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. 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." 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 , 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. 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. 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: 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 Solutions to the Exercises 387 Now, we illustrate the same solution in MATHEMATICA language. fl=x"[t]+83.3 x[t]+5.553 yl=DSolve[{fl==0,x[0]==.314,x'[0]==0},{x},{t}] The solution corresponds to (d). jl=Plot[Evaluate[x[t]/.yl],{t,0,.2},AxesLabel->{"t","x"}] j2=Plot[x= 0664,{t,0,.2}] Show[jl,j2] The curve in the graphic representation begins at the point 0.314 m. The horizon- tal lines jc* = -0.0664 m and x** = 0.157 - 0.0664 = 0.0906 m, in turn, indicate: x* is the zone where the point K does not reach because of the resistance torque T r (from 0 to -0.0664 m); it is the new zero point relative to which the value x** (the position of the point K after the rope is rewound for half a revolution) is denned and which is achieved at t= 0.103 second. FIGURE 3E-1.1 Displacement (of the point K) x versus time. 2 Solution to Exercise 3E-2 The solution is divided into two stages: stage a) the blade's movement from the initial point until it comes into contact with the wire (here we neglect the frictional resistance), and stage b) the cutting of the wire. Stage a) We use Equations (3.27) to (3.81). Thus, 388 Solutions to the Exercises which means that the weight mg of the blade aids its downward motion. From (a) we obtain where where CD is the natural frequency of the mechanism. The solution x is made up of two components: The homogeneous component is expressed as The partial solution is given by Substituting X into Equation (a) we find Thus, Substituting the initial conditions into (b), we obtain the coefficients A and B. When t= 0 and x = L 0 = 0.2 m, then 0.2= A + 0.002, and finally we obtain Differentiating (b), we obtain where t= 0, i= 0, and we thus obtain Finally, we obtain We now calculate the time 1 1 needed by the blade to reach the point at which it comes into contact with the wire, i.e., x = 0.1 m. From (d) we obtain or Solutions to the Exercises 389 The speed x { developed by the blade at this moment in time is calculated from (c): In MATHEMATICA language, we solve the above-derived equation as follows: f2=xl"[t]+5000xl[t] -10 y2=DSolve[{f2==0,xl[01==.2,xl'[0]==0},{xl},{t}] j21=Plot[Evaluate[xl[t]/.y2],{t,0,.025}, AxesLabel->{"t","xl"}] gl=Plot[{xl=.l},{t,0,.02}] Show[gl,j21] FIGURE 3E-2.1 Movement of the blade xjt] until it makes contact with the wire (x= 0.1 m). Stage b) We consider two ways to solve this stage. I. We begin with a simple physical estimation of the time needed for cutting the wire. The whole energy E (kinetic plus potential components) of the blade at the moment in time when it comes into contact with the wire is The work A that must be expended for cutting the wire is expressed as The saved energy E* after the cutting is accomplished is given by [...]... component t" is calculated from Expression (3. 127 ) We now need to determine the value of the pressure Pc: where ft = 1 (same pressures in the receiver and the filled volume), The total time t needed to fulfill the task is 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... the Exercises Solution to Exercise 3E- 5a) The time sought here is a sum of three components Component 1 The first component t0 is the time needed by the pressure wave to travel from the valve to the inlet of the volume: where Vs is the speed of sound Component 2 The second component ^ is the time needed to reach a value of pressure Prin the given volume Here, we consider the combination of supercritical... the time needed to reach critical pressure Pcr in the given volume Vc of the cylinder (supercritical regime), while the component t/' is the time needed subsequently to reach the pressure PC500 (subcritical regime) The value Then, from Expression (3. 120 ) we obtain for r/ Here, as in solution 3E- 5, Fp= 11.3 10'5 m2 The component t" is calculated from Expression (3. 127 ) We now need to determine the value... latter transforms the resistance force Q to the needed torque on the screw Ts which, respectively, is 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... to the Exercises and this energy (a remaining sum of kinetic and potential components after the cutting is accomplished) is given by where x* is the speed of the blade after cutting the wire From (e) we derive: We now express the loss of the momentum M as and, finally, the impulse of force Here, and, therefore, Thus, II Now let us solve this part of the problem describing the process of the blade's... j=Plot[s,{t,0,.l},AxesLabel->{"t","s"}] Solutions to the Exercises FIGURE 3E- 4a).l 8 401 Displacement of the piston-rod versus time Solution to Exercise 3E- 5 The time sought is a sum of three or four components Component 1 The first component t0 is the time needed by the pressure wave to travel from the valve to the inlet of the cylinder: r 0 = £ / V 5 = 10m/340m/secE0. 029 4sec, where Vs is the speed of sound Component 2. .. regimes The duration ^ consists, therefore, of two components; thus, ^ = f/ + t" The component £/ is the time needed to reach critical pressure Pcr in the given volume (supercritical regime), while the component t" is the time needed to subsequently reach the pressure Pc500 (subcritical regime) The value Then, from Expression (3. 120 ), we obtain for f/ Here, as in solution 3E- 5, Fp = 11.3 10'5 m2 The... moving parts with respect to the shaft of the brake requires calculation of the common kinetic energy E of the mechanism, which is where . 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 . 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 . obtain or Solutions to the Exercises 389 The speed x { developed by the blade at this moment in time is calculated from (c): In MATHEMATICA language, we solve the above-derived equation