linearly with coverage due to dsorbent\u2013adsorbate interaction. The adsorption is characterized\n0\n0.02\n0.04\n0.06\n0.08\n0.1\n2502001501005025\nR L\nCo (mg/L)\nig. 6. Effect of RBO3R initial concentration on dimensionless separation factor RL; 13 K (\u2666), 318 K ( ) and 333 K ( ).\nby a uniform distribution of binding energies, up to some maximum binding energy [17]. Temkin model is expressed as: qe = ( RT\nbT\n) ln(ACe) (7)\nwhere RT/bT = B (J/mol), which is the Temkin constant related to heat of sorption whereas A (l/g) is the equilibrium binding constant corresponding to the maximum binding energy. R (8.314 J/mol K) is the universal gas constant and T (K) is the absolute solution temperature. A plot of qe versus ln Ce gives a linear graph with B as the slope and B (ln AT) as the intercept (Fig. 8).\nTable 1 summarizes all the constants and R2 values obtained from the three isotherm models applied for adsorption of RBO3R on the CHAC. Langmuir model gave the highest R2 values which were greater than 0.99. Conformation of the experimental data into"
        ]
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure14.62-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure14.62-1.png",
        "caption": "Figure 14.62 (a) Kinematically indeterminate bar structure, is (b) changed into a two-hinged frame. (c) The support reactions of the statically indeterminate two-hinged frame if axial deformation is ignored; the line of force coincides with the bent member axis: there is no bending.",
        "texts": [
            "58a has been changed into a kinematically determinate truss. All the interior members are zero. As in contrast to the cable, the truss has the benefit that the shape does not change when the load changes. In Figure 14.61, the forces on the truss are shifted to the horizontal plane through the supports. The verticals are no longer zero members. These 682 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM trusses, in which all the members are subject to extension (or are zero members), can be an alternative for the beam subject to bending in Figure 14.56. Figure 14.62a shows the bar structure from Figure 14.58b, but now without tension member. This kinematically indeterminate structure can be made kinematically determinate not only by changing it into a truss, but also by replacing the hinged joints between the bars by rigid joints. The structure then becomes a bent member, recognisable in Figure 14.62b as a two-hinged frame. One problem is that the frame is statically indeterminate to the first degree. As such, it is not possible to determine the horizontal support reaction H directly from the equilibrium. The deformation of the frame also has to be taken into account. If the deformation by normal forces is ignored (as it was in the cable), it is possible to show that no bending occurs under the given load, and that the normal forces in the frame are equal to the forces in the two-force members in Figure 14.62a. In Figure 14.62c, the frame has been isolated and all the forces acting on it are shown. The line of force coincides everywhere with the bent member axis and there is no bending anywhere. In reality, there is always some axial deformation due to normal forces. As such, the horizontal support reactions are somewhat smaller and, because the vertical support reactions remain equal, the line of force no longer coincides with the bent member axis (see Figure 14.63). Axial deformation therefore induces bending in the two-hinged frame"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001351_j.ijfatigue.2019.03.025-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001351_j.ijfatigue.2019.03.025-Figure1-1.png",
        "caption": "Fig. 1. AM build chamber layout for vertical fatigue bars.",
        "texts": [
            " The asbuilt and cylindrical bars were placed closely together at the grip sections with rows of bars perpendicular to the recoating direction. The rows were staggered such that the recoater only contacted the edge of one component at any given time. Triangular supports were added to support the end of each row. In total, 15 as-built bars were included in the assembly. One of the bars with parameter set A was deleted midbuild, because of a recoater jam, so 14 as-built fatigue bars were fully fabricated. The build layout can be seen in Fig. 1 and the fatigue bar geometry is shown in Fig. 2. The fatigue bars were heat treated post-process in order to relieve the residual stress and homogenize the microstructure. This will allow consistent assessment of the as-built surface roughness. The bars were stress relieved at 1065 \u00b0C for 90min and furnace cooled. The bars were removed from the build plate using wire EDM and post-processed through the standard solution and aging heat treatment for AM Alloy 718 [24,25]. No additional heat treatment was used to reduce the porosity because it has been shown that the surface roughness will dominate the fatigue failure [12]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003836_0013-4686(81)85156-0-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003836_0013-4686(81)85156-0-Figure3-1.png",
        "caption": "Fig. 3 . Flow-through cell with RVC electrode[18] . A, Sample solution inlet. B, Sample solution outlet . C, RVC disk. D, Reference electrode. E, Lead to working electrode. F, Teflon spacer, G, Kel-F tube. H, O-rings (reproduced by permission of the American Chemical Society) .",
        "texts": [],
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            "Electrochemical flow-cells During the last decade there has been increased interest in flow-through porous electrodes for the detection of electroactive species in flowing streams (eg, liquid chromatography), for electroorganic synthesis, for the removal of metal ions from waste streams, and for various other industrial electrolytic processes. Among the various forms of carbon used as flow-through porous electrodes RVC appears to be very promising due to its hydrodynamic and electrochemical advantages (discussed above) . Strohl and Curran have evaluated a flow cell with a RVC cylinder of 10.8-cm long and 1 cm . dia (surface area -515cm2)[9] . The cell could operate in an amperometric and a coulometric mode. Currentpotential curves, obtained for the reduction of ferricyanide and oxidation of ascorbic acid, have indicated mass-transport controlled reactions. The cell was demonstrated to be a useful detector for flow injection analysis[13] . With amperometric operation, sampling rates as high as 264 samples per hour were achieved, the relative precision beigg about 0.5 %. Applications has been made for the detection of ascorbic acid, L-Dopa, epinephrine, and ferricyanide, with detection limits of few tenths of a nanogram. Figure 2 illustrates characteristic response of this detector for successive injections of 28.61 nanograms ephinephrine solutions . Several applications of this cell, based on controlledpotential coulometry, have been reported[14] ; these include the determination of submicrogram amounts of hydroquinone, catechol, ascorbic acid (in vitamin C tablets), and other analytes, with accuracy of a few parts-per-trillion . Strohl and Curran have found the RVC to be very useful for monitoring the acidity of solutions[10] . The mV-pH response shows that the material has promise as a pH indicator electrode . Although the mechanism of this pH response is unclear, it is estimated that the surface acidic functional groups are responsible for this behavior. Blaedel and Wang[15] carried out a general evaluation of a flow cell assembly composed of a variable number of RVC disks. The characteristics of this cell were found to be in accordance with the theory of the porous electrode reactor[16]. The dependence of the Epinephrie Reticulated vitreous carbon-A new versatile electrode material limiting current upon the flow rate was of the type it - U' with m increasing with increased electrode length or decreased flow rate. Application was made to the on-line measurement of submicromolar concentrations. Efficient discrimination against the high background current that accompanies the high analytical current was obtained by the sensitive stopped-flow voltammetry, resulting with a detection limit around i nM ferrocyanide . Similar hydrodynamic modulation approach, the rapid pulsed flow voltammetry, was incorporated with the RVC electrode for achieving subsecond response times when monitoring submicromolar concentration levels[17] . A unique design of an enzyme electrode based on the RVC material was studied by Blaedel and Wang( 18] . It is a mixed bed reactor composed of an immobilized enzyme gel (Sepharose-bound alcohol dehydrogenase) that has been packed inside the pores of a 60 ppi RVC electrodes. High sensitivity and rapid response (with ethanol as the analyte) acrue due to the large surface area and intimate contact between the enzyme and the electrode surface, respectively . Flow cell with a mercury coated RVC electrode was proposed for on-line anodic stripping voltammetry of trace metal ions[12] . Being composed of glassy carbon the RVC is ideal substrate for mercury film (although its surface cannot be polished to a mirror-like finish like solid glassy carbon). Exploratory work have 1723 1 724 JOSEPH WANG shown the dependence of the stripping peak current upon various experimental parameters . High deposition yields and high sensitivity, associated with the ROTATED SHAFT large surface area, have allowed measurements at the parts-per-billion concentration level using 5-min deposition. The electrode functioned in the normal fashion for about two weeks, after which electrode failure (ie, increased background) was observed . Smooth mercury coating has been achieved also by depositing mercury on nickel-plated RVC[6] . Schieffer suggested the use of RVC particles (obtained by crushing the staff with a spatula) for packing a coulometric cell, located upstream to an amperometric detector[19] . Such a dual cell system results with improved selectivity of electrochemical detection in high pressure liquid chromatography, with the RVC electrode held in lower potential than the amperometric electrode separation of two interfering peaks (with different oxidation potentials) is provided . Another interesting and efficient design of a RVC flow-through electrode is consisted of a rotated RVC disk[20] . Coupling the large surface area of the RVC with the efficient mass-transport of the rotating electrode has resulted with large analytical currents . Pulsed-rotation modulation technique has been used for compensating the background current, resulting with well-defined current-potential curves and low detection limits (at the nanomolar concentration level of dopamine and ascorbic acid) . Batch applications Although most reported applications of the RVC has been devoted to electrochemical flow cells, the RVC electrodes were first applied for different batch applications. Norvell and Mamantov used a thin slice of the RVC as an optically transparent electrode (OTE) [6] . Light transmittance through a thin RVC slice is comparable to that of wire mesh minigrids and is essentially constant over the entire ua-visible region . Due to the special structure of the RVC, the RVC-OTE has several characteristics which overcome some of the disadvantages of minigrid or metal oxide OTEs, while combining the advantages of both of these types (electrochemical properties of metal oxide with spectroscopic advantages of the minigrid OTE). Another contribution from the University of Tennessee involved the use of a stationary RVC anode for the coulometric electrolysis of uranocene in THF[21] . Metallic anodes, like Pt gauze, have been passivated under the conditions of the electrolysis, indicating great promise of the RVC for electrosynthetic purposes . Sensitive electroanalytical measurements have been obtained with a rotated RVC disk electrode (Fig . 4)[l 1]. The efficiency of mass-transport at the rotating RVC electrode approaches that at the rde, apparently due to the combination of the high rotational speed with the high void volume of the RVC that promotes circulation of the solution inside the electrode volume . A square root dependence (i, - cu\u00b0 \u2022 5) is obtained between the limiting current and the rotation speed . Pulsed-rotation voltammetry at the rotating RVC electrode have been demonstrated for micromolar and submicromolar analyses of NADH and ferrocyanide . Fig. 4. Rotating RVC disk electrode[11] (reproduced by permission of the American Chemical Society) . Coating the rotating RVC disk with thin mercury film has resulted with sensitive ASV indicator electrode . Special advantages have been gained with the differential pulse stripping approach that discriminates against the high charging current. Table 2 summarizes the various electrochemical applications and cell configurations of the RVC electrode."
        ]
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        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure7.28-1.png",
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        "caption": "Figure 7.28 A sketch of the cross-section of the door, with the water levels on both sides.",
        "texts": [
            " From the alternative approach, one can conclude the following: The resultant of the horizontal water pressure on the slide is independent of the shape of the slide and is exclusively determined by the height of the slide and the depth at which it is located under the water surface. Figure 7.26 shows the lines of action of Rx and Ry . The line of action of Rx can be found directly from the trapezoidal load diagram on AD.1 Example 3 At Hoek van Holland, near Rotterdam in the Netherlands, the Maeslantkering became operational in 1997. This storm barrier in the Nieuwe Waterweg consists of two sector doors with a radius r = 250 m (see Figure 7.27). The arc length of AB is 209.5 m. The door is 22.5 m in height. Figure 7.28 is a sketch of the longitudinal section of the door, with the water levels on both sides. The specific weight of water is \u03b3w = 10.25 kN/m3. To simplify the question, the part of the door within the parking dock is ignored. 1 The calculation is left to the reader. See Section 6.3.1, Example 1. 7 Gas Pressure and Hydrostatic Pressure 263 In addition, it is assumed that the water levels in front and behind the dam are present over the entire length of arc AB and that the pressure distribution on both sides is hydrostatic"
        ],
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    },
    {
        "image_filename": "designv10_0_0003728_1.3099008-Figure14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003728_1.3099008-Figure14-1.png",
        "caption": "Fig 14. Woodpecker (from Pfeiffer, 1984), windmill (from BOrger, 1980) and tippy top (from Magnus, 1971)",
        "texts": [],
        "surrounding_texts": [
            "Appl Mech Rev vol 51, no 5, May 1998\nthe operator could press down on the drill without burning the grip (Davison, 1957; Davison, 1957/58). Remains and drawings of such hand-fitted bearings are included in the above-mentioned Egyptian drills. Vitruvius recorded the usage of roller bearings fitted between a battering ram and a track during the siege of Cadiz around 330 BC. The credited designer was the Greek engineer Diades (Davison, 1957; Davison, 1957/58).\nLeonardo da Vinci (1452-1519) had sketched roller beatings in his notebooks (Leonardo da Vinci, 1939), as shown in Fig 11. The upper sketch shows bearings which allow a crank to turn under pressure. The drawing below includes axled rollers which keep the ball bearings separated. Inscriptions are in Leonardo's peculiar tight-to-left handwriting.\nThe first published drawings of roller bearings are believed to be those of Ramelli (1588), who had also sketched thrust beatings. One of Ramelli's bearing devices is shown in Fig 12. Among Ramelli's machines is the flywheel, which he designed to ease the efforts of the operator. In fact, flywheels go back at least to the Egyptians, who used them on drilling devices (Ferguson, in Ramelli (1588)).\nIt can be argued that friction played a motivational role in the domestication of animals for labor. Beasts of burden were better at pulling chariots and ploughs. Ploughing involves friction as well as tearing, lifting and turning of soil.\n\"The early plough was no more than a hoe drawn through the ground, perhaps first by man with a rope, but already in the Old Kingdom [of Egypt] by a pair of oxen\" (Drower, 1956).\nZeuner (1956) suggests that domestication came in four stages: 1) dogs, 2) reindeer, goat and sheep, 3) pigs and cattle, and 4) horses, asses and onagers. The first two stages probably were motivated by the availability o f food,\nFeeny et al: Historical review of dry friction and stick-slip phenomena 325\nwhile the latter stages may have been related to work. The domestication of pigs and cattle required a somewhat settied life, and is likely to have developed in the later neolithic period. The use of animals for transport and traction probably started with cattle, and were later extended to animals of the fourth group.\nChilde (1956a) suggests that \"oxen had been yoked to ploughs in many regions long before they or other animals began pulling carts or wagons.\" He notes further that \"the yoke was originally designed to fit the shoulders of oxen, and only secondarily transferred to onagers, horses, and asses--to their great discomfort\" (see Fig 13).\nWhile friction motivated technologies that eased labor, it was also important in enhancing play. Various types of mechanical toys require friction to achieve their entertaining effect. Among such toys are the tippy top, magic windmill, fiddlesticks, celt stones, mountain climber, and woodpecker. Some of these toys are discussed by BOrger (1980), and others by Walker (1975).\nThe tippy top, when motionless, rests with a rounded portion in contact with the ground. When spun, this configuration destabilizes. The top tips and over turns, spinning in the inverted position in a stable manner. Ultimately, dissipative effects dominate, the top slows down, flops over, and comes to rest in its original position. Friction had been attributed to the behavior by many interested parties (such as Jellett, 1872; Braams, 1952; Hugenholtz, 1952; and Del Campo, 1954). Fokker (1941) traced the motion of the tops by using carbon paper.\nThe magic windmill, or Indian mystery stick, is a notched stick with a propeller at the end. When the notches are rubbed, rotations are induced in the propeller. A clever operator can control the direction and speed of the propeller. Explanations for the behavior include friction (Laird, 1955) and polarized elliptic motion of the notched stick (Leonard, 1937; Bfirger, 1980). Bfirger suggests that the notched stick was born with Leonard in 1937. Laird cites Reg US Patent Office for the origin.\nCelt stones are elongated rigid bodies with specially shaped surfaces and skewed principle axes. When spun, they have a tendency to dip. When spun in one direction, a wobbly motion is induced, and then the wobbly motion is convened to a spin in the opposite direction. Friction is impor-\nDownloaded From: http://appliedmechanicsreviews.asmedigitalcollection.asme.org/ on 10/19/2017 Terms of Use: http://www.asme.org/about-asme/terms-of-use",
            "326 Feeny et al: Historical review of dry fdction and stick-slip phenomena Appl Mech Rev vol 51, no 5, May 1998\ntant in this reversal of momentum. Spinning in the opposite direction is stable.\nThe woodpecker consists of a rigid bird mounted to a slider through a soft spring. The slider slides on a vertical metal pole. The bird can rest statically such that the moment on the slider allows it to grip the pole. As the slider falls, and the rigid bird oscillates on its spring, there is an alternation between static gripping of the pole, and a sliding, falling state. This descent with an oscillation in the bird is stable, until the bird reaches the bottom. Den Hartog (1972) offered a partial analysis of this motion after assigning it as a finalexam problem.\nFiddlesticks are made up of poles with rings. As the rings slide down the poles, rolling whirls are induced, and the falling motion nearly halts (Walker, 1975). The mountain climber is a man made of wood, capable of bending at the waist. A string is strung through the feet and hands in a clever way. When the string is pulled, the man straightens and slides at the hands. When the tension is relaxed, he bends, and slips at the feet. Cyclic tension and relaxation causes the climber to traverse the string.\nA toy best known under the trade name of Lego employs a frictional grip between fitted plastic bricks, and other parts, so that children, and some adults, can build all sorts of fascinating things. Of course, a good billiards player uses friction to put english on the ball to influence impacts and ricochets (Bt.lrger, 1980).\nFriction has also played a role in sports. Ice hockey evolved from field hockey. North American Indians probably played a similar game on ice long ago. A 16th century picture shows Dutch skaters hitting a ball. The modern form of hockey was born in 1860 when British soldiers, veterans of the Crimean war stationed in Kingston, Ontario, used a flat disk on the ice. Racing on skates started in the low countries during the 17th century. Another ice sport is curling, in which a curling stone is slid across the ice toward a target. Spin is applied so that the stone may curve, and sweeping the ice reduces friction in front of the stone. Archeological digs have unearthed curling stones from bogs in Scotland. A painting by Pieter Breughel (1530-1569) shows a similar game being played by the Dutch (Heller, 1979).\nFrictional wear is exploited when writing or drawing with chalk or pencil. The ancient idea may have started with the use of charcoal. Cave art from the time of the famous works of Lascaux (some 15,000 years ago) may include charcoal scratchings. In reference to paleolithic painting, Leakey (1956) writes:\n\"The discovery of pointed crayons made of mineral colouring materials in cultural levels in some prehistoric sites, and the examination of actual paintings, make it probable that some paintings were carried out by means of a dry crayon technique.\"\nTraces of wear on such crayons suggest that they may have been used to draw rough sketches, \"which the artist could then go over with a wet brushing\" (Raspoli, 1978). Figure 15 comes from Rigaud et al (1988).\nThe first known modern pencil, a tube of wood with lead inside and a knobby eraser at the end, was sketched and described by Konrad Gesner of Zurich, Switzerland, in his book on fossils in 1565. Gesner died later that year of the plague (Petroski, 1990).\nThe pencil industry motivated friction and wear testing (Petroski, 1990). Around 1949, Eagle Pencil Company had been performing wear tests by dragging a loaded pencil on a drum of rolled paper. By measuring the distance as the circumference times the number of revolutions, Eagle could justify their advertising motto for the Mikado line: \"Thirty-five miles for a nickel.\" Eagle also measured smoothness from a pendulum rig, in which the number of oscillations indicated the amount of friction. After World War I1, India began developing a pencil industry of its own, and identified blackness, wear, and friction as critical properties. The Indian Standards Institution drafted its Spec!fications Jbr Black Lead Pencils in 1959 based on a set of tests. Wear tests were performed by measuring the number of millimeters of lead needed to draw a line of a given length on sandpaper. Friction was measured from the\nDownloaded From: http://appliedmechanicsreviews.asmedigitalcollection.asme.org/ on 10/19/2017 Terms of Use: http://www.asme.org/about-asme/terms-of-use",
            "Appl Mech Rev vol 51, no 5, May 1998 Feeny\nforce exerted while pulling a paper-covered cart under a loaded pencil. Blackness was quantified by reflecting light from a sheet of parallel lines into a photocell.\nAnother fascinating thing may take place when certain bodies are rubbed: attractive forces between the bodies may develop as electrons are shed: The Greek philosopher Thales (circa 636-546 BC) noted that when animal fur is rubbed on amber, a fossilized resin called elektron by the Greeks (and now famous for its proposed role in the cloning of dinosaurs), it would then attract other objects, such as feathers and bits of dried leaves (Asimov, 1966). Although Asimov (1966) suggests that this phenomenon was known to prehistoric people, Thales may have been the first to document what is now known as the electric charging of a body. Later it was found that this property belonged to other materials, such as diamond, sapphire, amethyst, carbuncle, jet, and rock crystal, as discovered by William Gilbert (1540-1603). In 1600, Gilbert suggested that such materials be called electrics. Otto yon Guericke (1602-1686) built the first electrical friction machine. It consisted of a sphere of sulfur which could be tumed by a crank. By placing a hand on it, it became \"electrified to a hitherto unprecedented extent.\" It was a shocking discovery. In 1729, Stephen Gray rubbed glass tubes and found that the corks at the ends became electrified. Thus he discovered electric f lu id in that electricity could move about (Asimov, 1966). Lightning is caused by the rubbing of airborne particles, and can occur in volcanic eruptions as charges develop due to the friction between volcanic ash particles and steam and other gases (Blong, 1984).\nModern technology Coming back to transportation, more recent developments have resulted from the automobile. One particular area is in braking. At first, braking took place on the rear Wheels alone. On loose dirt roads, the wheels typically would lock up. In 1902, the Automobile Club of America found that the average braking distance at a speed of 32 km/h was 18m. Fourwheel brakes were developed by the Aland Motor Car Company in 1916. Hydraulic brakes were then introduced by Duesenberg in 1921. Studebaker developed self-adjusting drum brakes on its 1946 models. Disc brakes were originally designed for airplanes. Attempts at incorporating disc brakes into automobiles were made by Crosely on its Hotshot sports car, and also by Chrysler on its 1949 Imperials. These brakes did not dissipate heat well enough. Later, Dunlop patented the modem disc brake in Britain. Drum brakes with cooling flanges were introduced by Oldsmobile in 1958. Oldsmobile and Studebaker debuted the master cylinder in 1962, so that if the brakes failed on one of the two-wheel sets, they would continue to function on the other. The rear wheel anti-lock brake was produced by Ford in 1969 (Yanik, 1996a; 1996b).\nThe automobile tire also went through several stages. Early tires where made of several layers of cloth covered with rubber. Attempts at adding convoluted treads for improved skid behavior began in 1908. Firestone introduced the \"balloon\" tire in 1922. Previous tires had pressures of 345- 380 kPa, which provided a poor \"footprint\" which was susceptible to skidding. The balloon tires reduced the air pressure to 240-275 kPa, improving the footprint, and hence the\net al: Historical review of dry friction and stick-slip phenomena 327\ntraction. Tubeless tires, which tumed blowouts into \"slowouts,\" were introduced by BF Goodrich in 1948, and became a standard by 1955. Goodrich made radials available to the domestic market in 1965 (Yanik, 1996a; 1996b).\nThe continuously variable transmission, a continuous gear system with infinitely many gear ratios, made its debut in 1886, when Daimer and Benz used a rubber V-belt in their first gasoline-powered automobile (Ashley, 1994).\nWhile the development of the wheel significantly reduced the friction in transport, levitated motion is \"an old dream of mankind\" (Schweitzer et al, 1994). This quest has been fulfilled to some extent, and has also fueled some interesting developments in physics. In fact, the original levitated vehicle is the ship (Moon and Schweitzer, 1994), brought off the hard ground in ancient times to haul heavy building materials over long distances, as in the construction of the Great Pyramids and Stonehenge. Sketches of more recent critical events can be found in Schweitzer et al (1994) and Chang (1991), and are summarized here. Eamshaw had shown in 1842 that stable levitation could not be achieved in six degrees of freedom by means of permanent magnets alone. However, diamagnetic materials can be used to levitate objects under some conditions (Braunbek, 1939). In 1937, Kemper received a patent for an active hovering system.\nUsing such an idea, Beams (1937) and Holmes (1937) levitated tiny beads and spun them to speeds of 18 million rpm! The centrifugal stress at such speeds caused the beads to explode. By 1976, in West Germany, the KOMET, a levitated test vehicle with an active magnetic suspension had reached speeds of up to 401 km/h. Today, the TRANSRAPID 07 is ready for intercity service with a maximum speed of 450 km/h. Tieste and Popp (1995) used a magnetically levitated linear guide for increased precision in the positioning of machine tools and for the active control of the cutting process for avoiding chatter.\nA permanent magnet was first suspended above a superconductor in 1945 by Arkadiev. This technology is used in Japanese levitated train, which first flew by 1976 (Yamamura, 1976). Levitation also led to a new kind of bearing. Magnetic beatings have the desireable feature of no contact, meaning virtually no friction or wear. This property makes magnetic beatings very suitable for applications involving high speeds and energy or wear concerns. Active magnetic bearings have been used for high-speed milling operations, flywheels for energy storage, and clean-room activities (Schweitzer et al, 1994). Superconducting magnetic beatings have also been applied. Moon and Chang (1990) achieved rotational speeds of 120,000 rpm in a rotor suspended in this way.\nFRICTION AND MECHANICS\nThere are ancient indications of the recognition of the role of friction in mechanics. Written in the fourth century BC, the Mo Ching, with hints at the idea of dissipation, states the equivalent of Newton's first law of motion: \"The cessation of motion is due to the opposing force. If there is no opposing force, the motion will never stop. This is as true as that an ox is not a horse\" (Dai, 1983). The Mo Ching were stored and forgotten until recently.\nDownloaded From: http://appliedmechanicsreviews.asmedigitalcollection.asme.org/ on 10/19/2017 Terms of Use: http://www.asme.org/about-asme/terms-of-use"
        ]
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    {
        "image_filename": "designv10_0_0000637_978-3-642-83006-8-Figure2.47-1.png",
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        "caption": "Fig. 2.47. Compensating movements for the double-support gait upon level ground for different ZHP laws and dura tion of double-support phase p=20%",
        "texts": [],
        "surrounding_texts": [
            "for the single-support phase, while for the double-support gait phase the relations (2.5.23) and (2.5.25) hold. In the latter case, thecal culation of the right-hand sides of differential equations is executed in two steps. First, 6 equations for the prescribed moments of the first chain are separated from the set of equations (2.5.23). Then, from. these equations is ca.lculated the vector a (ground reaction). The ~l as s L in k [k g 1 1 2 1 0. 0 2 1. 53 3 0. 0 4 3. 21 5 S. 41 6 0. 0 7 0. 0 S 6. 96 9 0. 0 10 0. 0 11 S. 41 T a b le 2 .4 . K in e m a ti c a n d d y n am ic p a ra m e te rs o f th e m ec h an is m M om en t of i n er ti a [k gm ~ D is ta nc e of t he a xe s ce nt re s of j o in ts f ro m th e J x J y Jz 1i nk c en tr e [m l 3 4 5 6 0. 0 0. 0 0. 0 7 T r 1 ,1 = ( 0, 0, 0. 00 01 ) ; 7 T r 1 ~ = ( 0 , 0, -0 .0 00 1) ,t . 0. 00 00 6 0. 00 05 5 0. 00 04 5 -: t T 7 T r~ 2 = ( 0, 0, 0. 03 0) ; r 2 ,3 = ( 0, 0, -0 .0 70 ) L , 0. 0 0. 0 0. 0 7 T r 3 ,3 = ( 0, 0, 0. 00 01 ) ; -: t T r 3 ,4 = ( 0 , 0, -0 .0 00 1) 0. 00 39 3 0. 00 39 3 0' 00 03 S 7 T r 4 ,4 = ( 0, 0, 0. 21 0) ; -: t T r 4 ,5 = ( 0 , 0, -0 .2 10 ) 0. 01 12 0 0. 01 20 0 0. 00 30 0 f 5 ,5 = ( 0, 0, O. 2 20 ) T ; -: t T r 5 ,6 = ( 0 , 0, -0 .2 20 ) 0. 0 0. 0 0. 0 7 T r 6 ,6 = ( 0, 0, 0. 00 01 ) ; -: t T r 6 ,7 = ( 0 , 0, - 0. 00 01 ) 0. 0 0. 0 0. 0 7 T r 7 ,7 = ( 0, 0, 0. 00 01 ) ; -: t T r 7 ,S = ( 0, 0, -0 .0 00 1) 0. 00 70 0 0. 00 56 5 0. 00 62 5 -: t T rS ,S = ( 0, 0. 13 5, 0. 1) ; 7: T rS ,9 = ( 0 , -0 .1 35 , 0. 1) 7 T r S ,1 5 = ( 0 , 0, -0 .0 5) 0. 0 -: t T 7: T 0. 0 0. 0 r9 9 = ( 0, 0, -0 .0 00 1) ; r9 ,1 0 = ( 0 , 0, 0. 00 01 ) , 0. 0 0. 0 0. 0 -: t T r 1 0 ,1 0 = ( 0 , 0, - 0. 00 01 ) ; 7: T r 1 0, 11 = ( 0 , 0, 0. 00 01 ) 0. 01 12 0 0. 01 20 0 0. 00 30 0 f 1 1 ,1 1 = ( 0, 0 , -0 .2 20 )T ; f 1 1 ,1 2 = ( 0 , 0, 0. 22 0) T - - - _ L . . - - - - - - - - - - - - - - - - - - - - - - - J.o i n t un it a xe s 7 -: t T e 1 = (l ,0 ,0 ) -: t T e 2 = (0 ,1 ,0 ) 7: T e 3 = (l ,D ,0 ) 7 T e 4 = ( 0 , 1, 0 ) -: t T e 5 = ( 0, 1, 0 ) 7 T e 6 = (O ,1 ,D ) 7 T e 7 = (1 ,0 ,0 ) 7 T e S = ( 1 , 0, 0 ) -: t T e 9 = ( 1 , 0, 0) 7: T e 1 0 = ( 0 , 0, 1) 7: T e 1 1 = (0 ,- 1 ,0 ) ~ I\\ ) (J '1 1 2 3 4 5 12 3. 21 0. 00 39 3 0. 00 39 3 0. 00 03 8 13 0. 0 0. 0 0. 0 0. 0 14 1. 5 3 0. 00 00 6 0. 00 05 5 0. 00 04 5 15 0. 0 0. 0 0. 0 0. 0 16 30 .8 5 0. 15 14 0 0. 13 70 0 0. 02 83 0 17 2. 07 0. 00 20 0 0. 00 20 0 0. 00 02 2 18 1. 1 4 0. 00 25 0 0. 00 42 5 0. 00 01 4 19 2. 07 0. 00 20 0 0. 00 20 0 0. 00 02 2 20 1. 14 0. 00 25 0 0. 00 42 5 0. 00 01 4 T a b le 2 .4 . C o n t. 6 ~ T r 1 2 ,1 2 = ( 0, 0, -0 .2 10 ) ; ~ T r 1 3 ,1 3 = ( 0, 0, -C .0 00 1) \u00b7; ? T r 1 4 ,1 4 = ( 0, 0, -0 .0 70 ) ? T r 1 5 ,1 6 = ( 0, 0, 0 .0 0 0 1 ); ~ T r 1 6 ,1 6 = ( 0, 0, 0. 34 ) ; ~ T r 1 6 ,1 9 = ( 0, -0 .2 , -0 .0 6) ~ T r 1 7 ,1 7 = ( 0, 0, -0 .1 54 ) ; ? T r 1 8 ,1 8 = ( 0, 0, -0 .1 32 ) ? T r 1 9 ,1 9 = ( 0, 0, -0 .1 54 ) ; ~ T r 2 0 ,2 0 = ( 0, 0, -0 .1 32 ) ? T r 1 2 ,1 3 = ( 0, 0, 0. 21 0) ? T r 1 3 ,1 4 = ( 0 , 0, G .0 00 1) ? T r 1 5 , 16 = ( 0, 0, -0 .0 00 1) ~ T r 1 6 ,1 7 = ( 0 , 0 .2 , -0 .0 6) ; :: T r 1 7 ,1 8 = ( 0 , 0, 0. 15 4) ~ T r J 9 ,2 0 = ( 0 , 0, 0. 15 4) 7 ? T e 1 2 = (0 ,- 1 ,0 ) ~ T e 1 3 = (0 ,- 1 ,0 ) ~ T e 1 4 = ( 1 , 0, 0) ? T e 1 5 = ( 0 , 1, 0) ~ T e 1 6 = ( 1 ,0 ,0 ) :: T e 1 7 = ( 1 ,0 ,0 ) ? T e 1 8 = ( 1 , 0, 0) ct - T e 1 9 = (- 1 ,0 ,0 ) ~ T e 2 0 = (- 1 ,0 ,0 ) ~ '\" '\" 127 sagittal (b) plane 128 second step is to calculate the right-hand sides of the differential equations describing the system's equilibrium in both the sagittal and frontal plane. 129 This procedure enables numerical integration of the system in both phases of the gait. As in the previous examples, an iterative procedu re is to be employed for the synthesis of nominal dynamics. Compensating movements for different duration periods p, and ZI1P law (Fig. 2.23) of the double-support phase are presented in Fig. 2.46-2.47. Example 5. This example is related to the modelling of a two-link foot Fig. 2.19. The mechanism consists of 19 links and 19 revolute jOints (Fig. frontal (2.49a) and sagittal (2.49b) plane, respectively. The right foot is modelled by links 2 and 3, the left one by links 13 and 14. The arms are fi xed, i.e. q16=q18=1.862253 [radj and q17=q19=1.5707 [rad]. The conditi ons q8: = q8_q 4 and q9: = q9+q 4 ensure the pelvis (link 8) is in the horizontal position and the legs are parallel. in the frontal plane. Other input data are: N=3; 11=1,12=8,1 3 =9; K1 =14, K2 =11, K3 =11,NT=19; t;i=o, i=1, \u2022\u2022\u2022 ,19; t;~=1, t;~5=1; value t;~iS1fOriE:{4,13} and 0 for other values of index i; r01 =(O, 0, -O.0001)T; ~1=(1, 0, O)T; the initial 4 15\u00b74 \u00b715 values of the open part of the system (q (0), q (0), q (0) and q (0)); the transformation matrix W for repeatibility conditions is the same as in Example 2, i.e. (2.8.4). (a) (b) Fig. 2.49. Compensating d.o.f. in the frontal (a) and sagittal (b) plane T a b le 2 .5 . K in e m a ti c a n d d y n am ic p a ra m e te rs o f th e m e c h a n is m t1 as s M om en t of in er ti a [k gm 2 ] D is ta nc e of t he a xe s ce n tr es o f jo in ts f ro m th e L in k [k g] li n k c en tr e [m ] Jo in t u n it a xe s J J Jz x y 1 2 3 4 5 6 7 7 T 7 T T T 1 0. 0 0. 0 0. 0 0. 0 r 1 ,1 = ( 0 ,0 ,0 .0 0 0 1 ) ; r 1 ~ = ( 0 , 0, -0 .0 00 1) e 1 = (1 ,0 ,0 ) ,t :. 2 1. 5 3 0. 00 00 6 0. 00 05 5 0. 00 04 5 T T r 2 ,2 = ( 0, 0, 0. 03 0) ; T T r 2 ,3 = ( 0 , 0, -0 .0 70 ) \"t T e 2 = (0 ,1 ,0 ) \"t T T T :: T 3 0. 0 0. 0 0. 0 0. 0 r 3 ,3 = ( 0, 0, 0. 00 01 ) ; r 3 ,4 = ( 0 , 0, -0 .0 00 1) e 3 = ( 0 , 1 , 0) I 4 0. 0 0. 0 0. 0 0. 0 T T r 4 ,4 = ( 0, 0, 0. 00 01 ) ; \"t T r 4 ,5 = ( 0 , 0, -0 .0 00 1) T T e 4 = (1 ,0 ,0 ) I 5 3. 21 0. 00 39 3 0. 00 39 3 0. 00 03 8 T T r 5 ,5 = ( 0, 0, 0. 21 0) ; 7 T r 5 ,6 = ( 0 , 0, -0 .2 10 ) T T e 5 = (0 ,1 ,0 ) 6 8. 41 0. 01 12 0 0. 01 20 0 0. 00 30 0 T T r 6 ,6 = ( 0, 0, 0. 22 0) ; T T r 6 ,7 = ( 0 , 0, -0 .2 20 ) \"t T e 6 = ( 0 , 1 , 0) 7 0. 0 0. 0 0. 0 0. 0 T T r 7 ,7 = ( 0, 0, 0. 00 01 ) ; T T r 7 ,8 = ( 0 , 0, -0 .0 00 1) T T e 7 = (0 ,1 ,0 ) 8 6. 96 0. 00 70 0 0. 00 56 5 0. 00 62 7 7 T r 8 ,8 = ( 0, 0. 13 5, 0. 1) ; T T r 8 ,9 = ( 0 , -0 .1 35 , 0. 1) ; T T e 8 = ( 1 , 0, 0) T T r 8 ,1 5 = ( 0 , 0, -0 .0 5) 9 0. 0 0. 0 0. 0 0. 0 T T r 9 ,9 = ( 0 , 0, -0 .0 00 1) ; :: T r 9 ,1 0 = ( 0 , 0, 0. 00 01 ) T T e 9 = (1 ,0 ,0 ) 10 8. 41 0. 01 12 0 0. 01 20 0 0. 00 30 0 f 1 D ,1 0 = ( 0, 0, -0 .2 20 )T ; T T r 1 0 ,1 1 = ( 0 , 0, 0. 22 0) T T e 1 0 = ( 0 , -1 , 0) 11 3. 21 0. 00 39 3 0. 00 39 3 0. 00 03 8 ~ T T T T T r 1 1 ,1 1 = ( 0 , 0, -0 .2 10 ) ; r 1 1 ,1 2 = ( 0 , 0, 0. 21 0) e 1 1 = (0 ,- 1 ,0 ) ~ ~ 1 2 3 4 5 12 0. 0 0. 0 0. 0 0. 0 13 0. 0 0. 0 0. 0 0. 0 14 1. 5 3 0. 00 00 6 0. 00 05 5 0. 00 04 5 15 30 .8 5 0. 15 14 0 0. 13 70 0 0. 02 83 0 16 2. 07 0. 00 20 0 0. 00 20 0 0. 00 02 2 17 1. 1 4 0. 00 25 0 0. 00 42 5 0. 00 01 4 18 2. 07 0. 00 20 0 0. 00 20 0 0. 00 02 2 19 1. 1 4 0. 00 25 0 0. 00 42 5 0. 00 01 4 T a b le 2 .5 . C o n t. 6 7 T r 1 2 ,1 2 = ( 0, 0, -0 .0 00 1) ; 7 T r 1 2 ,1 3 = ( 0 , 0, 0. 00 01 ) 7 T r 1 3 ,1 3 = ( 0 , 0, -0 .0 00 1) ; 7 T r 1 3 ,1 4 = ( 0, 0, 0. 00 01 ) ? T r 1 4 ,1 4 = ( 0, 0, -0 .0 70 ) 7 T r 1 5 ,1 5 = ( 0, 0, 0. 34 ) ; ~ T r 1 5 ,1 6 = ( 0, 0 .2 , -0 .0 6) ? T r 1 5 ,1 8 = ( 0 , -0 .2 , -0 .0 6) 7 T r 1 6 ,1 6 = ( 0 , 0, -0 .1 54 ) ; ? T r 1 6 ,1 7 = ( 0, 0, 0. 15 4) 7 T r 1 7 ,1 7 = ( 0, 0, -0 .1 32 ) 7 T r 1 8 ,1 8 = ( 0 , 0, -0 .1 54 ) ; ? T r 1 8 ,1 9 = ( 0 , 0, 0. 15 4) 7 T r 1 9 ,1 9 = ( 0 , 0, -0 .1 32 ) 7 7 T e 1 2 = (0 ,- 1 ,0 ) 7 T e 1 3 = (1 ,0 ,0 ) zT e 1 4 = ( 0 , -1 , 0) 7 T e 1 5 = (0 ,1 ,0 ) ? T e 1 6 = (1 ,0 ,0 ) 7 T e 1 7 = ( 1 ,0 ,0 ) ? T e 1 8 = (- 1 ,0 ,0 ) I ? T e 1 9 = (- 1 ,0 ,0 ) I W I\\ J 8=0.6 and different ZMP laws 133 134"
        ]
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure9.81-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure9.81-1.png",
        "caption": "Figure 9.81 The zero-force members in the truss. (a) to (d) represent the order in which the zero-force members can be found.",
        "texts": [],
        "surrounding_texts": [
            "We can often shorten the calculation that needs to be done by first looking for zero-force members in a truss. Zero-force members are members in which no forces are acting (N = 0) due to the present loading. 364 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM There are three situations of frequent occurrence in which zero-force members can be easily recognised: 1. If only two members meet in an unloaded joint, both are zero-force members (see Figure 9.71). 2. If three members meet in an unloaded joint of which two are in a direct line with one another, then the third is a zero-force member (see Figure 9.72). 3. If two members meet in an unloaded joint and the line of action of the load coincides with one of the members, the other member is a zero-force member (see Figure 9.73). These three rules are the direct consequence of the joint equilibrium, as shown below for each of the cases. Rule 1. Two members meet in unloaded joint A in Figure 9.71. The force in one of the members has a component normal to the direction of the other member. If we write down the equilibrium of joint A in the given (local) xy coordinate system, we find \u2211 Fx = N1 + N2 cos \u03b1 = 0,\u2211 Fy = N2 sin \u03b1 = 0 with the solution (because sin \u03b1 = 0):1 N1 = N2 = 0. Equilibrium is possible only if both member forces are zero. 1 In a kinematically determinate truss, members 1 and 2 cannot be an extension of one another, so that \u03b1 = 0 and \u03b1 = 180\u25e6. 9 Trusses 365 Rule 2. In Figure 9.72, three members meet in joint B, of which members 1 and 3 are in a direct line with one another. The force in member 2 has a component normal to members 1 and 3. There can be equilibrium only if this component is zero, or in other words, if N2 = 0. If we write down the equilibrium of joint B in the given (local) xy coordinate system, we find \u2211 Fx = N1 + N2 cos \u03b1 \u2212 N3 = 0,\u2211 Fy = N2 sin \u03b1 = 0 so that N2 = 0 and N1 = N3. In addition to the fact that member 2 is a zero-force member, the normal forces in the continuous members 1 and 3, which are in a direct line with one another, are equal. Rule 3. The situation in Figure 9.73 is clearly similar to that in Figure 9.71. The equations for the equilibrium of joint C are \u2211 Fx = N1 + N2 cos \u03b1 \u2212 F = 0,\u2211 Fy = N2 sin \u03b1 = 0 so that N2 = 0 and N2 = F. By using rules 1 to 3 to determine the zero-force members first, you can often shorten the required calculation. A fourth rule with which we can shorten the calculation relates to an unloaded joint, in which four members meet and in pairs are in a direct line with one another. This situation is 366 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM shown in Figure 9.74. For the given xy coordinate system, the equilibrium of joint D gives \u2211 Fx = N1 + N2 cos \u03b1 \u2212 N3 \u2212 N4 cos \u03b1 = 0,\u2211 Fy = N2 sin \u03b1 \u2212 N4 sin \u03b1 = 0 so that N1 = N3 and N2 = N4. Conclusion: Rule 4. If four members meet in an unloaded joint that in pairs are in a direct line with one another, these members can be considered crossing members as far as the transfer of forces is concerned. The three examples below show how it is possible to simplify the calculation with these four rules. Example 1 You are given the truss in Figure 9.75. Question: Which members are zero-force members for the given load? Solution: A is an unloaded joint in which two members meet (see Figure 9.76). Both members are zero-force members (rule 1), so that N1 = 0 and N2 = 0. B is an unloaded joint in which three members meet, and of which two are in a direct line with one another. The third member is therefore a zero-force 9 Trusses 367 member (rule 2), so that N9 = 0. C is a loaded joint where two members meet, and where the line of action of the load coincides with member 17. Thus (rule 3) N16 = 0. The zero-force members are shown in Figure 9.76 with a \u201c0\u201d through the member axis. Zero-force members do not participate in the force flow for the present load. When calculating the forces in the other members, you can leave out the zero-force members from the truss. If you leave out zero-force member 9 from the truss, you immediately notice that N8 = N10. If you leave out zero-force member 16, you see that N17 = \u2212F. That this (imaginary) omission of zero-force members can significantly reduce the effort in calculating is further emphasised in the following two examples. Example 2 You are given the truss in Figure 9.77. Question: Determine all the zero-force members for the given load. 368 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Solution: From the equilibrium in joint B, it follows that member 3 is a zero-force member (rule 2). With the given load, this member does not participate in the transfer of forces, and could therefore be omitted. In Figure 9.78a, the member is now shown by means of a dashed line. The equilibrium of joint C means that member 5 is also a zero-force member (rule 2 again) (see Figure 9.78b). If we continue, we notice that members 7 and 9 are also zero-force members (see Figures 9.78c and 9.78d). Since the support reaction in G is horizontal, member 11 is also a zero-force member (rule 3) (see Figure 9.78e). All the verticals and diagonals are zero-force members. The load is therefore fully transferred by the bottom and top chord members. For the (continuous) top chord members we find N1 = N4 = N8 For the (continuous) bottom chord members we find N2 = N6 = N10. When we talk about omitting zero-force members, this is done only to simplify the calculation. If the zero-force members are removed from the truss in reality, the truss becomes kinematically indeterminate. Zero-force members therefore have a genuine function in the truss. On the one hand they ensure the truss retains its shape, while on the other they can prevent buckling (in the plane of the structure) of (long) compressed members, such as the bottom chord in Figure 9.77, or the top chord in Figure 9.79. 9 Trusses 369 Example 3 You are given the truss in Figure 9.80. The diagonals are crossing members. Question: Determine all the zero-force members for the given load. Solution: In this truss, it is not possible to find a section across three members (that do not intersect in one point), nor is there a joint with less than two unknowns (member forces or support reactions). We therefore cannot determine the member forces with the method of sections, or with the method of joints, unless we first determine the support reactions. For determining the zero-force members in the truss, it is enough to know that the support reaction at the point of the roller is vertical, so that N2 = 0. This means that N7 = 0, and so forth (see Figures 9.81a\u20139.81d). We subsequently discover that members 10, 12 and 13 are zero-force members. Determining the other member forces is now a relatively simple task. Note that the force flow does not change when the crossing diagonals are joined at the point where they cross (rule 4)."
        ]
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.1-1.png",
        "caption": "FIGURE 7.1. A front-wheel-steering vehicle and the Ackerman condition.",
        "texts": [
            " A coupler point of an inverted slider-crank mechanism if \u03b82 = 30deg. Consider an inverted slider-crank mechanism with the following parameters. a = 10 cm d = 45 cm e = 5 cm c = 10 cm \u03b1 = 30deg Determine the coordinates of the coupler point if \u03b82 = 30deg. 7 To maneuver a vehicle we need a steering mechanism to turn wheels. Steering dynamics which we review in this chapter, introduces new requirements and challenges. 7.1 Kinematic Steering Consider a front-wheel-steering 4WS vehicle that is turning to the left, as shown in Figure 7.1. When the vehicle is moving very slowly, there is a kinematic condition between the inner and outer wheels that allows them to turn slip-free. The condition is called the Ackerman condition and is expressed by cot \u03b4o \u2212 cot \u03b4i = w l (7.1) where, \u03b4i is the steer angle of the inner wheel, and \u03b4o is the steer angle of the outer wheel. The inner and outer wheels are defined based on the turning center O. The distance between the steer axes of the steerable wheels is called the track and is shown by w"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure3.47-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure3.47-1.png",
        "caption": "FIGURE 3.47. A front view of a cambered tire and the generated camber force.",
        "texts": [
            "24 \u00b4 dy dx = 0.0144\u03c4yM . (3.150) If we calculate the lateral force Fy = 1000N by measuring the lateral acceleration, then the maximum lateral stress is \u03c4yM = Fz 0.014 4 = 69444Pa (3.151) and the lateral stress distribution over the tireprint is \u03c4y(x, y) = 69444 \u00b3 1\u2212 x 0.05 \u00b4\u00b5 1\u2212 x3 0.053 \u00b6 cos2 \u00b3 y\u03c0 0.24 \u00b4 Pa. (3.152) 3. Tire Dynamics 145 3.8 Camber Force Camber angle \u03b3 is the tilting angle of tire about the longitudinal x-axis. Camber angle generates a lateral force Fy called camber trust or camber force. Figure 3.47 illustrates a front view of a cambered tire and the generated camber force Fy. Camber angle is assumed positive \u03b3 > 0, when it is in the positive direction of the x-axis, measured from the z-axis to the tire. A positive camber angle generates a camber force along the \u2212y-axis. The camber force is proportional to \u03b3 at low camber angles, and depends directly on the wheel load Fz. Therefore, Fy = Fy j\u0302 (3.153) Fy = \u2212C\u03b3 \u03b3 (3.154) where C\u03b3 is called the camber stiffness of tire. C\u03b3 = lim \u03b3\u21920 \u2202 (\u2212Fy) \u2202\u03b3 (3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure6.21-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure6.21-1.png",
        "caption": "FIGURE 6.21. Kinematic model of an inverted slider-crank mechanism.",
        "texts": [
            " Hence, the inverted slider-crank is an inversion of a slider-crank mechanism. 340 6. Applied Mechanisms The angular position of the output slider \u03b84 and the length of the coupler link b are functions of the lengths of the links and the value of the input variable \u03b82. These variables are: b = \u00b1 p a2 + d2 \u2212 e2 \u2212 2ad cos \u03b82 (6.172) \u03b84 = \u03b83 + \u03c0 2 = 2 tan\u22121 \u00c3 \u2212H \u00b1 \u221a H2 \u2212 4GI 2G ! (6.173) where G = d\u2212 e\u2212 a cos \u03b82 (6.174) H = 2a sin \u03b82 (6.175) I = a cos \u03b82 \u2212 d\u2212 e. (6.176) Proof. We show the inverted slider-crank mechanism by a vector loop as shown in Figure 6.21. The direction of each vector is arbitrary, however, the angles should be associated to the vector\u2019s direction and be measured with positive direction of the x-axis. The links and their expression vectors are shown in Table 6.6. The vector loop is Gr1 + Gr2 \u2212 Gr3 \u2212 Gr4 = 0 (6.177) 6. Applied Mechanisms 341 which can be decomposed into sin and cos components. a sin \u03b82 \u2212 b sin \u00b3 \u03b84 \u2212 \u03c0 2 \u00b4 \u2212 e sin \u03b84 = 0 (6.178) \u2212d+ a cos \u03b82 \u2212 b sin \u00b3 \u03b84 \u2212 \u03c0 2 \u00b4 \u2212 e cos \u03b84 = 0 (6.179) To derive the relationship between the input angle \u03b82 and the output \u03b84, we eliminate b between Equations (6",
            "199) Assuming \u03b82 and \u03c92 are given values, and b, \u03b84 are known from Equations (6.172) and (6.173), we may solve Equations (6.197) and (6.198) for b\u0307 and \u03c94. b\u0307 = a b \u03c92 [b cos (\u03b84 \u2212 \u03b82)\u2212 e sin (\u03b84 \u2212 \u03b82)] (6.200) \u03c94 = \u03c93 = a b \u03c92 sin (\u03b82 \u2212 \u03b84) (6.201) 344 6. Applied Mechanisms Example 240 Velocity of moving joints for an inverted slider-crank mechanism. Having the coordinates \u03b82, \u03b84, b and velocities \u03c92, \u03c94, b\u0307 enables us to calculate the absolute and relative velocities of points A and B shown in Figure 6.21. The absolute and relative velocities of points A and B are GvA = G\u03c92 \u00d7 Gr2 = \u23a1\u23a3 0 0 \u03c92 \u23a4\u23a6\u00d7 \u23a1\u23a3 a cos \u03b82 a sin \u03b82 0 \u23a4\u23a6 = \u23a1\u23a3 \u2212a\u03c92 sin \u03b82a\u03c92 cos \u03b82 0 \u23a4\u23a6 (6.202) GvB4 = G\u03c94 \u00d7 Gr4 = \u23a1\u23a3 0 0 \u03c94 \u23a4\u23a6\u00d7 \u23a1\u23a3 e cos \u03b84 e sin \u03b84 0 \u23a4\u23a6 = \u23a1\u23a3 \u2212e\u03c94 sin \u03b84e\u03c94 cos \u03b84 0 \u23a4\u23a6 (6.203) GvB3/A = G\u03c93 \u00d7 \u00a1 \u2212Gr3 \u00a2 = \u23a1\u23a3 0 0 \u03c94 \u23a4\u23a6\u00d7 \u23a1\u23a3 \u2212b cos \u03b84\u2212b sin \u03b84 0 \u23a4\u23a6 = \u23a1\u23a3 b\u03c94 sin \u03b84 \u2212b\u03c94 cos \u03b84 0 \u23a4\u23a6 (6.204) GvB3 = GvB3/A + GvA = \u23a1\u23a3 b\u03c94 sin \u03b84 \u2212b\u03c94 cos \u03b84 0 \u23a4\u23a6+ \u23a1\u23a3 \u2212a\u03c92 sin \u03b82a\u03c92 cos \u03b82 0 \u23a4\u23a6 = \u23a1\u23a3 b\u03c94 sin \u03b84 \u2212 a\u03c92 sin \u03b82 a\u03c92 cos \u03b82 \u2212 b\u03c94 cos \u03b84 0 \u23a4\u23a6 (6"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000013_j.actamat.2016.07.019-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000013_j.actamat.2016.07.019-Figure2-1.png",
        "caption": "Fig. 2. Schematics of an EBM machine (from Murr et al. [30]), 1: electron gun, 2: lens system, 3: deflection lens, 4: powder cassettes with feedstock, 5: rake, 6: building component, 7: build table.",
        "texts": [
            " Nitrogen or argon is fed into the chamber to avoid undesired interactions of the metal powder with its environment and to protect the melt. Furthermore, secondary products of the process such as weld fume and weld spatter are removed by the inert gas flow around the work area [29]. In Electron BeamMelting (EBM) a powder bed is created similar to the LBM process. Therefore, metal powder is fed from a hopper and distributed by a rake across a build plate with a size of typically 200 mm 200mm in x-y-direction or 350mm in diameter [30], cf. Fig. 2. Usually, the powder layer thickness amounts to 50 mme200 mm [31]. Instead of a laser beam an electron beam functions as heat source to melt the powder as prescribed by the 3D CAD data. The electron beam is generated in an electron gun (Figs. 2 and 1) before it is accelerated with an acceleration voltage of 60 kV, focused by electromagnetic lenses (Fig. 2) and directed by a magnetic scan coil (Figs. 2 and 3) to the desired positions in the x-y plane on the build plate (Figs. 2 and 7) [32]. The power, focus and scan speed of the electron beam are generally determined by the choice of beam current, focus offset and speed function respectively [33]. At first, the powder bed is pre-heated by a defocused beamwhich scans the powder bed surface several times [34]. Using a high beam current of up to 30mA and a scan speed of about 104mm/s, temperatures of >700 C of the powder material are achieved for Ti-6Al-4V [8,35,36], while for e"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure2.10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure2.10-1.png",
        "caption": "Fig. 2.10. Coreless winding of a three-phase, eight-pole AFPM machine with twin external rotor.",
        "texts": [
            " Coreless stator windings are used in twin-rotor double-sided AFPM machines (Fig. 1.4d). For the ease of construction, the stator winding normally consists of a number of single layer trapezoidally shaped coils. The assembly of the stator is made possible by bending the ends of the coils by a certain angle, so that the active conductors lie evenly in the same plane and the end windings 2.2 Windings 37 38 2 Principles of AFPM Machines nest closely together. The windings are held together in position by using a composite material of epoxy resin and hardener. Fig. 2.10 shows the coreless stator winding of a three-phase, eight-pole AFPM machine. Obviously, the relations used in the slotted stator windings can be directly used for coreless trapezoidal stator winding with the exception that the term \u201cslot\u201d is replaced by the \u201ccoil side\u201d. Another coil profile that has been used in coreless stator AFPM machines is the rhomboidal coil. It has shorter end connections than the trapezoidal coils. The inclined arrangement of the coil\u2019s active sides makes it possible to place water cooling ducts inside the stator"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-FigureD.2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-FigureD.2-1.png",
        "caption": "FIGURE D.2 Two orthogonal wires.",
        "texts": [
            "6) where R1,2 = \u221a (x1 \u2212 x2)2 + (y1 \u2212 y2)2 + (z1 \u2212 z2)2 (D.7a) and \ud835\udcc11 = xe1 \u2212 xs1, \ud835\udcc12 = xe2 \u2212 xs2. (D.7b) Then, the closed form of this partial integral is Ip12 = 1 \ud835\udcc11\ud835\udcc12 rq 4\u2211 k=1 (\u22121)k+1 \u221a a2 k + (y2 \u2212 y1)2 + (z2 \u2212 z1)2 (D.8) with a1 = xe2 \u2212 xs1, a2 = xs2 \u2212 xs1 (D.9a) a3 = xs2 \u2212 xe1, a4 = xe2 \u2212 xe1. (D.9b) We treat these integrals in the same symmetrical way we do the other partial elements. D.1.3 1\u2215R3 Integral Ip12 for Two Orthogonal Filaments Unlike partial inductances, the potentials are coupled for the orthogonal conductors shown in Fig. D.2. The same is true for other coupling elements. 412 COMPUTATION OF PARTIAL COEFFICIENTS OF POTENTIAL The kernel for this case is of the 1\u2215R3 form and the integral to be solved is Ip12 = 1 \ud835\udcc11\ud835\udcc12 \u222b xe1 xs1 \u222b ye2 ys2 1 R3 1,2 dx2 dy1, (D.10) where R1,2 = \u221a (x1 \u2212 x2)2 + (y1 \u2212 y2)2 + (z1 \u2212 z2)2 (D.11a) and \ud835\udcc11 = xe1 \u2212 xs1, \ud835\udcc12 = ye2 \u2212 ys2. (D.11b) Then, the closed form of this partial integral is Ip12 = 1 4\ud835\udf0b\ud835\udf16 1 \ud835\udcc11\ud835\udcc12 (Ie \u2212 Is) (D.12) with Ii = |xe1 \u2212 x2| 2(xe1 \u2212 x2) Z [ sin\u22121 ( (a2 i \u2212 Z2) be \u2212 2 a2 i Z2 be|ai + Z| ) \u2212 sin\u22121 ( (a2 i \u2212 Z2) bs \u2212 2 a2 i Z2 bs |ai + Z| )] , (D"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001305_dnx004-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001305_dnx004-Figure4-1.png",
        "caption": "Fig. 4. Average absolute error as a function of the tuning parameters for a predefined-stable system with noisy measurements.",
        "texts": [
            "2) was simulated for several values of Tc and mq and with a Gaussian-distributed noise v. This noise was zero-mean and with a standard deviation of 0.1. The simulations were carried out through Euler\u2019s method with a sampling period of 0.01 time units and from the initial condition x(0) = 0. In each case, the average absolute error from t = 0 to t = 400, given by 1 400 \u222b 400 0 |x(t)| dt was calculated. The results for this average absolute error as a function of the tuning parameters are depicted in Fig. 4. As it was expected, the effect of the noise grows in magnitude as Tc decreases, since smaller values of Tc impose a faster and steeper convergence. The effect of mq, however, was not so clear since the several occurrences of this expression in the differential equation of the system have different influences. In particular, as mq increases, the factors 1 mq and 1 \u2016x+v\u2016mq decrease but exp \u2016x + v\u2016mq increases. The evidence in Fig. 4 suggests that this opposing effects come to the best balance when mq is near the middle of the interval (0, 1]. A basic problem in the design of feedback control systems is the stabilization and tracking in the presence of uncertainty caused by plant parameters variation and external perturbations. In order to deal with this problem, several approaches have been proposed. Most of them are based on Lyapunov stability theory and variable structure systems with SMs. The SM techniques are based on the idea of the sliding manifold, that is, an integral manifold with finite reaching time (Drakunov & Utkin, 1992), and have been widely used for the problems of control and observation of dynamical systems due to their characteristics of finite time convergence as well as robustness and insensitivity to uncertainties due to external bounded disturbances and parameters variation (Utkin, 1992; Utkin et al"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure4.34-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure4.34-1.png",
        "caption": "Figure 4.34 Examples of three-hinged frames.",
        "texts": [
            " Frames are planar, bent beams structures that are loaded in the plane of the structure. Such structures are often used to cover a space (warehouse, sports arena, and so forth). Figure 4.32 shows a number of simple examples of frames. In Figure 4.33, both fixed supports have been replaced by hinged supports, so that the structure is now referred to as a two-hinged frame. If the structure with hinged supports itself consists of two parts joined by a hinge, this is referred to as a three-hinged frame (see Figure 4.34). If the beam structure is not bent but arched, then the structure in Figure 4.35a is called a two-hinged arch, and the structure in Figure 4.35b a three-hinged arch. 130 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM It will be clear that a wide range of planar structures can be constructed using line elements. Two types of structure not mentioned in the earlier categories are shown here. The structure in Figure 4.36 is called a shored frame. The structures in Figure 4.37 go by the name of trussed beams"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000025_00207179308923053-Figure9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000025_00207179308923053-Figure9-1.png",
        "caption": "Figure 9.",
        "texts": [
            " Considering now the existence of the sliding mode on the manifold 0 - g( a) = 0 we have ~ (0 - g(a\u00bb = ij - g~(a)o = LuLua - g~g(a) + ~Lua, U dt au = c - g~(a)g(a) - Kasign(o - g(a\u00bb 1260 A. Levant The last term must dominate the first two terms but it can do so only when g;,(a)g(a) is bounded. That puts a restriction on g(a). Consider now the algorithm (16), (17) and (18). In this case (j = C + KUl - ;'pK _a_'_ lal 1 - p The last term here involves \u00b0and a gain which is unbounded when a-+ O. The real trajectory lies inside the set of the solution points of the associated inclusion equation (with lui < 1) (j E -[Kmtl\" - Co, KMtl\" + Co]signa - ;'p[Km, KM]-_o lal 1 - p Figure 9 depicts this case. I! follows from the accurate estimations that for p = 1/2 the sequence {OJ} of the intersection points with the axis a = 0 satisfies 10j+I!0;1,,;;; q < 1 which implies 10;1-+ 0 as i -+ co. For 0 < P < 1/2 the convergence to the origin is even faster. Also, the sum of the encircling time sequence is estimated by a geometric series. 6. Mathematical modelling We consider now the following example Xl = -5xj + 10X2 + 4X3 + Xl sin t + (u 2 - 1)(Xl - X2) X2 = 6XI - 3X2 - 2X3 + 3(xj + X2 + X3)COSt X3 = Xl + 3X3 + 4x2cos5t + 4sin5t + 10(1 + 0\u00b75cos IOt)JL(u) <P(x) where JL(U) = 3u - cos30tsinu - u2/4 } <P(x) = [xi + x~ + x~ + 1]1(2 (24) (25) Order and accuracy in sliding mode control 1261 Let a = X3/<P(X) be the constraint function"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000637_978-3-642-83006-8-Figure2.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000637_978-3-642-83006-8-Figure2.1-1.png",
        "caption": "Fig. 2.1. Joint coordinate of a revolute kinematic pair {C i - 1 , Ci }",
        "texts": [
            " ,s}, where s=6-j is the number of degrees of freedom, (d.o.f.) and j is the class of pair P ik . The definition of joint coordinates may be simplified if we consider fifth-class pairs only, i.e., pairs with a single d.o.f. of relative motion. This assumption does not reduce the generality of considerati on, since a pair of any class may be represented by superposition of virtual fifth-class pairs. It is therefore particularly significant to introduce, for such pairs a unique manner of defining the joint coor dinates. Fig. 2.1. shows a revolute kinematic pair whose joint angle i (coordinate) is denoted by q . Scalar quantities xek ' kE{1, ..\u2022 ,m}, which determine the position and (partially or completely) orientation of the n-th link of chain An with respect to a reference coordinate frame are said to be the externaZ (Cartesian)coordinates of the mechanism. An active mechanism represents a system comprising: a) a mechanical part which may be modelled by an appropriate kinematic chain, and b) a set of actuators by means of which driving torques (forces) are reali zed"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001394_tro.2010.2043757-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001394_tro.2010.2043757-Figure4-1.png",
        "caption": "Fig. 4. Virtual-linkage model for humanoid robots. We define a virtual-linkage model anchored at the contact CoP\u2019s. It enables the characterization of internal tensions and moments against contact surfaces. The virtual-linkage model abstracts the behavior of internal and CoM forces with respect to reaction forces. These characteristics make the virtual-linkage model a powerful tool for the analysis and efficient control of CoM maneuvers and contact interactions.",
        "texts": [
            " Note that placing contact CoP\u2019s at the centers of the links is not a necessary constraint. They can be placed at any position below the links in contact, but away from contact vertices and edges, to prevent rotations. Therefore, in this paper, we only consider flat-surface-contact interactions. Extensions to corner and edge contacts could also be explored using a similar methodology; however, we will leave this analysis open for the future. We associate a virtual-linkage model connecting all contact CoP\u2019s. In the scenario, which is shown in Fig. 4, each body in contact introduces a tension with respect to its neighboring nodes as well as normal and tangential moments. For contacts with ns > 2, we can independently specify 3(ns \u2212 2) tensions, ns normal moments, and 2ns tangential moments describing the CoP\u2019s. Any additional link in contact introduces three new tensions with respect to its neighbors and three more moments. No more than three tensions between neighboring nodes can be independently specified for a given contact. Internal tensions and normal moments characterize the behavior of contact bodies with respect to the friction cones and rotational friction properties of the surfaces in contact",
            " The control of the right hand to interact with the side panel is implemented using the force-control methods, as described in [22]. Therefore, a unified force/position operational task is created to control the right hand. Postural behavior is controlled by optimizing a criterion that minimizes the distance with respect to a human prerecorded posture using a method similar to the one described in [7]. During the multicontact phases (i.e., three or more supporting contacts) virtual-linkage models similar to that shown in Fig. 4 are implemented. During the biped phase, the special case of virtual-linkage model, as discussed in Fig. 11 in the Appendix, is used. Details on the algebraic expressions of the matrices involved in the virtual-linkage models are also given in the Appendix. The first two rows of the accompanying data correspond to the phase with three supporting contacts, where the right hand is controlled to clean the panel. In this phase, the horizontal position of the CoM is commanded to remain fixed. We display data on the tension forces occurring on the virtual-linkage model assigned between the left hand and the right foot, the moments about the CoP on the left hand, the normal force on the panel, and the sagittal and vertical trajectories of the CoM",
            " After applying the above spatial transformations, we select the tangential moments with respect to surface tangential coordinates using the matrix Scop =  000100 000010 [0]2\u00d76 \u00b7 \u00b7 \u00b7 [0]2\u00d76 . . . ... ... \u00b7 \u00b7 \u00b7 000100 000010  \u03b5R2ns \u00d76ns . (87) Next, we specify the matrices involved in the representation of the 3(ns \u2212 2) vector of tension forces. Based on (10), we first take the 9(ns \u2212 2) differences between pairs of linear forces. The following matrix shows an instance of a differential matrix for the virtual-linkage model of Fig. 4, where there exists four contact links and six tension forces: \u2206t =  I \u2212I 0 0 \u2223\u2223 0 0 0 0 I 0 \u2212I 0 \u2223\u2223 0 0 0 0 I 0 0 \u2212I \u2223\u2223 0 0 0 0 0 I \u2212I 0 \u2223\u2223 0 0 0 0 0 I 0 \u2212I \u2223\u2223 0 0 0 0 0 0 I \u2212I \u2223\u2223 0 0 0 0  \u03b5 R9(ns \u22122)\u00d76ns (88) where I is the identity matrix; in this case, it is of dimension 3 \u00d7 3. The first row corresponds to the differential of forces between the right hand and the left hand, the second row corresponds to the differential between the right hand and the right foot, the third row corresponds to the differential between the right hand and the left foot, etc., until we express the six tension differentials. The rotation matrix Rt takes differential forces expressed in global frame and rotates them along the links of the virtuallinkage model using the following expression for the case of Fig. 4: Rt =  R12 0 0 0 0 0 0 R13 0 0 0 0 0 0 R14 0 0 0 0 0 0 R23 0 0 0 0 0 0 R24 0 0 0 0 0 0 R34  (89) \u03b5R9(ns \u22122)\u00d79(ns \u22122) Rij =  x\u0302 T ij y\u0302 T ij z\u0302 T ij   x\u0302ij = Pi \u2212Pj \u2016Pi \u2212Pj \u2016 y\u0302ij = (\u2212x\u0302ij (2) x\u0302ij (1) 0 )T z\u0302ij = x\u0302ij \u00d7 y\u0302ij . (90) which result on the x-component aligned with the links. To obtain the tension forces, all is needed is to select the x-directions using the selection matrix St =  1 0 0 [0]1\u00d73 [0]1\u00d73 \u00b7 \u00b7 \u00b7 [0]1\u00d73 1 0 0 [0]1\u00d73 ... . . . [0]1\u00d73 \u00b7 \u00b7 \u00b7 [0]1\u00d73 1 0 0  \u03b5R3(ns \u22122)\u00d79(ns \u22122) . (91) Finally, we specify the matrices involved in the expression of normal moments of (11). We note that it involves a spatial transformation from global frame to surface local frames using the previous definition of Ts . However, this time we choose the normal-moment components using the following selection matrix also specify for the particular case of Fig. 4: Sn =  0 \u00b7 \u00b7 \u00b7 01 [0]1\u00d76 [0]1\u00d76 [0]1\u00d76 [0]1\u00d76 0 \u00b7 \u00b7 \u00b7 01 [0]1\u00d76 [0]1\u00d76 [0]1\u00d76 [0]1\u00d76 0 \u00b7 \u00b7 \u00b7 01 [0]1\u00d76 [0]1\u00d76 [0]1\u00d76 [0]1\u00d76 0 \u00b7 \u00b7 \u00b7 01  \u03b5Rns \u00d76ns . (92) The authors would like to thank T. Yoshikawa for his contributions and R. Philippsen and P. Fraisse for reviewing this paper. [1] C. O. Alford and S. M. Belyeu, \u201cCoordinated control of two robot arms,\u201d in Proc. IEEE Int. Conf. Robot. Autom., Mar. 1984, pp. 468\u2013473. [2] T. Bretl and S. Lall, \u201cTesting static equilibrium for legged robots,\u201d IEEE Trans"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure5.34-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure5.34-1.png",
        "caption": "FIGURE 5.34 Single loop over a ground plane and connected to it in a point.",
        "texts": [
            " Both loops have a length l = 5 cm and a width w = 3 cm. We assume that both copper loops have a cross-section of 0.1 mm \u00d7 0.1mm. Assume that a current source of 10 mA is applied between C and D. Make a inductance-resistance PEEC Matlab program to compute the voltage between the terminals A and B as a function of frequency. PROBLEMS 129 5.7 Approximate inductance of a loop over a ground plane Here we make the assumption that the inductance will be reduced if we place the loop over a lossless ground plane as shown in Fig. 5.34. We assume that all the loop dimensions are the same as the ones in the previous problem. We assume that the distance to the center of the wire from the ground plane is 1 mm. Assume that the influence of the ground plane can be taken into account with a closed loop at a distance of h = 2 mm. Compute the new inductance for the two-loop situation. Observe that the equivalent circuit is a two-loop transformer where the secondary loop is a short circuit. 5.8 Impact of lossy loop over lossy ground plane on impedance We use a different approximation in this example for the same loop dimensions and ground plane distances"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003860_ma034921g-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003860_ma034921g-Figure1-1.png",
        "caption": "Figure 1. Schematic of the clamping system used in the DMA setup.",
        "texts": [],
        "surrounding_texts": [
            "Materials and Techniques. 4-Hydroxybutyl acrylate inhibited with hydroquinone, 4,4\u2032-methylenebis(phenyl isocyanate) (MDI), and anhydrous toluene were obtained from Aldrich. Single-wall carbon nanotubes were purchased from Carbon Nanotechnologies, Inc. (Houston, TX). Prior to polymerization, hydroquinone was removed from 4-hydroxybutyl acrylate using an inhibitor removal column (Column DHR-4 from Scientific Polymer Products, Inc.). Analytical TLC was conducted on Whatman precoated silica gel 60-F254 plates. Molecular weight was determined by GPC using a Shimadzu LC-10A liquid chromatograph equipped with Plgel 5 \u00b5m Mixed-D column. THF was used as as the mobile phase and polystyrene as standard. GC/FID sample analysis was carried out on a HP 6890 instrument equipped with a 30 m DB-5 column, with an oven program of 50 \u00b0C (1 min) to 300 \u00b0C (10 min) at a rate of 8 \u00b0C/min. 1H NMR spectra were recorded on a Bruker DRX-400 spectrometer. All spectra were run in CDCl3 solution. Thermal Analysis. A Perkin-Elmer differential scanning calorimeter DSC 7 equipped with a CCA 7 liquid nitrogencooling accessory was used to study the nematic-isotropic phase transition of the monomers and their corresponding polymers. Scans were done at 2 \u00b0C/min heating and cooling rate over a temperature range of 0-110 \u00b0C. The melting temperature of indium was used as a standard for temperature calibration, which agreed to within 1 deg of the expected value. Mechanical Studies. All measurements were done on a TA Instrument DMA 2980 with tension clamp for thermoelastic and stress/strain measurements. The clamp assembly consists of a fixed upper clamp and a mobile lower clamp activated by an air bearing under pressure. The fiber was positioned between the two clamps and held under a preload of 0.001 N. The force was increased at the rate of 0.001 N/min, and the corresponding stress-strain values were measured. Thermoelastic experiments follow the length change as a function of temperature. The % strain vs temperature data were collected while the temperature was ramped up (heating cycle) and down (cooling cycle). Isostrain experiments involve heating the fiber while holding it under a predetermined strain. The stress values are measured while the fiber is taken through the nematic-isotropic transition at a constant of 0.5 \u00b0C/min. Synthesis. The syntheses of the monomers MAOC4 and MACC5 were described elsewhere.19 But here we discuss the synthetic steps used to prepare the terpolymer and the subsequent cross-linking using MDI material. Polymerization of (4\u2032\u2032-Acryloxybutyl) 2,5-Di(4\u2032-butyloxybenzoyloxy)benzoate (MAOC4) and (4\u2032\u2032-Acryloxyloxybutyl) 2,5- Di(4\u2032-pentylcyclohexylcarboxyloxy)benzoate (MACC5) with 4-Hydroxybutyl Acrylate. A mixture of MAOC4 (0.25 g, 0.4 mmol), MACC5 (0.38 g, 0.6 mmol), 4-hydroxybutyl acrylate (0.014 g, 0.10 mmol, 10%), and 1.56 \u00d7 10-3 g (0.0094 mmol, 1%) of azobis(isobutyronitrile) (AIBN) were dissolved in 8 mL of toluene. The mixture was purged with nitrogen for 30 min to remove oxygen from the solution. The flask was immersed in an oil bath preset at temperature of 65 \u00b0C. After heating overnight, the solution was poured in methanol. The polymer was purified by successive reprecipitation from THF solutions into methanol until TLC analysis shows no traces of unreacted monomers. The polymer was dried in a vacuum oven for 24 h to yield 0.5 g of white waxy material. 1H NMR (CDCl3): 0.8- 2.6 (m, 77H aliph), 3.42-4.1 (m, 16H, 8CH2-O), 6.84-8.0 (m, 14H, ArH). The signal from olefinic hydrogens, which indicate the presence of the acrylic group, had completely disappeared from the NMR spectrum. The molecular mass determined from GPC (THF, polystyrene standard) was found to be Mw ) 30 000 mol/g and Mn ) 19 200 mol/g. The density of the hydroxybutyl group present in the polymer was indirectly determined by using the GC technique. A small amount of the mixture before and after polymerization was taken and injected into a GC instrument. The mixture consists of MAOC4, MACC5, hydroxybutyl acrylate, toluene, and cyclohexanone (reference standard). Comparison of the area under the peaks for cyclohexanone and hydroxybutyl acrylate compounds before and after polymerization shows that all the cross-linker has been consumed. Preparation of Liquid Crystal Elastomer (LCE) Fibers. Fibers were drawn from a melt mixture of the polymer and MDI cross-linker. In a typical experiment, a small amount of the polymer (0.04 g) is heated to 80 \u00b0C in a microscope slide placed on a thermal stage (Figure 2). A known amount of MDI (0.25/1 mole ratio of MDI relative to the hydroxyl unit present in the polymer backbone) was added at this temperature. Using a concentration higher than 0.25 mol of MDI led to a brittle and inhomogeneous fiber. The temperature was then dropped to around 60 \u00b0C, and the sample was mixed well at this temperature (about a minute) until the mixture appears homogeneous. At this point, the mixture becomes viscous, indicating that the cross-linking has started to occur.The fibers were drawn by dipping the tip of a metallic tweezer and pulling the mixture with it as quickly as possible as shown schematically in Figure 2. The fibers were left at room temperature for about 120 h, time over which the cross-linking reaction is completed. The average diameter of the fiber drawn this way was about 300 \u00b5m. Carbon Nanotubes Doped Fiber. A solution of singlewalled nanotubes (5 mg) in dichloromethane, terpolymer (30 mg), and MDI (0.62 mg) was stirred for 2 h at room temperature. The solvent was evaporated under vacuum, and the fiber was drawn from a melt at 60 \u00b0C as described above."
        ]
    },
    {
        "image_filename": "designv10_0_0001189_j.ijmachtools.2005.11.005-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001189_j.ijmachtools.2005.11.005-Figure1-1.png",
        "caption": "Fig. 1. Model geometry for the laser-assisted machining of a cylindrical workpiece [10].",
        "texts": [
            " The objectives for this study on LAM are to (i) apply a thermal model for LAM to accurately predict and characterize the thermal fields of Inconel 718 undergoing LAM, (ii) experimentally verify such predictions using surface temperature measurements for a wide range of operating parameters, (iii) utilize this validated model to discern the underlying physical phenomena, which include machinability, material removal mechanism, tool wear mechanism, and tool wear rate, and (iv) find optimized operating ranges of process parameters for LAM of Inconel 718. A transient, three-dimensional thermo-mechanical model of a rotating workpiece undergoing laser heating and material removal was first developed by Rozzi et al. [10\u201312] for an opaque homogenous ceramic workpiece. The schematics of workpiece geometry with the complete heat transfer problem is shown in Fig. 1. Assuming isotropic thermal conductivity, the heat equation governing this phenomenon is given by ro qh qf \u00fe rV z qh qz|fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl} advection \u00fe r qh qt|ffl{zffl} storage \u00bc 1 r q qz rk qT qr \u00fe 1 r2 q qf k qT qf \u00fe q qz k qT qz |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} conduction \u00fe q000|{z} generation , \u00f01\u00de where r is the density in kg/m3, o the rotational speed in rad/s, h the enthalpy in J/kg, k the thermal conductivity in W/mK, and r, f, and z are cylindrical coordinates in m"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003438_s0921-5093(01)01179-0-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003438_s0921-5093(01)01179-0-Figure2-1.png",
        "caption": "Fig. 2. Deposition patterns for beam substrates.",
        "texts": [
            " The effect of creep was not included in the model since the material was at a high temperature for a relatively short time [16,17]. These assumptions were made to simplify the model and reduce the computation time. While these simplifications may introduce errors, all calculated results were compared to experimental values to confirm the finite element results. This FEM was used to investigate two different deposition patterns on a beam substrate, a long raster pattern and a short raster pattern (Fig. 2). Figs. 3 and 4 show the FEM calculated principal stresses in the x and y directions on the top surface of the beam after the first line has been remelted and cooled and after the last line has been remelted and cooled. Only 1/4 of the beam is shown due to the two symmetry planes used in the analysis. Two observations may be made from these stress plots. First, the highest principal stresses were found in the direction of the long axis of the deposition line. This is demonstrated in Fig. 3a and Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-FigureD.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-FigureD.1-1.png",
        "caption": "FIGURE D.1 Partial potential for two filament wires.",
        "texts": [
            " For this reason, we also attempt to compute the partial elements with sufficient accuracy for large aspect ratio cells. A mixed, analytic, and numerical approach we call multifunction is presented in Appendix E. D.1 PARTIAL POTENTIAL COEFFICIENTS FOR ORTHOGONAL GEOMETRIES D.1.1 Pp12 for Two Parallel Wires One of the simplest coefficients is the wire-to-wire (filament-to-filament) partial potential, which we call Pp12. Of course, this result is singular if the wires touch. The integrals for the wire-to-wire partial potential Pp12 for the geometry shown in Fig. D.1 is Pp12 = 1 4\ud835\udf0b\ud835\udf16 1 \ud835\udcc11\ud835\udcc12 \u222b xe1 xs1 \u222b xe2 xs2 1 R1,2 dx2 dx1, (D.2) where R1,2 = \u221a (x1 \u2212 x2)2 + (y1 \u2212 y2)2 + (z1 \u2212 z2)2 (D.3a) and the wire lengths are given by \ud835\udcc11 = xe1 \u2212 xs1, \ud835\udcc12 = xe2 \u2212 xs2. (D.3b) PARTIAL POTENTIAL COEFFICIENTS FOR ORTHOGONAL GEOMETRIES 411 The closed-form solution for (D.2) is easy to obtain. Pp12 = 1 4\ud835\udf0b\ud835\udf16 1 \ud835\udcc11\ud835\udcc12 4\u2211 k=1 (\u22121)k+1[ak log (ak + rk) \u2212 rk ] (D.4) with a1 = xs2 \u2212 xe1, a2 = xe2 \u2212 xe1 (D.5a) a3 = xe2 \u2212 xs1, a4 = xs2 \u2212 xs1 (D.5b) rk = \u221a a2 k + (y2 \u2212 y1)2 + (z2 \u2212 z1)2. (D.5c) This exact equation is used extensively for wire-type models and also for numerical integration over cross-sectional-type geometries. D.1.2 1\u2215R3 Integral Ip12 for Two Parallel Wires For some applications, such as the surface formulation in Section 11.4.2, we also need integrals with an 1\u2215R3 kernel. The geometry is the same as the one in the previous section in Fig. D.1. Hence, the integral to be solved is Ip12 = 1 \ud835\udcc11\ud835\udcc12 \u222b xe1 xs1 \u222b xe2 xs2 1 R3 1,2 dx2 dx1, (D.6) where R1,2 = \u221a (x1 \u2212 x2)2 + (y1 \u2212 y2)2 + (z1 \u2212 z2)2 (D.7a) and \ud835\udcc11 = xe1 \u2212 xs1, \ud835\udcc12 = xe2 \u2212 xs2. (D.7b) Then, the closed form of this partial integral is Ip12 = 1 \ud835\udcc11\ud835\udcc12 rq 4\u2211 k=1 (\u22121)k+1 \u221a a2 k + (y2 \u2212 y1)2 + (z2 \u2212 z1)2 (D.8) with a1 = xe2 \u2212 xs1, a2 = xs2 \u2212 xs1 (D.9a) a3 = xs2 \u2212 xe1, a4 = xe2 \u2212 xe1. (D.9b) We treat these integrals in the same symmetrical way we do the other partial elements. D.1.3 1\u2215R3 Integral Ip12 for Two Orthogonal Filaments Unlike partial inductances, the potentials are coupled for the orthogonal conductors shown in Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000233_978-1-4614-0460-6-Figure9.9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000233_978-1-4614-0460-6-Figure9.9-1.png",
        "caption": "Fig. 9.9 Flexible coupling of components I and II to form assembly III. The force F2b is applied to the assembly at coordinate X2b in order to determine H2b2b, H2a2b, and H12b",
        "texts": [
            " F2a F2a H2a2a \u00bc h2a2a h2a2a h2a2a \u00fe h2b2b \u00fe 1 k 1 h2a2a \u00f09:34\u00de H2b2a \u00bc X2b F2a \u00bc x2b F2a \u00bc h2b2bf2b F2b \u00bc h2b2b h2a2a \u00fe h2b2b \u00fe 1 k 1 h2a2aF2a F2a H2b2a \u00bc h2b2b h2a2a \u00fe h2b2b \u00fe 1 k 1 h2a2a \u00f09:35\u00de H12a \u00bc X1 F2a \u00bc x1 F2a \u00bc h12af2a F2a \u00bc h12a 1 h2a2a \u00fe h2b2b \u00fe 1 k 1 h2a2a ! F2a F2a H12a \u00bc h12a h12a h2a2a \u00fe h2b2b \u00fe 1 k 1 h2a2a \u00f09:36\u00de IN A NUTSHELL While these computations are straightforward, they are more complicated than the rigid connection case. Increased accuracy often comes at the expense of increased computational complexity. The essence of engineering is to determine the required accuracy of the measurement/model/desired outcome and proceeding accordingly. Our final scenario for the two-component flexible coupling is shown in Fig. 9.9. Here, we apply the external force F2b to coordinate X2b to obtain the direct and cross-assembly receptances H2b2b, H2a2b, and H12b. The component displacements are the same as the previous case: x1 \u00bc h12af2a and x2a \u00bc h2a2af2a for substructure I and x2b \u00bc h2b2bf2b for substructure II. However, the equilibrium condition is modified to be f2a \u00fe f2b \u00bc F2b and the compatibility condition is rewritten as: k x2a x2b\u00f0 \u00de \u00bc f2a: (9.37) 332 9 Receptance Coupling replacing f2b with F2a f2a. k h2a2af2a h2b2bF2b \u00fe h2b2bf2a\u00f0 \u00de \u00bc f2a kh2a2af2a kh2b2bF2b \u00fe kh2b2bf2a \u00bc f2a h2a2a \u00fe h2b2b \u00fe 1 k f2a \u00bc h2b2bF2b f2a \u00bc h2a2a \u00fe h2b2b \u00fe 1 k 1 h2b2bF2b \u00f09:38\u00de Again using the equilibrium condition we find the equation for f2b. f2b \u00bc F2b f2a \u00bc 1 h2a2a \u00fe h2b2b \u00fe 1 k 1 h2b2b ! F2b (9.39) The direct and cross receptances for the case shown in Fig. 9.9 are given in Eqs. 9.40 through 9.42. H2b2b \u00bc X2b F2b \u00bc x2b F2b \u00bc h2b2bf2b F2b \u00bc h2b2b 1 h2a2a \u00fe h2b2b \u00fe 1 k 1 h2b2b ! F2b F2b H2b2b \u00bc h2b2b h2b2b h2a2a \u00fe h2b2b \u00fe 1 k 1 h2b2b \u00f09:40\u00de H2a2b \u00bc X2a F2b \u00bc x2a F2b \u00bc h2a2af2a F2b \u00bc h2a2a h2a2a \u00fe h2b2b \u00fe 1 k 1 h2b2bF2b F2b H2a2b \u00bc h2a2a h2a2a \u00fe h2b2b \u00fe 1 k 1 h2b2b \u00f09:41\u00de f2b f2a x2b x2a II I x1 III 9.3 Two-Component Flexible Coupling 333 H12b \u00bc X1 F2b \u00bc x1 F2b \u00bc h12af2a F2a \u00bc h12a h2a2a \u00fe h2b2b \u00fe 1 k 1 h2b2bF2b F2b H12b \u00bc h12a h2a2a \u00fe h2b2b \u00fe 1 k 1 h2b2b \u00f09:42\u00de As we discussed in Sect"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000730_s0022112006002631-Figure17-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000730_s0022112006002631-Figure17-1.png",
        "caption": "Figure 17. The geometry for the trajectories of squirmers 1 and 2. \u03b80 is the angle between e1 and e2. Squirmer 1 is always placed 10 units apart from squirmer 2 in the x-direction, and the x-coordinate of point A relative to the centre of squirmer 2 is equal to the distance between point A and the centre of squirmer 1. The relative positions of A and A\u2032 from the centre of squirmer 1 can be uniquely determined as functions of \u03b80. \u03b4y is the distance between the centre of squirmer 2 and point A\u2032.",
        "texts": [
            " Some sample lines are shown in table 1 for the case with \u03b2 = 1; the total number of lines in this case is 54 481. The full database is available as a supplement to the online version of the paper. In this section, we will introduce some interesting features of the trajectories. We begin by considering two-dimensional configurations, in which the centres and orientation vectors of the two squirmers are in the same plane. Let the x-direction be e2, and let the centre of squirmer 2 be placed \u03b4y apart from a point A\u2032 in the y-direction (see figure 17). Let point A be the intersection between the extension of e1 H y d ro dy n a m ic in tera ctio n o f tw o sw im m in g m o d el m icro -o rg a n ism s 1 4 7 from the centre of squirmer 1 and the extension of e2 from point A\u2032. We restrict the configurations so that the x-coordinate of point A relative to the centre of squirmer 2 is equal to the distance between point A and the centre of squirmer 1. Initially, squirmer 1 is always placed a distance of 10 units from squirmer 2 in the x-direction, i"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000930_s0022112057000713-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000930_s0022112057000713-Figure4-1.png",
        "caption": "Figure 4. Transition curves from spherical nose to cones of semi-angles 45\" and 30\". Bodies of revolution with these meridian sections would have constant surface pressure behind the spherical portion, according to the Ivey, Klunker & Bowen formula (67).",
        "texts": [
            " Here p , is the pressure (to the first approximation) at the point r = r; on the surface. For example, if one takes the pressure to be its value on a sphere of radius a up to a certain value of r, andthereafter constant and equal to that on a cone of semi-angle xo, in other words if It is to be noted how rapidly x --f xo as r increases ; the transition from sphere to cone is made fast, although without the pressure dropping below its final value. T h e shape is plotted for x0 = 45\" and xo = 30\" in figure 4. These are probably rather suitable ' transition curves ' from the spherical to the conical shape, since the fact that the pressure formula overestimates the centrifugal effect can only mean that the tendency to pressure rise, already eliminated by the shape chosen, is less marked even than was assumed in the calculation. 3.8. Shape of shock wave beyond where it separates from the body Both for bodies from which the shock wave separates at the base, and for those from which it is thrown off (as it were) by centrifugal force at a point ahead of the base, it is of interest to investigate the shape of the shock wave beyond where it separates"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000206_978-0-387-74315-8-Figure4-1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000206_978-0-387-74315-8-Figure4-1-1.png",
        "caption": "Figure 4-1: Robot connectivity",
        "texts": [
            " The dynamics algorithms in this and the following chapters are described using a system model of the robot. This is a description of the robot in terms of the number of links, their inertias, and the joints connecting them. The first step in defining a system model is to labeI the various components (joints and links) and specify which is connected to which. Normally the connectivity9 has to be given explicitly, but for the moment we will consider only robots consisting of a single un-branched open-Ioop kinematic chain, as shown in Figure 4-1, for which the connectivity is 9Connectivity is used here in the topological sense. 72 implicit in the nl1mbering scheme. More general robot mechanisms will be considered in Chapters 7 and 9. Let there be n movable links, numbered l..n, and n one-degree-of freedom joints, also numbered l..n, such that joint i connects link i to link i-l. Joint 1 connects link 1 to the immovable base member, which will be called link O for convenience. The next step is to define the geometry of the linkage, the types of joint and the inertia parameters (see Figure 4-2)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure3.36-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure3.36-1.png",
        "caption": "Fig. 3.36: The free-body diagram of a manipulator link for dynamics.",
        "texts": [
            " Both problems have practical interest when they are solved with efficient computational algorithms. Direct problem is fundamental for simulating the robot operation before building or operating them. Inverse problem is useful for regulating the robot operation during a given manipulation task. In addition, the solutions of both problems can be used and are used in control algorithms for high performance robot motion. The model of Fig.3.34 can be extended to the case of dynamics by adding the consideration of inertia forces and torques, as shown in Fig. 3.36. The inertia forces and torques are expressed as Gii in i m aF \u2212= (3.6.3) i Gi iii Gi i in i xIIT \u2212\u2212= (3.6.4) in which inertia characteristics are evaluated in terms of mass mi and inertia tensor Ii Gi of the i-th link. Indeed, inertia characteristics are determined when both mass distribution and kinematic status of a link are known. Chapter 3: Fundamentals of the Mechanics of Robots154 Mass distribution is characterized by the mass of a link and inertia tensor. The mass mi of a link can be evaluated by computing \u03c1= Vi i dxdydzm (3",
            " Similarly to mass of a body, the elements of the inertia tensor can be difficult to be evaluated, mainly for built manipulators. Therefore, numerical approximation can be obtained by again using a discrete sum of known volumes with simple shapes in combination with application of the Huygens Theorem. Experimental determination can also be carried out by using the pendulum test to evaluate the moment of inertia about the axis of the used joint through the formula of Eq. (3.6.6). The Newton\u2013Euler formulation can use the model of Fig. 3.36 to deduce the dynamic equations for a link. Thus, one can write the Newton\u2013Euler equations to express the dynamic equilibrium with respect to the i-th frame in the form of the D'Alembert Principle by summing forces acting on the body link in the form 0R i1i1iii in ii =+\u2212+ ++ FffF (3.6.11) and summing torques about the center of mass in the form 0xRxR iii1i1ii1iii1i1iii in ii =+\u2212+\u2212+ +++++ FGOfOOTttT (3.6.12) in which the inertia force and torque of Eqs (3.6.3) and (3.6.4) are expressed with respect to the i-th frame"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000880_6.2004-4900-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000880_6.2004-4900-Figure5-1.png",
        "caption": "Figure 5. Linear Model for Straight-line Following Case",
        "texts": [
            " In this section, linear analyses are performed for the following three cases: \u2022 Case 1: following a straight line \u2022 Case 2: following a non-straight line which is a perturbation from a straight line \u2022 Case 3: following a circular path and significant features are pointed out. An important design choice in the guidance logic is the distance L1 between the vehicle and the reference point. This value can be chosen with the help of a linear system analysis. Case 1: Following a Straight-line and Selection of L1 Figure 5 defines the notation used in the linearization. L1 is the distance from the vehicle to the reference point, d is the cross-track error, and V is a vehicle nominal speed. Assuming \u03b7 is small in magnitude sin \u03b7 \u2248 \u03b7 = \u03b71 + \u03b72 and \u03b71 \u2248 d L1 , \u03b72 \u2248 d\u0307 V 4 of 16 American Institute of Aeronautics and Astronautics Combining the above with the guidance formula leads to ascmd = 2 V 2 L1 sin \u03b7 \u2248 2 V L1 ( d\u0307 + V L1 d ) (3) Hence, linearization of the nonlinear guidance logic yields a PD (proportional and derivative) controller for the cross-track error"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000876_j.surfcoat.2004.06.029-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000876_j.surfcoat.2004.06.029-Figure6-1.png",
        "caption": "Fig. 6. (a) Sixty-millimeter diameter C45 bar with experimental laser tracks clustered in groups of five tracks with the same feeding rate parameter F. Scanning speed S as well as feeding rate F increases from the left to the right. The visible black pen lines mark the place with the laser power at level of 362.5 and 1500 W during the power grade experiment around the whole bar. Detached laser tracks as well as segments without cladding are visible. (b) Part of the experimental bar after the cutting for laser tracks inspection.",
        "texts": [
            " Area IV may be characterized as an area of low efficiency laser cladding. Solid powder particles reach the liquid surface with relatively high velocities, which leads to the solid-particle/ liquid-surface interaction with a high repelling force [4]. The velocity of clad particles was not varied in our experiments. However, the relatively steep dependence of Pmp on particle velocity is a good assumption that for efficient coaxial cladding an optimal particle speed can be found for a wide interval of powder sizes and properties. As Fig. 6 clearly demonstrates, all 25 laser tracks were cross-sectioned in four cuts, which provided a total of 175 usable data points corresponding to different combinations of S, F and P values. The laser cladding process is based on heat transfer among the laser beam, the substrate and the powder and mass transfer between the powder flow and the molten surface. Therefore, the investigation of the process can be performed based on the so-called combined parameters [18]. Two quantities are fundamental: The amount of powder provided per unit length of the laser track F/S and the total heat input per unit length of the laser track P/S"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001467_j.mechmachtheory.2010.05.001-Figure9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001467_j.mechmachtheory.2010.05.001-Figure9-1.png",
        "caption": "Fig. 9. The position and direction of $T1 relative to $O1.",
        "texts": [
            " dmax denotes the potential maximal length of the common perpendicular between the axes of the unit screws $T1 and $O1; where and hT1 and hO1 represent the pitches of $T1 and $O1, respectively. Based on Eqs. (31) and (33), the numerator in Eq. (36) can be obtained as j$T1\u2218$O1j = j hT1 + hO1\u00f0 \u00decos \u03b8\u2212dsin\u03b8j = jm1;4k1;1 + m1;5k1;2 + m1;6k1;3 + m1;1k1;4 + m1;2k1;5 + m1;3k1;6j \u00f037a\u00de Since the transmission wrench in each leg is a pure force, the pitch of the TWS vanishes, i.e., hT1 = 0 \u00f037b\u00de The pitch of the unit twist screw $O1 can be obtained as hO1 = k1;1k1;4 + k1;2k1;5 + k1;3k1;6 \u00f037c\u00de As shown in Fig. 9, the center of S joint (S1) is the intersection point of the axis of $T1 and the moving platform, and its position relative to the axis of $O1 is constant. The line segment CD represents the distance between the axes of the unit screws $T1 and $O1. As the position and direction of $T1 relative to $O1 changes, the position of the line segment CD will be different. Considering that the axis of $T1 is always passing through point S1, when point C coincideswith point S1, length d of line segment CD has its maximal value dmax"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002124_j.optlastec.2016.07.012-Figure13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002124_j.optlastec.2016.07.012-Figure13-1.png",
        "caption": "Fig. 13. The schematic of direction for microhardness measurements.",
        "texts": [
            " The slice thickness mainly has effect on the remelting times of the molten pool while the slice thickness is less than the minimum of the melting depth. As remelting times increase, the gaseous bubbles formed in the last rapid melting and solidification have more opportunities to escape from the molten pool. Thus the quantity and size of the pores increase with the increase of the slice thickness. As a result, the density increases with the decreasing of the slice thickness. Fig. 14 presents the microhardness profile in the directions (as shown in Fig. 13) of x, y and 45\u00b0 on both cross section and vertical section of the densest SLMed sample. The microhardness of all the rest samples (Porosityr2%) was measured in one direction and it is an average value of the ten data on this direction. According to Fig. 14, the average microhardness of the densest sample is 259 HV. Meanwhile, the microhardness measured from all of the six directions yielded the uniform and similar value, indicating that microhardness is directionally independent. The relationship between scan velocity and microhardness is shown in Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000554_978-3-540-85629-0-Figure7.4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000554_978-3-540-85629-0-Figure7.4-1.png",
        "caption": "Fig. 7.4. A one degree of freedom model with backslash.",
        "texts": [
            " The parameters d, \u03b1 and \u03b2 represent the viscous, the quadratic and the Coulomb damping respectively. The dynamics of the system is described now by 270 7 Dynamic Analysis of Trajectories mz\u0308 + dz\u0307 + kz = \u2212my\u0308 \u2212 \u03b1|z\u0307|z\u0307 \u2212 \u03b2 z\u0307 |z\u0307| . where z = x \u2212 y. This equation may also be written as z\u0308 + 2\u03b4\u03c9nz\u0307 + \u03c92 nz = \u2212\u03c92 n\u03b6 with \u03b6 = y\u0308 \u03c92 n + \u03b1 m\u03c92 n |z\u0307|z\u0307 + \u03b2 m\u03c92 n z\u0307 |z\u0307| . If a backslash effect has to be considered, it is necessary to define an additional nonlinear term, such as the one represented in Fig. 7.4. In this case, with a proper choice of the origin of the reference frames, the \u2018contact\u2019 between motor and load occurs only if y \u2212 x > x0. The equations describing the dynamics are now { mx\u0308 + kx = 0, y \u2212 x < x0 mx\u0308 + dx\u0307 + kx = ky + dy\u0307, y \u2212 x > x0. 7.1.4 Nonlinear model with n degrees of freedom A n-dimensional model with nonlinear passive parameters is shown in Fig. 7.5. Let x1, x2, . . . , xn be the positions of the masses m1, m2, . . . , mn, and y the input motion. By using the state variables z1 = x1\u2212y, z2 = x2\u2212x1, "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000857_j.matdes.2014.07.043-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000857_j.matdes.2014.07.043-Figure1-1.png",
        "caption": "Fig. 1. Example of overhanging structures: (a) downward sloping face, (b) and (c) downw building platform.",
        "texts": [
            " To do this, it is necessary to analyze, first of all, the structures that are typically classified as critical during the building process, as the overhang structures. The samples were fabricated in AlSi10Mg and Ti6Al4V alloys, two of the most common employed metal materials for AM. As mentioned above, to reduce the support structures, it is necessary to determine the optimal orientation of the part on the building platform. This can be done starting from the identification of critical geometries for the SLM process. These are generically called overhanging structures. In AM process, it is considered an overhanging structure (Fig. 1) a part of a component that is not supported during building, by solidified material or a substrate on the bottom side. Consequently, the melt pool created by the heat input from the laser is supported by powder material. From this definition, it is clear that whether a part of a component is an overhanging structure or not also depends on the orientation given to the part while building it (Fig. 1d). Therefore, analyzing the limits of construction of the overhanging structures, it is possible to determine the areas that require support structures and then derive the optimal orientation. In any building orientation, the part is defined with its base on the xy-plane and the building direction is along the z axis. Unfortunately, overhanging structures cannot always be avoided. The main defects that frequently happen during their fabrication are staircase effect, dross formation and warp. SLM process starts with the creation of a three-dimensional CAD-model of an object"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003483_s0890-6955(02)00163-3-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003483_s0890-6955(02)00163-3-Figure1-1.png",
        "caption": "Fig. 1. Powder delivery rate sensor. (a) Schematic of the powder delivery rate sensor. (b) Setup of the powder delivery rate sensor.",
        "texts": [
            " Due to the low sampling frequency of the feedback, a steady delivery rate can only be realized if the delivery rate is averaged in a long period of time after the feed screw reaches a stable rotational speed. In a smaller time scale, fluctuations in the powder delivery can be observed. An optoelectronic sensor is developed to sense the metal powder delivery rate in real-time. The sensor consists of a laser diode, a photo diode, and a glass window. The components are installed in such a way that the laser beam emitted from the laser diode passes through the powder stream flowing inside the glass chamber and is received by the photo diode (Fig. 1 (a)). The carrier gas and the metal powder are mixed well so the powder particles distribute uniformly in the carrier gas. Because of the diffusion, absorption, and reflection of the powder particles to the laser beam, the laser energy received by the photo diode decreases if the powder delivery rate increases, which means there is higher percentage of powder particles in the carrier gas. The laser diode emits a red light with a wavelength of 600\u2013710 nm and a power less than 500 mW. The photo diode is characterized with a good linearity between the illumination energy received by the diode and the output current that is converted later to a voltage signal through a current signal pick-up circuit. Fig. 1 (b) shows the setup of the metal powder delivery rate sensor. A group of experiments are set up to test and calibrate the sensor. By setting the feed screw of the powder feeder at different rotational speeds, a series of powder delivery rates can be obtained. The feed screw runs at the preset rotational speed for about 2 min before any measurement is taken to ensure that a static delivery status has been reached. The averaged powder delivery rate is then calibrated by measuring the time period and the weight of the powder delivered during that time period"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001344_j.addma.2018.11.010-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001344_j.addma.2018.11.010-Figure7-1.png",
        "caption": "Figure 7: A CAD rendering of the completed overhang test artifact. Note the FoV of the high speed camera.",
        "texts": [
            " Data from a further three parameter combinations (PV #27, 28, and #29) were inadvertently recorded using an incorrect exposure time and were therefore not used for training. This experiment was designed to investigate melt pool behavior during the fusion of an unsupported overhang. Specially, the goal was to observe any melt pool morphological changes triggered by the differing thermal conditions present due to the low thermal conductivity of the unfused powder [47,48] below the overhang. As shown in Figure 7, the camera\u2019s FoV is centered on an overhang spanning a gap of 5 mm. To decrease the turnaround time between the experiments, a channel was milled into several of the sub-size plates \u2013 thereby reducing the build time and allowing additional test artifacts to be fabricated in the event of a build or data capture failure. Pads 160 \u00b5m in height were built using nominal parameters (PV #1) and a hatch spacing of 110 \u00b5m on either side of the pre-cut channel in order to ensure appropriate bonding between the overhanging layers of interest and the sub-size plate"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003881_mssp.2001.1414-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003881_mssp.2001.1414-Figure3-1.png",
        "caption": "Figure 3. Finite element model of the spur gears in mesh.",
        "texts": [
            " In this paper, the matrix capability of the Simulink environment was used to solve the equations, resulting in a faster solution than that provided by the MATLAB ode solvers. Most of the previously published \"nite element analysis gear models have involved only partial gear body models. In an investigation of gear transmission errors using the variational torsional mesh sti!ness, the whole body of the gears in mesh should be modelled. A \"nite element model of the spur gears in mesh, without any tooth damage, is shown in Fig. 3. The gears were modelled using quadratic two-dimensional (2-D) plane strain elements and the contact e!ect was modelled using 2-D line-to-line general contact elements which included elastic Coulomb frictional e!ects. The torsional mesh sti!ness of gears in mesh at particular positions throughout the mesh cycle was generated by rotating both solid gears, then creating a \"nite element model in that particular position. The torsional mesh sti!ness of the gears in mesh, K m was obtained from the FEA model by introducing a prescribed input torque to the coupled nodes of the input gear hub, while restraining all nodes of the output gear hub"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000903_tcyb.2016.2518300-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000903_tcyb.2016.2518300-Figure1-1.png",
        "caption": "Fig. 1. Schematic of electromechanical system.",
        "texts": [
            " If r > 0, r1 > 0, \u03bb > 0, kj, ci > 0, and Q are large and \u03c4 > 0 is small, the control energy is large. Therefore, in practical applications, the design parameters should be adjusted carefully for achieving suitable transient performance and control action. V. SIMULATION RESULTS In this section, a practical simulation example (electromechanical system) is presented, which is borrowed from [22] and [40], and it is used to verify the effectiveness and advantage of our method. Example 1: We consider the electromechanical system which is shown by Fig. 1 [22], [40]. The dynamics of the electromechanical system is described by the following differential equations: { Mq\u0308 + Bq\u0307 + N sin(q) = I LI\u0307 = V0 \u2212 RI \u2212 KBq\u0307 (84) where M = (J/K\u03c4 )+(mL2 0/3K\u03c4 )+(M0L2 0/K\u03c4 )+(2M0R2 0/5K\u03c4 ), N = (mL0G/2K\u03c4 ) + (M0L0G/K\u03c4 ), and B = (B0/K\u03c4 ). J is the rotor inertia, m is the link mass, M0 is the load mass, L0 is the link length, R0 is the radius of the load, G is the gravity coefficient, B0 is the coefficient of viscous friction at the joint, q(t) is the angular motor position (and hence the position of the load), I(t) is the motor armature current, and K\u03c4 is the coefficient which characterizes the electromechanical conversion of armature current to torque"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure9.15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure9.15-1.png",
        "caption": "FIGURE 9.15. A model for a 2R planar manipulator.",
        "texts": [
            "313) where m is the mass and I is the moment of inertia of the pendulum about O. The Lagrangean of the manipulator is L = K \u2212 V = 1 2 I\u03b8\u0307 2 +mg cos \u03b8 (9.314) and therefore, the equation of motion of the pendulum is M = d dt \u00b5 \u2202L \u2202\u03b8\u0307 \u00b6 \u2212 \u2202L \u2202\u03b8 = I \u03b8\u0308 +mgl sin \u03b8. (9.315) The generalized force M is the contribution of the motor torque Q and the viscous friction torque \u2212c\u03b8\u0307. Hence, the equation of motion of the manipulator is Q = I \u03b8\u0308 + c\u03b8\u0307 +mgl sin \u03b8. (9.316) Example 370 An ideal 2R planar manipulator dynamics. An ideal model of a 2R planar manipulator is illustrated in Figure 9.15. It is called ideal because we assume the links are massless and there is no friction. The masses m1 and m2 are the mass of the second motor to run 564 9. Applied Dynamics the second link and the load at the endpoint. We take the absolute angle \u03b81 and the relative angle \u03b82 as the generalized coordinates to express the configuration of the manipulator. The global positions of m1 and m2 are\u2219 X1 Y2 \u00b8 = \u2219 l1 cos \u03b81 l1 sin \u03b81 \u00b8 (9.317)\u2219 X2 Y2 \u00b8 = \u2219 l1 cos \u03b81 + l2 cos (\u03b81 + \u03b82) l1 sin \u03b81 + l2 sin (\u03b81 + \u03b82) \u00b8 (9"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure9.70-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure9.70-1.png",
        "caption": "Figure 9.70 The isolated joints G and H.",
        "texts": [
            " Solution: After first determining the support reactions from the equilibrium of the truss as a whole, we can determine the forces in members 1 to 6 from the equilibrium of joints A, B and C respectively. In the situation shown in Figure 9.69a we get stuck, as more than two member forces are unknown in both D and E. Since members 8 and 12, and 10 and 13 are in a direct line with one another, we can determine the forces in the members 11 and 9 from the equilibrium of joints H and G. The vertical equilibrium of joint H in Figure 9.70 gives N11 = 2F. We now have the situation as shown in Figure 9.69b. From the equilibrium in G in the direction normal to members 10 and 13 we find next (see Figure 9.70): N9 + 1 2N11 \u221a 2 = N9 + 1 2 \u00d7 2F \u00d7 \u221a 2 = 0 \u21d2 N9 = \u2212F \u221a 2. Now that N9 is known (see Figure 9.69c), we can find the remaining member forces by consecutively elaborating the equilibrium of joints D, E, G, H and L. The order in which we handle the joints is therefore A \u21d2 B \u21d2 C \u21d2 H \u21d2 G \u21d2 D \u21d2 E \u21d2 G \u21d2 H \u21d2 L. 9 Trusses 363 Instead of determining N9 and N11 from the equilibrium of joints H and G, it is far easier to revert to the method of sections. With the section shown in Figure 9.69d across members 12, 13 and 14, we can determine the force in member 14 from \u2211 Tz|K = 0"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002277_j.prostr.2016.02.039-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002277_j.prostr.2016.02.039-Figure7-1.png",
        "caption": "Fig. 7 \u2013 Stress distribution of the original component (left) and the optimized component (right) for Load Case 3.",
        "texts": [],
        "surrounding_texts": [
            "The mesh has great influence in the final solution. Highly refined meshes give very different topologies from less refined meshes. In the final design domain, a more controlled method for meshing was used in order to ensure its high quality. The control parameters also have great influence not only in the solution convergence degree, but also in the computing time, thus a convergence study on these parameters was run. In this optimisation there were two control parameters which were studied, the Relative Convergence Criterion (RCC) and the Discreteness Parameter (DP) which is the equivalent for penalty factor in TO theory. The objective function was the weighted compliance in order to consider the three load cases in the topology optimisation [HyperWorks Guide]. This response is given by Equation 1 \ud835\udc36\ud835\udc36\ud835\udc64\ud835\udc64 = \u2211\ud835\udc64\ud835\udc64\ud835\udc56\ud835\udc56\ud835\udc36\ud835\udc36\ud835\udc56\ud835\udc56 (1) where wi is the weight and Ci is the compliance of load case i which is given by Equation 2 Ci = 1 2 ui Tfi (2) where ui and fi are the displacement and force vectors, respectively, corresponding to load case i. The objective was the minimization of the weighted compliance and each load case was given the same weight. There were two constraints defined in the optimisation. The first one was regarding the volume fraction of the design domain. The second was a symmetry constraint, forcing the optimised solution to be symmetric with respect to the component\u2019s mid vertical plane as the original component is. Figure 3 and Figure 4 illustrate the TO boundary conditions and final solution pseudo-density distribution. Fig. 3 \u2013 Final Design Domain (blue) and boundary conditions. Fig. 4 \u2013 TO solution with element pseudo-density distribution. Author name / Structural Integrity Procedia 00 (2016) 000\u2013000 5"
        ]
    },
    {
        "image_filename": "designv10_0_0001514_j.engfailanal.2015.06.004-Figure13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001514_j.engfailanal.2015.06.004-Figure13-1.png",
        "caption": "Fig. 13. Schematic for plastic inclination of gear tooth due to tooth root crack [45].",
        "texts": [
            " 11), and compared the difference of the TVMS under two crack propagation paths: a straight line and multiple straight lines. Based on the a curved crack propagation path obtained using FRANC (see Fig. 12), Zhao et al. [44] proposed a potential energy method to calculate the TVMS of a cracked gear pair. Considering the combined effects of tooth crack and the plastic inclination, Shao and Chen [45] developed a tooth plastic inclination model for a spur gear pair by considering the cracked tooth as a variable cross-section cantilever beam (see Fig. 13). Considering the effects of the gear tooth root crack which propagates into the gear-body, Zhou et al. [46] presented a revised mathematical model based on the assumption of a linear propagation path (see Fig. 14). For a spur gear pair, Saxena et al. [47] established an analytical model by taking the influences of the shaft misalignment and the friction of tooth surface into account. Omar et al. [48] presented a TVMS expression for a gear pair with tooth root crack, which can be written as follows: km \u00bc XN 1 i\u00bc1 ki m \u00fe kcracked m ; \u00f01\u00de where ki m \u00bc f k xt 2p i 1 N \u00fe n 2p x i 1 N \u00fe n 6 t 6 2p x i 1 N \u00fe n \u00fe hf x 0 elsewhere ( ; \u00f02\u00de kcracked m is evaluated using Eq"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002049_j.ymssp.2021.107945-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002049_j.ymssp.2021.107945-Figure7-1.png",
        "caption": "Fig. 7. Bending and stretching dominated behaviors in cellular materials: (a) influence of relative density on the mechanical properties of a bending cellular structure, and (b) relative density of a stretching-dominated cellular architecture.",
        "texts": [
            " Tancogne-Dejean et al. [126] designed plate-lattices by placing plates along the closest-packed planes of crystal structures and discussed their applications for heat-exchange, thermal insulation, acoustics, and biomedical engineering. Alfouneh et al. [127] designed multi-cellular cores for sandwich panels by topology optimization to attenuate harmonic excitations. Cellular structures are typically classified into bending and stretching dominated structures in terms of deformation, as shown in Fig. 7 [48]. Due to the effective energy absorption ability of the cellular structures, they have been used in engineering for vibration attenuation and sound absorption. For a typical cellular structure, stress\u2013strain curves under compression shown in Fig. 8 may be characterized by three regimes [128]: 1) a linear elastic regime with cell edge bending or face stretching; 2) a stress plateau associated with progressive cell collapse by elastic buckling, plastic yielding or brittle crushing, depending on the mechanical properties of the base materials; and 3) densification, corresponding to the collapse of the cells throughout the material and subsequent loading of the cell edges and faces against one another"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001422_tsmc.2016.2586108-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001422_tsmc.2016.2586108-Figure5-1.png",
        "caption": "Fig. 5. Blueprint of electromechanical system.",
        "texts": [
            " 2 shows trajectories of adaptive laws \u03b81 and \u03b82, and Figs. 3 and 4 demonstrate the trajectory of designed control signal v and control input signal with saturation u, respectively. As can be seen from them, signals of the result- ing closed-loop systems are bounded and system states are regulated to the origin, and the practical control input u satisfies the input constraint, which confirms the effectiveness of the proposed method. Example 2: To further validate the effectiveness of the proposed approach, the electromechanical system shown in Fig. 5 is considered and its governing equation is as follows: Mq\u0308 + Bq\u0307 + N sin(q) = I V\u03b5 \u2212 RI \u2212 KBq\u0307 = LI\u0307 where and detailed symbols meaning can be found in Table I. Values for the above parameters are the same as those in [62]. Referring to the method used in [62], and considering unmodeled dynamics, the above equation can be transformed into the following expressions: z\u0307 = \u22122z + 0.25x2 1 x\u03071 = x2 + (sin(x1)+ z)2 x\u03072 = x3 \u2212 N M sin(x1)\u2212 B M x2 + x2z sin(x1) x\u03073 = u \u2212 KB LM x2 \u2212 R LM x3 where u(v) = sat(v) = { sign(v)\u00d7 5, |v| \u2265 5 v, |v| < 5"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.59-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.59-1.png",
        "caption": "FIGURE 7.59. A multi-link steering mechanism tha must be optimized by varying x.",
        "texts": [
            " Determine the coordinates of turning center of a 4WS vehicle in terms of outer steer angles \u03b4of and \u03b4or. 13. F Turning radius. Determine the turning radius of a 4WS vehicle in terms of \u03b4r. cot \u03b4r = 1 2 (cot \u03b4ir + cot \u03b4or) 14. F A three-axle car. 454 7. Steering Dynamics Consider a three-axle off-road pick-up car. Assume a1 = 1100mm a2 = 1240mm a3 = 1500mm w = 1457mm and determine \u03b4o, R1, Rf , and R if \u03b4i = 10deg. 15. F Steering mechanism optimization. Find the optimum length x for the multi-link steering mechanism shown in Figure 7.59 to operate as close as possible to the kinematic steering condition. The vehicle has a track w = 2.64m and a wheelbase l = 3.84m, and must be able to turn the front wheels within a working range equal to \u221222 deg \u2264 \u03b4 \u2264 22 deg. 8 The suspension is what links the wheels to the vehicle body and allows relative motion. This chapter covers the suspension mechanisms, and discusses the possible relative motions between the wheel and the vehicle body. The wheels, through the suspension linkage, must propel, steer, and stop the vehicle, and support the associated forces"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure6.3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure6.3-1.png",
        "caption": "Fig. 6.3. Construction of a three-phase, 8-pole AFPM brushless machines with PMs arranged in Halbach array: (a) PM ring; (b) stator winding; (c) one half of the twin rotor; (d) stator winding and complete twin rotor.",
        "texts": [],
        "surrounding_texts": [
            "Fig. 6.6 shows the results of the 2D FEM modelling of the magnetic field in the air gap of a coreless AFPM brushless machine. NdFeB magnets with remanent flux density Br = 1.2 T and coercivity Hc = 950 kA/m have been 196 6 AFPM Machines Without Stator and Rotor Cores 6.3 Air Gap Magnetic Flux Density 197 considered. The thickness of each PM has been assumed 6 mm, the coreless stator winding thickness is 10 mm and one-sided air gap thickness equals to 1 mm. With the aid of Halbach array a high peak value (over 0.6 T) of the normal component of the magnetic flux density has been excited. This value is sufficient to obtain a high electromagnetic torque. The peak value of the flux density can be even higher for the optimized magnetic circuit of an AFPM machine. In practice, 60o and 45o Halbach arrays produce similar peak magnetic flux density (Fig. 3.22). The peak value of the normal component of the magnetic flux density is higher than that in a standard arrangement of surface PMs. The backing ferromagnetic discs added to the double-sided PM structure with a large nonmagnetic gap do not increase the flux density as much as Halbach array excitation does. 198 6 AFPM Machines Without Stator and Rotor Cores"
        ]
    },
    {
        "image_filename": "designv10_0_0000741_iros.2003.1248880-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000741_iros.2003.1248880-Figure1-1.png",
        "caption": "Figure 1: A model of humanoid robot",
        "texts": [
            " In Section 4, we explain the detailed calculation of inertia matrix which is necessary to our method. In Section 5, using a humanoid robot HW-1, kicking and walking motions are generated and tested. Section 6 concludes the paper and states our future plan. 2 Momentum equation 2.1 Momentum and joint velocities We represent a humanoid robot as a mechanism of tree structure whose root is a f ieflying base link 07803-7860-1103l517.00 Q 2003 IEEE 1644 (pelvis), a rigid body having 6 D.0.F in 3D space (Figure 1). We define the frame Cg embedded in the base link whose translational velocity and angular velocity are ug and W B , respec!ively. In addition, we define a n x 1 column vector 0 which contains velocities of all joints as its elements where n is the total number of joints. The linear momentum P (3 x 1) and the angular momentum L (3 x 1) of the whole mechanism are given by where is the total mass of the robot, E is am identity matrix of 3 x 3, TB-: is the 3 x 1 vector from the base link to the total center of mass (CoM) and i is the 3 x 3 inertia matrix with respect to the COW"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001205_s10846-016-0442-0-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001205_s10846-016-0442-0-Figure2-1.png",
        "caption": "Fig. 2 Vehicle view and common nomenclature used in vehicle Modelling",
        "texts": [
            " Usually known as handling model, important properties usually observed are the lateral and yaw motions. For handling model, vertical motions are always considered constant by assuming flat road surface where ride properties and vertical forces from suspension is negligible. Most common vehicle model adopt bicycle concept where the vehicle is reduced to a two-tire configuration at the front and rear by assuming same behaviour for left and right tires. With this assumption, the right and left tire is merged and represented by one tire at front and rear, each. Figure 2 shows the common representation of vehicle model usually used in handling studies depicting the full vehicle model. Usual configurations consists of geometric, kinematic and dynamic model. Geometrical dimensions of the vehicle are considered in geometric model with no regard on the kinematic (acceleration and its derivatives) and dynamic (forces, inertia, and energy) properties. Similarly, kinematic model consider the motion of the vehicle in terms of its acceleration, velocity and position related to the geometric of the whole vehicle"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure7.20-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure7.20-1.png",
        "caption": "Fig. 7.20. Small coreless AFPM brushless motor. Photo courtesy of the University of Stellenbosch, South Africa.",
        "texts": [
            " A low pass filter (LPF) is furthermore used to extract from the dq currents the fundamental current components iad and iaq, which are necessary for the fundamental current controller of the drive (Fig. 7.13). The frequency of the injected voltage signals must be much higher than the fundamental frequency, but also much lower than the switching frequency of the inverter, and is typically between 500 Hz and 2 kHz. Numerical Example 7.1 A small, 35-W, 4300 r/min, 12 pole, Y-connected air-cored, trapezoidal AFPM brushless motor is used as part of a reaction wheel for a micro-satellite (see Fig. 7.20). Reaction or momentum wheels are used on satellites to keep the satellite steady and to orientate it. Orientation control is necessary as, e.g. the photovoltaic panels must be pointed at all times to the sun for maximum power generation. The motor has a stator phase inductance of Lp = 3.2 \u00b5H and a phase resistance ofR1 = 22 m\u2126. The motor is fed by a 3-phase MOSFET inverter with Vd = 14 V, and is under square-wave current control as described in Section 7.1.3. Assuming the voltage drop across a MOSFET switches during switching-on as 1 V, calculate: (a) the switching frequency fs of the inverter; (b) the external phase inductance required for a maximum peak\u2013to\u2013peak rip- ple current less than \u2206irmax = 0"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001786_j.mechatronics.2011.02.007-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001786_j.mechatronics.2011.02.007-Figure4-1.png",
        "caption": "Fig. 4. Prototype quadrotor.",
        "texts": [
            " We choose f2 much larger than f1 in order to force the control weights close to the alternate weights when the learning error is small. The term aCTs helps guarantee bounded signals in the Lyapunov-derivation of Appendix C, where all signals are shown to be uniformly ultimately bounded. To conduct control experiments, one can modify the electronics on commercially-available quadrotors like Dragonflyer from RC Toy and EADS Quattrocopter MAV12 or obtain vehicles designed for researchers like The X-4 Flyer. However, building our own quadrotor prototype provided low-cost control algorithm testing (Fig. 4). Moreover, the quadrotor design ensures modeling assumptions are met, especially the rigid frame, and the electronics are suitable for accomplishing the proposed control method. The rigid Q4 Dragster frame from lipoly.de consists of a lightweight carbon fiber electronics cage with four aluminum arms. The arms are 23.2 cm long from center to tip, and the total mass Fig. 3. Proposed method of the prototype (including battery and electronics) measures 753 g. Testing established the four rotors lift 960 g when running at 3500 RPM (366 rad/s), and speeds of close to 6000 RPM are possible"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001252_j.ymssp.2007.12.001-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001252_j.ymssp.2007.12.001-Figure8-1.png",
        "caption": "Fig. 8. Thirty-four-DOF dynamic model of the gear test rig (vertical direction y aligned with the Line of Action of the gears).",
        "texts": [
            " In order to study the dynamics of this system, the equations of motion are described next for the rotational and translational along the Line of Action (LoA) using Eqs. (30) and (31), respectively. mi \u20acyi \u00fe ksiyi cm\u00f0r2 _y2 r1 _y1 _x2 \u00fe _x1 \u00fe _et\u00de kmri\u00f0r2y2 r1y1 x2 \u00fe x1 \u00fe et\u00de \u00bc 0, (30) ji \u20acy\u00fe cmri\u00f0r2 _y2 r1 _y1 _x2 \u00fe _x1 \u00fe _et\u00de \u00fe kmri\u00f0r2y2 r1y1 x2 \u00fe x1 \u00fe et\u00de \u00bc Ti, (31) where i \u00bc 1, 2 (1 for gear, 2 for pinion). These equations are solved iteratively using Matlabs/Simulinks. There are 34 DOF in the new model as opposed to the 16 DOF in the previous model [1]. This is illustrated in the schematic diagram of Fig. 8. The extra 18 DOF are due to the inclusion of the five-DOF bearing model (Fig. 3), and to the fact that translational DOF are now considered both along the Line of Action (LOA) and perpendicular to it (Figs. 3 and 7). The main assumptions, on which the new model is based, are as follows. (1) Shaft mass and inertia are lumped at the bearings or at the gears. (2) Translational DOF are considered along the LoA and perpendicular to it, with the LoA aligned vertically. (3) Two resonances of the gear casing are considered"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003650_s0167-6911(02)00342-0-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003650_s0167-6911(02)00342-0-Figure5-1.png",
        "caption": "Fig. 5. Flexible manipulator for Example 3.",
        "texts": [
            " If P is in7nite, the analysis breaks down because the switching can no longer be guaranteed to stop and stability results for the case of persistent switching in nonlinear systems are lacking. However, simulations indicate that the system still displays the desired behavior. A typical trajectory for the case P=[0:1; 2]\u00d7[0:1; 3] is depicted in Fig. 4. A parking movie generated with MATLAB/Simulink which illustrates the corresponding motion of the robot is available from the web [25]. Example 3. Consider the Lexible manipulator shown in Fig. 5. For small bending, this system can be modeled by the following PDE: Ty(x; t) + EI y\u2032\u2032\u2032\u2032(x; t) =\u2212x T (t); (6) with boundary conditions y(0; t) = y\u2032(0; t) = 0; y\u2032\u2032(L; t) = y\u2032\u2032\u2032(L; t) + mt y\u2032\u2032\u2032\u2032(L; t) = 0; T (t) = IH T (t)\u2212 EIy\u2032\u2032(0; t); where E denotes elasticity of the beam, I the inertia of a transversal slice, IH the axis\u2019 inertia, L the beam\u2019s length, the beam\u2019s mass density, and mt the load mass at the tip of the manipulator. Four measurements are 6 Notable exceptions include the work reported in [7,20]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000832_j.addma.2016.05.007-Figure7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000832_j.addma.2016.05.007-Figure7-1.png",
        "caption": "Figure 7. Temperature distribution plot for all scanning strategy cases.",
        "texts": [
            " The reason is that the scanning strategy changes scanning direction for neighboring island and rotates scanning direction for the same island on the subsequent layer. Therefore, the directional deflection difference has been reduced. The localized temperature evolution is considered as the main cause of the deformation defect in SLM process. Thus, the temperature distribution contours for laser scanning on layer 3 have been collected at the time point which is half of total time needed for one layer scanning, and shown in Figure 7. The minimum temperature of part substrate did not maintain at room temperature, it has reached around 300 oC when laser beam traveled on the third layer for all cases due to small part size and high beam power. It is noticed that the five line or rotate line scanning cases have comparatively lower minimum substrate temperature while the island scanning case has the highest minimum substrate temperature. The island scanning case needs comparatively more time to complete a layer, thus more heat is introduced to the substrate"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure5.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure5.1-1.png",
        "caption": "Figure 5.1 (a) A light mast fixed at A with (b) the support reactions.",
        "texts": [
            " For this reason, in addition to self-contained structures, we will also look at compound and related structures, such as hinged beams, three-hinged frames (with or without tie-rods), shored structures and trussed beams. The loading remains limited to a few point loads. In one case, the structure is loaded by a concentrated couple. 154 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM In this section, we will use five examples to show how, for statically determinate self-contained structures, it is possible to determine the support reactions and interaction forces directly from the equilibrium. Example 1 The light mast ABC in Figure 5.1a is fixed at A and is loaded at C by a vertical force of 6 kN. Question: Draw the support reactions at A as they are expected to act and determine them. Solution: No horizontal loading is being exerted on the mast. The horizontal support reaction at A is therefore zero. The vertical support reaction at A must generate an equilibrium with the vertical force of 6 kN, and will therefore be pointed upwards. In order to determine the fixed-end moment, the isolated structure is considered to be pinned at A. The load causes a clockwise rotation about A. The fixed-end moment has to prevent this rotation and will therefore act counter-clockwise. The support reactions are shown in Figure 5.1b. The equilibrium equations are \u2211 Fx = Ah = 0,\u2211 Fy = \u2212(6 kN) + Av = 0,\u2211 Tz|A = \u2212(6 kN) \u00d7 (1.5 m) + Am = 0. The solution is Ah = 0, Av = 6 kN and Am = 9 kNm. 5 Calculating Support Reactions and Interaction Forces 155 The fact that the solutions found are positive confirms the correctness of the directions assumed for these support reactions. Note that the support reaction Av and the force of 6 kN at C together form a couple that is in equilibrium with the fixed-end moment Am. Example 2 In Figure 5"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure10.9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure10.9-1.png",
        "caption": "Figure 10.9 The positive directions of the section forces N , V and M in different coordinate systems.",
        "texts": [
            " This sign convention has already been used in trusses (see Section 9.3). \u2022 A shear force V is positive when it acts on a positive sectional plane in the positive z direction, and on a negative sectional plane in the negative z direction. \u2022 A bending moment M is positive when it causes tension (tensile stresses) at the positive z side of the x axis, and causes compression (compressive stresses) at the negative z side. 392 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM In an xy coordinate system, the positive/negative section forces are defined in the same way. Figure 10.9 shows the positive directions of the section forces in various coordinate systems.1 The sign convention given here for the section forces N , V and M is associated with the sign convention for stresses in the cross-section. We look at this in more detail in Section 10.1.2. On a positive sectional plane, consider a small area A. Let F be the resultant of all the small forces that are transferred by the matter via that small area. F is built up by the contributions of a large number of interactions between the particles of matter"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.42-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.42-1.png",
        "caption": "FIGURE 8.42. Illustration of trust angle.",
        "texts": [
            " Example 324 Castor versus camber. Camber doesn\u2019t improve turn-in as the positive caster does. Camber is not generally good for tire wear. Camber in one wheel does not improve directional stability. Camber adversely affects braking and acceleration efforts. 8.4.4 F Trust Angle The trust angle \u03c5 is the angle between vehicle\u2019s centerline and perpendicular to the rear axle. It compares the direction that the rear axle is aimed with the centerline of the vehicle. A nonzero angle configuration is shown in Figure 8.42. Zero angle confirms that the rear axle is parallel to the front axle, and the wheelbase on both sides of the vehicle are the same. A reason for nonzero 484 8. Suspension Mechanisms 8. Suspension Mechanisms 485 trust angle would have unequal toe-in or toe-out on both sides of the axle. Example 325 Torque reaction. There are two kinds of torque reactions in rear-whel-drive: 1\u2212 the reaction of the axle housing to rotate in the opposite direction of the crown wheel rotation, and 2\u2212 the reaction of axle housing to spin about its own center, opposite to the direction of pinion\u2019s rotation"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure9.13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure9.13-1.png",
        "caption": "Figure 9.13 Model of (a) a steel garden gate with falling diagonal (tension diagonal) and (b) a wooden garden gate with rising diagonal (compression diagonal).",
        "texts": [
            " The rising diagonals in case (b) shorten and are loaded by compressive forces. 9 Trusses 327 In steel trusses, falling diagonals (tension diagonals) are used most frequently, as (usually slender) steel members subject to compression run the risk of buckling. Preferably apply them as tension members. In contrast, rising diagonals (compression diagonals) are most often used in wooden trusses, as in general a wooden joint is more suitable to transfer compressive forces rather than tensile forces. An example close to home is the simple garden gate in Figure 9.13, with (a) a steel version (falling diagonal) and (b) a wooden version (rising diagonal). The wooden lock-gate in Figure 9.14 is another example. The wooden diagonal strut is a rising diagonal and acts as a compression member under influence of the dead weight of the gate. The wooden planking is facing the same way as the diagonal strut. The steel falling diagonal is a tension bar. The joints in a truss are numbered or lettered (see Figure 9.15). The numbers or letters used to indicate the joints can be used as an index"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003684_ias.1997.643010-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003684_ias.1997.643010-Figure2-1.png",
        "caption": "Figure 2 The stator and rotor flux linkages in different",
        "texts": [
            " ANALYSIS OF THE TORQUE IN THE STATOR FLUX REFERENCE FRAME The basic idea of the DTC is in the control of the stator flux vector (in amplitude and angular position) with feedback of observed torque and stator flux from measured voltages and currents from motor terminals. (Using modern devices, computations necessary for these two quantities can now be carried out in real time). The observers can be set up in the stationary reference frame, which leads eventually to the elimination of the rotor position sensor. The stator flux linkage vector, qs, and rotor flux linkage vector, qf, can be drawn in the rotor flux (dq-), stator flux (xy-) and stationary (DQ-) reference frames as in figure 2. The angle between the stator and rotor flux linkages, S, is the torque angle. In the steady state, 6 is constant corresponding to a load torque and both stator and rotor flux rotate at the synchronous speed. In transient operation, 6 varies and the stator and rotor flux rotate at different speeds. Since the electrical time-constant is normally much smaller than the mechanical time-constant, the rotating speed of stator flux with respect to the rotor flux can be easily changed. The stator flux linkage and the torque angle 6 are given by (2) and ( 3 ) "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000757_tsmcb.2007.910740-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000757_tsmcb.2007.910740-Figure8-1.png",
        "caption": "Fig. 8. (a) Outlook of the PIP device. (b) Profile of the PIP.",
        "texts": [
            " 1\u20133, although the tracking is achieved, the control-input chattering with SMC is severe. When the ASMC is applied, the control input is much smooth, and the tracking performance is excellent, as shown in Figs. 4\u20136. The control input is much smaller, and the switching control part is small once the sliding layer is entered. The convergence of \u0393\u0302 is confirmed in Fig. 7. Notably, the knowledge of the upper bounds on the uncertainties is not required. A planetary gear-type inverted-pendulum (PIP) mechanism [16] is utilized to verify the effectiveness of the proposed ASMC. Fig. 8 shows the outlook of the PIP device. The angular acceleration of the star gear causes the pendulum to swing. An important advantage of the PIP mechanism is that the planet gear encircles the star gear, avoiding the winding wire problem. Define the state vector as x = [x1, x2] T, where x1 = \u03b80 and x2 = \u03b8\u03070 are the angle and the angular velocity of the pendulum, respectively. According to the Lagrange method, the state-space equation of the PIP control system can be represented as x\u03071 =x2 x\u03072 = p(x) + q\u03c4 (21) where p(x) = \u2212 ( \u03b52I1+I2 )( r1 \u03b5(r1+r2)I1 ) \u00d7 ( m0 ( l1 2 + l2 ) + m1(r1 + r2) + m3(l1 + l2) + m4l2 2 ) \u00b7g \u00b7sinx1 \u00b7 ( \u03b5 ( r1+r2 r1 ) I1\u2212 ( \u03b52I1+I2 )( r1 \u03b5(r1+r2)I1 ) \u00b7 ( 4I1 + m0(l1 + l2) 3 + m1(r1 + r2) 2 + m3(l1 + l2) 2+ m4l 2 2 3 \u2212 l32 3l1 ))\u22121 (22) q = ( \u03b5 ( r1+r2 r1 ) I1\u2212 ( \u03b52I1+I2 )( r1 \u03b5(r1+r2)I1 ) \u00b7 ( 4I1 + m0(l1 + l2) 3 + m1(r1 + r2) 2 + m3(l1 + l2) 2 + m4l 2 2 3 \u2212 l32 3l1 ))\u22121 (23) and \u03c4 is the control torque of the motor"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000946_j.addma.2014.10.003-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000946_j.addma.2014.10.003-Figure5-1.png",
        "caption": "Fig. 5. The half-symmetry finite element mesh of the substrate, wall, and the fi",
        "texts": [
            " The properties of an element are witched from \u201cquiet\u201d to active when any Gauss point of the lement is consumed by the heat source volume (Eq. (3)). In ddition to the properties being switched, the temperature of the ctivated element is reset to the ambient temperature to prevent rroneous heating of the element. The free surfaces of the part re re-assessed whenever an element is switch from \u201cquiet\u201d to ctive to ensure that convection and radiation are applied proprly to the evolving part surface, which includes the interfaces etween the \u201cquiet\u201d and active elements. .2. Finite element mesh Fig. 5 presents the 3-dimensional half-symmetry mesh that is sed for both the thermal and mechanical analysis. The black surace represents the symmetry plane. The mesh comprises 23,295 odes and 15,627 elements. There are 11,904 elements in the 5.2 mm tall, 38.1 mm long, 1.5 mm thick wall. The aluminum J.C. Heigel et al. / Additive Manufacturing 5 (2015) 9\u201319 13 c i s a 1 s m c p % w b t t 5 c u m s i d o h c T u m T ( q d h s i r w t 2 t p r l i a T b ( v f e m h xture\u2019s aluminum clamp. lamp used to hold the substrate in the measurement fixture is ncluded in the mesh to account for the heat transfer from the subtrate into it"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure9.43-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure9.43-1.png",
        "caption": "Fig. 9.43. Axial flux brushless machine with bulk HTS rotor developed by Kitano Seiki , Japan: (a) computer 3D image; (b) motor and propeller on test bed. 1 \u2014 HTS bulk magnet, 2 \u2014 armature coil, 3 \u2014 magnetic seal unit, 4 \u2014 vacuum pump, 5 \u2014 liquid nitrogen inlet, 6 \u2014 armature winding terminal, 7 \u2014 coil cooling layer.",
        "texts": [
            " With YBCO wire the current density is considerably improved. Another Japanese industry-academia group3 has developed a 15 kW, 720 rpm, coreless axial flux HTS motor, similar to that shown in Figs 9.40 and 9.41. Kitano Seiki is marketing this motor [153]. The axial gap superconductive motor uses GdBaCuO bulk HTS for rotor excitation system and HTS tapes for the stator (armature) winding. The rotor system is in the central position of the machine and the twin stator (armature) is on both sides of the rotor (Fig. 9.43). The rotor has 8 poles 26-mm diameter each. The armature winding has 6 poles. The rotor assembly is 0.3 m long and is 0.5 m in diameter. The magnetic flux density in the air gap is 3 T at 77K under field cool magnetization. The bulk HTS poles are magnetized inside the machine using a pulsed-copper coils. The armature system is stationary so that there are no slip rings, only armature leads. The rotor field excitation system with bulk HTS does not require any leads. No electric power supply is needed to maintain the magnetic field"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001538_j.jmatprotec.2011.09.020-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001538_j.jmatprotec.2011.09.020-Figure2-1.png",
        "caption": "Fig. 2. Specimen arrange",
        "texts": [
            " Specimen manufacture For the purposes of this report, a number of sets of specimens were manufactured as shown in Table 1. Cylindrical specimens of cellular structure of diameter (15 mm) and height (30 mm) were designed and manufactured to a nominal porosity of 65%, with each part being individually identified. The structure consisted of octahedral wire frames contained within imaginary unit cells of 600 m cell size created using Manipulator\u00a9 Version 4.7.1 software (The University of Liverpool, UK). 25 of these samples were built on all substrate plates in a 5 \u00d7 5 array (45 mm spacing in X and Y directions) (Fig. 2). This particular style of build was chosen for a number of reasons. Firstly the structures represent typical orthopaedic bone in-growth structures (the primary concern of a larger project) and secondly the parts being produced can only be built within a narrow process window owing to physical and mechanical requirements of the parts. Consequently, if any minor disturbance to the operating settings or environment is encountered, the affect of such disturbances will be amplified and may cause build failure or tolerance issues",
            " It is clear that in comparison to the original as delivered design, significant improvement in the uniformity of the flow has been achieved. The flow is by no means fully optimised as a variation of up to 50% in the flow velocity across the process area was observed. To verify the results of the CFD analysis, a new gas rail was manufactured using Fused Deposition Modelling (FDM) and installed in the chamber. Tests showed that the samples at the back of the substrate (A1, B1, C1, D1 and E1, as shown in Fig. 2) were under poor flow conditions. This was later confirmed by the examination of a statistically significant number of samples built using this design gas rail. As uniformity of gas flow was shown to significantly affect the strength of the parts another iteration of gas rail design was undertaken to further improve the gas flow, this became the gas rail for the completed condition and can be seen in Fig. 7. This rail design includes a series of individual adjustable nozzles and a diffuser (offset diamond formation, four diamonds \u00d7 four rows) at the inlet of the gas flow rail to ensure even distribution to the channels"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.12-1.png",
        "caption": "FIGURE 8.12. A Watt suspension mechanisms with a Panhard arm.",
        "texts": [],
        "surrounding_texts": [
            "The suspension is what links the wheels to the vehicle body and allows relative motion. This chapter covers the suspension mechanisms, and discusses the possible relative motions between the wheel and the vehicle body. The wheels, through the suspension linkage, must propel, steer, and stop the vehicle, and support the associated forces. 8.1 Solid Axle Suspension The simplest way to attach a pair of wheels to a vehicle is to mount them at opposite ends of a solid axle, such as the one that is shown in Figure 8.1. The solid axle must be attached to the body such that an up and down motion in the z-direction, as well as a roll rotation about the x-axis, is possible. So, no forward and lateral translation, and also no rotation about the axle and the z-axis, is allowed. There are many combinations of links and springs that can provide the kinematic and dynamic requirements. The simplest design is to clamp the axle to the middle of two leaf springs with their ends tied or shackled to the vehicle frame as shown schematically in Figure 8.1. A side view of a multi-leaf spring and solid axle is shown in Figure 8.2. A suspension with a solid connection between the left and right wheels is called dependent suspension. 456 8. Suspension Mechanisms The performance of a solid axle with leaf springs suspension can be improved by adding a linkage to guide the axle kinematically and provide dynamic support to carry the non z-direction forces. The solid axle with leaf spring combination came to vehicle industry from horse-drawn vehicles. Example 299 Hotchkiss drive. When a live solid axle is connected to the body with nothing but two leaf springs, it is called the Hotchkiss drive, which is the name of the car that used it first. The main problems of a Hotchkiss drive, which is shown in Figure 8.2, are locating the axle under lateral and longitudinal forces, and having a low mass ratio \u03b5 = ms/mu, where ms is the sprung mass and mu is the unsprung mass. Sprung mass refers to all masses that are supported by the spring, such as vehicle body. Unsprung mass refers to all masses that are attached to and not supported by the spring, such as wheel, axle, or brakes. Example 300 Leaf spring suspension and flexibility problem. The solid axle suspension systems with longitudinal leaf springs have many drawbacks. The main problem lies in the fact that springs themselves act as locating members. Springs are supposed to flex under load, but their flexibility is needed in only one direction. However, it is the nature of leaf springs to twist and bend laterally and hence, flex also in planes other than the tireplane. Leaf springs are not suited for taking up the driving and braking traction forces. These forces tend to push the springs into an S-shaped profile, as shown in Figure 8.3. The driving and braking flexibility of leaf springs, generates a negative caster and increases instability. Long springs provide better ride. However, long sprigs exaggerate their bending and twisting under different load conditions. Example 301 Leaf spring suspension and flexibility solution. To reduce the effect of a horizontal force and S-shaped profile appearance in a solid axle with leaf springs, the axle may be attached to the chassis by a longitudinal bar as Figure 8.4(a) shows. Such a bar is called an anti-tramp 8. Suspension Mechanisms 457 bar, and the suspension is the simplest cure for longitudinal problems of a Hotchkiss drive. A solid axle with an anti-tramp bar may be kinematically approximated by a four-bar linkage, as shown in Figure 8.4(b). Although an anti-tramp bar may control the shape of the leaf spring, it introduces a twisting angle problem when the axle is moving up and down, as shown in Figure 8.5. Twisting the axle and the wheel about the axle is called caster. The solid axle is frequently used to help keeping the wheels perpendicular to the road. Example 302 Leaf spring location problem. The front wheels need room to steer left and right. Therefore, leaf springs cannot be attached close to the wheel hubs, and must be placed closer to the middle of the axle. That gives a narrow spring-base, which means that a small side force can sway or tilt the body relative to the axle through a considerable roll angle due to weight transfer. This is uncomfortable for the vehicle passengers, and may also produce unwanted steering. The solid axle positively prevents the camber change by body roll. The wheels remain upright and hence, do not roll on a side. However, a solid axle shifts laterally from its static plane and its center does not remain on the vehicle\u2019s longitudinal axis under a lateral force. 458 8. Suspension Mechanisms A solid axle produces bump-camber when single-wheel bump occurs. If the right wheel goes over a bump, the axle is raised at its right end, and that tilts the left wheel hub, putting the left wheel at a camber angle for the duration of deflection. Example 303 Triangular linkage. A triangulated linkage, as shown in Figure 8.6, may be attached to a solid axle to provide lateral and twist resistance during acceleration and braking. Example 304 Panhard arm. High spring rate is a problem of leaf springs. Reducing their stiffness by narrowing them and using fewer leaves, reduces the lateral stiffness and increases the directional stability of the suspension significantly. A Panhard arm is a bar that attaches a solid axle suspension to the chassis laterally. Figure 8.7 illustrates a solid axle and a Panhard arm to guide the axle. Figure 8.8 shows a triangular linkage and a Panhard arm combination for guiding a solid axle. A double triangle mechanism, as shown in Figure 8.9, is an alternative design to guide the axle and support it laterally. Example 305 Straight line linkages. There are many mechanisms that can provide a straight line motion. The simplest mechanisms are four-bar linkages with a coupler point moving straight. Some of the most applied and famous linkages are shown in Figure 8.10. By having proper lengths, the Watt, Robert, Chebyshev, and Evance linkages can make the coupler point C move on a straight line vertically. Such a mechanism and straight motion may be used to guide a solid axle. Two Watt suspension mechanisms with a Panhard arm are shown in Figures 8.11 and 8.12. 460 8. Suspension Mechanisms 8. Suspension Mechanisms 461 Figures 8.13, 8.14, and 8.15 illustrate three combinations of Robert suspension linkages equipped with a Panhard arm. Example 306 Solid axle suspension and unsprung mass problem. A solid axle is counted as an unsprung member, and hence, the unsprung mass is increased where using solid axle suspension. A heavy unsprung mass ruins both, the ride and handling of a vehicle. Lightening the solid axle makes it weaker and increases the most dangerous problem in vehicles: axle breakage. The solid axle must be strong enough to make sure it will not break under any loading conditions at any age. As a rough estimate, 90% of the leaf spring mass may also be counted as unsprung mass, which makes the problem worse. The unsprung mass problem is worse in front, and it is the main reason that they are no longer used in street cars. However, front solid axles are still common on trucks and buses. These are heavy vehicles and solid axle 8. Suspension Mechanisms 463 464 8. Suspension Mechanisms suspension does not reduce the mass ratio \u03b5 = ms/mu very much. When a vehicle is rear-wheel-drive and a solid axle suspension is used in the back, the suspension is called live axle. A live axle is a casing that contains a differential, and two drive shafts. The drive shafts are connected to the wheel hubs. A live axle can be three to four times heavier than a dead I-beam axle. It is called live axle because of rotating gears and shafts inside the axle. Example 307 Solid axle and coil spring. To decrease the unsprung mass and increase vertical flexibility of solid axle suspensions, it is possible to equip them with coil springs. A sample of a solid axle suspension with coil spring is shown in Figure 8.16. The suspension mechanism is made of four longitudinal bars between the axle and chassis. The springs may have some lateral or longitudinal angle to introduce some lateral or longitudinal compliance. Example 308 De Dion axle. When a solid axle is a dead axle with no driving wheels, the connecting beam between the left and right wheels may have different shapes to do different jobs, usually to give the wheels independent flexibility. We may also modify the shape of a live axle to attach the differential to the chassis and reduce the unsprung mass. De Dion design is a modification of a beam axle that may be used as a dead axle or to attach the differential to the chassis and transfer the driving power to the drive wheels by employing universal joints and split shafts. Figure 8.17 illustrates a De Dion suspension. 8.2 Independent Suspension Independent suspensions is introduced to let a wheel to move up and down without affecting the opposite wheel. There are many forms and designs of independent suspensions. However, double A-arm and McPherson strut suspensions are the simplest and the most common designs. Figure 8.18 illustrates a sample of a double A-arm and Figure 8.19 shows a McPherson suspension. Kinematically, a double A-arm suspension mechanism is a four-bar linkage with the chassis as the ground link, and coupler as the wheel carrying link. A McPherson suspension is an inverted slider mechanism that has the chassis as the ground link and the coupler as the wheel carrying link. A double A-arm and a McPherson suspension mechanism on the left and right wheels are schematically shown in Figures 8.20 and 8.21 respectively. Double A-arm, is also called double wishbone, or short/long arm suspension. McPherson also may be written as MacPherson. Example 309 Double A-arm suspension and spring position. Consider a double A-arm suspension mechanism. The coil spring may be between the lower arm and the chassis, as shown in Figure 8.18. It is also possible to install the spring between the upper arm and the chassis, or between the upper and lower arms. In either case, the lower or the upper arm, which supports the spring, is made stronger and the other arm acts as a connecting arm. Example 310 Multi-link suspension mechanism. When the two side bars of an A-arm are attached to each other with a joint, as shown in Figure 8.22, then the double A-arm is called a multi-link mechanism. A multi-link mechanism is a six-bar mechanism that may have a better coupler motion than a double A-arm mechanism. However, multi- 466 8. Suspension Mechanisms 8. Suspension Mechanisms 467 link suspensions are more expensive, less reliable, and more complicated compare to a double A-arm four-bar linkage. There are vehicles with more than six-link suspension with possibly better kinematic performance. Example 311 Swing arm suspension. An independent suspension may be as simple as a triangle shown in Figure 8.23. The base of the triangle is jointed to the chassis and the wheel to the tip point. The base of the triangle is aligned with the longitudinal axis of the vehicle. Such a suspension mechanism is called a swing axle or swing arm. The variation in camber angle for a swing arm suspension is maximum, compared to the other suspension mechanisms. Example 312 Trailing arm suspension. Figure 8.24 illustrates a trailing arm suspension that is a longitudinal arm with a lateral axis of rotation. The camber angle of the wheel, supported by a trailing arm, will not change during the up and down motion. Trailing arm suspension has been successfully using in a variety of frontwheel-drive vehicles, to suspend their rear wheels. Example 313 Semi-trailing arm Semi-trailing arm suspension, as shown in Figure 8.25, is a compromise between the swing arm and trailing arm suspensions. The joint axis may have any angle, however an angle not too far from 45 deg is more applied. Such suspensions have acceptable camber angle change, while they can handle both, the lateral and longitudinal forces. Semi-trailing design has successfully applied to a series of rear-wheel-drive cars for several decades. Example 314 Antiroll bar and roll stiffness. Coil springs are used in vehicles because they are less stiff with better 470 8. Suspension Mechanisms ride comfort compared to leaf springs. Therefore, the roll stiffness of the vehicle with coil springs is usually less than in vehicles with leaf springs. To increase the roll stiffness of such suspensions, an antiroll bar must be used. Leaf springs with reduced layers, uni-leaf, trapezoidal, or nonuniform thickness may also need an antiroll bar to compensate for their reduced roll stiffness. The antiroll bar is also called a stabilizer. Figure 8.26 illustrates an anti-roll bare attached to a solid axle with coil springs. Example 315 Need for longitudinal compliance. A bump is an obstacle on the road that opposes the forward motion of a wheel. When a vehicle goes over a bump, the first action is a force that tends to push the wheel backward relative to the rest of the vehicle. So, the lifting force has a longitudinal component, which will be felt inside the vehicle unless the suspension system has horizontal compliance. There are situations in which the horizontal component of the force is even higher than the vertical component. Leaf springs can somewhat absorb this horizontal force by flattening out and stretching the distance from the forward spring anchor and the axle. Such a stretch is usually less than 1/2 in \u2248 1 cm. 8.3 Roll Center and Roll Axis The roll axis is the instantaneous line about which the body of a vehicle rolls. Roll axis is found by connecting the roll center of the front and rear suspensions of the vehicle. Assume we cut a vehicle laterally to disconnect the front and rear half of the vehicle. Then, the roll center of the front or rear suspension is the instantaneous center of rotation of the body with respect to the ground. 8. Suspension Mechanisms 471 472 8. Suspension Mechanisms Figure 8.27 illustrates a sample of the front suspensions of a car with a double A-arm mechanism. To find the roll center of the body with respect to the ground, we analyze the two-dimensional kinematically equivalent mechanism shown in Figure 8.28. The center of tireprint is the instant center of rotation of the wheel with respect to the ground, so the wheels are jointed links to the ground at their center of tireprints. The instant center I18 is the roll center of the body with respect to the ground. To find I18, we apply the Kennedy theorem and find the intersection of the line I12I28 and I13I38 as shown in Figure 8.29. The point I28 and I38 are the instant center of rotation for the wheels with respect to the body. The instant center of rotation of a wheel with respect to the body is called suspension roll center. So, to find the roll center of the front or rear half of a car, we should determine the suspension roll centers, and find the intersection of the lines connecting the suspension roll centers to the center of their associated tireprints. The Kennedy theorem states that the instant center of every three relatively moving objects are colinear. Example 316 McPherson suspension roll center. A McPherson suspension is an inverted slider crank mechanism. The instant centers of an example of inverted slider crank mechanism are shown in Figure 8.30. In this figure, the point I12 is the suspension roll center, which is the instant center of rotation for the wheel link number 2 with respect to the chassis link number 1. A car with a McPherson suspension system is shown in Figure 8.31. The kinematic equivalent mechanism is depicted in Figure 8.32. Suspension roll centers along with the body roll center are shown in Figure 8.33. To 8. Suspension Mechanisms 473 find the roll center of the front or rear half of a car, we determine each suspension roll center and then find the intersection of the lines connecting the suspension roll centers to the center of the associated tireprint. Example 317 Roll center of double A-arm suspension. The roll center of an independent suspension such as a double A-arm can be internal or external. The kinematic model of a double A-arm suspension for the front left wheel of a car is illustrated in Figure 8.34. The suspension roll center in Figure 8.34(a) is internal, and in Figure 8.34(b) is external. An internal suspension roll center is toward the vehicle body, while an external suspension roll center goes away from the vehicle body. A suspension roll center may be on, above, or below the road surface, as shown in Figure 8.35(a)-(c) for an external suspension roll center. When the suspension roll center is on the ground, above the ground, or below the ground, the vehicle roll center would be on the ground, below the ground, and above the ground, respectively. Example 318 F Camber variation of double A-arm suspension. When a wheel moves up and down with respect to the vehicle body, de- 474 8. Suspension Mechanisms 8. Suspension Mechanisms 475 pending on the suspension mechanism, the wheel may camber. Figure 8.36 illustrates the kinematic model for a double A-arm suspension mechanism. The mechanism is equivalent to a four-bar linkage with the ground link as the vehicle chasis. The wheel is always attached to a coupler point C of the mechanism. We set a local suspension coordinate frame (x, y) with the x-axis indicating the ground link MN . The x-axis makes a constant angle \u03b80 with the vertical direction. The suspension machanism has a length a for the upper A-arm, b for the coupler link, c for the lower A-arm, and d for the ground link. The configuration of the suspension is determined by the angles \u03b82, \u03b83, and \u03b84, all measured from the positive direction of the x-axis. When the suspension is at its equilibrium position, the links of the double A-arm suspension make initial angles \u03b820 \u03b830, and \u03b840 with the x-axis. The equilibrium position of a suspension is called the rest position. To determine the camber angle during the fluctuation of the wheel, we should determine the variation of the coupler angle \u03b83, as a function of vertical motion z of the coupler point C. Using \u03b82 as a parameter, we can find the coordinates (xC , yC) of the 476 8. Suspension Mechanisms coupler point C in the suspension coordinate frame (x, y) as xC = a cos \u03b82 + e cos (p\u2212 q + \u03b1) (8.1) yC = a sin \u03b82 + e sin (p\u2212 q + \u03b1) (8.2) where, q = tan\u22121 a sin \u03b82 d\u2212 a cos \u03b82 (8.3) p = tan\u22121 q 4b2f2 \u2212 (b2 + f2 \u2212 c2) 2 b2 + f2 \u2212 c2 (8.4) f = p a2 + d2 \u2212 2ad cos \u03b82. (8.5) The position vector of the coupler point is uC uC = xC \u0131\u0302+ yC j\u0302 (8.6) and the unit vector in the z-direction is u\u0302z = \u2212 cos \u03b80\u0131\u0302\u2212 sin \u03b80j\u0302. (8.7) Therefore, the displacement z in terms of xC and yC is: z = uC \u00b7 u\u0302z = \u2212xC cos \u03b80 \u2212 yC sin \u03b80 (8.8) 8. Suspension Mechanisms 477 The initial coordinates of the coupler point C and the initial value of z are: xC0 = a cos \u03b820 + e cos (p0 \u2212 q0 + \u03b1) (8.9) yC0 = a sin \u03b820 + e sin (p0 \u2212 q0 + \u03b1) (8.10) z0 = \u2212xC0 cos \u03b80 \u2212 yC0 sin \u03b80 (8.11) and hence, the vertical displacement of the wheel center can be calculated by h = z \u2212 z0 (8.12) The initial angle of the coupler link with the vertical direction is \u03b80 \u2212 \u03b830. Therefore, the camber angle of the wheel would be \u03b3 = (\u03b80 \u2212 \u03b83)\u2212 (\u03b80 \u2212 \u03b830) = \u03b830 \u2212 \u03b83 (8.13) The angle of the coupler link with the x-direction is equal to \u03b83 = 2 tan \u22121 \u00c3 \u2212E \u00b1 \u221a E2 \u2212 4DF 2D ! (8.14) where, D = J5 \u2212 J1 + (1 + J4) cos \u03b82 (8.15) E = \u22122 sin \u03b82 (8.16) F = J5 + J1 \u2212 (1\u2212 J4) cos \u03b82 (8.17) and J1 = d a (8.18) J2 = d c (8.19) J3 = a2 \u2212 b2 + c2 + d2 2ac (8.20) J4 = d b (8.21) J5 = c2 \u2212 d2 \u2212 a2 \u2212 b2 2ab . (8.22) Substituting (8.14) and (8.13), and then, eliminating \u03b82 between (8.13) and (8.8) provides the relationship between the vertical motion of the wheel, z, and the camber angle \u03b3. 478 8. Suspension Mechanisms Example 319 F Camber angle and wheel fluctuations. Consider the double A-arm suspension that is shown in Figure 8.36. The dimensions of the equivalent kinematic model are: a = 22.4 cm b = 22.1 cm c = 27.3 cm d = 17.4 cm \u03b80 = 24.3 deg (8.23) The coupler point C is at: e = 14.8 cm \u03b1 = 54.8 deg (8.24) If the angle \u03b82 at the rest position is at \u03b820 = 121.5 deg (8.25) then the initial angle of the other links are: \u03b830 = 18.36 deg \u03b840 = 107.32 deg (8.26) At the rest position, the coupler point is at: xC0 = \u221222.73 cm yC0 = 9.23 cm z0 = 16.92 cm (8.27) We may calculate h and \u03b3 by varying the parameter \u03b82. Figure 8.37 illustrates h as a function of the camber angle \u03b3. For this suspension mechanism, the wheel gains a positive camber when the wheel moves up, and gains a negative camber when the it moves down. The mechanism is shown in Figure 8.38, when the wheel is at the rest position and has a positive or a negative displacement. 8.4 F Car Tire Relative Angles There are four major wheel alignment parameters that affect vehicle dynamics: toe, camber, caster, and trust angle. 8. Suspension Mechanisms 479 8.4.1 F Toe When a pair of wheels is set so that their leading edges are pointed toward each other, the wheel pair is said to have toe-in. If the leading edges point away from each other, the pair is said to have toe-out. Toe-in and toe-out front wheel configurations of a car are illustrated in Figure 8.39. The amount of toe can be expressed in degrees of the angle to which the wheels are not parallel. However, it is more common to express the toein and toe-out as the difference between the track widths as measured at the leading and trailing edges of the tires. Toe settings affect three major performances: tire wear, straight-line stability, and corner entry handling. For minimum tire wear and power loss, the wheels on a given axle of a car should point directly ahead when the car is running in a straight line. Excessive toe-in causes accelerated wear at the outboard edges of the tires, while too much toe-out causes wear at the inboard edges. Toe-in increases the directional stability of the vehicle, and toe-out increases the steering response. Hence, a toe-in setting makes the steering function lazy, while a toe-out makes the vehicle unstable. With four wheel independent suspensions, the toe may also be set at the rear of the car. Toe settings at the rear have the same effect on wear, directional stability, and turn-in as they do on the front. However, we usually do not set up a rear-drive race car toed out in the rear, because of excessive instability. When driving torque is applied to the wheels, they pull themselves forward and try to create toe-in. Furthermore, when pushed down the road, a non-driven wheel or a braking wheel will tend to toe-out. Example 320 Toe-in and directional stability. Toe settings have an impact on directional stability. When the steering 480 8. Suspension Mechanisms 8. Suspension Mechanisms 481 wheel is centered, toe-in causes the wheels to tent to move along paths that intersect each other in front of the vehicle. However, the wheels are in balance and no turn results. Toe-in setup can increase the directional stability caused by little steering fluctuations and keep the car moving straight. Steering fluctuations may be a result of road disturbances. If a car is set up with toe-out, the front wheels are aligned so that slight disturbances cause the wheel pair to assume rolling directions that approach a turn. Therefore, toe-out encourages the initiation of a turn, while toe-in discourages it. Toe-out makes the steering quicker. So, it may be used in vehicles for a faster response. The toe setting on a particular car becomes a trade-off between the straight-line stability afforded by toe-in and the quick steering response by toe-out. Toe-out is not desirable for street cars, however, race car drivers are willing to drive a car with a little directional instability, for sharper turn-in to the corners. So street cars are generally set up with toe-in, while race cars are often set up with toe-out. Example 321 Toe-in and toe-out in the front and rear axles. Front toe-in: slower steering response, more straight-line stability, greater wear at the outboard edges of the tires. Front toe-zero: medium steering response, minimum power loss, minimum tire wear. Front toe-out: quicker steering response, less straight-line stability, greater wear at the inboard edges of the tires. Rear toe-in: straight-line stability, traction out of the corner, more steerability, higher top speed. 482 8. Suspension Mechanisms 8.4.2 F Caster Angle Caster is the angle to which the steering pivot axis is tilted forward or rearward from vertical, as viewed from the side. Assume the wheel is straight to have the body frame and the wheel frame coincident. If the steering axis is turned about the wheel yw-axis then the wheel has positive caster. If the steering axis is turned about the wheel \u2212yw-axis, then the wheel has negative caster. Positive and negative caster configurations on the front wheel of a car are shown in Figure 8.40. Negative caster aids in centering the steering wheel after a turn and makes the front tires straighten quicker. Most street cars are made with 4\u22126 deg negative caster. Negative caster tends to straighten the wheel when the vehicle is traveling forward, and thus is used to enhance straight-line stability. Example 322 Negative caster of shopping carts. The steering axis of a shopping cart wheel is set forward of where the wheel contacts the ground. As the cart is pushed forward, the steering axis pulls the wheel along, and because the wheel drags along the ground, it falls directly in line behind the steering axis. The force that causes the wheel to follow the steering axis is proportional to the distance between the steering axis and the wheel-to-ground contact point, if the caster is small. This distance is referred to as trail. The cars\u2019 steering axis intersects the ground at a point in front of the tireprint, and thus the same effect as seen in the shopping cart casters is achieved. While greater caster angles improves straight-line stability, they also cause an increase in steering effort. Example 323 Characteristics of caster in front axle. Zero castor provides: easy steering into the corner, low steering out of the corner, low straight-line stability. 8. Suspension Mechanisms 483 Negative caster provides: low steering into the corner, easy steering out of the corner, more straight-line stability, high tireprint area during turn, good turn-in response, good directional stability, good steering feel. When a castered wheel rotates about the steering axis, the wheel gains camber. This camber is generally favorable for cornering. 8.4.3 F Camber Camber is the angle of the wheel relative to vertical line to the road, as viewed from the front or the rear of the car. Figure 8.41 illustrates the wheel number 1 of a vehicle. If the wheel leans in toward the chassis, it is called negative camber and if it leans away from the car, it is called positive camber. The cornering force that a tire can develop is highly dependent on its angle relative to the road surface, and so wheel camber has a major effect on the road holding of a car. A tire develops its maximum lateral force at a small camber angle. This fact is due to the contribution of camber thrust, which is an additional lateral force generated by elastic deformation as the tread rubber pulls through the tire/road interface. To optimize a tire\u2019s performance in a turn, the suspension should provide a slight camber angle in the direction of rotation. As the body rolls in a turn, the suspension deflects vertically. The wheel is connected to the chassis by suspension mechanism, which must rotate to allow for the wheel deflection. Therefore, the wheel can be subject to large camber changes as the suspension moves up and down. So, the more the wheel must deflect from its static position, the more difficult it is to maintain an ideal camber angle. Thus, the relatively large wheel travel and soft roll stiffness needed to provide a smooth ride in passenger cars presents a difficult design challenge, while the small wheel travel and high roll stiffness inherent in racing cars reduces the problem. Example 324 Castor versus camber. Camber doesn\u2019t improve turn-in as the positive caster does. Camber is not generally good for tire wear. Camber in one wheel does not improve directional stability. Camber adversely affects braking and acceleration efforts. 8.4.4 F Trust Angle The trust angle \u03c5 is the angle between vehicle\u2019s centerline and perpendicular to the rear axle. It compares the direction that the rear axle is aimed with the centerline of the vehicle. A nonzero angle configuration is shown in Figure 8.42. Zero angle confirms that the rear axle is parallel to the front axle, and the wheelbase on both sides of the vehicle are the same. A reason for nonzero 484 8. Suspension Mechanisms 8. Suspension Mechanisms 485 trust angle would have unequal toe-in or toe-out on both sides of the axle. Example 325 Torque reaction. There are two kinds of torque reactions in rear-whel-drive: 1\u2212 the reaction of the axle housing to rotate in the opposite direction of the crown wheel rotation, and 2\u2212 the reaction of axle housing to spin about its own center, opposite to the direction of pinion\u2019s rotation. The first reaction leads to a lifting force in the differential causing a wind-up in springs. The second reaction leads to a lifting force on the right wheels. 8.5 Suspension Requirements and Coordinate Frames The suspension mechanism should allow a relative motion between the wheel and the vehicle body. The relative motions are needed to pass the road irregularities and steering. To function properly, a suspension mechanism should have some kinematic and dynamics requirements. 8.5.1 Kinematic Requirements To express the motions of a wheel, we attach a wheel coordinate system W (oxwywzw) to the center of the wheel. A wheel, as a rigid body, has six degrees-of-freedom with respect to the vehicle body: three translations and three rotations, as shown in Figure 8.43. 486 8. Suspension Mechanisms The axes xw, yw, and zw indicate the direction of forward, lateral, and vertical translations and rotations. In the position shown in the figure, the rotation about the xw-axis is the camber angle, about the yw-axis is the spin, and about the zw-axis is the steer angle. Consider a non-steerable wheel. Translation in zw-direction and spin about the yw-axis are the only two DOF allowed for such a wheel. So, we need to take four DOF. If the wheel is steerable, then translation in the zw-direction, spin about the yw-axis, and steer rotation about the zw-axis are the three DOF allowed. So, we must take three DOF of a steerable wheel. Kinematically, non-steerable and steerable wheels should be supported as shown in Figures 8.44 and 8.45 respectively. Providing the required freedom, as well as eliminating the taken DOF, are the kinematic requirements of a suspension mechanism. 8.5.2 Dynamic Requirements Wheels should be able to propel, steer, and stop the vehicle. So, the suspension system must transmit the driving traction and deceleration braking forces between the vehicle body and the ground. The suspension members must also resist lateral forces acting on the vehicle. Hence, the wheel suspension system must make the wheel rigid for the taken DOF. However, there must also be some compliance members to limit the untaken DOF. The most important compliant members are spring and dampers to provide returning and resistance forces in the z-direction. 8. Suspension Mechanisms 487 8.5.3 Wheel, wheel-body, and tire Coordinate Frames Three coordinate frames are employed to express the orientation of a tire and wheel with respect to the vehicle: the wheel frame W , wheel-body frame C, and tire frame T . A wheel coordinate frame W (xw, yw, zw) is attached to the center of a wheel. It follows every translation and rotation of the wheel except the spin. Hence, the xw and zw axes are always in the tire-plane, while the yw-axis is always along the spin axis. A wheel coordinate frame is shown in Figure 8.43. When the wheel is straight and the W frame is parallel to the vehicle coordinate frame, we attach a wheel-body coordinate frame C (xc, yc, zc) at the center of the wheel parallel to the vehicle coordinate axes. The wheelbody frame C is motionless with respect to the vehicle coordinate and does not follow any motion of the wheel. The tire coordinate frame T (xt, yt, zt) is set at the center of the tireprint. The zt-axis is always perpendicular to the ground. The xt-axis is along the intersection line of the tire-plane and the ground. The tire frame does not follows the spin and camber rotations of the tire however, it follows the steer angle rotation about the zc-axis. Figure 8.46 illustrates a tire and a wheel coordinate frames. Example 326 Visualization of the wheel, tire, and wheel-body frames. Figure 8.47 illustrates the relative configuration of a wheel-body frame C, a tire frame T , and a wheel frame W . If the steering axis is along the zc-axis then, the rotation of the wheel about the zc-axis is the steer angle \u03b4. Rotation about the xt-axis is the camber angle \u03b3. 488 8. Suspension Mechanisms Generally speaking, the steering axis may have any angle and may go through any point of the ground plane. Example 327 Wheel to tire coordinate frame transformation. If TdW indicates the T -expression of the position vector of the wheel frame origin relative to the tire frame origin, then having the coordinates of a point P in the wheel frame, we can find its coordinates in the tire frame using the following equation. T rP = TRW W rP + TdW (8.28) If W rP indicates the position vector of a point P in the wheel frame, W rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 (8.29) then the coordinates of the point P in the tire frame T rP are T rP = TRW W rP + Td = TRW W rP + TRW W T dW = \u23a1\u23a3 xP yP cos \u03b3 \u2212Rw sin \u03b3 \u2212 zP sin \u03b3 Rw cos \u03b3 + zP cos \u03b3 + yP sin \u03b3 \u23a4\u23a6 (8.30) 8. Suspension Mechanisms 489 where, WT dW is the W -expression of the position vector of the wheel frame in the tire frame, Rw is the radius of the tire, and TRW is the rotation matrix to go from the wheel frame W to the tire frame T . TRW = \u23a1\u23a3 1 0 0 0 cos \u03b3 \u2212 sin \u03b3 0 sin \u03b3 cos \u03b3 \u23a4\u23a6 (8.31) W T dW = \u23a1\u23a3 0 0 Rw \u23a4\u23a6 . (8.32) As an example, the center of the wheel W rP = W ro = 0 is the origin of the wheel frame W , that is at T ro = TdW = TRW W T dW = \u23a1\u23a3 0 \u2212Rw sin \u03b3 Rw cos \u03b3 \u23a4\u23a6 (8.33) in the tire coordinate frame T . 490 8. Suspension Mechanisms Example 328 F Tire to wheel coordinate frame transformation. If rP indicates the position vector of a point P in the tire coordinate frame, T rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 (8.34) then the position vector W rP of the point P in the wheel coordinate frame is W rP = WRT T rP \u2212 W T dW (8.35) = \u23a1\u23a3 xP yP cos \u03b3 + zP sin \u03b3 zP cos \u03b3 \u2212Rw \u2212 yP sin \u03b3 \u23a4\u23a6 because WRT = \u23a1\u23a3 1 0 0 0 cos \u03b3 sin \u03b3 0 \u2212 sin \u03b3 cos \u03b3 \u23a4\u23a6 (8.36) WdT = \u23a1\u23a3 0 0 Rw \u23a4\u23a6 (8.37) and we may multiply both sides of Equation (8.28) by TRT W to get TRT W T rP = W rP + TRT W TdW (8.38) = W rP + W T dW W rP = WRT T rP \u2212 W T dW . (8.39) As an example, the center of tireprint in the wheel frame is at W rP = \u23a1\u23a3 1 0 0 0 cos \u03b3 \u2212 sin \u03b3 0 sin \u03b3 cos \u03b3 \u23a4\u23a6T \u23a1\u23a3 0 0 0 \u23a4\u23a6\u2212 \u23a1\u23a3 0 0 Rw \u23a4\u23a6 = \u23a1\u23a3 0 0 \u2212Rw \u23a4\u23a6 . (8.40) Example 329 F Wheel to tire homogeneous transformation matrices. The transformation from the wheel to tire coordinate frame may also be expressed by a 4\u00d7 4 homogeneous transformation matrix TTW , T rP = TTW W rP = \u2219 TRW TdW 0 1 \u00b8 W rP (8.41) where TTW = \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 cos \u03b3 \u2212 sin \u03b3 \u2212Rw sin \u03b3 0 sin \u03b3 cos \u03b3 Rw cos \u03b3 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 . (8.42) 8. Suspension Mechanisms 491 The corresponding homogeneous transformation matrix WTT from the tire to wheel frame would be WTT = \u2219 WRT WdT 0 1 \u00b8 = \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 cos \u03b3 \u2212 sin \u03b3 0 0 sin \u03b3 cos \u03b3 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 . (8.43) It can be checked that WTT = TT\u22121W , using the inverse of a homogeneous transformation matrix rule. TT\u22121W = \u2219 TRW TdW 0 1 \u00b8\u22121 = \u2219 TRT W \u2212 TRT W TdW 0 1 \u00b8 = \u2219 WRT \u2212WRT TdW 0 1 \u00b8 (8.44) Example 330 F Tire to wheel-body frame transformation. The origin of the tire frame is at CdT in the wheel-body frame. CdT = \u23a1\u23a3 0 0 \u2212Rw \u23a4\u23a6 (8.45) The tire frame can steer about the zc-axis with respect to the wheel-body frame. The associated rotation matrix is CRT = \u23a1\u23a3 cos \u03b4 \u2212 sin \u03b4 0 sin \u03b4 cos \u03b4 0 0 0 1 \u23a4\u23a6 (8.46) Therefore, the transformation between the tire and wheel-body frames can be expressed by Cr = CRT T r+ CdT (8.47) or equivalently, by a homogeneous transformation matrix CTT . CTT = \u2219 CRT CdT 0 1 \u00b8 = \u23a1\u23a2\u23a2\u23a3 cos \u03b4 \u2212 sin \u03b4 0 0 sin \u03b4 cos \u03b4 0 0 0 0 1 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 (8.48) As an example, the wheel-body coordinates of the point P on the tread of a 492 8. Suspension Mechanisms negatively steered tire at the position shown in Figure 8.48, are: Cr = CTT T rP = \u23a1\u23a2\u23a2\u23a3 cos\u2212\u03b4 \u2212 sin\u2212\u03b4 0 0 sin\u2212\u03b4 cos\u2212\u03b4 0 0 0 0 1 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 \u23a1\u23a2\u23a2\u23a3 Rw 0 Rw 1 \u23a4\u23a5\u23a5\u23a6 = \u23a1\u23a2\u23a2\u23a3 Rw cos \u03b4 \u2212Rw sin \u03b4 0 1 \u23a4\u23a5\u23a5\u23a6 (8.49) The homogeneous transformation matrix for tire to wheel-body frame TTC is: TTC = CT\u22121T = \u2219 CRT CdT 0 1 \u00b8\u22121 = \u2219 CRT T \u2212CRT T CdT 0 1 \u00b8 = \u2219 CRT T \u2212 T CdT 0 1 \u00b8 = \u23a1\u23a2\u23a2\u23a3 cos \u03b4 sin \u03b4 0 0 \u2212 sin \u03b4 cos \u03b4 0 0 0 0 1 Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 (8.50) Example 331 F Cycloid. Assume that the wheel in Figure 8.48 is turning with angular velocity \u03c9 and has no slip on the ground. If the point P is at the center of the tireprint 8. Suspension Mechanisms 493 when t = 0, MrP = \u23a1\u23a3 0 0 \u2212Rw \u23a4\u23a6 (8.51) then we can find its position in the wheel frame at a time t by employing another coordinate frame M . The frame M is called the rim frame and is stuck to the wheel at its center. Because of spin, the M frame turns about the yw-axis, and therefore, the rotation matrix to go from the rim frame to the wheel frame is: WRM = \u23a1\u23a3 cos\u03c9t 0 sin\u03c9t 0 1 0 \u2212 sin\u03c9t 0 cos\u03c9t \u23a4\u23a6 (8.52) So the coordinates of P in the wheel frame are: W rP = WRM MrP = \u23a1\u23a3 \u2212Rw sin t\u03c9 0 \u2212Rw cos t\u03c9 \u23a4\u23a6 (8.53) The center of the wheel is moving with speed vx = Rw\u03c9 and it is at Gr =\u00a3 vxt 0 Rw \u00a4 in the global coordinate frame G on the ground. Hence, the coordinates of point P in the global frame G, would be GrP = W rP + \u23a1\u23a3 vxt 0 Rw \u23a4\u23a6 = \u23a1\u23a3 Rw (\u03c9t\u2212 sin t\u03c9) 0 Rw (1\u2212 cos t\u03c9) \u23a4\u23a6 . (8.54) The path of motion of point P in the (X,Z)-plane can be found by eliminating t between X and Z coordinates. However, it is easier to expressed the path by using \u03c9t as a parameter. Such a path is called cycloid. In general case, point P can be at any distance from the center of the rim frame. If the point is at a distance d 6= Rw, then its path of motion is called the trochoid. A trochoid is called a curtate cycloid when d < Rw and a prolate cycloid when d > Rw. Figure 8.49(a)-(c) illustrate a cycloid, curtate cycloid, and prolate cycloid respectively. Example 332 F Wheel to wheel-body frame transformation. The homogeneous transformation matrix CTW to go from the wheel frame 494 8. Suspension Mechanisms to the wheel-body frame can be found by combined transformation. CTW = CTT TTW (8.55) = \u23a1\u23a2\u23a2\u23a3 c\u03b4 \u2212s\u03b4 0 0 s\u03b4 c\u03b4 0 0 0 0 1 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 c\u03b3 \u2212s\u03b3 \u2212Rw sin \u03b3 0 s\u03b3 c\u03b3 Rw cos \u03b3 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 = \u23a1\u23a2\u23a2\u23a3 cos \u03b4 \u2212 cos \u03b3 sin \u03b4 sin \u03b3 sin \u03b4 Rw sin \u03b3 sin \u03b4 sin \u03b4 cos \u03b3 cos \u03b4 \u2212 cos \u03b4 sin \u03b3 \u2212Rw cos \u03b4 sin \u03b3 0 sin \u03b3 cos \u03b3 Rw cos \u03b3 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 If rP indicates the position vector of a point P in the wheel coordinate frame, W rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 (8.56) then the homogeneous position vector CrP of the point P in the wheel-body 8. Suspension Mechanisms 495 coordinate frame is: CrP = CTW W rP = \u23a1\u23a2\u23a2\u23a3 xP cos \u03b4 \u2212 yP cos \u03b3 sin \u03b4 + (Rw + zP ) sin \u03b3 sin \u03b4 xP sin \u03b4 + yP cos \u03b3 cos \u03b4 \u2212 (Rw + zP ) cos \u03b4 sin \u03b3 \u2212Rw + (Rw + zP ) cos \u03b3 + yP sin \u03b3 1 \u23a4\u23a5\u23a5\u23a6 (8.57) The position of the wheel center W r = 0, for a cambered and steered wheel is at Cr = CTW W r = \u23a1\u23a2\u23a2\u23a3 Rw sin \u03b3 sin \u03b4 \u2212Rw cos \u03b4 sin \u03b3 \u2212Rw(1\u2212 cos \u03b3) 1 \u23a4\u23a5\u23a5\u23a6 (8.58) The zc = Rw (cos \u03b3 \u2212 1) indicates how much the center of the wheel comes down when the wheel cambers. If the wheel is not steerable, then \u03b4 = 0 and the transformation matrix CTW reduces to CTW = \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 cos \u03b3 \u2212 sin \u03b3 \u2212Rw sin \u03b3 0 sin \u03b3 cos \u03b3 Rw (cos \u03b3 \u2212 1) 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 (8.59) that shows CrP = CTW W rP = \u23a1\u23a2\u23a2\u23a3 xP yP cos \u03b3 \u2212Rw sin \u03b3 \u2212 zP sin \u03b3 zP cos \u03b3 + yP sin \u03b3 +Rw (cos \u03b3 \u2212 1) 1 \u23a4\u23a5\u23a5\u23a6 (8.60) Example 333 F Tire to vehicle coordinate frame transformation. Figure 8.50 illustrates the first and fourth tires of a 4-wheel vehicle. There is a body coordinate frame B (x, y, z) attached to the mass center C of the vehicle. There are also two tire coordinate frames T1 (xt1 , yt1 , zt1) and T4 (xt4 , yt4 , zt4) attached to the tires 1 and 4 at the center of their tireprints. The origin of the tire coordinate frame T1 is at Bd1 BdT1 = \u23a1\u23a3 a1 \u2212b1 \u2212h \u23a4\u23a6 (8.61)"
        ]
    },
    {
        "image_filename": "designv10_0_0001261_j.mprp.2016.12.062-Figure9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001261_j.mprp.2016.12.062-Figure9-1.png",
        "caption": "FIGURE 9 Fan case produced by adding features with AM (laser aided directed energy deposition) to a forged preform (courtesy: Jim Sears).",
        "texts": [
            " Adding features to a forged or cast preform as opposed to machining of such features can provide the most cost-effective manufacturing option, where a significant reduction of the preform size and weight can be effected through the elimination of the need for a machining allowance. Examples are various casings, housings etc. in jet engines where flanges, bosses, etc. can be added on cast or forged cylindrical preforms. This is demonstrated for a feature addition on a titanium fan casing for an aero-engine (Fig. 9). One of the best application areas suited for Directed Energy Deposition techniques is remanufacturing and repair of damaged, Please cite this article in press as: B. Dutta, F.H.S. Froes, Met. Powder Rep. (2017), http://dx 6 worn-out, corroded parts. Due to their ability to add metal on select locations on 3D surfaces, these technologies can be used to rebuild lost material on various components [27\u201329]. Closed loop technologies, such as DMD offer particular benefit of minimum heat affected zone (HAZ) in the repaired part and helps to retain the integrity of the part"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure11.16-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure11.16-1.png",
        "caption": "Figure 11.16 (a) A cantilever beam with a uniformly distributed load, and the associated (b) V diagram and (c) M diagram. The bending moment M is extreme where the shear force V is zero or changes sign.",
        "texts": [
            " end condition at C: \u221220 \u00d7 72 + C (2) 1 \u00d7 7 + C (2) 2 = 0. 454 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Or more neatly put C (1) 2 = 0, 5C (1) 1 + C (1) 2 \u2212 5C (2) 1 \u2212 C (2) 2 = 0, C (2) 1 = 280, 7C (2) 1 + C (2) 2 = 980. The solution to the set is C (1) 1 = 84 kN, C (1) 2 = 0, C (2) 1 = 280 kN, C (2) 2 = \u2212980 kNm. From this it follows for field (1) with (0 m)<x<(5 m) that V (1) = (\u221240x + 84) kN, M(1) = (\u221220x2 + 84x) kNm, and for field (2) with (5 m)<x <(7 m) that V (2) = (\u221240x + 280) kN, M(2) = (\u221220x2 + 280x \u2212 980) kNm, Figure 11.16 shows the V and M diagrams. At A, B and C, the tangents to the M diagram are also shown. These intersect at the middle of each field. Note that p1 = 1 8 \u00d740 \u00d752 = 125 kNm and p2 = 1 8 \u00d740 \u00d722 = 20 kNm, or in other words, for each field: \u201cp = 1 8q 2\u201d. 11 Mathematical Description of the Relationship between Section Forces and Loading 455 The bending moment in field (1) is a maximum where the tangent to the M diagram is horizontal, or where dM dx = V = \u221240x + 84 = 0 \u21d2 x = 2.1 m. The maximum bending moment therefore occurs to the left of the middle of AB"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure3.10-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure3.10-1.png",
        "caption": "Fig. 3.10. Diagram of a PM for finding the origin of the recoil line and operating point.",
        "texts": [
            " toroid, this PM does not show any magnetic properties in the external space, in spite of the fact that there is the magnetic flux \u03a6r = BrSM = BrwM lM (3.16) corresponding to the remanent flux density Br inside the PM. A PM previously magnetized and placed alone in an open space, as in Fig. 3.9a, generates a magnetic field. To sustain a magnetic flux in the external open space, an MMF developed by the magnet is necessary. The state of the PM is characterized by the point K on the demagnetization curve (Fig. 3.10). The location of the point K is at the intersection of the demagnetization curve with a straight line representing the permeance of the external magnetic circuit (open space): Gext = \u03a6K FK , tan\u03b1ext = \u03a6K/\u03a6r FK/Fc = Gext Fc \u03a6r (3.17) The permeance Gext corresponds to the flux \u03a6\u2014MMF coordinate system and is referred to as MMF at the ends of the PM. In the \u03a6\u2014MMF coordinate 100 3 Materials and Fabrication system the remanent flux \u03a6r is according to eqn (3.16) and the MMF corresponding to the coercivity Hc is Fc = HchM (3.18) The magnetic energy per unit produced by the PM in the external space is wK = BKHK/2. This energy is proportional to the rectangle limited by the coordinate system and lines perpendicular to the \u03a6 and F coordinates projected from the point K. It is obvious that the maximum magnetic energy is for BK = Bmax and for HK = Hmax. If the poles are furnished with pole shoes (Fig. 3.9b) the permeance of the external space increases. The point which characterizes a new state of the PM in Fig. 3.10 moves along the recoil line from the point K to the point A. The recoil line KGM is the same as the internal permeance of the PM, i.e. GM = \u00b5rec wM lM hM = \u00b5rec SM hM (3.19) The point A is the intersection of the recoil line KGM and the straight line OGA representing the leakage permeance of the PM with pole shoes, i.e. GA = \u03a6A FA , tan\u03b1A = GA Fc \u03a6r (3.20) The energy produced by the PM in the external space decreases as compared with the previous case, i.e. wA = BAHA/2. 3.2 Rotor Magnetic Circuits 101 The next stage is to place the PM in an external ferromagnetic circuit as shown in Fig. 3.9c. The resultant permeance of this system is GP = \u03a6P FP , tan\u03b1P = GP Fc \u03a6r (3.21) which meets the condition GP > GA > Gext. For an external magnetic circuit without any electric circuit carrying the armature current, the magnetic state of the PM is characterized by the point P (Fig. 3.10), i.e. the intersection of the recoil line KGM and the permeance line OGP . When the external magnetic circuit is furnished with an armature winding and when this winding is fed with a current which produces an MMF magnetizing the PM (Fig. 3.9d), the magnetic flux in the PM increases to the value \u03a6N . The d-axis MMF F \u2032ad of the external (armature) field acting directly on the PM corresponds to \u03a6N . The magnetic state of the PM is described by the point N located on the recoil line on the right-hand side of the origin of the coordinate system",
            " To obtain this point it is necessary to lay off the distance OF \u2032ad and to draw a line GP from the point F \u2032ad inclined by the angle \u03b1P to the F -axis. The intersection of the recoil line and the permeance line GP gives the point N . If the exciting current in the external armature winding is increased further, the point N will move further along the recoil line to the right, up to the saturation of the PM. When the excitation current is reversed, the external armature magnetic field will demagnetize the PM. In this case it is necessary to lay off the distance OF \u2032ad from the origin of the coordinate system to the left (Fig. 3.10). The line GP drawn from the point F \u2032ad with the slope \u03b1P intersects the demagnetization curve at the point K \u2032. This point can be above or below the point K (for the PM alone in the open space). The point K \u2032 is the origin of a new recoil line K \u2032G\u2032M . Now if the armature exciting current decreases, the operating point will move along the new recoil line K \u2032G\u2032M to the right. If the armature current drops down to zero, the operating point takes the position P \u2032 (intersection of the new recoil line K \u2032G\u2032M with the permeance line GP drawn from the origin of the coordinate system). On the basis of Fig. 3.10 the energies wP \u2032 = BP \u2032HP \u2032/2, wP = BPHP /2, and wP \u2032 < wP . The location of the origin of the recoil line, as well as the location of the operating point, determine the level of utilization of the energy produced by the PM. A PM behaves differently than a d.c. electromagnet : the energy of a PM is not constant if the permeance and exciting current of the external armature changes. The location of the origin of the recoil line is determined by the minimum value of the permeance of the external magnetic circuit or the demagnetization action of the external field"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000348_pnas.0406724102-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000348_pnas.0406724102-Figure1-1.png",
        "caption": "Fig. 1. Construct for quantitative analysis of bacterial dynamics at a slanting interface. (A) Apparatus for simultaneous dark-field and fluorescence imaging of fluid drops: laser (L), shutter (S), beam-expander lenses (L1 and L2), dichroic beam splitter (D), long-pass filter (LPF), fiber-optic ring light (RL), and digital camera (C). The chamber has antireflection areas (AR) to reduce glare and hydrophobic areas (H) to pin the drop. (B) Boundary conditions and coordinates in the wedge (described in Mathematical Model).",
        "texts": [
            " The 5-h culture was then slowly added to the chamber with a syringe (21-gauge needle). The 20\u00b0 stock was prepared by adding spores on sand to 10 ml of TB at room temperature in a Petri dish and allowing for 24 h of growth. One milliliter of 24-h stock was added to 50 ml of TB and incubated for 18 h. One milliliter of the 18-h culture was added to 50 ml of TB and incubated for 5 h. Aliquots composed of 0.75 ml of 5-h culture mixed gently with 0.25 ml of glycerin were placed in cold storage. The chambers (Fig. 1A) were constructed from microscope slides cemented together with UV curing adhesive (Norland, Cranbury, NJ); care was taken to remove excess adhesive to avoid interference with visualizations near the contact line and to seal the chamber adequately to avoid evaporation and associated flows. The space between the vertical slides that form the front and back of the chamber was typically 1 mm. Imaging was achieved with a digital charge-coupled device camera [Hamamatsu (Ichinocho, Japan) C7300; 1,024 1,024 pixels; 12 bit] under computer control (National Instruments, Austin, TX) attached to a macrophotography bellows (PB-6, Nikon) with a 105-mm f 2",
            " Subject to incompressibility, u 0, the fluid equations (Eq. 3) use the Boussinesq approximation in which the density variations caused by bacteria appear only in the buoyant forcing, with b , where Vb and b are the bacterial volume and density, respectively, and is the pure fluid density. In the Appendix we provide a nondimensionalization of these equations convenient for numerical studies. The boundary conditions on c, n, and u are central to the global f lows and possible singularities. As shown in Fig. 1B, there is no flux (of cells or oxygen) through the substrate, under the drop and to its side, and no cell f lux through the fluid\u2013air interface. That interface is stress-free, whereas the fluid\u2013glass boundary has no slip. Because the oxygen-diffusion coefficient in the air is three orders of magnitude larger than that in the fluid, the oxygen concentration outside the drop can be assumed equal to its saturation value cs inside the fluid and on the free surface. Oxygen diffusion in the drop is equivalent to that in the enlarged space, the bottom of which is the reflection of the meniscus across the fluid\u2013glass interface",
            " Close enough to the contact line the numerics indicate that the velocity along the upper free surface tends to zero, so advective contributions vanish asymptotically there. The simplest steady-state oxygen-diffusion problem, for which 2c 0 inside the drop and c cs on the free surface, is mathematically equivalent to the equation defining the electrostatic potential in that same region, the exterior of the drop being a conductor (5). This mapping reveals (14) that a singularity arises from a solution of the form c r, cs m 1 amrm /2 cos m 2 , [5] with polar coordinates (r, ) as shown in Fig. 1B. This form enforces the condition c cs on the drop surface ( ) and the no-flux condition on the glass substrate ( 0). After successive differentiation, the potentially most singular term is m 1, and we expect a1 0 because c cs in the drop interior. Noting that  c r /2 1 and 2c r /2 2, singularities appear in two cases: 1. 2, an overhanging meniscus. This case is like the familiar \u2018\u2018lightning-rod\u2019\u2019 effect. Both the Fick\u2019s law oxygen flux, Dc c, and the chemotactic f lux, rn c, diverge as r3 0, leading to a singular accumulation at the contact line"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002348_rpj-06-2014-0075-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002348_rpj-06-2014-0075-Figure8-1.png",
        "caption": "Figure 8 - Stress distribution around cooling channel for varying pitch and depths from surface. Common distributed load applied to top surface. Where is the channel depth from the mould surface to top of the channel, is cooling channel pitch and d is channel diameter.",
        "texts": [
            " The stress concentration effect is dependent on the diameter of the cooling channel, the depth of the channel from the mould surface, and the pitch spacing between adjacent channels. A static-structural Finite Element (FE) analysis was conducted in ANSYS simulation software to quantify the stress concentration effect for standard channels with varying channel diameter (d), depth (y) and pitch (x). A twodimensional model representative of a solid mould section was used which featured a series of cooling channels (Figure 8). A distributed unit load was applied to the upper edge of the model, with the lower edge constrained against vertical displacement. The stress concentration factor associated with the channels was determined by comparison of the resulting maximum stress to the effective stress in the absence of cooling channels. Figure 8 shows an example stress contour for a single evaluation. D ow nl oa de d by U ni te d A ra b E m ir at es U ni ve rs ity A t 1 0: 09 2 2 M ar ch 2 01 6 (P T ) Figure 10 summarizes stress concentration factors for a range of associated parameters where the axes correspond to channel spacing, the marker to channel diameter, and the marker colour to the resultant stress concentration. A low stress concentration level of 4 is indicated at the low end of the colour map. The analysis shows that magnitude of the stress concentration increases with an increase in channel diameter and with a decrease in channel depth and pitch"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003781_978-94-011-3396-8_5-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003781_978-94-011-3396-8_5-Figure2-1.png",
        "caption": "Fig. 2. Cyclic voltammograms (current vs. potential plots) for electrodes made with (a) native glucose oxidase with or without glucose; (b) glucose oxidase modified by covalent bonding of 12 electron-relays per enzyme molecule in the absence of glucose; (c) as in (b), but at 0.8 mM glucose and (d) as in (b) but at 5 mM glucose. Scan rate 2mV/sec.",
        "texts": [
            " (Figures 1,2) The current was produced by a sequence of electron transfer steps, where glucose reduced the enzyme's FAD centers to FADH2; the FADH2 centers were reoxidized to FAD upon transferring electrons to the ferricinium relays, reducing these to ferrocene; ferrocene transferred electrons to the electrode, and was reoxidized to ferricinium. The greater the number of relays we bound to the enzyme, the better its FADH2 centers communicated with electrodes.5 This was also the case in D-amino acid oxidase,4,5 The current in relay-modified glucose oxidase electrodes increased when glucose was added to the solution (Figure 2) but did not change when sucrose, other hexoses, or pentoses were added.5 ELECTRICAL WIRING OF REDOX ENZYMES 73 Bot tom: Flow of electrons through a relay from a redox center of an enzyme (glucose oxidase) to an electrode. A current is observed when the substrate (glucose) transfers a pair of electrons to an FAD center of the enzyme that, in tum, transfers these either to a relay or to a molecular wire, which then transfers the electrons to the electrode. 74 A.HELLER Another effective glucose oxidase-bound relay that we and Bartlett and Whitaker,5,45 studied was ferrocene acetic acid, covalently bound to glucose oxidase lysines"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure5.6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure5.6-1.png",
        "caption": "FIGURE 5.6. Second Euler angle.",
        "texts": [
            " To find the Euler angles rotation matrix to go from the global frame G(OXY Z) to the final body frame B(Oxyz), we employ a body frame B0(Ox0y0z0) as shown in Figure 5.5 that before the first rotation coincides with the global frame. Let there be at first a rotation \u03d5 about the z0-axis. Because Z-axis and z0-axis are coincident, by our theory B0 r = B0 RG Gr (5.72) B0 RG = Rz,\u03d5 = \u23a1\u23a3 cos\u03d5 sin\u03d5 0 \u2212 sin\u03d5 cos\u03d5 0 0 0 1 \u23a4\u23a6 . (5.73) Next we consider the B0(Ox0y0z0) frame as a new fixed global frame and introduce a new body frame B00(Ox00y00z00). Before the second rotation, the two frames coincide. Then, we execute a \u03b8 rotation about x00-axis as shown in Figure 5.6. The transformation between B0(Ox0y0z0) and B00(Ox00y00z00) is B00 r = B00 RB0 B0 r (5.74) B00 RB0 = Rx,\u03b8 = \u23a1\u23a3 1 0 0 0 cos \u03b8 sin \u03b8 0 \u2212 sin \u03b8 cos \u03b8 \u23a4\u23a6 . (5.75) Finally, we consider the B00(Ox00y00z00) frame as a new fixed global frame and consider the final body frame B(Oxyz) to coincide with B00 before the third rotation. We now execute a \u03c8 rotation about the z00-axis as shown in 232 5. Applied Kinematics 5. Applied Kinematics 233 Figure 5.7. The transformation between B00(Ox00y00z00) and B(Oxyz) is Br = BRB00 B 00 r (5"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure13.7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure13.7-1.png",
        "caption": "FIGURE 13.7 Example loop problem geometry for passivity study.",
        "texts": [
            " The result is shown in Fig. 13.6, where both sides of the antenna patches divided with six and seven subdivision cells. The large impact of the cell subdivisions for frequencies above fmax is apparent. Next, we consider the frequency domain response for a loop using a scattering response again with the same port connection. The frequency domain reflection coefficient \ud835\udf0c(s) or scattering response S11 is computed by using (13.12). We show results for a loop problem PASSIVITY ASSESSMENT OF SOLUTION 339 shown in Fig. 13.7, which is a longer version (xl = 0.53 mm, yl = 50 \u03bcm, w=10 \u03bcm) of an example used in Ref. [16] and in Chapter 9. Here, we first present the solution without passivity enhancement. First, we consider the real part of the input impedance of the loop as a function of frequency. The real part of the loop input impedance is given in Fig. 13.8. It is important that the response must be computed far above the maximum frequency of interest fmax. It is clear that the real part of the high-frequency response prevents strict passivity due to this response"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure9.5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure9.5-1.png",
        "caption": "Figure 9.5 The structure of a simple truss bridge.",
        "texts": [
            " These structures behave as trusses, and can be calculated as such only if the structure remains kinematically determinate when all the rigid joints are replaced by hinged joints. Figures 9.4e to 9.4h show the same structures as in (a) to (d), but now with hinged joints. After applying hinges, structures (a) and (b) are kinematically determinate and can therefore be considered trusses. With structures (c) and (d), a mechanism is formed after introducing hinged joints. They are now kinematically indeterminate and cannot be calculated as trusses. The force flow in these structures occurs mainly by bending. The simple truss bridge in Figure 9.5 shows how to ensure that the load on the bridge ends up at the joints of the truss. The bridge consists of two main beams constructed as plane trusses. Cross beams have been introduced 1 The proof for this cannot be given at this stage, but is based on the characteristic that the members in a truss are relatively weak with respect to bending, and relatively stiff with respect to extension (changing length). 9 Trusses 323 between the main beams, which are supported at the joints of the truss. Between the cross beams, stringers carry the deck (not shown)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure3.15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure3.15-1.png",
        "caption": "Fig. 3.15 Definition of Variables for Calculation of ZMP by Multiple 1 Axis Force Sensor",
        "texts": [
            "14 shows the humanoid robot H5 [70]. To make the foot light, the ZMP is measured by using twelve force sensing registers: FSR and sorbothane sandwiched by two aluminum planes (Fig. 3.14(b)). Since the electric resistance changes according to the applied force, the FSR can be used as a one dimensional force sensor to measure the vertical component of ground reaction force. 3.2 Measurement of ZMP 81 To measure the ZMP, the x and y components of the force are set to be 0 in (3.24) and (3.25). As shown in Fig.3.15, when there are N one-dimensional force sensors, the ZMP can be obtained by px = \u2211N j=1 pjxfjz\u2211N j=1 fjz (3.28) py = \u2211N j=1 pjyfjz\u2211N j=1 fjz . (3.29) Figure 3.16 shows the humanoid robot Morph3 and its foot [129, 120]. Morph3 measures the ZMP by using four 3 axis force sensors attached at each foot (Fig. 3.16(b)). The 3 axis force sensor measures the 3 dimensional forces applied to the sole split into four parts. By using this measuring system, we can obtain measurement on the point of contact. To calculate the ZMP 82 3 ZMP and Dynamics of each foot, the element of moment, \u03c4jx and \u03c4jy , in (3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003586_s0731-7085(98)00056-9-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003586_s0731-7085(98)00056-9-Figure3-1.png",
        "caption": "Fig. 3. Design of the disposable sensor-array cartridge used in the i-STAT portable clinical analyzer (courtesy of i-STAT).",
        "texts": [
            " Such fabrication technology couples various processes, used in the manufacture of electronic circuits, including film deposition, photolithographic patterning, and etching. It offers a greater resolution (down to submicrometer structures) at higher capital and manufacturing costs, and is thus particularly suitable for the production of sensor arrays. One successful example is the i-STAT portable (hand-held) clinical analyzer that simultaneously performs eight common clinical tests on a 60 ml patient blood sample in about 90 s ([34], Fig. 3). Upon completion of the assay, the results are displayed on a liquid-crystal screen, along with the patient identification number and time. Thin-film biosensors can be readily integrated with miniaturized total analytical systems (mTAS), produced by silicon micromachining technology. Miniaturized analytical systems based on fluid- handling silicon microstructures, offer an improved efficiency with respect to sample size, response time and reagent consumption. A recent example is the development of a mTAS system for on-line monitoring of lactate and glucose in biological systems, based on integrated microdialysis sampling and photolithographically-prepared lactate and glucose electrodes ([35], Fig"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure4.21-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure4.21-1.png",
        "caption": "Figure 4.21 A steel hinged support, previously used in smaller bridges. The horizontal movement is prevented by the pin.",
        "texts": [
            " For the hinged support in Figure 4.19 with the coordinate system shown, the following applies for motion at A: ux;A = 0 (prescribed motion), uy;A = 0 (prescribed motion), \u03d5z;A = unknown (free motion). and for the forces at A: Fx;A = unknown (free force), Fy;A = unknown (free force), Tz;A = 0 (prescribed force). A hinged support therefore has one degree of freedom (a rotation) and generates two support reactions (the two components of a force). 4 Structures 125 Figure 4.20 is a good example of a hinged support. Figure 4.21 shows how a steel hinged support can be used in small bridges. Horizontal motion is prevented by a pin. The steel hinged support in Figure 4.22 can transfer large forces and is an example of what is used in larger bridges. Like roller supports, hinged supports can be made from materials other than steel, or from a combination of materials Although the supports in Figures 4.21 and 4.22 can transfer only compressive forces, it is assumed below that hinged supports can also transfer tensile forces"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000492_02783649922066376-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000492_02783649922066376-Figure1-1.png",
        "caption": "Fig. 1. The sketch of a 3-D extended rigid-body biped robot (left), and a view with its support foot artificially disconnected from the shank to show the intervening forces (right). The CoP, GCoM, and the FRI point are denoted by P , C, and F , respectively.",
        "texts": [
            " We address the mechanics of foot rotation, and do not concern ourselves with the formulation or implementation of any control law. However, since the real interest in this area re- sults from control problems, a brief description of the control issues is included for completeness (in Section 5). Please note that whenever the context permits, we loosely use \u201cforce\u201d to mean \u201cforce/torque.\u201d 2. FRI Point of a General 3-D Biped Robot To formally introduce the FRI point, we first treat the entire biped robot\u2014a general n-segment extended rigid-body kinematic chain (see the sketch in Fig. 1 left)\u2014as a system, and determine its response to external force/torque. We may employ Newton\u2019s or d\u2019Alembert\u2019s principle for this purpose. The external forces acting on the robot are the resultant groundreaction force/torques, R and M , acting at the CoP (denoted by P ; see Fig. 1, right), and gravity. The equation for rotational dynamic equilibrium1 is obtained by noting that the sum of the external moments on the robot, computed either at its GCoM or at any stationary reference point, is equal to the sum of the rates of change of angular momentum of the individual segments about the same point. Taking moments at the origin O, we have M + OP \u00d7 R + \u2211 OGi \u00d7 mig = \u2211 H\u0307Gi + \u2211 OGi \u00d7 miai , (1) where mi is the mass, Gi is the CoM location, ai is the CoM linear acceleration, and HGi is the angular momentum about CoM, of the ith segment. An important aspect of our approach is to treat the stance foot as the focus of attention of our analysis. Indeed, as the only robot segment interacting with the ground, the stance foot is a \u201cspecial\u201d segment subjected to joint forces, gravity forces, and the ground-reaction forces. Viewing from the stance foot, the dynamics of the rest of the robot may be completely represented by the ankle force/torque \u2212R1 and \u2212\u03c41 (negative signs are for convention). Figure 1 (right) artificially disconnects the support foot from the shank to clearly show the forces in action at that joint. The dynamic equilibrium equation of the foot (segment 1) is M + OP \u00d7 R + OG1 \u00d7 m1g \u2212 \u03c41 \u2212 OO1 \u00d7 R1 = H\u0307G1 + OG1 \u00d7 m1a1. (2) The equations for static equilibrium of the foot are obtained by setting the dynamic terms (in the right-hand side) in eq. (2) to zero: M + OP \u00d7 R + OG1 \u00d7 m1g \u2212 \u03c41 \u2212 OO1 \u00d7 R1 = 0. (3) Recall that to derive eq. (3) we could compute the moments at any other stationary reference point",
            " (9) Carrying out the operation, we may finally obtain OFx = m1OG1yg + n\u2211 i=2 miOGiy(aiz + g) m1g + n\u2211 i=2 mi(aiz + g) \u2212 n\u2211 i=2 miOGizaiy + n\u2211 i=2 H\u0307Gix m1g + n\u2211 i=2 mi(aiz + g) , (10) OFy = m1OG1xg + n\u2211 i=2 miOGix(aiz + g) m1g + n\u2211 i=2 mi(aiz + g) \u2212 n\u2211 i=2 miOGizaix \u2212 n\u2211 i=2 H\u0307Giy m1g + n\u2211 i=2 mi(aiz + g) . (11) 2.1. Properties of the FRI Point Some useful properties of the FRI point which may be exploited in gait planning include the following: 1. The FRI point indicates the occurrence of foot rotation, as already described. 2. The location of the FRI point indicates the magnitude of the unbalanced moment on the foot. The total moment MI A due to the impressed forces about a point A on the support-polygon boundary (Fig. 1, right) is MI A = AF \u00d7 (m1g \u2212 R1), (12) which is proportional to the distance between A and F . If F is situated inside the support polygon, MI A is counteracted by the moment due to R and is precisely compensated; see Figure 4 (left) for a planar example. Otherwise, MI A is the uncompensated moment that causes the foot to rotate; see Figure 4 (right). at UNIVERSITY OF BRIGHTON on July 15, 2014ijr.sagepub.comDownloaded from 3. The FRI point indicates the direction of foot rotation. This we derive from eq",
            " It may be shown that for an idealized rigid foot the CoP is situated at a boundary point unless the foot is in stable equilibrium. Since the position of the CoP cannot distinguish between the marginal state of static equilibrium and a complete loss of equilibrium of the foot (in both cases it is situated at the support boundary), its utility in gait planning is limited. The FRI point, on the other hand, may exit the physical boundary of the support polygon, and it does so whenever the foot is subjected to a net rotational moment. The GCoM, represented by C in Figure 1, satisfies CG \u00d7 \u2211 mig = 0, (18) where G is the center of mass of the entire robot and\u2211 mi = M is the total robot mass. Noting that CG \u2211 mi =\u2211 CGimi and CGi = CP +PGi , we can rewrite eq. (18) as CP \u00d7 \u2211 mig + \u2211 PGi \u00d7 mig = 0. (19) Substituting in eq. (1), we get M \u2212 CP \u00d7 \u2211 mig = \u2211 H\u0307Gi + \u2211 PGi \u00d7 miai . (20) From above, P and C coincide if ( \u2211 H\u0307Gi + \u2211 PGi \u00d7 miai ) t = 0, which is possible if the robot is stationary or has uniform linear and angular velocities in all the joints. The objective of this section is to elucidate the idea behind the FRI point by means of four simple examples, depicted in Figures 6 and 7"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002216_j.ijfatigue.2021.106317-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002216_j.ijfatigue.2021.106317-Figure1-1.png",
        "caption": "Fig. 1. Schematics showing (a) the scan strategy for the LPBF process used to manufacture the AlSi10Mg alloy wall and (b) the orientation of the test-pieces relative to the build direction (BD).",
        "texts": [
            ", Ltd) AM forming equipment was used to build the AlSi10Mg alloy sample block (110 \u00d7 55 \u00d7 120 mm3). The AlSi10Mg powder particles used here were ranged from 15 to 45 \u03bcm in diameter (based on the sieving method and tested by SEM) and chemical composition is shown in Table 1. The following optimal LPBF parameters were used: laser power ~ 360\u2013400 W, scanning speed ~ 1200\u20131500 mm/s, hatch spacing ~ 0.13\u20130.16 mm, using a layer thickness of 50 \u03bcm and a scan strategy employing a 67\u25e6 angle between successive layers (see Fig. 1(a)). After AM but prior to machining specimens, the printed block was heated to 300 \u25e6C for about 2 h (SR1 condition according to ASTM F3318-18 standard) to relieve residual stress after which was naturally aged for 2 weeks. Nomenclature Symbols \u221aareaeff Murakami\u2019s defect size parameter \u221aarea Actual size of defect a, b, c Principal semi-axes of the ellipsoid C, D Intrinsic factor without defect and sensitive factor of defects d Shortest distance between two adjacent defects e Elongation f Loading frequency Fw Murakami geometry factor h Shortest distance between specimen free surface and defect edge hs Thickness of surface area Nf Fatigue life R Stress ratio (\u03c3min/\u03c3max) RA Reduction in area Sd Surface area of defect V Prospective volume under examination V0 Inspection sub-volume",
            " normal to the build direction H-T Tensile specimen loaded normal to the build direction H-C(T) Compact tension specimen loaded normal to the build direction H-HCF High cycle fatigue specimen loaded normal to the build direction HCF High cycle fatigue ISIA Improved backward statistical inference approach K-T Kitagawa-Takahashi LOF Lack of fusion MPB Melt pool boundary ML Machine learning POT Peaks-over-threshold SIF Stress intensity factor SEM Scanning electron microscopy LPBF Laser powder bed fusion V Vertical direction, i.e. parallel to the build direction V-T Tensile specimen loaded along the build direction V-C(T) Compact tension specimen loaded along the build direction V-HCF High cycle fatigue specimen loaded along the build direction CT Computed tomograph Z. Wu et al. International Journal of Fatigue 151 (2021) 106317 Fig. 1 presents the schematics of LPBF process strategy and the specimen extraction along with the coordinate system used in this work. As illustrated the compact tension (C(T)) samples were cut in two orientations. Firstly, so that we could assess the crack propagation threshold in the X direction with the Mode I driving force applied relative to the vertical build direction (Z), often referred to the Z-X configuration, and secondly, the threshold of crack propagation in the build direction (Z) with Mode I driving force applied relative to the horizontal direction (X), often referred to as the X-Z configuration"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000538_acs.chemrev.8b00172-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000538_acs.chemrev.8b00172-Figure5-1.png",
        "caption": "Figure 5. Screen-printed electrode. (1) Co-phthalocyanine mediated screen-printed electrode (a) and the electrode region (b). Reproduced with permission from ref 106. Copyright 2014 Springer. (2) Electrochemical sensor (ECS) modules, ECS1 and ECS2. Reproduced with permission from ref 107. Copyright 2017 Wiley-VCH Verlag GmbH & Co.",
        "texts": [
            " Electrodes in microfluidic systems can be produced with thin-film or thick-film techniques, respectively. Thick-film technology, with screen printing as its most established technique, in general is significantly cheaper than thin-film technology of noble metals and is highly compatible with the fabrication of carbon electrodes. Screen printing of carbon electrodes can be performed on various substrates, e.g. alumina ceramic plate,101 PDMS,102 and thermoplastics,103\u2212105 and can successfully be integrated into microfluidic systems. Also, the technique is popular to provide Ag/AgCl REs (Figure 5).106,107 In fact, for the integration of electrodes into paper-based LOC platforms, screen printing of the electrodes is the most common technique. Additionally, inkjet printing has been employed as an efficient technique to pattern electrodes on flexible substrates. For example, inkjet printing of silver electrodes on PDMS is reported by Wu et al.108 (Figure 6-1). Inkjet-printed gold electrodes on paper are reported as well.109 Common electrode materials for the WE and CE are carbon, copper, gold, or aluminum ink, while the RE is usually fabricated with Ag/AgCl inks"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002358_j.arcontrol.2018.10.009-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002358_j.arcontrol.2018.10.009-Figure5-1.png",
        "caption": "Fig. 5. Examples of modular aerial robots: (a) a planar three ducted-fan ( Naldi, Forte, & Marconi, 2011 ), (b) four distributed flight array modules ( Oung, 2013 ).",
        "texts": [
            " However, due to the system complexity, only over flight condition was implemented and tested. One challenge n the modular approach is to build an accurate dynamic model of he system. A bigger challenge is to design control algorithms and stimation strategies for a system that consists of tens or maybe undreds of individual elements interacting dynamically. This can ead to a very unregular and unexpected behavior. Thus, the major ocus of their work was on developing distributed and decentralzed control systems that are capable of controlling such compliated systems. Fig. 5 shows the examples of introducing system edundancy in modular aerial robots. System redundancy can also be achieved by simply adding exra actuators to a quadrotor frame. A hexacopter, a six-rotor UAV, s an example of vehicles with motor redundancy. Thanks to its implicity, this approach is often the preferred solution for actutor failure. For example, a modification on the standard design f hexagon shaped hexacopter was done in Giribet, S\u00e1nchez-Pe\u00f1a, nd Ghersin (2016) by tilting the rotors towards the vehicle\u2019s verical axis"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure3.75-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure3.75-1.png",
        "caption": "Fig. 3.75: A numerical approximation for workspace evaluation of the CaPaMan manipulator in terms of: a) volume of the position workspace; a) volume of the orientation workspace.",
        "texts": [
            "110) can be used conveniently as objective function for prescribing singularity-free conditions. Stiffness analysis of CaPaMan has been formulated by modeling each leg of CaPaMan as explained previously according to the scheme of Fig. 3.73. The lumped stiffness parameters can be assumed as kbk=kdk=2.625 \u00d7 106 N/m and kTk= 58.4 \u00d7 103 Nm/rad. Position and orientation workspace volumes have been conveniently evaluated by using the kinematics of the CaPaMan manipulator and they have been characterized by using the schemes of Fig. 3.75 for the approximation in the optimization numerical process. Thus, the design problem for the CaPaMan manipulator has been formulated as minimizing a function F (X), which is the weighted sum of the objective functions as in Eqs (3.9.26) but grouping those criteria for facilitating the numerical procedure to have 'V *V 1)(f pos pos 1 \u2212=X 'V *V 1)(f or or 2 \u2212=X +\u2212+\u2212+\u2212= g d g d g d 3 z z 1 y y 1 x x 1)(f X (3.9.111) \u2212+\u2212+ \u03d5 \u03d5 \u2212+ g d g d g d 111 8G/1)(f 4 \u2212=X with the constraints ' maxmax xx \u2264 ; ' maxmax yy \u2264 ; ' maxmax zz \u2264 ' maxmax \u03d5\u2264\u03d5 ; ' maxmax \u03c8\u2264\u03c8 ; (3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003933_robot.1989.99999-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003933_robot.1989.99999-Figure8-1.png",
        "caption": "Figure 8: 3R Manipulator Null Spaces",
        "texts": [
            " Remark: The null space of the Jacobian matrix, evaluated at a particular joint configuration, B o , is the tangent to the self-motion manifold at go. The collective null spaces of one self-motion is the tangent bund le3 of the self-motion manifold. The set of all Jacobian null spaces over a fixed end-effector location will be tlie union of all self-motion tangeiit bundles. For example, tlie null space of the maiiipulator in Figure 4 at go is a line, Et1, tangent to the self-motion manifold at go. The collection of all null spaces for one self-motion of this manipulator is diffeomorpliic to S' x R', a cylinder. Figure 8 shows the tangent spaces of the self-motion manifold over point 2 in Figure 4. For an n-dimensional manifold X , the tangent bundle of X, denoted T(X'), is the subset of X x R\" defined by T ( S ) = {(.,.) 6 X x R\": 2) E Tz(.z)}. I n other words, the tangcnt bundle of S is a n 271- dimensioiial nianifold formed by \"tacking 011\" to every X ~ S a copy of the tangent space a t 1'. 268 7. Self-motion Homotopy Classes In the example of Figure 4, the self-motion manifolds over point 1 included a 2.rr joint rotation, while the self-motions over point 2 did not",
            " Such x are termed coregular, and the self-motions in the preimage of such points are termed coregular self-motions However, a coregular preimage does not properly form a manifold. The structure of coregular self-motions can be investigated by considering the 312 planar manipulator self-motions as the end-effector nears a singularity (which is not a workspace boundary singularity). At the singularity, two self-motions (which are diffeomorphic to circles) join at a point, to form a \u201cbouquet of circles,\u201d or a curve which is diffeomorphic to a figure 8. Only the waist of the figure 8 is a critical point, whereas the other points in the figure 8 are coregular. The preimage of an entire critical value manifold (which is formed by sweeping f-\u2019(x) along the critical value manifold) is a continuous surface, termed a coregular surface. Coregular surfaces are not manifolds, but by virtne of their dimension, they can separate the configuration into disjoint regions. Figure 9 illustrates all of the . . coregular surfaces for Two maps, fo: X i Y and f l : X - Y are homotopic if there exists a smooth map, F : X x I - Y such that F ( z , O ) = fo(z) and F ( z , 1) = fI(z)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002652_j.jmapro.2019.12.009-Figure14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002652_j.jmapro.2019.12.009-Figure14-1.png",
        "caption": "Fig. 14. Broken gear tooth repairing process: (a) slices of the broken tooth model, (b) the shape of the first layer of the broken tooth model slices, (c) remanufactured tooth using laser cladding, (d) remanufactured tooth after post-processing.",
        "texts": [
            " In the CDRS, the scanning path of the second layer is perpendicular to the first layer, which means it will not make the defect worse. Therefore, the surface of the block formed with the CDRS is smoother than that with SDRS. After comparing the forming qualities of the two blocks, CDRS was applied in the remanufacturing process. Based on the systematical investigation, the broken tooth CAD model was constructed and the optimal remanufacturing parameters were confirmed. The tooth model should be sliced before repairing, as shown in Fig. 14(a). The stratification of 3D model is based on the method of creating datum planes. The distance between each two datum planes is the height of the optimal Z-axis increment, which is 0.3 mm, obtained by the experiment. After stratification, the length and width of each slice need to be measured to design the scanning path. Fig. 14(b) shows the shape of the first layer of the broken tooth model slices. It can be seen that the top view of the first layer is rectangle. The short side of the rectangle L1 is used as the length of each cladding track. In order to form the first layer, the number of the cladding track is calculated using Eq. (5). = + \u2212 \u2212L W N \u03b7 W( 1)(1 )2 = \u23a1 \u23a2\u23a2 \u2212 \u2212 + \u23a4 \u23a5\u23a5 N L W W \u03b7(1 ) 12 (5) In Eq. (5), L2 is the length of the long side of the rectangle, W is the width of the cladding track, \u03b7 is the optimal overlapping rate which equals 20 % in this paper, N is the total number of the cladding track when forming this layer. The forming strategy of other layers is similar to that of the first layer. Other remanufacturing parameters such as scanning strategy and multi-layer accumulation pattern have been confirmed in Section 3.2.3 and Section 3.2.4. Finally, the broken tooth is remanufactured by laser cladding as shown in Fig. 14(c). It can be seen the remanufactured tooth has basically recovered its appearance before breakage. But it also can be seen that the surface quality of the remanufactured tooth is very poor, because there are a lot of sticky powders on the surface. Also the meshing lines of the tooth are not accurate. Therefore, a secondary machining is conducted on the remanufactured tooth to form a smooth surface and accurate meshing lines, as shown in Fig. 14(d). Fig. 15 shows the typical microstructure of the as-deposited laser cladding H13 steel. It can be seen from the Fig. 15(a) that the microstructure of the cladding layer is mainly equiaxed crystal and the grain size increases with the distance from the substrate. As shown in Fig. 15(b), columnar dendrites show and grow epitaxially in the region of molten pool, perpendicular to the substrate surface. In the process of laser remanufacturing for gear, the main factor affecting the dendrite growth direction is the direction of heat flux [22]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000406_j.actamat.2015.06.036-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000406_j.actamat.2015.06.036-Figure2-1.png",
        "caption": "Fig. 2. (a) Schematic illustration of the horizontal Ti\u20136Al\u20134V ELI block (100 mm 10 mm 30 mm) built by EBM. (b) Machinery of the block into standard tensile testpieces. (c) Layout of the four tensile testpieces. The build direction is indicated by an arrow.",
        "texts": [
            "0e 3 mBar, controlled by using high-purity helium as regulating gas in order to prevent powder charging. Recycling of non-melted and/or sintered powder was achieved via the powder recovery system (PRS) and a vibrating sieve (mesh size 6150 lm). In order to investigate the consistency of microstructure and mechanical properties of EBM-built parts, two horizontal blocks (100 mm 10 mm 30 mm) were fabricated. Four tensile testpieces were wire-cut one-by-one from a horizontal block. They are termed 10 mm-1, 10 mm-2, 10 mm-3 and 10 mm-4 from bottom to top (as illustrated in Fig. 2). Optical microscopy (OM; ZEISS Axioskop 2 MAT), scanning electron microscopy (SEM; JEOL JMS-6700F; 10 kV), X-ray diffraction (XRD; PANalytical Empyrean; Cu Ka; step size of 0.01 ) and transmission electron microscopy (TEM; JEOL-2010; 200 kV) were used to examine the microstructure of as-built Ti\u20136Al\u20134V. OM and SEM samples were etched in Kroll\u2019s reagent (1\u20133% HF, 2\u20136% HNO3, and 91\u201397% H2O) for 10 s. Quantitative image analysis was carried out by using Image J software. TEM samples were prepared using ion milling with a beam voltage of 3",
            " Tensile tests were performed on Instron Static Tester (series 5569) using subsize specimens with a gauge dimension of 25 mm 6 mm 6 mm according to ASTM E8 standard at a strain rate of 3.33e 4 s 1. Yield strength (YS, 0.2% offset method), elongation at break (% EL) and ultimate tensile strength (UTS) were measured from the engineering stress\u2013strain curves. Vickers microhardness (HV) (1 kg f, 15 s hold) measurements were conducted on the metallographic samples using a Future Tech FM-300e microhardness tester. At least 5 microhardness measurements were performed for each sample. Fig. 3 shows the XRD peaks of EBM-built Ti\u20136Al\u20134V sample taken from the 10 mm-thick block in Fig. 2a. From Safdar et al. [16], it is known that most of the peaks originates from a phase. Only a weak (110) peak of b phase could be observed after zooming in, which indicates that a very small amount of b phase retained in the final microstructure. It is thus confirmed that the microstructure of EBM-built Ti\u20136Al\u20134V mainly consists of a and b phases. The lattice constants of hcp a phase in the EBM-built Ti\u20136Al\u20134V can be determined to be a = 0.293 nm and c = 0.467 nm respectively, with c/a ratio of 1",
            " Moreover, it was found to be prevalent in all of our EBM-built Ti\u20136Al\u20134V parts. This interface layer appears to be quite similar with the a/b interface phase observed in the previous studies [6\u20138]. The difference is that the interface layer was not located symmetrically on both sides of b phase. SAED results demonstrate that it is the L phase with an fcc crystal structure of a = 0.441 nm [10]. In order to quantitatively study the a/b interface, APT was used to analyze the specimen cut from the EBM-built Ti\u20136Al\u20134V block in Fig. 2a. According to the mass-to-charge spectrum illustrated in Fig. 6, the form of the interstitial elements, e.g. O, Fe, N, C, H etc. that exist in the initial pre-alloyed powder could be directly known. O mainly exists as TiO and it also combine with Al and H to form AlO and H2O. Otherwise, the H2O detected in Fig. 6 might also be from the absorbed moisture. Fe and C will not react with other elements. N can exist either as element or compound of TiN and Ti2N. H exists in the form of TiH and VH but it may mainly come from the analysis chamber of APT"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000768_cdc.2006.377588-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000768_cdc.2006.377588-Figure1-1.png",
        "caption": "Fig. 1. A quadrotor configuration",
        "texts": [
            ". INTRODUCTION A quadrotor is a four rotor helicopter and its concept has appeared some time ago. The first reported quadrotor helicopter Gyroplane No. 1 was built in 1907 by the Breguet Brothers. The configuration of a quadrotor helicopter is shown in Fig. 1. It consists of a body fixed frame and four rotors or propellers which can generate four independent thrusts Fi\u2019s (i = 1, 2, 3, 4). The two pairs of rotors (1,3) and (2,4) turn in opposite directions: the rotation direction of 1 and 3 are clockwise while 2 and 4 are counterclockwise, in order to balance the moments and produce yaw motions as needed. By varying the rotor speeds, one can change the lift forces and create motion. Thus, increasing or decreasing the four propeller\u2019s speeds together generates vertical motion"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002362_j.ymssp.2020.106640-Figure17-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002362_j.ymssp.2020.106640-Figure17-1.png",
        "caption": "Fig. 17. Geared rotor dynamic model.",
        "texts": [
            " 16 (tooth pair C is in Fig. 10). The shadowed area represents the spalling region. It can be concluded that the spalling faults occur more frequently in single tooth contact region due to the high contact stress. It is remarkable that the multiple teeth spalling is taken into consideration (tooth pair D and E). However, spalling defect only occurs on one tooth surface in traditional models. In order to take the flexibility of the shafts into account, the shafts are modeled using Timoshenko beam elements (see Fig. 17). The two shafts are supported by four identical rolling bearings, which are simplified as linear springs. The mesh stiffness and the non-loaded static transmission error are imported into the geared rotor dynamic model to acquire the dynamic response. The non-loaded static transmission error (also known as geometric transmission error) includes these four parts: nlste \u00bc erunout \u00fe etooth \u00fe eerror \u00fe espall \u00f06\u00de where erunout, etooth, eerror and espall represent the run-out error, the tooth profile error, the random error and the error caused by the spalling defect",
            " The expressions of the run-out error, the tooth profile error, the random error are as follows [41]: erunout \u00bc 30 10 6sin 2pt Tr \u00fe / \u00f07\u00de where Tr represents the rotational period of the shaft. u denotes the phase difference (in this paper, u = 0). etooth \u00bc 15 10 6sin 2pZt Tr \u00fe / \u00f08\u00de where Z is the number of teeth. eerror \u00bc 0:2 10 6randn \u00f09\u00de where randn denotes the random generator of standard normal distribution. The contact relationship of the spur gear pair is modeled as meshing element. The general displacement vector of the gear pair can be expressed as: X12 \u00bc \u00bdx1 y1 z1 hx1 hy1 hz1 x2 y2 z2 hx2 hy2 hz2 \u00f010\u00de The definition of x1, x2, y1, y2, hz1 and hz2 can be found in Fig. 17. The relative distance between the gear pair can be expressed as: p12 \u00bc \u00f0x1sina0 x2sina0 y1cosa0 \u00fe y2cosa0 \u00fe rb1hz1 \u00fe rb2hz2\u00de nlste \u00f011\u00de where a0 is the pressure angle of the reference circle. The equations of motion of the geared rotor system are: M\u20acu\u00fe \u00f0C\u00fe G\u00de _u\u00fe Ku \u00bc Fu \u00f012\u00de where u is the displacement vector. Fu is the load vector. M, C, G, K represent mass matrix, damping matrix, gyroscopic matrix and stiffness matrix, respectively. The detailed information of these matrixes is elaborately expressed in Appendix A and Refs"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure8.12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure8.12-1.png",
        "caption": "Fig. 8.12. AFPM machine cooled by heat pipes.",
        "texts": [
            " The high temperature and the corresponding high pressure in this region cause the vapour to flow to the cooler condenser region, where the vapour condenses, dissipating its latent heat of vaporisation. The capillary forces existing in the wicking structure then pump the liquid back to the evaporator. The wick structure thus ensures that the heat pipe can transfer heat whether the heat source is below the cooled end or above the cooled end. A heat pipe presents an alternative means for removing the heat from the AFPM machine. The heat pipe in an AFPM machine may be configured as shown in Fig. 8.12. Heat is transferred into the atmosphere through the finned 268 8 Cooling and Heat Transfer surface. The finned surface is cooled by air moving over the fins. The heat loss removed by the heat pipe, \u2206Php, is given by [235]: \u2206Php = \u03d1hot \u2212 \u03d1cold 1 hhotAhot + 1 hcoldAcold + 1 \u03b7finhfinAfin (8.29) where \u03d1hot is the average temperature of the elements that surround the heat pipe in the stator, \u03d1cold is the average temperature of the air cooling the finned surface, hhot is the convective heat transfer coefficient on the inside wall of the heat pipe in the stator, Ahot is the exposed area of the heat pipe in the stator, hcold is the convection heat transfer coefficient on the inside wall of the heat pipe in the finned area, Acold is the exposed area of the heat pipe at the finned surface, \u03b7fin is the efficiency of the finned surface, hfin is the convection heat transfer coefficient on the surface of the fins and Afin is the total exposed area of the finned surface",
            " Due to the temperature dependency of Rr1, a simple computer program using Gauss Seidel iteration has been created to find the solutions of the equations. The results are given in Table 8.4. Numerical Example 8.2 A totally enclosed AFPM brushless machine has a power loss of \u2206P1W =2500 W in the stator winding at continuous duty. The machine\u2019s outer and inner diameters are Dout = 0.72 m and Din = 0.5 m respectively. To remove the heat from the stator, use is made of heat pipes for direct cooling of the motor as shown in Fig. 8.12. The convection heat transfer coefficients on the inside wall of heat pipes in the stator and in the finned area are assumed to be h = 1000 W/(m2 oC). The average convective heat transfer coefficient on the fin surface is taken as hfin = 50 W/(m2 oC). The finned surface has an overall area of Afin = 1.8 m2 and an efficiency of \u03b7fin = 92%. The length of heat pipe embedded in the finned surface is lfin = 1.5 m. Find: (a) Steady-state temperature of the stator winding if the heat pipe with a diameter Dhp = 9 mm is placed along the average radius of the stator 8"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000554_978-3-540-85629-0-Figure7.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000554_978-3-540-85629-0-Figure7.1-1.png",
        "caption": "Fig. 7.1. Linear model with one degree of freedom.",
        "texts": [
            " The order of presentation starts from simple to more complex models, able to better describe the dynamics of the real mechanical systems, with the drawback of an increased computational complexity. For notational ease, all the models are described considering linear motions and forces. The discussion is obviously valid also in case of rotational movements and torques. 7.1.1 Linear model with one degree of freedom Let us consider the lumped parameters model composed by a mass m, a spring with elastic coefficient k and a dumper d that takes into consideration the friction and then dissipates energy, see Fig. 7.1. Let x be the position of the mass m, and y the input position of the mechanism, i.e the position of the actuation system. The dynamics of the system is described by the differential equation mx\u0308 + d x\u0307 + k x = d y\u0307 + k y. (7.1) If the difference z = x\u2212y due to the elastic element is considered, one obtains mz\u0308 + d z\u0307 + k z = \u2212my\u0308 or z\u0308 + 2 \u03b4 \u03c9nz\u0307 + \u03c92 nz = \u2212y\u0308 (7.2) where \u03c9n = \u221a k m , \u03b4 = d 2m\u03c9n are the natural frequency and the dumping coefficient respectively of the considered dynamic model. The second order differential equation (7",
            "28) can be estimated by computing the integral in a closed form or in a numerical way, or, equivalently, by considering the output of the stable linear filter G\u2212 0 (s) with the profile q(t) as input. 9 The double S trajectory considered in all the examples of this section is obtained with the conditions q0 = 0, q1 = 10, vmax = 10, amax = 50, jmax = 100. The total duration of the trajectory is T = 1.585 s. 334 7 Dynamic Analysis of Trajectories Example 7.8 Given the one-dof elastic system already considered in Sec. 7.1.1 and described by the transfer function G(s) = d s + k ms2 + ds + k where the meaning of the parameters m, d, k is explained in Fig. 7.1, the feedforward filter is F (s) = G\u22121(s) = [ m d s + ( 1 \u2212 mk d2 )] + G\u2212 0 (s) with G\u2212 0 (s) = mk2 d3 1 s + k/d . Therefore, the signal y(t) which guarantees a perfect tracking of the trajectory q(t) is composed by two terms. The former is represented by a linear combination of q(t) and q\u0307(t), while the latter is given by the trajectory q(t) filtered by G\u2212 0 (s) (see Fig. 7.63): y(t) = [ m d q\u0307(t) + ( 1 \u2212 mk d2 ) q(t) ] + \u222b t 0 mk2 d3 e\u2212 k d (t\u2212\u03c4)q(\u03c4)d\u03c4. (7.29) Differently from Example 7.7, in this case a continuous input function y(t) can be obtained if the trajectory q(t) is continuous up to the first derivative"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure9.35-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure9.35-1.png",
        "caption": "FIGURE 9.35 A rectangular bar with thickness.",
        "texts": [
            "15 and the shorted ends as in Fig. 9.16. Consider the dimensions xs1 = xs2 = 0, xe1 = xe2 = 2 cm, ys1 = 0, ye1 = 0.1 cm, ys2 = ye1 + 0.2 cm, ye2 = ys2 + 0.1 cm, zs1 = zs2 = 0, ze1 = ze2 = 50 \u03bcm. The result will be a frequency-dependent 2\u00d72 impedance matrix. PROBLEMS 245 9.2 Construct a Single-conductor VFI skin-effect model Construct a single-conductor skin-effect model similar to the previous problem. Compute the impedance from end to end of the rectangular conductor (copper \ud835\udf0e = 5.8 \u00d7 107 S\u2215m), illustrated in Fig. 9.35. Consider xs = 0, xe = 5 mm, ys = 0, ye = 20 \u03bcm, zs = 0 \u03bcm, ze = 10 \u03bcm. Compute the length impedance in the x-direction of the conductor at 1 MHz, 1 GHz, and 10 GHz. Use the model in Fig. 9.13 with subdivisions along the length x. 9.3 Skin effect using 1D model in Section 9.2.2 Compare the impedance of the previous problem with a skin effect with a model that is constructed with the high-frequency 1D skin-effect model in Section 9.2.2. Hint: Assume that all four side surfaces are represented with such a 1D impedance model"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.38-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.38-1.png",
        "caption": "FIGURE 7.38. Bicycle model for a negative 4WS vehicle.",
        "texts": [
            " To find the vehicle\u2019s turning radius R, we may define equivalent bicycle models as shown in Figure 7.37 and 7.38 for positive and negative 4WS 420 7. Steering Dynamics vehicles. The radius of rotation R is perpendicular to the vehicle\u2019s velocity vector v at the mass center C. Let\u2019s examine the positive 4WS situation in Figure 7.37. Using the geometry shown in the bicycle model, we have R2 = (a2 + c2) 2 +R21 (7.131) cot \u03b4f = R1 c1 = 1 2 (cot \u03b4if + cot \u03b4of ) (7.132) and therefore, R = q (a2 + c2) 2 + c21 cot 2 \u03b4f . (7.133) Examining Figure 7.38 shows that the turning radius of a negative 4WS vehicle can be determined from the same equation (7.133). Example 289 F FWS and 4WS comparison. The turning center of a FWS car is always on the extension of the rear axel, and its steering length ls is always equal to 1. However, the turning center of a 4WS car can be: 7. Steering Dynamics 421 1\u2212 ahead of the front axle, if ls < \u22121 2\u2212 for a FWS car, if \u22121 < ls < 1 or 3\u2212 behind the rear axle, if 1 < ls A comparison among the different steering lengths is illustrated in Figure 7"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.26-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.26-1.png",
        "caption": "FIGURE 8.26. An anti-roll bar attached to a solid axle with coil springs.",
        "texts": [
            " Coil springs are used in vehicles because they are less stiff with better 470 8. Suspension Mechanisms ride comfort compared to leaf springs. Therefore, the roll stiffness of the vehicle with coil springs is usually less than in vehicles with leaf springs. To increase the roll stiffness of such suspensions, an antiroll bar must be used. Leaf springs with reduced layers, uni-leaf, trapezoidal, or nonuniform thickness may also need an antiroll bar to compensate for their reduced roll stiffness. The antiroll bar is also called a stabilizer. Figure 8.26 illustrates an anti-roll bare attached to a solid axle with coil springs. Example 315 Need for longitudinal compliance. A bump is an obstacle on the road that opposes the forward motion of a wheel. When a vehicle goes over a bump, the first action is a force that tends to push the wheel backward relative to the rest of the vehicle. So, the lifting force has a longitudinal component, which will be felt inside the vehicle unless the suspension system has horizontal compliance. There are situations in which the horizontal component of the force is even higher than the vertical component"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000556_bfb0036073-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000556_bfb0036073-Figure2-1.png",
        "caption": "Fig. 2. Generalized coordinates of a unicycle",
        "texts": [
            " 2 M o d e l i n g a n d a n a l y s i s o f t h e c a r - l i k e r o b o t In this section, we shall first derive the kinematic equations of a car-like robot and then analyze the fundamental properties of the corresponding system from a control viewpoint. 2.1 K inema t i c mode l ing The main feature of the kinematic model of wheeled mobile robots is the presence of nonholonomic constraints due to the rolling without slipping condition between the wheels and the ground. The case of a single wheel is analyzed first. Consider a wheel that rolls on a plane while keeping its body vertical, as shown in Fig. 2. This kind of system is also referred to as a unicycle. Its configuration can be described by a vector q of three generalized coordinates, namely the position coordinates x, y of the point of contact with the ground in a fixed frame and the angle ~ measuring the wheel orientation with respect to the x axis. The system generalized velocities q cannot assume independent values; in particular, they must satisfy the constraint [ sin ~ - cos 8 0 ] -= 0, (2) entailing that the linear velocity of the wheel center lies in the body plane of the wheel (zero lateral velocity)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure4.38-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure4.38-1.png",
        "caption": "Figure 4.38 In a plane, a rigid body has three degrees of freedom; (b) due to the bar support at A this number is reduced to two; (c) two bar supports act as a hinged support at the centre of rotation RC, the point of intersection of the two bars; there is only one degree of freedom left: the rotation about RC; (d) with three bar supports, the body is immovable or kinematically determinate.",
        "texts": [
            " If there are too many unknown forces to be able to determine them from the equilibrium, the structure is said to be statically indeterminate. In order to determine the forces in a statically indeterminate structure, the deformation of the structure must be taken into account. 4 Structures 131 A dimensionally stable structure or self-contained structure is a structure that, isolated from its supports, retains its shape. If we neglect the deformations that occur, a self-contained structure can be seen as a rigid body. In a plane, a rigid body has three degrees of freedom: two components of a translation and one rotation, see Figure 4.38a. In Figure 4.38b, the block is supported by a bar (two-force member) and is free to move in the direction perpendicular to the bar (on the condition that the movements remain small, see Section 4.3.1) and can rotate about A. The bar support at A reduces the three degrees of freedom of the body to two. The freedom of movement can be limited further with a second bar support, for example at B (see Figure 4.38c). The movement that the body can now perform, with (minor) movement at A and B perpendicular to the bars, can be interpreted as a rotation about the so-called (instantaneous) centre of rotation RC, which is located on the intersection of the two bars.1 With two bar supports the number of degrees of freedom of the block has been reduced to one. The last degree of freedom, the rotation about RC, can be removed with a third bar support, for example at C (see Figure 4.38d). The three bars now prevent all possible movement. However the body is pulled or pushed, it remains where it is. This is referred to as the body having an immovable or kinematically determinate support. Three bar supports (at least) are required for an immovable or kinematically 1 The fact that the centre of rotation RC is a fixed point is true only if the rotation is still small. When considering Figure 4.42, one should not be confused by the fact that the displacements have been drawn to a large scale with respect to the dimensions of the structure",
            " In the support in Figure 4.39a, all bars intersect at the rotation centre RC, allowing the body to rotate. This support is movable or kinematically indeterminate. The support in Figure 4.39b, in which all bars are parallel to one another (intersect at a point at infinite distance), is also kinematically indeterminate, as the body is free to move in the direction perpendicular to the bar supports. The similarities between bar supports and roller supports were repeatedly pointed out in Section 4.2. Figure 4.38c also shows that two bar supports act as a hinged support at the rotation centre RC, the intersection of the two bars. An immovable support of a rigid body is therefore also possible with three roller supports, as in Figure 4.40a, or with a roller and hinged support, as in Figure 4.40b. It should be clear that a fixed support of a rigid body, as in Figure 4.40c, also is an immovable support. Instead of investigating the freedom of movement of a body, it is possible to determine also how many support reactions would be needed to keep the body in equilibrium under all imaginable loading conditions",
            " The displacement of BE with respect to AG consists of a rotation about RC(BE), the centre of rotation of BE, that coincides with 1 The concept self-contained was covered earlier in Section 4.5.1. 2 The open circles for hinged joints are consistently omitted (see Section 9.1.1). 9 Trusses 329 Figure 9.18 Based on a simple triangle, we can repeatedly create a new joint by adding two members. Figure 9.19 Simple trusses constructed in the way shown in Figure 9.18. In (c) and (d) we can start with the dark triangle in the middle. For all these trusses it holds that s = 2k \u2212 3. the intersection of two-force members AB and EG. See also Section 4.5.1 and Figure 4.38c. When looking at the deformed quadrangle once more, it is important to note that the displacements are depicted large in the figure as compared to the length of the members. The simplest way of constructing self-contained trusses is to start with a triangle, and, as in Figure 9.18, repeatedly create a new joint with two members. To retain its shape the truss does not have to consist only of triangles. For example, the quadrangle ABEG from Figure 9.17 is found again in the self-contained truss in Figure 9"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002097_978-1-4939-3017-3-Figure9.1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002097_978-1-4939-3017-3-Figure9.1-1.png",
        "caption": "Figure 9.1.1. The pendulum on a cart on an inclined plane.",
        "texts": [
            " The unifying idea is that control inputs are used to alter (and prescribe) the form of the Lagrangian governing the behavior of the q-variables. Hence both the design and analysis of control laws may be carried out using the powerful machinery of analytical mechanics. Equations (9.1.2)\u2013(9.1.3) provide mathematical descriptions of a wide and interesting class of controlled mechanical systems, examples of which will be studied in detail. 9.1.2 Example (Cart with Pendulum on an Inclined Plane). Consider the free pendulum on a cart on an inclined plane. This is depicted in Figure 9.1.1. Let s denote the position of the cart along the incline, and let \u03b8 denote the angle of the pendulum with respect to the downward pointing vertical. The configuration space for this system is Q = R 1 \u00d7 S1, with the first factor being the cart position s and the second factor being the pendulum angle \u03b8. We assume that s is directly controlled, while \u03b8 is controlled only indirectly through dynamic interactions within the mechanism. Clearly, 470 9. Energy-Based Methods for Stabilization this can be cast in the form (9",
            ", there is no g\u03c1 modification needed), but we can also choose the metric g\u03c3 to modify the original metric g only in the group directions by a scalar factor \u03c3. In this case, the controlled Lagrangian takes the form L\u03c4,\u03c3(v) = L(v + [\u03c4(v)]Q(q)) + \u03c3 2 g([\u03c4(v)]Q, [\u03c4(v)]Q). (9.2.7) The Inverted Pendulum on a Cart The system we consider is the inverted pendulum on a cart that we described in Chapter 1. We remind the reader of the setup here: The Lagrangian. Recall that we let s denote the position of the cart on the s-axis and let \u03b8 denote the angle of the pendulum with the vertical, as in Figure 9.1.1. In the notation depicted in Figure 9.1.1, \u03c8 = 0. The configuration space for this system is Q = S \u00d7 G = S1 \u00d7 R. For notational convenience we rewrite the Lagrangian as L(\u03b8, s, \u03b8\u0307, s\u0307) = 1 2 (\u03b1\u03b8\u03072 + 2\u03b2 cos \u03b8s\u0307\u03b8\u0307 + \u03b3s\u03072) +D cos \u03b8, (9.2.8) where \u03b1 = ml2, \u03b2 = ml, \u03b3 = M + m, and D = mgl are constants. The momentum conjugate to \u03b8 is p\u03b8 = \u2202L \u2202\u03b8\u0307 = \u03b1\u03b8\u0307 + \u03b2 (cos \u03b8) s\u0307, and the momentum conjugate to s is ps = \u2202L \u2202s\u0307 = \u03b3s\u0307+ \u03b2 (cos \u03b8) \u03b8\u0307. The relative equilibrium defined by \u03b8 = \u03c0, \u03b8\u0307 = 0, and s\u0307 = 0 is unstable, since D > 0. 476 9. Energy-Based Methods for Stabilization The equations of motion for the pendulum\u2013cart system with a control force u acting on the cart (and no direct forces acting on the pendulum) are, as we saw earlier, d dt \u2202L \u2202\u03b8\u0307 \u2212 \u2202L \u2202\u03b8 = 0, d dt \u2202L \u2202s\u0307 = u, that is, d dt p\u03b8 + \u03b2 sin \u03b8s\u0307\u03b8\u0307 +D sin \u03b8 = 0; that is, d dt (\u03b1\u03b8\u0307 + \u03b2 cos \u03b8s\u0307) + \u03b2 sin \u03b8s\u0307\u03b8\u0307 +D sin \u03b8 = 0 (9"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003581_jsvi.1997.1031-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003581_jsvi.1997.1031-Figure3-1.png",
        "caption": "Figure 3. The bearing geometry for an inner race defect.",
        "texts": [
            " (19) Therefore, the frequency spectrum of the displacement for both radial and axial load will have components at Zjvc (j=1, 2, 3, . . .); i.e., the outer race defect frequency and its multiples. The amplitude of the component at Zjvc is YZjvc =0si P(j)Xi (j) Miv 2 i 1ZFZj . (20) For an axial load, P(j) is constant for any position of the defect, but for a radial load, P(j) changes with the change in position of the defect. When a defect on the inner race is struck by the rolling elements, the response obtained is mainly due to the vibration of inner race. The relevant geometry is shown in Figure 3. The frequency of impact of a particular rolling element with a defect on the inner race is (vc \u2212vs ). Since the inner race is not stationary, the ball/roller load is a function of time. The generalized force due to the impact of inner race defect with a rolling element is Q'i =P(vct)F(t) g 2p 0 Xi (f)d[f\u2212(vc \u2212vs )t] df =P(vct)F(t)Xi [(vc \u2212vs )t]. (21) The pulse F(t) has a period of 2p/(vc \u2212vs ) and, therefore, F(t) can be expressed as F(t)=F0 + s s Fs cos s(vc \u2212vs )t. (22) 3.2.1. Axial loading In the case of axial loading, the generalized force due to the striking of the defect by the kth rolling element is Q'i =PtFo [sin i(vct\u2212vst\u2212 uk )+ cos i(vct\u2212vst\u2212 uk )] + s s PtFs 2 [sin (i+ s)(vct\u2212vst\u2212 uk )+ sin (i\u2212 s)(vct\u2212vst\u2212 uk ) + cos (i+ s)(vct\u2212vst\u2212 uk )+ cos (i\u2212 s)(vct\u2212vst\u2212 uk )]"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002381_j.matdes.2017.12.031-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002381_j.matdes.2017.12.031-Figure1-1.png",
        "caption": "Figure 1: Integrated modeling framework to reproduce the major procedures of SEBM.",
        "texts": [
            " Keywords: Electron Beam, Additive Manufacturing, Thermal-fluid Flow, Discrete Element Method, Multiple Layers, Scan Path Preprint submitted to Journal Name December 19, 2017 AC C EP TE D M AN U SC R IP T Powder bed based Additive Manufacturing (AM) technologies for metallic components, including Selective Laser Melting (SLM) and Selective Electron Beam Melting (SEBM), have drawn increasing attention over the past decade. In addition to manufacturing components with complex geometry, these powder-based AM technologies are opening new avenues for locally manipulating chemical compositions and mechanical properties. For example, Ge et al. [1, 2, 3] manufactured functionally graded Ti-TiAl materials using SEBM. The SLM/SEBM process mainly consists of two repeated procedures [1], as shown in Fig.1. 1. Apply one layer of powder on a preheated platform or previously deposited layers. 2. Selectively melt the powder bed along the designated scan path. Although the basic principle of SLM/SEBM is rather straightforward, the actual processes consists of multiple physical phenomena, e.g., powder particle packing, heat transfer, phase transformation and fluid flow, where a number of factors have large influence over the process and fabrication quality [4]. There are a considerable number of fundamental physical mechanisms affecting each fabrication procedure, complicating understanding of the process",
            " Most powder melting models only simulate the melting process of a single track, which has not shed light into how successive tracks and layers interact with each other. Moreover, few studies have been done to comprehensively incorporate the procedure of powder spreading. In this study, we develop an integrated modeling framework to simulate the two repeated procedures of SEBM: spreading the powder layer and selectively melting the newly applied powder. This approach consists of a powder spreading model using the DEM and a thermal-fluid flow model for powder AC C EP TE D M AN U SC R IP T melting, as shown in Fig.1. The powder spreading model simulates the frictional contact and collisions both between particles and with the rake and substrate, and the powder melting model captures the material evolution process as a result of input energy, including heat transfer, phase change, and molten pool flow. The 3D geometry model of the powder bed used in the powder melting model is generated by the powder spreading model, with the powder size distribution taken from experimental measurements. After simulating the melting and solidification process, the solidified geometry is transferred back to the powder spreading model to apply a fresh powder layer"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001383_025403-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001383_025403-Figure4-1.png",
        "caption": "Figure 4. Simplified schematic of the powder\u2014melt-pool interaction in coaxial regime (melt-pool 1 is included in the powder stream whereas melt pool 2 is longer than the melt-pool length). The dotted line corresponds to the interaction area between melt-pool 2 and the powder distribution (based upon Toyserkani et al [4]).",
        "texts": [
            " Prediction of widths and deposition heights. The main assumption of our model is that powder feeding contributes to direct manufacturing provided that: (1) powder grains arriving at average temperature, TP and with a local mass rate D\u2217 m interact with the molten pool and (2) the local energy Qij (J ) inside the melt pool is high enough to melt the incoming powder. Considering those two necessary factors for growing matter, our model first considers a simple local interaction between melt pool and powder distribution (figure 4), then applies a simplified thermal model to limit layer growth by comparing the energy required to melt the powder stream with the energy present in the melt pool. The main issue here is to calculate the interaction area between different meltpool sizes and a unique powder distribution area (figure 4). When the melt-pool size is totally included into the powder distribution (melt-pool 1 on figure 4), the melt-pool area forms the interaction area, whereas when the melt-pool length is longer than the powder stream radius (melt-pool 2), one has to consider a specific interaction area, smaller than the melt-pool and restricted by the powder area. The melt-pool geometry was calculated using steady state calculations on COMSOL Multiphysics\u2122 software (figure 5(a)), using the Eulerian formalism where matter flows through a fixed mesh matrix (Lagrange linear tetrahedrons elements), simultaneously irradiated by a stationary laser source"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure2.4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure2.4-1.png",
        "caption": "Fig. 2.4 Arm Local Coordinates \u03a3a and the Lower Arm Local Coordinates \u03a3b",
        "texts": [
            " The Homogeneous Transformation T a converts points described in arm local coordinates to world coordinates [ p 1 ] = T a [ ap 1 ] . Note that the point ap can be any point in the left arm, thus the arm shape can be represented by its aggregation. On the other hand, the arm configuration (position and orientation) is separately represented by the matrix T a. In general, we can say that a Homogeneous Transformation in itself describes the position and attitude of an object. It is possible to define a Local Coordinate System which has another Local Coordinate System as it\u2019s parent. In Fig. 2.4(a) we show a Local Coordinate System \u03a3b which has \u03a3a as it\u2019s parent. \u03a3b moves together with the lower arm and is set so that the axes are in-line with the axes of \u03a3a when the elbow is straight. Let us define the rotation angle of the elbow joint as \u03b8. Let us also define the unit vectors of which are the axes x, y, z of \u03a3b as aebx, aeby, aebz respectively. They are defined as follows, 2 The same matrix is referred to in computer graphics as an Affine transformation matrix. There is a difference as in an Affine transformation, Ra is not necessarily a simple rotation matrix (\u2192 Section 2.2) and may include scaling and shearing as well. 24 2 Kinematics aebx = \u23a1 \u23a3 cos \u03b8 0 sin \u03b8 \u23a4 \u23a6 , aeby = \u23a1 \u23a3 0 1 0 \u23a4 \u23a6 , aebz = \u23a1 \u23a3 \u2212 sin \u03b8 0 cos \u03b8 \u23a4 \u23a6 . (2.8) Since the elbow rotates around the y axis, only the vectors aebx and aebz change by \u03b8. The vectors are defined in \u03a3a space and thus they have the indicator a on the top left. Let us define a matrix aRb as a combination of the three unit vectors. aRb \u2261 [aebx aeby aebz] (2.9) The conversion of bph (in Fig. 2.4(a)) which is in \u03a3b space, to aph (in Fig. 2.4 (b)) in \u03a3a space becomes as follows [ aph 1 ] = aT b [ bph 1 ] . (2.10) Here we defined a homogeneous transformation aT b as follows aT b \u2261 [ aRb apb 0 0 0 1 ] . Note that apb is the origin of \u03a3b viewed from \u03a3a. By combining the results of (2.10) with (2.7) we get the equation which converts a point defined in \u03a3b space to world coordinates. The example below converts the manipulator end point in \u03a3b [ ph 1 ] = T a aT b [ bph 1 ] . (2.11) 2.2 Characteristics of Rotational Motion 25 Let us define the product of homogeneous transformation on the right hand side of the equation as one matrix and call it T b T b \u2261 T a aT b"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure4.50-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure4.50-1.png",
        "caption": "Fig. 4.50: A linkage mechanism for an articulated finger with 1 d.o.f.",
        "texts": [
            "49b) refers to a planar grasp that is obtained by the thumb and index finger, whose motion is properly described through the corresponding joint angles \u03b81, \u03b82, and \u03b83. Similar schemes can be determined for each of the many grasping configurations that a human hand can adapt to different objects. The kinematics of a finger motion can also be studied as referring to a scheme of the finger as a planar 3R manipulator. Fundamentals of the Mechanics of Robotic Manipulation 301 Chapter 4 Fundamentals of the Mechanics of Grasp302 Alternatively, planar mechanism can be used even to obtain 1 d.o.f. articulated fingers, as in the case of the diagram in Fig. 4.50. The 1 d.o.f. mechanism solution for fingers is of great interest since it permits the obtaining of simple low-cost designs, but it requires special attention for feasible operation. In particular the scheme of Fig. 4.50 can be used to obtain specific size of the linkages so that the phalanx bodies can move with a good approximation, as in a human finger. An example is shown in Fig. 4.51, in which the closing movement of the index finger is reproduced with a specific design of the linkage mechanism of Fig. 4.50. Fundamentals of the Mechanics of Robotic Manipulation 303 Fig. 4.50, as built in Cassino. Fig. 4.52 shows a prototype of a 1 d.o.f. articulated finger that has been built by using the mechanism layout of Fig. 4.50 with a specific dimensional synthesis to obtain a compact design in which the linkages are hidden in the finger body. The mechanism solution for an articulated finger of Fig. 4.50 can also be considered as convenient because it permits the application of features and requirements for design and operation performances that can be deduced similarly to those for grippers. An alternative design that is widely used, is shown in Fig. 4.53 in which the architecture of a planar 3R manipulator can be easily recognized. In the scheme the problem of multiple grasping contacts is also indicated by means of the determination of the contact points P1, P2, P3 . The action of a finger grasping an object through the forces F1, F2, F3 at the contact points P1, P2, P3 can be evaluated in a straightforward way by applying the Principle of Virtual work, when friction forces are assumed as negligible, in the form 332211332211 vFvFvF ++=\u03b8\u03c4+\u03b8\u03c4+\u03b8\u03c4 (4"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000252_j.automatica.2006.10.008-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000252_j.automatica.2006.10.008-Figure4-1.png",
        "caption": "Fig. 4. Performance of the controller (28), (25), (26) with noisy measurements.",
        "texts": [],
        "surrounding_texts": [
            "The controller includes a real-time second-order differentiator (Levant, 2003) z\u03070 = v0 = \u221213.4|z0 \u2212 |2/3 sign(z0 \u2212 ) + z1, (29) z\u03071 = v1 = \u221226.0|z0 \u2212 v0|1/2 sign(z0 \u2212 v0) + z2, (30) z\u03072 = \u2212330 sign(z2 \u2212 v1), = x \u2212 xc, (31) where z0, z1, z2 are real-time estimations of , \u0307, \u0308, respectively. Differentiator (29)\u2013(31) is exact for input signals with the third derivative not exceeding 300 in absolute value. The initial values z0(0) = (0), z1(0) = z2(0) = 0 were taken. Consider a controller of the form (19), (23) which establishes in finite time the 2-sliding mode = \u0307 = 0, where = + \u0307. Substituting the estimations z0, z1, z2 of the derivatives , \u0307, \u0308 obtain the controller u\u0307 = min{30, max[\u221230, \u221240(ln(|s1|/|s0|1/2 + 1) \u00b7 sign s1 + ln 2 \u00b7 sign s0)]} with |u| < 10, |s1| < 100|s0|1/2, (32) u\u0307 = \u221230 sign s1 with |u| < 10, |s1| 100|s0|1/2, (33) u\u0307 = \u2212u with |u| 10, s0 = z0 + z1, s1 = z1 + z2, (34) where the function (\u2217) = ln(\u2217 + 1) is chosen, u(0) = 0. Without changing the control values, (33) is constructed so that the overflow be avoided during the computer simulation (or practical implementation). In the absence of noises the controller provides in finite time for keeping + \u0307 \u2261 0. As a result the asymptotically stable 3-sliding mode =\u0307=\u0308=0 is established. Note that since (19) is quasi-continuous, the applied controller (29)\u2013(34) produces the control u, whose derivative u\u0307 remains continuous until the entrance into the 2-sliding mode =\u0307=0. The graph of u\u0307 is very similar to Fig. 3e and is omitted. The graphs of 3-sliding deviations , \u0307, \u0308 and the control u are demonstrated in Figs. 5a,c, respectively. The 2-sliding convergence in the plane , \u0307 is demonstrated in Fig. 5b. It is seen from Fig. 5d that + \u0307 \u2261 0 is kept in 2-sliding mode. Due to this equality the accuracies achieved at t = 10 are of the same order: | | = |x \u2212 xc| 4.0\u00d710\u22124, |\u0307|=|x\u0307\u2212 x\u0307c| 4.0\u00d710\u22124, |\u0308|=|x\u0308\u2212 x\u0308c| 4.0 \u00d7 10\u22124."
        ]
    },
    {
        "image_filename": "designv10_0_0003517_978-1-4020-2110-7-Figure3.11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003517_978-1-4020-2110-7-Figure3.11-1.png",
        "caption": "Fig. 3.11: A descriptive view of workspace generation for serial manipulators with revolute joints.",
        "texts": [
            " Fundamentals of Mechanics of Robotic Manipulation 89 The workspace is defined as generated by a reference point H on the extremity of the manipulator chain that is moved to reach all possible positions because of the mobility ranges of the joints. Workspace geometry of manipulators with revolute joints has been recognized as a ring topology, and this can be conveniently used to analytically describe the workspace. Thus, the ring geometry can be used with the following modeling and reasoning by referring to Fig. 3.11 to characterize the generation of the manipulator workspace. Revolving a torus about an axis generates a ring. Therefore, a ring volume W3R(H) can be thought as the union of the points swept by the revolving torus TR2R3(H), due to the mobility in R2 and R3 joints, during the \u03b81 revolution about Z1 axis. This can be expressed as ( ) ( ) \u03c0 =\u03d1 = 2 0 RR3R 1 32 HTHW (3.1.27) Chapter 3: Fundamentals of the Mechanics of Robots90 Alternatively, a ring W3R(H) can be considered as the union of the tori TR1R2(H), which are due to the mobility in R1 and R2 joints and are traced by all parallel circles which can be cut on generating torus TR2R3(H) so that a ring workspace can be formulated as ( ) ( ) \u03c0 =\u03d1 = 2 0 RRR3 3 21 HTHW ",
            "28) Thus, the boundary \u2202W3R(H) of a ring can be thought as the envelope of torus surfaces generated by revolution of the generating torus or, alternatively, it can be obtained by an envelope of torus surfaces that are traced from the parallel circles of the generating torus. The latter procedure can be expressed according to Eq. (3.1.28) in the form ( ) ( )HTHW 21 3 RR 2 0 R3 env \u03c0 =\u03d1 =\u2202 (3.1.29) where \u2018env\u2019 is an envelope operator performing an envelope process. A ring geometry has been recognized as topologically common also to hyper-rings as a \u2018solid hollow ring\u2019 shape on the basis of a consideration of an iterative revolving process for hyper-ring generation in N-R manipulators. In fact, referring to Fig. 3.11 a (N-j+1)R hyper-ring W(N-j+1)R (H) is generated by revolving a (N-j)R hyper-ring W(N-j)R (H) about Zj axis. The (N-j+1)R hyper-ring, which is traced by a point H on the last link of an open chain with N-j+1 consecutive revolute pairs, is the volume swept by the (Nj)R hyper-ring during its revolution. A basic element is a generating parallel circle, which is cut on the revolving W(N-j)R (H) because of a revolution about Zj+1 axis and whose revolution about Zj generates a torus TRjRj+1 (H). The envelope of the tori of the family will give the boundary \u2202W(N-j+1)R (H) of the hyper-ring"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure12.11-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure12.11-1.png",
        "caption": "FIGURE 12.11. Equivalent mass-spring vibrator for a pendulum.",
        "texts": [
            " The mass ms represents one quarter of the car\u2019s body, which is mounted on a suspension made of a spring ks and a damper cs. When ms is vibrating at a position such as in Figure 12.10(b), its free body diagram is as Figure 12.10(c) shows. Applying Newton\u2019s method, the equation of motion would be msx\u0308 = \u2212ks (xs \u2212 y)\u2212 cs (x\u0307s \u2212 y\u0307) (12.41) which can be simplified to the following equation, when we separate the input y and output x variables. msx\u0308+ csx\u0307s + ksxs = ksy + csy\u0307. (12.42) Example 428 Equivalent mass and spring. Figure 12.11(a) illustrates a pendulum made by a point mass m attached to a massless bar with length l. The coordinate \u03b8 shows the angular position of the bar. The equation of motion for the pendulum can be found by using the Euler equation and employing the free-body-diagram shown in Figure 12.11(b). ml2\u03b8\u0308 = \u2212mgl sin \u03b8 (12.43) Simplifying the equation of motion and assuming a very small swing angle shows that l\u03b8\u0308 + g\u03b8 = 0. (12.44) 12. Applied Vibrations 741 This equation is equivalent to an equation of motion for a mass-spring system made by a mass m \u2261 l, and a spring with stiffness k \u2261 g. The displacement of the mass would be x \u2261 \u03b8. Figure 12.11(c) depicts such an equivalent mass-spring system. Example 429 Force proportionality. The equation of motion for a vibrating system is a balance between four different forces. A force proportional to displacement, \u2212kx, a force proportional to velocity, \u2212cv, a force proportional to acceleration, ma, and an applied external force f (x, v, t), which can be a function of displacement, velocity, and time. Based on Newton\u2019s method, the force proportional to acceleration, ma, is always equal to the sum of all the other forces"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001618_j.actamat.2016.11.018-Figure6-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001618_j.actamat.2016.11.018-Figure6-1.png",
        "caption": "Fig. 6. a) Transverse cross section with boundary and domain indicated and b) Longitudinal (scan direction) cross-section with boundary and domains.",
        "texts": [
            " Hence, the multi scale approach should be used to predict microstructure development during dendritic solidification. Phase field analysis was carried out on undercooled mushy region, since the high temperature inside melt pool does not allow nucleation of the dendrites. The numerical solution in 3D is very expensive and time consuming; therefore the problemwas divided into two plane problems in two orthogonal microstructural cross sections correspondingly. The first cross-section is in the transverse direction (y-z plane in Fig. 6a). By solving this \u201ctransverse\u201d plane problem we assume that all cross-sections have the same solidification pattern. Such an assumption allows quasi-steady state formulation, which does not include the laser source velocity. The second cross-section is in the scan velocity direction or in longitudinal plane (x-z plane in Fig. 6b). This problem requires direct account for the effect of moving heat source. In simulations of the longitudinal cross-section, the bottom and left (far end from the melt pool) boundaries remained at solidus temperature to replicate the conditions of moving mushy zone while the mushy zone still remained hot due to release of latent heat. First, the model for the stationary position of the heat source is presented for transverse cross section in Fig. 9a. The model adopted here is based on Kobayashi [19] and Boettinger et al"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.41-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.41-1.png",
        "caption": "FIGURE 8.41. A positive and negative camber on the front wheel of a car.",
        "texts": [
            " Suspension Mechanisms 483 Negative caster provides: low steering into the corner, easy steering out of the corner, more straight-line stability, high tireprint area during turn, good turn-in response, good directional stability, good steering feel. When a castered wheel rotates about the steering axis, the wheel gains camber. This camber is generally favorable for cornering. 8.4.3 F Camber Camber is the angle of the wheel relative to vertical line to the road, as viewed from the front or the rear of the car. Figure 8.41 illustrates the wheel number 1 of a vehicle. If the wheel leans in toward the chassis, it is called negative camber and if it leans away from the car, it is called positive camber. The cornering force that a tire can develop is highly dependent on its angle relative to the road surface, and so wheel camber has a major effect on the road holding of a car. A tire develops its maximum lateral force at a small camber angle. This fact is due to the contribution of camber thrust, which is an additional lateral force generated by elastic deformation as the tread rubber pulls through the tire/road interface"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure9.76-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure9.76-1.png",
        "caption": "Figure 9.76 The zero-force members in the truss.",
        "texts": [
            " For the given xy coordinate system, the equilibrium of joint D gives \u2211 Fx = N1 + N2 cos \u03b1 \u2212 N3 \u2212 N4 cos \u03b1 = 0,\u2211 Fy = N2 sin \u03b1 \u2212 N4 sin \u03b1 = 0 so that N1 = N3 and N2 = N4. Conclusion: Rule 4. If four members meet in an unloaded joint that in pairs are in a direct line with one another, these members can be considered crossing members as far as the transfer of forces is concerned. The three examples below show how it is possible to simplify the calculation with these four rules. Example 1 You are given the truss in Figure 9.75. Question: Which members are zero-force members for the given load? Solution: A is an unloaded joint in which two members meet (see Figure 9.76). Both members are zero-force members (rule 1), so that N1 = 0 and N2 = 0. B is an unloaded joint in which three members meet, and of which two are in a direct line with one another. The third member is therefore a zero-force 9 Trusses 367 member (rule 2), so that N9 = 0. C is a loaded joint where two members meet, and where the line of action of the load coincides with member 17. Thus (rule 3) N16 = 0. The zero-force members are shown in Figure 9.76 with a \u201c0\u201d through the member axis. Zero-force members do not participate in the force flow for the present load. When calculating the forces in the other members, you can leave out the zero-force members from the truss. If you leave out zero-force member 9 from the truss, you immediately notice that N8 = N10. If you leave out zero-force member 16, you see that N17 = \u2212F. That this (imaginary) omission of zero-force members can significantly reduce the effort in calculating is further emphasised in the following two examples"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002581_j.matdes.2019.107866-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002581_j.matdes.2019.107866-Figure5-1.png",
        "caption": "Fig. 5. (a) Thermal distribution on powder layer surface, (b) top view of powder layer surface, (c) dimensions of simulated melt-pool cross-section, and (d) experimental dimensions of melt-pool cross-section [25].",
        "texts": [
            "0119 is the contact size ratio between neighboring powder particles [1]. 4. Verification of simulation model with experimental data in [22] The validity of the thermal-fluid simulation model described above was confirmed by comparing the simulation results for the melt-pool dimensions (depth and width) with the experimental results reported by Kamath et al. [25] under identical processing conditions (e.g., scanning speed V, laser power P, beam diameter D, and powder layer thickness (see Table 1), and powder size distribution (Section 3.2). Fig. 5(a) and (b) show the temperature distribution on the top surface of the CL20ES powder bed given a scanning speed of V = 1800 mm/s and a laser power of P = 300 W. Note that the green regions of Fig. 5(a) and (b) correspond to the temperature above 1690 K (the melting temperature of the SS316 material) and therefore correspond to themelt-pool. Fig. 5(c) and (d) present the simulation and experimental results for the melt-pool cross-section, respectively. It is seen that a reasonable qualitative agreement exists between the dimensions of the twomelt-pools. Thus, the basic validity of the thermal-fluid model is confirmed. Fig. 6 compares the simulation results obtained for themelt-pool depth and melt-pool width, respectively, with the experimental results presented in [22] for a laser power of P = 200 W and a scanning velocity V in the range of 500\u20131500 mm/s"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000972_j.msea.2018.01.103-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000972_j.msea.2018.01.103-Figure2-1.png",
        "caption": "Fig. 2. Building strategy of tensile specimens (series A, B, C and D).",
        "texts": [
            " If the scan vectors in one layer were parallel to the x axis, in the subsequent layer they were parallel to the y axis. According to the XY scanning strategy (Fig. 1b), each layer was scanned two times. During the first scanning with use of alternating scan vectors parallel to the x axis, melting of powder occurred. During the second scanning with use of alternating scan vectors parallel to the y axis, the same layer was remelted only. The tensile specimens were built in four series, differing in orientation of their axes with respect to the build direction z (Fig. 2): series A (0\u00b0) \u2013 tension axis parallel to z, series B (90\u00b0) \u2013 tension axis normal to z, series C (45\u00b0) \u2013 tension axis at 45\u00b0 to z, series D (45\u00b0 x 45\u00b0) \u2013 tension axis at 45\u00b0 to z and also at 45\u00b0 to both scanning directions x and y. The tensile specimens were CNC finished to remove residues of support structures and to achieve smooth, notch-free surfaces. For each series, five specimens were used. Both kinds of specimens (cuboid-shaped and tensile specimens) were prepared in the standard procedure for microstructure observation",
            " This should be taken into account at evaluation of susceptibility of 316 L in as-built condition to stress corrosion cracking (SCC) and at selection of parameters of post-process heat treatment. Susceptibility to SCC is significant when hardness is higher than 300 HV1. From the viewpoint of grain refinement by recrystallization annealing, dislocation and highangle boundaries densities are low enough that annealing at a temperature over 1000 \u00b0C can lead only to grain growth and twinning. Tensile testing was carried-out on specimens prepared with scan strategy according to variant I, in four series (A, B, C, and D) differing in building strategy (see Fig. 2). Examinations of all series of the specimens made it possible to check repeatability of the properties determined for various orientations of tension direction in respect to build direction, i.e. in respect to grain growth direction, positions of tracks and layers. Only the variant I scanning strategy was selected for A-D series of specimens. The specimens were characterised by the largest amount of \u03b4-ferrite in comparison to the other variants and by the highest level of residual stresses due to the highest cooling rate"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001199_b978-0-08-100433-3.00013-0-Figure13.7-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001199_b978-0-08-100433-3.00013-0-Figure13.7-1.png",
        "caption": "Figure 13.7 The first titanium bracket connector produced using additive manufacturing on board the Airbus A350 XWB. Reproduced from Airbus. Cabin bracket produced on Concept Laser\u2019s LaserCUSING machine.",
        "texts": [
            " The new design features have more intricate cooling pathways and support ligaments, which result in five times the service life compared with the predecessor part made by conventional manufacturing. The AM process allows engineers to use a simpler design that reduces the number of brazes and welds from 25 to only 5; the number of parts used to make the nozzle was reduced from 18 to 1. In addition, the weight of these nozzles is 25% lighter than that of the predecessor part. Concept Laser [24] adopted the SLM process to fabricate aircraft components with a \u201cbionic\u201d design using their LaserCUSING system. An example of the bracket connector used in the Airbus A350 XWB is shown in Fig. 13.7. The bone-like porous structure significantly reduces the weight. The component was previously a milled part made of aluminum. Now it is made of titanium using AM and is significantly lighter. The new structure design is more than 30% lighter than conventional designs realized using casting or milling processes. The considerable weight reduction results in lower fuel consumption and the potential to increase the load capacity of the aircraft. On the other hand, toolless fabrication with the Concept Laser SLM process saves money and shortens the development time by up to 75%: Component development time was previously 6 months but is now down to 1 month"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure9.24-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure9.24-1.png",
        "caption": "FIGURE 9.24. A simplified models of a trebuchet.",
        "texts": [],
        "surrounding_texts": [
            "Dynamics of a rigid vehicle may be considered as the motion of a rigid body with respect to a fixed global coordinate frame. The principles of Newton and Euler equations of motion that describe the translational and rotational motion of the rigid body are reviewed in this chapter. 9.1 Force and Moment In Newtonian dynamics, the forces acting on a system of connected rigid bodied can be divided into internal and external forces. Internal forces are acting between connected bodies, and external forces are acting from outside of the system. An external force can be a contact force, such as the traction force at the tireprint of a driving wheel, or a body force, such as the gravitational force on the vehicle\u2019s body. External forces and moments are called load, and a set of forces and moments acting on a rigid body, such as forces and moments on the vehicle shown in Figure 9.1, is called a force system. The resultant or total force F is the sum of all the external forces acting on a body, and the resultant or 522 9. Applied Dynamics total moment M is the sum of all the moments of the external forces. F = X i Fi (9.1) M = X i Mi (9.2) Consider a force F acting on a point P at rP . The moment of the force about a directional line l passing through the origin is Ml = lu\u0302 \u00b7 (rP \u00d7F) (9.3) where u\u0302 is a unit vector on l. The moment of the force F, about a point Q at rQ is MQ = (rP \u2212 rQ)\u00d7F (9.4) so, the moment of F about the origin is M = rP \u00d7F. (9.5) The moment of a force may also be called torque or moment. The effect of a force system is equivalent to the effect of the resultant force and resultant moment of the force system. Any two force systems are equivalent if their resultant forces and resultant moments are equal. If the resultant force of a force system is zero, the resultant moment of the force system is independent of the origin of the coordinate frame. Such a resultant moment is called couple. When a force system is reduced to a resultant FP andMP with respect to a reference point P , we may change the reference point to another point Q and find the new resultants as FQ = FP (9.6) MQ = MP + (rP \u2212 rQ)\u00d7FP = MP + QrP \u00d7FP . (9.7) The momentum of a moving rigid body is a vector quantity equal to the total mass of the body times the translational velocity of the mass center of the body. p = mv (9.8) The momentum p is also called translational momentum or linear momentum. Consider a rigid body with momentum p. The moment of momentum, L, about a directional line l passing through the origin is Ll = lu\u0302 \u00b7 (rC \u00d7 p) (9.9) 9. Applied Dynamics 523 where u\u0302 is a unit vector indicating the direction of the line, and rC is the position vector of the mass center C. The moment of momentum about the origin is L = rC \u00d7 p. (9.10) The moment of momentum L is also called angular momentum. A bounded vector is a vector fixed at a point in space. A sliding or line vector is a vector free to slide on its line of action. A free vector is a vector that may move to any point as long as it keeps its direction. Force is a sliding vector and couple is a free vector. However, the moment of a force is dependent on the distance between the origin of the coordinate frame and the line of action. The application of a force system is emphasized by Newton\u2019s second and third laws of motion. The second law of motion, also called the Newton\u2019s equation of motion, states that the global rate of change of linear momentum is proportional to the global applied force. GF = Gd dt Gp = Gd dt \u00a1 mGv \u00a2 (9.11) The third law of motion states that the action and reaction forces acting between two bodies are equal and opposite. The second law of motion can be expanded to include rotational motions. Hence, the second law of motion also states that the global rate of change of angular momentum is proportional to the global applied moment. GM = Gd dt GL (9.12) Proof. Differentiating from angular momentum (9.10) shows that Gd dt GL = Gd dt (rC \u00d7 p) = \u00b5 GdrC dt \u00d7 p+ rC \u00d7 Gdp dt \u00b6 = GrC \u00d7 Gdp dt = GrC \u00d7 GF = GM. (9.13) Kinetic energy K of a moving body point P with mass m at a position GrP , and having a velocity GvP , is K = 1 2 mGv2P = 1 2 m \u00b3 Gd\u0307B + BvP + B G\u03c9B \u00d7 BrP \u00b42 . (9.14) 524 9. Applied Dynamics The work done by the applied force GF on m in moving from point 1 to point 2 on a path, indicated by a vector Gr, is 1W2 = Z 2 1 GF \u00b7 dGr. (9.15) However, Z 2 1 GF \u00b7 dGr = m Z 2 1 Gd dt Gv \u00b7 Gvdt = 1 2 m Z 2 1 d dt v2dt = 1 2 m \u00a1 v22 \u2212 v21 \u00a2 = K2 \u2212K1 (9.16) that shows 1W2 is equal to the difference of the kinetic energy between terminal and initial points. 1W2 = K2 \u2212K1 (9.17) Equation (9.17) is called principle of work and energy. Example 342 Position of center of mass. The position of the mass center of a rigid body in a coordinate frame is indicated by BrC and is usually measured in the body coordinate frame. BrC = 1 m Z B Br dm (9.18) \u23a1\u23a3 xC yC zC \u23a4\u23a6 = \u23a1\u23a2\u23a2\u23a3 1 m R B x dm 1 m R B y dm 1 m R B z dm \u23a4\u23a5\u23a5\u23a6 (9.19) Applying the mass center integral on the symmetric and uniform L-section rigid body with \u03c1 = 1 shown in Figure 9.2 provides the position of mass center C of the section. The x position of C is xC = 1 m Z B xdm = 1 A Z B x dA = \u2212b 2 + ab\u2212 a2 4ab+ 2a2 (9.20) and because of symmetry, we have yC = \u2212xC = b2 + ab\u2212 a2 4ab+ 2a2 . (9.21) 9. Applied Dynamics 525 When a = b, the position of C reduces to yC = \u2212xC = 1 2 b. (9.22) Example 343 F Every force system is equivalent to a wrench. The Poinsot theorem states: Every force system is equivalent to a single force, plus a moment parallel to the force. Let F andM be the resultant force and moment of a force system. We decompose the moment into parallel and perpendicular components,Mk andM\u22a5, to the force axis. The force F and the perpendicular moment M\u22a5 can be replaced by a single force F0 parallel to F. Therefore, the force system is reduced to a force F0 and a moment Mk parallel to each other. A force and a moment about the force axis is called a wrench. The Poinsot theorem is similar to the Chasles theorem that states: Every rigid body motion is equivalent to a screw, which is a translation plus a rotation about the axis of translation. Example 344 F Motion of a moving point in a moving body frame. The velocity and acceleration of a moving point P as shown in Figure 5.12 are found in Example 200. GvP = Gd\u0307B + GRB \u00a1 BvP + B G\u03c9B \u00d7 BrP \u00a2 (9.23) GaP = Gd\u0308B + GRB \u00a1 BaP + 2 B G\u03c9B \u00d7 BvP + B G\u03c9\u0307B \u00d7 BrP \u00a2 +GRB \u00a1 B G\u03c9B \u00d7 \u00a1 B G\u03c9B \u00d7 BrP \u00a2\u00a2 (9.24) Therefore, the equation of motion for the point mass P is GF = mGaP = m \u00b3 Gd\u0308B + GRB \u00a1 BaP + 2 B G\u03c9B \u00d7 BvP + B G\u03c9\u0307B \u00d7 BrP \u00a2\u00b4 +m GRB \u00a1 B G\u03c9B \u00d7 \u00a1 B G\u03c9B \u00d7 BrP \u00a2\u00a2 . (9.25) 526 9. Applied Dynamics Example 345 Newton\u2019s equation in a rotating frame. Consider a spherical rigid body, such as Earth, with a fixed point that is rotating with a constant angular velocity. The equation of motion for a moving point vehicle P on the rigid body is found by setting Gd\u0308B = B G\u03c9\u0307B = 0 in the equation of motion of a moving point in a moving body frame (9.25) BF = mBaP +mB G\u03c9B \u00d7 \u00a1 B G\u03c9B \u00d7 BrP \u00a2 + 2mB G\u03c9B \u00d7 B r\u0307P (9.26) 6= mBaP which shows that the Newton\u2019s equation of motion F = ma must be modified for rotating frames. Example 346 Coriolis force. The equation of motion of a moving vehicle point on the surface of the Earth is BF = mBaP +mB G\u03c9B \u00d7 \u00a1 B G\u03c9B \u00d7 BrP \u00a2 + 2mB G\u03c9B \u00d7 BvP (9.27) which can be rearranged to BF\u2212mB G\u03c9B \u00d7 \u00a1 B G\u03c9B \u00d7 BrP \u00a2 \u2212 2mB G\u03c9B \u00d7 BvP = mBaP . (9.28) Equation (9.28) is the equation of motion for an observer in the rotating frame, which in this case is an observer on the Earth. The left-hand side of this equation is called the effective force Feff , Feff = BF\u2212mB G\u03c9B \u00d7 \u00a1 B G\u03c9B \u00d7 BrP \u00a2 \u2212 2mB G\u03c9B \u00d7 BvP (9.29) because it seems that the particle is moving under the influence of this force. The second term is negative of the centrifugal force and pointing outward. The maximum value of this force on the Earth is on the equator r\u03c92 = 6378.388\u00d7 103 \u00d7 \u00b5 2\u03c0 24\u00d7 3600 366.25 365.25 \u00b62 = 3.3917\u00d7 10\u22122m/ s2 (9.30) which is about 0.3% of the acceleration of gravity. If we add the variation of the gravitational acceleration because of a change of radius from R = 6356912m at the pole to R = 6378388m on the equator, then the variation of the acceleration of gravity becomes 0.53%. So, generally speaking, a sportsman such as a pole-vaulter who has practiced in the north pole can show a better record in a competition held on the equator. The third term is called the Coriolis force or Coriolis effect, FC, which is perpendicular to both \u03c9 and BvP . For a mass m moving on the north hemisphere at a latitude \u03b8 towards the equator, we should provide a 9. Applied Dynamics 527 lateral eastward force equal to the Coriolis effect to force the mass, keeping its direction relative to the ground. FC = 2mB G\u03c9B \u00d7 Bvm = 1.4584\u00d7 10\u22124 Bpm cos \u03b8 kgm/ s2 (9.31) The Coriolis effect is the reason why the west side of railways, roads, and rivers wears. The lack of providing the Coriolis force is the reason for turning the direction of winds, projectiles, flood, and falling objects westward. Example 347 Work, force, and kinetic energy in a unidirectional motion. A vehicle with mass m = 1200 kg has an initial kinetic energy K = 6000 J. The mass is under a constant force F = F I\u0302 = 4000I\u0302 and moves from X(0) = 0 to X(tf ) = 1000m at a terminal time tf . The work done by the force during this motion is W = Z r(tf ) r(0) F \u00b7 dr = Z 1000 0 4000 dX = 4\u00d7 106Nm = 4MJ (9.32) The kinetic energy at the terminal time is K(tf ) =W +K(0) = 4006000 J (9.33) which shows that the terminal speed of the mass is v2 = r 2K(tf ) m \u2248 81.7m/ s. (9.34) Example 348 Direct dynamics. When the applied force is time varying and is a known function, then, F(t) = m r\u0308. (9.35) The general solution for the equation of motion can be found by integration. r\u0307(t) = r\u0307(t0) + 1 m Z t t0 F(t)dt (9.36) r(t) = r(t0) + r\u0307(t0)(t\u2212 t0) + 1 m Z t t0 Z t t0 F(t)dt dt (9.37) This kind of problem is called direct or forward dynamics. 528 9. Applied Dynamics 9.2 Rigid Body Translational Dynamics Figure 9.3 depicts a moving body B in a global coordinate frame G. Assume that the body frame is attached at the mass center of the body. Point P indicates an infinitesimal sphere of the body, which has a very small mass dm. The point mass dm is acted on by an infinitesimal force df and has a global velocity GvP . According to Newton\u2019s law of motion we have df = GaP dm. (9.38) However, the equation of motion for the whole body in a global coordinate frame is GF = mGaB (9.39) which can be expressed in the body coordinate frame as BF = mB GaB +m B G\u03c9B \u00d7 BvB (9.40)\u23a1\u23a3 Fx Fy Fz \u23a4\u23a6 = \u23a1\u23a3 max +m (\u03c9yvz \u2212 \u03c9zvy) may \u2212m (\u03c9xvz \u2212 \u03c9zvx) maz +m (\u03c9xvy \u2212 \u03c9yvx) \u23a4\u23a6 . (9.41) In these equations, GaB is the acceleration vector of the body mass center C in the global frame,m is the total mass of the body, and F is the resultant of the external forces acted on the body at C. 9. Applied Dynamics 529 Proof. A body coordinate frame at the mass center is called a central frame. If frame B is a central frame, then the center of mass, C, is defined such that Z B Brdm dm = 0. (9.42) The global position vector of dm is related to its local position vector by Grdm = GdB + GRB Brdm (9.43) where GdB is the global position vector of the central body frame, and therefore, Z B Grdm dm = Z B GdB dm+ GRB Z m Brdm dm = Z B GdB dm = GdB Z B dm = mGdB. (9.44) A time derivative of both sides shows that mGd\u0307B = mGvB = Z B Gr\u0307dm dm = Z B Gvdm dm (9.45) and another derivative is mGv\u0307B = mGaB = Z B Gv\u0307dm dm. (9.46) However, we have df = Gv\u0307P dm and therefore, mGaB = Z B df . (9.47) The integral on the right-hand side accounts for all the forces acting on the body. The internal forces cancel one another out, so the net result is the vector sum of all the externally applied forces, F, and therefore, GF = m GaB = m Gv\u0307B . (9.48) In the body coordinate frame we have BF = BRG GF = m BRG GaB = m B GaB = m BaB +m B G\u03c9B \u00d7 BvB. (9.49) 530 9. Applied Dynamics The expanded form of the Newton\u2019s equation in the body coordinate frame is then equal to BF = m BaB +m B G\u03c9B \u00d7 BvB\u23a1\u23a3 Fx Fy Fz \u23a4\u23a6 = m \u23a1\u23a3 ax ay az \u23a4\u23a6+m \u23a1\u23a3 \u03c9x \u03c9y \u03c9z \u23a4\u23a6\u00d7 \u23a1\u23a3 vx vy vz \u23a4\u23a6 = \u23a1\u23a3 max +m (\u03c9yvz \u2212 \u03c9zvy) may \u2212m (\u03c9xvz \u2212 \u03c9zvx) maz +m (\u03c9xvy \u2212 \u03c9yvx) \u23a4\u23a6 . (9.50) 9.3 Rigid Body Rotational Dynamics The rigid body rotational equation of motion is the Euler equation BM = Gd dt BL = BL\u0307+B G\u03c9B \u00d7 BL = BI B G\u03c9\u0307B + B G\u03c9B \u00d7 \u00a1 BI B G\u03c9B \u00a2 (9.51) where L is the angular momentum BL = BI B G\u03c9B (9.52) and I is the moment of inertia of the rigid body. I = \u23a1\u23a3 Ixx Ixy Ixz Iyx Iyy Iyz Izx Izy Izz \u23a4\u23a6 (9.53) The elements of I are functions of the mass distribution of the rigid body and may be defined by Iij = Z B \u00a1 r2i \u03b4mn \u2212 ximxjn \u00a2 dm , i, j = 1, 2, 3 (9.54) where \u03b4ij is Kronecker\u2019s delta. \u03b4mn = \u00bd 1 if m = n 0 if m 6= n (9.55) The expanded form of the Euler equation (9.51) is Mx = Ixx\u03c9\u0307x + Ixy\u03c9\u0307y + Ixz\u03c9\u0307z \u2212 (Iyy \u2212 Izz)\u03c9y\u03c9z \u2212Iyz \u00a1 \u03c92z \u2212 \u03c92y \u00a2 \u2212 \u03c9x (\u03c9zIxy \u2212 \u03c9yIxz) (9.56) 9. Applied Dynamics 531 My = Iyx\u03c9\u0307x + Iyy\u03c9\u0307y + Iyz\u03c9\u0307z \u2212 (Izz \u2212 Ixx)\u03c9z\u03c9x \u2212Ixz \u00a1 \u03c92x \u2212 \u03c92z \u00a2 \u2212 \u03c9y (\u03c9xIyz \u2212 \u03c9zIxy) (9.57) Mz = Izx\u03c9\u0307x + Izy\u03c9\u0307y + Izz\u03c9\u0307z \u2212 (Ixx \u2212 Iyy)\u03c9x\u03c9y \u2212Ixy \u00a1 \u03c92y \u2212 \u03c92x \u00a2 \u2212 \u03c9z (\u03c9yIxz \u2212 \u03c9xIyz) . (9.58) which can be reduced to M1 = I1\u03c9\u03071 \u2212 (I2 \u2212 I2)\u03c92\u03c93 M2 = I2\u03c9\u03072 \u2212 (I3 \u2212 I1)\u03c93\u03c91 (9.59) M3 = I3\u03c9\u03073 \u2212 (I1 \u2212 I2)\u03c91\u03c92 in a special Cartesian coordinate frame called the principal coordinate frame. The principal coordinate frame is denoted by numbers 123 to indicate the first, second, and third principal axes. The parameters Iij , i 6= j are zero in the principal frame. The body and principal coordinate frame sit at the mass center C. Kinetic energy of a rotating rigid body is K = 1 2 \u00a1 Ixx\u03c9 2 x + Iyy\u03c9 2 y + Izz\u03c9 2 z \u00a2 \u2212Ixy\u03c9x\u03c9y \u2212 Iyz\u03c9y\u03c9z \u2212 Izx\u03c9z\u03c9x (9.60) = 1 2 \u03c9 \u00b7 L (9.61) = 1 2 \u03c9T I \u03c9 (9.62) that in the principal coordinate frame reduces to K = 1 2 \u00a1 I1\u03c9 2 1 + I2\u03c9 2 2 + I3\u03c9 2 3 \u00a2 . (9.63) Proof. Let mi be the mass of the ith particle of a rigid body B, which is made of n particles and let ri = Bri = \u00a3 xi yi zi \u00a4T (9.64) be the Cartesian position vector of mi in a central body fixed coordinate frame Oxyz. Assume that \u03c9 = B G\u03c9B = \u00a3 \u03c9x \u03c9y \u03c9z \u00a4T (9.65) is the angular velocity of the rigid body with respect to the ground, expressed in the body coordinate frame. 532 9. Applied Dynamics The angular momentum of mi is Li = ri \u00d7mir\u0307i = mi [ri \u00d7 (\u03c9 \u00d7 ri)] = mi [(ri \u00b7 ri)\u03c9 \u2212 (ri \u00b7 \u03c9) ri] = mir 2 i\u03c9 \u2212mi (ri \u00b7 \u03c9) ri. (9.66) Hence, the angular momentum of the rigid body would be L = \u03c9 nX i=1 mir 2 i \u2212 nX i=1 mi (ri \u00b7 \u03c9) ri. (9.67) Substitution for ri and \u03c9 gives us L = \u00b3 \u03c9x \u0131\u0302+ \u03c9y j\u0302+ \u03c9z k\u0302 \u00b4 nX i=1 mi \u00a1 x2i + y2i + z2i \u00a2 \u2212 nX i=1 mi (xi\u03c9x + yi\u03c9y + zi\u03c9z) \u00b7 \u00b3 xi\u0131\u0302+ yij\u0302+ zik\u0302 \u00b4 (9.68) and therefore, L = nX i=1 mi \u00a1 x2i + y2i + z2i \u00a2 \u03c9x \u0131\u0302 + nX i=1 mi \u00a1 x2i + y2i + z2i \u00a2 \u03c9y j\u0302 + nX i=1 mi \u00a1 x2i + y2i + z2i \u00a2 \u03c9z k\u0302 \u2212 nX i=1 mi (xi\u03c9x + yi\u03c9y + zi\u03c9z)xi\u0131\u0302 \u2212 nX i=1 mi (xi\u03c9x + yi\u03c9y + zi\u03c9z) yij\u0302 \u2212 nX i=1 mi (xi\u03c9x + yi\u03c9y + zi\u03c9z) zik\u0302 (9.69) or L = nX i=1 mi \u00a3\u00a1 x2i + y2i + z2i \u00a2 \u03c9x \u2212 (xi\u03c9x + yi\u03c9y + zi\u03c9z)xi \u00a4 \u0131\u0302 + nX i=1 mi \u00a3\u00a1 x2i + y2i + z2i \u00a2 \u03c9y \u2212 (xi\u03c9x + yi\u03c9y + zi\u03c9z) yi \u00a4 j\u0302 + nX i=1 mi \u00a3\u00a1 x2i + y2i + z2i \u00a2 \u03c9z \u2212 (xi\u03c9x + yi\u03c9y + zi\u03c9z) zi \u00a4 k\u0302 (9.70) 9. Applied Dynamics 533 which can be rearranged as L = nX i=1 \u00a3 mi \u00a1 y2i + z2i \u00a2\u00a4 \u03c9x\u0131\u0302 + nX i=1 \u00a3 mi \u00a1 z2i + x2i \u00a2\u00a4 \u03c9y j\u0302 + nX i=1 \u00a3 mi \u00a1 x2i + y2i \u00a2\u00a4 \u03c9zk\u0302 \u2212 \u00c3 nX i=1 (mixiyi)\u03c9y + nX i=1 (mixizi)\u03c9z ! \u0131\u0302 \u2212 \u00c3 nX i=1 (miyizi)\u03c9z + nX i=1 (miyixi)\u03c9x ! j\u0302 \u2212 \u00c3 nX i=1 (mizixi)\u03c9x + nX i=1 (miziyi)\u03c9y ! k\u0302. (9.71) By introducing the moment of inertia matrix I with the following elements, Ixx = nX i=1 \u00a3 mi \u00a1 y2i + z2i \u00a2\u00a4 (9.72) Iyy = nX i=1 \u00a3 mi \u00a1 z2i + x2i \u00a2\u00a4 (9.73) Izz = nX i=1 \u00a3 mi \u00a1 x2i + y2i \u00a2\u00a4 (9.74) Ixy = Iyx = \u2212 nX i=1 (mixiyi) (9.75) Iyz = Izy = \u2212 nX i=1 (miyizi) (9.76) Izx = Ixz = \u2212 nX i=1 (mizixi) . (9.77) we may write the angular momentum L in a concise form Lx = Ixx\u03c9x + Ixy\u03c9y + Ixz\u03c9z (9.78) Ly = Iyx\u03c9x + Iyy\u03c9y + Iyz\u03c9z (9.79) Lz = Izx\u03c9x + Izy\u03c9y + Izz\u03c9z (9.80) 534 9. Applied Dynamics or in a matrix form L = I \u00b7 \u03c9 (9.81)\u23a1\u23a3 Lx Ly Lz \u23a4\u23a6 = \u23a1\u23a3 Ixx Ixy Ixz Iyx Iyy Iyz Izx Izy Izz \u23a4\u23a6\u23a1\u23a3 \u03c9x \u03c9y \u03c9z \u23a4\u23a6 . (9.82) For a rigid body that is a continuous solid, the summations must be replaced by integrations over the volume of the body as in Equation (9.54). The Euler equation of motion for a rigid body is BM = Gd dt BL (9.83) where BM is the resultant of the external moments applied on the rigid body. The angular momentum BL is a vector defined in the body coordinate frame. Hence, its time derivative in the global coordinate frame is GdBL dt = BL\u0307+B G\u03c9B \u00d7 BL. (9.84) Therefore, BM = dL dt = L\u0307+ \u03c9 \u00d7 L = I\u03c9\u0307 + \u03c9\u00d7 (I\u03c9) (9.85) or in expanded form BM = (Ixx\u03c9\u0307x + Ixy\u03c9\u0307y + Ixz\u03c9\u0307z) \u0131\u0302 +(Iyx\u03c9\u0307x + Iyy\u03c9\u0307y + Iyz\u03c9\u0307z) j\u0302 +(Izx\u03c9\u0307x + Izy\u03c9\u0307y + Izz\u03c9\u0307z) k\u0302 +\u03c9y (Ixz\u03c9x + Iyz\u03c9y + Izz\u03c9z) \u0131\u0302 \u2212\u03c9z (Ixy\u03c9x + Iyy\u03c9y + Iyz\u03c9z) \u0131\u0302 +\u03c9z (Ixx\u03c9x + Ixy\u03c9y + Ixz\u03c9z) j\u0302 \u2212\u03c9x (Ixz\u03c9x + Iyz\u03c9y + Izz\u03c9z) j\u0302 +\u03c9x (Ixy\u03c9x + Iyy\u03c9y + Iyz\u03c9z) k\u0302 \u2212\u03c9y (Ixx\u03c9x + Ixy\u03c9y + Ixz\u03c9z) k\u0302 (9.86) and therefore, the most general form of the Euler equations of motion for a rigid body in a body frame attached to C are Mx = Ixx\u03c9\u0307x + Ixy\u03c9\u0307y + Ixz\u03c9\u0307z \u2212 (Iyy \u2212 Izz)\u03c9y\u03c9z \u2212Iyz \u00a1 \u03c92z \u2212 \u03c92y \u00a2 \u2212 \u03c9x (\u03c9zIxy \u2212 \u03c9yIxz) (9.87) My = Iyx\u03c9\u0307x + Iyy\u03c9\u0307y + Iyz\u03c9\u0307z \u2212 (Izz \u2212 Ixx)\u03c9z\u03c9x \u2212Ixz \u00a1 \u03c92x \u2212 \u03c92z \u00a2 \u2212 \u03c9y (\u03c9xIyz \u2212 \u03c9zIxy) (9.88) Mz = Izx\u03c9\u0307x + Izy\u03c9\u0307y + Izz\u03c9\u0307z \u2212 (Ixx \u2212 Iyy)\u03c9x\u03c9y \u2212Ixy \u00a1 \u03c92y \u2212 \u03c92x \u00a2 \u2212 \u03c9z (\u03c9yIxz \u2212 \u03c9xIyz) . (9.89) 9. Applied Dynamics 535 Assume that we are able to rotate the body frame about its origin to find an orientation that makes Iij = 0, for i 6= j. In such a coordinate frame, which is called a principal frame, the Euler equations reduce to M1 = I1\u03c9\u03071 \u2212 (I2 \u2212 I2)\u03c92\u03c93 (9.90) M2 = I2\u03c9\u03072 \u2212 (I3 \u2212 I1)\u03c93\u03c91 (9.91) M3 = I3\u03c9\u03073 \u2212 (I1 \u2212 I2)\u03c91\u03c92. (9.92) The kinetic energy of a rigid body may be found by the integral of the kinetic energy of a mass element dm, over the whole body. K = 1 2 Z B v\u03072dm = 1 2 Z B (\u03c9 \u00d7 r) \u00b7 (\u03c9 \u00d7 r) dm = \u03c92x 2 Z B \u00a1 y2 + z2 \u00a2 dm+ \u03c92y 2 Z B \u00a1 z2 + x2 \u00a2 dm+ \u03c92z 2 Z B \u00a1 x2 + y2 \u00a2 dm \u2212\u03c9x\u03c9y Z B xy dm\u2212 \u03c9y\u03c9z Z B yz dm\u2212 \u03c9z\u03c9x Z B zx dm = 1 2 \u00a1 Ixx\u03c9 2 x + Iyy\u03c9 2 y + Izz\u03c9 2 z \u00a2 \u2212Ixy\u03c9x\u03c9y \u2212 Iyz\u03c9y\u03c9z \u2212 Izx\u03c9z\u03c9x (9.93) The kinetic energy can be rearranged to a matrix multiplication form K = 1 2 \u03c9T I \u03c9 (9.94) = 1 2 \u03c9 \u00b7 L. (9.95) When the body frame is principal, the kinetic energy will simplify to K = 1 2 \u00a1 I1\u03c9 2 1 + I2\u03c9 2 2 + I3\u03c9 2 3 \u00a2 . (9.96) Example 349 A tilted disc on a massless shaft. Figure 9.4 illustrates a disc with mass m and radius r, mounted on a massless shaft. The shaft is turning with a constant angular speed \u03c9. The disc is attached to the shaft at an angle \u03b8. Because of \u03b8, the bearings at A and B must support a rotating force. We attach a principal body coordinate frame at the disc center as shown in the figure. The angular velocity vector in the body frame is B G\u03c9B = \u03c9 cos \u03b8 \u0131\u0302+ \u03c9 sin \u03b8 j\u0302 (9.97) 536 9. Applied Dynamics and the mass moment of inertia matrix is BI = \u23a1\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3 mr2 2 0 0 0 mr2 4 0 0 0 mr2 4 \u23a4\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6 . (9.98) Substituting (9.97) and (9.98) in (9.90)-(9.92), with 1 \u2261 x, 2 \u2261 y, 3 \u2261 z, provides that Mx = 0 (9.99) My = 0 (9.100) Mz = mr2 4 \u03c9 cos \u03b8 sin \u03b8. (9.101) Therefore, the bearing reaction forces FA and FB are FA = \u2212FB = \u2212Mz l = \u2212mr2 4l \u03c9 cos \u03b8 sin \u03b8. (9.102) Example 350 Steady rotation of a freely rotating rigid body. The Newton-Euler equations of motion for a rigid body are GF = mGv\u0307 (9.103) BM = I B G\u03c9\u0307B + B G\u03c9B \u00d7 BL. (9.104) 9. Applied Dynamics 537 Consider a situation in which the resultant applied force and moment on the body are zero. GF = BF = 0 (9.105) GM = BM = 0 (9.106) Based on the Newton\u2019s equation, the velocity of the mass center will be constant in the global coordinate frame. However, the Euler equation reduces to \u03c9\u03071 = I2 \u2212 I3 I1 \u03c92\u03c93 (9.107) \u03c9\u03072 = I3 \u2212 I1 I22 \u03c93\u03c91 (9.108) \u03c9\u03073 = I1 \u2212 I2 I3 \u03c91\u03c92 (9.109) that show the angular velocity can be constant if I1 = I2 = I3 (9.110) or if two principal moments of inertia, say I1 and I2, are zero and the third angular velocity, in this case \u03c93, is initially zero, or if the angular velocity vector is initially parallel to a principal axis. Example 351 Angular momentum of a two-link manipulator. A two-link manipulator is shown in Figure 9.5. Link A rotates with angular velocity \u03d5\u0307 about the z-axis of its local coordinate frame. Link B is attached to link A and has angular velocity \u03c8\u0307 with respect to A about the xA-axis. We assume that A and G were coincident at \u03d5 = 0, therefore, the rotation matrix between A and G is GRA = \u23a1\u23a3 cos\u03d5(t) \u2212 sin\u03d5(t) 0 sin\u03d5(t) cos\u03d5(t) 0 0 0 1 \u23a4\u23a6 . (9.111) Frame B is related to frame A by Euler angles \u03d5 = 90deg, \u03b8 = 90deg, and \u03c8, hence, ARB = \u23a1\u23a3 c\u03c0c\u03c8 \u2212 c\u03c0s\u03c0s\u03c8 \u2212c\u03c0s\u03c8 \u2212 c\u03c0c\u03c8s\u03c0 s\u03c0s\u03c0 c\u03c8s\u03c0 + c\u03c0c\u03c0s\u03c8 \u2212s\u03c0s\u03c8 + c\u03c0c\u03c0c\u03c8 \u2212c\u03c0s\u03c0 s\u03c0s\u03c8 s\u03c0c\u03c8 c\u03c0 \u23a4\u23a6 \u23a1\u23a3 \u2212 cos\u03c8 sin\u03c8 0 sin\u03c8 cos\u03c8 0 0 0 \u22121 \u23a4\u23a6 (9.112) 538 9. Applied Dynamics and therefore, GRB = GRA ARB (9.113) = \u23a1\u23a3 \u2212 cos\u03d5 cos\u03c8 \u2212 sin\u03d5 sin\u03c8 cos\u03d5 sin\u03c8 \u2212 cos\u03c8 sin\u03d5 0 cos\u03d5 sin\u03c8 \u2212 cos\u03c8 sin\u03d5 cos\u03d5 cos\u03c8 + sin\u03d5 sin\u03c8 0 0 0 \u22121 \u23a4\u23a6 . The angular velocity of A in G, and B in A are G\u03c9A = \u03d5\u0307K\u0302 (9.114) A\u03c9B = \u03c8\u0307\u0131\u0302A. (9.115) Moment of inertia matrices for the arms A and B can be defined as AIA = \u23a1\u23a3 IA1 0 0 0 IA2 0 0 0 IA3 \u23a4\u23a6 (9.116) BIB = \u23a1\u23a3 IB1 0 0 0 IB2 0 0 0 IB3 \u23a4\u23a6 . (9.117) These moments of inertia must be transformed to the global frame GIA = GRB AIA GRT A (9.118) GIB = GRB BIB GRT B . (9.119) 9. Applied Dynamics 539 The total angular momentum of the manipulator is GL = GLA + GLB (9.120) where GLA = GIA G\u03c9A (9.121) GLB = GIB G\u03c9B = GIB \u00a1 G A\u03c9B + G\u03c9A \u00a2 . (9.122) Example 352 Poinsot\u2019s construction. Consider a freely rotating rigid body with an attached principal coordinate frame. HavingM = 0 provides a motion under constant angular momentum and constant kinetic energy L = I \u03c9 = cte (9.123) K = 1 2 \u03c9T I \u03c9 = cte. (9.124) Because the length of the angular momentum L is constant, the equation L2 = L \u00b7 L = L2x + L2y + L2z = I21\u03c9 2 1 + I22\u03c9 2 2 + I23\u03c9 2 3 (9.125) introduces an ellipsoid in the (\u03c91, \u03c92, \u03c93) coordinate frame, called the momentum ellipsoid. The tip of all possible angular velocity vectors must lie on the surface of the momentum ellipsoid. The kinetic energy also defines an energy ellipsoid in the same coordinate frame so that the tip of the angular velocity vectors must also lie on its surface. K = 1 2 \u00a1 I1\u03c9 2 1 + I2\u03c9 2 2 + I3\u03c9 2 3 \u00a2 (9.126) In other words, the dynamics of moment-free motion of a rigid body requires that the corresponding angular velocity \u03c9(t) satisfy both Equations (9.125) and (9.126) and therefore lie on the intersection of the momentum and energy ellipsoids. For clarity, we may define the ellipsoids in the (Lx, Ly, Lz) coordinate system as L2x + L2y + L2z = L2 (9.127) L2x 2I1K + L2y 2I2K + L2z 2I3K = 1. (9.128) Equation (9.127) is a sphere and Equation (9.128) defines an ellipsoid with\u221a 2IiK as semi-axes. To have a meaningful motion, these two shapes must intersect. The intersection may form a trajectory, as shown in Figure 9.6. 540 9. Applied Dynamics It can be deduced that for a certain value of angular momentum there are maximum and minimum limit values for acceptable kinetic energy. Assuming I1 > I3 > I3 (9.129) the limits of possible kinetic energy are Kmin = L2 2I1 (9.130) Kmax = L2 2I3 (9.131) and the corresponding motions are turning about the axes I1 and I3 respectively. Example 353 F Alternative derivation of Euler equations of motion. Assume that the moment of the small force df is shown by dm and a mass element is shown by dm, then, dm = Grdm \u00d7 df = Grdm \u00d7 Gv\u0307dm dm. (9.132) The global angular momentum dl of dm is equal to dl = Grdm \u00d7 Gvdm dm (9.133) and according to (9.12) we have dm = Gd dt dl (9.134) Grdm \u00d7 df = Gd dt \u00a1 Grdm \u00d7 Gvdm dm \u00a2 . (9.135) 9. Applied Dynamics 541 Integrating over the body results inZ B Grdm \u00d7 df = Z B Gd dt \u00a1 Grdm \u00d7 Gvdm dm \u00a2 = Gd dt Z B \u00a1 Grdm \u00d7 Gvdm dm \u00a2 . (9.136) However, utilizing Grdm = GdB + GRB Brdm (9.137) where GdB is the global position vector of the central body frame, can simplify the left-hand side of the integral toZ B Grdm \u00d7 df = Z B \u00a1 GdB + GRB Brdm \u00a2 \u00d7 df = Z B GdB \u00d7 df + Z B G Brdm \u00d7 df = GdB \u00d7 GF+ GMC (9.138) where MC is the resultant external moment about the body mass center C. The right-hand side of Equation (9.136) is Gd dt Z B \u00a1 Grdm \u00d7 Gvdm dm \u00a2 = Gd dt Z B \u00a1\u00a1 GdB + GRB Brdm \u00a2 \u00d7 Gvdm dm \u00a2 = Gd dt Z B \u00a1 GdB \u00d7 Gvdm \u00a2 dm+ Gd dt Z B \u00a1 G Brdm \u00d7 Gvdm \u00a2 dm = Gd dt \u00b5 GdB \u00d7 Z B Gvdmdm \u00b6 + Gd dt LC = Gd\u0307B \u00d7 Z B Gvdmdm+ GdB \u00d7 Z B Gv\u0307dmdm+ d dt LC . (9.139) We use LC for angular momentum about the body mass center. Because the body frame is at the mass center, we haveZ B Grdm dm = mGdB = mGrC (9.140)Z B Gvdmdm = mGd\u0307B = mGvC (9.141)Z B Gv\u0307dmdm = mGd\u0308B = mGaC (9.142) and therefore, Gd dt Z B \u00a1 Grdm \u00d7 Gvdm dm \u00a2 = GdB \u00d7 GF+ Gd dt GLC . (9.143) 542 9. Applied Dynamics Substituting (9.138) and (9.143) in (9.136) provides the Euler equation of motion in the global frame, indicating that the resultant of externally applied moments about C is equal to the global derivative of angular momentum about C. GMC = Gd dt GLC . (9.144) The Euler equation in the body coordinate can be found by transforming (9.144) BMC = GRT B GMC = GRT B Gd dt LC = Gd dt GRT B LC = Gd dt BLC = BL\u0307C + B G\u03c9B \u00d7 BLC . (9.145) 9.4 Mass Moment of Inertia Matrix In analyzing the motion of rigid bodies, two types of integrals arise that belong to the geometry of the body. The first type defines the center of mass and is important when the translation motion of the body is considered. The second is the moment of inertia that appears when the rotational motion of the body is considered. The moment of inertia is also called centrifugal moments, or deviation moments. Every rigid body has a 3 \u00d7 3 moment of inertia matrix I, which is denoted by I = \u23a1\u23a3 Ixx Ixy Ixz Iyx Iyy Iyz Izx Izy Izz \u23a4\u23a6 . (9.146) The diagonal elements Iij , i = j are called polar moments of inertia Ixx = Ix = Z B \u00a1 y2 + z2 \u00a2 dm (9.147) Iyy = Iy = Z B \u00a1 z2 + x2 \u00a2 dm (9.148) Izz = Iz = Z B \u00a1 x2 + y2 \u00a2 dm (9.149) and the off-diagonal elements Iij , i 6= j are called products of inertia Ixy = Iyx = \u2212 Z B xy dm (9.150) 9. Applied Dynamics 543 Iyz = Izy = \u2212 Z B yz dm (9.151) Izx = Ixz = \u2212 Z B zx dm. (9.152) The elements of I for a rigid body, made of discrete point masses, are defined in Equation (9.54). The elements of I are calculated about a body coordinate frame attached to the mass center C of the body. Therefore, I is a frame-dependent quantity and must be written like BI to show the frame it is computed in. BI = Z B \u23a1\u23a3 y2 + z2 \u2212xy \u2212zx \u2212xy z2 + x2 \u2212yz \u2212zx \u2212yz x2 + y2 \u23a4\u23a6 dm (9.153) = Z B \u00a1 r2I\u2212 r rT \u00a2 dm (9.154) = Z B \u2212r\u0303 r\u0303 dm. (9.155) Moments of inertia can be transformed from a coordinate frame B1 to another coordinate frame B2, both installed at the mass center of the body, according to the rule of the rotated-axes theorem B2I = B2RB1 B1I B2RT B1 . (9.156) Transformation of the moment of inertia from a central frame B1 located at B2rC to another frame B2, which is parallel to B1, is, according to the rule of parallel-axes theorem, B2I = B1I +mr\u0303C r\u0303TC . (9.157) If the local coordinate frame Oxyz is located such that the products of inertia vanish, the local coordinate frame is called the principal coordinate frame and the associated moments of inertia are called principal moments of inertia. Principal axes and principal moments of inertia can be found by solving the following equation for I:\u00af\u0304\u0304\u0304 \u00af\u0304 Ixx \u2212 I Ixy Ixz Iyx Iyy \u2212 I Iyz Izx Izy Izz \u2212 I \u00af\u0304\u0304\u0304 \u00af\u0304 = 0 (9.158) det ([Iij ]\u2212 I [\u03b4ij ]) = 0. (9.159) Since Equation (9.159) is a cubic equation in I, we obtain three eigenvalues I1 = Ix I2 = Iy I3 = Iz (9.160) 544 9. Applied Dynamics that are the principal moments of inertia. Proof. Two coordinate frames with a common origin at the mass center of a rigid body are shown in Figure 9.7. The angular velocity and angular momentum of a rigid body transform from the frame B1 to the frame B2 by vector transformation rule B2\u03c9 = B2RB1 B1\u03c9 (9.161) B2L = B2RB1 B1L. (9.162) However, L and \u03c9 are related according to Equation (9.52) B1L = B1I B1\u03c9 (9.163) and therefore, B2L = B2RB1 B1I B2RT B1 B2\u03c9 = B2I B2\u03c9 (9.164) which shows how to transfer the moment of inertia from the coordinate frame B1 to a rotated frame B2 B2I = B2RB1 B1I B2RT B1 . (9.165) Now consider a central frame B1, shown in Figure 9.8, at B2rC , which rotates about the origin of a fixed frame B2 such that their axes remain parallel. The angular velocity and angular momentum of the rigid body transform from frame B1 to frame B2 by B2\u03c9 = B1\u03c9 (9.166) B2L = B1L+ (rC \u00d7mvC) . (9.167) 9. Applied Dynamics 545 Therefore, B2L = B1L+mB2rC \u00d7 \u00a1 B2\u03c9\u00d7B2rC \u00a2 = B1L+ \u00a1 m B2 r\u0303C B2 r\u0303TC \u00a2 B2\u03c9 = \u00a1 B1I +m B2 r\u0303C B2 r\u0303TC \u00a2 B2\u03c9 (9.168) which shows how to transfer the moment of inertia from frame B1 to a parallel frame B2 B2I = B1I +mr\u0303C r\u0303TC . (9.169) The parallel-axes theorem is also called the Huygens-Steiner theorem. Referring to Equation (9.165) for transformation of the moment of inertia to a rotated frame, we can always find a frame in which B2I is diagonal. In such a frame, we have B2RB1 B1I = B2I B2RB1 (9.170) or \u23a1\u23a3 r11 r12 r13 r21 r22 r23 r31 r32 r33 \u23a4\u23a6\u23a1\u23a3 Ixx Ixy Ixz Iyx Iyy Iyz Izx Izy Izz \u23a4\u23a6 = \u23a1\u23a3 I1 0 0 0 I2 0 0 0 I3 \u23a4\u23a6\u23a1\u23a3 r11 r12 r13 r21 r22 r23 r31 r32 r33 \u23a4\u23a6 (9.171) which shows that I1, I2, and I3 are eigenvalues of B1I. These eigenvalues can be found by solving the following equation for \u03bb:\u00af\u0304\u0304\u0304 \u00af\u0304 Ixx \u2212 \u03bb Ixy Ixz Iyx Iyy \u2212 \u03bb Iyz Izx Izy Izz \u2212 \u03bb \u00af\u0304\u0304\u0304 \u00af\u0304 = 0. (9.172) 546 9. Applied Dynamics The eigenvalues I1, I2, and I3 are principal moments of inertia, and their associated eigenvectors are called principal directions. The coordinate frame made by the eigenvectors is the principal body coordinate frame. In the principal coordinate frame, the rigid body angular momentum is\u23a1\u23a3 L1 L2 L3 \u23a4\u23a6 = \u23a1\u23a3 I1 0 0 0 I2 0 0 0 I3 \u23a4\u23a6\u23a1\u23a3 \u03c91 \u03c92 \u03c93 \u23a4\u23a6 . (9.173) Example 354 Principal moments of inertia. Consider the inertia matrix I I = \u23a1\u23a3 20 \u22122 0 \u22122 30 0 0 0 40 \u23a4\u23a6 . (9.174) We set up the determinant (9.159)\u00af\u0304\u0304\u0304 \u00af\u0304 20\u2212 \u03bb \u22122 0 \u22122 30\u2212 \u03bb 0 0 0 40\u2212 \u03bb \u00af\u0304\u0304\u0304 \u00af\u0304 = 0 (9.175) which leads to the following characteristic equation. (20\u2212 \u03bb) (30\u2212 \u03bb) (40\u2212 \u03bb)\u2212 4 (40\u2212 \u03bb) = 0 (9.176) Three roots of Equation (9.176) are I1 = 30.385, I2 = 19.615, I3 = 40 (9.177) and therefore, the principal moment of inertia matrix is I = \u23a1\u23a3 30.385 0 0 0 19.615 0 0 0 40 \u23a4\u23a6 . (9.178) Example 355 Principal coordinate frame. Consider the inertia matrix I I = \u23a1\u23a3 20 \u22122 0 \u22122 30 0 0 0 40 \u23a4\u23a6 (9.179) the direction of a principal axis xi is established by solving\u23a1\u23a3 Ixx \u2212 Ii Ixy Ixz Iyx Iyy \u2212 Ii Iyz Izx Izy Izz \u2212 Ii \u23a4\u23a6\u23a1\u23a3 cos\u03b1i cos\u03b2i cos \u03b3i \u23a4\u23a6 = \u23a1\u23a3 0 0 0 \u23a4\u23a6 (9.180) 9. Applied Dynamics 547 for direction cosines, which must also satisfy cos2 \u03b1i + cos 2 \u03b2i + cos 2 \u03b3i = 1. (9.181) For the first principal moment of inertia I1 = 30.385 we have\u23a1\u23a3 20\u2212 30.385 \u22122 0 \u22122 30\u2212 30.385 0 0 0 40\u2212 30.385 \u23a4\u23a6\u23a1\u23a3 cos\u03b11 cos\u03b21 cos \u03b31 \u23a4\u23a6 = \u23a1\u23a3 0 0 0 \u23a4\u23a6 (9.182) or \u221210.385 cos\u03b11 \u2212 2 cos\u03b21 + 0 = 0 (9.183) \u22122 cos\u03b11 \u2212 0.385 cos\u03b21 + 0 = 0 (9.184) 0 + 0 + 9.615 cos \u03b31 = 0 (9.185) and we obtain \u03b11 = 79.1 deg (9.186) \u03b21 = 169.1 deg (9.187) \u03b31 = 90.0 deg . (9.188) Using I2 = 19.615 for the second principal axis\u23a1\u23a3 20\u2212 19.62 \u22122 0 \u22122 30\u2212 19.62 0 0 0 40\u2212 19.62 \u23a4\u23a6\u23a1\u23a3 cos\u03b12 cos\u03b22 cos \u03b32 \u23a4\u23a6 = \u23a1\u23a3 0 0 0 \u23a4\u23a6 (9.189) we obtain \u03b12 = 10.9 deg (9.190) \u03b22 = 79.1 deg (9.191) \u03b32 = 90.0 deg . (9.192) The third principal axis is for I3 = 40\u23a1\u23a3 20\u2212 40 \u22122 0 \u22122 30\u2212 40 0 0 0 40\u2212 40 \u23a4\u23a6\u23a1\u23a3 cos\u03b13 cos\u03b23 cos \u03b33 \u23a4\u23a6 = \u23a1\u23a3 0 0 0 \u23a4\u23a6 (9.193) which leads to \u03b13 = 90.0 deg (9.194) \u03b23 = 90.0 deg (9.195) \u03b33 = 0.0 deg . (9.196) 548 9. Applied Dynamics Example 356 Moment of inertia of a rigid rectangular bar. Consider a homogeneous rectangular link with mass m, length l, width w, and height h, as shown in Figure 9.9. The local central coordinate frame is attached to the link at its mass center. The moments of inertia matrix of the link can be found by the integral method. We begin with calculating Ixx Ixx = Z B \u00a1 y2 + z2 \u00a2 dm = Z v \u00a1 y2 + z2 \u00a2 \u03c1dv = m lwh Z v \u00a1 y2 + z2 \u00a2 dv = m lwh Z h/2 \u2212h/2 Z w/2 \u2212w/2 Z l/2 \u2212l/2 \u00a1 y2 + z2 \u00a2 dx dy dz = m 12 \u00a1 w2 + h2 \u00a2 (9.197) which shows Iyy and Izz can be calculated similarly Iyy = m 12 \u00a1 h2 + l2 \u00a2 (9.198) Izz = m 12 \u00a1 l2 + w2 \u00a2 . (9.199) Since the coordinate frame is central, the products of inertia must be zero. To show this, we examine Ixy. Ixy = Iyx = \u2212 Z B xy dm = Z v xy\u03c1dv = m lwh Z h/2 \u2212h/2 Z w/2 \u2212w/2 Z l/2 \u2212l/2 xy dxdy dz = 0 (9.200) 9. Applied Dynamics 549 Therefore, the moment of inertia matrix for the rigid rectangular bar in its central frame is I = \u23a1\u23a3 m 12 \u00a1 w2 + h2 \u00a2 0 0 0 m 12 \u00a1 h2 + l2 \u00a2 0 0 0 m 12 \u00a1 l2 + w2 \u00a2 \u23a4\u23a6 . (9.201) Example 357 Translation of the inertia matrix. The moment of inertia matrix of the rigid body shown in Figure 9.10, in the principal frame B(oxyz) is given in Equation (9.201). The moment of inertia matrix in the non-principal frame B0(ox0y0z0) can be found by applying the parallel-axes transformation formula (9.169). B0 I = BI +m B0 r\u0303C B0 r\u0303TC (9.202) The mass center is at B0 rC = 1 2 \u23a1\u23a3 l w h \u23a4\u23a6 (9.203) and therefore, B0 r\u0303C = 1 2 \u23a1\u23a3 0 \u2212h w h 0 \u2212l \u2212w l 0 \u23a4\u23a6 (9.204) that provides B0 I = \u23a1\u23a3 1 3h 2m+ 1 3mw2 \u221214 lmw \u221214hlm \u221214 lmw 1 3h 2m+ 1 3 l 2m \u221214hmw \u221214hlm \u2212 14hmw 1 3 l 2m+ 1 3mw2 \u23a4\u23a6 . (9.205) 550 9. Applied Dynamics Example 358 Principal rotation matrix. Consider a body inertia matrix as I = \u23a1\u23a3 2/3 \u22121/2 \u22121/2 \u22121/2 5/3 \u22121/4 \u22121/2 \u22121/4 5/3 \u23a4\u23a6 . (9.206) The eigenvalues and eigenvectors of I are I1 = 0.2413 , \u23a1\u23a3 2.351 1 1 \u23a4\u23a6 (9.207) I2 = 1.8421 , \u23a1\u23a3 \u22120.8511 1 \u23a4\u23a6 (9.208) I3 = 1.9167 , \u23a1\u23a3 0 \u22121 1 \u23a4\u23a6 . (9.209) The normalized eigenvector matrix W is equal to the transpose of the required transformation matrix to make the inertia matrix diagonal W = \u23a1\u23a3 | | | w1 w 2 w 3 | | | \u23a4\u23a6 = 2RT 1 = \u23a1\u23a3 0.856 9 \u22120.515 6 0.0 0.364 48 0.605 88 \u22120.707 11 0.364 48 0.605 88 0.707 11 \u23a4\u23a6 . (9.210) We may verify that 2I \u2248 2R1 1I 2RT 1 =WT 1I W = \u23a1\u23a3 0.2413 \u22121\u00d7 10\u22124 0.0 \u22121\u00d7 10\u22124 1.842 1 \u22121\u00d7 10\u221219 0.0 0.0 1.916 7 \u23a4\u23a6 . (9.211) Example 359 F Relative diagonal moments of inertia. Using the definitions for moments of inertia (9.147), (9.148), and (9.149) it is seen that the inertia matrix is symmetric, andZ B \u00a1 x2 + y2 + z2 \u00a2 dm = 1 2 (Ixx + Iyy + Izz) (9.212) and also Ixx + Iyy \u2265 Izz (9.213) Iyy + Izz \u2265 Ixx (9.214) Izz + Ixx \u2265 Iyy. (9.215) 9. Applied Dynamics 551 Noting that (y \u2212 z) 2 \u2265 0 it is evident that \u00a1 y2 + z2 \u00a2 \u2265 2yz and therefore Ixx \u2265 2Iyz (9.216) and similarly Iyy \u2265 2Izx (9.217) Izz \u2265 2Ixy. (9.218) Example 360 F Coefficients of the characteristic equation. The determinant (9.172)\u00af\u0304\u0304\u0304 \u00af\u0304 Ixx \u2212 \u03bb Ixy Ixz Iyx Iyy \u2212 \u03bb Iyz Izx Izy Izz \u2212 \u03bb \u00af\u0304\u0304\u0304 \u00af\u0304 = 0 (9.219) for calculating the principal moments of inertia, leads to a third-degree equation for \u03bb, called the characteristic equation. \u03bb3 \u2212 a1\u03bb 2 + a2\u03bb\u2212 a3 = 0 (9.220) The coefficients of the characteristic equation are called the principal invariants of [I]. The coefficients of the characteristic equation can directly be found from the following equations: a1 = Ixx + Iyy + Izz = tr [I] (9.221) a2 = IxxIyy + IyyIzz + IzzIxx \u2212 I2xy \u2212 I2yz \u2212 I2zx = \u00af\u0304\u0304\u0304 Ixx Ixy Iyx Iyy \u00af\u0304\u0304\u0304 + \u00af\u0304\u0304\u0304 Iyy Iyz Izy Izz \u00af\u0304\u0304\u0304 + \u00af\u0304\u0304\u0304 Ixx Ixz Izx Izz \u00af\u0304\u0304\u0304 = 1 2 \u00a1 a21 \u2212 tr \u00a3 I2 \u00a4\u00a2 (9.222) a3 = IxxIyyIzz + IxyIyzIzx + IzyIyxIxz \u2212 (IxxIyzIzy + IyyIzxIxz + IzzIxyIyx) = IxxIyyIzz + 2IxyIyzIzx \u2212 \u00a1 IxxI 2 yz + IyyI 2 zx + IzzI 2 xy \u00a2 = det [I] (9.223) 552 9. Applied Dynamics Example 361 F The principal moments of inertia are coordinate invariants. The roots of the inertia characteristic equation are the principal moments of inertia. They are all real but not necessarily different. The principal moments of inertia are extreme. That is, the principal moments of inertia determine the smallest and the largest values of Iii. Since the smallest and largest values of Iii do not depend on the choice of the body coordinate frame, the solution of the characteristic equation is not dependent of the coordinate frame. In other words, if I1, I2, and I3 are the principal moments of inertia for B1I, the principal moments of inertia for B2I are also I1, I2, and I3 when B2I = B2RB1 B1I B2RT B1 . We conclude that I1, I2, and I3 are coordinate invariants of the matrix [I], and therefore any quantity that depends on I1, I2, and I3 is also coordinate invariant. The matrix [I] has only three independent invariants and every other invariant can be expressed in terms of I1, I2, and I3. Since I1, I2, and I3 are the solutions of the characteristic equation of [I] given in (9.220), we may write the determinant (9.172) in the form (\u03bb\u2212 I1) (\u03bb\u2212 I2) (\u03bb\u2212 I3) = 0. (9.224) The expanded form of this equation is \u03bb3 \u2212 (I1 + I2 + I3)\u03bb 2 + (I1I2 + I2I3 + I3I1) a2\u03bb\u2212 I1I2I3 = 0. (9.225) By comparing (9.225) and (9.220) we conclude that a1 = Ixx + Iyy + Izz = I1 + I2 + I3 (9.226) a2 = IxxIyy + IyyIzz + IzzIxx \u2212 I2xy \u2212 I2yz \u2212 I2zx = I1I2 + I2I3 + I3I1 (9.227) a3 = IxxIyyIzz + 2IxyIyzIzx \u2212 \u00a1 IxxI 2 yz + IyyI 2 zx + IzzI 2 xy \u00a2 = I1I2I3. (9.228) Being able to express the coefficients a1, a2, and a3 as functions of I1, I2, and I3 determines that the coefficients of the characteristic equation are coordinate-invariant. Example 362 F Short notation for the elements of inertia matrix. Taking advantage of the Kronecker\u2019s delta (5.138) we may write the el- 9. Applied Dynamics 553 ements of the moment of inertia matrix Iij in short notation forms Iij = Z B \u00a1\u00a1 x21 + x22 + x23 \u00a2 \u03b4ij \u2212 xixj \u00a2 dm (9.229) Iij = Z B \u00a1 r2\u03b4ij \u2212 xixj \u00a2 dm (9.230) Iij = Z B \u00c3 3X k=1 xkxk\u03b4ij \u2212 xixj ! dm (9.231) where we utilized the following notations: x1 = x x2 = y x3 = z. (9.232) Example 363 F Moment of inertia with respect to a plane, a line, and a point. The moment of inertia of a system of particles may be defined with respect to a plane, a line, or a point as the sum of the products of the mass of the particles into the square of the perpendicular distance from the particle to the plane, line, or point. For a continuous body, the sum would be definite integral over the volume of the body. The moments of inertia with respect to the xy, yz, and zx-plane are Iz2 = Z B z2dm (9.233) Iy2 = Z B y2dm (9.234) Ix2 = Z B x2dm. (9.235) The moments of inertia with respect to the x, y, and z axes are Ix = Z B \u00a1 y2 + z2 \u00a2 dm (9.236) Iy = Z B \u00a1 z2 + x2 \u00a2 dm (9.237) Iz = Z B \u00a1 x2 + y2 \u00a2 dm (9.238) and therefore, Ix = Iy2 + Iz2 (9.239) Iy = Iz2 + Ix2 (9.240) Iz = Ix2 + Iy2 . (9.241) 554 9. Applied Dynamics The moment of inertia with respect to the origin is Io = Z B \u00a1 x2 + y2 + z2 \u00a2 dm = Ix2 + Iy2 + Iz2 = 1 2 (Ix + Iy + Iz) . (9.242) Because the choice of the coordinate frame is arbitrary, we can say that the moment of inertia with respect to a line is the sum of the moments of inertia with respect to any two mutually orthogonal planes that pass through the line. The moment of inertia with respect to a point has similar meaning for three mutually orthogonal planes intersecting at the point. 9.5 Lagrange\u2019s Form of Newton\u2019s Equations of Motion Newton\u2019s equation of motion can be transformed to d dt \u00b5 \u2202K \u2202q\u0307r \u00b6 \u2212 \u2202K \u2202qr = Fr r = 1, 2, \u00b7 \u00b7 \u00b7n (9.243) where Fr = nX i=1 \u00b5 Fix \u2202fi \u2202q1 + Fiy \u2202gi \u2202q2 + Fiz \u2202hi \u2202qn \u00b6 . (9.244) Equation (9.243) is called the Lagrange equation of motion, where K is the kinetic energy of the n degree-of-freedom (DOF ) system, qr, r = 1, 2, \u00b7 \u00b7 \u00b7 , n are the generalized coordinates of the system, F = \u00a3 Fix Fiy Fiz \u00a4T is the external force acting on the ith particle of the system, and Fr is the generalized force associated to qr. Proof. Let mi be the mass of one of the particles of a system and let (xi, yi, zi) be its Cartesian coordinates in a globally fixed coordinate frame. Assume that the coordinates of every individual particle are functions of another set of coordinates q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, and possibly time t. xi = fi(q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, t) (9.245) yi = gi(q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, t) (9.246) zi = hi(q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, t) (9.247) If Fxi, Fyi, Fzi are components of the total force acting on the particle mi, then the Newton equations of motion for the particle would be Fxi = mix\u0308i (9.248) Fyi = miy\u0308i (9.249) Fzi = miz\u0308i. (9.250) 9. Applied Dynamics 555 We multiply both sides of these equations by \u2202fi \u2202qr \u2202gi \u2202qr \u2202hi \u2202qr respectively, and add them up for all the particles to have nX i=1 mi \u00b5 x\u0308i \u2202fi \u2202qr + y\u0308i \u2202gi \u2202qr + z\u0308i \u2202hi \u2202qr \u00b6 = nX i=1 \u00b5 Fxi \u2202fi \u2202qr + Fyi \u2202gi \u2202qr + Fzi \u2202hi \u2202qr \u00b6 (9.251) where n is the total number of particles. Taking a time derivative of Equation (9.245), x\u0307i = \u2202fi \u2202q1 q\u03071 + \u2202fi \u2202q2 q\u03072 + \u2202fi \u2202q3 q\u03073 + \u00b7 \u00b7 \u00b7+ \u2202fi \u2202qn q\u0307n + \u2202fi \u2202t (9.252) we find \u2202x\u0307i \u2202q\u0307r = \u2202 \u2202q\u0307r \u00b5 \u2202fi \u2202q1 q\u03071 + \u2202fi \u2202q2 q\u03072 + \u00b7 \u00b7 \u00b7+ \u2202fi \u2202qn q\u0307n + \u2202fi \u2202t \u00b6 = \u2202fi \u2202qr . (9.253) and therefore, x\u0308i \u2202fi \u2202qr = x\u0308i \u2202x\u0307i \u2202q\u0307r = d dt \u00b5 x\u0307i \u2202x\u0307i \u2202q\u0307r \u00b6 \u2212 x\u0307i d dt \u00b5 \u2202x\u0307i \u2202q\u0307r \u00b6 . (9.254) However, x\u0307i d dt \u00b5 \u2202x\u0307i \u2202q\u0307r \u00b6 = x\u0307i d dt \u00b5 \u2202fi \u2202qr \u00b6 = x\u0307i \u00b5 \u22022fi \u2202q1\u2202qr q\u03071 + \u00b7 \u00b7 \u00b7+ \u22022fi \u2202qn\u2202qr q\u0307n + \u22022fi \u2202t\u2202qr \u00b6 = x\u0307i \u2202 \u2202qr \u00b5 \u2202fi \u2202q1 q\u03071 + \u2202fi \u2202q2 q\u03072 + \u00b7 \u00b7 \u00b7+ \u2202fi \u2202qn q\u0307n + \u2202fi \u2202t \u00b6 = x\u0307i \u2202x\u0307i \u2202qr (9.255) and we have x\u0308i \u2202x\u0307i \u2202q\u0307r = d dt \u00b5 x\u0307i \u2202x\u0307i \u2202q\u0307r \u00b6 \u2212 x\u0307i \u2202x\u0307i \u2202qr (9.256) 556 9. Applied Dynamics which is equal to x\u0308i x\u0307i q\u0307r = d dt \u2219 \u2202 \u2202q\u0307r \u00b5 1 2 x\u03072i \u00b6\u00b8 \u2212 \u2202 \u2202qr \u00b5 1 2 x\u03072i \u00b6 . (9.257) Now substituting (9.254) and (9.257) in the left-hand side of (9.251) leads to nX i=1 mi \u00b5 x\u0308i \u2202fi \u2202qr + y\u0308i \u2202gi \u2202qr + z\u0308i \u2202hi \u2202qr \u00b6 = nX i=1 mi d dt \u2219 \u2202 \u2202q\u0307r \u00b5 1 2 x\u03072i + 1 2 y\u03072i + 1 2 z\u03072i \u00b6\u00b8 \u2212 nX i=1 mi \u2202 \u2202qr \u00b5 1 2 x\u03072i + 1 2 y\u03072i + 1 2 z\u03072i \u00b6 = 1 2 nX i=1 mi d dt \u2219 \u2202 \u2202q\u0307r \u00a1 x\u03072i + y\u03072i + z\u03072i \u00a2\u00b8 \u22121 2 nX i=1 mi \u2202 \u2202qr \u00a1 x\u03072i + y\u03072i + z\u03072i \u00a2 . (9.258) where 1 2 nX i=1 mi \u00a1 x\u03072i + y\u03072i + z\u03072i \u00a2 = K (9.259) is the kinetic energy of the system. Therefore, the Newton equations of motion (9.248), (9.249), and (9.250) are converted to d dt \u00b5 \u2202K \u2202q\u0307r \u00b6 \u2212 \u2202K \u2202qr = nX i=1 \u00b5 Fxi \u2202fi \u2202qr + Fyi \u2202gi \u2202qr + Fzi \u2202hi \u2202qr \u00b6 . (9.260) Because of (9.245), (9.246), and (9.247), the kinetic energy is a function of q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn and time t. The left-hand side of Equation (9.260) includes the kinetic energy of the whole system and the right-hand side is a generalized force and shows the effect of changing coordinates from xi to qj on the external forces. Let us assume that the coordinate qr alters to qr+ \u03b4qr while the other coordinates q1, q2, q3, \u00b7 \u00b7 \u00b7 , qr\u22121, qr+1, \u00b7 \u00b7 \u00b7 , qn and time t are unaltered. So, the coordinates of mi are changed to xi + \u2202fi \u2202qr \u03b4qr (9.261) yi + \u2202gi \u2202qr \u03b4qr (9.262) zi + \u2202hi \u2202qr \u03b4qr (9.263) 9. Applied Dynamics 557 Such a displacement is called virtual displacement. The work done in this virtual displacement by all forces acting on the particles of the system is \u03b4W = nX i=1 \u00b5 Fxi \u2202fi \u2202qr + Fyi \u2202gi \u2202qr + Fzi \u2202hi \u2202qr \u00b6 \u03b4qr. (9.264) Because the work done by internal forces appears in opposite pairs, only the work done by external forces remains in Equation (9.264). Let\u2019s denote the virtual work by \u03b4W = Fr (q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, t) \u03b4qr. (9.265) Then we have d dt \u00b5 \u2202K \u2202q\u0307r \u00b6 \u2212 \u2202K \u2202qr = Fr (9.266) where Fr = nX i=1 \u00b5 Fxi \u2202fi \u2202qr + Fyi \u2202gi \u2202qr + Fzi \u2202hi \u2202qr \u00b6 . (9.267) Equation (9.266) is the Lagrange form of equations of motion. This equation is true for all values of r from 1 to n. We thus have n second-order ordinary differential equations in which q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn are the dependent variables and t is the independent variable. The coordinates q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn are called generalized coordinates and can be any measurable parameters to provide the configuration of the system. The number of equations and the number of dependent variables are equal, therefore, the equations are theoretically sufficient to determine the motion of all mi. Example 364 Equation of motion for a simple pendulum. A pendulum is shown in Figure 9.11. Using x and y for the Cartesian position of m, and using \u03b8 = q as the generalized coordinate, we have x = f(\u03b8) = l sin \u03b8 (9.268) y = g(\u03b8) = l cos \u03b8 (9.269) K = 1 2 m \u00a1 x\u03072 + y\u03072 \u00a2 = 1 2 ml2\u03b8\u0307 2 (9.270) and therefore, d dt \u00b5 \u2202K \u2202\u03b8\u0307 \u00b6 \u2212 \u2202K \u2202\u03b8 = d dt (ml2\u03b8\u0307) = ml2\u03b8\u0308. (9.271) The external force components, acting on m, are Fx = 0 (9.272) Fy = mg (9.273) 558 9. Applied Dynamics and therefore, F\u03b8 = Fx \u2202f \u2202\u03b8 + Fy \u2202g \u2202\u03b8 = \u2212mgl sin \u03b8. (9.274) Hence, the equation of motion for the pendulum is ml2\u03b8\u0308 = \u2212mgl sin \u03b8. (9.275) Example 365 A pendulum attached to an oscillating mass. Figure 9.12 illustrates a vibrating mass with a hanging pendulum. The pendulum can act as a vibration absorber if designed properly. Starting with coordinate relationships xM = fM = x (9.276) yM = gM = 0 (9.277) xm = fm = x+ l sin \u03b8 (9.278) ym = gm = l cos \u03b8 (9.279) we may find the kinetic energy in terms of the generalized coordinates x and \u03b8. K = 1 2 M \u00a1 x\u03072M + y\u03072M \u00a2 + 1 2 m \u00a1 x\u03072m + y\u03072m \u00a2 = 1 2 Mx\u03072 + 1 2 m \u00b3 x\u03072 + l2\u03b8\u0307 2 + 2lx\u0307\u03b8\u0307 cos \u03b8 \u00b4 (9.280) Then, the left-hand side of the Lagrange equations are d dt \u00b5 \u2202K \u2202x\u0307 \u00b6 \u2212 \u2202K \u2202x = (M +m)x\u0308+ml\u03b8\u0308 cos \u03b8 \u2212ml\u03b8\u0307 2 sin \u03b8 (9.281) d dt \u00b5 \u2202K \u2202\u03b8\u0307 \u00b6 \u2212 \u2202K \u2202\u03b8 = ml2\u03b8\u0308 +mlx\u0308 cos \u03b8. (9.282) 9. Applied Dynamics 559 The external forces acting on M and m are FxM = \u2212kx (9.283) FyM = 0 (9.284) Fxm = 0 (9.285) Fym = mg. (9.286) Therefore, the generalized forces are Fx = FxM \u2202fM \u2202x + FyM \u2202gM \u2202x + Fxm \u2202fm \u2202x + Fym \u2202gm \u2202x = \u2212kx (9.287) F\u03b8 = FxM \u2202fM \u2202\u03b8 + FyM \u2202gM \u2202\u03b8 + Fxm \u2202fm \u2202\u03b8 + Fym \u2202gm \u2202\u03b8 = \u2212mgl sin \u03b8 (9.288) and finally the Lagrange equations of motion are (M +m)x\u0308+ml\u03b8\u0308 cos \u03b8 \u2212ml\u03b8\u0307 2 sin \u03b8 = \u2212kx (9.289) ml2\u03b8\u0308 +mlx\u0308 cos \u03b8 = \u2212mgl sin \u03b8. (9.290) Example 366 Kinetic energy of the Earth. Earth is approximately a rotating rigid body about a fixed axis. The two motions of the Earth are called revolution about the sun, and rotation about an axis approximately fixed in the Earth. The kinetic energy of the Earth due to its rotation is K1 = 1 2 I\u03c921 = 1 2 2 5 \u00a1 5.9742\u00d7 1024 \u00a2\u00b56356912 + 6378388 2 \u00b62\u00b5 2\u03c0 24\u00d7 3600 366.25 365.25 \u00b62 = 2.5762\u00d7 1029 J 560 9. Applied Dynamics and the kinetic energy of the Earth due to its revolution is K2 = 1 2 Mr2\u03c922 = 1 2 \u00a1 5.9742\u00d7 1024 \u00a2 \u00a1 1.49475\u00d7 1011 \u00a22\u00b5 2\u03c0 24\u00d7 3600 1 365.25 \u00b62 = 2.6457\u00d7 1033 J where r is the distance from the sun and \u03c92 is the angular speed about the sun. The total kinetic energy of the Earth is K = K1 +K2. However, the ratio of the revolutionary to rotational kinetic energies is K2 K1 = 2.6457\u00d7 1033 2.5762\u00d7 1029 \u2248 10000. Example 367 F Explicit form of Lagrange equations. Assume the coordinates of every particle are functions of the coordinates q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn but not the time t. The kinetic energy of the system made of n massive particles can be written as K = 1 2 nX i=1 mi \u00a1 x\u03072i + y\u03072i + z\u03072i \u00a2 = 1 2 nX j=1 nX k=1 ajkq\u0307j q\u0307k (9.291) where the coefficients ajk are functions of q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn and ajk = akj . (9.292) The Lagrange equations of motion d dt \u00b5 \u2202K \u2202q\u0307r \u00b6 \u2212 \u2202K \u2202qr = Fr r = 1, 2, \u00b7 \u00b7 \u00b7n (9.293) are then equal to d dt nX m=1 amrq\u0307m \u2212 1 2 nX j=1 nX k=1 ajk \u2202qr q\u0307j q\u0307k = Fr (9.294) or nX m=1 amrq\u0308m + nX k=1 nX n=1 \u0393rk,nq\u0307kq\u0307n = Fr (9.295) where \u0393ij,k is called the Christoffel operator \u0393ij,k = 1 2 \u00b5 \u2202aij \u2202qk + \u2202aik \u2202qj \u2212 \u2202akj \u2202qi \u00b6 . (9.296) 9. Applied Dynamics 561 9.6 Lagrangian Mechanics Assume for some forces F = \u00a3 Fix Fiy Fiz \u00a4T there is a function V , called potential energy, such that the force is derivable from V F = \u2212\u2207V. (9.297) Such a force is called potential or conservative force. Then, the Lagrange equation of motion can be written as d dt \u00b5 \u2202L \u2202q\u0307r \u00b6 \u2212 \u2202L \u2202qr = Qr r = 1, 2, \u00b7 \u00b7 \u00b7n (9.298) where L = K \u2212 V (9.299) is the Lagrangean of the system and Qr is the nonpotential generalized force. Proof. Assume the external forces F = \u00a3 Fxi Fyi Fzi \u00a4T acting on the system are conservative. F = \u2212\u2207V (9.300) The work done by these forces in an arbitrary virtual displacement \u03b4q1, \u03b4q2, \u03b4q3, \u00b7 \u00b7 \u00b7 , \u03b4qn is \u2202W = \u2212\u2202V \u2202q1 \u03b4q1 \u2212 \u2202V \u2202q2 \u03b4q2 \u2212 \u00b7 \u00b7 \u00b7 \u2202V \u2202qn \u03b4qn (9.301) then the Lagrange equation becomes d dt \u00b5 \u2202K \u2202q\u0307r \u00b6 \u2212 \u2202K \u2202qr = \u2212 \u2202V \u2202q1 r = 1, 2, \u00b7 \u00b7 \u00b7n. (9.302) Introducing the Lagrangean function L = K \u2212 V converts the Lagrange equation to d dt \u00b5 \u2202L \u2202q\u0307r \u00b6 \u2212 \u2202L \u2202qr = 0 r = 1, 2, \u00b7 \u00b7 \u00b7n (9.303) for a conservative system. The Lagrangean is also called kinetic potential. If a force is not conservative, then the virtual work done by the force is \u03b4W = nX i=1 \u00b5 Fxi \u2202fi \u2202qr + Fyi \u2202gi \u2202qr + Fzi \u2202hi \u2202qr \u00b6 \u03b4qr = Qr \u03b4qr (9.304) and the equation of motion would be d dt \u00b5 \u2202L \u2202q\u0307r \u00b6 \u2212 \u2202L \u2202qr = Qr r = 1, 2, \u00b7 \u00b7 \u00b7n (9.305) where Qr is the nonpotential generalized force doing work in a virtual displacement of the rth generalized coordinate qr. 562 9. Applied Dynamics Example 368 Spherical pendulum. A pendulum analogy is utilized in modeling of many dynamical problems. Figure 9.13 illustrates a spherical pendulum with mass m and length l. The angles \u03d5 and \u03b8 may be used as describing coordinates of the system. The Cartesian coordinates of the mass as a function of the generalized coordinates are \u23a1\u23a3 X Y Z \u23a4\u23a6 = \u23a1\u23a3 r cos\u03d5 sin \u03b8 r sin \u03b8 sin\u03d5 \u2212r cos \u03b8 \u23a4\u23a6 (9.306) and therefore, the kinetic and potential energies of the pendulum are K = 1 2 m \u00b3 l2\u03b8\u0307 2 + l2\u03d5\u03072 sin2 \u03b8 \u00b4 (9.307) V = \u2212mgl cos \u03b8. (9.308) The kinetic potential function of this system is then equal to L = 1 2 m \u00b3 l2\u03b8\u0307 2 + l2\u03d5\u03072 sin2 \u03b8 \u00b4 +mgl cos \u03b8 (9.309) which leads to the following equations of motion: \u03b8\u0308 \u2212 \u03d5\u03072 sin \u03b8 cos \u03b8 + g l sin \u03b8 = 0 (9.310) \u03d5\u0308 sin2 \u03b8 + 2\u03d5\u0307\u03b8\u0307 sin \u03b8 cos \u03b8 = 0. (9.311) Example 369 Controlled compound pendulum. A massive arm is attached to a ceiling at a pin joint O as illustrated in Figure 9.14. Assume that there is viscous friction in the joint where an ideal motor can apply a torque Q to move the arm. The rotor of an ideal motor has no moment of inertia by assumption. 9. Applied Dynamics 563 The kinetic and potential energies of the manipulator are K = 1 2 I\u03b8\u0307 2 = 1 2 \u00a1 IC +ml2 \u00a2 \u03b8\u0307 2 (9.312) V = \u2212mg cos \u03b8 (9.313) where m is the mass and I is the moment of inertia of the pendulum about O. The Lagrangean of the manipulator is L = K \u2212 V = 1 2 I\u03b8\u0307 2 +mg cos \u03b8 (9.314) and therefore, the equation of motion of the pendulum is M = d dt \u00b5 \u2202L \u2202\u03b8\u0307 \u00b6 \u2212 \u2202L \u2202\u03b8 = I \u03b8\u0308 +mgl sin \u03b8. (9.315) The generalized force M is the contribution of the motor torque Q and the viscous friction torque \u2212c\u03b8\u0307. Hence, the equation of motion of the manipulator is Q = I \u03b8\u0308 + c\u03b8\u0307 +mgl sin \u03b8. (9.316) Example 370 An ideal 2R planar manipulator dynamics. An ideal model of a 2R planar manipulator is illustrated in Figure 9.15. It is called ideal because we assume the links are massless and there is no friction. The masses m1 and m2 are the mass of the second motor to run 564 9. Applied Dynamics the second link and the load at the endpoint. We take the absolute angle \u03b81 and the relative angle \u03b82 as the generalized coordinates to express the configuration of the manipulator. The global positions of m1 and m2 are\u2219 X1 Y2 \u00b8 = \u2219 l1 cos \u03b81 l1 sin \u03b81 \u00b8 (9.317)\u2219 X2 Y2 \u00b8 = \u2219 l1 cos \u03b81 + l2 cos (\u03b81 + \u03b82) l1 sin \u03b81 + l2 sin (\u03b81 + \u03b82) \u00b8 (9.318) and therefore, the global velocity of the masses are\u2219 X\u03071 Y\u03071 \u00b8 = \u2219 \u2212l1\u03b8\u03071 sin \u03b81 l1\u03b8\u03071 cos \u03b81 \u00b8 (9.319) \u2219 X\u03072 Y\u03072 \u00b8 = \u23a1\u23a3 \u2212l1\u03b8\u03071 sin \u03b81 \u2212 l2 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 sin (\u03b81 + \u03b82) l1\u03b8\u03071 cos \u03b81 + l2 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 cos (\u03b81 + \u03b82) \u23a4\u23a6 . (9.320) The kinetic energy of this manipulator is made of kinetic energy of the masses and is equal to K = K1 +K2 = 1 2 m1 \u00b3 X\u03072 1 + Y\u0307 2 1 \u00b4 + 1 2 m2 \u00b3 X\u03072 2 + Y\u0307 2 2 \u00b4 = 1 2 m1l 2 1\u03b8\u0307 2 1 + 1 2 m2 \u00b5 l21\u03b8\u0307 2 1 + l22 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b42 + 2l1l2\u03b8\u03071 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 cos \u03b82 \u00b6 . (9.321) 9. Applied Dynamics 565 The potential energy of the manipulator is V = V1 + V2 = m1gY1 +m2gY2 = m1gl1 sin \u03b81 +m2g (l1 sin \u03b81 + l2 sin (\u03b81 + \u03b82)) . (9.322) The Lagrangean is then obtained from Equations (9.321) and (9.322) L = K \u2212 V (9.323) = 1 2 m1l 2 1\u03b8\u0307 2 1 + 1 2 m2 \u00b5 l21\u03b8\u0307 2 1 + l22 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b42 + 2l1l2\u03b8\u03071 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 cos \u03b82 \u00b6 \u2212 (m1gl1 sin \u03b81 +m2g (l1 sin \u03b81 + l2 sin (\u03b81 + \u03b82))) . which provides the required partial derivatives as follows: \u2202L \u2202\u03b81 = \u2212 (m1 +m2) gl1 cos \u03b81 \u2212m2gl2 cos (\u03b81 + \u03b82) (9.324) \u2202L \u2202\u03b8\u03071 = (m1 +m2) l 2 1\u03b8\u03071 +m2l 2 2 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 +m2l1l2 \u00b3 2\u03b8\u03071 + \u03b8\u03072 \u00b4 cos \u03b82 (9.325) d dt \u00b5 \u2202L \u2202\u03b8\u03071 \u00b6 = (m1 +m2) l 2 1\u03b8\u03081 +m2l 2 2 \u00b3 \u03b8\u03081 + \u03b8\u03082 \u00b4 +m2l1l2 \u00b3 2 \u03b8\u03081 + \u03b8\u03082 \u00b4 cos \u03b82 \u2212m2l1l2\u03b8\u03072 \u00b3 2\u03b8\u03071 + \u03b8\u03072 \u00b4 sin \u03b82 (9.326) \u2202L \u2202\u03b82 = \u2212m2l1l2\u03b8\u03071 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 sin \u03b82 \u2212m2gl2 cos (\u03b81 + \u03b82) (9.327) \u2202L \u2202\u03b8\u03072 = m2l 2 2 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 +m2l1l2\u03b8\u03071 cos \u03b82 (9.328) d dt \u00b5 \u2202L \u2202\u03b8\u03072 \u00b6 = m2l 2 2 \u00b3 \u03b8\u03081 + \u03b8\u03082 \u00b4 +m2l1l2\u03b8\u03081 cos \u03b82 \u2212m2l1l2\u03b8\u03071\u03b8\u03072 sin \u03b82 (9.329) Therefore, the equations of motion for the 2R manipulator are Q1 = d dt \u00b5 \u2202L \u2202\u03b8\u03071 \u00b6 \u2212 \u2202L \u2202\u03b81 = (m1 +m2) l 2 1\u03b8\u03081 +m2l 2 2 \u00b3 \u03b8\u03081 + \u03b8\u03082 \u00b4 +m2l1l2 \u00b3 2 \u03b8\u03081 + \u03b8\u03082 \u00b4 cos \u03b82 \u2212m2l1l2\u03b8\u03072 \u00b3 2\u03b8\u03071 + \u03b8\u03072 \u00b4 sin \u03b82 +(m1 +m2) gl1 cos \u03b81 +m2gl2 cos (\u03b81 + \u03b82) (9.330) 566 9. Applied Dynamics Q2 = d dt \u00b5 \u2202L \u2202\u03b8\u03072 \u00b6 \u2212 \u2202L \u2202\u03b82 = m2l 2 2 \u00b3 \u03b8\u03081 + \u03b8\u03082 \u00b4 +m2l1l2\u03b8\u03081 cos \u03b82 \u2212m2l1l2\u03b8\u03071\u03b8\u03072 sin \u03b82 +m2l1l2\u03b8\u03071 \u00b3 \u03b8\u03071 + \u03b8\u03072 \u00b4 sin \u03b82 +m2gl2 cos (\u03b81 + \u03b82) . (9.331) The generalized forces Q1 and Q2 are the required forces to drive the generalized coordinates. In this case, Q1 is the torque at the base motor and Q2 is the torque of the motor at m1. The equations of motion can be rearranged to have a more systematic form Q1 = \u00a1 (m1 +m2) l 2 1 +m2l2 (l2 + 2l1 cos \u03b82) \u00a2 \u03b8\u03081 +m2l2 (l2 + l1 cos \u03b82) \u03b8\u03082 \u22122m2l1l2 sin \u03b82 \u03b8\u03071\u03b8\u03072 \u2212m2l1l2 sin \u03b82 \u03b8\u0307 2 2 +(m1 +m2) gl1 cos \u03b81 +m2gl2 cos (\u03b81 + \u03b82) (9.332) Q2 = m2l2 (l2 + l1 cos \u03b82) \u03b8\u03081 +m2l 2 2\u03b8\u03082 +m2l1l2 sin \u03b82 \u03b8\u0307 2 1 +m2gl2 cos (\u03b81 + \u03b82) . (9.333) Example 371 Mechanical energy. If a system of masses mi are moving in a potential force field Fmi = \u2212\u2207iV (9.334) their Newton equations of motion will be mir\u0308i = \u2212\u2207iV i = 1, 2, \u00b7 \u00b7 \u00b7n. (9.335) The inner product of equations of motion with r\u0307i and adding the equations nX i=1 mir\u0307i \u00b7 r\u0308i = \u2212 nX i=1 r\u0307i \u00b7\u2207iV (9.336) and then, integrating over time 1 2 nX i=1 mir\u0307i \u00b7 r\u0307i = \u2212 Z nX i=1 ri \u00b7\u2207iV (9.337) shows that K = \u2212 Z nX i=1 \u00b5 \u2202V \u2202xi xi + \u2202V \u2202yi yi + \u2202V \u2202zi zi \u00b6 = \u2212V +E (9.338) 9. Applied Dynamics 567 where E is the constant of integration. E is called the mechanical energy of the system and is equal to kinetic plus potential energies. E = K + V (9.339) Example 372 Falling wheel. Figure 9.16 illustrates a wheel turning, without slip, over a cylindrical hill. We may use the conservation of mechanical energy to find the angle at which the wheel leaves the hill. At the initial instant of time, the wheel is at point A. We assume the initial kinetic and potential, and hence, the mechanical energies are zero. When the wheel is turning over the hill, its angular velocity, \u03c9, is \u03c9 = v r (9.340) where v is the speed at the center of the wheel. At any other point B, the wheel achieves some kinetic energy and loses some potential energy. At a certain angle, where the normal component of the weight cannot provide more centripetal force, mg cos \u03b8 = mv2 R+ r . (9.341) the wheel separates from the surface. Employing the conservation of energy, we have EA = EB (9.342) KA + VA = KB + VB. (9.343) The kinetic and potential energy at the separation point B are KB = 1 2 mv2 + 1 2 IC\u03c9 2 (9.344) VB = \u2212mg (R+ r) (1\u2212 cos \u03b8) (9.345) 568 9. Applied Dynamics where IC is the mass moment of inertia for the wheel about its center. Therefore, 1 2 mv2 + 1 2 IC\u03c9 2 = mg (R+ r) (1\u2212 cos \u03b8) (9.346) and substituting (9.340) and (9.341) provides\u00b5 1 + IC mr2 \u00b6 (R+ r) g cos \u03b8 = 2g (R+ r) (1\u2212 cos \u03b8) (9.347) and therefore, the separation angle is \u03b8 = cos\u22121 2mr2 IC + 3mr2 . (9.348) Let\u2019s examine the equation for a disc wheel with IC = 1 2 mr2. (9.349) and find the separation angle. \u03b8 = cos\u22121 4 7 (9.350) \u2248 0.96 rad \u2248 55.15 deg Example 373 Turning wheel over a step. Figure 9.17 illustrates a wheel of radius R turning with speed v to go over a step with height H < R. We may use the principle of energy conservation and find the speed of the wheel after getting across the step. Employing the conservation of energy, 9. Applied Dynamics 569 we have EA = EB (9.351) KA + VA = KB + VB (9.352) 1 2 mv21 + 1 2 IC\u03c9 2 1 + 0 = 1 2 mv22 + 1 2 IC\u03c9 2 2 +mgH (9.353)\u00b5 m+ IC R2 \u00b6 v21 = \u00b5 m+ IC R2 \u00b6 v22 + 2mgH (9.354) and therefore, v2 = vuutv21 \u2212 2gH 1 + IC mR2 . (9.355) The condition for having a real v2 is v1 > vuut 2gH 1 + IC mR2 . (9.356) The second speed (9.355) and the condition (9.356) for a solid disc are v2 = r v21 \u2212 4 3 Hg (9.357) v1 > r 4 3 Hg (9.358) because we assumed that IC = 1 2 mR2. (9.359) Example 374 Trebuchet. A trebuchet, shown schematically in Figure 9.18, is a shooting weapon of war powered by a falling massive counterweight m1. A beam AB is pivoted to the chassis with two unequal sections a and b. The figure shows a trebuchet at its initial configuration. The origin of a global coordinate frame is set at the pivot point. The counterweight m1 is at (x1, y1) and is hinged at the shorter arm of the beam at a distance c from the end B. The mass of the projectile is m2 and it is at the end of a massless sling with a length l attached to the end of the longer arm of the beam. The three independent variable angles \u03b1, \u03b8, and \u03b3 describe the motion of the device. We consider the parameters a, b, c, d, l, m1, and m2 constant, and determine the equations of motion by the Lagrange method. Figure 9.19 illustrates the trebuchet when it is in motion. The position coordinates of masses m1 and m2 are x1 = b sin \u03b8 \u2212 c sin (\u03b8 + \u03b3) (9.360) y1 = \u2212b cos \u03b8 + c cos (\u03b8 + \u03b3) (9.361) 570 9. Applied Dynamics and x2 = \u2212a sin \u03b8 \u2212 l sin (\u2212\u03b8 + \u03b1) (9.362) y2 = \u2212a cos \u03b8 \u2212 l cos (\u2212\u03b8 + \u03b1) . (9.363) Taking a time derivative provides the velocity components x\u03071 = b\u03b8\u0307 cos \u03b8 \u2212 c \u00b3 \u03b8\u0307 + \u03b3\u0307 \u00b4 cos (\u03b8 + \u03b3) (9.364) y\u03071 = b\u03b8\u0307 sin \u03b8 \u2212 c \u00b3 \u03b8\u0307 + \u03b3\u0307 \u00b4 sin (\u03b8 + \u03b3) (9.365) x\u03072 = l (c\u2212 \u03b1\u0307) cos (\u03b1\u2212 \u03b8)\u2212 a\u03b8\u0307 cos (\u03b8) (9.366) y\u03072 = a\u03b8\u0307 sin \u03b8 \u2212 l \u00b3 \u03b8\u0307 \u2212 \u03b1\u0307 \u00b4 sin (\u03b1\u2212 \u03b8) . (9.367) 9. Applied Dynamics 571 which shows that the kinetic energy of the system is K = 1 2 m1v 2 1 + 1 2 m2v 2 2 = 1 2 m1 \u00a1 x\u030721 + y\u030721 \u00a2 + 1 2 m2 \u00a1 x\u030722 + y\u030722 \u00a2 = 1 2 m1 \u00b3\u00a1 b2 + c2 \u00a2 \u03b8\u0307 2 + c2\u03b3\u03072 + 2c2\u03b8\u0307\u03b3\u0307 \u00b4 \u2212m1bc\u03b8\u0307 \u00b3 \u03b8\u0307 + \u03b3\u0307 \u00b4 cos \u03b3 + 1 2 m2 \u00b3\u00a1 a2 + l2 \u00a2 \u03b8\u0307 2 + l2\u03b1\u03072 \u2212 2l2\u03b8\u0307\u03b1\u0307 \u00b4 \u2212m2al\u03b8\u0307 \u00b3 \u03b8\u0307 \u2212 \u03b1\u0307 \u00b4 cos (2\u03b8 \u2212 \u03b1) . (9.368) The potential energy of the system can be calculated by y-position of the masses. V = m1gy1 +m2gy2 = m1g (\u2212b cos \u03b8 + c cos (\u03b8 + \u03b3)) +m2g (\u2212a cos \u03b8 \u2212 l cos (\u2212\u03b8 + \u03b1)) (9.369) Having the energies K and V , we can set up the Lagrangean L. L = K \u2212 V (9.370) Using the Lagrangean, we are able to find the three equations of motion. d dt \u00b5 \u2202L \u2202\u03b8\u0307 \u00b6 \u2212 \u2202L \u2202\u03b8 = 0 (9.371) d dt \u00b5 \u2202L \u2202\u03b1\u0307 \u00b6 \u2212 \u2202L \u2202\u03b1 = 0 (9.372) d dt \u00b5 \u2202L \u2202\u03b3\u0307 \u00b6 \u2212 \u2202L \u2202\u03b3 = 0. (9.373) The trebuchet appeared in 500 to 400 B.C. China and was developed by Persian armies around 300 B.C. It was used by the Arabs against the Romans during 600 to 1200 A.D. The trebuchet is also called the manjaniq, catapults, or onager. The \"Manjaniq\" is the root of the words \"machine\" and \"mechanic\". 9.7 Summary The translational and rotational equations of motion for a rigid body, expressed in the global coordinate frame, are GF = Gd dt Gp (9.374) GM = Gd dt GL (9.375) 572 9. Applied Dynamics where GF and GM indicate the resultant of the external forces and moments applied on the rigid body, measured at the mass center C. The vector Gp is the momentum and GL is the moment of momentum for the rigid body at C p = mv (9.376) L = rC \u00d7 p. (9.377) The expression of the equations of motion in the body coordinate frame are BF = Gp\u0307+ B G\u03c9B \u00d7 Bp = m BaB +m B G\u03c9B \u00d7 BvB (9.378) BM = BL\u0307+B G\u03c9B \u00d7 BL = BI B G\u03c9\u0307B + B G\u03c9B \u00d7 \u00a1 BI B G\u03c9B \u00a2 (9.379) where I is the moment of inertia for the rigid body. I = \u23a1\u23a3 Ixx Ixy Ixz Iyx Iyy Iyz Izx Izy Izz \u23a4\u23a6 . (9.380) 9. Applied Dynamics 573 The elements of I are functions of the mass distribution of the rigid body and are defined by Iij = Z B \u00a1 r2i \u03b4mn \u2212 ximxjn \u00a2 dm , i, j = 1, 2, 3 (9.381) where \u03b4ij is Kronecker\u2019s delta. Every rigid body has a principal body coordinate frame in which the moment of inertia is in the form BI = \u23a1\u23a3 I1 0 0 0 I2 0 0 0 I3 \u23a4\u23a6 . (9.382) The rotational equation of motion in the principal coordinate frame simplifies to M1 = I1\u03c9\u03071 \u2212 (I2 \u2212 I2)\u03c92\u03c93 M2 = I2\u03c9\u03072 \u2212 (I3 \u2212 I1)\u03c93\u03c91 (9.383) M3 = I3\u03c9\u03073 \u2212 (I1 \u2212 I2)\u03c91\u03c92. The equations of motion for a mechanical system having n DOF can also be found by the Lagrange equation d dt \u00b5 \u2202L \u2202q\u0307r \u00b6 \u2212 \u2202L \u2202qr = Qr r = 1, 2, \u00b7 \u00b7 \u00b7n (9.384) L = K \u2212 V (9.385) where L is the Lagrangean of the system, K is the kinetic energy, V is the potential energy, and Qr is the nonpotential generalized force. Qr = nX i=1 \u00b5 Qix \u2202fi \u2202q1 +Qiy \u2202gi \u2202q2 +Qiz \u2202hi \u2202qn \u00b6 (9.386) The parameters qr, r = 1, 2, \u00b7 \u00b7 \u00b7 , n are the generalized coordinates of the system, Q = \u00a3 Qix Qiy Qiz \u00a4T is the external force acting on the ith particle of the system, and Qr is the generalized force associated to qr. When (xi, yi, zi) are Cartesian coordinates in a globally fixed coordinate frame for the particle mi, then its coordinates may be functions of another set of generalized coordinates q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn and possibly time t. xi = fi(q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, t) (9.387) yi = gi(q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, t) (9.388) zi = hi(q1, q2, q3, \u00b7 \u00b7 \u00b7 , qn, t) (9.389) 574 9. Applied Dynamics 9.8 Key Symbols a, b, w, h length a acceleration C mass center d position vector of the body coordinate frame df infinitesimal force dm infinitesimal mass dm infinitesimal moment E mechanical energy F force FC Coriolis force g gravitational acceleration H height I moment of inertia matrix I1, I2, I3 principal moment of inertia K kinetic energy l directional line L moment of momentum L = K \u2212 V Lagrangean m mass M moment p momentum P,Q points in rigid body r radius of disc r position vector R radius R rotation matrix t time u\u0302 unit vector to show the directional line v \u2261 x\u0307, v velocity V potential energy w eigenvector W work W eigenvector matrix x, y, z, x displacement \u03b4ij Kronecker\u2019s delta \u0393ij,k Christoffel operator \u03bb eigenvalue \u03d5, \u03b8, \u03c8 Euler angles \u03c9,\u03c9 angular velocity k parallel \u22a5 orthogonal 9. Applied Dynamics 575 Exercises 1. Kinetic energy of a rigid link. Consider a straight and uniform bar as a rigid bar. The bar has a mass m. Show that the kinetic energy of the bar can be expressed as K = 1 6 m (v1 \u00b7 v1 + v1 \u00b7 v2 + v2 \u00b7 v2) where v1 and v2 are the velocity vectors of the endpoints of the bar. 2. Discrete particles. There are three particles m1 = 10kg, m2 = 20 kg, m3 = 30 kg, at r1 = \u23a1\u23a3 1 \u22121 1 \u23a4\u23a6 r1 = \u23a1\u23a3 \u22121\u22123 2 \u23a4\u23a6 r1 = \u23a1\u23a3 2 \u22121 \u22123 \u23a4\u23a6 . Their velocities are v1 = \u23a1\u23a3 2 1 1 \u23a4\u23a6 v1 = \u23a1\u23a3 \u221210 2 \u23a4\u23a6 v1 = \u23a1\u23a3 3 \u22122 \u22121 \u23a4\u23a6 . Find the position and velocity of the system at C. Calculate the system\u2019s momentum and moment of momentum. Calculate the system\u2019s kinetic energy and determine the rotational and translational parts of the kinetic energy. 3. Newton\u2019s equation of motion in the body frame. Show that Newton\u2019s equation of motion in the body frame is\u23a1\u23a3 Fx Fy Fz \u23a4\u23a6 = m \u23a1\u23a3 ax ay az \u23a4\u23a6+ \u23a1\u23a3 0 \u2212\u03c9z \u03c9y \u03c9z 0 \u2212\u03c9x \u2212\u03c9y \u03c9x 0 \u23a4\u23a6\u23a1\u23a3 vx vy vz \u23a4\u23a6 . 4. Work on a curved path. A particle of mass m is moving on a circular path given by GrP = cos \u03b8 I\u0302 + sin \u03b8 J\u0302 + 4 K\u0302. Calculate the work done by a force GF when the particle moves from \u03b8 = 0 to \u03b8 = \u03c0 2 . (a) GF = z2 \u2212 y2 (x+ y) 2 I\u0302 + y2 \u2212 x2 (x+ y) 2 J\u0302 + x2 \u2212 y2 (x+ z) 2 K\u0302 576 9. Applied Dynamics (b) GF = z2 \u2212 y2 (x+ y) 2 I\u0302 + 2y x+ y J\u0302 + x2 \u2212 y2 (x+ z) 2 K\u0302 5. Principal moments of inertia. Find the principal moments of inertia and directions for the following inertia matrices: (a) [I] = \u23a1\u23a3 3 2 2 2 2 0 2 0 4 \u23a4\u23a6 (b) [I] = \u23a1\u23a3 3 2 4 2 0 2 4 2 3 \u23a4\u23a6 (c) [I] = \u23a1\u23a3 100 20 \u221a 3 0 20 \u221a 3 60 0 0 0 10 \u23a4\u23a6 6. Rotated moment of inertia matrix. A principal moment of inertia matrix B2I is given as [I] = \u23a1\u23a3 3 0 0 0 5 0 0 0 4 \u23a4\u23a6 . The principal frame was achieved by rotating the initial body coordinate frame 30 deg about the x-axis, followed by 45 deg about the z-axis. Find the initial moment of inertia matrix B1I. 7. Rotation of moment of inertia matrix. Find the required rotation matrix that transforms the moment of inertia matrix [I] to an diagonal matrix. [I] = \u23a1\u23a3 3 2 2 2 2 0.1 2 0.1 4 \u23a4\u23a6 8. F Cubic equations. The solution of a cubic equation ax3 + bx2 + cx+ d = 0 9. Applied Dynamics 577 where a 6= 0, can be found in a systematic way. Transform the equation to a new form with discriminant 4p3 + q2, y3 + 3py + q = 0 using the transformation x = y \u2212 b 3a , where, p = 3ac\u2212 b2 9a2 q = 2b3 \u2212 9abc+ 27a2d 27a3 . The solutions are then y1 = 3 \u221a \u03b1\u2212 3 p \u03b2 y2 = e 2\u03c0i 3 3 \u221a \u03b1\u2212 e 4\u03c0i 3 3 p \u03b2 y3 = e 4\u03c0i 3 3 \u221a \u03b1\u2212 e 2\u03c0i 3 3 p \u03b2 where, \u03b1 = \u2212q + p q2 + 4p3 2 \u03b2 = \u2212q + p q2 + 4p3 2 . For real values of p and q, if the discriminant is positive, then one root is real, and two roots are complex conjugates. If the discriminant is zero, then there are three real roots, of which at least two are equal. If the discriminant is negative, then there are three unequal real roots. Apply this theory for the characteristic equation of the matrix [I] and show that the principal moments of inertia are real. 9. Kinematics of a moving car on the Earth. The location of a vehicle on the Earth is described by its longitude \u03d5 from a fixed meridian, say, the Greenwich meridian, and its latitude \u03b8 from the equator, as shown in Figure 9.20. We attach a coordinate frame B at the center of the Earth with the x-axis on the equator\u2019s plane and the y-axis pointing to the vehicle. There are also two coordinate frames E and G where E is attached to the Earth and G is the global coordinate frame. Show that the angular velocity of B and the velocity of the vehicle are B G\u03c9B = \u03b8\u0307 \u0131\u0302B + (\u03c9E + \u03d5\u0307) sin \u03b8 j\u0302B + (\u03c9E + \u03d5\u0307) cos \u03b8 k\u0302 B GvP = \u2212r (\u03c9E + \u03d5\u0307) cos \u03b8 \u0131\u0302B + r\u03b8\u0307 k\u0302. Calculate the acceleration of the vehicle. 578 9. Applied Dynamics \u03b8 X Y Z x y z r \u03c9E Z (a) (b) G B Y z x y E \u03d5 \u03b8 y z \u03c9E P P 9. Applied Dynamics 579 (c) the pivot O has a uniform motion on a circle rO = R cos\u03c9t I\u0302 +R sin\u03c9t J\u0302. 13. Equations of motion from Lagrangean. Consider a physical system with a Lagrangean as L = 1 2 m (ax\u0307+ by\u0307) 2 \u2212 1 2 k (ax+ by) 2 . and find the equations of motion. The coefficients m, k, a, and b are constant. 14. Lagrangean from equation of motion. Find the Lagrangean associated to the following equations of motions: (a) mr2\u03b8\u0308 + k1l1\u03b8 + k2l2\u03b8 +mgl = 0 580 9. Applied Dynamics (b) r\u0308 \u2212 r \u03b8\u0307 2 = 0 r2 \u03b8\u0308 + 2r r\u0307 \u03b8\u0307 = 0 15. Trebuchet. Derive the equations of motion for the trebuchet shown in Figure 9.18. 16. Simplified trebuchet. Three simplified models of a trebuchet are shown in Figures 9.23 to 9.25. Derive and compare their equations of motion. 9. Applied Dynamics 581 10"
        ]
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure12.26-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure12.26-1.png",
        "caption": "Figure 12.26 Simply supported beam with the V and M diagrams due to (a) the force and (b) the couple. (c) The V and M diagrams due to the force and the couple together, found by superposing the V and M diagrams from (a) and (b). (d) The V and M diagrams from (c) transferred to a horizontal axis.",
        "texts": [
            " If several loads are acting on a structure, the separate influences of the various loads on the support reactions and section forces can be added together.1 The validity of this so-called principle of superposition is a result of the linear relationships between the loads, section forces and support reactions. 1 In Section 6.3.1, we showed that distributed loads can be split and that the individual influences on the support reactions can be added together. 506 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Example 1 In Figure 12.26, the principle of superposition has been applied to determine the V and M diagrams for an 8-metre beam that is loaded by a force of 40 kN and a couple of 80 kNm. In Figure 12.26a, the V and M diagrams have been calculated due to the force only. In Figure 12.26b, the V and M diagrams have been calculated due to the couple only. The final V and M diagrams with concurrent loading by the force and the couple is shown in Figure 12.26c. To draw these V and M diagrams, one of the two diagrams to be superposed has been reflected with respect to the horizontal axis to simplify the graphics. In areas with opposite deformation symbols that overlap one another, the combined contribution to the section force is zero. The remaining areas have been filled and form the final V and M diagrams. In Figure 12.26d, these diagrams have been transferred to a horizontal axis, but this is generally not necessary. Of course the superposition can also be performed by determining the ordinates at a number of points and adding them together. Example 2 The second example relating to the principle of superposition concerns further analysis of the force flow in segment BC of beam AD in Figure 12.27, with a uniformly distributed load over BC. The V and M diagrams for this beam were calculated in Section 12.1.3, Example 2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure7.13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure7.13-1.png",
        "caption": "FIGURE 7.13 Orthogonal coplanar quadrilateral for partial mutual term.",
        "texts": [
            " Remember that the original integrals have 1\u2215R singularities for the self-terms. In this section, we are using global coordinates for the evaluation of the integrals since an analytic result can be obtained in global coordinates for this case. The original work was presented in Ref. [23], whereas the new solution was led by D. Romano in a recent work. The evaluation of the partial potential coefficient using local coordinates is given by (7.24). A self-term for a quadrilateral cell is shown in Fig. 7.12 and the coupling situation for two quadrilateral cells is given in Fig. 7.13. We observe that the key difference between the partial coefficients of potential and the partial inductances is a different multiplication factor as well as the dot-product between the current directions. We have to approximate the dot-product for partial inductances by using average current directions. However, the accuracy is reduced if the current direction changes considerably over at least one of the cells. This may require further subdivisions of the cells. The geometry of the problem of interest is shown in Fig. 7.13 for the general case for two quadrilaterals, whereas Fig. 7.12 represents the special case for the self-term. In this computation, we have to assume that both surfaces are located in the z = 0 plane in the general coordinate system (x, y, z). The two zero thickness quadrilateral cells 1 and 2 are specified using the local corner coordinates 0\u20133, with the associated global coordinates x, y, z with the surfaces 1 for quadrilateral 1 and 2 for 2 corresponding to the primed coordinates. The integral to be evaluated for the zero thickness Pp or Lp is IT = \u222b2 \u222b1 1|r \u2212 r\u2032| d d \u2032 = \u222b2 \u222b1 1 R d d \u2032 (7"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure5.32-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure5.32-1.png",
        "caption": "Figure 5.32 The forces acting on joint A and frame half AS.",
        "texts": [
            " 176 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Check: With the hinge forces calculated, the left-hand frame part AS must also be in force equilibrium. d. The following forces are acting on joint A: \u2022 the support reactions Ah = 40 kN and Av = 50 kN; \u2022 the force N exerted by the tie-rod AB, with components Nh = 40 kN and Nv = 20 kN; \u2022 the forces exerted by the left-hand frame part AS. The last-mentioned forces can be found from the force equilibrium of joint A. All the forces on the joint are shown in Figure 5.32. Check: The part AS isolated from joint A has to be in force equilibrium. Note that here the horizontal load of 40 kN is transferred via a long detour to the support at A. 5.5 The portal-like structure in Figure 5.33a, with only hinged joints, is kinematically indeterminate. The structure can tilt. To prevent this, one can fix one or more of the columns (Figure 5.33b). Or one can replace one or more of the hinges between column and beam by rigid connections (Figure 5.33c). It is also possible to prevent the construction from tilting by applying so-called shoring bars, indicated in Figure 5"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002642_jas.2020.1003057-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002642_jas.2020.1003057-Figure3-1.png",
        "caption": "Fig. 3. Construction of joint capacity region for small cells.",
        "texts": [
            " Intuitively, this unified solution should be located on the boundary of the capacity regions of both links. In other words, the boundaries of both links will intersect at the optimal \u03b1. Consequently, the outcome boundary of the joint capacity region can be achieved by tracing intersection points for different values of \u03b1. Fig. 2 shows the boundaries of the capacity region across both links for a two-user scenario. As it can be seen, the intersection points of the corresponding boundaries (with the same values of \u03b1) form the boundary of the joint capacity region in thick solid line. Fig. 3 shows the capacity regions of the same scenario for a uniform quantized set of \u03b1 (with 100 levels). As shown, the joint capacity region can be constructed as the union of capacity regions for different values of \u03b1. 5) Subgradient Method With Small-Cell Grouping: As shown, the optimality conditions for both subproblems A.1 and A.2 follow a generic optimality conditions according to (17) and (19) per link where tuning parameter \u03bcn in conjunction with duration of transmission balance the operation across both links"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003549_robot.1996.509185-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003549_robot.1996.509185-Figure4-1.png",
        "caption": "Fig. 4: 3D Force-Angle stability measure.",
        "texts": [
            " For a consistent formillation ithe pi are numbered in ascending order following a right-hand rule convention where the thumb is directed downwards along the gravity vector, i.e. points are numbered in clockwise order when viewed from ,above. The lines which join the ground contact ]points are the candidate tipover mode axes, ai , and the set of these lines will be referred to as the support pattern. The ith tipover mode axis is given by ai = pi+l -pi i = (1, . . . , n - I} ( 3 ) a, = p1 -Pn (4) as shown in Fig. 4. The ground contact point numbering convention was required in order to obtain a set of tipover axes whose directions all coincide with that of stabilizing moments. A natural (i.e. untripped) tipover of the .vehicle will always occur about a tipover mode axis ai. A tripped tipover of the vehicle occurs when one of the ground contact points encounters an (obstacle or a sudden change in the ground conditions. In a tripped tipover the vehicle undergoes a rotation about an axis which is some linear combination of the tipover mode axes associated with the single remaining ground contact point"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001217_s41586-020-03123-5-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001217_s41586-020-03123-5-Figure1-1.png",
        "caption": "Fig. 1 | Design, programming and mechanical response of a single m-bit. a, Photograph of an m-bit comprising a bistable elastic shell (i), a magnetic cap (ii), four curved columns (iii), two stoppers (iv) and a top lid (v). b, The m-bit can be programmed using electromagnetic coils. When the current impulse activates the field, a force is exerted on the magnetic cap, which in turn switches the state of the bistable shell. c, Schematic diagram illustrating the two states of an m-bit. The magnetic cap and the stoppers are lowered (raised)",
        "texts": [
            " First, just as a digital bit can sustain many cycles of rewriting, the state change during programming of an m-bit must be elastic (reversible). Second, as in non-volatile storage, the state changes induced through writing must be indefinitely stable. Third, the resolution of programming should be at the level of an individual m-bit, independently of any interactions with neighbouring m-bits. Similarly to memory access, reading or physical loading of the m-bit should not change its state. Given the above requirements, our m-bit design (Fig.\u00a01a) consists of: (i) a bistable, elastic and conic shell that possesses two geometric states26,27, (ii) a magnetic cap that drives the shell, (iii) two sets of naturally curved columns, (iv) two stoppers that amplify the difference in the mechanical behaviour of the two states and (v) a top lid (see\u00a0Methods sections \u2018Design of m-bits\u2019 and \u2018Fabrication of m-bits\u2019 for the detailed design process). Each m-bit is cubic with an edge length of l\u00a0=\u00a030\u00a0mm. The magnetic cap is cast using an elastomer composite made of vinyl polysiloxane (VPS) mixed with NdFeB particles, which are subsequently magnetized (see\u00a0Methods section \u2018Characterization of silicone materials\u2019). All other parts are cast using pure VPS. The m-bit is fabricated in its ON state. Here we refer to \u2018programming\u2019 as the switching of the two stable equilibria of the bistable shell, which is achieved through magnetic interaction between two electromagnetic coils and the magnetic cap atop the conical shell22 (Fig.\u00a01b). The polarity of the current in the coils dictates the direction of the vertical force on the magnetic cap. This magnetic loading raises or lowers the magnetic cap if the magnetic force is sufficiently large to trigger the snap-through of the bistable shell (Fig.\u00a01c). Two different classes of force-displacement behaviour are observed for the ON or OFF states under compression (Fig.\u00a01d). In both cases, the bottom surface is fixed, and the top one is compressed uniformly (see Supplementary Video\u00a01). In the ON state, the force-displacement behaviour has a well defined ultimate strength. By contrast, in the OFF state, an initial compliant behaviour is followed by stiffening (see\u00a0Methods sections \u2018Experimental https://doi.org/10.1038/s41586-020-03123-5 Received: 14 August 2020 Accepted: 26 November 2020 Check for updates 1Flexible Structures Laboratory, Institute of Mechanical Engineering, \u00c9cole Polytechnique Fe\u0301de\u0301rale de Lausanne, Lausanne, Switzerland. 2Geometric Computing Laboratory, Institute of Computer and Communication Sciences, \u00c9cole Polytechnique Fe\u0301de\u0301rale de Lausanne, Lausanne, Switzerland. \u2709e-mail: mark.pauly@epfl.ch; pedro.reis@epfl.ch Nature | Vol 589 | 21 January 2021 | 387 protocols\u2019 and \u2018Finite-element simulations\u2019 for details). Owing to their cubic geometry, multiple m-bits can be tessellated into a square-lattice configuration (Fig.\u00a0 1e) to produce a planar-array structure28,29. Programming is performed by moving the entire array on the x\u2013y plane so that the targeted m-bit is positioned at the centre of the electromagnetic coils. The mechanical response is characterized by applying a displacement-controlled cyclic compression to the top while fixing the bottom surface. The mechanical response of an m-bit, in both its ON and OFF states, is characterized through cyclic compression testing. Three separate actions enable the substantial differences between these two states",
            " Second, in the ON state, stoppers are confined by the two columns, thereby creating an axial load path (Fig.\u00a02a; Cp,2 and Cs,2). Third, when the m-bit is in the OFF state, the curved columns undergo pure bending until they come into contact with the stoppers at much larger\u00a0deformation of \u03b5\u00a0\u2248\u00a00.15 (Fig.\u00a02a; Cp,3 and Cs,3). Consequently, the ON state is dominated by axial compression, contributing to a stiffer and\u00a0stronger behaviour. Conversely, the OFF state is dominated by the bending of the columns before contact. We extract the following quantities from the force-displacement curves (Fig.\u00a01d): (1) the effective stiffness modulus k, (2) the effective yield strength \u03c32% and (3) the\u00a0stored strain energy density U\u0302. In the ON state, we define k as the maximum slope of each loading curve before the onset of softening. In the OFF state, k corresponds to the bending stiffness of the columns before the contact-induced stiffening (see\u00a0Methods section \u2018Parametric study of the curved columns\u2019). Column\u2013column contact is designed to occur at \u03b5\u00a0\u2248\u00a00.15 by tuning the initial amplitude of the curved columns (Fig",
            " Furthermore, we can select two types of qualitatively different mechanical behaviour that can be switched with distinct nonlinear elastic responses. This on-demand control achieves a tenet of metamaterial programmability and offers a powerful strategy for future designs. \u2018Reading\u2019 of an m-bit is interpreted as the realization of the desired force responses when a displacement is applied. The m-bit exhibits reversible force-displacement responses in both states over cyclic loading, indicating that the programmed \u2018data\u2019 are not corrupted (Fig.\u00a01d). This largely elastic recovery is crucial but not inherently achieved, given that the bistable shell experiences compression in both the programming and the testing phases. We note that in the OFF state, the curved columns undergo pure bending due to compression, whereas there is contact with the stoppers in the ON state. Consequently, the downward displacement when the m-bit is in the OFF state, POFF, has slower downward velocity than that for the ON state, PON (Fig.\u00a02c). This incompatibility results in\u00a0the stoppers exerting a downward force through tangential contact with the columns",
            " X-ray micro-computed tomography (\u03bcCT) scans at these five representative stages (Fig.\u00a03c) confirm the desired geometrical changes during programming and demonstrate that the shell\u2019s boundary does not exhibit sizable deformation. Views of the scanned geometries are shown in Supplementary Video\u00a02; the scanning and testing procedures are detailed in Methods section \u2018Experimental protocols\u2019. The force required to trigger bistability is provided through the interaction between the magnetic cap and the field generated by a pair of electromagnetic coils (Fig.\u00a01b); see\u00a0Methods section \u2018Parametric study of the bistable shell\u2019. These two coils (radius R) are set in the Maxwell configuration (separation R3 ) to generate a magnetic field along the centre line, Bz(r\u00a0=\u00a00), with a constant gradient dBz(r\u00a0=\u00a00)/dz (Fig. 3b). A permanent magnet placed in this field experiences a vertical force, fm, proportional to the gradient of the field. The experimental measurements of the field agree with predictions from the Biot\u2013Savart law (see\u00a0Methods section \u2018Magnetic actuation\u2019)",
            " The MRE was prepared by adding part A of the silicone and the particles in the appropriate ratio. This composite solution was mixed using a Thinky mixer for 2\u00a0min, then cooled to \u221220 \u00b0C to reduce viscosity. Curing does not yet occur during this step because part B of the silicone has not yet been introduced. Part B was then added and mixed for an additional 1\u00a0min, followed by 20\u00a0s of defoaming before casting. The mechanical properties of the polymers mentioned above were characterized under uniaxial tension. Three dog-bone specimens (Extended Data Fig.\u00a01a) were fabricated for each material and tested following ASTM D412 using a static testing machine (5940 Single Column Tabletop Testing System, Instron). One week elapsed before testing to allow full curing of the silicone (Extended Data Fig.\u00a01b). Because the Mullins effect37 can be observed with the MRE, cyclic testing was done to obtain both the initial and the steady-state stress\u2013strain curves. The 6th load cycle, up to a maximum strain of \u03b5\u00a0=\u00a00.125, was used to determine the material parameters. We note that, in contrast to the rest of the m-bit, the MRE magnetic cap experiences negligible strain (Fig.\u00a02a). The material behaviour of the various silicone polymers was characterized using a second-order (N\u00a0=\u00a02) Ogden hyperelastic model38. From the strain energy density of this model, the stress from uniaxial tension experiments can be derived as \u03c3 \u03bc \u03bb \u03bb= \u2211 ( \u2212 )p N p \u03b1 \u03b1 =1 \u2212 /2p p , where \u03bb is the axial stretch of the specimen within the gauge. The parameters \u03bcp and \u03b1p (Extended Data Fig.\u00a01c) were obtained from the experimental uniaxial testing data using the parameter-fitting tool for hyperelastic materials in the commercial package Simulia Abaqus FEA (v6.14, Dassault Syst\u00e8mes). We followed a systematic and iterative process in the design of the m-bit. The process begins by stating the core functionalities that the m-bits must possess. Then, each component is designed sequentially through a systematic parametric exploration. Finally, the assembly is simulated and tested, and the results are used to inform the next iteration of the component-wise design",
            " Five different scans are performed, each with a different vertical displacement corresponding to the critical points on the force-displacement curve shown in Fig.\u00a03a. The scans are reconstructed in three dimensions using ImageJ40, and the resulting STL files are post-processed using McNeel Rhinoceros 3D (Supplementary Video\u00a02). Finite-element modelling was used to compute the mechanical response of the m-bit in both the ON and OFF states. Ogden hyperelastic material models were implemented in Abaqus 6.14 using the parameters shown in Extended Data Fig.\u00a01c. Quadratic tetrahedral elements (C3D10H with four corner nodes and six intermediate nodes) were used to mesh the m-bit finite-element model. The element edge length was set at ~0.5\u00a0mm, as determined by a mesh sensitivity analysis. A multi-step geometrically nonlinear static analysis was performed for both the ON and OFF states. In the ON state, gravity was applied to the whole structure in the first step; then, a compressive displacement was applied to the top surface while the bottom surface was pinned",
            " Correspondence and requests for materials should be addressed to M.P. or P.M.R. Peer review information Nature thanks Corentin Coulais and the other, anonymous, reviewer(s) for their contribution to the peer review of this work. Reprints and permissions information is available at http://www.nature.com/reprints. Article 115.0 6.0 25.0 25 a b c VPS8 + NdFeb VPS32 VPS16 \u03bc0 [kg mm-2] 7.94 1.98 0.528 \u03bc1 [kg mm-2] 1137.9 395.06 165.85 \u03b10 25.000 14.372 9.6244 \u03b11 -3.4244 -7.3476 1.7163 10 mm Extended Data Fig. 1 | Characterizations of silicone materials. a, Dimensions of the silicone cast dog-bone specimens. The gauge length is indicated. All dimensions are in millimetres. b, Stress\u2013strain curves obtained from uniaxial testing of VPS32, VPS16 and the MRE (the VPS8-NdFeB composite). The corresponding Ogden hyperelastic model results using the parameters listed in c are overlaid as solid lines. c, Table listing the material parameters obtained from fitting the experimental data to the Ogden hyperelastic model for each of the silicone materials used in the fabrication of an m-bit"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001766_j.cirp.2019.05.004-Figure34-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001766_j.cirp.2019.05.004-Figure34-1.png",
        "caption": "Fig. 34. Artefact geometries proposed for standardization by ASTM F42/ISO TC (a) linear accuracy, (b) circular accuracy, (c) resolution pins, (d) resolution hole resolution slots, (f) resolution ribs, (g) surface texture [188]. Note that the sca different for each geometry.",
        "texts": [
            " To solve the need for standardisation, the ASTM F42/ISO TC 261 Joint Group for standard test artefacts (Joint Group 52) is proposing a new approach to avoid the problem that one single artefact might not address the many different needs of different users. The Joint Group 52 decided that the specification standard under development will describe a suite of test geometries, each testing a different characteristic, and users will be allowed to configure the individual geometries to whatever best fits the specific needs [188]. Fig. 34 shows the proposed artefact geometries which are still being debated and subject to change in future reviews. Examples of artefacts developed to include surface texture specific features are shown in Fig. 20 (see Section 4.2). Besides artefacts for testing the performance of AM systems or XCT measuring systems, in recent years a number of artefacts have been developed for testing measuring systems and establishing traceability in specific dimensional measurements of metal AM products [42,130]. An example of such artefacts, recently proposed by Hermanek et al"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure5.4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure5.4-1.png",
        "caption": "FIGURE 5.4. Local roll-pitch-yaw angles.",
        "texts": [
            " The rotation matrix for a body point P (x, y, z) after rotation Rz,\u03d5 followed by Rx,\u03b8 and Ry,\u03c8 is BRG = Ry,\u03c8Rx,\u03b8Rz,\u03d5 = \u23a1\u23a3 c\u03d5c\u03c8 \u2212 s\u03b8s\u03d5s\u03c8 c\u03c8s\u03d5+ c\u03d5s\u03b8s\u03c8 \u2212c\u03b8s\u03c8 \u2212c\u03b8s\u03d5 c\u03b8c\u03d5 s\u03b8 c\u03d5s\u03c8 + s\u03b8c\u03c8s\u03d5 s\u03d5s\u03c8 \u2212 c\u03d5s\u03b8c\u03c8 c\u03b8c\u03c8 \u23a4\u23a6 . (5.70) Example 155 Local roll-pitch-yaw angles Rotation about the x-axis of the local frame is called roll or bank, rotation about y-axis of the local frame is called pitch or attitude, and rotation about the z-axis of the local frame is called yaw, spin, or heading. The local roll-pitch-yaw angles are shown in Figure 5.4. The local roll-pitch-yaw rotation matrix is BRG = Rz,\u03c8Ry,\u03b8Rx,\u03d5 = \u23a1\u23a3 c\u03b8c\u03c8 c\u03d5s\u03c8 + s\u03b8c\u03c8s\u03d5 s\u03d5s\u03c8 \u2212 c\u03d5s\u03b8c\u03c8 \u2212c\u03b8s\u03c8 c\u03d5c\u03c8 \u2212 s\u03b8s\u03d5s\u03c8 c\u03c8s\u03d5+ c\u03d5s\u03b8s\u03c8 s\u03b8 \u2212c\u03b8s\u03d5 c\u03b8c\u03d5 \u23a4\u23a6 . (5.71) Note the difference between roll-pitch-yaw and Euler angles, although we show both utilizing \u03d5, \u03b8, and \u03c8. 5. Applied Kinematics 231 5.5 F Euler Angles The rotation about the Z-axis of the global coordinate is called precession, the rotation about the x-axis of the local coordinate is called nutation, and the rotation about the z-axis of the local coordinate is called spin"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure14.42-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure14.42-1.png",
        "caption": "FIGURE 14.42. A double A-arm suspension and its equivalent vibrating system.",
        "texts": [
            " Its equivalent vibrating system is shown in Figure 14.41(b). (a) Determine keq and ceq if a = 22 cm b = 45 cm k = 10000N/m c = 1000N s/m \u03b1 = 12deg . (b) Determine the stiffness k such that the natural frequency of the vibrating system is fn = 1Hz, if a = 22 cm b = 45 cm m = 1000/4 kg \u03b1 = 12deg . 928 14. Suspension Optimization (c) Determine the damping c such that the damping ratio of the vibrating system is \u03be = 0.4, if a = 22 cm b = 45 cm m = 1000/4 kg \u03b1 = 12deg fn = 1Hz. 2. Equivalent double A-arm suspension parameters. Figure 14.42(a) illustrates a double A-arm suspension. Its equivalent vibrating system is shown in Figure 14.42(b). (a) Determine keq and ceq if a = 32 cm b = 45 cm k = 8000N/m c = 1000N s/m \u03b1 = 10deg . (b) Determine the stiffness k such that the natural frequency of the 14. Suspension Optimization 929 vibrating system is fn = 1Hz, if a = 32 cm b = 45 cm m = 1000/4 kg \u03b1 = 10deg . (c) Determine the damping c such that the damping ratio of the vibrating system is \u03be = 0.4, if a = 32 cm b = 45 cm m = 1000/4 kg \u03b1 = 10deg fn = 1Hz. 3. Road excitation frequency. A car is moving on a wavy road. What is the wave length d1 if the excitation frequency is fn = 5Hz and (a) v = 30km/h (b) v = 60km/h (c) v = 100km/h"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001966_j.jallcom.2016.02.183-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001966_j.jallcom.2016.02.183-Figure2-1.png",
        "caption": "Fig. 2. Laser scanning strategy (cross-hatching technique).",
        "texts": [
            " The morphology under a scanning electronmicroscope (SEM, PhenomG2), and the size distribution of the powder are showed in Fig. 1. Nearly all of the particles are spherical and most particles have diameter no larger than 45 mm (D90: 31 mm, D50: 19 mm, D10: 10 mm). To obtain an even denser specimen, laser power was set at the highest value of 95 W, layer thickness of 25 mm, track width of 77 mm, and laser scanning speed range of 300 mm/s to 2400 mm/s was performed to build the specimen. Meanwhile, as shown in Fig. 2, linear scanning strategy was used and laser scanning pattern between layers was rotated by 90 (so-called cross-hatching technique) [13]. The density of the specimen was measured on the Sartorius BSA224S. The relation between the density and laser scanning speed (shown in Fig. 3) indicates that the densest specimen is the one of 900 mm/s. Therefore, the optimum laser scanning speed of 900 mm/s was chosen to build the specimen for this study. The b-transus temperature (Tb) of the as-received Tie6Ale4V alloy is 995 \u00b1 5 C, firstly determined via metallographic method"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001568_s11661-014-2722-2-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001568_s11661-014-2722-2-Figure2-1.png",
        "caption": "Fig. 2\u2014The 3 XZ-planes chosen for data collection with the respective x, y, and z locations. (a) The 5 mm 9 10 mm 9 15 mm DLMS part. (b) The 5 mm 9 10 mm 9 20 mm EBM part.",
        "texts": [
            ", near 90 deg to obtain a nearly cubic gauge volume). A gauge volume of 2 9 2 9 2 mm was selected for DLMS, but 3 9 3 9 3 mm was selected for EBM because of the larger grain size to ensure adequate particle statistics. Three xz-planes at y\u20141.4, 2.5, and 3.6\u2014were chosen for data collection on the DLMS unsectioned sample with 20 points in each plane. Only one xz-plane at y = 2.5 was chosen for the EBM sample with 30 points because of the larger gauge volume. These data locations for both the unsectioned DLMS and EBM samples are shown in Figure 2. The quality of METALLURGICAL AND MATERIALS TRANSACTIONS A the counting statistics was confirmed by observing the intensity of the diffracted peak as the sample was translated through the gauge volume. The samples were initially aligned optically to a precision of 0.2 mm and selected locations were confirmed using x, y, and z-axis edge scans of the sample though the beam/gauge volume to avoid \u2018\u2018pseudostrains\u2019\u2019 due to partial gauge volume burial.[20] The strain can be determined with the interplanar spacing, dhkl, as follows: ehkl \u00bc dhkl dhkl0 dhkl0 : \u00bd1 Using Hooke\u2019s law and the measured strain in three orthogonal directions, the residual stresses, rii, in three orthogonal directions (labeled as x, y and z in Figure 1) were calculated as follows: r11 \u00bc E \u00f01\u00fe m\u00de\u00f01 2m\u00de e11 1 m\u00f0 \u00de \u00fe m\u00f0e22 \u00fe e33\u00de\u00bd ; \u00bd2 r22 \u00bc E \u00f01\u00fe m\u00de\u00f01 2m\u00de e22 1 m\u00f0 \u00de \u00fe m\u00f0e11 \u00fe e33\u00de\u00bd ; \u00bd3 r33 \u00bc E \u00f01\u00fe m\u00de\u00f01 2m\u00de e33 1 m\u00f0 \u00de \u00fe m\u00f0e11 \u00fe e22\u00de\u00bd ; \u00bd4 where E and m are the Young\u2019s modulus and Poisson\u2019s ratio (m) were taken to be 200 MPa and 0.29 deg, respectively. e11, e22, and e33 are the orthogonal strain vectors determined using Eq. [1]. Based on the build geometry and knowledge of the thermal history, the principal directions of strain were assumed to be along the x, y, and z axes in Figure 2. The shear strains were not measured and were assumed to be negligible. As discussed by Hutchings et al. 2005,[8] an alternate approach to determine d0 involves balancing the force METALLURGICAL AND MATERIALS TRANSACTIONS A and moment across one or more cross-sections of the sample, as required from mechanical equilibrium. In the present case, an equilibrium approach was used wherein the individual force components are related to the stress by the area perpendicular to the component given by Eq",
            " In our case, we disregard the area because the area for which the forces are acting upon is a constant (i.e., the gauge volume). Therefore, we will be balancing the residual stresses. The sum of all residual stresses in each orthogonal direction must be equal to zero (Eq. [6]). Similarly, the sum of the moments of inertia in a give plane must be equal to 0 (Eq. [7]). rij \u00bc Fij=A, \u00bd5 X r11 \u00bc 0; X r22 \u00bc 0; X r33 \u00bc 0; \u00bd6 X r11L3 \u00bc 0; X Rr22L3 \u00bc 0; X Rr33L1 \u00bc 0; \u00bd7 where L3 represents the distance from the origin with z equal to zero in Figure 2 and L1 represents the distance from the origin with x = 0. All three planes (XZ, YZ, XY) were used for moment balance, whereas for only the XZ-planes were selected for force balance in both the EBM part and DLMS part because these planes included the most extensive coverage of experimental data. Wolfram Mathematica Version 9 was used to determine the solution for Eq. [5] using a built-in function (\u2018\u2018Err (x)\u2019\u2019) of Wolfram Mathematica v.9 to find the accurate d0 interplanar spacings for the x, y and z directions"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure9.14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure9.14-1.png",
        "caption": "Figure 9.14 closed position. (b) Interior view of the left-hand lock door, as found in older wooden mitre gates, with a diagonal strutt (compression diagonal) and a steel tension bar (tension diagonal).",
        "texts": [
            " 9 Trusses 327 In steel trusses, falling diagonals (tension diagonals) are used most frequently, as (usually slender) steel members subject to compression run the risk of buckling. Preferably apply them as tension members. In contrast, rising diagonals (compression diagonals) are most often used in wooden trusses, as in general a wooden joint is more suitable to transfer compressive forces rather than tensile forces. An example close to home is the simple garden gate in Figure 9.13, with (a) a steel version (falling diagonal) and (b) a wooden version (rising diagonal). The wooden lock-gate in Figure 9.14 is another example. The wooden diagonal strut is a rising diagonal and acts as a compression member under influence of the dead weight of the gate. The wooden planking is facing the same way as the diagonal strut. The steel falling diagonal is a tension bar. The joints in a truss are numbered or lettered (see Figure 9.15). The numbers or letters used to indicate the joints can be used as an index. For example, x4; y4 gives the x and y coordinates of joint 4, and Fx;C is the x component of force F on joint C"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002604_tie.2019.2902834-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002604_tie.2019.2902834-Figure1-1.png",
        "caption": "Fig. 1. A quadrotor with a suspended payload",
        "texts": [
            " Section II presents the dynamic model of the system of a quadrotor with a suspended payload. In Section III, an energy based adaptive control strategy is developed. And in Section IV, the equilibrium of quadrotor suspended-payload system is proven to be asymptotically stable via a Lyapunov-based analysis. In Section V the real-time flight experimental results are provided. Finally, Some conclusion remarks are given in Section VI. The schematic of a composite system which contains a quadrotor and a cable-connected payload is shown in Fig. 1. The mass of the payload, and the quadrotor UAV are denoted by mL and mQ respectively. The length of the cable is denoted by L, and the acceleration of gravity is denoted by g. Let I = {xI , yI , zI} represent a right hand inertial reference frame with zI being the vertical direction towards the ground, and the body fixed frame denoted by B = {xB , yB , zB} , is located at the center of gravity of the quadrotor. The rotation matrix from B to I is represented by R(t) \u2208 R3\u00d73 as follows [3] R = c\u03c6c\u03b8 s\u03c6s\u03b8c\u03c8 \u2212 s\u03c8c\u03c6 c\u03c6s\u03b8c\u03c8 + s\u03c6s\u03c8 s\u03c6c\u03b8 s\u03c6s\u03b8s\u03c8 + c\u03c6c\u03c8 c\u03c6s\u03b8s\u03c8 \u2212 c\u03c8s\u03c6 \u2212s\u03b8 c\u03b8s\u03c6 c\u03b8c\u03c6  (1) where s\u2022 and c\u2022 represent sin(\u00b7) and cos(\u00b7) respectively, \u03c8(t), \u03b8(t), and \u03c6(t) are the roll, pitch, and yaw angles of the quadrotor with respect to the inertia frame I, respectively",
            " Finally, real-time experimental results illustrate the good performance of the proposed control design by comparison with a PID controller and a terminal sliding mode controller in both stabilization control and regulation control. The future work will focus on conducting outdoor experiments, and developing a dynamic planning methodology based trajectory generation design to reduce the maximum swing angle of the payload during the transportation procedure. To obtain the dynamic model in (7)-(12), the kinetic energy K(t) and potential energy P (t) of the quadrotor payload composite system illustrated in Fig. 1 are defined as K (t) = 1 2 mQP\u0307 T Q (t) P\u0307Q (t) + 1 2 mLP\u0307 T L (t) P\u0307L (t) , (64) P (t) =\u2212mQP T Q (t)G\u2212mLP T L (t)G. (65) The Lagrangian of the composite system is given by \u039b (t) = K (t)\u2212 P (t) . (66) Then, the partial derivatives of \u039b (t) with respect to PQ (t) , P\u0307Q (t) , \u03b3 (t) , and \u03b3\u0307 (t) can be obtained as follows: \u2202\u039b (t) \u2202PQ (t) = (mQ +mL)G, (67) \u2202\u039b (t) \u2202P\u0307Q (t) = (mQ +mL) P\u0307Q (t) +mLLq\u0307 (t) , (68) \u2202\u039b (t) \u2202\u03b3 (t) = +mLL \u2202q\u0307 (t) \u2202\u03b3T (t) T ( P\u0307Q (t) + Lq\u0307 (t) ) +mLL \u2202q (t) \u2202\u03b3T (t) T G, (69) \u2202\u039b (t) \u2202\u03b3\u0307 (t) =mLL \u2202q\u0307 (t) \u2202\u03b3\u0307T (t) T ( P\u0307Q (t) + Lq\u0307 (t) ) ",
            " (74) By substituting q\u0307, q\u0308, \u2202q\u0307(t) \u2202\u03b3\u0307T (t) , \u2202q(t) \u2202\u03b3T (t) , \u2202q\u0307(t) \u2202\u03b3T (t) , and d dt ( \u2202q\u0307(t) \u2202\u03b3\u0307T (t) ) into (73) and (74), the Lagrangian equation can be rewritten as follows: F (t) = (mQ +mL) P\u0308Q +mLL  cxcy \u2212sxsy 0 cy \u2212sxcy cxsy  \u03b3\u0308 \u2212mLL [ S1 S2 ] \u2212 (mQ +mL)G, (75) 0 =mLL  cxcy \u2212sxsy 0 cy \u2212cysx cxsy T P\u0308Q +mLLg [ sxcy cxsy ] +mLL 2 [ cos2 \u03b3y 0 0 1 ] \u03b3\u0308 \u2212mLL 2cysy [ \u03b3\u0307y \u03b3\u0307x \u2212\u03b3\u0307x 0 ] . (76) where S1 =  \u03b3\u0307xsxcy + \u03b3\u0307ycxsy 0 \u03b3\u0307xcxcy \u2212 \u03b3\u0307ysxsy  , (77) S2 =  \u03b3\u0307xcxsy + \u03b3\u0307ysxcy \u03b3\u0307ysy \u03b3\u0307xsxsy + \u03b3\u0307ycxcy  . (78) Thus, the dynamic model for the composite system illustrated in Fig. 1 can be obtained as M (p) p\u0308+ C (p, p\u0307) p\u0307+ Vm (p) = \u03c4 (t) . (79) with M(p), C(p, p\u0307), and Vm(p) being defined in (8)-(12). [1] X. Zhang, B. Xian, B. Zhao, and Y. Zhang, \u201cAutonomous flight control of a nano quadrotor helicopter in a gps-denied environment using onboard vision,\u201d IEEE Transactions on Industrial Electronics, vol. 62, no. 10, pp. 6392\u20136403, 2015. 0278-0046 (c) 2018 IEEE. Personal use is permitted, but republication/redistribution requires IEEE permission. See http://www.ieee.org/publications_standards/publications/rights/index"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000366_tcst.2004.825052-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000366_tcst.2004.825052-Figure2-1.png",
        "caption": "Fig. 2. The PVTOL aircraft.",
        "texts": [
            " This should be done while keeping the total thrust constant. In view of its configuration, the four-rotor rotorcraft in Fig. 1 has some similarities with planar vertical take off and landing (PVTOL) aircraft problem [2], [6], [7]. The PVTOL is a mathematical model of a flying object that evolves in a vertical plane. It has three degrees of freedom ( , , and ) corresponding to its position and orientation in the plane. The PVTOL is composed of two independent thrusters that produce a force and a moment on the flying machine, see Fig. 2. The PVTOL is an underactuated system since it has three degrees of freedom and only two inputs. The PVTOL is a very interesting nonlinear control problem. Furthermore, the four-rotor rotorcraft reduces to a PVTOL when the roll and yaw angles are set to zero. In a way, the four-rotor rotorcraft can be seen as two PVTOL connected such that their axes are orthogonal. In this paper, we present the model of a four-rotor rotorcraft whose dynamical model is obtained via a Lagrange approach [5]. A control strategy is proposed having in mind that the four-rotor rotorcraft can be seen as the interconnection of two PVTOL aircraft"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure12.12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure12.12-1.png",
        "caption": "Figure 12.12 (a) The shear forces directly to the left and right of joint B are in equilibrium with the point load of 40 kN on joint B. (b) The bending moments directly to the left and right of joint C are in equilibrium with the load due to the concentrated couple of 80 kNm on joint C.",
        "texts": [
            " (d) Bending moment diagram. A bend occurs at the point of application of the concentrated force, and a step change occurs where the couple is applied. 12 Bending Moment, Shear Force and Normal Force Diagrams 485 7). We cannot however see from the V diagram that the M diagram also has extreme values directly to the left and to the right of C, the point where the couple is applied. The maximum bending moment in an absolute sense occurs in the section directly to the right of C and is 60 kNm. b. In Figure 12.12a, joint B has been isolated. In the sections, only the shear forces are shown. The shear forces directly to the left and right of B have to be in equilibrium with the vertical force on the joint. From this, it follows that the step change in the V diagram must be equal to the force on the joint. In Figure 12.12b, joint C has been isolated. In the sections, only the bending moments are shown. From the moment equilibrium of joint C it follows that the bending moments directly to the left and to the right of C have to be in equilibrium with the couple at C. The magnitude of the step change in the M diagram must therefore be equal to the magnitude of the couple. c. Rule 11, states that the change M of the bending moment M is equal to the area of the V diagram. This rule is still valid as long as no couples are acting in the field that is considered"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003540_s0006-3495(93)81129-9-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003540_s0006-3495(93)81129-9-Figure1-1.png",
        "caption": "FIGURE 1 Definition of variables associated with the unbounded and half-space kernels.",
        "texts": [
            " These may be expressed as boundary integral equations in terms of the velocity field by the use of Betti's Reciprocal Work Theorem (see Banerjee and Butterfield (42), Brebbia et al. (43), or Kim and Karrila (41)): Cij (X)uj(X)= fD u*(x, X) t (X) dS(X) aD - ft* (x,)uj(X)dS(X), (2) where AD is the Liapunov smooth boundary of the solution domain D, X E AD, uj(X) is the j component of the velocity at X, tj(X) is the j component of the boundary traction at X, u* (x, X) is the i component of velocity field at x due to a Stokeslet in the j direction at X, and t*} (x, X) is its associated traction (see Fig. 1). Here, cij(x) depends upon the location of x and the local geometry; it is given by 128ij X E aD smooth at x cii =J i xEED 0 x(4D Expressions for u * (x, X) and t*} (x, X) depend on whether the flow domain D is an unbounded fluid or a half-space due to the presence of a zero-displacement boundary and are discussed in Appendix 1. For a boundary discretized intoM elements over which the boundary solution is approximated by a piecewise continuous polynomial (for example, the velocity over an element may be given by uj = Q()un)), Eq",
            " This tends to prevent fluid drift as well as provide a good depth of fluid (a fluid film thickness of up to two orders of magnitude greater than the largest dimension of the organism is attainable), thereby greatly reducing the likelihood of significant hydrodynamic interactions. APPENDIX 1: UNBOUNDED AND HALF-SPACE KERNELS For an unbounded fluid, Brebbia et al. (43) give the Kelvin solution for the fundamental velocity and traction kernels uK and tK', respectively, as K 1( r + rirj and tK = 3 rijmm (13) where r = x - X (see Fig. 1), x is the position vector of the field point, and X is the point where the Stokeslet is situated. Here, nj is a unit vector pointing out of the flow domain, and 8ij is the usual Kronecker delta. To account for a half-space, the complementary terms uCj and tc as well as the extra terms uE and tE are included such that (see Phan-Thien (65)) U = -uK_ + UE1 and ty = tK tC + tE1 (14) where K (8 1~1\\ K R.R.R n~(~Uii8S(k+R3) and tK= _3 (15) U 4 R3 { 8j3R + 8i3Rj-28j38j3R3 [ R. 1]~R +x3 [2883 -8ij + 3- (Rj -2823R) (16) E - 3 I 38Ri 27rRs5 38i3RRmnm -3R3ni(25j3R3 - Rj) - 3x3{ 8jiRmnm + Rinj - 28j3(8i3Rmnn + Rjn3) + (5R12 - (283R3 - Rj)1} 283(8i3Rmfn + Rin3) + (5 RiRmm - ni )(2j2R3- Rj)]} (17) and R = x - X* (see Fig. 1). The complementary terms are identical to those of the Kelvin solution, except that the Stokeslet is taken at the point X* (rather than X), which is the reflection of X about the half-space. APPENDIX 2: DISCRETIZATION SCHEMES The primary nodes, needed to define the vertices of the triangular elements, on the sphere of radius a and center (0, 0, 0), are generated by beginning with the vertices of the relevant octahedron: (\u00b1a,0,0), (0, \u00b1a,O), (00,O\u00b1a). By extending the midpoint of each edge to the surface of the sphere, the remaining necessary 12 vertices are found to be ( a a \u00b10,1\u00b1 a a-+- , 09S\u00b0 -+^0 / a a X The extra midside nodes are found in much the same manner"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003549_robot.1996.509185-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003549_robot.1996.509185-Figure5-1.png",
        "caption": "Fig. 5: Use of equivalent force couple to replace moment at c.g.",
        "texts": [
            " For a given tipoeer axis d, we are only concerned with those components off , and n, which act about the tipover axis, so we let and Angular Loads Since the IForce-Angle staioiliity measure is based on the computation of the angle between the net force vector and each of the tipover axis normals, it is necessary to replace the moment ni with an equivalent force couple f,, for each tipover axis. The equivalent force couple must, necessarily lie in the plane normal to the moment ni. The most judicious choice of the infinite possible force iiouple locations and directions in this plane, is that pair of minimum magnitude where one member of the couple passes through the centerof-mass and the other through the h e of the tipover axis. As shown in Fig. 5, the member of the force couple acting on the center-of-mass is then given by 31 13 where ith tipover axis is thus = 1/ 11111. The new net force vector f;' for the li x ni Illill fi* = fi + - Force-Angle Stability Measure Letting f * = f*/I/f*II, the candidate angles for the Force-Angle stability measure are then simply given by where -T 5 Qi 5 T ' The sign of Bi is determined by ri as follows where i = (1, . . . ! n} . Note that the appropriate sign of the angle measure associated with each tipover axis is determined by establishing whether or not the net force vector lies inside the support pattern"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002281_j.mechmachtheory.2020.103838-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002281_j.mechmachtheory.2020.103838-Figure5-1.png",
        "caption": "Fig. 5. Gear tooth deformation due to the coupling effect of gear foundation structure.",
        "texts": [
            " [40] are used to calculate the gear body structure coupling effect. The gear body structure coupling deformation 1/ K f ij is defined as the displacement of the intersection point of the i th tooth profile and the line of action when a unit force is applied to the intersection point of the j th tooth profile and the line of action. The direction for both the displacement and the unit force is along the line of action. For example, when the gear pair is in the double-tooth mesh region shown in Fig. 5 , the 1/ K f12 denotes the displacement of the point where F 1 ( F 1 = 0) is applied to due to the gear foundation deflection when F 2 = 1, and the 1/ K f21 denotes the displacement of the point where F 2 ( F 2 = 0) is applied to due to the gear foundation deflection when F 1 = 1. They can be calculated by [40] , 1 K f21 = 1 W E cos \u03b11 cos \u03b12 [ L 1 ( u 1 u 2 S 2 )2 + ( tan \u03b12 M 1 + P 1 ) u 1 S + ( tan \u03b11 Q 1 + R 1 ) u 2 S + ( tan \u03b11 S 1 + T 1 ) tan \u03b12 + U 1 tan \u03b11 + V 1 ] (8) 1 K f12 = 1 W E cos \u03b11 cos \u03b12 [ L 2 ( u 1 u 2 S 2 )2 + ( tan \u03b11 M 2 + P 2 ) u 2 S + ( tan \u03b12 Q 2 + R 2 ) u 1 S + ( tan \u03b12 S 2 + T 2 ) tan \u03b11 + U 2 tan \u03b12 + V 2 ] (9) where, the symbols u i and \u03b1i ( i = 1,2 denoting the tooth number) has the same meaning as the symbol u and \u03b11 in Eq",
            " The intentional tooth profile modification is an effective approach that has been widely used for vibration and noise reduction of gear transmissions. However, it will inevitably complicate the dynamic interactions between the teeth and the tooth pairs in mesh. The gear mesh results, namely the mesh stiffness, loaded static transmission error, and load sharing ratio of the spur gear pair with different tooth tip relief are calculated by using the proposed analytical method in Section 2 and shown in Figs. 8 and 9 . The definitions of the tooth tip relief are the same as that in [29] and shown in Fig. 5 , where the normalized relief parameters are calculated as: C n = C a /0.02 m, L n = L a /0.6 m. Here, the symbol m represents the module of the gear. It can be seen that the double-tooth region will reduce and the slope of the transition region between single- and double-tooth mesh will become greater as the amplitude increase of the tooth tip relief with a fixed length of tooth profile modification. The loaded static transmission error curves have the similar variation trend but with an inverse shape compared with the mesh stiffness curves"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure3.17-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure3.17-1.png",
        "caption": "Fig. 3.17 Definition of Variables for ZMP for Both Feet Contact",
        "texts": [],
        "surrounding_texts": [
            "From the previous discussion, we can express the ground reaction force acting upon a humanoid robot by using the ZMP, the linear force, and the moment about a vertical line passing the ZMP. In this section, we discuss the relationship between the ground reaction force and the robot\u2019s motion. After showing basic equations, we explain the principle of it. Lastly, we show some calculation algorithms."
        ]
    },
    {
        "image_filename": "designv10_0_0001446_j.oceaneng.2016.06.041-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001446_j.oceaneng.2016.06.041-Figure3-1.png",
        "caption": "Fig. 3. Rudder saturation.",
        "texts": [
            " Consider the following Lyapunov function candidate: \u03c1 \u03bc \u03c9 = + \u02dc \u02dc + \u02dc + \u02dc + \u02dc + \u02dc + \u02dc + \u02dc ( ) \u22a4 V s k h h k b b k k g k k D 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 31y 2 01 02 1 1 2 03 2 04 2 05 2 06 2 2 Its time derivative is \u03c1\u03c1 \u03bc\u03bc \u03c9 \u03c9 \u03c1\u03c1 \u03bc\u03bc \u03c9 \u03c9 \u03c9 \u0307 = \u0307 + \u02dc \u02dc \u0307 + \u02dc \u02dc \u0307 + \u02dc \u02dc \u0307 + \u02dc \u02dc \u0307 + \u02dc \u02dc \u0307 + \u02dc \u02dc \u0307 + \u02dc \u02dc \u0307 = \u0307 + \u02dc ^ \u0307 + \u02dc ^ \u0307 + \u02dc ^ \u0307 + \u02dc ^ \u0307 + \u02dc ^ \u0307 + \u02dc ^ \u0307 + \u02dc ( ^ \u0307 \u2212 \u0307 ) ( ) \u22a4 \u22a4 V ss k h h k b b b k k gg k k DD ss k h h k b b b k k gg k k DD 1 1 1 1 1 1 1 1 1 1 1 1 32 y y y y y 01 02 1 1 1 03 04 05 06 01 02 1 1 1 03 04 05 06 Substituting (29) and (30) into (32), we have \u03c4 \u03c1 \u03c3 \u03c9 \u03a6 \u03c1 \u03c9 \u03bc\u03c9 \u03c9 \u03c9 \u03c9 \u03a6 \u03bc \u03c9 \u03c4 \u03c9 \u03c4 \u0307 = \u2212 \u2212 | | \u2212 ^| | \u2212 ^ \u2212 \u02dc + \u2212 \u02dc \u02dc + \u02dc| | + \u02dc \u2212 \u02dc \u02dc \u2212 \u02dc \u02dc \u2212 \u02dc( ^ \u2212 ) + \u02dc \u02dc + \u02dc + \u02dc + \u02dc \u2212 \u02dc \u2264 \u2212 \u2212 | | \u2212 \u02dc \u2264 \u2212 \u2212 | | ( ) \u22a4 \u22a4\u23a1 \u23a3\u23a2 \u23a4 \u23a6\u23a5 V s k s s gs Ds c s h s gs g u D h u g D k s k s k s k s 33 y y y y y y y y 2 3 07 2 07 2 2 Thus, \u0307 <V 0 for all \u2260s 0. Based on Barbalat's lemma (Slotine and Li, 1991), \u2192s 0 as \u2192 \u221et . This implies that the system achieves asymptotic stability and that the error converges to zero asymptotically. This completes the proof.\u25a1 4.4. Control design with input saturation Now, we consider the case with actuator saturation. When the computed control input reaches its bound, we will design an antiwindup compensator to reduce the influence of saturation. The control input with actuator saturation is shown in Fig. 3. We can describe the saturation as \u03b4 = ( ) ( )usat 34r where u is the designed control input, \u03b4r is the actual control input generated by the actuator, and (\u00b7)sat denotes the nonlinear saturation characteristic, which can be written as \u03b4 \u03b4 \u03b4 \u03b4 \u03b4 \u03b4 \u03b4 = ( ) = > \u2212 \u2264 \u2264 \u2212 < \u2212 ( ) \u23a7 \u23a8\u23aa \u23a9\u23aa u u u u u sat , , , 35 r rm rm rm rm rm rm where \u03b4rm is the upper bound of the actuator. To solve the problem of attitude control subjected to actuator saturation, we employ an anti-windup compensator as (Chen et al"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000684_s0022-0728(80)80003-9-Figure12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000684_s0022-0728(80)80003-9-Figure12-1.png",
        "caption": "Fig. 12. V o l t a m m o g r a m a n d b a c k g r o u n d c u r r e n t fo r a t h i n l aye r o n a c a r b o n e l e c t r o d e , c = 10 -6 m o l 1-1 . Supe r f i c i a l c o n c e n t r a t i o n 2.0 \u00d7 10 - 1 \u00b0 m o l c m -2 ; p H 6 .6 ; v = 0.1 V s -1 .",
        "texts": [
            " On a carbon electrode, the determination of the exact shape of the peak is complicated by the fact that after the peak the bacJ~ground current is a little higher than the current observed in the absence of reactant; as we have ascertained by examining the curves for a non-adsorbable substance at the same 20 Fig. 13. I n f l u e n c e o f t h e s can ra te o n t h e c a t h o d i c p e a k fo r m u l t i l a y e r a d s o r p t i o n at pH 6 .6 ; v (V s -1 ): (a) 0 .01 , (b ) 0.1, (c) 1. T h e b a c k g r o u n d c u r r e n t was s u b t r a c t e d f r o m t h e curves . concentrat ion (1 methyl 1,5-naphthyridinium iodide), this is in part due to a contribution of convection or diffusion. The voltammetric peaks do not have quite the theoretical width (Fig. 12), probably because of the influence of the heterogeneities of the electrode [ 5]. Their shape is discussed in section (V.2). However, by applying the theory for a monolayer [4], we determined values of 21 ks ex, which roughly indicate that the reaction becomes much slower on carbon than on mercury, but that its rate still depends on the pH (Fig. l l b ) . (V.2) Multilayer adsorption A few examples of voltammograms are shown in Figs. 13 and 14. The influence of the scan rate can be seen in Fig. 13",
            " For the peak potentials, the variations for a multilayer parallel those for a thin layer. The variations of the peak currents (Fig. 16) conform to the predictions of the theory. The increased width at large values of v (Fig. 17) reflects the appearance of the tailing. At the highest sweep rates, where the current for the multitayers can be large, an influence of the ohmic drop on the data of Figs. 15--17 cannot be excluded. 22 In contrast with that observed on mercury , some tailing is obtained, even for the thin layers on carbon (Fig. 12) for which the surface concentrat ions correspond to about one layer, if it is assumed that the actual area is equal to the geometrical area. This could be due to a heterogeneous deposit, in which certain parts of the surface would be covered with more than one layer of molecules. The case of a catalysis will be examined in Part III of this series. A C K N O W L E D G E M E N T We thank Mme M.T. Compain for her technical assistance. 23 1 E Lav~xon, J . E l e e t r o a n a l . C h e m , , 1 1 2 ( 1 9 8 0 ) 1 ( th l s Issue) "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000683_978-3-642-82997-0-Figure3.24-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000683_978-3-642-82997-0-Figure3.24-1.png",
        "caption": "Fig. 3.24. Manipulator UMS-2",
        "texts": [
            " By including the dynamics of actuators which power particular mechanism joints we can establish the complete dynamic robot model which enables us to calculate its exact dynamics in a perturbed working regime. This perturbed regime comes from mechanical vibrations which can be present as noticed in Sect. 3.3.1, by \"white noise\". Using the complete dynamic model for a real manipulation robot, the manipulator deviation from nominal trajectory can be calculated, the results of which are presented in the next example. Example We observe the motion of the manipulator UMS-2 which is shown in Fig. 3.24. The geometric manipulator parameters are presented in Table 3.1, the dynamic parameters in Table 3.2, and the actuator parameters in Table 3.3. 88 3 Dynamics and Dynamic Analysis of Manipulation Robots 3.3 Dynamics of Manipulation Robots in Conditions of Mechanical Vibrations Impact 89 The manipulator motion is observed in internal coordinates whose initial and final positions are presented in Table 3.4. The motion between the initial and final positions is performed over a straight-line segment with an adopted parabolic velocity profile"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure4.22-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure4.22-1.png",
        "caption": "Figure 4.22 A steel hinged support as still found in many bridges today.",
        "texts": [
            "19 with the coordinate system shown, the following applies for motion at A: ux;A = 0 (prescribed motion), uy;A = 0 (prescribed motion), \u03d5z;A = unknown (free motion). and for the forces at A: Fx;A = unknown (free force), Fy;A = unknown (free force), Tz;A = 0 (prescribed force). A hinged support therefore has one degree of freedom (a rotation) and generates two support reactions (the two components of a force). 4 Structures 125 Figure 4.20 is a good example of a hinged support. Figure 4.21 shows how a steel hinged support can be used in small bridges. Horizontal motion is prevented by a pin. The steel hinged support in Figure 4.22 can transfer large forces and is an example of what is used in larger bridges. Like roller supports, hinged supports can be made from materials other than steel, or from a combination of materials Although the supports in Figures 4.21 and 4.22 can transfer only compressive forces, it is assumed below that hinged supports can also transfer tensile forces. A fixed support is an infinitely stiff or rigid joint between a body and its environment, see also Section 4.2.2. Figure 4.23a is a model of a fixed support (the dotted line is generally omitted)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002770_tii.2021.3089340-Figure4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002770_tii.2021.3089340-Figure4-1.png",
        "caption": "Fig. 4. The mass spring damping system for a rolling bearing.",
        "texts": [
            " (1) and (4), we can obtain the time-varying displacement excitations 1( )H t and 2 ( )H t . Then we can calculate the total contact deformation of the ith rolling element at any angle by [30]  (t)=x cos sini i iy H t       (8) where  is the radial clearance of rolling bearing.  H t is taken as 1( )H t in the stage of defect initiation, while it is taken as 2 ( )H t in other two stages. Next, in order to obtain the life-cycle dynamic response of rolling bearing, we can simplify the bearing as a mass spring damping system, as shown in the Fig. 4. By substituting Eq. (8) into the two-degree of freedom dynamic model of rolling bearing [18], the life-cycle bearing dynamic equations are derived as: 2 3/2 2 1 2 3/ 2 2 1 ( ) cos ( ) sin                    n i i i x i n i i i y i d x dx m c K t W dtdt d y dy m c K t W dtdt (9) where m is the total mass of the inner race and the shaft; c is the damping coefficient; x and y respectively represent the bearing vibration displacement along the X and Y directions; xW and yW denote the radial force acting on the shaft in X and Y di- rections; K is the total contact stiffness that can be calculated by Harris's method [35]; and i represents the valid contact area parameter of the ith rolling element that is written as: 0 0 1 0        i i i (10) To sum up, the complete steps for constructing the life-cycle dynamic model of rolling bearing are listed in Table II"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure13.23-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure13.23-1.png",
        "caption": "FIGURE 13.23. Two similar double pendulums, expressed by absolute and relative coordinates.",
        "texts": [
            " Vehicle Vibrations 877 n number of DOF p momentum qi, Qi generalized force r = \u03c9 \u03c9n frequency ratio r,R radius S quadrature t time T period uij jth element of the ith mode shape u mode shape, eigenvector ui ith eigenvector v,v, x\u0307, x\u0307 velocity V potential energy w track x absolute displacement X steady-state amplitude of x y base excitation displacement Y steady-state amplitude of y z relative displacement Z steady-state amplitude of z Zi short notation parameter \u03b4 deflection \u03be = c 2 \u221a km damping ratio \u03bb eigenvalue \u03bbi ith eigenvalue \u03c9 = 2\u03c0f angular frequency [ rad/ s] \u03c9n natural frequency \u03c9i ith natural frequency Subscript d driver f front r rear s sprung u unsprung 878 13. Vehicle Vibrations Exercises 1. Equation of motion of a multiple DOF system. Figure 13.22 illustrates a two DOF vibrating system. (a) Determine K, V , and D functions. (b) Determine the equations of motion using the Lagrange method. (c) F Rewrite K, V , and D in quadrature form. (d) Determine the natural frequencies and mode shapes of the system. 2. Absolute and relative coordinates. Figure 13.23 illustrates two similar double pendulums. We express the motion of the left one using absolute coordinates \u03b81 and \u03b82, and express the motion of the right one with absolute coordinate \u03b81 and relative coordinate \u03b82. (a) Determine the equation of motion of the absolute coordinate double pendulum. (b) Determine the equation of motion of the relative coordinate double pendulum. (c) Compare their mass and stiffness matrices. 13. Vehicle Vibrations 879 3. One-eight car model. Consider a one-eight car model as a base excited one DOF system"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure5.21-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure5.21-1.png",
        "caption": "Fig. 5.21 Definition of Falling Angle \u03b8",
        "texts": [
            " First we divided the fall itself into five states and the robots motion was controlled differently in each state. 1. SQUATTING State This is the initial state of falling where the robot\u2019s center of mass deviates too far from the supporting polygon. All control is stopped and the fall control takes over. It starts to bend the robot\u2019s knees and also starts to bend the neck, waist and arms so that the robot falls on its hips. 2. EXTEND1 State The fall proceeds. When the angle made by the ground and the line that connects the point the robot\u2019s heels touch the ground and the hip landing point \u03b8(Fig. 5.21) goes below a certain value, the robot starts to extend it\u2019s knees to decrease the velocity at which the hips hit the ground. This also makes sure that the robot hits the ground with it\u2019s hips. 3. TOUCHDOWN State The fall proceeds to the next stage. When angle \u03b8 goes below another value, the robot prepares for landing and stops servo control of the joints. 4. EXTEND2 State When a certain amount of time elapses after the fall, the robot extends it\u2019s legs to prevent rolling on it\u2019s head due to the momentum of the fall"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure8.5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure8.5-1.png",
        "caption": "Fig. 8.5. Leakage flow in an AFPM machine (not to scale).",
        "texts": [
            " For approximately radial blades, the Stanitz slip factor ks (80o < \u03b22 < 90o) is ks = 1\u2212 0.63\u03c0/nb (8.23) where \u03b22 is the blade angle at exit and nb is the number of the blades. When applying a slip factor, the pressure relation (8.22) becomes \u2206p = \u03c1\u21262(ksR 2 out \u2212R2 in) + \u03c1 2 ( 1 A2 1 \u2212 1 A2 2 )Q2 (8.24) 8.3 Cooling of AFPM Machines 261 Shock, leakage and friction Energy losses due to friction, separation of the boundary layer (shock loss) and leakage should also be considered in the flow analysis. As illustrated in Fig. 8.5, if the total volumetric flow rate through the PM channel is Qt, the pressure difference between the PM exit and the entrance will cause a leakage or recirculation of a volume of fluid Ql, thus reducing the flow rate at outlet to Q = Qt\u2212Ql. The Ql is a function of mass flow rate and discharge and leakage path resistances. The leakage flow reaches its maximum when the main outlet flow is shut. These losses can be accounted for by introducing a pressure loss term \u2206pl in eqn (8.24) as follows [232] \u2206p = \u03c1\u21262(ksR 2 out \u2212R2 in) + \u03c1 2 ( 1 A2 1 \u2212 1 A2 2 )Q2 \u2212\u2206pl (8"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure14.49-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure14.49-1.png",
        "caption": "Figure 14.49 (a) Support reactions and line of action figure. (b) Bending moment diagram. (c) Normal force diagram.",
        "texts": [
            " For N \u21920 it always holds that ez \u2192\u221e, and the centre of force is at infinity. \u2022 If the bending moment is zero, ez = 0 applies, and the centre of force is on the member axis. This means that in structures with hinges, the line of force always passes through the hinges, as the bending moment is zero there. \u2022 Where the line of force intersects the member axis (or coincides with it), ez = 0 and the bending moment is zero. Example 1 Figure 14.48 shows a three-hinged portal frame, loaded by a vertical force F = 50 kN at E. Figure 14.49a shows the support reactions and Figures 14.49b and 14.49c shows the M and N diagrams. The calculation is left to the reader. In Figure 14.49a, the lines of action of force F and the support reactions at A and B are shown. These lines of action intersect in one point. This offers a graphical check of the moment equilibrium of the three-hinged frame (see Section 5.3, Example 1). Questions: a. Determine the centre of force in cross-section G of post AD, 1.50 m above support A. b. Determine the line of force for the parts AD, THE, DSC and BC. 676 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM Solution: a. Figure 14.50a shows the bending moment and the normal force at crosssection (cs) G",
            " The centre of force is indeed to the left of the member axis. b. In Figure 14.51, the part AG has been isolated. The support reactions act at A. If there is a equilibrium, a force has to act at cross-section G that has the same magnitude as the resulting force at A, and has the same line of action, but has to act in the opposite direction. In other words, the centre of 14 Cables, Lines of Force and Structural Shapes 677 force at cross-section G is located on the line of action (a) of the resulting force at A (see also Figure 14.49a). This leads to the following statement: the centre of force is the intersection of the cross-section with the line of action of the resultant of all forces that the cross-section has to transfer. Since all the cross-sections in AD have to transfer the same resulting force at A, the centres of force in those cross-sections are on the same line of action (a). The line of force for AD therefore coincides with line of action (a) (see Figure 14.52a). Since the normal force in AC is a compressive force, the line of force is a line of pressure",
            " Note: If the normal force in a beam is constant, the figure that is enclosed between the line of force and the member axis has the same shape as the M diagram. This is not surprising, for1 M = Ne. 1 Without taking into account the coordinate system and signs. 678 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM The scale factor is N . If N is a tensile force, the line of force is on the same side of the member axis as the M diagram. If N is a compressive force, the line of force and the M diagram are not on the same side. It is up to the reader to verify this, using the bending moment diagram in Figure 14.49b and the lines of force in Figure 14.52. Example 2 The structure ABCD in Figure 14.53a is fixed at A, and loaded by a horizontal force F at C and a vertical force F at D. Question: Determine the lines of force for AB, BC and BD. Solution: The lines of force are shown in Figure 14.53b. All cross-sections between C and B have to transfer the horizontal force F . The line of force for BC therefore coincides with the line of action of this horizontal force. All cross-sections between D and B have to transfer the vertical force F "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.52-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.52-1.png",
        "caption": "FIGURE 7.52. A car pulling a one-axle trailer.",
        "texts": [
            " Steering Dynamics compare with the optimal multi-link mechanism. The optimality of x = \u22120.824m may be more clear in Figure 7.49 that shows the difference \u2206 = \u03b42 \u2212 \u03b4Ac for different values of x. The optimal multi-link steering mechanism along with the length of its links is shown in Figure 7.50. The mechanism and the meaning of negative value for x are shown in Figure 7.51 where the mechanism is in a positive turning position. 7.7 F Trailer-Truck Kinematics Consider a car pulling a one-axle trailer, as shown in Figure 7.52. We may normalize the dimensions such that the length of the trailer is 1. The 7. Steering Dynamics 435 positions of the car at the hinge point and the trailer at the center of its axle are shown by vectors r and s. Assuming r is a given differentiable function of time t, we would like to examine the behavior of the trailer by calculating s, and predict jackknifing. When the car is moving forward, we say the car and trailer are jackknifed if r\u0307 \u00b7 z < 0 (7.185) where z = r\u2212 s. (7.186) A jackknifed configuration is shown in Figure 7.53, while Figure 7.52 is showing an unjackknifed configuration. Mathematically, we want to know if the truck-trailer will jackknife for a 436 7. Steering Dynamics given path of motion r = r(t) and what conditions we must impose on r(t) to prevent jackknifing. The velocity of the trailer can be expressed by s\u0307 = c (r\u2212 s) (7.187) where c = r\u0307 \u00b7 z (7.188) and the unjackknifing condition is c > 0. (7.189) Assume the twice continuously differentiable function r is the path of car motion. If |z| = 1, and r has a radius of curvature R(t) > 1, and r\u0307(0) \u00b7 z(0) > 0 (7",
            "218) and the trailer moves in an unstable configuration. Any deviation from \u03b8 = 0 ends up to change the situation and leads to the stable limiting solution (7.217). If \u03b8 = \u03c0, then C1 = \u221e C2 = \u221e xt = t\u2212 1 yt = 0 (7.219) and the trailer follows the car in an stable configuration. Any deviation from \u03b8 = 0 will disappear after a while. 7. Steering Dynamics 441 Example 297 F Straight car motion with different initial \u03b8. Consider a car moving on an x-axis with constant speed. The car is pulling a trailer, which is initially at \u03b8 such as shown in Figure 7.52. Using a normalized length, we assume the distance between the center of the trailer axle and the hinge is the length of trailer, and is equal to 1. If we show the absolute position of the car at hinge by r = \u00a3 xc yc \u00a4T and the absolute position of the trailer by s = \u00a3 xt yt \u00a4T then the position of the trailer is a function of the car\u2019s motion. When the position of the car is given by a time-dependent vector function r = \u2219 xc(t) yc(t) \u00b8 (7.220) the trailer position can be found by solving two coupled differential equation",
            "227) When the trailer starts from s = \u00a3 cos \u03b8 sin \u03b8 \u00a4T , the constants of integral would be equal to Equations (7.215) and (7.216) and therefore, xt = t+ e\u22122t (cos \u03b8 + 1)\u2212 cos \u03b8 + 1 e\u22122t (cos \u03b8 + 1) + cos \u03b8 \u2212 1 (7.228) yt = 2e\u2212t sin \u03b8 e\u22122t (cos \u03b8 + 1) + cos \u03b8 \u2212 1 . (7.229) 442 7. Steering Dynamics Figures 7.55, 7.56, and 7.57 illustrate the behavior of the trailer starting from \u03b8 = 45deg, \u03b8 = 90deg, \u03b8 = 135deg. Example 298 F Circular motion of the car with constant velocity. Consider a car pulling a trailer such as Figure 7.52 shows. The car is traveling along a circle of radius R > 1, based on a normalized length in which the length of the trailer is 1. In a circular motion with a normalized angular velocity \u03c9 = 1 and period T = 2\u03c0, the position of the car is given by the following time-dependent vector function: r = \u2219 xc(t) yc(t) \u00b8 = \u2219 R cos(t) R sin(t) \u00b8 . (7.230) The initial position of the trailer must lie on a unit circle with a center at r(0) = \u00a3 xc(0) yc(0) \u00a4T . s(0) = \u2219 xt(0) yt(0) \u00b8 = \u2219 xc(0) yc(0) \u00b8 + \u2219 cos \u03b8 sin \u03b8 \u00b8 (7"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003558_978-1-4020-2249-4-Figure9-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003558_978-1-4020-2249-4-Figure9-1.png",
        "caption": "Figure 9. Family ROIF JQ",
        "texts": [
            " Thus, all manipulators in domain 1 have the same workspace topology, namely, 0 node, 0 cusp and a hole inside their workspace. This workspace topology is referred to as wn (Workspace Topology #1). Domain 2: Figures 4 and 2 show two distinct workspace topologies of manipulators in domain 2, which feature 2 nodes and 0 node and which we call WT2 and WT3 , respectively. Transition between these two workspace topologies is one such that the two lateral segments of the internal boundary meet tangentially (Fig. 9). Equation of this transition can be derived geometrically and the following equation is found (Baili, 2003): d4 =(A-B)/2 (10) As noted in section 3.2, a third topology exists in this domain, where the internal boundary exhibits a '2-tail fish'. This workspace topology, which we call WT4 , features two nodes like in Fig. 4, but these nodes do not play the same role. They coincide with two isolated singular points, which are associated with the two singularity lines defined by B.3=\u00b1arccos(-d3/ c4) (the operation point lies on the second joint axis and the inverse kinematics admits infinitely many solutions)",
            " With the set of parameters {a,s,d,t,f3=20.91\u00b0},h(cp)=acoscp-ssincp and the components 1\\ = kcosf3 + hsinf3 + (kcosf3 - hsinf3)/ 2,12 = (kcosf3 - hSinf3)..J3 /2 and 13 = 2hcosf3 of the gusset vector t. the distance k(cp) can be found from the condition I~ + t~ + Iff = 12 and therewith the variable length of the dodecahedral edges is given by: L( cp) = 2~-5h( cp)2 + 212 - 3h( cpi cos2f3 /(cosf3..J6)+ 6(a sincp + scoscp). For the mechanism in Fig. 8 we find F = 6 + 3- 6 x 1 = 3 and for the Do decahedral Zig-Zag Linkage (Fig.9): F = I./; -61 = (20x6+30)-6x31 = -46. 359 For thkosahedral Zig-Zag Linkage we use the Flat Ring Articula tion. With the {a,s,d,t,{3 = arccos[(-3+,j5)/~(20-8,j5) -n/2 - 31, 71\u00b0}and h(cp) = acoscp-ssincp the edge length ofthe icosahedron can be calculated: L( cp) \"\" 2[(1 I 2) + h( cp )cos( n I 5)]/[ cos{3 sine n I 5)] + 6(a sin cp + s cos cp). The mobility of the mechanism shown in Fig.10 (Flat Ring Articulation and 5 Nuremberg scissors) is found by F = ~/; - 6/ .. (10+ 5)- 6x 1 = 9. For the Ikosahedral Zig-Zag Linkage (Figs",
            " CO J is already described by Karouia and Herve, 2002b. 5.2 Family ROJIO J G In a limb ROflO JpP RO, the second RO pair may be perpendicular to the plane of PP. Then the sequence PP RO becomes PPRo, which is equivalent to a planar pair G. We obtain a new family ROflO jG employing the equivalencies of G. In a limb of the family ROflO JG the H pair axis passes through the fixed point 0 and is parallel to the plane of G. The limbs of the family ROflO JG, are deducted by replacing G by equivalent generators of planar gliding motion (figure 9). 379 5.3 Family ROCo Jp HD J and family Cop HD JRO In a limb of the family RORopP EP J, the sequence ROE. may become (RO l!J equivalent to QO J and, we obtain the subfamily ROQO J E. EP J (figure 10 left) and its special subfamily ROEP JpP EP J (figure 10 right) derived from RO(EP I\u00a3)E.EP J(figure 10 midle). By the same way, a limb of the type ROPPEP JRo, may be (ROE.) E.EP JRo, which leads to the subfamily QO JE.EP JRo (figure 11). The paper describes new special SPM limbs with no inactive pair nor passive motion"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000349_978-1-4020-8227-6-Figure9.32-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000349_978-1-4020-8227-6-Figure9.32-1.png",
        "caption": "Fig. 9.32. Construction of computer HDDs with AFPM brushless motors: (a) single-sided motor; (b) double-sided motor. 1 \u2014 stator coil, 2 \u2014 PM, 3 \u2014 shaft, 4 \u2014 bearing, 5 \u2014 hub, 6 \u2014 rotor ferromagnetic yoke, 7 \u2014 stator ferromagnetic yoke, 8 \u2014 base plate, 9 \u2014 pole fingers, 10 \u2014 nonmagnetic ring.",
        "texts": [
            " Special design features of computer HDD motors are their high starting torque, limited current supply, low vibration and noise, physical constraints on volume and shape, protection against contamination and scaling problems. A high starting torque, 10 to 20 times the running torque is required, since 9.11 Ventricular Assist Devices 313 the read/write head tends to stick to the disc when not moving. The starting current is limited by the computer power supply unit, which in turns, can severely limit the starting torque. AFPM brushless motors (Fig. 9.32) can develop higher starting torque than their radial field counterparts. The additional advantage is zero cogging torque. The drawback of the single-sided AFPM HDD motor shown in Fig. 9.32a is the high normal attractive force between the stator with ferromagnetic yoke and rotor mounted PMs. In double-sided HDD AFPM motors the stator does not have any ferromagnetic core and no normal attractive force is produced at zero-current state. The stator has a three-phase winding with coreless coils fabricated with the aid of a lithographic method. In a typical design, there are two coils per phase for 2p = 8 PM poles and three coils per phase for 2p = 12 PM poles. To reduce the air gap and increase the magnetic flux density, the so called pole fingers are created in the lower part of the hub and the rotor magnetic circuit is bended by pressing it towards the stator coil centres (Fig. 9.32b). According to [140], for a HDD AFPM brushless motor with a 51-mm stator outer diameter, N1 = 180 turns per phase and torque constant kT = 6.59\u00d710\u22123 Nm/A the current consumption is 0.09 A at 13 900 rpm and no load condition. The acoustic noise of HDD brushless motors with ball bearings is usually below 30 dB(A) and projected mean time between failures MTBF = 100 000 hours. HDD spindle motors are now changing from ball bearing to fluid dynamic bearing (FDB) motors. Contact-free FDBs produce less noise and are serviceable for an extended period of time"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002262_asjc.1758-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002262_asjc.1758-Figure1-1.png",
        "caption": "Fig. 1. Quadrotor configuration with body and earth reference frames.",
        "texts": [
            " Several projects on quadrotor UAV by various researchers have been accomplished since the last decade. Quadrotor has recently been used for surveillance, aerial photography, building exploration, climate forecasting, bridge inspection, 3D mapping and swarm missions etc. [2\u20135]. The quadrotor UAV holds very simple mechanical structure, while on the contrary, conventional helicopter normally retains variable pitch rotor and extremely complex mechanical control structure. Quadrotor has cross configuration and there are four fixed-pitch rotors at each end. Referring to Fig. 1, rotors 1 and 3 constitute an odd pair while rotors 2 and 4 constitute an even pair. Both pairs of rotors rotate contrarily to balance the total rotating moment of the body. The output states to be controlled are more than available inputs in a quadrotor, therefore, it constitutes an underactuated system (fewer actuators having more DOF to be controlled) and embraces inherent instability. Six output states which need to be controlled are three Cartesian position states (x, y, and z) and three Euler angles (roll, pitch, and yaw)",
            " These studies have expressed the quadrotor dynamics globally on the special Euclidean group and the geometric approach guarantees the complex quadrotor maneuvers by resolving singularity issues. A complete and precise quadrotor mathematical model is difficult to build due to a number of complex aerodynamic forces which act during flight. The more complete mathematical model helps in designing more accurate and robust flight controller [1,20\u201323]. Quadrotor model presented in this study takes an account of gyroscopic effect which occurs due to the rigid body and propeller rotation [8,24]. As shown in Fig. 1, two reference frames are used to illustrate the position and orientation of the quadrotor. One is the earth-fixed frame and the other is body-fixed frame. Earth-fixed frame has also been referred as inertial reference frame in some studies and can be represented as E = [ex ey ez], while body-fixed frame can be characterized by B \u00bc bx by bz\u00bd . Euler angles (i.e. attitude angles) roll, pitch and yaw are symbolized by \u03d5, \u03b8 and \u03c8 correspondingly. From body-fixed frame to earth-fixed frame, the position of quadrotor in space can be transformed using rotation matrix R, while R \u2208 SO3"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001766_j.cirp.2019.05.004-Figure31-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001766_j.cirp.2019.05.004-Figure31-1.png",
        "caption": "Fig. 31. Geometrical model of the \u201cNIST part\u201d [191].",
        "texts": [
            " The same review points out the need for a standardised test artefact and builds upon the findings of previous work by Richter and Jacobs [225], which identified the following \u201crules\u201d for test artefacts: should be large enough to allow testing of the whole build area, should include many small, medium, and large features, should include in- and out-wards pointing features, should allow a short build time, should allow low material consumption, should be easy to measure, should include \u201creal part\u201d features. Based on these rules and on rigorous design criteria, NIST developed a test artefact intended for standardisation (see Fig. 31) [191]. The \u201cNIST part\u201d incorporates many of the necessary structures and features for the evaluation of the process and machine capabilities, allows the assessment of systematic errors at the same time, and contains datum features for alignment and measurement. Despite this and other standardisation efforts, a large number of different test artefacts exist and still continue to be developed by individual users to address very specific requirements. For example, the \u201cNIST part\u201d was used by Bauza et"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003820_b415051a-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003820_b415051a-Figure1-1.png",
        "caption": "Fig. 1 Cyclic voltammograms of SWNT modified electrodes where the tubes were cut for 2 hours and either randomly dispersed or vertically aligned. CVs are recorded relative to an Ag/AgCl reference electrode in 1 mM K3Fe(CN)6 in a background of 0.05 M KH2PO4 and 0.05 M KCl at pH 7.0 at 100 mV s21. The y-axis shows the current normalised scale relative to the anodic peak current.",
        "texts": [
            " In the first method, droplets of nanotubes dispersed in ethanol were applied to the cysteamine modified electrode to give a bed of randomly orientated nanotubes whilst in the second, the nanotubes were vertically aligned normal to the surface by selfassembly as confirmed via atomic force microscopy (AFM){ images described previously.11 Cyclic voltammograms in the presence of 1 mM K3Fe(CN)6 in a background of 0.05 M KH2PO4 and 0.05 M KCl at pH 7.0 at an electrode modified with randomly dispersed nanotubes cut for 2 hours and aligned carbon nanotubes cut for 2 hours are shown in Fig. 1 with the current scale normalised relative to the anodic peak current to account for the different current magnitudes due to different active electrode areas. A comparison between the CVs shows that the difference in potential between the anodic and cathodic peaks for ferricyanide (DEp), is 105 mV for the randomly dispersed SWNTs and 72 mV for the aligned SWNTs. At a bare gold electrode, DEp is 93 mV. As DEp is a function of the rate of electron transfer, the results suggest that vertical alignment of the cut nanotubes improves the electrochemical properties of the modified electrode",
            " View Article Online / Journal Homepage / Table of Contents for this issue dispersed SWNTs, Compton and co-workers have shown that the electrochemistry is dominated by the ends of the tubes.6 Therefore we propose that with the aligned SWNTs there is a greater presentation of the ends per unit electrode area to the ferricyanide in solution compared with the randomly dispersed SWNTs where it is mostly walls that are presented. Hence in the former case, DEp is expected to be smaller, as observed in Fig. 1. If this hypothesis is correct for electrodes modified with randomly dispersed nanotubes, the shorter the nanotubes, the greater the contribution of the ends to the ferricyanide electrochemistry and the lower the DEp. Furthermore, DEp should be insensitive to cutting time for the aligned SWNT modified electrode which should display the lowest DEp of all the nanotube modified electrodes. Table 1 summarises the DEp values as a function of the cutting time, and hence length of the SWNTs, for both the randomly dispersed and aligned SWNT modified electrodes"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure7.15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure7.15-1.png",
        "caption": "FIGURE 7.15. By increasing the speed at a turn, parallel or reverse steering is needed instead of Ackerman steering.",
        "texts": [
            " Under these conditions the inner front wheel of a kinematic steering vehicle would be at a higher slip angle than required for maximum lateral force. 7. Steering Dynamics 393 394 7. Steering Dynamics Therefore, the inner wheel of a vehicle in a high speed turn must operate at a lower steer angle than kinematic steering. Reducing the steer angle of the inner wheel reduces the difference between steer angles of the inner and outer wheels. For race cars, it is common to use parallel or reverse steering. Ackerman, parallel, and reverse Ackerman steering are illustrated in Figure 7.15. The correct steer angle is a function of the instant wheel load, road condition, speed, and tire characteristics. Furthermore, the vehicle must also be able to turn at a low speed under an Ackerman steering condition. Hence, there is no ideal steering mechanism unless we control the steer angle of each steerable wheel independently using a smart system. Example 267 F Speed dependent steering system. There is a speed adjustment idea that says it is better to have a harder steering system at high speeds"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003933_robot.1989.99999-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003933_robot.1989.99999-Figure2-1.png",
        "caption": "Figure 2: Configuration Space of a 2R Manipulator",
        "texts": [
            " The configuration space, C, of an n-revolute-jointed manipulator is a product space formed by the n-times product of the individual joint manifolds: where Tn is an n-torus, which is a compact n-dimensional manifold. Each of the circles that make up the torus is termed a generator of the torus, and is physically equivalent to a 27r rotation of one joint. There is a one-to-one correspondence between each point in the n-torus configuration space manifold and a distinct manipulator configuration. For example, the 2R planar manipulator in Figure 2(a) has a 2-torus configuration space (Figure 2(b)). While the torus geometry properly captures the manifold structure of C, there are times when other representations of the torus are useful. For example, C in Figure 2 can be presented as a square with dimension 2n by \u201ccutting\u201d the torus along two generators (Figure 3). The 3-torus configuration space of a 3R manipulator can not be directly viewed in a 3-dimensional space, but it can be presented as a cube of dimensions 2n by cutting along the 3-tori generators, and the cubes in the following figures are 3-ton sliced along their generators. Attach a frame to the manipulator end-effector. The manipulator\u2019s workspace manifold, W , is the set of all possible locations and orientations of this frame as the manipulator joints are swept through all points of C"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000880_6.2004-4900-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000880_6.2004-4900-Figure8-1.png",
        "caption": "Figure 8. Circular Path Following Case",
        "texts": [
            "(s) = \u03c92 ne\u03c4s s2 + 2\u03b6\u03c9ns + \u03c92 n where \u03b6 = 0.707, \u03c9n = \u221a 2V L1 , \u03c4 \u2248 L1/V (8) Bode diagrams for this system, as a function of L1/Lp, are shown in Figure 7. The plot clearly shows the improvement in phase response near the system bandwidth (when L1/Lp \u2248 1/4.4 = 0.23) that results from the anticipation (e\u03c4s). Furthermore, if Lp is the wavelength of the highest frequency content in the desired path, then L1 must be chosen to be less than about Lp/4.4 if the vehicle is to accurately follow the desired path. Case 3: Following a Circular Path Figure 8 shows a diagram for a case of following a circular desired path. In this analysis \u03b71 and \u03b72 are assumed to be small, but \u03b73 is not necessarily small. \u03b71 \u2248 0, \u03b72 \u2248 0, |\u03b73| \u00c0 0 (9) It should be noticed that the angle \u03b73 is associated with the local circular segment as shown in the diagram. The current position of the vehicle is specified by r = R + d and \u03b8. \u03c8 indicates the velocity direction. \u03b72 is 6 of 16 American Institute of Aeronautics and Astronautics an angle that is created by the difference between the velocity direction and the current tangent line to the circular desired path"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003482_rob.4620120607-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003482_rob.4620120607-Figure3-1.png",
        "caption": "Figure 3. Simple model for an inchworm robot.",
        "texts": [
            " Briefly, some momentum is conserved along a trajectory that is horizontal with respect to this connection. A more complete description may be found in Bloch et a1.* 3.3. Other Systems Other types of locomotion systems can be modeled, at least approximately, using connections on princi- pal bundles. We require that the motion of such systems be determined exclusively by their kinematics, without dynamic terms. We illustrate with the example below and those in section 5. Example 3. Inchworm Robot Consider a simple model for the linear locomotion of a snakelike robot of unit length, as depicted in Figure 3. This system is visually reminiscent of a crawling inchworm, although we make no attempt here to construct a biologically accurate model of legless locomotion (such as that presented in Keller and Falkovitz9). The variable I , denotes the length of the rear segment in contact with the ground, sh the arclength of the raised segment, or hump, and l h the longitudinal span of the hump. Segments in contact with the ground share a constant uniform thickness. The robot is aligned with the x axis. If we assume viscous friction where the robot comes in contact with the plane, sliding motion of a segment of the robot will be opposed by a force proportional to both the sliding velocity and the length of the contact segment"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000233_978-1-4614-0460-6-Figure3.33-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000233_978-1-4614-0460-6-Figure3.33-1.png",
        "caption": "Fig. 3.33 By the Numbers 3.4 \u2013 Base motion example for automobile suspension (not to scale)",
        "texts": [
            "41) The corresponding phase lag of the mass motion with respect to the base motion is: f \u00bc tan 1 Im Re \u00bc tan 1 2zr3 1\u00fe 4z2 1 r2 ! : (3.42) The suspension for an automobile can be modeled as a single degree of freedom spring\u2013mass\u2013damper as shown in Fig. 3.32 with a natural frequency of 3 Hz and a damping ratio of 0.5. The natural frequency is based on the combination of the car\u2019s mass and the suspension spring\u2019s stiffness. The damping comes primarily from the shock absorber. If y\u00f0t\u00de represents the motion of the wheel center as it follows the oscillating road surface depicted in Fig. 3.33, determine the magnitude of the automobile\u2019s motion, X, when the car is traveling at 60 mph. The first step in the solution is to determine the forcing frequency, f (in Hz), due to the automobile\u2019s motion across the wavy road. The car\u2019s speed, v, is: v \u00bc 60 min hr 1 3; 600 hr s 5;280 ft min 12 in: ft 0:0254 m in: \u00bc 26:8 m s : 3.6 Base Motion 115 The forcing frequency is the speed divided by the spatial wavelength of the road\u2019s surface: f \u00bc 26:8 8 \u00bc 3:35Hz: The frequency ratio is therefore r \u00bc f fn \u00bc 3:35 3 \u00bc 1:12"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001073_b978-0-12-384050-9.50007-6-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001073_b978-0-12-384050-9.50007-6-Figure5-1.png",
        "caption": "Fig. 5. C l a w grip in arboreal m a m m a l s . ( A ) C l a w e d digit gr ipping flat surface , showing angle of penetration (a). T h e size of the e m b e d d e d part of the claw is exaggerated here. ( B ) C l a w e d m a m m a l clinging to cylindrical support . Sol id circle, actual support surface tangent to animal's p a d s ; dashed circle, effective support sur face tangent to volar aspect of e m b e d d e d claws; 0 3 , actual central angle; 0 2 , effective central angle.",
        "texts": [],
        "surrounding_texts": [
            "(Petter and Peyrieras, 1970).* The specialized pointed nails of the needleclawed galago, Galago elegantulus, allow it to eat sap and resin from large smooth trunks and branches, which other galagos do not climb on (Charles-Dominique, 1971). Similar marked specializations for clinging to large vertical supports appear in Phaner and in Microcebus coquereli, which also feed on sap and resin (Petter et al, 1971). The claws of the marmosets Callithrix and Saguinus facilitate several squirrel-like loco motor habits, including running spirally up and down large tree trunks (Bates, 1863; Krieg, 1930; Thorington, 1968; Hershkovitz, 1969). * \"Ses griffes puissantes . . . lui permettent en outre de gr imper aux gros troncs ou de s 'agripper fortement dans des positions impossibles \u00e0 adopter pour les L \u00e9 m u r s lorsqu'il a t taque le bois avec ses dents.\""
        ]
    },
    {
        "image_filename": "designv10_0_0000233_978-1-4614-0460-6-Figure3.29-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000233_978-1-4614-0460-6-Figure3.29-1.png",
        "caption": "Fig. 3.29 A rotating shaft supported by bearing. Large magnitude rotating unbalance could damage the bearings",
        "texts": [
            " Xj j \u00bc m M er2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 r2\u00f0 \u00de2 \u00fe 2zr\u00f0 \u00de2 q (3.27) f \u00bc tan 1 2zr 1 r2 (3.28) MX me was obtained by movingM,m, and e from the numerator of the right-hand side to the left-hand side in Eq. 3.27. Responses are provided for z \u00bc 0.01, 0.05, and 0.1. In some cases, we wish to maintain a vibration magnitude below a certain level to avoid damage to the system, such as bearing damage for a rotating shaft; see Fig. 3.29. In this case, we may need to select a range of acceptable rotating speeds so that this maximum vibration magnitude is not exceeded. In Fig. 3.30, the acceptable rotating speed ranges are o<o1 and o>o2. The problem with the second range is that the speed must pass through resonance (o \u00bc on) in order reach o2. This would only be an acceptable alternative if short-term, large vibration magnitudes will not harm the system. 110 3 Single Degree of Freedom Forced Vibration 3.5 Rotating Unbalance 111 IN A NUTSHELL Designers of rotating systems use Eq"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure2.5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure2.5-1.png",
        "caption": "FIGURE 2.5. A parked car on inclined pavement.",
        "texts": [
            " The height h of C can be found by taking a moment of the forces about the tireprint of the front tire. h = Fz2 (a1 + a2) mg \u2212 a1 cos \u00b5 sin\u22121 H a1 + a2 \u00b6 + Rf +Rr 2 (2.35) Example 45 Statically indeterminate. A vehicle with more than three wheels is statically indeterminate. To determine the vertical force under each tire, we need to know the mechanical properties and conditions of the tires, such as the value of deflection at the center of the tire, and its vertical stiffness. When a car is parked on an inclined pavement as shown in Figure 2.5, the normal force, Fz, under each of the front and rear wheels, Fz1 , Fz2 , is: 2. Forward Vehicle Dynamics 45 Fz1 = 1 2 mg a2 l cos\u03c6+ 1 2 mg h l sin\u03c6 (2.36) Fz2 = 1 2 mg a1 l cos\u03c6\u2212 1 2 mg h l sin\u03c6 (2.37) l = a1 + a2 where, \u03c6 is the angle of the road with the horizon. The horizon is perpendicular to the gravitational acceleration g. Proof. Consider the car shown in Figure 2.5. Let us assume the parking brake forces are applied on only the rear tires. It means the front tires are free to spin. Applying the planar static equilibrium equationsX Fx = 0 (2.38)X Fz = 0 (2.39)X My = 0 (2.40) shows that 2Fx2 \u2212mg sin\u03c6 = 0 (2.41) 2Fz1 + 2Fz2 \u2212mg cos\u03c6 = 0 (2.42) \u22122Fz1a1 + 2Fz2a2 \u2212 2Fx2h = 0. (2.43) 46 2. Forward Vehicle Dynamics These equations provide the brake force and reaction forces under the front and rear tires. Fz1 = 1 2 mg a2 l cos\u03c6\u2212 1 2 mg h l sin\u03c6 (2.44) Fz2 = 1 2 mg a1 l cos\u03c6+ 1 2 mg h l sin\u03c6 (2"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000595_1.3209132-Figure3-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000595_1.3209132-Figure3-1.png",
        "caption": "Fig. 3 Stress history at a subsurface point in a Hertzian line contact",
        "texts": [
            " For the derivation of the subsurface stresses for line and point Hertzian contact, please refer to Refs. 17\u201324 . 2. Contrary to most classical fatigue phenomena, rolling contact fatigue is typically a multiaxial fatigue mechanism. In the past few decades, several multiaxial fatigue criteria have been developed and verified with experiments 25\u201331 . 3. Contrary to classical fatigue, the loading history at a point below the surface is nonproportional; i.e., the stress components do not rise and fall with time in the same proportion to each other; for example, as shown in Fig. 3, the stress history for a point located at the depth where the orthogonal shear stress xz is maximum. As seen, there is a complete reversal of the shear stress xz, while the normal stresses x and z always remain compressive. Also, the peaks of the two normal stresses do not coincide with the peaks for the shear stress. 4. There is a high hydrostatic stress component present in the case of nonconformal contacts, which is absent in classical tension-compression or bending fatigue. 5. The principal axes in nonconformal contacts constantly change in direction during a stress cycle due to which the planes of maximum shear stress also keep changing"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure15.12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure15.12-1.png",
        "caption": "Figure 15.12 The components of force F at point i.",
        "texts": [
            " VOLUME 1: EQUILIBRIUM In a plane1 there are three equilibrium equations, two for the force equilibrium: \u2211 Fx = 0,\u2211 Fy = 0 and one for the moment equilibrium: \u2211 Tz = 0. There is equilibrium when all three conditions are satisfied. The requirement G = \u03bb1 \u2211 Fx + \u03bb2 \u2211 Fy + \u03bb3 \u2211 Tz = 0 for all arbitrary combinations of \u03bb1; \u03bb2; \u03bb3 (not all concurrently zero) is an alternative for the three equilibrium equations: \u2022 The combination \u03bb1 = 0; \u03bb2 = 0; \u03bb3 = 0 leads to \u2211 Fx = 0. \u2022 The combination \u03bb1 = 0; \u03bb2 = 0; \u03bb3 = 0 leads to \u2211 Fy = 0. \u2022 The combination \u03bb1 = 0; \u03bb2 = 0; \u03bb3 = 0 leads to \u2211 Tz = 0. Assume a number of forces are acting on the body (see Figure 15.12): Fxi ; Fyi are the components of Fi at point i with coordinates xi ; yi ; Fxj ; Fyj are the components of Fj at point j with coordinates xj ; yj ; etc. 1 For simplicity we will look only at rigid bodies in the xy plane. 15 Virtual Work 721 To keep the picture simple, Figure 15.12 includes only the components of the force at point i. The body is in equilibrium when the following condition is satisfied: G = \u03bb1 \u2211 i Fxi + \u03bb2 \u2211 i Fyi + \u03bb3 \u2211 i (xiFyi \u2212 yiFxi) = 0 for each arbitrary choice of \u03bb1; \u03bb2; \u03bb3 (not all concurrently zero). The displacement of a rigid body in a plane is defined by the displacement of, for example, point O to O\u2032 and a rotation about O. Assume that the components of the translation (displacement) are ux0; uy0 and the rotation is \u03d5z0. Instead of using its coordinates xi ; yi , we can define the location of an arbitrary point i also by its angle \u03b1i and the radius ri "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000576_j.ijfatigue.2016.03.014-Figure1-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000576_j.ijfatigue.2016.03.014-Figure1-1.png",
        "caption": "Fig. 1. (a) Schematic of SLM process [1], (b) SEM image of 17-4 PH SS powder used i horizontally-orientated cylindrical bars.",
        "texts": [
            " Selective Laser Melting (SLM), a conventional L-PBF process, involves depositing a thin layer of metallic powder on a substrate and then moving a focused laser in a specified pattern along the powder bed. Due to the high heat flux imposed by the laser beam, the irradiated powder melts forming a micro-sized melt pool (i.e. molten pool). Upon removal of the laser, the melt pool rapidly solidifies, thus forming individual tracks of solid material with length and direction governed by the laser scan pattern. The sum of individual tracks within a plane then forms a layer and this process is repeated until the part(s) are completed, as shown schematically in Fig. 1(a). The building process is typically conducted in an inert atmosphere (e.g. argon) to reduce part oxidization while at elevated temperature. Compared to DLD methods, which involve injected powder (rather than a powder bed), SLM generally allows for higher precision, powder efficiency and smoother surface finishes. Still, the surface finish of SLM parts is relatively rough due to partially melted/sintered particles existing along the periphery of the manufactured parts. Post-SLM machining/polishing processes can be utilized to clean and smooth the as-built surfaces, if required",
            " Next, the monotonic and cyclic deformation, as well as the fatigue behavior and failure mechanisms of various SLM 17-4 PH SS specimens, are discussed. This is followed by a discussion on requirements for a meaningful process parameter optimization to enhance mechanical behavior of AM products. Finally, the main conclusions drawn from this study are summarized. Commercially-available, gas-atomized (argon) 17-4 PH SS powder (Phenix Systems) with 80% of its particle size distribution below 22 lm was utilized in this study. Scanning electron microscopy (SEM) images of the typical, as-received powder morphology is shown in Fig. 1(b). It may be seen that the majority of the asreceived powder possessed a spherical shape. Process parameters (i.e. laser power, scanning speed, layer thickness, and hatching pitch) were optimized to obtain an acceptable level of final part density using a design of experiments (DOE) methodology [15]. The optimized process parameters, which remained constant with time (i.e. un-controlled parameters), are summarized in Table 1 and were utilized for fabricating two different groups of 17-4 PH SS samples within an argon-purged SLM machine (ProXTM 100). The first group contained vertically oriented cylindrical rods, while the second group contained horizontally orientated cylindrical rods, as depicted in Fig. 1(c) and (d), respectively. The cylindrical rods from each set possessed an 8 mm diameter and 75 mm height (all dimensions nominal and refer to actual CAD drawing). It has been found that precipitation hardening does not occur via direct aging of 17-4 PH SS [8]. Therefore, half of the as-built samples, from both the \u2018vertical\u2019 and \u2018horizontal\u2019 sets, underwent post-SLM heat treatment consisting of (in order): solution annealing for 30 min at approximately 1040 C, air cooling (AC) to room temperature (ConditionA), precipitation hardening for 1 h at 482 C, then AC to room temperature (Condition H900 or peak-aging)"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000801_j.finel.2014.04.003-Figure14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000801_j.finel.2014.04.003-Figure14-1.png",
        "caption": "Fig. 14. Wall build using inactive element layers, quiet elements within layer (1C, case wiq2).",
        "texts": [
            " In this approach, equation numbering and solver initialization is repeated only when each layer is activated resulting into slightly faster computer run times of about 976 s wall (6284 s using 1 core) per case. In cases wq2, wi2, wiq2, and wiq3, the temperature of quiet or inactive elements before activation n 1T is set to the initial temperature 0T , while in cases wq1, wi1, wiq1, it is not. The temperature result during the deposition of the 61st layer for case wq2 is illustrated in Fig. 12. Fig. 13 illustrates the temperature result for case wq2 at the same time instance with all quiet element removed. The temperature result during the deposition of the 61st layer for case wi2 is illustrated in Fig. 14. Temperature results along line AA are extracted and plotted in Fig. 15. Results from cases wq2 and wi3 are also plotted in the same figure. Using case wiq2 as reference, the maximum % error along the line is listed in Table 5. As seen in Table 5, neglecting to reset n 1T upon element activation leads to errors (3.5323\u20135.3056%). The error is higher in the inactive element method. When n 1T is reset, all methods result into nearly equivalent results (maximum error .5%). In all cases, except case wi3, surface and convection and radiation are applied on the interface between active and inactive (or quiet) elements"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure1.20-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure1.20-1.png",
        "caption": "FIGURE 1.20. The difference between aluminum, magnesium, and steel rims in regaining road contact after a jump.",
        "texts": [
            " Tire and Rim Fundamentals cling. Magnesium is about 30% lighter than aluminum, and is excellent for size stability and impact resistance. However, magnesium is more expensive and it is used mainly for luxury or racing cars. The corrosion resistance of magnesium is not as good as aluminum. Titanium is much stronger than aluminum with excellent corrosion resistance. However, titanium is expensive and hard to be machine processed. The difference between aluminum, magnesium, and steel rims is illustrated in Figure 1.20. Light weight wheels regain contact with the ground quicker than heavier wheels. Example 37 Spare tire. Road vehicles typically carry a spare tire, which is already mounted on a rim ready to use in the event of flat tire. After 1980, some cars have been equipped with spare tires that are smaller than normal size. These spare tires are called doughnuts or space-saver spare tires. Although the doughnut spare tire is not very useful or popular, it can help to save a little space, 1. Tire and Rim Fundamentals 25 weight, cost, and gas mileage"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.20-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.20-1.png",
        "caption": "FIGURE 8.20. A double A-arm suspension mechanism on the left and right wheels",
        "texts": [],
        "surrounding_texts": [
            "The suspension is what links the wheels to the vehicle body and allows relative motion. This chapter covers the suspension mechanisms, and discusses the possible relative motions between the wheel and the vehicle body. The wheels, through the suspension linkage, must propel, steer, and stop the vehicle, and support the associated forces. 8.1 Solid Axle Suspension The simplest way to attach a pair of wheels to a vehicle is to mount them at opposite ends of a solid axle, such as the one that is shown in Figure 8.1. The solid axle must be attached to the body such that an up and down motion in the z-direction, as well as a roll rotation about the x-axis, is possible. So, no forward and lateral translation, and also no rotation about the axle and the z-axis, is allowed. There are many combinations of links and springs that can provide the kinematic and dynamic requirements. The simplest design is to clamp the axle to the middle of two leaf springs with their ends tied or shackled to the vehicle frame as shown schematically in Figure 8.1. A side view of a multi-leaf spring and solid axle is shown in Figure 8.2. A suspension with a solid connection between the left and right wheels is called dependent suspension. 456 8. Suspension Mechanisms The performance of a solid axle with leaf springs suspension can be improved by adding a linkage to guide the axle kinematically and provide dynamic support to carry the non z-direction forces. The solid axle with leaf spring combination came to vehicle industry from horse-drawn vehicles. Example 299 Hotchkiss drive. When a live solid axle is connected to the body with nothing but two leaf springs, it is called the Hotchkiss drive, which is the name of the car that used it first. The main problems of a Hotchkiss drive, which is shown in Figure 8.2, are locating the axle under lateral and longitudinal forces, and having a low mass ratio \u03b5 = ms/mu, where ms is the sprung mass and mu is the unsprung mass. Sprung mass refers to all masses that are supported by the spring, such as vehicle body. Unsprung mass refers to all masses that are attached to and not supported by the spring, such as wheel, axle, or brakes. Example 300 Leaf spring suspension and flexibility problem. The solid axle suspension systems with longitudinal leaf springs have many drawbacks. The main problem lies in the fact that springs themselves act as locating members. Springs are supposed to flex under load, but their flexibility is needed in only one direction. However, it is the nature of leaf springs to twist and bend laterally and hence, flex also in planes other than the tireplane. Leaf springs are not suited for taking up the driving and braking traction forces. These forces tend to push the springs into an S-shaped profile, as shown in Figure 8.3. The driving and braking flexibility of leaf springs, generates a negative caster and increases instability. Long springs provide better ride. However, long sprigs exaggerate their bending and twisting under different load conditions. Example 301 Leaf spring suspension and flexibility solution. To reduce the effect of a horizontal force and S-shaped profile appearance in a solid axle with leaf springs, the axle may be attached to the chassis by a longitudinal bar as Figure 8.4(a) shows. Such a bar is called an anti-tramp 8. Suspension Mechanisms 457 bar, and the suspension is the simplest cure for longitudinal problems of a Hotchkiss drive. A solid axle with an anti-tramp bar may be kinematically approximated by a four-bar linkage, as shown in Figure 8.4(b). Although an anti-tramp bar may control the shape of the leaf spring, it introduces a twisting angle problem when the axle is moving up and down, as shown in Figure 8.5. Twisting the axle and the wheel about the axle is called caster. The solid axle is frequently used to help keeping the wheels perpendicular to the road. Example 302 Leaf spring location problem. The front wheels need room to steer left and right. Therefore, leaf springs cannot be attached close to the wheel hubs, and must be placed closer to the middle of the axle. That gives a narrow spring-base, which means that a small side force can sway or tilt the body relative to the axle through a considerable roll angle due to weight transfer. This is uncomfortable for the vehicle passengers, and may also produce unwanted steering. The solid axle positively prevents the camber change by body roll. The wheels remain upright and hence, do not roll on a side. However, a solid axle shifts laterally from its static plane and its center does not remain on the vehicle\u2019s longitudinal axis under a lateral force. 458 8. Suspension Mechanisms A solid axle produces bump-camber when single-wheel bump occurs. If the right wheel goes over a bump, the axle is raised at its right end, and that tilts the left wheel hub, putting the left wheel at a camber angle for the duration of deflection. Example 303 Triangular linkage. A triangulated linkage, as shown in Figure 8.6, may be attached to a solid axle to provide lateral and twist resistance during acceleration and braking. Example 304 Panhard arm. High spring rate is a problem of leaf springs. Reducing their stiffness by narrowing them and using fewer leaves, reduces the lateral stiffness and increases the directional stability of the suspension significantly. A Panhard arm is a bar that attaches a solid axle suspension to the chassis laterally. Figure 8.7 illustrates a solid axle and a Panhard arm to guide the axle. Figure 8.8 shows a triangular linkage and a Panhard arm combination for guiding a solid axle. A double triangle mechanism, as shown in Figure 8.9, is an alternative design to guide the axle and support it laterally. Example 305 Straight line linkages. There are many mechanisms that can provide a straight line motion. The simplest mechanisms are four-bar linkages with a coupler point moving straight. Some of the most applied and famous linkages are shown in Figure 8.10. By having proper lengths, the Watt, Robert, Chebyshev, and Evance linkages can make the coupler point C move on a straight line vertically. Such a mechanism and straight motion may be used to guide a solid axle. Two Watt suspension mechanisms with a Panhard arm are shown in Figures 8.11 and 8.12. 460 8. Suspension Mechanisms 8. Suspension Mechanisms 461 Figures 8.13, 8.14, and 8.15 illustrate three combinations of Robert suspension linkages equipped with a Panhard arm. Example 306 Solid axle suspension and unsprung mass problem. A solid axle is counted as an unsprung member, and hence, the unsprung mass is increased where using solid axle suspension. A heavy unsprung mass ruins both, the ride and handling of a vehicle. Lightening the solid axle makes it weaker and increases the most dangerous problem in vehicles: axle breakage. The solid axle must be strong enough to make sure it will not break under any loading conditions at any age. As a rough estimate, 90% of the leaf spring mass may also be counted as unsprung mass, which makes the problem worse. The unsprung mass problem is worse in front, and it is the main reason that they are no longer used in street cars. However, front solid axles are still common on trucks and buses. These are heavy vehicles and solid axle 8. Suspension Mechanisms 463 464 8. Suspension Mechanisms suspension does not reduce the mass ratio \u03b5 = ms/mu very much. When a vehicle is rear-wheel-drive and a solid axle suspension is used in the back, the suspension is called live axle. A live axle is a casing that contains a differential, and two drive shafts. The drive shafts are connected to the wheel hubs. A live axle can be three to four times heavier than a dead I-beam axle. It is called live axle because of rotating gears and shafts inside the axle. Example 307 Solid axle and coil spring. To decrease the unsprung mass and increase vertical flexibility of solid axle suspensions, it is possible to equip them with coil springs. A sample of a solid axle suspension with coil spring is shown in Figure 8.16. The suspension mechanism is made of four longitudinal bars between the axle and chassis. The springs may have some lateral or longitudinal angle to introduce some lateral or longitudinal compliance. Example 308 De Dion axle. When a solid axle is a dead axle with no driving wheels, the connecting beam between the left and right wheels may have different shapes to do different jobs, usually to give the wheels independent flexibility. We may also modify the shape of a live axle to attach the differential to the chassis and reduce the unsprung mass. De Dion design is a modification of a beam axle that may be used as a dead axle or to attach the differential to the chassis and transfer the driving power to the drive wheels by employing universal joints and split shafts. Figure 8.17 illustrates a De Dion suspension. 8.2 Independent Suspension Independent suspensions is introduced to let a wheel to move up and down without affecting the opposite wheel. There are many forms and designs of independent suspensions. However, double A-arm and McPherson strut suspensions are the simplest and the most common designs. Figure 8.18 illustrates a sample of a double A-arm and Figure 8.19 shows a McPherson suspension. Kinematically, a double A-arm suspension mechanism is a four-bar linkage with the chassis as the ground link, and coupler as the wheel carrying link. A McPherson suspension is an inverted slider mechanism that has the chassis as the ground link and the coupler as the wheel carrying link. A double A-arm and a McPherson suspension mechanism on the left and right wheels are schematically shown in Figures 8.20 and 8.21 respectively. Double A-arm, is also called double wishbone, or short/long arm suspension. McPherson also may be written as MacPherson. Example 309 Double A-arm suspension and spring position. Consider a double A-arm suspension mechanism. The coil spring may be between the lower arm and the chassis, as shown in Figure 8.18. It is also possible to install the spring between the upper arm and the chassis, or between the upper and lower arms. In either case, the lower or the upper arm, which supports the spring, is made stronger and the other arm acts as a connecting arm. Example 310 Multi-link suspension mechanism. When the two side bars of an A-arm are attached to each other with a joint, as shown in Figure 8.22, then the double A-arm is called a multi-link mechanism. A multi-link mechanism is a six-bar mechanism that may have a better coupler motion than a double A-arm mechanism. However, multi- 466 8. Suspension Mechanisms 8. Suspension Mechanisms 467 link suspensions are more expensive, less reliable, and more complicated compare to a double A-arm four-bar linkage. There are vehicles with more than six-link suspension with possibly better kinematic performance. Example 311 Swing arm suspension. An independent suspension may be as simple as a triangle shown in Figure 8.23. The base of the triangle is jointed to the chassis and the wheel to the tip point. The base of the triangle is aligned with the longitudinal axis of the vehicle. Such a suspension mechanism is called a swing axle or swing arm. The variation in camber angle for a swing arm suspension is maximum, compared to the other suspension mechanisms. Example 312 Trailing arm suspension. Figure 8.24 illustrates a trailing arm suspension that is a longitudinal arm with a lateral axis of rotation. The camber angle of the wheel, supported by a trailing arm, will not change during the up and down motion. Trailing arm suspension has been successfully using in a variety of frontwheel-drive vehicles, to suspend their rear wheels. Example 313 Semi-trailing arm Semi-trailing arm suspension, as shown in Figure 8.25, is a compromise between the swing arm and trailing arm suspensions. The joint axis may have any angle, however an angle not too far from 45 deg is more applied. Such suspensions have acceptable camber angle change, while they can handle both, the lateral and longitudinal forces. Semi-trailing design has successfully applied to a series of rear-wheel-drive cars for several decades. Example 314 Antiroll bar and roll stiffness. Coil springs are used in vehicles because they are less stiff with better 470 8. Suspension Mechanisms ride comfort compared to leaf springs. Therefore, the roll stiffness of the vehicle with coil springs is usually less than in vehicles with leaf springs. To increase the roll stiffness of such suspensions, an antiroll bar must be used. Leaf springs with reduced layers, uni-leaf, trapezoidal, or nonuniform thickness may also need an antiroll bar to compensate for their reduced roll stiffness. The antiroll bar is also called a stabilizer. Figure 8.26 illustrates an anti-roll bare attached to a solid axle with coil springs. Example 315 Need for longitudinal compliance. A bump is an obstacle on the road that opposes the forward motion of a wheel. When a vehicle goes over a bump, the first action is a force that tends to push the wheel backward relative to the rest of the vehicle. So, the lifting force has a longitudinal component, which will be felt inside the vehicle unless the suspension system has horizontal compliance. There are situations in which the horizontal component of the force is even higher than the vertical component. Leaf springs can somewhat absorb this horizontal force by flattening out and stretching the distance from the forward spring anchor and the axle. Such a stretch is usually less than 1/2 in \u2248 1 cm. 8.3 Roll Center and Roll Axis The roll axis is the instantaneous line about which the body of a vehicle rolls. Roll axis is found by connecting the roll center of the front and rear suspensions of the vehicle. Assume we cut a vehicle laterally to disconnect the front and rear half of the vehicle. Then, the roll center of the front or rear suspension is the instantaneous center of rotation of the body with respect to the ground. 8. Suspension Mechanisms 471 472 8. Suspension Mechanisms Figure 8.27 illustrates a sample of the front suspensions of a car with a double A-arm mechanism. To find the roll center of the body with respect to the ground, we analyze the two-dimensional kinematically equivalent mechanism shown in Figure 8.28. The center of tireprint is the instant center of rotation of the wheel with respect to the ground, so the wheels are jointed links to the ground at their center of tireprints. The instant center I18 is the roll center of the body with respect to the ground. To find I18, we apply the Kennedy theorem and find the intersection of the line I12I28 and I13I38 as shown in Figure 8.29. The point I28 and I38 are the instant center of rotation for the wheels with respect to the body. The instant center of rotation of a wheel with respect to the body is called suspension roll center. So, to find the roll center of the front or rear half of a car, we should determine the suspension roll centers, and find the intersection of the lines connecting the suspension roll centers to the center of their associated tireprints. The Kennedy theorem states that the instant center of every three relatively moving objects are colinear. Example 316 McPherson suspension roll center. A McPherson suspension is an inverted slider crank mechanism. The instant centers of an example of inverted slider crank mechanism are shown in Figure 8.30. In this figure, the point I12 is the suspension roll center, which is the instant center of rotation for the wheel link number 2 with respect to the chassis link number 1. A car with a McPherson suspension system is shown in Figure 8.31. The kinematic equivalent mechanism is depicted in Figure 8.32. Suspension roll centers along with the body roll center are shown in Figure 8.33. To 8. Suspension Mechanisms 473 find the roll center of the front or rear half of a car, we determine each suspension roll center and then find the intersection of the lines connecting the suspension roll centers to the center of the associated tireprint. Example 317 Roll center of double A-arm suspension. The roll center of an independent suspension such as a double A-arm can be internal or external. The kinematic model of a double A-arm suspension for the front left wheel of a car is illustrated in Figure 8.34. The suspension roll center in Figure 8.34(a) is internal, and in Figure 8.34(b) is external. An internal suspension roll center is toward the vehicle body, while an external suspension roll center goes away from the vehicle body. A suspension roll center may be on, above, or below the road surface, as shown in Figure 8.35(a)-(c) for an external suspension roll center. When the suspension roll center is on the ground, above the ground, or below the ground, the vehicle roll center would be on the ground, below the ground, and above the ground, respectively. Example 318 F Camber variation of double A-arm suspension. When a wheel moves up and down with respect to the vehicle body, de- 474 8. Suspension Mechanisms 8. Suspension Mechanisms 475 pending on the suspension mechanism, the wheel may camber. Figure 8.36 illustrates the kinematic model for a double A-arm suspension mechanism. The mechanism is equivalent to a four-bar linkage with the ground link as the vehicle chasis. The wheel is always attached to a coupler point C of the mechanism. We set a local suspension coordinate frame (x, y) with the x-axis indicating the ground link MN . The x-axis makes a constant angle \u03b80 with the vertical direction. The suspension machanism has a length a for the upper A-arm, b for the coupler link, c for the lower A-arm, and d for the ground link. The configuration of the suspension is determined by the angles \u03b82, \u03b83, and \u03b84, all measured from the positive direction of the x-axis. When the suspension is at its equilibrium position, the links of the double A-arm suspension make initial angles \u03b820 \u03b830, and \u03b840 with the x-axis. The equilibrium position of a suspension is called the rest position. To determine the camber angle during the fluctuation of the wheel, we should determine the variation of the coupler angle \u03b83, as a function of vertical motion z of the coupler point C. Using \u03b82 as a parameter, we can find the coordinates (xC , yC) of the 476 8. Suspension Mechanisms coupler point C in the suspension coordinate frame (x, y) as xC = a cos \u03b82 + e cos (p\u2212 q + \u03b1) (8.1) yC = a sin \u03b82 + e sin (p\u2212 q + \u03b1) (8.2) where, q = tan\u22121 a sin \u03b82 d\u2212 a cos \u03b82 (8.3) p = tan\u22121 q 4b2f2 \u2212 (b2 + f2 \u2212 c2) 2 b2 + f2 \u2212 c2 (8.4) f = p a2 + d2 \u2212 2ad cos \u03b82. (8.5) The position vector of the coupler point is uC uC = xC \u0131\u0302+ yC j\u0302 (8.6) and the unit vector in the z-direction is u\u0302z = \u2212 cos \u03b80\u0131\u0302\u2212 sin \u03b80j\u0302. (8.7) Therefore, the displacement z in terms of xC and yC is: z = uC \u00b7 u\u0302z = \u2212xC cos \u03b80 \u2212 yC sin \u03b80 (8.8) 8. Suspension Mechanisms 477 The initial coordinates of the coupler point C and the initial value of z are: xC0 = a cos \u03b820 + e cos (p0 \u2212 q0 + \u03b1) (8.9) yC0 = a sin \u03b820 + e sin (p0 \u2212 q0 + \u03b1) (8.10) z0 = \u2212xC0 cos \u03b80 \u2212 yC0 sin \u03b80 (8.11) and hence, the vertical displacement of the wheel center can be calculated by h = z \u2212 z0 (8.12) The initial angle of the coupler link with the vertical direction is \u03b80 \u2212 \u03b830. Therefore, the camber angle of the wheel would be \u03b3 = (\u03b80 \u2212 \u03b83)\u2212 (\u03b80 \u2212 \u03b830) = \u03b830 \u2212 \u03b83 (8.13) The angle of the coupler link with the x-direction is equal to \u03b83 = 2 tan \u22121 \u00c3 \u2212E \u00b1 \u221a E2 \u2212 4DF 2D ! (8.14) where, D = J5 \u2212 J1 + (1 + J4) cos \u03b82 (8.15) E = \u22122 sin \u03b82 (8.16) F = J5 + J1 \u2212 (1\u2212 J4) cos \u03b82 (8.17) and J1 = d a (8.18) J2 = d c (8.19) J3 = a2 \u2212 b2 + c2 + d2 2ac (8.20) J4 = d b (8.21) J5 = c2 \u2212 d2 \u2212 a2 \u2212 b2 2ab . (8.22) Substituting (8.14) and (8.13), and then, eliminating \u03b82 between (8.13) and (8.8) provides the relationship between the vertical motion of the wheel, z, and the camber angle \u03b3. 478 8. Suspension Mechanisms Example 319 F Camber angle and wheel fluctuations. Consider the double A-arm suspension that is shown in Figure 8.36. The dimensions of the equivalent kinematic model are: a = 22.4 cm b = 22.1 cm c = 27.3 cm d = 17.4 cm \u03b80 = 24.3 deg (8.23) The coupler point C is at: e = 14.8 cm \u03b1 = 54.8 deg (8.24) If the angle \u03b82 at the rest position is at \u03b820 = 121.5 deg (8.25) then the initial angle of the other links are: \u03b830 = 18.36 deg \u03b840 = 107.32 deg (8.26) At the rest position, the coupler point is at: xC0 = \u221222.73 cm yC0 = 9.23 cm z0 = 16.92 cm (8.27) We may calculate h and \u03b3 by varying the parameter \u03b82. Figure 8.37 illustrates h as a function of the camber angle \u03b3. For this suspension mechanism, the wheel gains a positive camber when the wheel moves up, and gains a negative camber when the it moves down. The mechanism is shown in Figure 8.38, when the wheel is at the rest position and has a positive or a negative displacement. 8.4 F Car Tire Relative Angles There are four major wheel alignment parameters that affect vehicle dynamics: toe, camber, caster, and trust angle. 8. Suspension Mechanisms 479 8.4.1 F Toe When a pair of wheels is set so that their leading edges are pointed toward each other, the wheel pair is said to have toe-in. If the leading edges point away from each other, the pair is said to have toe-out. Toe-in and toe-out front wheel configurations of a car are illustrated in Figure 8.39. The amount of toe can be expressed in degrees of the angle to which the wheels are not parallel. However, it is more common to express the toein and toe-out as the difference between the track widths as measured at the leading and trailing edges of the tires. Toe settings affect three major performances: tire wear, straight-line stability, and corner entry handling. For minimum tire wear and power loss, the wheels on a given axle of a car should point directly ahead when the car is running in a straight line. Excessive toe-in causes accelerated wear at the outboard edges of the tires, while too much toe-out causes wear at the inboard edges. Toe-in increases the directional stability of the vehicle, and toe-out increases the steering response. Hence, a toe-in setting makes the steering function lazy, while a toe-out makes the vehicle unstable. With four wheel independent suspensions, the toe may also be set at the rear of the car. Toe settings at the rear have the same effect on wear, directional stability, and turn-in as they do on the front. However, we usually do not set up a rear-drive race car toed out in the rear, because of excessive instability. When driving torque is applied to the wheels, they pull themselves forward and try to create toe-in. Furthermore, when pushed down the road, a non-driven wheel or a braking wheel will tend to toe-out. Example 320 Toe-in and directional stability. Toe settings have an impact on directional stability. When the steering 480 8. Suspension Mechanisms 8. Suspension Mechanisms 481 wheel is centered, toe-in causes the wheels to tent to move along paths that intersect each other in front of the vehicle. However, the wheels are in balance and no turn results. Toe-in setup can increase the directional stability caused by little steering fluctuations and keep the car moving straight. Steering fluctuations may be a result of road disturbances. If a car is set up with toe-out, the front wheels are aligned so that slight disturbances cause the wheel pair to assume rolling directions that approach a turn. Therefore, toe-out encourages the initiation of a turn, while toe-in discourages it. Toe-out makes the steering quicker. So, it may be used in vehicles for a faster response. The toe setting on a particular car becomes a trade-off between the straight-line stability afforded by toe-in and the quick steering response by toe-out. Toe-out is not desirable for street cars, however, race car drivers are willing to drive a car with a little directional instability, for sharper turn-in to the corners. So street cars are generally set up with toe-in, while race cars are often set up with toe-out. Example 321 Toe-in and toe-out in the front and rear axles. Front toe-in: slower steering response, more straight-line stability, greater wear at the outboard edges of the tires. Front toe-zero: medium steering response, minimum power loss, minimum tire wear. Front toe-out: quicker steering response, less straight-line stability, greater wear at the inboard edges of the tires. Rear toe-in: straight-line stability, traction out of the corner, more steerability, higher top speed. 482 8. Suspension Mechanisms 8.4.2 F Caster Angle Caster is the angle to which the steering pivot axis is tilted forward or rearward from vertical, as viewed from the side. Assume the wheel is straight to have the body frame and the wheel frame coincident. If the steering axis is turned about the wheel yw-axis then the wheel has positive caster. If the steering axis is turned about the wheel \u2212yw-axis, then the wheel has negative caster. Positive and negative caster configurations on the front wheel of a car are shown in Figure 8.40. Negative caster aids in centering the steering wheel after a turn and makes the front tires straighten quicker. Most street cars are made with 4\u22126 deg negative caster. Negative caster tends to straighten the wheel when the vehicle is traveling forward, and thus is used to enhance straight-line stability. Example 322 Negative caster of shopping carts. The steering axis of a shopping cart wheel is set forward of where the wheel contacts the ground. As the cart is pushed forward, the steering axis pulls the wheel along, and because the wheel drags along the ground, it falls directly in line behind the steering axis. The force that causes the wheel to follow the steering axis is proportional to the distance between the steering axis and the wheel-to-ground contact point, if the caster is small. This distance is referred to as trail. The cars\u2019 steering axis intersects the ground at a point in front of the tireprint, and thus the same effect as seen in the shopping cart casters is achieved. While greater caster angles improves straight-line stability, they also cause an increase in steering effort. Example 323 Characteristics of caster in front axle. Zero castor provides: easy steering into the corner, low steering out of the corner, low straight-line stability. 8. Suspension Mechanisms 483 Negative caster provides: low steering into the corner, easy steering out of the corner, more straight-line stability, high tireprint area during turn, good turn-in response, good directional stability, good steering feel. When a castered wheel rotates about the steering axis, the wheel gains camber. This camber is generally favorable for cornering. 8.4.3 F Camber Camber is the angle of the wheel relative to vertical line to the road, as viewed from the front or the rear of the car. Figure 8.41 illustrates the wheel number 1 of a vehicle. If the wheel leans in toward the chassis, it is called negative camber and if it leans away from the car, it is called positive camber. The cornering force that a tire can develop is highly dependent on its angle relative to the road surface, and so wheel camber has a major effect on the road holding of a car. A tire develops its maximum lateral force at a small camber angle. This fact is due to the contribution of camber thrust, which is an additional lateral force generated by elastic deformation as the tread rubber pulls through the tire/road interface. To optimize a tire\u2019s performance in a turn, the suspension should provide a slight camber angle in the direction of rotation. As the body rolls in a turn, the suspension deflects vertically. The wheel is connected to the chassis by suspension mechanism, which must rotate to allow for the wheel deflection. Therefore, the wheel can be subject to large camber changes as the suspension moves up and down. So, the more the wheel must deflect from its static position, the more difficult it is to maintain an ideal camber angle. Thus, the relatively large wheel travel and soft roll stiffness needed to provide a smooth ride in passenger cars presents a difficult design challenge, while the small wheel travel and high roll stiffness inherent in racing cars reduces the problem. Example 324 Castor versus camber. Camber doesn\u2019t improve turn-in as the positive caster does. Camber is not generally good for tire wear. Camber in one wheel does not improve directional stability. Camber adversely affects braking and acceleration efforts. 8.4.4 F Trust Angle The trust angle \u03c5 is the angle between vehicle\u2019s centerline and perpendicular to the rear axle. It compares the direction that the rear axle is aimed with the centerline of the vehicle. A nonzero angle configuration is shown in Figure 8.42. Zero angle confirms that the rear axle is parallel to the front axle, and the wheelbase on both sides of the vehicle are the same. A reason for nonzero 484 8. Suspension Mechanisms 8. Suspension Mechanisms 485 trust angle would have unequal toe-in or toe-out on both sides of the axle. Example 325 Torque reaction. There are two kinds of torque reactions in rear-whel-drive: 1\u2212 the reaction of the axle housing to rotate in the opposite direction of the crown wheel rotation, and 2\u2212 the reaction of axle housing to spin about its own center, opposite to the direction of pinion\u2019s rotation. The first reaction leads to a lifting force in the differential causing a wind-up in springs. The second reaction leads to a lifting force on the right wheels. 8.5 Suspension Requirements and Coordinate Frames The suspension mechanism should allow a relative motion between the wheel and the vehicle body. The relative motions are needed to pass the road irregularities and steering. To function properly, a suspension mechanism should have some kinematic and dynamics requirements. 8.5.1 Kinematic Requirements To express the motions of a wheel, we attach a wheel coordinate system W (oxwywzw) to the center of the wheel. A wheel, as a rigid body, has six degrees-of-freedom with respect to the vehicle body: three translations and three rotations, as shown in Figure 8.43. 486 8. Suspension Mechanisms The axes xw, yw, and zw indicate the direction of forward, lateral, and vertical translations and rotations. In the position shown in the figure, the rotation about the xw-axis is the camber angle, about the yw-axis is the spin, and about the zw-axis is the steer angle. Consider a non-steerable wheel. Translation in zw-direction and spin about the yw-axis are the only two DOF allowed for such a wheel. So, we need to take four DOF. If the wheel is steerable, then translation in the zw-direction, spin about the yw-axis, and steer rotation about the zw-axis are the three DOF allowed. So, we must take three DOF of a steerable wheel. Kinematically, non-steerable and steerable wheels should be supported as shown in Figures 8.44 and 8.45 respectively. Providing the required freedom, as well as eliminating the taken DOF, are the kinematic requirements of a suspension mechanism. 8.5.2 Dynamic Requirements Wheels should be able to propel, steer, and stop the vehicle. So, the suspension system must transmit the driving traction and deceleration braking forces between the vehicle body and the ground. The suspension members must also resist lateral forces acting on the vehicle. Hence, the wheel suspension system must make the wheel rigid for the taken DOF. However, there must also be some compliance members to limit the untaken DOF. The most important compliant members are spring and dampers to provide returning and resistance forces in the z-direction. 8. Suspension Mechanisms 487 8.5.3 Wheel, wheel-body, and tire Coordinate Frames Three coordinate frames are employed to express the orientation of a tire and wheel with respect to the vehicle: the wheel frame W , wheel-body frame C, and tire frame T . A wheel coordinate frame W (xw, yw, zw) is attached to the center of a wheel. It follows every translation and rotation of the wheel except the spin. Hence, the xw and zw axes are always in the tire-plane, while the yw-axis is always along the spin axis. A wheel coordinate frame is shown in Figure 8.43. When the wheel is straight and the W frame is parallel to the vehicle coordinate frame, we attach a wheel-body coordinate frame C (xc, yc, zc) at the center of the wheel parallel to the vehicle coordinate axes. The wheelbody frame C is motionless with respect to the vehicle coordinate and does not follow any motion of the wheel. The tire coordinate frame T (xt, yt, zt) is set at the center of the tireprint. The zt-axis is always perpendicular to the ground. The xt-axis is along the intersection line of the tire-plane and the ground. The tire frame does not follows the spin and camber rotations of the tire however, it follows the steer angle rotation about the zc-axis. Figure 8.46 illustrates a tire and a wheel coordinate frames. Example 326 Visualization of the wheel, tire, and wheel-body frames. Figure 8.47 illustrates the relative configuration of a wheel-body frame C, a tire frame T , and a wheel frame W . If the steering axis is along the zc-axis then, the rotation of the wheel about the zc-axis is the steer angle \u03b4. Rotation about the xt-axis is the camber angle \u03b3. 488 8. Suspension Mechanisms Generally speaking, the steering axis may have any angle and may go through any point of the ground plane. Example 327 Wheel to tire coordinate frame transformation. If TdW indicates the T -expression of the position vector of the wheel frame origin relative to the tire frame origin, then having the coordinates of a point P in the wheel frame, we can find its coordinates in the tire frame using the following equation. T rP = TRW W rP + TdW (8.28) If W rP indicates the position vector of a point P in the wheel frame, W rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 (8.29) then the coordinates of the point P in the tire frame T rP are T rP = TRW W rP + Td = TRW W rP + TRW W T dW = \u23a1\u23a3 xP yP cos \u03b3 \u2212Rw sin \u03b3 \u2212 zP sin \u03b3 Rw cos \u03b3 + zP cos \u03b3 + yP sin \u03b3 \u23a4\u23a6 (8.30) 8. Suspension Mechanisms 489 where, WT dW is the W -expression of the position vector of the wheel frame in the tire frame, Rw is the radius of the tire, and TRW is the rotation matrix to go from the wheel frame W to the tire frame T . TRW = \u23a1\u23a3 1 0 0 0 cos \u03b3 \u2212 sin \u03b3 0 sin \u03b3 cos \u03b3 \u23a4\u23a6 (8.31) W T dW = \u23a1\u23a3 0 0 Rw \u23a4\u23a6 . (8.32) As an example, the center of the wheel W rP = W ro = 0 is the origin of the wheel frame W , that is at T ro = TdW = TRW W T dW = \u23a1\u23a3 0 \u2212Rw sin \u03b3 Rw cos \u03b3 \u23a4\u23a6 (8.33) in the tire coordinate frame T . 490 8. Suspension Mechanisms Example 328 F Tire to wheel coordinate frame transformation. If rP indicates the position vector of a point P in the tire coordinate frame, T rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 (8.34) then the position vector W rP of the point P in the wheel coordinate frame is W rP = WRT T rP \u2212 W T dW (8.35) = \u23a1\u23a3 xP yP cos \u03b3 + zP sin \u03b3 zP cos \u03b3 \u2212Rw \u2212 yP sin \u03b3 \u23a4\u23a6 because WRT = \u23a1\u23a3 1 0 0 0 cos \u03b3 sin \u03b3 0 \u2212 sin \u03b3 cos \u03b3 \u23a4\u23a6 (8.36) WdT = \u23a1\u23a3 0 0 Rw \u23a4\u23a6 (8.37) and we may multiply both sides of Equation (8.28) by TRT W to get TRT W T rP = W rP + TRT W TdW (8.38) = W rP + W T dW W rP = WRT T rP \u2212 W T dW . (8.39) As an example, the center of tireprint in the wheel frame is at W rP = \u23a1\u23a3 1 0 0 0 cos \u03b3 \u2212 sin \u03b3 0 sin \u03b3 cos \u03b3 \u23a4\u23a6T \u23a1\u23a3 0 0 0 \u23a4\u23a6\u2212 \u23a1\u23a3 0 0 Rw \u23a4\u23a6 = \u23a1\u23a3 0 0 \u2212Rw \u23a4\u23a6 . (8.40) Example 329 F Wheel to tire homogeneous transformation matrices. The transformation from the wheel to tire coordinate frame may also be expressed by a 4\u00d7 4 homogeneous transformation matrix TTW , T rP = TTW W rP = \u2219 TRW TdW 0 1 \u00b8 W rP (8.41) where TTW = \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 cos \u03b3 \u2212 sin \u03b3 \u2212Rw sin \u03b3 0 sin \u03b3 cos \u03b3 Rw cos \u03b3 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 . (8.42) 8. Suspension Mechanisms 491 The corresponding homogeneous transformation matrix WTT from the tire to wheel frame would be WTT = \u2219 WRT WdT 0 1 \u00b8 = \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 cos \u03b3 \u2212 sin \u03b3 0 0 sin \u03b3 cos \u03b3 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 . (8.43) It can be checked that WTT = TT\u22121W , using the inverse of a homogeneous transformation matrix rule. TT\u22121W = \u2219 TRW TdW 0 1 \u00b8\u22121 = \u2219 TRT W \u2212 TRT W TdW 0 1 \u00b8 = \u2219 WRT \u2212WRT TdW 0 1 \u00b8 (8.44) Example 330 F Tire to wheel-body frame transformation. The origin of the tire frame is at CdT in the wheel-body frame. CdT = \u23a1\u23a3 0 0 \u2212Rw \u23a4\u23a6 (8.45) The tire frame can steer about the zc-axis with respect to the wheel-body frame. The associated rotation matrix is CRT = \u23a1\u23a3 cos \u03b4 \u2212 sin \u03b4 0 sin \u03b4 cos \u03b4 0 0 0 1 \u23a4\u23a6 (8.46) Therefore, the transformation between the tire and wheel-body frames can be expressed by Cr = CRT T r+ CdT (8.47) or equivalently, by a homogeneous transformation matrix CTT . CTT = \u2219 CRT CdT 0 1 \u00b8 = \u23a1\u23a2\u23a2\u23a3 cos \u03b4 \u2212 sin \u03b4 0 0 sin \u03b4 cos \u03b4 0 0 0 0 1 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 (8.48) As an example, the wheel-body coordinates of the point P on the tread of a 492 8. Suspension Mechanisms negatively steered tire at the position shown in Figure 8.48, are: Cr = CTT T rP = \u23a1\u23a2\u23a2\u23a3 cos\u2212\u03b4 \u2212 sin\u2212\u03b4 0 0 sin\u2212\u03b4 cos\u2212\u03b4 0 0 0 0 1 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 \u23a1\u23a2\u23a2\u23a3 Rw 0 Rw 1 \u23a4\u23a5\u23a5\u23a6 = \u23a1\u23a2\u23a2\u23a3 Rw cos \u03b4 \u2212Rw sin \u03b4 0 1 \u23a4\u23a5\u23a5\u23a6 (8.49) The homogeneous transformation matrix for tire to wheel-body frame TTC is: TTC = CT\u22121T = \u2219 CRT CdT 0 1 \u00b8\u22121 = \u2219 CRT T \u2212CRT T CdT 0 1 \u00b8 = \u2219 CRT T \u2212 T CdT 0 1 \u00b8 = \u23a1\u23a2\u23a2\u23a3 cos \u03b4 sin \u03b4 0 0 \u2212 sin \u03b4 cos \u03b4 0 0 0 0 1 Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 (8.50) Example 331 F Cycloid. Assume that the wheel in Figure 8.48 is turning with angular velocity \u03c9 and has no slip on the ground. If the point P is at the center of the tireprint 8. Suspension Mechanisms 493 when t = 0, MrP = \u23a1\u23a3 0 0 \u2212Rw \u23a4\u23a6 (8.51) then we can find its position in the wheel frame at a time t by employing another coordinate frame M . The frame M is called the rim frame and is stuck to the wheel at its center. Because of spin, the M frame turns about the yw-axis, and therefore, the rotation matrix to go from the rim frame to the wheel frame is: WRM = \u23a1\u23a3 cos\u03c9t 0 sin\u03c9t 0 1 0 \u2212 sin\u03c9t 0 cos\u03c9t \u23a4\u23a6 (8.52) So the coordinates of P in the wheel frame are: W rP = WRM MrP = \u23a1\u23a3 \u2212Rw sin t\u03c9 0 \u2212Rw cos t\u03c9 \u23a4\u23a6 (8.53) The center of the wheel is moving with speed vx = Rw\u03c9 and it is at Gr =\u00a3 vxt 0 Rw \u00a4 in the global coordinate frame G on the ground. Hence, the coordinates of point P in the global frame G, would be GrP = W rP + \u23a1\u23a3 vxt 0 Rw \u23a4\u23a6 = \u23a1\u23a3 Rw (\u03c9t\u2212 sin t\u03c9) 0 Rw (1\u2212 cos t\u03c9) \u23a4\u23a6 . (8.54) The path of motion of point P in the (X,Z)-plane can be found by eliminating t between X and Z coordinates. However, it is easier to expressed the path by using \u03c9t as a parameter. Such a path is called cycloid. In general case, point P can be at any distance from the center of the rim frame. If the point is at a distance d 6= Rw, then its path of motion is called the trochoid. A trochoid is called a curtate cycloid when d < Rw and a prolate cycloid when d > Rw. Figure 8.49(a)-(c) illustrate a cycloid, curtate cycloid, and prolate cycloid respectively. Example 332 F Wheel to wheel-body frame transformation. The homogeneous transformation matrix CTW to go from the wheel frame 494 8. Suspension Mechanisms to the wheel-body frame can be found by combined transformation. CTW = CTT TTW (8.55) = \u23a1\u23a2\u23a2\u23a3 c\u03b4 \u2212s\u03b4 0 0 s\u03b4 c\u03b4 0 0 0 0 1 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 c\u03b3 \u2212s\u03b3 \u2212Rw sin \u03b3 0 s\u03b3 c\u03b3 Rw cos \u03b3 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 = \u23a1\u23a2\u23a2\u23a3 cos \u03b4 \u2212 cos \u03b3 sin \u03b4 sin \u03b3 sin \u03b4 Rw sin \u03b3 sin \u03b4 sin \u03b4 cos \u03b3 cos \u03b4 \u2212 cos \u03b4 sin \u03b3 \u2212Rw cos \u03b4 sin \u03b3 0 sin \u03b3 cos \u03b3 Rw cos \u03b3 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 If rP indicates the position vector of a point P in the wheel coordinate frame, W rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 (8.56) then the homogeneous position vector CrP of the point P in the wheel-body 8. Suspension Mechanisms 495 coordinate frame is: CrP = CTW W rP = \u23a1\u23a2\u23a2\u23a3 xP cos \u03b4 \u2212 yP cos \u03b3 sin \u03b4 + (Rw + zP ) sin \u03b3 sin \u03b4 xP sin \u03b4 + yP cos \u03b3 cos \u03b4 \u2212 (Rw + zP ) cos \u03b4 sin \u03b3 \u2212Rw + (Rw + zP ) cos \u03b3 + yP sin \u03b3 1 \u23a4\u23a5\u23a5\u23a6 (8.57) The position of the wheel center W r = 0, for a cambered and steered wheel is at Cr = CTW W r = \u23a1\u23a2\u23a2\u23a3 Rw sin \u03b3 sin \u03b4 \u2212Rw cos \u03b4 sin \u03b3 \u2212Rw(1\u2212 cos \u03b3) 1 \u23a4\u23a5\u23a5\u23a6 (8.58) The zc = Rw (cos \u03b3 \u2212 1) indicates how much the center of the wheel comes down when the wheel cambers. If the wheel is not steerable, then \u03b4 = 0 and the transformation matrix CTW reduces to CTW = \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 cos \u03b3 \u2212 sin \u03b3 \u2212Rw sin \u03b3 0 sin \u03b3 cos \u03b3 Rw (cos \u03b3 \u2212 1) 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 (8.59) that shows CrP = CTW W rP = \u23a1\u23a2\u23a2\u23a3 xP yP cos \u03b3 \u2212Rw sin \u03b3 \u2212 zP sin \u03b3 zP cos \u03b3 + yP sin \u03b3 +Rw (cos \u03b3 \u2212 1) 1 \u23a4\u23a5\u23a5\u23a6 (8.60) Example 333 F Tire to vehicle coordinate frame transformation. Figure 8.50 illustrates the first and fourth tires of a 4-wheel vehicle. There is a body coordinate frame B (x, y, z) attached to the mass center C of the vehicle. There are also two tire coordinate frames T1 (xt1 , yt1 , zt1) and T4 (xt4 , yt4 , zt4) attached to the tires 1 and 4 at the center of their tireprints. The origin of the tire coordinate frame T1 is at Bd1 BdT1 = \u23a1\u23a3 a1 \u2212b1 \u2212h \u23a4\u23a6 (8.61)"
        ]
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure7.12-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure7.12-1.png",
        "caption": "Figure 7.12 The forces acting on the ring beam.",
        "texts": [
            " Here too, the (distributed) tensile forces are independent of the aperture angle \u03b1. They are, however, half as large as the circumferential tensile forces in the circular cylindrical pneu from the previous example. 252 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM In this example n = 1 2pr = 1 2 \u00d7 (268 N/m2)(2200 m) = 295 kN/m. The aperture angle \u03b1 is (see Figure 7.11) \u03b1 = arccos ( 1960 m 2200 m ) = 27\u25e6. At the ring beam, the horizontal and vertical components of n are nh = n cos \u03b1 = (295 kN/m) \u00d7 cos 27\u25e6 = 263 kN/m, nv = n sin \u03b1 = (295 kN/m) \u00d7 sin 27\u25e6 = 134 kN/m. Figure 7.12 shows all the forces acting on the ring beam. These are the distributed force n, which the pneu exerts on the ring beam, and the dead weight qdw of the ring beam (also a force per length). The vertical component nv tries to lift the ring beam. In order to prevent this, the dead weight qdw has to be larger than nv = 134 kN/m. If the ring beam is made of concrete, with a specific weight of 24 kN/m3, then the cross-section A of the beam has to obey qdw = A \u00d7 (24 kN/m3) \u2265 nv = 134 kN/m \u21d2 A \u2265 134 kN/m 24 kN/m3 = 5"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000457_j.sna.2017.12.028-Figure8-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000457_j.sna.2017.12.028-Figure8-1.png",
        "caption": "Fig. 8. Schematic diagram of the formation of stretchable graphene nano-papers [178].",
        "texts": [
            " / Sensors and Actuators A 270 (2018) 177\u2013194 185 t c w t i a d d f he electrical resistance with the movement of the throat musles. After the response of the sensor for every English phoneme as analyzed to differentiate its output, different words were tried o validate the developed sensor further. There were other studes reported on GWF where even a G.F. factor up to \u223c105 was chieved. Graphene, in the form of nano-papers, were also used to evelop strain sensors [24,177,178]. The sensor patch was used to etect human movements. Fig. 8 gives the schematic diagram of the abrication steps of the nano-cellulose based sensor patch. Crum- pled graphene and nano-cellulose fibril were mixed at a weight ratio of 1:1 on a polycarbonate membrane. Following filtration of the embedded nanocomposite, the thin film was peeled off and impregnated with PDMS to develop the stretchable nano-papers. The movement of the fingers caused a 3D movement of the sensor leading to a maximum strain exceeding 50%. Table 6 depicts the some of the techniques used to develop the strain sensors with graphene electrodes along with their strain percent and gauge factor"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-FigureC.4-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-FigureC.4-1.png",
        "caption": "FIGURE C.4 Exact partial mutual inductance for filament and sheet.",
        "texts": [
            " The result for the tube conductor is given by 388 COMPUTATION OF PARTIAL INDUCTANCES Lp11 = \ud835\udf070\ud835\udcc1 4 [( k2 480 + k4 1280 + 1 3600 ) \ud835\udf0b 3 + ( 1 18 \u2212 k2 24 ) \ud835\udf0b + ( 2 \u2212 2 log(\ud835\udcc1) + 6 log(2) + 2 log(a) \u2212 4 log(k\ud835\udf0b) ) 1 \ud835\udf0b + 8 a \ud835\udcc1 \ud835\udf0b2 ] , (C.17) where k = 2a\u2215\ud835\udcc1. We should note that the equation is also used in equation (9.29) for skin-effect models [9]. C.1.3 Lp12 for Filament and Current Sheet Another important partial mutual inductance represents the case where one conductor is approximated with a filament and the other with a zero thickness sheet as shown in Fig. C.4. Of course, the current in the sheet is in the same x-direction as in the wire as is shown in Fig. C.4 and the integral is given by Lp12 = 1 (ye1 \u2212 ys1) \ud835\udf07 4\ud835\udf0b \u222b ye1 ys1 \u222b xe1 xs1 \u222b xe2 xs2 1 R1,2 dx2 dx1 dy1. (C.18) The analytical solution for the above integrals is Lp12 = \ud835\udf070 4\ud835\udf0b 1 w1 4\u2211 k=1 2\u2211 m=1 (\u22121)k+m [ a2 k \u2212 Z2 2 log(bm + rkm + \ud835\udf16) + ak bm log(ak + rkm + \ud835\udf16) \u2212 ak Z tan\u22121 ( ak bm Z rkm ) \u2212 bm 2 rkm ] (C.19) with a1 = xs2 \u2212 xe1, a2 = xe2 \u2212 xe1 (C.20) a3 = xe2 \u2212 xs1, a4 = xs2 \u2212 xs1 (C.21) PARTIAL INDUCTANCE FORMULAS FOR ORTHOGONAL GEOMETRIES 389 b1 = y2 \u2212 ys1, b2 = y2 \u2212 ye1 (C.22) Z = z2 \u2212 z1, rkm = \u221a a2 k + b2 m + Z2 "
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003628_91.705510-Figure15-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003628_91.705510-Figure15-1.png",
        "caption": "Fig. 15. Structure of a ball-beam system.",
        "texts": [
            " In the simulation, the following parameters are used: m, m, m/s , kg, kg, kg, , , , , , , . The initial values are: , , , , . Figs. 11\u201314 show the simulation runs with . The angles of both poles converge to zero eventually. One point needs special attention: although the poles can be balanced, the cart was not forced back to the origin. Nevertheless, Fig. 14 shows that , indicating the cart was not experiencing any external force when the poles were stabilized. Consider a ball-beam system as the one depicted in Fig. 15. The mathematical expression of this system can be written as (39) where angle of the beam with respect to the horizontal axis; angular velocity of the beam with respect to the horizontal axis; position of the ball; velocity of the ball; ; moment of inertia of the ball; mass of the ball; radius of the ball; acceleration of gravity. The center of rotation is assumed to be frictionless and the ball is free to roll along the beam. It is required that the ball remains in contact with the beam and that the rolling occurs without slipping"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001146_978-3-642-54536-8-Figure6.21-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001146_978-3-642-54536-8-Figure6.21-1.png",
        "caption": "Fig. 6.21 Two links in space",
        "texts": [
            " The second, and more intrinsic problem is to solve (6.39), a huge simultaneous linear equations. Although the efficient algorithms like Gauss-Jordan elimination or LU decomposition are known [136], it always takes time proportional to (n+ 6)3 and it becomes a serious bottle-neck for faster forward dynamics calculation. In the next section, we introduce the method to solve this problem. 6.4 Dynamics of Link System 203 Suppose we have two links connected by a revolutional joint floating in 3D space as shown in Fig. 6.21. Assuming that we have already known the spatial acceleration of link1, we want to derive the joint acceleration q\u03082 for a given torque u2. The spatial acceleration of the link2 is given by \u03be\u03072 = \u03be\u03071 + s\u03072q\u03072 + s2q\u03082. (6.43) From the acceleration, we can calculate the necessary force and moment by using the equation of motion. [ f2 \u03c4 2 ] = IS 2 \u03be\u03072 + \u03be2 \u00d7 IS 2 \u03be2 (6.44) Moreover, the joint torque u2 is a projection of this force and moment u2 = sT2 [ f2 \u03c4 2 ] . (6.45) Substituting (6.43) and (6"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure6.28-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure6.28-1.png",
        "caption": "FIGURE 6.28. The roll center of a double A-arm suspension and its equivalent kinematic model.",
        "texts": [
            "218) The four instant centers of rotation for a four-bar linkage, a slider-crank, and an inverted slider-crank mechanisms are shown in Figure 6.26. The instant center of rotation for two links that slide on each other is at infinity, 350 6. Applied Mechanisms on a line normal to the common tangent. So, I14 in Figure 6.26(b) is on a line perpendicular to the ground, and I34 in Figure 6.26(c) is on perpendicular to the link 3. Figure 6.27 depicts the 15 instant centers for a six-link mechanism. Example 244 Application of instant center of rotation in vehicles. Figure 6.28 illustrates a double A-arm suspension and its equivalent kinematic model. The wheel will be fastened to the coupler link AB, witch connects the upper A-arm BN to the lower A-arm AN . The A-arms are connected to the body with two revolute joints at N and M . The body of the vehicle acts as the ground link for the suspension mechanism, which is a four-bar linkage. Points N and M are, respectively, the instant centers of rotation for the upper and lower arms with respect to the body. The intersection point of the extension line for the upper and lower A-arms indicates the instant center of rotation for the coupler with respect to the body"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure6.37-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure6.37-1.png",
        "caption": "FIGURE 6.37. A slider-crank mechanism and a coupler point at C.",
        "texts": [
            " When the wheel moves up and down, the wheel center moves on a the couple point curve shown in the figure. The wheel\u2019s center of proper suspension mechanism is supposed to move vertically, however, the wheel center of the suspension moves on a high curvature path and generates an undesired camber. A small motion of the kinematic model of suspension is shown in Figure 6.35, and the actual suspension and wheel configurations are shown in Figure 6.36. 6. Applied Mechanisms 359 360 6. Applied Mechanisms 6.5.2 Coupler Point Curve for a Slider-Crank Mechanism Figure 6.37 illustrates a slider-crank mechanism and a coupler point at C. When the mechanism moves, coupler point C will move on a coupler point curve with the following parametric equation: xC = a cos \u03b82 + c cos (\u03b1\u2212 \u03b3) (6.262) yC = a sin \u03b82 + c sin (\u03b1\u2212 \u03b3) (6.263) The angle \u03b82 is the input angle and acts as a parameter, and angle \u03b3 can be calculated from the following equation. \u03b3 = sin\u22121 a sin \u03b82 \u2212 e b (6.264) Proof. We attach a planar Cartesian coordinate frame to the ground link at M . The x-axis is parallel to the ground indicated by the sliding surface, as shown in Figure 6.37. Drawing a line l through A and parallel to the ground shows that \u03b2 = \u03b1\u2212 \u03b3 (6.265) where \u03b3 is the angle between the coupler link and the ground. The coordinates (xC , yC) for point C are xC = a cos \u03b82 + c cos\u03b2 (6.266) yC = a sin \u03b82 + c sin\u03b2. (6.267) 6. Applied Mechanisms 361 To calculate the angle \u03b3, we examine 4AEB and find sin \u03b3 = AE AB = a sin \u03b82 \u2212 e b (6.268) that finishes Equation (6.264). Therefore, the coordinates xC and yC can be calculated as two parametric functions of \u03b82 for a given set of a, b, c, e, and \u03b1"
        ],
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    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure12.13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure12.13-1.png",
        "caption": "FIGURE 12.13. An inverted pendulum with a tip mass m and two supportive springs.",
        "texts": [
            "48) using any other assumption, such as xs < xu > y, xs > xu < y, or xs < xu < y. However, having an assumption helps to make a consistent free body diagram. We usually rearrange the equations of motion for a linear system in a matrix form [M ] x\u0307+ [c] x\u0307+ [k]x = F (12.49) to take advantage of matrix calculus. Rearrangement of Equations (12.47) and (12.48) results in the following set of equations:\u2219 ms 0 0 mu \u00b8 \u2219 x\u0308s x\u0308u \u00b8 + \u2219 cs \u2212cs \u2212cs cs + cu \u00b8 \u2219 x\u0307s x\u0307u \u00b8 +\u2219 ks \u2212ks \u2212ks ks + ku \u00b8 \u2219 xs xu \u00b8 = \u2219 0 kuy + cuy\u0307 \u00b8 (12.50) Example 431 F Inverted pendulum. Figure 12.13(a) illustrates an inverted pendulum with a tip mass m and a length l. The pendulum is supported by two identical springs attached to point B at a distance a < l from the pivot A. A free body diagram of the pendulum is shown in Figure 12.13(b). The equation of motion may be found by taking a moment about A. X MA = IA\u03b8\u0308 (12.51) mg (l sin \u03b8)\u2212 2ka\u03b8 (a cos \u03b8) = ml2\u03b8\u0308 (12.52) 12. Applied Vibrations 743 To derive Equation (12.52) we assumed that the springs are long enough to remain almost straight when the pendulum oscillates. Rearrangement and assuming a very small \u03b8 shows that the nonlinear equation of motion (12.52) can be approximated by ml2\u03b8\u0308 + \u00a1 mgl \u2212 2ka2 \u00a2 \u03b8 = 0 (12.53) which is equivalent to a linear oscillator meq \u03b8\u0308 + keq\u03b8 = 0 (12"
        ],
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    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure10.14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure10.14-1.png",
        "caption": "FIGURE 10.14. (a)- A 180 deg, fast turning maneuver from reverse. (b)- A 180deg, fast turning maneuver from forward.",
        "texts": [
            " To make a fast 180 deg turn without stopping, the driver may follow these steps: 1\u2212 The driver should push the gas pedal to gain enough speed, 2\u2212 free the gas pedal and put the gear in neutral, 3\u2212 cut the steering wheel sharply around 90 deg, 4\u2212 change the gear to drive, and 5\u2212 push the gas pedal and return the steering wheel to 0 deg after the car has completed the 180 deg turn. The backward speed before step 2 may be around 20m/ s \u2248 70 km/h \u2248 45mi/h. Steps 2 to 4 should be done fast and almost simultaneously. Figure 10.14(a) illustrates the 180 deg fast turning maneuver from reverse. This example should never be performed by the reader of this book. Example 389 F The race care 180 deg quick turn from forward. You have seen that race car drivers can turn a 180 deg quickly when the car is moving forward. Here is how they do this. The driver moves forward when the car is in drive or a forward gear. To make a fast 180 deg turn without stopping, the driver may follow these steps: 1\u2212 The driver should push the gas pedal to gain enough speed, 2\u2212 free the gas pedal and put the gear in neutral, 3\u2212 cut the steering wheel sharply around 90 deg while 10. Vehicle Planar Dynamics 619 620 10. Vehicle Planar Dynamics you pull hard the hand brake, 4\u2212 while the rear swings around, return the steering wheel to 0 deg and put the gear into drive, and 5\u2212 push the gas pedal after the car has completed the 180 deg turn. Figure 10.14(b) illustrates this maneuver. The forward speed before step 2 may be around 20m/ s \u2248 70 km/h \u2248 45mi/h. Steps 2 to 4 should be done fast and almost simultaneously. The 180 deg fast turning from forward is more difficult than backward and can be done because the hand brakes are connected to the rear wheels. It can be done better when the rear of a car is lighter than the front to slides easier. Road condition, nonuniform friction, slippery surface can cause flipping the car and spinning out of control"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001514_j.engfailanal.2015.06.004-Figure19-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001514_j.engfailanal.2015.06.004-Figure19-1.png",
        "caption": "Fig. 19. FE model of crack gear [56]: (a) meshing gear, (b) constraint and loading of the gear model.",
        "texts": [
            " [42] established a FE model of a cracked tooth in NX8 software, and calculated the mesh stiffness of the single tooth by applying a single force on the theoretical contact point (see Fig. 18), then obtained the TVMS of the gear pair by considering the effects of alternating single-tooth and double-tooth contact. Chaari et al. [30] established a 3D FE model of a spur gear with one tooth to improve the computational efficiency, and evaluated the TVMS of a cracked gear pair by applying the concentrated force. Based on a 3D quasi-static FE model (see Fig. 19), Zouari et al. [56] calculated the TVMS of a spur gear pair with a tooth root crack by applying a constant load along the tooth width, and analyzed the influence of the crack sizes and crack propagation directions along the tooth root on the TVMS. Based on the static analysis of MSC.MARC software and Loaded Tooth Contact Analysis (LTCA) [57], Endo et al. [58\u201360] established the FE models of gears with spalling and tooth crack in which the elastic deformations of the engaging teeth and the gear body as well as the Hertzian deformation are considered"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0003586_s0731-7085(98)00056-9-Figure2-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0003586_s0731-7085(98)00056-9-Figure2-1.png",
        "caption": "Fig. 2. Schematic of an implantable needle-type glucose sensor (reproduced from Zhang and Wilson [14]).",
        "texts": [
            " Such monitoring requires special adaptation of biosensors in term of miniaturization, biocompatability (involving both the tissue and blood response), enzyme stability, cofactor dependency (oxygen deficit), drift, in vivo calibration, safety, or convenience. In spite of these major challenges, significant progress has been made towards the continuous monitoring of physiologically important molecules, such as glucose, lactate or glutamate. Most of the attention has been given to the probing of blood glucose levels as an aid to diabetes therapy [12,13]. In particular, subcutaneously implantable needle-type amperometric glucose sensors, being developed in various laboratories (e.g. Fig. 2 [14]), should offer hypo- or hyperglycemia alarm capabilities to enable swift and appropriate corrective action (through a closed-loop insulin delivery system, i.e. an artificial pancreas). While success in this research has reached the level of short-term human implantation, a continuously functioning implantable glucose sensor possessing long-term stability has not been realized. Besides the obvious biocompatabil- ity (biofouling) challenge, such implantable glucose sensors may be prone to errors due to low oxygen tension or electroactive interferences"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0000199_978-0-387-74244-1-Figure8.21-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0000199_978-0-387-74244-1-Figure8.21-1.png",
        "caption": "FIGURE 8.21. A McPherson suspension mechanism on the left and right wheels.",
        "texts": [],
        "surrounding_texts": [
            "The suspension is what links the wheels to the vehicle body and allows relative motion. This chapter covers the suspension mechanisms, and discusses the possible relative motions between the wheel and the vehicle body. The wheels, through the suspension linkage, must propel, steer, and stop the vehicle, and support the associated forces. 8.1 Solid Axle Suspension The simplest way to attach a pair of wheels to a vehicle is to mount them at opposite ends of a solid axle, such as the one that is shown in Figure 8.1. The solid axle must be attached to the body such that an up and down motion in the z-direction, as well as a roll rotation about the x-axis, is possible. So, no forward and lateral translation, and also no rotation about the axle and the z-axis, is allowed. There are many combinations of links and springs that can provide the kinematic and dynamic requirements. The simplest design is to clamp the axle to the middle of two leaf springs with their ends tied or shackled to the vehicle frame as shown schematically in Figure 8.1. A side view of a multi-leaf spring and solid axle is shown in Figure 8.2. A suspension with a solid connection between the left and right wheels is called dependent suspension. 456 8. Suspension Mechanisms The performance of a solid axle with leaf springs suspension can be improved by adding a linkage to guide the axle kinematically and provide dynamic support to carry the non z-direction forces. The solid axle with leaf spring combination came to vehicle industry from horse-drawn vehicles. Example 299 Hotchkiss drive. When a live solid axle is connected to the body with nothing but two leaf springs, it is called the Hotchkiss drive, which is the name of the car that used it first. The main problems of a Hotchkiss drive, which is shown in Figure 8.2, are locating the axle under lateral and longitudinal forces, and having a low mass ratio \u03b5 = ms/mu, where ms is the sprung mass and mu is the unsprung mass. Sprung mass refers to all masses that are supported by the spring, such as vehicle body. Unsprung mass refers to all masses that are attached to and not supported by the spring, such as wheel, axle, or brakes. Example 300 Leaf spring suspension and flexibility problem. The solid axle suspension systems with longitudinal leaf springs have many drawbacks. The main problem lies in the fact that springs themselves act as locating members. Springs are supposed to flex under load, but their flexibility is needed in only one direction. However, it is the nature of leaf springs to twist and bend laterally and hence, flex also in planes other than the tireplane. Leaf springs are not suited for taking up the driving and braking traction forces. These forces tend to push the springs into an S-shaped profile, as shown in Figure 8.3. The driving and braking flexibility of leaf springs, generates a negative caster and increases instability. Long springs provide better ride. However, long sprigs exaggerate their bending and twisting under different load conditions. Example 301 Leaf spring suspension and flexibility solution. To reduce the effect of a horizontal force and S-shaped profile appearance in a solid axle with leaf springs, the axle may be attached to the chassis by a longitudinal bar as Figure 8.4(a) shows. Such a bar is called an anti-tramp 8. Suspension Mechanisms 457 bar, and the suspension is the simplest cure for longitudinal problems of a Hotchkiss drive. A solid axle with an anti-tramp bar may be kinematically approximated by a four-bar linkage, as shown in Figure 8.4(b). Although an anti-tramp bar may control the shape of the leaf spring, it introduces a twisting angle problem when the axle is moving up and down, as shown in Figure 8.5. Twisting the axle and the wheel about the axle is called caster. The solid axle is frequently used to help keeping the wheels perpendicular to the road. Example 302 Leaf spring location problem. The front wheels need room to steer left and right. Therefore, leaf springs cannot be attached close to the wheel hubs, and must be placed closer to the middle of the axle. That gives a narrow spring-base, which means that a small side force can sway or tilt the body relative to the axle through a considerable roll angle due to weight transfer. This is uncomfortable for the vehicle passengers, and may also produce unwanted steering. The solid axle positively prevents the camber change by body roll. The wheels remain upright and hence, do not roll on a side. However, a solid axle shifts laterally from its static plane and its center does not remain on the vehicle\u2019s longitudinal axis under a lateral force. 458 8. Suspension Mechanisms A solid axle produces bump-camber when single-wheel bump occurs. If the right wheel goes over a bump, the axle is raised at its right end, and that tilts the left wheel hub, putting the left wheel at a camber angle for the duration of deflection. Example 303 Triangular linkage. A triangulated linkage, as shown in Figure 8.6, may be attached to a solid axle to provide lateral and twist resistance during acceleration and braking. Example 304 Panhard arm. High spring rate is a problem of leaf springs. Reducing their stiffness by narrowing them and using fewer leaves, reduces the lateral stiffness and increases the directional stability of the suspension significantly. A Panhard arm is a bar that attaches a solid axle suspension to the chassis laterally. Figure 8.7 illustrates a solid axle and a Panhard arm to guide the axle. Figure 8.8 shows a triangular linkage and a Panhard arm combination for guiding a solid axle. A double triangle mechanism, as shown in Figure 8.9, is an alternative design to guide the axle and support it laterally. Example 305 Straight line linkages. There are many mechanisms that can provide a straight line motion. The simplest mechanisms are four-bar linkages with a coupler point moving straight. Some of the most applied and famous linkages are shown in Figure 8.10. By having proper lengths, the Watt, Robert, Chebyshev, and Evance linkages can make the coupler point C move on a straight line vertically. Such a mechanism and straight motion may be used to guide a solid axle. Two Watt suspension mechanisms with a Panhard arm are shown in Figures 8.11 and 8.12. 460 8. Suspension Mechanisms 8. Suspension Mechanisms 461 Figures 8.13, 8.14, and 8.15 illustrate three combinations of Robert suspension linkages equipped with a Panhard arm. Example 306 Solid axle suspension and unsprung mass problem. A solid axle is counted as an unsprung member, and hence, the unsprung mass is increased where using solid axle suspension. A heavy unsprung mass ruins both, the ride and handling of a vehicle. Lightening the solid axle makes it weaker and increases the most dangerous problem in vehicles: axle breakage. The solid axle must be strong enough to make sure it will not break under any loading conditions at any age. As a rough estimate, 90% of the leaf spring mass may also be counted as unsprung mass, which makes the problem worse. The unsprung mass problem is worse in front, and it is the main reason that they are no longer used in street cars. However, front solid axles are still common on trucks and buses. These are heavy vehicles and solid axle 8. Suspension Mechanisms 463 464 8. Suspension Mechanisms suspension does not reduce the mass ratio \u03b5 = ms/mu very much. When a vehicle is rear-wheel-drive and a solid axle suspension is used in the back, the suspension is called live axle. A live axle is a casing that contains a differential, and two drive shafts. The drive shafts are connected to the wheel hubs. A live axle can be three to four times heavier than a dead I-beam axle. It is called live axle because of rotating gears and shafts inside the axle. Example 307 Solid axle and coil spring. To decrease the unsprung mass and increase vertical flexibility of solid axle suspensions, it is possible to equip them with coil springs. A sample of a solid axle suspension with coil spring is shown in Figure 8.16. The suspension mechanism is made of four longitudinal bars between the axle and chassis. The springs may have some lateral or longitudinal angle to introduce some lateral or longitudinal compliance. Example 308 De Dion axle. When a solid axle is a dead axle with no driving wheels, the connecting beam between the left and right wheels may have different shapes to do different jobs, usually to give the wheels independent flexibility. We may also modify the shape of a live axle to attach the differential to the chassis and reduce the unsprung mass. De Dion design is a modification of a beam axle that may be used as a dead axle or to attach the differential to the chassis and transfer the driving power to the drive wheels by employing universal joints and split shafts. Figure 8.17 illustrates a De Dion suspension. 8.2 Independent Suspension Independent suspensions is introduced to let a wheel to move up and down without affecting the opposite wheel. There are many forms and designs of independent suspensions. However, double A-arm and McPherson strut suspensions are the simplest and the most common designs. Figure 8.18 illustrates a sample of a double A-arm and Figure 8.19 shows a McPherson suspension. Kinematically, a double A-arm suspension mechanism is a four-bar linkage with the chassis as the ground link, and coupler as the wheel carrying link. A McPherson suspension is an inverted slider mechanism that has the chassis as the ground link and the coupler as the wheel carrying link. A double A-arm and a McPherson suspension mechanism on the left and right wheels are schematically shown in Figures 8.20 and 8.21 respectively. Double A-arm, is also called double wishbone, or short/long arm suspension. McPherson also may be written as MacPherson. Example 309 Double A-arm suspension and spring position. Consider a double A-arm suspension mechanism. The coil spring may be between the lower arm and the chassis, as shown in Figure 8.18. It is also possible to install the spring between the upper arm and the chassis, or between the upper and lower arms. In either case, the lower or the upper arm, which supports the spring, is made stronger and the other arm acts as a connecting arm. Example 310 Multi-link suspension mechanism. When the two side bars of an A-arm are attached to each other with a joint, as shown in Figure 8.22, then the double A-arm is called a multi-link mechanism. A multi-link mechanism is a six-bar mechanism that may have a better coupler motion than a double A-arm mechanism. However, multi- 466 8. Suspension Mechanisms 8. Suspension Mechanisms 467 link suspensions are more expensive, less reliable, and more complicated compare to a double A-arm four-bar linkage. There are vehicles with more than six-link suspension with possibly better kinematic performance. Example 311 Swing arm suspension. An independent suspension may be as simple as a triangle shown in Figure 8.23. The base of the triangle is jointed to the chassis and the wheel to the tip point. The base of the triangle is aligned with the longitudinal axis of the vehicle. Such a suspension mechanism is called a swing axle or swing arm. The variation in camber angle for a swing arm suspension is maximum, compared to the other suspension mechanisms. Example 312 Trailing arm suspension. Figure 8.24 illustrates a trailing arm suspension that is a longitudinal arm with a lateral axis of rotation. The camber angle of the wheel, supported by a trailing arm, will not change during the up and down motion. Trailing arm suspension has been successfully using in a variety of frontwheel-drive vehicles, to suspend their rear wheels. Example 313 Semi-trailing arm Semi-trailing arm suspension, as shown in Figure 8.25, is a compromise between the swing arm and trailing arm suspensions. The joint axis may have any angle, however an angle not too far from 45 deg is more applied. Such suspensions have acceptable camber angle change, while they can handle both, the lateral and longitudinal forces. Semi-trailing design has successfully applied to a series of rear-wheel-drive cars for several decades. Example 314 Antiroll bar and roll stiffness. Coil springs are used in vehicles because they are less stiff with better 470 8. Suspension Mechanisms ride comfort compared to leaf springs. Therefore, the roll stiffness of the vehicle with coil springs is usually less than in vehicles with leaf springs. To increase the roll stiffness of such suspensions, an antiroll bar must be used. Leaf springs with reduced layers, uni-leaf, trapezoidal, or nonuniform thickness may also need an antiroll bar to compensate for their reduced roll stiffness. The antiroll bar is also called a stabilizer. Figure 8.26 illustrates an anti-roll bare attached to a solid axle with coil springs. Example 315 Need for longitudinal compliance. A bump is an obstacle on the road that opposes the forward motion of a wheel. When a vehicle goes over a bump, the first action is a force that tends to push the wheel backward relative to the rest of the vehicle. So, the lifting force has a longitudinal component, which will be felt inside the vehicle unless the suspension system has horizontal compliance. There are situations in which the horizontal component of the force is even higher than the vertical component. Leaf springs can somewhat absorb this horizontal force by flattening out and stretching the distance from the forward spring anchor and the axle. Such a stretch is usually less than 1/2 in \u2248 1 cm. 8.3 Roll Center and Roll Axis The roll axis is the instantaneous line about which the body of a vehicle rolls. Roll axis is found by connecting the roll center of the front and rear suspensions of the vehicle. Assume we cut a vehicle laterally to disconnect the front and rear half of the vehicle. Then, the roll center of the front or rear suspension is the instantaneous center of rotation of the body with respect to the ground. 8. Suspension Mechanisms 471 472 8. Suspension Mechanisms Figure 8.27 illustrates a sample of the front suspensions of a car with a double A-arm mechanism. To find the roll center of the body with respect to the ground, we analyze the two-dimensional kinematically equivalent mechanism shown in Figure 8.28. The center of tireprint is the instant center of rotation of the wheel with respect to the ground, so the wheels are jointed links to the ground at their center of tireprints. The instant center I18 is the roll center of the body with respect to the ground. To find I18, we apply the Kennedy theorem and find the intersection of the line I12I28 and I13I38 as shown in Figure 8.29. The point I28 and I38 are the instant center of rotation for the wheels with respect to the body. The instant center of rotation of a wheel with respect to the body is called suspension roll center. So, to find the roll center of the front or rear half of a car, we should determine the suspension roll centers, and find the intersection of the lines connecting the suspension roll centers to the center of their associated tireprints. The Kennedy theorem states that the instant center of every three relatively moving objects are colinear. Example 316 McPherson suspension roll center. A McPherson suspension is an inverted slider crank mechanism. The instant centers of an example of inverted slider crank mechanism are shown in Figure 8.30. In this figure, the point I12 is the suspension roll center, which is the instant center of rotation for the wheel link number 2 with respect to the chassis link number 1. A car with a McPherson suspension system is shown in Figure 8.31. The kinematic equivalent mechanism is depicted in Figure 8.32. Suspension roll centers along with the body roll center are shown in Figure 8.33. To 8. Suspension Mechanisms 473 find the roll center of the front or rear half of a car, we determine each suspension roll center and then find the intersection of the lines connecting the suspension roll centers to the center of the associated tireprint. Example 317 Roll center of double A-arm suspension. The roll center of an independent suspension such as a double A-arm can be internal or external. The kinematic model of a double A-arm suspension for the front left wheel of a car is illustrated in Figure 8.34. The suspension roll center in Figure 8.34(a) is internal, and in Figure 8.34(b) is external. An internal suspension roll center is toward the vehicle body, while an external suspension roll center goes away from the vehicle body. A suspension roll center may be on, above, or below the road surface, as shown in Figure 8.35(a)-(c) for an external suspension roll center. When the suspension roll center is on the ground, above the ground, or below the ground, the vehicle roll center would be on the ground, below the ground, and above the ground, respectively. Example 318 F Camber variation of double A-arm suspension. When a wheel moves up and down with respect to the vehicle body, de- 474 8. Suspension Mechanisms 8. Suspension Mechanisms 475 pending on the suspension mechanism, the wheel may camber. Figure 8.36 illustrates the kinematic model for a double A-arm suspension mechanism. The mechanism is equivalent to a four-bar linkage with the ground link as the vehicle chasis. The wheel is always attached to a coupler point C of the mechanism. We set a local suspension coordinate frame (x, y) with the x-axis indicating the ground link MN . The x-axis makes a constant angle \u03b80 with the vertical direction. The suspension machanism has a length a for the upper A-arm, b for the coupler link, c for the lower A-arm, and d for the ground link. The configuration of the suspension is determined by the angles \u03b82, \u03b83, and \u03b84, all measured from the positive direction of the x-axis. When the suspension is at its equilibrium position, the links of the double A-arm suspension make initial angles \u03b820 \u03b830, and \u03b840 with the x-axis. The equilibrium position of a suspension is called the rest position. To determine the camber angle during the fluctuation of the wheel, we should determine the variation of the coupler angle \u03b83, as a function of vertical motion z of the coupler point C. Using \u03b82 as a parameter, we can find the coordinates (xC , yC) of the 476 8. Suspension Mechanisms coupler point C in the suspension coordinate frame (x, y) as xC = a cos \u03b82 + e cos (p\u2212 q + \u03b1) (8.1) yC = a sin \u03b82 + e sin (p\u2212 q + \u03b1) (8.2) where, q = tan\u22121 a sin \u03b82 d\u2212 a cos \u03b82 (8.3) p = tan\u22121 q 4b2f2 \u2212 (b2 + f2 \u2212 c2) 2 b2 + f2 \u2212 c2 (8.4) f = p a2 + d2 \u2212 2ad cos \u03b82. (8.5) The position vector of the coupler point is uC uC = xC \u0131\u0302+ yC j\u0302 (8.6) and the unit vector in the z-direction is u\u0302z = \u2212 cos \u03b80\u0131\u0302\u2212 sin \u03b80j\u0302. (8.7) Therefore, the displacement z in terms of xC and yC is: z = uC \u00b7 u\u0302z = \u2212xC cos \u03b80 \u2212 yC sin \u03b80 (8.8) 8. Suspension Mechanisms 477 The initial coordinates of the coupler point C and the initial value of z are: xC0 = a cos \u03b820 + e cos (p0 \u2212 q0 + \u03b1) (8.9) yC0 = a sin \u03b820 + e sin (p0 \u2212 q0 + \u03b1) (8.10) z0 = \u2212xC0 cos \u03b80 \u2212 yC0 sin \u03b80 (8.11) and hence, the vertical displacement of the wheel center can be calculated by h = z \u2212 z0 (8.12) The initial angle of the coupler link with the vertical direction is \u03b80 \u2212 \u03b830. Therefore, the camber angle of the wheel would be \u03b3 = (\u03b80 \u2212 \u03b83)\u2212 (\u03b80 \u2212 \u03b830) = \u03b830 \u2212 \u03b83 (8.13) The angle of the coupler link with the x-direction is equal to \u03b83 = 2 tan \u22121 \u00c3 \u2212E \u00b1 \u221a E2 \u2212 4DF 2D ! (8.14) where, D = J5 \u2212 J1 + (1 + J4) cos \u03b82 (8.15) E = \u22122 sin \u03b82 (8.16) F = J5 + J1 \u2212 (1\u2212 J4) cos \u03b82 (8.17) and J1 = d a (8.18) J2 = d c (8.19) J3 = a2 \u2212 b2 + c2 + d2 2ac (8.20) J4 = d b (8.21) J5 = c2 \u2212 d2 \u2212 a2 \u2212 b2 2ab . (8.22) Substituting (8.14) and (8.13), and then, eliminating \u03b82 between (8.13) and (8.8) provides the relationship between the vertical motion of the wheel, z, and the camber angle \u03b3. 478 8. Suspension Mechanisms Example 319 F Camber angle and wheel fluctuations. Consider the double A-arm suspension that is shown in Figure 8.36. The dimensions of the equivalent kinematic model are: a = 22.4 cm b = 22.1 cm c = 27.3 cm d = 17.4 cm \u03b80 = 24.3 deg (8.23) The coupler point C is at: e = 14.8 cm \u03b1 = 54.8 deg (8.24) If the angle \u03b82 at the rest position is at \u03b820 = 121.5 deg (8.25) then the initial angle of the other links are: \u03b830 = 18.36 deg \u03b840 = 107.32 deg (8.26) At the rest position, the coupler point is at: xC0 = \u221222.73 cm yC0 = 9.23 cm z0 = 16.92 cm (8.27) We may calculate h and \u03b3 by varying the parameter \u03b82. Figure 8.37 illustrates h as a function of the camber angle \u03b3. For this suspension mechanism, the wheel gains a positive camber when the wheel moves up, and gains a negative camber when the it moves down. The mechanism is shown in Figure 8.38, when the wheel is at the rest position and has a positive or a negative displacement. 8.4 F Car Tire Relative Angles There are four major wheel alignment parameters that affect vehicle dynamics: toe, camber, caster, and trust angle. 8. Suspension Mechanisms 479 8.4.1 F Toe When a pair of wheels is set so that their leading edges are pointed toward each other, the wheel pair is said to have toe-in. If the leading edges point away from each other, the pair is said to have toe-out. Toe-in and toe-out front wheel configurations of a car are illustrated in Figure 8.39. The amount of toe can be expressed in degrees of the angle to which the wheels are not parallel. However, it is more common to express the toein and toe-out as the difference between the track widths as measured at the leading and trailing edges of the tires. Toe settings affect three major performances: tire wear, straight-line stability, and corner entry handling. For minimum tire wear and power loss, the wheels on a given axle of a car should point directly ahead when the car is running in a straight line. Excessive toe-in causes accelerated wear at the outboard edges of the tires, while too much toe-out causes wear at the inboard edges. Toe-in increases the directional stability of the vehicle, and toe-out increases the steering response. Hence, a toe-in setting makes the steering function lazy, while a toe-out makes the vehicle unstable. With four wheel independent suspensions, the toe may also be set at the rear of the car. Toe settings at the rear have the same effect on wear, directional stability, and turn-in as they do on the front. However, we usually do not set up a rear-drive race car toed out in the rear, because of excessive instability. When driving torque is applied to the wheels, they pull themselves forward and try to create toe-in. Furthermore, when pushed down the road, a non-driven wheel or a braking wheel will tend to toe-out. Example 320 Toe-in and directional stability. Toe settings have an impact on directional stability. When the steering 480 8. Suspension Mechanisms 8. Suspension Mechanisms 481 wheel is centered, toe-in causes the wheels to tent to move along paths that intersect each other in front of the vehicle. However, the wheels are in balance and no turn results. Toe-in setup can increase the directional stability caused by little steering fluctuations and keep the car moving straight. Steering fluctuations may be a result of road disturbances. If a car is set up with toe-out, the front wheels are aligned so that slight disturbances cause the wheel pair to assume rolling directions that approach a turn. Therefore, toe-out encourages the initiation of a turn, while toe-in discourages it. Toe-out makes the steering quicker. So, it may be used in vehicles for a faster response. The toe setting on a particular car becomes a trade-off between the straight-line stability afforded by toe-in and the quick steering response by toe-out. Toe-out is not desirable for street cars, however, race car drivers are willing to drive a car with a little directional instability, for sharper turn-in to the corners. So street cars are generally set up with toe-in, while race cars are often set up with toe-out. Example 321 Toe-in and toe-out in the front and rear axles. Front toe-in: slower steering response, more straight-line stability, greater wear at the outboard edges of the tires. Front toe-zero: medium steering response, minimum power loss, minimum tire wear. Front toe-out: quicker steering response, less straight-line stability, greater wear at the inboard edges of the tires. Rear toe-in: straight-line stability, traction out of the corner, more steerability, higher top speed. 482 8. Suspension Mechanisms 8.4.2 F Caster Angle Caster is the angle to which the steering pivot axis is tilted forward or rearward from vertical, as viewed from the side. Assume the wheel is straight to have the body frame and the wheel frame coincident. If the steering axis is turned about the wheel yw-axis then the wheel has positive caster. If the steering axis is turned about the wheel \u2212yw-axis, then the wheel has negative caster. Positive and negative caster configurations on the front wheel of a car are shown in Figure 8.40. Negative caster aids in centering the steering wheel after a turn and makes the front tires straighten quicker. Most street cars are made with 4\u22126 deg negative caster. Negative caster tends to straighten the wheel when the vehicle is traveling forward, and thus is used to enhance straight-line stability. Example 322 Negative caster of shopping carts. The steering axis of a shopping cart wheel is set forward of where the wheel contacts the ground. As the cart is pushed forward, the steering axis pulls the wheel along, and because the wheel drags along the ground, it falls directly in line behind the steering axis. The force that causes the wheel to follow the steering axis is proportional to the distance between the steering axis and the wheel-to-ground contact point, if the caster is small. This distance is referred to as trail. The cars\u2019 steering axis intersects the ground at a point in front of the tireprint, and thus the same effect as seen in the shopping cart casters is achieved. While greater caster angles improves straight-line stability, they also cause an increase in steering effort. Example 323 Characteristics of caster in front axle. Zero castor provides: easy steering into the corner, low steering out of the corner, low straight-line stability. 8. Suspension Mechanisms 483 Negative caster provides: low steering into the corner, easy steering out of the corner, more straight-line stability, high tireprint area during turn, good turn-in response, good directional stability, good steering feel. When a castered wheel rotates about the steering axis, the wheel gains camber. This camber is generally favorable for cornering. 8.4.3 F Camber Camber is the angle of the wheel relative to vertical line to the road, as viewed from the front or the rear of the car. Figure 8.41 illustrates the wheel number 1 of a vehicle. If the wheel leans in toward the chassis, it is called negative camber and if it leans away from the car, it is called positive camber. The cornering force that a tire can develop is highly dependent on its angle relative to the road surface, and so wheel camber has a major effect on the road holding of a car. A tire develops its maximum lateral force at a small camber angle. This fact is due to the contribution of camber thrust, which is an additional lateral force generated by elastic deformation as the tread rubber pulls through the tire/road interface. To optimize a tire\u2019s performance in a turn, the suspension should provide a slight camber angle in the direction of rotation. As the body rolls in a turn, the suspension deflects vertically. The wheel is connected to the chassis by suspension mechanism, which must rotate to allow for the wheel deflection. Therefore, the wheel can be subject to large camber changes as the suspension moves up and down. So, the more the wheel must deflect from its static position, the more difficult it is to maintain an ideal camber angle. Thus, the relatively large wheel travel and soft roll stiffness needed to provide a smooth ride in passenger cars presents a difficult design challenge, while the small wheel travel and high roll stiffness inherent in racing cars reduces the problem. Example 324 Castor versus camber. Camber doesn\u2019t improve turn-in as the positive caster does. Camber is not generally good for tire wear. Camber in one wheel does not improve directional stability. Camber adversely affects braking and acceleration efforts. 8.4.4 F Trust Angle The trust angle \u03c5 is the angle between vehicle\u2019s centerline and perpendicular to the rear axle. It compares the direction that the rear axle is aimed with the centerline of the vehicle. A nonzero angle configuration is shown in Figure 8.42. Zero angle confirms that the rear axle is parallel to the front axle, and the wheelbase on both sides of the vehicle are the same. A reason for nonzero 484 8. Suspension Mechanisms 8. Suspension Mechanisms 485 trust angle would have unequal toe-in or toe-out on both sides of the axle. Example 325 Torque reaction. There are two kinds of torque reactions in rear-whel-drive: 1\u2212 the reaction of the axle housing to rotate in the opposite direction of the crown wheel rotation, and 2\u2212 the reaction of axle housing to spin about its own center, opposite to the direction of pinion\u2019s rotation. The first reaction leads to a lifting force in the differential causing a wind-up in springs. The second reaction leads to a lifting force on the right wheels. 8.5 Suspension Requirements and Coordinate Frames The suspension mechanism should allow a relative motion between the wheel and the vehicle body. The relative motions are needed to pass the road irregularities and steering. To function properly, a suspension mechanism should have some kinematic and dynamics requirements. 8.5.1 Kinematic Requirements To express the motions of a wheel, we attach a wheel coordinate system W (oxwywzw) to the center of the wheel. A wheel, as a rigid body, has six degrees-of-freedom with respect to the vehicle body: three translations and three rotations, as shown in Figure 8.43. 486 8. Suspension Mechanisms The axes xw, yw, and zw indicate the direction of forward, lateral, and vertical translations and rotations. In the position shown in the figure, the rotation about the xw-axis is the camber angle, about the yw-axis is the spin, and about the zw-axis is the steer angle. Consider a non-steerable wheel. Translation in zw-direction and spin about the yw-axis are the only two DOF allowed for such a wheel. So, we need to take four DOF. If the wheel is steerable, then translation in the zw-direction, spin about the yw-axis, and steer rotation about the zw-axis are the three DOF allowed. So, we must take three DOF of a steerable wheel. Kinematically, non-steerable and steerable wheels should be supported as shown in Figures 8.44 and 8.45 respectively. Providing the required freedom, as well as eliminating the taken DOF, are the kinematic requirements of a suspension mechanism. 8.5.2 Dynamic Requirements Wheels should be able to propel, steer, and stop the vehicle. So, the suspension system must transmit the driving traction and deceleration braking forces between the vehicle body and the ground. The suspension members must also resist lateral forces acting on the vehicle. Hence, the wheel suspension system must make the wheel rigid for the taken DOF. However, there must also be some compliance members to limit the untaken DOF. The most important compliant members are spring and dampers to provide returning and resistance forces in the z-direction. 8. Suspension Mechanisms 487 8.5.3 Wheel, wheel-body, and tire Coordinate Frames Three coordinate frames are employed to express the orientation of a tire and wheel with respect to the vehicle: the wheel frame W , wheel-body frame C, and tire frame T . A wheel coordinate frame W (xw, yw, zw) is attached to the center of a wheel. It follows every translation and rotation of the wheel except the spin. Hence, the xw and zw axes are always in the tire-plane, while the yw-axis is always along the spin axis. A wheel coordinate frame is shown in Figure 8.43. When the wheel is straight and the W frame is parallel to the vehicle coordinate frame, we attach a wheel-body coordinate frame C (xc, yc, zc) at the center of the wheel parallel to the vehicle coordinate axes. The wheelbody frame C is motionless with respect to the vehicle coordinate and does not follow any motion of the wheel. The tire coordinate frame T (xt, yt, zt) is set at the center of the tireprint. The zt-axis is always perpendicular to the ground. The xt-axis is along the intersection line of the tire-plane and the ground. The tire frame does not follows the spin and camber rotations of the tire however, it follows the steer angle rotation about the zc-axis. Figure 8.46 illustrates a tire and a wheel coordinate frames. Example 326 Visualization of the wheel, tire, and wheel-body frames. Figure 8.47 illustrates the relative configuration of a wheel-body frame C, a tire frame T , and a wheel frame W . If the steering axis is along the zc-axis then, the rotation of the wheel about the zc-axis is the steer angle \u03b4. Rotation about the xt-axis is the camber angle \u03b3. 488 8. Suspension Mechanisms Generally speaking, the steering axis may have any angle and may go through any point of the ground plane. Example 327 Wheel to tire coordinate frame transformation. If TdW indicates the T -expression of the position vector of the wheel frame origin relative to the tire frame origin, then having the coordinates of a point P in the wheel frame, we can find its coordinates in the tire frame using the following equation. T rP = TRW W rP + TdW (8.28) If W rP indicates the position vector of a point P in the wheel frame, W rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 (8.29) then the coordinates of the point P in the tire frame T rP are T rP = TRW W rP + Td = TRW W rP + TRW W T dW = \u23a1\u23a3 xP yP cos \u03b3 \u2212Rw sin \u03b3 \u2212 zP sin \u03b3 Rw cos \u03b3 + zP cos \u03b3 + yP sin \u03b3 \u23a4\u23a6 (8.30) 8. Suspension Mechanisms 489 where, WT dW is the W -expression of the position vector of the wheel frame in the tire frame, Rw is the radius of the tire, and TRW is the rotation matrix to go from the wheel frame W to the tire frame T . TRW = \u23a1\u23a3 1 0 0 0 cos \u03b3 \u2212 sin \u03b3 0 sin \u03b3 cos \u03b3 \u23a4\u23a6 (8.31) W T dW = \u23a1\u23a3 0 0 Rw \u23a4\u23a6 . (8.32) As an example, the center of the wheel W rP = W ro = 0 is the origin of the wheel frame W , that is at T ro = TdW = TRW W T dW = \u23a1\u23a3 0 \u2212Rw sin \u03b3 Rw cos \u03b3 \u23a4\u23a6 (8.33) in the tire coordinate frame T . 490 8. Suspension Mechanisms Example 328 F Tire to wheel coordinate frame transformation. If rP indicates the position vector of a point P in the tire coordinate frame, T rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 (8.34) then the position vector W rP of the point P in the wheel coordinate frame is W rP = WRT T rP \u2212 W T dW (8.35) = \u23a1\u23a3 xP yP cos \u03b3 + zP sin \u03b3 zP cos \u03b3 \u2212Rw \u2212 yP sin \u03b3 \u23a4\u23a6 because WRT = \u23a1\u23a3 1 0 0 0 cos \u03b3 sin \u03b3 0 \u2212 sin \u03b3 cos \u03b3 \u23a4\u23a6 (8.36) WdT = \u23a1\u23a3 0 0 Rw \u23a4\u23a6 (8.37) and we may multiply both sides of Equation (8.28) by TRT W to get TRT W T rP = W rP + TRT W TdW (8.38) = W rP + W T dW W rP = WRT T rP \u2212 W T dW . (8.39) As an example, the center of tireprint in the wheel frame is at W rP = \u23a1\u23a3 1 0 0 0 cos \u03b3 \u2212 sin \u03b3 0 sin \u03b3 cos \u03b3 \u23a4\u23a6T \u23a1\u23a3 0 0 0 \u23a4\u23a6\u2212 \u23a1\u23a3 0 0 Rw \u23a4\u23a6 = \u23a1\u23a3 0 0 \u2212Rw \u23a4\u23a6 . (8.40) Example 329 F Wheel to tire homogeneous transformation matrices. The transformation from the wheel to tire coordinate frame may also be expressed by a 4\u00d7 4 homogeneous transformation matrix TTW , T rP = TTW W rP = \u2219 TRW TdW 0 1 \u00b8 W rP (8.41) where TTW = \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 cos \u03b3 \u2212 sin \u03b3 \u2212Rw sin \u03b3 0 sin \u03b3 cos \u03b3 Rw cos \u03b3 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 . (8.42) 8. Suspension Mechanisms 491 The corresponding homogeneous transformation matrix WTT from the tire to wheel frame would be WTT = \u2219 WRT WdT 0 1 \u00b8 = \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 cos \u03b3 \u2212 sin \u03b3 0 0 sin \u03b3 cos \u03b3 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 . (8.43) It can be checked that WTT = TT\u22121W , using the inverse of a homogeneous transformation matrix rule. TT\u22121W = \u2219 TRW TdW 0 1 \u00b8\u22121 = \u2219 TRT W \u2212 TRT W TdW 0 1 \u00b8 = \u2219 WRT \u2212WRT TdW 0 1 \u00b8 (8.44) Example 330 F Tire to wheel-body frame transformation. The origin of the tire frame is at CdT in the wheel-body frame. CdT = \u23a1\u23a3 0 0 \u2212Rw \u23a4\u23a6 (8.45) The tire frame can steer about the zc-axis with respect to the wheel-body frame. The associated rotation matrix is CRT = \u23a1\u23a3 cos \u03b4 \u2212 sin \u03b4 0 sin \u03b4 cos \u03b4 0 0 0 1 \u23a4\u23a6 (8.46) Therefore, the transformation between the tire and wheel-body frames can be expressed by Cr = CRT T r+ CdT (8.47) or equivalently, by a homogeneous transformation matrix CTT . CTT = \u2219 CRT CdT 0 1 \u00b8 = \u23a1\u23a2\u23a2\u23a3 cos \u03b4 \u2212 sin \u03b4 0 0 sin \u03b4 cos \u03b4 0 0 0 0 1 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 (8.48) As an example, the wheel-body coordinates of the point P on the tread of a 492 8. Suspension Mechanisms negatively steered tire at the position shown in Figure 8.48, are: Cr = CTT T rP = \u23a1\u23a2\u23a2\u23a3 cos\u2212\u03b4 \u2212 sin\u2212\u03b4 0 0 sin\u2212\u03b4 cos\u2212\u03b4 0 0 0 0 1 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 \u23a1\u23a2\u23a2\u23a3 Rw 0 Rw 1 \u23a4\u23a5\u23a5\u23a6 = \u23a1\u23a2\u23a2\u23a3 Rw cos \u03b4 \u2212Rw sin \u03b4 0 1 \u23a4\u23a5\u23a5\u23a6 (8.49) The homogeneous transformation matrix for tire to wheel-body frame TTC is: TTC = CT\u22121T = \u2219 CRT CdT 0 1 \u00b8\u22121 = \u2219 CRT T \u2212CRT T CdT 0 1 \u00b8 = \u2219 CRT T \u2212 T CdT 0 1 \u00b8 = \u23a1\u23a2\u23a2\u23a3 cos \u03b4 sin \u03b4 0 0 \u2212 sin \u03b4 cos \u03b4 0 0 0 0 1 Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 (8.50) Example 331 F Cycloid. Assume that the wheel in Figure 8.48 is turning with angular velocity \u03c9 and has no slip on the ground. If the point P is at the center of the tireprint 8. Suspension Mechanisms 493 when t = 0, MrP = \u23a1\u23a3 0 0 \u2212Rw \u23a4\u23a6 (8.51) then we can find its position in the wheel frame at a time t by employing another coordinate frame M . The frame M is called the rim frame and is stuck to the wheel at its center. Because of spin, the M frame turns about the yw-axis, and therefore, the rotation matrix to go from the rim frame to the wheel frame is: WRM = \u23a1\u23a3 cos\u03c9t 0 sin\u03c9t 0 1 0 \u2212 sin\u03c9t 0 cos\u03c9t \u23a4\u23a6 (8.52) So the coordinates of P in the wheel frame are: W rP = WRM MrP = \u23a1\u23a3 \u2212Rw sin t\u03c9 0 \u2212Rw cos t\u03c9 \u23a4\u23a6 (8.53) The center of the wheel is moving with speed vx = Rw\u03c9 and it is at Gr =\u00a3 vxt 0 Rw \u00a4 in the global coordinate frame G on the ground. Hence, the coordinates of point P in the global frame G, would be GrP = W rP + \u23a1\u23a3 vxt 0 Rw \u23a4\u23a6 = \u23a1\u23a3 Rw (\u03c9t\u2212 sin t\u03c9) 0 Rw (1\u2212 cos t\u03c9) \u23a4\u23a6 . (8.54) The path of motion of point P in the (X,Z)-plane can be found by eliminating t between X and Z coordinates. However, it is easier to expressed the path by using \u03c9t as a parameter. Such a path is called cycloid. In general case, point P can be at any distance from the center of the rim frame. If the point is at a distance d 6= Rw, then its path of motion is called the trochoid. A trochoid is called a curtate cycloid when d < Rw and a prolate cycloid when d > Rw. Figure 8.49(a)-(c) illustrate a cycloid, curtate cycloid, and prolate cycloid respectively. Example 332 F Wheel to wheel-body frame transformation. The homogeneous transformation matrix CTW to go from the wheel frame 494 8. Suspension Mechanisms to the wheel-body frame can be found by combined transformation. CTW = CTT TTW (8.55) = \u23a1\u23a2\u23a2\u23a3 c\u03b4 \u2212s\u03b4 0 0 s\u03b4 c\u03b4 0 0 0 0 1 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 c\u03b3 \u2212s\u03b3 \u2212Rw sin \u03b3 0 s\u03b3 c\u03b3 Rw cos \u03b3 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 = \u23a1\u23a2\u23a2\u23a3 cos \u03b4 \u2212 cos \u03b3 sin \u03b4 sin \u03b3 sin \u03b4 Rw sin \u03b3 sin \u03b4 sin \u03b4 cos \u03b3 cos \u03b4 \u2212 cos \u03b4 sin \u03b3 \u2212Rw cos \u03b4 sin \u03b3 0 sin \u03b3 cos \u03b3 Rw cos \u03b3 \u2212Rw 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 If rP indicates the position vector of a point P in the wheel coordinate frame, W rP = \u23a1\u23a3 xP yP zP \u23a4\u23a6 (8.56) then the homogeneous position vector CrP of the point P in the wheel-body 8. Suspension Mechanisms 495 coordinate frame is: CrP = CTW W rP = \u23a1\u23a2\u23a2\u23a3 xP cos \u03b4 \u2212 yP cos \u03b3 sin \u03b4 + (Rw + zP ) sin \u03b3 sin \u03b4 xP sin \u03b4 + yP cos \u03b3 cos \u03b4 \u2212 (Rw + zP ) cos \u03b4 sin \u03b3 \u2212Rw + (Rw + zP ) cos \u03b3 + yP sin \u03b3 1 \u23a4\u23a5\u23a5\u23a6 (8.57) The position of the wheel center W r = 0, for a cambered and steered wheel is at Cr = CTW W r = \u23a1\u23a2\u23a2\u23a3 Rw sin \u03b3 sin \u03b4 \u2212Rw cos \u03b4 sin \u03b3 \u2212Rw(1\u2212 cos \u03b3) 1 \u23a4\u23a5\u23a5\u23a6 (8.58) The zc = Rw (cos \u03b3 \u2212 1) indicates how much the center of the wheel comes down when the wheel cambers. If the wheel is not steerable, then \u03b4 = 0 and the transformation matrix CTW reduces to CTW = \u23a1\u23a2\u23a2\u23a3 1 0 0 0 0 cos \u03b3 \u2212 sin \u03b3 \u2212Rw sin \u03b3 0 sin \u03b3 cos \u03b3 Rw (cos \u03b3 \u2212 1) 0 0 0 1 \u23a4\u23a5\u23a5\u23a6 (8.59) that shows CrP = CTW W rP = \u23a1\u23a2\u23a2\u23a3 xP yP cos \u03b3 \u2212Rw sin \u03b3 \u2212 zP sin \u03b3 zP cos \u03b3 + yP sin \u03b3 +Rw (cos \u03b3 \u2212 1) 1 \u23a4\u23a5\u23a5\u23a6 (8.60) Example 333 F Tire to vehicle coordinate frame transformation. Figure 8.50 illustrates the first and fourth tires of a 4-wheel vehicle. There is a body coordinate frame B (x, y, z) attached to the mass center C of the vehicle. There are also two tire coordinate frames T1 (xt1 , yt1 , zt1) and T4 (xt4 , yt4 , zt4) attached to the tires 1 and 4 at the center of their tireprints. The origin of the tire coordinate frame T1 is at Bd1 BdT1 = \u23a1\u23a3 a1 \u2212b1 \u2212h \u23a4\u23a6 (8.61)"
        ]
    },
    {
        "image_filename": "designv10_0_0002235_9781119078388-Figure7.14-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002235_9781119078388-Figure7.14-1.png",
        "caption": "FIGURE 7.14 Relevant geometry for the double-line integrals.",
        "texts": [
            " If the two sides are perpendicular where k = u\u0302 \u22c5 u\u0302\u2032 = 0, then the computation of Ip is zero. Hence, removing the evaluation saves compute time. Of course, the x and y axes are exchanged if the integration is set up in the wrong direction. The analytical evaluation of Ip is considered in the following section for the general case as well as the special case when two sides coincide is considered. It is clear from the previous section that to evaluate IT , we have to solve the integrals Ip between two segments l and l\u2032 as shown in Fig. 7.14. Hence, the integrals involve two lines l with the end points P1(x1, y1), P2(x2, y2) and l\u2032 with the end points P3(x3, y3) and P4(x4, y4), which can substitute from (7.35) in the global reference system (x, y, z) for l as y = mx + q m = y2 \u2212 y1 x2 \u2212 x1 , q = y1 \u2212 mx1, dl = \u221a 1 + m2dx, (7.37) and for l\u2032 as y\u2032 = m\u2032x\u2032 + q\u2032 m\u2032 = y4 \u2212 y3 x4 \u2212 x3 , q\u2032 = y3 \u2212 m\u2032x3, dl\u2032 = \u221a 1 + m\u20322dx\u2032. (7.38) Substituting (7.31b), (7.37), and (7.38) in (7.36) one obtains Ip = \u222bl\u2032\u222bl \u221a (x \u2212 x\u2032)2 + (y \u2212 y\u2032)2 dl dl\u2032. (7.39) EVALUATION OF PARTIAL ELEMENTS FOR NONORTHOGONAL PEEC CIRCUITS 175 I = \u222bl \u221a (x \u2212 x\u2032)2 + (y \u2212 y\u2032)2 dl = \u221a 1 + m2 \u222b x2 x1 \u221a cx2 + bx + c dx = \u221a 1 + m2 [ (2cx2 + b) 4c \u221a cx2 2 + bx2 + a + \u0394 8c \u221a c ln ( b + 2cx2 2 \u221a c + \u221a cx2 2 + bx2 + a ) \u2212 (2cx1 + b) 4c \u221a cx2 1 + bx1 + a \u2212 \u0394 8c \u221a c ln ( b + 2cx1 2 \u221a c + \u221a cx2 1 + bx1 + a )] , (7"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0002667_j.matpr.2020.08.415-Figure5-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0002667_j.matpr.2020.08.415-Figure5-1.png",
        "caption": "Fig. 5. Selective laser melting [9].",
        "texts": [
            " Sometimes CO2 laser beam is used [16]. In metal SLS process a mixture powder is used. It consists of one powder having low melting point and another high melting point. During SLS process only low melting point powders can partially melt and forms the bond with unmelt high melting point powders. 2.2. Selective Laser melting (SLM) process The working of SLM process is similar to that of SLS process. It melts alloy powder particles instead of sintering [1,2,6,7,8,9,10,12,15,17,18]. It is shown in Fig. 5. The manufacturing cycle time of complex part production is rapidly reduced. Finer microstructure can be observed at high cooling rate compared to conventional manufacturing methods. SLM parts have excellent mechanical properties. Aerospace, automobile, and medical applications parts are produced by SLM. Lattice structure can be pro- Fig. 6 showing the DMLS process [1,2,7,8,9,10,12], which is similar to that of SLM process. Functional prototypes, short-run components, and functionally graded materials (FGM) are manufactured by direct metal laser sintering"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001179_978-1-4020-5483-9-Figure6.26-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001179_978-1-4020-5483-9-Figure6.26-1.png",
        "caption": "Figure 6.26 All the forces acting on the isolated beam. The distributed load is composed of the roof load and a dead weight of 6 kN/m. The support reactions of 46.4 kN have to be provided by the columns.",
        "texts": [
            " If one takes an arbitrary strip of the roof of 1-metre width, the total load on the strip would be (4 m)(1 m)(2.8 kN/m2) = 11.2 kN. Each beam carries half of this, or in other words, 5.6 kN over a 1-metre length, (see Figure 6.25b). Over the full length, the beam is therefore loaded by a uniformly distributed load of 5.6 kN/m. We also have to include the dead weight of the beam. If we assume a dead weight of 6 kN/m, the total 232 ENGINEERING MECHANICS. VOLUME 1: EQUILIBRIUM load on the beam is (see Figure 6.26) q = (5.6 kN/m) + (6 kN/m) = 11.6 kN/m. The beam is simply supported. The support reactions, which have to be provided by the columns, amount to 1 2 \u00d7 (11.6 kN/m)(8 m) = 46.4 kN. Equal and opposite forces are acting on the columns. Figure 6.27a shows the cross-sectional dimensions of the columns. With mass density \u03c1 = 2400 kg/m, the specific weight of concrete is \u03c1g = (2400 kg/m3) \u00d7 (10 N/kg) = 24000 N/m3) = 24 kN/m3. For the cross-sectional dimensions of the column in Figure 6.27a, the dead weight per length is (0"
        ],
        "surrounding_texts": []
    },
    {
        "image_filename": "designv10_0_0001270_jproc.2020.3046112-Figure13-1.png",
        "original_path": "designv10-0/openalex_figure/designv10_0_0001270_jproc.2020.3046112-Figure13-1.png",
        "caption": "Fig. 13. PCB busbar for 135-kW SiC-based traction inverter.",
        "texts": [
            " The power modules are positioned such that the copper traces connect directly onto the module pins. Because of the high bus voltage, the PCB design should adhere to various standards (such as UL796 and IPC-2221a specifications) to withstand voltages in the range of several kilo-Volts between the conductive elements on the board. Furthermore, thermal stress needs to be considered so that the temperature rise during operation does not exceed the capability of the PCB material. In the PCB busbar design shown in Fig. 13, minimization of the loop inductance is addressed by selecting a vertical stacking architecture with the power planes distributed in a specific order [36]. For the dc side, the positive and negative power planes are stacked in pairs with maximum overlapping between the positive and negative layers. Designing the busbar in this way fully uses the magnetic coupling existing between the PCB conducting layers, thus enabling effective flux cancellation. In the case of ac planes, the phases do not have any vertical overlap with each other"
        ],
        "surrounding_texts": []
    }
]