Robotics 2 E Part 15 pdf

410
Solutions
to the
Exercises
or
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).
fl=.01
v"[t]+.5
(Cos[.51])*2
v"[t]-
.125(Sin[t])v'[t] l
jl=NDSolvel{fl==0,vIO]==0,v'[0]==0},{v[t]},{t,0,2}l
bl=Plot[Evaluate[v[t]/.jl],{t,0,2}^xesLabel->{"t","v"}]
£2=.01
w'ltl+.S
(Cos[.5
t])A2
w'W-
.125(Sin[t])w[t] l
j2=NDSolve[{f2==0,w[0]==0},{w[t]},{t,0,2}]
b2=Plot[Evaluate[w[t]/.j2]
,{t,0,2},AxesLabel->{"t","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
mecha-
nism.
The
driving torque
Tmust
be
equalized
by
inertia torques,
in
keeping with expres-
sion (3.165):

where
/ is the
common moment
of
inertia
of the
disc
/
0
and the
moving
mass
I
r
.
Obviously:
Solutions
to the
Exercises
411
Therefore,
For
a DC
motor with
a
characteristic
T=T^-
r
0
0,

the
Expression
(a)
becomes
For
the
same
data
as in
solution 3E-6,
and
where
7\
=
0.1
Nm and
T
0
=
0.025 Nm/sec
and
RQ
= 0.5 m, the
following
solution using
MATHEMATICA
is
given.
The
equation

is
0.010+0.5
(cos
0.51]
2
0-0.125
sin
10-0.1+0.025
0 = 0.
f01=.01
v"[t]+.5
(Cos[.5
t])
A
2
v"[t] 125
(Sin[t])
v'[t]
1
+0.025
v'[t]
j01=NDSolve[{f01==0,v[0]==oy[0]==0},{v[t]},{t,0,2}]
b01=Plot[Evaluate[v[t]/.j01],{t,0,2}^VxesLabel->{"t","v"}]
f02=.01
w1t]+.5
(Cos[.5
t])A2
w'[t] 125
(Sin[tl)
w[t]

1
+0.025 w[t]
j02=NDSolve[{f02==0,w[0]==0,w'[0]==0},{w[t]},{t,0,2}]
b02=Plot[Evaluate[w[t]
/.J02]
,{t,0,2},AxesLabel->{"t","w"}]
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
W
s
= 2 is
doubled,
we
have
the
following expressions
for
the
response time:
412
Solutions
to the
Exercises
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]:
Solutions
to the
Exercises
413
From
(c) and the
Relationship (4.4)
we
obtain
Substituting
the
numerical data into
(c) and
(d),
we
obtain
Case

b)
From
the
geometry
of the
given mechanism,
we
have
AD =
CE.
Then,
the
motion
function
Yl(x)
is
defined
as
follows:
y
=
n(</0
=
AOsin0
=
0.2sm30°
=
0.1m.
Thus,
y

=
n'(0)0
=
0.2cos30°
5
=
0.866
m /
sec,
and
y
=
n"(0)0
2
=-0.2sin30°
5
2
=
-2.5m/sec
2
.
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
414
Solutions
to the
Exercises
To
find the
angle
0
corresponding
to the
maximum pressure angle
a
max
,

we
differ-
entiate (b):
From
(c),
it
follows
that
2-0.08
cos
(4-0)+fecos(4-0)-Mcos
2
(4-0)+sin
2
(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
of
Newton,
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
the
form
Solutions
to the
Exercises
415
where
A
is the
vibrational
amplitude. Assuming
that

the
vibrations
S
have
the
form:
Then
the
accelerations
are
Thus,
Case
a)
Condition
(a) can
then
be
rewritten
in a
form
that
takes into account that
ft>
=
2flrf=2*:50
=
3141/sec:
From
(a)
follows

or
Case
b)
From
the
condition
(a) and
taking into account
(b) of the
previous case,
it
follows
also
for
this case that
Now
we
obtain
or
and
416
Solutions
to the
Exercises
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
O
become
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
Solutions
to the
Exercises
417
and
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
418
Solutions
to the
Exercises
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<w
2
sina£
=
-0.0001-314
2
sin314tm/sec

2
.
The
maximal value
of the
acceleration
S
max
is
Smax
=
aa>
2
=
0.0001-314
2
=
10
ml
sec
2
.
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 oscil-
lations. Thus,
and
The
latter expression means
that
during
the
second half-period
of
oscillations slide
conditions 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. There-
fore,

there
is no
rebound
in the
discussed case.
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 the
groove
lags behind
the
item,
is
defined
as
Solutions
to the
Exercises
419
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 def-
inition
to the
form
F=
//mg.
This
force
causes deceleration:
W
L
=
-ju
=

-0.6
• 9.8
=
5.88
m
/
sec
2
.
This
assumption gives
a
lower estimation
of the
displacement.
The
following
gives
the
upper estimation:
This
condition exists during time
t
2
,
which
is
defined
as
The

displacement
S
1
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
compu-
tation
in
MATHEMATICA
language. Results
are
given
in
Figure 7E-2. (For convenience
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]
420
Solutions
to the
Exercises
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 the
device.
However,
its
characteristics remain
the
same.
Therefore,
we
have
the
same
values
of the
critical accelerations
S
cr
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 the
acceleration
of the
bowl
S
max
becomes higher,
due to the
increased oscillations ampli-
tude
"a."
In
this case,
we

have
Smax
=a-a)
2
=0.00015-314
2
=15m/sec.
We
follow
the
same procedure
as in the
previous problem,
and for the
value
of the
speed
at the
moment
in
time that
the
slide begins,
we
obtain
and
Solutions
to the
Exercises
421

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
>
d
l
>
-0.000044
m.
The
total displacement
S
3
during
one
period
of the
bowl's vibrations
is
obviously
6
3
=S
l
+S
2
=0.00021

+
0.000044
=
0.000166m
and
<5
3
'
=
81
+
S
2
'
=
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 the
MATHEMATICA
language.
The
result
is
shown
in
Figure 7E-3.
FIGURE
7E-3.1
Body
movement

on the
tray
of the
vibrofeeder.
422
Solutions
to the
Exercises
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 the
dimensions).
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
different
positions
on the
tray
are
possible. This results
are
presented
in the

following
table.
H
= 2h
H*2h
A
= B = C
3
12
A*B
= C
6
12
A*B*C
12
12
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
differ-
ent
positions
on the
tray
are
possible. Because
of its
internal asymmetry, this body
requires special means
for its
orientation. These means are,
for
instance,
a)
utilization
of
the
location
of the
asymmetrical mass center,

and b)
means
of
electrodynamic
or
magnetic 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
424
Recommended
Readings
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
Chapter
1
L
distance, length
P
productivity
of a
machine
T
pure processing time
^
duration
of a
movement
t
z
duration
of a

pause
V
speed
of the
product
in the
machine
T
time
losses
Chapter
2
none
Chapter
3
A, B, C, D
integration constants
«j,
02,
b
lt
b
2
,
q
constants
a
linear acceleration
c
stiffness

of a
spring
or
elastic link
F,P
force
/
dry
friction
coefficient
g
acceleration
of
gravitation
I
moment
of
inertia
of a
link
425
426
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
T
torque
t
time
a
angular acceleration
«!,
a

2
constants
d
clearance
or gap
0
inclination angle, angular displacement
(o
frequency
of
oscillations, angular speed
Chapter
4
a,
D, e, h,
I,
r
geometrical dimensions
b,
c
damping
coefficients
c,
k
stiffness
of
springs
or
elastic links
E

Young
modulus
F
force
/
dry
friction
coefficient
/
moment
of
inertia
of
a
cross-section
of a
link
/
moment
of
inertia
of
a
massive body
m
mass
q
deflection
of an
elastic link

s,
x, y
linear displacement
T
torque
r
time
V
speed
Z-L,
z
2
,
z
3
,
z
4
number
of
teeth
a, ft, 7
geometrical angular dimensions
n
symbol
of
the
position
function
0,

y
angular displacements
CD
angular speed,
frequency
of
oscillations
Chapter
5
a, d,
I
geometrical linear dimensions
c
capacitance
E
electromotive
force
/
area
of
a
pipe's cross section
List
of
Main Symbols
427
H,
h
height, pressure
L

inductance
R
electrical resistance
Afl
increment
of
electrical resistance
s,
x
linear displacements
or
distance
T
temperature
t
time
V
speed
v
voltage
Ay
increment
of
voltage
W
energy
w
number
of
winds

Xi
inductive resistance
z
complete electrical resistance
a
angular displacement
«!,
«
2
coefficient
of
flow
rates
in
pipe sections
1 and 2,
respectively
e
dielectric
permitiviry
//
magnetic permeability
<£
magnetic
flow
p
density
of flowing
liquid
CD

frequency
of
oscillations
Chapter
6
A
linear acceleration
g
gravitation acceleration
L
geometrical dimensions
m
mass
P
force
T
time, period
V
speed
x
displacement
z
number
of
slots
or
teeth
fj,
friction
coefficient

0
angular displacement
Q,
co
angular speed
Chapter
7
a
linear acceleration
a, b, c, d, D, h,
I,
L
geometrical dimensions
428
List
of
Main Symbols
g
acceleration
of
gravity
F,N,P,Q
forces
m
mass
n
revolutions
per
minute
s,

x, y
displacements
T
torque
t
time
At
value
of a
clearance
V
speed
W
energy
a,
f},
j
angles
of
inclination
A
value
of a
geometrical
gap
JLI
friction coefficient
p
friction
angle

a>
frequency
of
oscillations
Chapter
8
none
Chapter
9
a
linear acceleration
A,
B,
C
components
of
opposed
forces,
constants
of
integration
A.
A
2
,
W
energy
a,
b,
c, h,

I,
L
geometrical dimensions
of
links
C
number
of
combinations
c
stiffness
of
elastic elements
F
force
f
A
,
f
B
,
f
c
friction forces
corresponding
to the
opposing
forces
/
moment

of
inertia
K
constant
M,
m
moving mass
N
power
P
weight
of an
element
R
radius
of an
element
r(t)
variable distance
of a
rotating mass
from
the
rotation
axis
5
displacement
of a
gripper's
jaw

T
torque
t,
T
time
List
of
Main
Symbols
429
MI,
u
2
control
functions
V
velocity
of a
moving body
x
displacement
of a
body
x
lt
x
2
, x
3
,

x^
auxiliary variables
X,
Y,
Z
components
of
forces
z(t]
variable heights
a
angular acceleration
a
v
,
fi
v
binding angles
of the
links
of the
v-th
leg
0(f)
variable azimuth
A0,
Ayf
increments
of
angles

V(t)
variable angle
of a
link
co
frequency
of
oscillations
This page intentionally left blank
Index
A
clepsydra,
4
absorber
of
energy,
161
concepts
copying,
45
accuracy,
148,362
conveyor
air-cushion,
369
chain-like,
211
amplifiers
vibrating,
223,245

electromechanical,
135,140,144
cylindrical manipulator,
15,315
hydraulic,
138
pneumatic,
173
D
android,
3
damping
of
vibrations,
157
Architas
of
Tarentum,
3
dextrous hand,
355
artificial
muscle,
339
drilling head,
310
assembling
by
electromagnetic means,
298

drives
automatic assembly,
284,312
electric,
75
azimuth,
316
electromagnetic,
71
hydraulic,
88
B
mechanical,
64
Bacon,
Roger,
3
pneumatic,
91
Boucher,
Guillaume,
3
brakes,
99 E
bridge, electrical,
176
electromotors
bending heads,
309
alternate current,

76
direct current,
75
C
stepper,
79,139
camshaft,
59,123,151
synchronous,
78
cards, perforated,
135
exoskeleton,
29
Cartesian-type manipulators,
15,16,324,359
Chapek,
Karel,
1 F
Chebyshev,
29,377
feeding
clamping,
232
bins,
242
431
432
Index
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
rolling,
366
manipulators,
2,12,315,
326
functional
devices,
25,352
manufacturing processes,
37
mechanisms
G
cam,
123
gauges Geneva,
117

pneumatic,
139,183
master controller,
135
strain,
195
miscellaneous,
307
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.),
1
Hero
of
Alexandria,
3
Hitchcock,
H. K., 49 O
hoppers,
18,237
optimal-time trajectory,
317
box,
239
orientation devices,
227,254
horizontal,
240
active,
266
vertical,
239
logical,
271

knife,
244
passive,
259
tray,
237
orientation
by
nonmechanical
means,
274
I
P
inspection, systems,
300
Pilkington,
62
interferometer,
181
Pitot device,
191
indexing tables,
217
position function,
116
processes
J
manufacturing,
37
Jacquard,

4,141
continuous,
23
Jacqnet-Droz,
3
periodic,
21
prostheses,
32
K
punching heads,
307
kinematic layout,
55
R
L
reliability
of
assembly processes,
294
Lanchester damper,
161
robot's
arm
vibration,
342
lathe,
4
robots
Lawton,

Tolbart,
5
bang-bang,
2,
8,35,131
layout
fixed
stop,
2
kinematic,
55
limited
degree-of-freedom,
3
Index
433
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
acceleration,
193
epicyclic,
220
capacitance,
180
linear,
206
displacement,
175
rotational,
217
electric,
175
vibrational,
223

flowrate,
190
force,
193 V
induction,
178
Vancanson,
de
Jacques,
3
item presence,
202
variable moment
of
inertia
and
mass,
103
photoelectric,
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,
315,324,326,359

Theophilus,
3