robotics Designing the Mechanisms for Automated Machinery Part 13 docx

9.6 Mobile and Walking Robots 377 By substituting the latter into (9.66), the position of the vehicle is determined. This algorithm can be used both when programming and controlling the trajectory of the vehicle and when analyzing the trajectory of the vehicle, for instance, for indicating the location of a car during its travel. The problem of travelling across rugged terrain has prompted investigations in the field of walking vehicles. There is no real substitute for legs, which can step over local obstacles, find the optimum point on a surface for reliable support, change the length of the stride, and change the pace from walking to running and jumping or from an amble to a trot to a gallop. It seems that investigators have covered all possibilities in seeking the optimal walking machine. In Figures 1.30 and 1.31 of Chapter 1 we briefly described the purely mechanical walking machine designed by the famous mathematician P. Chebyshev. This device can walk in an optimal way; however, it requires a flat surface and cannot make turns. Therefore, it does not fulfill the hopes a walking device has to realize. (By the way, this device clearly shows how cumbersome legs are in comparison with a wheel. The only reason the Creator (or evolution) used legs in his design of walking animals is the simplicity in providing reliable connection of blood vessels, nerves, and muscles. Otherwise we would have wheels, and shoes would be designed like tires.) We begin our consideration with some six-legged devices, for which insects have served as a prototype. Figure 9.58a) shows an example. (This design was influenced by the article by A. P. Bessonovand N. V. Umnov, Mechanism and Machine Theory, Insti- tute for the Study of Machines, Vol. 18, No. 4, pp. 261-265, Moscow, U.S.S.R, 1983.) The skeleton of the machine consists of rigid frame 1 to which links 3 are connected via joints 2. Another two pairs of joints 4 and rods 5 and 6 close the kinematic chains, cre- ating two parallelograms. Hydrocylinders 7 and 8 control the position of these two par- allelograms relative to frame 1. This arrangement serves for steering. Slide bushings 9 are mounted on rods 3 and frame 1. These bushings are driven along the rods by means of hydraulic cylinders 10. To each bushing 9 is fastened one of the six legs. Each leg includes a hydraulic cylinder 11 which is responsible for the height of the leg's foot. This is a design with 14 degrees of freedom. Figure 9.58b) explains the movement of a foot. By combining the movement of the cylinders in a certain time sequence, a step- over movement of the foot is carried out. Say, the foot at the beginning of the cycle is in its upper position I due to cylinder 11; then cylinder 10 moves it horizontally to point II. At this moment cylinder 11 lowers the foot to point III, from which cylinder 10 pulls the foot back (relative to the frame) to point IV, thus moving the "insect" one step leftward. Other six-legged walker concepts using other leg kinematics are also possible. For instance, a leg with three degrees of freedom is illustrated in Figure 9.59a). Shaft 1 with joint 2 represents the hip, to which the thigh 3 with knee 4 is attached. The latter serves for connecting to the shin 5. These links can be driven by hydraulic cylinders. Cylin- der 6 drives shaft 1 via rack-gear transmission 7. Cylinder 8 controls thigh 3 while cylin- der 9 is responsible for the movement of shin 5. We can imagine three of this kind of leg placed on each side of a rectangular frame, simulating an insect. In this case the TEAM LRN 378 Manipulators FIGURE 9.58 a) Six-legged walking machine with 12 degrees of freedom; b) Design of one leg. system possesses 18 degrees of freedom. (This kind of machine was created by Ivan E. Sutherland (see Scientific American, January, 1983).) It can also be made in the form presented in Figure 9.59b), where the hip-shafts 1 are installed along radii 0-1, O-II, 0- III, 0-IV, 0-V, and 0-VI. Thus, the axially symmetrical device is supported by at least three legs (black points in Figure 9.59c) at any moment. To move along vector V the sequence of lifting and lowering the legs is as shown in this figure when the circle moves rightward. Obviously, no preferred direction exists for this concept; it does not have a front or back. It can begin its travel in any direction and provides high maneu- FIGURE 9.59 a) Leg with three degrees of freedom; b) Circular form of a vehicle; c) Walking tracks of this vehicle. TEAM LRN 9.6 Mobile and Walking Robots 379 verability, including rotation in place. For this purpose foot I moves to point I', foot II moves to point II', and so on. Using the leg with three degrees of freedom shown in Figure 9.59a, one can design four-legged devices, three-legged vehicles (A. Seireg, R. M. Peterson, University of Wis- consin, Madison, Wisconsin, U.S.A.) and, of course, try to create a two- or one-legged walker. Here there are problems with balance. Marc H. Ralbert and his colleagues at Carnegie-Mellon University have built and demonstrated a machine that hops on a single leg and runs like a kangaroo. This device consists of two main parts (Figure 9.60): body 1 and leg 2. The body carries the energy-distribution and control units. The leg can change its length, due to pressure cylinder 3, and pivots with respect to the body around the X- and Y-axes due to spherical hinge 4. A gyrosystem keeps the body hor- izontal while the leg bounces on the pneumatic spring-pneumatic cylinder 3. The pres- sure in the cylinder is controlled to dictate the bouncing dynamics. Another pneumosystem controls the pivoting of the leg during its rebound. A feedback system provides the pivoting required for balance when landing. The balance problem is very important in walking devices. Solving it will permit four- and two-legged robots to be built. (The six-legged devices are always balanced on three supporting feet.) To solve the balancing problems a mathematical descrip- tion of the device must be analyzed. Let us consider the model of the two-legged walker shown in Figure 9.61, which presumably reflects the main dynamic properties of natural two-legged creatures. (The following discussion is taken from the book Two-legged Walking by V. V. Beletzky, Moscow, Nauka, 1984 (in Russian).) Different levels of com- plexity can be considered for the model; however, it is always true that: • The device possesses a certain mass, which is subject to gravitation; • The device is controlled, which means that its links are driven and torques act in every joint; (Here we consider an additional degree of freedom between the shin and the foot, compared with the leg shown in Figure 9.59a).) • Movement across a surface entails reactive forces against the feet. All walking machines can act in two regimes: • Walking—at any moment, at least one leg touches the ground; • Running—there are moments when no contact between the ground and feet exists. FIGURE 9.60 One-legged hopper. TEAM LRN 380 Manipulators The scheme shown in Figure 9.61 permits expressing the coordinates of the mass center C or any other point, through the parameters of the device. For instance, for the posi- tion in the figure we have for point O the coordinates are where v = index of the supporting leg, and z v and x v are components of the r v vector (position of the foot's contact point). Expression (9.73) allows, for instance, answering the question of what the kine- matic requirements are for providing constant height of point O above the ground (for better energy saving). This condition states and from (9.73) it then follows that The opposite kinematic problem can arise: for a given location of point 0 what are the corresponding angles a and/?? Assuming that a = b=l (the links have equal lengths) and denoting i »-» Obviously, the speeds and accelerations of the links can be calculated if the functions z(t} and x(f) are known. FIGURE 9.61 Mathematical model of a two- legged anthropoid walking machine. TEAM LRN 9.6 Mobile and Walking Robots 381 The dynamics of a two-legged robot present a complex problem for the general case. For the designations in Figure 9.61, the equations can be written in vector form as follows: Here, M= the mass of the system; r c = radius vector of the mass center C from the coordinates' initial point N; P = gravity force; R = reaction force; r T = radius vector of the point where the force R crosses the supporting surface; K= angular momentum of the system (moment of momentum). Let the reader try to solve Equations (9.76) and (9.77). Here the simplified model shown in Figure 9.62—a two-legged "spider"—will be considered. This spider consists of a massive particle and two three-link legs. Then vector R is the reactive force acting at the initial point. Gravity force P acts at point O where mass center C is also located. For this case Equations (9.76) and (9.77) take the following scalar form: Here x and z are coordinates of particle C, and M is the mass of the particle. It is supposed that at every moment the spider is supported by only one leg. The exchange of legs happens instantly and mass center C does not change its height. Thus, z = h = const and therefore z= 0. For periodical walking with step length L and dura- tion 2T, the following limit conditions are valid: FIGURE 9.62 Simplified model of a two- legged walker or "spider." TEAM LRN 382 Manipulators For these assumptions it follows from the second Equation (9.78) that and the first equation can be rewritten in the form (taking 9.80 into account) The solution for (9.81) for the conditions (9.79) has the following form: Thus, V 0 , L, and Tare related by Expression (9.83). The average marching speed yean be calculated from The work A that the reaction force's horizontal component produces is determined with the help of the following formula: Obviously, the power required for walking can be obtained from (9.86) and expressed as follows: Now let us consider running for this spider model. We assume momentary foot contact with the support surface. Between these contact instants, #=0 (there is no reaction force), and the spider, in essence, flies. The trajectory of the particle in this case consists of parabolic sections, as shown in Figure 9.63. In this figure V Q = initial speed of the particle, JC Q = horizontal component of speed V 0 , TEAM LRN 9.6 Mobile and Walking Robots 383 z Q = vertical component of speed V 0 , T= flying time, and L = flying distance. These concepts are related by known formulas where V- average speed of translational movement. At the end of the flying period, the vertical speed component ^ = -z 0 . To acceler- ate the particle again to speed z 0 , an amount of energy A l must be expended by the foot: In stopping at the end of the flying period, the amount of energy A 2 expended by the other foot is Thus, the energy A expended for one step is Substituting Expression (9.88) into (9.89), we obtain Obviously, as follows from (9.90), the power spent in running can be calculated from the expression Comparing the latter with Expression (9.87), we discover that when V > ^igh, it is worthwhile to run instead of walk; V < ^gh , it is worthwhile to walk instead of run. TEAM LRN 384 Manipulators Above, we considered the energy consumption of the walking or running body. However, to be more accurate, one must also take into account the power spent for moving the feet. This power can be estimated with the formula Here ju - the relation of foot mass m to particle mass M. Thus, together with (9.87), we have an expression for the total power expended for walking: The latter formula (9.93a and b) enables the value of the optimum length L 0 of a walking steo to be derived: The computation model presented here can give a rough estimation of power con- sumption in multi-legged vehicles by simply multiplying the results and distributing the mass of the moving body among all the pairs of legs. Using the derived formulas we can recommend that the reader walk with an optimum step which is, for an average person (h= 1 m, // = 0.2, V= 1.25 m/sec), L 0 = 0.7 m. Then he or she will expend about 150 watts (0.036 kcal/sec) of power. We also recommend changing from walking to running when a speed of 11.3 km/hr is reached. However, if the reader is overweight, let him or her continue to walk with higher speed (more energy will be expended). The speed record for walking is about 15.5 km/hr. On this optimistic tone we finish this chapter, the final one in the book. TEAM LRN Solutions to the Exercises 1 Solution to Exercise 3E-1 The first step is to reduce the given mechanism to a single-mass system. The resis- tance torque T r on drum 1, obviously, varies in inverse proportion to the ratio i - 1:3. Thus, The procedure of reducing inertia 7 2 of drum 2 to the axes of drum 1 requires cal- culation of the common kinetic energy of the mechanism, which is where co^ and co 2 are the angular velocities of drums 1 and 2, respectively. (The inertia of the gears and the shafts is neglected.) The kinetic energy of the reduced system with moment of inertia / is: The motion equation may then be written as follows: TEAM LRN 386 Solutions to the Exercises where x is the displacement of point K on the rope (see Figure 3E-1.1). Substituting the numerical data into (a) we obtain The solution x is made up of two components: x = x l + x 2 . The homogeneous component is sought in the form: Xi = Acoskt + Bsmkt, where k is the natural frequency of the system. Here, obviously, The partial solution x 2 , as follows from (a), is sought in the form of a constant X: From the initial conditions given in the formulation of the problem, it follows that for time t = 0, the spring is stretched for x 0 = 2nR = 2 • n • 0.05 = 0.314 m, while the speed JC G = 0. Thus, from (b) we derive Differentiating (b) in terms of speed, we obtain Substituting the initial conditions, we obtain 9.13 B = 0 or B = 0. Finally, from (b) we obtain the following expression for the solution: To answer the question formulated in the problem, we find t from (d), substituting the value X' (location of the point K after the rope had been rewound half a perime- ter around drum 1). Obviously, And from (d), it follows that TEAM LRN [...]... single-mass system The resistance torque Tr on the axes of the electromotor, obviously, varies in inverse proportion to the ratio i= 1:4 Thus, The procedure of reducing the inertia of all the moving parts with respect to the axes of the motor requires calculation of the common kinetic energy E of the mechanism, which is where CD is the angular speed of the shaft of the motor The kinetic energy of the reduced... for AC motor drive; b) Characteristic of the electromotor 5 Solution to Exercise 3E-3b) The first step is to reduce the given mechanism to a single-mass system The resistance torque Ton the axes of the electromotor, obviously, varies in inverse proportion to the total ratio of the gears (/= 1:3) and of the screw-nut transmission The latter transforms the resistance force Q to the needed torque on the. .. LRN Solutions to the Exercises 397 Thus, The procedure of reducing the inertia of all moving parts with respect to the axes of the motor requires calculation of the common kinetic energy E of the mechanism, which is where co is the angular speed of the shaft of the motor The kinetic energy of the reduced system with a moment of inertia / respective 1 the axes of the motor is or or The differential... supposing that all the pressure Pr acts during the entire lifting process Of course, this will be the lower estimate for the time the lifting will take Thus, the acting force Fa is then: and the acceleration a is then and the lifting time t2 is The total time 1500 is TEAM LRN 404 9 Solutions to the Exercises Solution to Exercise 3E-5a) The time sought here is a sum of three components Component 1 The first... we obtain for the coefficients fi either or A comparison of these results shows that the first belongs to a supercritical air flow regime while the second belongs to a subcritical regime A We begin with the supercritical case Let us calculate the second time component t^ To do so, we use the Formula (3.120): where Vc is the volume of the cylinder at the beginning of the action, and Fp is the cross-sectional... where Vs is the speed of sound Component 2 The second component ^ is the time needed to reach a pressure Pcin the cylinder volume that develops a force equal to the resisting force acting on the piston-rod (in our case this is the weight mg of the lifted mass m) Obviously, and for the problem under consideration this is either or TEAM LRN 402 Solutions to the Exercises Corresponding to these two situations,... pressures in the receiver and the filled volume), The total time t needed to fulfill the task is TEAM LRN Solutions to the Exercises 10 405 Solution to Exercise 3E-5b) The time sought here is comprised of three components Component 1 The first component f0 is the time needed by the pressure wave to travel from the valve to the inlet of the cylinder of the jig: where Vs is the speed of sound Component 2 The. .. The partial solution is sought as a constant a>2 = Q = const Substituting this constant into Equation (a) yields Thus, From the initial conditions, we find the coefficient A For the moment t= 0, the speed CD = 0 Therefore, and finally, To answer the question formulated in the problem, we substitute t = 0.1 sec into Expression (c): Integrating Expression (c), we find the angle of rotation . expressing the coordinates of the mass center C or any other point, through the parameters of the device. For instance, for the posi- tion in the figure we have for point O the . with the formula Here ju - the relation of foot mass m to particle mass M. Thus, together with (9.87), we have an expression for the total power expended for walking: The . Manipulators For these assumptions it follows from the second Equation (9.78) that and the first equation can be rewritten in the form (taking 9.80 into account) The solution for (9.81)