Robotics 2 E Part 4 ppt

Substituting Equation 3.55 into Equation 3.54 we obtainor 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 F

on the design and dimensions of the motors These parameters usually change in thefollowing range:

Torque: TL =22.5 -1125 kg m, Inertia of the rotor: Ir = 1.2-10000 g cm2,Maximum pulse rate: S = 150 -5- 50000 pps

A typical torque-versus-time characteristic for stepper motors is presented in Figure3.18 The point that should be stressed is that changes in the torque are different fordifferent pulse rates The lower the value of the pulse rate (i.e., the duration of onepulse is longer), the higher the torque at the beginning of the switching and the lowerthe 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 loadcan be run bi-directionally without losing a step This condition occurs by a speed ofabout 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 Crepresents the detent-like torque which is typical only for motors with permanentmagnet rotors At this point even a nonenergized stator resists the movement of therotor to move The motor "remembers" its position Curves 1 and 2 represent the behav-

FIGURE 3.18 Torque-versus-time dependencefor a stepper motor

FIGURE 3.19 Torque-versus-pulse-ratedependence for stepper motors

3.3 Electric Drives 81ior 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 motors as drives for automatic machines and systems, some other comparative fea-tures of these electromotors must be discussed

electro-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 formers) The advantage of AC motors (both one- and three-phase) is that they operate

trans-on the standard voltage available at any industrial site In addititrans-on, a three-phase tion motor with a squirrel-cage rotor is cheaper than any other type of motor of thesame power For accurate positioning both DC and AC motors require feedbacks Incontrast, 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)

induc-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 sion which is proportional to the speed of rotation of the machine Thus we can writefor Tr :

expres-Obviously, a l andor2 are constants The physical meaning of the value a 1 is the initialresistance 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 resistancetorque 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 tothe case under design

Substituting Equations (3.46) and (3.49) into Equation (3.41 ) we obtain a linearequation in the form

After simplification we obtain

where C = (oc 2 + «2) /I and 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 we have

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 a2 = 0 we obtain

3.3 Electric Drives 83

When there is no resistance torque, i.e., av = a2 = 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 Forthis purpose we must rewrite this expression as follows:

Integrating this dependence termwise, we obtain:

com-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

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 85When 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 (2Tm 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 coislant 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 tfsame way as are asynchronous motors Therefore, the calculations are of the same soiand 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 assumptioFirst, 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 trfollowing equation of the movement of the machine:

Rewriting this expression, we obtain

The solution consists of two components, co = col + (O 2 For the solution of the

homo-geneous equation we have

and for the particular solution we have

Substituting these solutions in the homogeneous form of Equation (3.83) and in itscomplete 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 bydirect integration of solution (3.88):

or

For the second assumption, we introduce into the excitation torque a "saw"-likeperiodic 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 straightlines, 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 Figure3.18) of a stepper motor by inclinedstraight lines

3.3 Electric Drives 87and 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 formsand 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

Obviously, the calculations shown above answer the following questions:

• How long does it take for the drive to reach the desired angle or (using sponding transmission) displacement?

corre-• 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 desiredangle, displacement, or speed in a specific time?

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 leftand 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 89Let us now try to define the operation time of the piston in the hydromechanismunder 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(£) Werewrite correspondingly:

and

Now, finally, we obtain

Integrating this, latter expression we obtain the required formula in the following form:

An example in MATHEMATICA language is given Let us suppose that a device responding to Figure 3.22a) is described by the following parameters:

cor-M = 1000 kg, p = 700 N/cm2, Q = 5000 N, F = 75 cm2,

¥=100Nsec2/m2

Then ft = 4.36 I/sec, m = 0.2 1/m, A = 47.5 ml sec 2

The solution for this specific example is:

s[t] = = 2/.2 Log[Cosh[4.36/2 t]l

j = Plot[2/.2 Log[Cosh[4.36/2 t]],{t,0,.l},AxesLabel->{"t","s"},

PlotRange->AU,Frame->True,GridLines->Automaticl

It is more difficult to solve the problem for a case in which the value A varies, say,

a function of the piston's displacement Thus: A(s) For this purpose we rearrange tion (3.100) and substitute y = s 2 in that expression We can then rewrite the equation

Equa-in the form

(Note: If s2 = y, then dy/dt=2ss, which gives s = dy/2ds.}

Equation (3.105) is linear with respect to y, and thus, in accordance with the

super-position principle, the solution must be expressed as the following sum:

where

y0 = the solution of the homogeneous equation,

y l = the partial solution for A in the right-hand side of the equation.

We seek y0 in the form

Substituting Equation (3.106) into Equation (3.105), we find that

FIGURE 3.22a) Solution: piston displacement versus

time for the above-given mechanism

3.5 Pneumodrive 91

Supposing, for example, that A(s) = a^ + a 2 s; we seek the solution y l in the form

Substituting this expression into Equation (3.105) and comparing the coefficients onboth sides of the equation, we find that

in in, For initial conditions t= 0 and V= 0, we also have y= 0 (Remember: y = 2V.V} This condition gives the following expression for Y:

Finally, the complete solution can be written as

Substituting back the meaning of y we obtain

3.5 Pneumodrive

In general, the dynamics of a pneumomechanism may be described by a system

of differential equations which depict the movement of the pneumatically driven massand the changes in the air parameters in the working volume The work of a pneu-momechanism differs from that of a hydraulic mechanism in the nature of the outflow

of the air through the orifices and the process of filling up the cylinder volume Let usconsider the mechanism for which the layout is given in Figure 3.23 Let us suppose

the processes of outflow and filling up are adiabatic, and the pressure p r in the receiver

1 is constant From thermodynamics we know that the rate of flow of the air throughthe pipeline 2 may be described by the formula

where

G = the rate of flow,

a = coefficient of aerodynamic resistance,

F p = cross-sectional area of pipe 2 (m2),

p r = air pressure in the receiver 1,

T = absolute temperature of the air in the receiver,

FIGURE 3.23 Layout of apneumatic drive.

R = gas constant,

ft = p/p r = ratio of the pressure in cylinder 3 to that in the receiver,

k = adiabatic exponent (k= 1.41).

At this stage, we must distinguish between supercritical and subcritical regimes If

we denote p cr as the critical pressure, then

(for airftcr = 0.528) for the supercritical regime pr > p/fi crt and for the subcritical regime,

p r < piP cr , where p = pressure in the cylinder.

To explain the meaning of the concept of the 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 from zero to one (when p becomes equal to pr) The air sumption through the connecting pipe develops as shown in Figure 3.25: from/? = 0 up

con-loft =ft cr it stays constant until ftcr = 0.528, while for the range ft cr <ft =£ 1 the tion changes from Gcr to zero as the pressure in the right volume becomes equal to pr

consump-FIGURE 3.24 Layout of gas transferfrom one volume into another

FIGURE 3.25 Air consumption in theconnecting pipe of the layout shown inFigure 3.24 versus the pressure ratio inthe connected volumes

3.5 Pneumodrive 93

For the supercritical regime, the pressure p l in the cylinder input orifice is constant,i.e.,

Substituting p l from Equation (3.113) into Equation (3.111), we obtain, for the

super-critical flow rate G cr

Under standard conditions (T r = 293°K) we have

Now let us find the time required to fill the cylinder from the initial value of the

pres-sure p 0 to the final value p l (p r < 0.528^) while the air temperature remains constant For a cylinder volume V c we have

where

Gc = instantaneous value of the air weight, and

T= the air temperature.

During an infinitesimal time period dt the pressure will change over a value dp, and the amount of air will change by G cr dt Thus,

Substituting Equation (3.116) from Equation (3.117), we obtain

By integration we obtain

By substituting Equation (3.115) into Equation (3.119), we finally reach

When time ^ has passed, two outcomes are possible:

1 The piston begins its movement earlier than ^ (i.e., the movement begins before

pressure p l appears in the cylinder volume)

2 The pressure in the cylinder is still not sufficient to initiate the movement ofthe piston

For the first case we have to integrate Equation (3.118) in the limits from p0 to some pressure p* which is less than that at ^ Thus, in place of Equation (3.120) we obtain

For the second case, we have to continue our investigation for the subcriticalregimes For a subcritical regime,

Now, in Equation (3.111) the value of/3 varies from the initial value of 0.528 to 1 In

this case we must substitute in the differential Equation (3.118) G, which is not stant and is defined by Equation (3.111) Thus,

con-Since p=fip r ,

Therefore,

To integrate this equation, we introduce an auxiliary function,

which gives

After substituting Equations (3.124) and (3.125) in Equation (3.123), we obtain

The limits of integration are determined by the initial value of/?0 and the critical value

of ft cr Thus,

In the 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