Robotics 2 E Part 14 potx

two-For these assumptions it follows from the second Equation 9.78 thatand the first equation can be rewritten in the form taking 9.80 into account The solution for 9.81 for the conditio

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 tion in the figure we have for point O the coordinates are

posi-where v = index of the supporting leg, and zv and xv are components of the rv vector

(position of the foot's contact point)

Expression (9.73) allows, for instance, answering the question of what the matic requirements are for providing constant height of point O above the ground (forbetter energy saving) This condition states

kine-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 legged anthropoid walking machine

two-The dynamics of a two-legged robot present a complex problem for the generalcase 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 Theexchange 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 tion 2T, the following limit conditions are valid:

dura-FIGURE 9.62 Simplified model of a legged walker or "spider."

two-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 footcontact with the support surface Between these contact instants, #=0 (there is noreaction force), and the spider, in essence, flies The trajectory of the particle in thiscase consists of parabolic sections, as shown in Figure 9.63 In this figure

V Q = initial speed of the particle,

JC = horizontal component of speed V ,

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 ^ = -z0 To

acceler-ate the particle again to speed z 0 , an amount of energy A l must be expended by thefoot:

In stopping at the end of the flying period, the amount of energy A 2 expended by theother 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 fromthe 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.

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 formoving 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 forwalking:

The latter formula (9.93a and b) enables the value of the optimum length L0 of awalking steo to be derived:

The computation model presented here can give a rough estimation of power sumption in multi-legged vehicles by simply multiplying the results and distributingthe mass of the moving body among all the pairs of legs

con-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),

L0 = 0.7 m Then he or she will expend about 150 watts (0.036 kcal/sec) of power Wealso recommend changing from walking to running when a speed of 11.3 km/hr isreached However, if the reader is overweight, let him or her continue to walk withhigher speed (more energy will be expended) The speed record for walking is about15.5 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 72 of drum 2 to the axes of drum 1 requires culation of the common kinetic energy of the mechanism, which is

cal-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 withmoment of inertia / is:

The motion equation may then be written as follows:

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

JCG = 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

Now, we illustrate the same solution in MATHEMATICA language.

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 theinitial point until it comes into contact with the wire (here we neglect the frictionalresistance), and stage b) the cutting of the wire

Stage a)

We use Equations (3.27) to (3.81) Thus,

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

We now calculate the time 11 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

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:

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

and this energy (a remaining sum of kinetic and potential components after thecutting 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

motion by a differential equation This latter is

or

and

The solution jc is made up of two components:

The homogeneous component is given by

The partial solution is expressed as

Substituting X into Equation (f), we find:

Thus,

Differentiating (g), we obtain

Substituting the initial conditions into (g), we obtain the coefficients 7 and A When

t = 0 and x = L l = Q.lm, then the speed is x=-12.17 m/sec, and we obtain from (g) and

FIGURE 3E-2.2 Movement of the blade x z [t] during the cutting

process (from xa = 0.1 m to x 1 = 0.096 m).

3 Solution to Exercise 3E-3

In this case, the total moment of inertia 7 of the masses driven by the tor is calculated from

electromo-The differential equation, according to equation (3.41), takes the following form:

Substituting the numerical data into this equation, we rewrite it as

or

The solution is sought as a sum (0 = 0)^ + co 2 , where ^ is the homogeneous solution

in the form co\ = Ae at

Substituting this expression into Equation (a), we obtain

The partial solution is then sought as a constant co 2 = Q = const.

Substitution of this solution into Equation (a) yields

Substituting the desired speed co = 10 I/sec into (d), we may rewrite this expression as

Integrating Expression (b), we find the rotational angle 0(t) of the motor:

Thus,

Substituting t- 0.134 sec into (e), we obtain

0 = 3l|l/2.9exp[-2.9 0.134] + 0.134-1/2.9J = 0.776rad = 0.123rev.

Taking into account the radius r of the drum, we obtain the height h that the mass

FIGURE 3E-3.1 Rotational speed of the motor versus time.

4 Solution to Exercise 3E-3a)

The first step is to reduce the given mechanism to a single-mass system The

resis-tance torque T r on the axes of the electromotor, obviously, varies in inverse

propor-tion 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

mecha-nism, which is

where CD is the angular speed of the shaft of the motor The kinetic energy of the reduced

system with moment of inertia / with respect to the axle of the motor is

or

or

The differential equation according to Expression (3.41) takes the following form:

where the characteristic of the motor gives T= T + T^co - 4 - 0.1 CD.

Substituting the numerical data into this equation, we may rewrite it as

or

The solution is sought as a sum co = co r + co 2 , where o) l is the homogeneous solution

in the form: co^ = Ae at

Substituting this expression into Equation (a), we obtain

The partial solution is sought as a constant a> 2 = Q = const.

Substituting this constant into Equation (a) yields

The distance mass m l travels then is 3.6 • 0.02/4 = 0.018m.

Now let us consider the case with an AC electromotor as the driving source In thiscase the driving torque Tis

The equation taking into the account the given characteristic, (as it is shown in the

Expression (3.48)) and considering the definition of the concept s is as follows:

We solve it using the MATHEMATICA package

5 Solution to Exercise 3E-3b)

The first step is to reduce the given mechanism to a single-mass system The

resis-tance 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 screw Ts which,

respec-tively, is

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 1the axes of the motor is

or

or

The differential equation, according to Expression (3.41), takes the following form:

where the characteristic of the motor gives T= 7\ + T0 = 4 + 0.1 CD.

Substituting the numerical data into this equation, we may rewrite it as

or

The solution is sought as a sum co = (ol + co 2 , where ^ is the homogeneous solution

in the form

Substituting this expression into Equation (a), we obtain

The partial solution is sought as a constant co 2 = Q = const.

Substituting this constant into Equation (a) yields

Substituting t= 0.5 sec into (d), we obtain

and the travelling distance s is derived from the explanation given above as

6 Solution to Exercise 3E-4

To solve this problem, we use Expression (3.103), which is derived from Equationsand Formulas (3.99), 3.100), 3.101), and (3.102) Substituting the given parameters intothe above-mentioned expression, we obtain

Now let us take Expression (3.104) and rewrite it in the following form:

Obviously, from (a), the distance 5 the piston travels during time t can be derived, but, the opposite function, i.e., the time t needed to accomplish a certain travelling distance s is much easier to obtain by computerized means Using MATHEMATICA language, we build a graphic representation for the function s(t) that answers both

questions

s=1.333 Log[Cosh[8.661]]

j=Plot[s,{t,0,.l},AxesLabel->{"t","s"}]

FIGURE 3E-4.1 Displacement of the piston rod versus time

7 Solution to Exercise 3E-4a)

The first step is to reduce the given mechanism to a single-mass system The force

of resistance Q on the piston-rod, obviously, varies in inverse proportion to the ratio

R/r= 1:2.5 Thus,

The procedure of reducing inertia / of the wheels with respect to the piston-rod

requires calculation of the common kinetic energy E of the mechanism, which is

where Vis the speed of the piston-rod The mass M of the reduced system is

or

The active area F of the piston when lifting the mass m2 is, obviously,

To solve this problem, we use Expression (3.103), which is derived from Equationsand Formulas (3.99), 3.100), 3.101), and (3.102) Substituting the given parameters intothe above-mentioned expression, we obtain

Now the travel distance can be re-estimated by using the following formula:

Substituting into the latter formula the values for time t = 0.134 sec and

ft = 3.66 I/sec, we obtain the distance s travelled by the piston during this time:

Using MATHEMATICA language, we build a graphic representation for the

func-tion s(f):

s=4.54 Log[Cosh[1.83 t]]

j=Plot[s,{t,0,.l},AxesLabel->{"t","s"}]