Robotics 2 E Part 15 pdf

From these formulas, it follows that when the voltage is doubled, u s = 2, we havethe following expressions for the response time: From these formulas it follows that when the mass of th

For a DC motor with a characteristic T= T^-To^, Expression (a) becomes

A solution using MATHEMATICA language follows The numerical data given in the

problem are substituted by symbols (here, for reasons of convenience, we denote <f> = v and0= w).

FIGURE 3E-6.1 a) Rotation angle v; b) Speed w of the column versus time.

13 Solution to Exercise 3E-6a)

The rotation angle 0 of the disc depends upon the torques acting on the nism The driving torque Tmust be equalized by inertia torques, in keeping with expres-sion (3.165):

mecha-where / is the common moment of inertia of the disc /0 and the moving mass Ir

Obviously:

For a DC motor with a characteristic T=T^- r00, the Expression (a) becomes

For the same data as in solution 3E-6, and where 7\ = 0.1 Nm and T0 = 0.025 Nm/sec and RQ = 0.5 m, the following solution using MATHEMATICA is given The equation is

FIGURE 3E-6a).l a) Rotation angle v; b) Speed w of the column versus time.

14 Solution to Exercise 3E-7

To answer the questions we use Formula (3.39) Thus, from this formula, it follows

that when the number of winds Ws = 2 is doubled, we have the following expressions

for the response time:

From these formulas, it follows that when the voltage is doubled, u s = 2, we have

the following expressions for the response time:

From these formulas it follows that when the mass of the armature is doubled,

m s = 2, we have the following expressions for the response time:

15 Solution to Exercise 4E-1

Case a)

From geometrical considerations, the motion function n(jc) becomes

Differentiating (a), we obtain

Thus,

Substituting the given data into (c), we obtain for y

By differentiating (b), we obtain the following dependence from Expression (4.3)

[the case where x = 0]:

From (c) and the Relationship (4.4) we obtain

Substituting the numerical data into (c) and (d), we obtain

16 Solution to Exercise 4E-2

From the Formula (4.24) and its derivatives, we have:

Here, it follows from the description of the problem that

and therefore

Thus, from (a) we obtain

To find the angle 0 corresponding to the maximum pressure angle amax, we entiate (b):

differ-From (c), it follows that

2-0.08 cos (4-0)+fecos(4-0)-Mcos2(4-0)+sin2(4-0)] = 0

or

On the other hand, from (b), we have

Substituting tana = tan 20° = 0.324 into (e), and from (d), we obtain

Solving Equation (f) by any method (for instance, graphically, by the method ofNewton, or by computer) we obtain

which from (d) gives for h

The solution in MATHEMATICA language is

al=(2 Sin[8 f]-.364 (Cos[4 f]+(Cos[4 f])A2))/(l-Cos[4 f])-.364

bl=FindRoot[al==0,{f,.5}]

{f-> 0.369625}

17 Solution to Exercise 6E-1

Condition (6.17) states that horizontal component A/, of the acceleration takes theform

A is the vibrational amplitude Assuming that the vibrations S have the form:

Then the accelerations are

18 Solution to Exercise 7E-1

Here we use Equation System (7.1) Since the two levers press the strip from both

sides (upper and lower), the mechanism must develop a friction force P = F/2 at every

contact point Thus, the equations for forces and torques with respect to point Obecome

and

Here, R x and R y are the reaction forces in hinge 0; Nis the normal force at the contact

point between the strip and the lever From (a), we express the normal force Nas

From the Equation (b), we express the force Q developed by the spring as

Reactions R x and Ry are, respectively,

19 Solution to Exercise 7E-la)

Here we use equation system (7.6) Since the two rollers press the strip from both

sides (upper and lower), the mechanism must develop a friction force F b = QI2 at every

point of contact with the strip Thus, the equations for forces and torques with respect

to point O become

From the Equation (a), it follows that

From the Equation (b), it follows that

or

From the Equation (c) and the given mechanism it follows that

and finally

20 Solution to Exercise 7E-lb

We continue to use Equation System (7.1) Since the two levers press the strip from

both sides (right and left), the mechanism must develop a friction force F= Q/2 at

every contact point Thus, the equations for forces and torques with respect to point

O become

and

21 Solution to Exercise 7E-2

The angular frequency co of the oscillations of the bowl is

CQ = 2nf=2x5Q = 3l4 I/sec.

The motion S of the bowl is: S = 0.0001 sin 314 t m The acceleration S of the bowl

obviously is

S = -0<w2sina£ = -0.0001-3142sin314tm/sec2.The maximal value of the acceleration Smax is

Smax = aa> 2 = 0.0001-3142 = 10 ml sec2

The angle/3 = y-oc = 30° - 2° = 28° From Expressions (7.33-7.34), we calculate the

values of critical accelerations for the half-periods of both positive and negative lations Thus,

oscil-and

The latter expression means that during the second half-period of oscillations slideconditions practically do not occur for the body on the tray By applying Expression(7.35), we check whether rebound conditions exist on the tray, a situation that occurs

when the acceleration exceeds the value S r Thus,

At any point of movement, no point of the bowl reaches this acceleration value fore, there is no rebound in the discussed case

There-We can now proceed to calculate the displacement of the items From the curves

in Figure 7.25 it follows that the time t v at which the slide begins (section EM) and thegroove lags behind the item, is defined as

At this moment in time, the speed V 0 of the item (and the bowl) is defined as

The slide begins with this speed and is under the influence of the friction force

F = -fj.m(g + y ) acting backwards For our engineering purposes, we simplify this inition to the form F= //mg This force causes deceleration:

def-W L = -ju = -0.6 • 9.8 = 5.88 m / sec2.This assumption gives a lower estimation of the displacement The following gives theupper estimation:

This condition exists during time t2 , which is defined as

The displacement S1 is then

or

or

0.000053m < 81 < 0.000083m.

It is interesting to observe the influence of the friction coefficient ju on the values

of the critical accelerations for both oscillation directions We show here the tation in MATHEMATICA language Results are given in Figure 7E-2 (For convenience

compu-in MATHEMATICA we use m for the friction coefficient.)

gl=Plot[9.8 (Sin[2 Degree]+m Cos[2 Degree])/

(m Sin [30 Degree]+Cos [30 Degree]),

{m,.2,l},AxesLabel->{"m","s""}]

g2=Plot[9.8 (Sin[2 Degree]-m Cos[2 Degree])/

(m Sin[30 Degree]-Cos [30 Degree]),

{m,.2,l},AxesLabel->{"m","s""}]

Show[gl,g2]

FIGURE 7E-2.1 Dependence "critical acceleration s"

versus friction coefficient "m" for the specific design

of the vibrofeeder as in this and next exercises

This displacement takes place 50 times every second Therefore, the total

dis-placement H during one second is

0.00265m <H V < 0.0041 m [e]

22 Solution to Exercise 7E-3

When increasing the vibrations amplitude "a" to 0.00015 m for the same

vibro-feeder as in the previous Exercise 7E-2, obviously, we change the dynamics of thedevice However, its characteristics remain the same Therefore, we have the same

values of the critical accelerations Scr and S'cr as before The changes in the dynamic

behavior of the device take place because of the fact that the maximal value of theacceleration of the bowl Smax becomes higher, due to the increased oscillations ampli-

tude "a." In this case, we have

Smax =a-a) 2 =0.00015-3142 =15m/sec

We follow the same procedure as in the previous problem, and for the value of thespeed at the moment in time that the slide begins, we obtain

and

The conditions of this exercise result in the appearance of a backslide in the domain

EK To calculate the back displacement 8 2 , we apply the ideas of the forward

dis-placement in formulas (b), (c), (d), and (e):

or

- 0.000087 m > dl > -0.000044 m.

The total displacement S3 during one period of the bowl's vibrations is obviously

6 3 =S l +S 2 =0.00021 + 0.000044 = 0.000166mand

<53' = 81 + S2 ' = 0.000165-0.000087 = 0.000078 m.

Finally,

0.0039 m <#! < 0.0083 m.

And, for the last time, we put forward an illustration of the displacement of a body

on the tray of the feeder calculated in keeping with Equation (7.34) by means of theMATHEMATICA language The result is shown in Figure 7E-3

FIGURE 7E-3.1 Body movement on the tray of the vibrofeeder

23 Solution to Exercise 7E-4

To answer the question we use Figure 7.28 We begin with the simplest case—a

cube with a hole drilled symmetrically in the middle of it (A =B = C and H=2h) This

case is analogous to case 4 in the figure (the hole makes a difference to one of thedimensions) Therefore, it has three different positions on the tray When a right par-

allelepiped with a symmetrically located hole (H= 2h)—for both cases: A ±B = C and

A±B± C—is considered, we have a body possessing three planes of symmetry—line

3 in the figure This gives six different positions of the body on the tray Finally, the

most common case, when the hole is located so that H± 2h, fits line 2 in the figure for

both cases The body possesses two planes of symmetry and, therefore, 12 differentpositions on the tray are possible This results are presented in the following table

H = 2h

H*2h

A = B = C

312

A * B = C

612

A * B * C

1212

24 Solution to Exercise 7E-5

To answer the question we use Figure 7.28 This is the case that corresponds to line

2 in the figure The body possesses two planes of symmetry and, therefore, 12 ent positions on the tray are possible Because of its internal asymmetry, this bodyrequires special means for its orientation These means are, for instance, a) utilization

differ-of the location differ-of the asymmetrical mass center, and b) means differ-of electrodynamic ormagnetic orientation

Recommended Readings

Fu, K S., R C Gonzales, and C S G Lee, Robotics: Control, Sensing, Vision and Intelligence,

McGraw-Hill Book Company, New York, 1987

Pessen, D W., Industrial Automation, John Wiley & Sons, New York, 1989.

Ogata, Katsuhiko, System Dynamics: Second Edition, Prentice-Hall, Englewood Cliffs, New

Jersey, 1992

Dieter, George, Engineering Design: A Materials and Processing Approach: Second Edition,

McGraw-Hill, Inc., New York, 1991

Schey, John A., Introduction to Manufacturing Processes: Second Edition, McGraw-Hill

Inter-national Editions, New York, 1987

Powers Jr., John H., Computer-Automated Manufacturing, McGraw-Hill International

Edi-tions, New York, 1987

Critchlow, Arthur J., Introduction to Robotics, Macmillan Publishing Company, New York,

Collier Macmillan Publishers, London, 1985

Bradley, D A., D Dawson, N C Burd, and A J Loader, Mechatronics: Electronics in Products

and Processes, Chapman & Hall, London, 1996.

Slocum, Alexander H., Precision Machine Design, Prentice-Hall, Englewood Cliffs, New Jersey,

1992

Erdman, Arthur G., George N Sandor, Mechanism Design: Analyses and Syntheses,

Prentice-Hall International, Inc., Simon & Schuster/ A Viacom Company, Upper Saddle River,New Jersey, 1997

Rampersad, Hubert K., Integrated and Simultaneous Design for Robotic Assembly, John Wiley

& Sons, Chichester, New York, 1993

Groover, Mikell P., Fundamentals of Modern Manufacturing: Materials, Processes and Systems,

Prentice-Hall International, Inc., Simon & Schuster, Upper Saddle River, New Jersey,1996

Lindberg, Roy A., Processes and Materials of Manufacture: Fourth Edition, Allyn and Bacon,

Boston, 1990

Krar, S E, J W Oswald, and J E St Amand, Technology of Machine Tools: Third Edition,

McGraw-Hill International Editions, New York, 1984

423

Miu, Denny K., Mechatronics: Electromechanics and Contromechanics, Springer-Verlag, New

York, Berlin, 1992

Groover, Mikell R, Automation, Production Systems, and Computer Integrated Manufacturing,

Prentice-Hall International, Inc., Simon & Schuster, Englewood Cliffs, New Jersey, 1987

Brown, James, Modern Manufacturing Processes, Industrial Press Inc., New York, 1991.

Fawcett, J N., J S Burdess, Basic Mechanics with Engineer ing Applications, Edward Arnold, A

division of Hodder and Stoughton, London, New York, 1988

Birmingham, R., G Cleland, R Driver, and D Maffin, Understanding Engineering Design:

Context, Theory and Practice, Prentice-Hall, London, New York, 1997.

Mabie, Hamilton H., Charles F Reinhoholtz, Mechanisms and Dynamics of Machinery, John

Wiley & Sons, New York, 1987

Meriam, J L., and L G Kraige, Engineering Mechanics: Dynamics, SI Version, vol 2, John Wiley

& Sons, Inc., New York, 1993

Askeland, Donald R., The Science and Engineering of Materials: Third Edition, PWS Publishing

Company, Boston, 1994

List of Main Symbols

L distance, length

M, m mass

p pressure in a hydraulic or pneumatic system

r, r(f) radius or variable distance of a rotating mass

s slip in a synchronous electromotor

s, x linear displacement, deflection

0 inclination angle, angular displacement

(o frequency of oscillations, angular speed

/ dry friction coefficient

/ moment of inertia of a cross-section of a link/ moment of inertia of a massive body

a, ft, 7 geometrical angular dimensions

n symbol of the position function

H, h height, pressure

L inductance

R electrical resistance

Afl increment of electrical resistance

s, x linear displacements or distance

A value of a geometrical gap

JLI friction coefficient

MI, u 2 control functions

V velocity of a moving body

A0, Ayf increments of angles

V(t) variable angle of a link

co frequency of oscillations

artificial muscle, 339 drilling head, 310

assembling by electromagnetic means, 298 drives

automatic assembly, 284,312 electric, 75

bridge, electrical, 176 electromotors

bending heads, 309 alternate current, 76

feeding, continued processing, 37

continuous, 227 length compensator, 207, 208

of granular materials, 229 Leshot, Jean-Frederic, 3

interrupted, 227 levitation, 370

of liquids, 228 limit switches, 147,208

of oriented parts, 235 loom, Jacquard's, 4,141

of rods, ribbons, strips, wires, 231

vibrofeeders, 223,245 M

friction magazines, 18,238

"dry," 362 Magnus, Albertus, 4

lubricated, 360, 371 Mandsley, Henry, 4

Geneva mechanism, 117,211 one-revolution, 130, 215

Goertz, 12 Mergenthaler, Ottomar, 5

grippers, 350 mobile robots, 3, 372

guides, 358 Muller, Johann, 3

Gutenberg, Johannes, 5 muscle, artificial, 339

H N

Heal, W E., 48 National Bureau of Standards (U.S.), 1Hero of Alexandria, 3

Hitchcock, H K., 49 O

hoppers, 18,237 optimal-time trajectory, 317

box, 239 orientation devices, 227,254

inspection, systems, 300 Pilkington, 62

interferometer, 181 Pitot device, 191

indexing tables, 217 position function, 116

Lawton, Tolbart, 5 bang-bang, 2, 8,35,131

layout fixed stop, 2

kinematic, 55 limited degree-of-freedom, 3

mobile, 3,372 threading head, 310

pick and place, 2 time

rolling supports, 366 auxilliary, 53

running, 382 operating, 53

timing diagram

S linear approach, 53

Sendzemir process, 44 circular approach, 53

sensors transporting devices

induction, 178 Vancanson, de Jacques, 3

item presence, 202 variable moment of inertia and mass, 103photoelectric, 182 vibrations

pneumatic, 183 automatic damping, 162,166

pressure, 198 damping, 157

speed, 188, 207 dynamic damping, 159,162

tactile, 355 free, 159

temperature, 200 of rotating shafts, 166

serpent-like manipulators, 317 vibrators, 252

spherical manipulators, 14, 328 vibrofeeding, 245

Stewart platform, 347 vibrotransportation, 223

subcritical air flow regime, 92

supercritical air flow regime, 92, 95 W

"waiter," automatic, 28, 375

T walking, 377

tables walking machines, 29, 377

indexing, 15,117, 217,218 Watt, James, 4

X-Fcoordinate, 15,315, 324, 326, 359

tachogenerator, 207 X

tactile sensors, 207 X-Fcoordinate tables, 15, 315,324,326,359Theophilus, 3