Robotics 2 E Part 4 ppt

80 Dynamic Analysis of Drives on the design and dimensions of the motors. These parameters usually change in the following range: Torque: T L =22.5 -1125 kg m, Inertia of the rotor: I r = 1.2-10000 g cm 2 , Maximum pulse rate: S = 150 -5- 50000 pps. A typical torque-versus-time characteristic for stepper motors is presented in Figure 3.18. The point that should be stressed is that changes in the torque are different for different pulse rates. The lower the value of the pulse rate (i.e., the duration of one pulse is longer), the higher the torque at the beginning of the switching and the lower the torque at the end of it. In more detail, the dependence torque versus pulse rate is shown in Figure 3.19. (These data are taken from Machine Design, April 29, 1976, p. 36.) Point A represents the conditions that ensure the maximum speed at which a load can be run bi-directionally without losing a step. This condition occurs by a speed of about five steps per second. Point B indicates the so-called stall torque. At this point, the stator windings being energized, for all kinds of motors, resist movement. Point C represents the detent-like torque which is typical only for motors with permanent magnet rotors. At this point even a nonenergized stator resists the movement of the rotor to move. The motor "remembers" its position. Curves 1 and 2 represent the behav- FIGURE 3.18 Torque-versus-time dependence for a stepper motor. FIGURE 3.19 Torque-versus-pulse-rate dependence for stepper motors. 3.3 Electric Drives 81 ior of motors provided with permanent magnet rotors, while curve 3 represents that of variable reluctance motors. Now that we have learned the characteristics of the most frequently used electro- motors as drives for automatic machines and systems, some other comparative fea- tures of these electromotors must be discussed. The advantage of DC motors lies in the ease of speed control, whereas speed control in AC motors requires the installation of sophisticated equipment (frequency trans- formers). The advantage of AC motors (both one- and three-phase) is that they operate on the standard voltage available at any industrial site. In addition, a three-phase induc- tion motor with a squirrel-cage rotor is cheaper than any other type of motor of the same power. For accurate positioning both DC and AC motors require feedbacks. In contrast, stepper motors, although more expensive, are suitable for accurate posi- tioning (almost always without any feedback) and speed control. Such motors are con- venient for engagement with digital means (computers). Let us now analyze Equation (3.41) for the case when T d is described by Equations (3.46). Let us suppose that the resistance torque T r is also described by a linear expres- sion which is proportional to the speed of rotation of the machine. Thus we can write for T r : Obviously, a l andor 2 are constants. The physical meaning of the value a 1 is the initial resistance of the driven system. Until the drive has developed this value of the driving torque, the system will not move. The value of a 2 controls the rate of the resistance torque during the speed increase of the accelerated system. The problem is how to estimate the values of a^ and a 2 . We feel that the only way to do this is to measure the resistance torques of existing machines and interpolate or extrapolate the results to the case under design. Substituting Equations (3.46) and (3.49) into Equation (3.41 ) we obtain a linear equation in the form After simplification we obtain where C = (oc 2 + « 2 ) /I an d B=(a l - a^) /I. Remembering that 0 = CD, we can rewrite Equation (3.51) to obtain The solution of this equation has the form where ^ is the solution of the homogeneous equation For co l we have 82 Dynamic Analysis of Drives Substituting Equation (3.55) into Equation (3.54) we obtain or For a> 2 we seek a solution in the form Substituting Equation (3.58) into Equation (3.52) we obtain or From Equation (3.53) we derive For the initial conditions at time t= 0 and speed CD = 0, we can rewrite Equation (3.61) and extract the unknown constant A in the following way: or And finally we obtain the solution: To calculate the time needed to achieve some speed of rotation co^ we derive the fol- lowing equation from Equation (3.64): From Expression (3.65) some particular cases can be obtained. For a constant resis- tance torque a 2 = 0 we obtain 3.3 Electric Drives 83 When there is no resistance torque, i.e., a v = a 2 = 0 we obtain For the case when a^ - 0 we derive from Expression (3.65) the following formula: Expression (3.64) allows us to find the dependence of the rotation angle 0 on time. For this purpose we must rewrite this expression as follows: Integrating this dependence termwise, we obtain: and Now for the particular case when oc 2 - 0: For For In Figure 3.20 the graphic representation of this dependence is shown to be com- posed of three components. This graph provides us with a tool for determining the time ^ needed to achieve rotation for the angle 0j. Let us consider the case where the drive is supplied by a series DC motor with the characteristics given by Expression (3.47). Because of analytical difficulties, we will 84 Dynamic Analysis of Drives FIGURE 3.20 Angular displacement of the DC motor's rotor (composed of three components) versus time as a solution of Equation 3.50. discuss here the simplest case, i.e., when in Equation (3.41) T r = 0, which means that resistance is negligible. Thus, we obtain or From Equation (3.75) it follows that and finally that Thus, for the dependence <p(t) we obtain or 3.3 Electric Drives 85 When the drive is supplied by an asynchronous induction motor, we substitute Equa- tion (3.48) into Equation (3.41). Here again, we will discuss the simplest case when T r = 0. Thus, we rewrite Equation (3.41) in the form Remembering the definition of slip given above, we obtain, instead of Equation (3.79), the expression Denoting (2T m s m co^ jl=A and s^cvo =B,we rewrite Equation (3.80) in the form After obvious transformation, the final result can be obtained in the form: For a synchronous motor the driving speed (as was explained above) remains coi slant over a certain range of torques until the motor stops. Thus, Q) = a> 0 - constant. To reach the speed a> 0 from a state of rest when CD = 0, an infinitely large acceleratic must be developed. To overcome this difficulty, synchronous motors are started in tf same way as are asynchronous motors. Therefore, the calculations are of the same soi and they may be described by Equations (3.79-3.82), which were previously applic to asynchronous drives. For the drive means of stepper motors, we must make two levels of assumptio First, we assume that the stepper motor develops a constant driving torqu Td=T 0 = constant (the higher the pulse rate, the more valid the assumption), which the average value of the torque for the "saw"-like form of the characteristic. Then, fro] the basic Equations (3.41) and (3.49), we obtain for the given torque characteristic tr following equation of the movement of the machine: Rewriting this expression, we obtain The solution consists of two components, co = co l + (O 2 . For the solution of the homo- geneous equation we have 86 Dynamic Analysis of Drives and for the particular solution we have Substituting these solutions in the homogeneous form of Equation (3.83) and in its complete form, respectively, we obtain Using the initial conditions that for t = 0 the speed co = 0, we obtain for the constant A Thus, the complete solution has the form The next step is to calculate the 0(Z) dependence. This can obviously be achieved by direct integration of solution (3.88): or For the second assumption, we introduce into the excitation torque a "saw"-like periodic component. To do so we must express this "saw" in a convenient form, i.e., describe it in terms of a Fourier series. Let us approximate this "saw" by inclined straight lines, as shown in Figure 3.21 (the reader can make another choice for the approxi- mation form). Then, this periodic torque component T p can be described analytically by the expression FIGURE 3.21 Approximation of the "saw"-like characteristic (see Figure 3.18) of a stepper motor by inclined straight lines. 3.3 Electric Drives 87 and its expansion into a Fourier series becomes where r is the torque amplitude. Thus, Equation (3.83) for this case can be rewritten in the form and its solution will be composed of three components: The solutions co l and co 2 are found as in the previous case for the corresponding forms and may be expressed as Here we show the solution a> 3 only for one first term of the series, namely, Substituting it into Equation (3.92) for the corresponding case, we obtain the form After rearrangement of the members and comparison between the left and right sides of this equation, we obtain Introducing the initial conditions, namely for t= 0, CD = 0, we derive Finally, the solution of the full Equation (3.92) is From Equation (3.97) we obtain the dependence 88 Dynamic Analysis of Drives Obviously, the calculations shown above answer the following questions: • How long does it take for the drive to reach the desired angle or (using corre- sponding transmission) displacement? • How long does it take for the drive to reach the desired speed? • What angle, displacement, or speed can be reached during a specific time interval? • Which parameters of the motor must be taken into account to reach the desired angle, displacement, or speed in a specific time? 3.4 Hydraulic Drive Let us now learn how to estimate the displacement time of a mass driven by a hydro - mechanism. Let us consider the hydromechanism presented schematically in Figure 3.22. This device consists of a cylinder 1, a piston 2, a piston rod 3 with a driven mass M, and a piping system 4 for pressure supply. We can describe the movement of the mass Mby the differential equation where 5 = the displacement of the driven mass, p = the pressure at the input of the cylinder, F= the area of the piston, Q = the useful and detrimental forces, ^ = F 3 p/2a 2 f 2 = the coefficient of hydraulic friction of the liquid flow in the cylinder, where p = density of the liquid, /= the area of the inlet-pipe cross section, a = the coefficient of the inlet hydraulic resistance. For movement of the piston to the right, the hydraulic friction is directed to the left and thus sgn 5=1. Denoting we can rewrite Equation (3.99) in the form The excitation A causes the movement of the mass M. FIGURE 3.22 Layout of a hydraulic drive. 3.4 Hydraulic Drive 89 Let us now try to define the operation time of the piston in the hydromechanism under consideration. For this purpose, we will rewrite Equation (3.100) in the form where Vis the speed of the mass. Let us assume that A can be taken as a constant value. Then Equation (3.101) can be rearranged as Integrating Equation (3.102), we obtain where C is the constant of the integration. The initial conditions are that when t = 0, V = 0; thus, C = 0, and we can finally write where From Equation (3.103) we obtain the following expression for the speed: From equation (3.104) we derive an expression describing the dependence s(£). We rewrite correspondingly: and Now, finally, we obtain [...]... three equations differ in the temperatures of the acting air In Equations (fl), ( 2) , and (f3), the absolute temperatures are taken as Tr = 29 3° K, T2 = 340 ° K and T3 = 40 0° K, respectively The higher the value of the temperature, the faster the piston moves, and the longer the distance 5 it travels during equal time intervals For instance (see Figure 3 .26 a)), in the considered example during 0.8 seconds... general case, the time t* required to reach a pressure sufficient to move the piston and overcome the load and the forces of resistance may be written in the form 3.5 Pneumodrive 95 There is an additional time component t0, which is the time needed by the pressure wave to travel from the valve to the orifice This time can be estimated as follows: where L is the length of the pipe from the valve to the... the cylinder, and Vs is the sound speed Vs = 340 m/sec In Figure 3 .26 we show the pressure development versus time To calculate the movement of the piston, we must deduce the differential equation for its displacement This requires some intermediate steps The thermodynamic equation for the air in the volume of the cylinder has the following form: (The subscript c indicates values belonging to the cylinder... rotated for one revolution, stretching the spring; thereafter, at a particular time, the system was freed Calculate the time needed by the drum 1 to complete 0.5 of a revolution under the influence of the spring overcoming the torque Tr FIGURE 3E- 1 Exercises 109 Exercise 3E -2 A blade with mass m = 1 kg driven by a spring is shown in Figure 3E -2 In the beginning, the spring is compressed by a distance... given temperatures, displacements of the piston correspondingly are T! = 29 3°K T2 = 340 °K T3 = 40 0°K ^ = 2. 12 meters, s2 = 2. 40 meters, s3 = 2. 76 meters [f 1 ] [f2] [f3] 3.5 Pneumodrive 97 FIGURE 3 .26 a) Displacement of the piston s versus time For a mechanism shown in Figure 3 .23 , for different air temperatures: 29 3, 340 , and 40 0°K fl = 40 0 s"[t]-(. 023 5 28 .7 29 3 00 02 5000001 +.05 100000 01)/s[t]+ 40 00+100... cylinder volume.) For the volume Vc, we can substitute the obvious expression where 5 = the displacement of the piston, and Fc = the cylinder's cross-sectional area After differentiating Equation (3. 129 ), we obtain or FIGURE 3 .26 Pressure development versus time in a real pneumatic drive Here: t0 is determined from (3. 128 ); ta from (3. 120 ); fc from (3. 127 ); p' and p" are intermediate pressure values 96... critical regime, let us make use of the example shown in Figure 3 . 24 In the left volume a pressure pr, which is much higher than pressure p, is created and maintained permanently At some moment of time the valve is opened and the gas from the left volume begins to flow into the right volume Because we stated in the beginning that p < pr, we have a situation where the ratio p / p r = f t grows theoretically... the geometrical axis of the column may be varied, thereby changing the moment of inertia of the whole system The case given in Figure 3 .29 b) is another illustration of the same case The drives providing the rotating speed a> in both cases are influenced by the variable moments of inertia of these systems We will now consider some simplified calculation examples corresponding to the cases described... of the driven mass.) Thus, from Equation (3. 140 ) we obtain and for T~ constant or Finally, we rewrite Equation (3.1 42 ) in the form or which gives 3.6 Brakes In this section we consider a special type of drive, one which must reduce the speed of a moving element until complete cessation of movement of the element is achieved, i .e. , a brake system Such a mechanism must be able to facilitate speed reduction... and 4kg In keeping with these parameters and for each of the chosen mass values, we write the needed expressions and obtain the solutions in graphic form as follows The calculations, as was mentioned above, were carried out for three different mass values This fact is reflected in the three curves on each graph shown in Figure 3.30a) and b) The upper ones belong to the smallest mass value (in our case . controls the rate of the resistance torque during the speed increase of the accelerated system. The problem is how to estimate the values of a^ and a 2 . We feel that the only . write where From Equation (3.103) we obtain the following expression for the speed: From equation (3.1 04) we derive an expression describing the dependence s(£). We rewrite correspondingly: and Now, . section we consider a special type of drive, one which must reduce the speed of a moving element until complete cessation of movement of the element is achieved, i .e. , a brake system.