robotics Designing the Mechanisms for Automated Machinery Part 13 docx

9.6
Mobile
and
Walking Robots
377
By
substituting
the
latter into
(9.66),
the
position
of the
vehicle
is
determined. This
algorithm
can be
used both when programming
and
controlling
the
trajectory
of the
vehicle
and
when analyzing
the
trajectory
of the
vehicle,

for
instance,
for
indicating
the
location
of a car
during
its
travel.
The
problem
of
travelling across rugged terrain
has
prompted investigations
in the
field
of
walking vehicles. There
is no
real substitute
for
legs, which
can
step
over local
obstacles,
find the
optimum point

on a
surface
for
reliable support, change
the
length
of
the
stride,
and
change
the
pace
from
walking
to
running
and
jumping
or
from
an
amble
to a
trot
to a
gallop.
It
seems
that

investigators have covered
all
possibilities
in
seeking
the
optimal
walking machine.
In
Figures 1.30
and
1.31
of
Chapter
1 we
briefly
described
the
purely
mechanical walking machine designed
by the
famous
mathematician
P.
Chebyshev.
This device
can
walk
in an
optimal way; however,

it
requires
a flat
surface
and
cannot
make turns.
Therefore,
it
does
not
fulfill
the
hopes
a
walking device
has to
realize.
(By
the
way, this device clearly shows
how
cumbersome legs
are in
comparison with
a
wheel.
The
only reason
the

Creator
(or
evolution) used legs
in his
design
of
walking
animals
is the
simplicity
in
providing reliable connection
of
blood vessels, nerves,
and
muscles. Otherwise
we
would have wheels,
and
shoes would
be
designed like tires.)
We
begin
our
consideration with some six-legged devices,
for
which insects have
served
as a

prototype. Figure 9.58a) shows
an
example. (This design
was
influenced
by the
article
by
A. P.
Bessonovand
N.
V.
Umnov,
Mechanism
and
Machine
Theory,
Insti-
tute
for the
Study
of
Machines,
Vol.
18, No. 4, pp.
261-265,
Moscow,
U.S.S.R,
1983.)
The

skeleton
of the
machine consists
of
rigid
frame
1 to
which links
3 are
connected
via
joints
2.
Another
two
pairs
of
joints
4 and
rods
5 and 6
close
the
kinematic chains, cre-
ating
two
parallelograms. Hydrocylinders
7 and 8
control
the

position
of
these
two
par-
allelograms relative
to
frame
1.
This arrangement serves
for
steering. Slide bushings
9
are
mounted
on
rods
3 and
frame
1.
These bushings
are
driven along
the
rods
by
means
of
hydraulic cylinders
10. To

each bushing
9 is
fastened
one of the six
legs. Each
leg
includes
a
hydraulic cylinder
11
which
is
responsible
for the
height
of the
leg's
foot.
This
is a
design with
14
degrees
of
freedom.
Figure 9.58b) explains
the
movement
of a
foot.

By
combining
the
movement
of the
cylinders
in a
certain time sequence,
a
step-
over
movement
of the
foot
is
carried out. Say,
the
foot
at the
beginning
of the
cycle
is
in
its
upper position
I due to
cylinder
11;
then

cylinder
10
moves
it
horizontally
to
point
II. At
this moment cylinder
11
lowers
the
foot
to
point III,
from
which cylinder
10
pulls
the
foot
back (relative
to the
frame)
to
point
IV,
thus moving
the
"insect"

one
step
leftward.
Other
six-legged walker concepts using other
leg
kinematics
are
also possible.
For
instance,
a leg
with three degrees
of
freedom
is
illustrated
in
Figure
9.59a).
Shaft
1
with
joint
2
represents
the
hip,
to
which

the
thigh
3
with knee
4 is
attached.
The
latter serves
for
connecting
to the
shin
5.
These links
can be
driven
by
hydraulic cylinders.
Cylin-
der
6
drives
shaft
1 via
rack-gear transmission
7.
Cylinder
8
controls thigh
3

while
cylin-
der 9 is
responsible
for the
movement
of
shin
5. We can
imagine three
of
this
kind
of
leg
placed
on
each side
of a
rectangular
frame,
simulating
an
insect.
In
this case
the
TEAM LRN
378
Manipulators

FIGURE
9.58
a)
Six-legged
walking machine with
12
degrees
of
freedom;
b)
Design
of one
leg.
system possesses
18
degrees
of
freedom. (This kind
of
machine
was
created
by
Ivan
E.
Sutherland (see
Scientific
American,
January,
1983).)

It can
also
be
made
in the
form
presented
in
Figure 9.59b), where
the
hip-shafts
1 are
installed along radii
0-1,
O-II,
0-
III,
0-IV,
0-V,
and
0-VI.
Thus,
the
axially
symmetrical device
is
supported
by at
least
three legs (black points

in
Figure 9.59c)
at any
moment.
To
move along vector
V the
sequence
of
lifting
and
lowering
the
legs
is as
shown
in
this
figure
when
the
circle
moves
rightward.
Obviously,
no
preferred direction exists
for
this
concept;

it
does
not
have
a
front
or
back.
It can
begin
its
travel
in any
direction
and
provides high maneu-
FIGURE
9.59
a) Leg
with three degrees
of
freedom;
b)
Circular
form
of a
vehicle;
c)
Walking
tracks

of
this
vehicle.
TEAM LRN
9.6
Mobile
and
Walking Robots
379
verability,
including rotation
in
place.
For
this purpose
foot
I
moves
to
point
I',
foot
II
moves
to
point
II',
and so on.
Using
the leg

with three degrees
of
freedom
shown
in
Figure
9.59a,
one can
design
four-legged
devices, three-legged vehicles
(A.
Seireg,
R. M.
Peterson, University
of
Wis-
consin, Madison, Wisconsin,
U.S.A.)
and,
of
course,
try to
create
a
two-
or
one-legged
walker.
Here there

are
problems with balance. Marc
H.
Ralbert
and his
colleagues
at
Carnegie-Mellon
University have built
and
demonstrated
a
machine that hops
on a
single
leg and
runs like
a
kangaroo. This device consists
of
two
main parts
(Figure
9.60):
body
1 and leg 2. The
body carries
the
energy-distribution
and

control units.
The leg
can
change
its
length,
due to
pressure cylinder
3, and
pivots with respect
to the
body
around
the X- and
Y-axes
due to
spherical hinge
4. A
gyrosystem
keeps
the
body hor-
izontal while
the leg
bounces
on the
pneumatic spring-pneumatic cylinder
3. The
pres-
sure

in the
cylinder
is
controlled
to
dictate
the
bouncing dynamics. Another
pneumosystem controls
the
pivoting
of the leg
during
its
rebound.
A
feedback system
provides
the
pivoting required
for
balance when landing.
The
balance problem
is
very important
in
walking devices. Solving
it
will permit

four-
and
two-legged robots
to be
built. (The six-legged devices
are
always balanced
on
three supporting
feet.)
To
solve
the
balancing problems
a
mathematical descrip-
tion
of
the
device must
be
analyzed.
Let us
consider
the
model
of
the
two-legged walker
shown

in
Figure
9.61,
which presumably
reflects
the
main dynamic properties
of
natural
two-legged
creatures.
(The following
discussion
is
taken
from
the
book
Two-legged
Walking
by
V.
V.
Beletzky,
Moscow, Nauka, 1984
(in
Russian).)
Different
levels
of

com-
plexity
can be
considered
for the
model; however,
it is
always true that:
• The
device
possesses
a
certain mass, which
is
subject
to
gravitation;
• The
device
is
controlled, which means
that
its
links
are
driven
and
torques
act
in

every joint;
(Here
we
consider
an
additional degree
of
freedom between
the
shin
and the
foot,
compared with
the leg
shown
in
Figure
9.59a).)
•
Movement across
a
surface entails reactive
forces
against
the
feet.
All
walking machines
can act in two
regimes:

•
Walking—at
any
moment,
at
least
one leg
touches
the
ground;
•
Running—there
are
moments when
no
contact between
the
ground
and
feet
exists.
FIGURE
9.60
One-legged
hopper.
TEAM LRN
380
Manipulators
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
posi-
tion
in the figure we
have
for
point
O the
coordinates
are
where

v =
index
of the
supporting leg,
and
z
v
and
x
v
are
components
of the
r
v
vector
(position
of the
foot's
contact point).
Expression
(9.73)
allows,
for
instance, answering
the
question
of
what
the

kine-
matic requirements
are for
providing constant height
of
point
O
above
the
ground
(for
better
energy saving). This condition states
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
two-
legged
anthropoid
walking
machine.
TEAM LRN
9.6
Mobile
and
Walking Robots
381
The
dynamics
of a
two-legged robot present
a
complex problem
for the
general
case.
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.
The
exchange
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
dura-
tion
2T, the

following
limit conditions
are
valid:
FIGURE
9.62
Simplified
model
of a
two-
legged
walker
or
"spider."
TEAM LRN
382
Manipulators
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
foot
contact with
the
support
surface.
Between these contact instants,
#=0
(there
is no
reaction
force),
and the
spider,
in
essence,
flies. The
trajectory
of the
particle
in
this
case consists
of
parabolic sections,
as

shown
in
Figure 9.63.
In
this
figure
V
Q
=
initial speed
of the
particle,
JC
Q
=
horizontal component
of
speed
V
0
,
TEAM LRN
9.6
Mobile
and
Walking Robots
383
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
^

=
-z
0
.
To
acceler-
ate the
particle again
to
speed
z
0
,
an
amount
of
energy
A
l
must
be
expended
by the
foot:
In
stopping
at the end of the flying
period,
the
amount

of
energy
A
2
expended
by the
other
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
from
the
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.
TEAM LRN
384
Manipulators
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
for
moving
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
for
walking:
The
latter
formula
(9.93a
and b)
enables
the
value
of the

optimum length
L
0
of a
walking
steo
to be
derived:
The
computation model presented here
can
give
a
rough estimation
of
power con-
sumption
in
multi-legged vehicles
by
simply multiplying
the
results
and
distributing
the
mass
of the
moving body among
all the

pairs
of
legs.
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),
L
0
= 0.7 m.
Then
he or she
will expend about
150
watts
(0.036
kcal/sec)

of
power.
We
also recommend changing
from
walking
to
running when
a
speed
of
11.3
km/hr
is
reached. However,
if the
reader
is
overweight,
let him or her
continue
to
walk with
higher
speed (more energy will
be
expended).
The
speed record
for

walking
is
about
15.5 km/hr.
On
this optimistic tone
we finish
this chapter,
the final one in the
book.
TEAM LRN
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
7
2
of
drum
2 to the
axes
of
drum
1
requires cal-
culation
of the

common kinetic energy
of the
mechanism, which
is
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 with
moment
of
inertia

/
is:
The
motion equation
may
then
be
written
as
follows:
TEAM LRN
386
Solutions
to the
Exercises
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
JC
G
= 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
TEAM LRN
Solutions
to the
Exercises
387
Now,
we
illustrate
the
same solution
in
MATHEMATICA

language.
fl=x"[t]+83.3
x[t]+5.553
yl=DSolve[{fl==0,x[0]==.314,x'[0]==0},{x},{t}]
The
solution corresponds
to
(d).
jl=Plot[Evaluate[x[t]/.yl],{t,0,.2},AxesLabel->{"t","x"}]
j2=Plot[x= 0664,{t,0,.2}]
Show[jl,j2]
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
the
initial point until
it
comes into contact with
the
wire (here
we
neglect
the
frictional
resistance),
and
stage
b) the
cutting
of the
wire.
Stage
a)
We
use

Equations
(3.27)
to
(3.81). Thus,
TEAM LRN
388
Solutions
to the
Exercises
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
Thus,
Substituting
the
initial conditions into (b),
we
obtain
the
coefficients
A and
B.
When
t= 0 and x =
L
0

= 0.2 m,
then
0.2=
A
+
0.002,
and finally we
obtain
Differentiating
(b),
we
obtain
where
t=
0,
i=
0, and we
thus obtain
Finally,
we
obtain
We
now
calculate
the
time
1
1
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
TEAM LRN
Solutions
to the
Exercises
389
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:
f2=xl"[t]+5000xl[t]
-10
y2=DSolve[{f2==0,xl[01==.2,xl'[0]==0},{xl},{t}]
j21=Plot[Evaluate[xl[t]/.y2],{t,0,.025},
AxesLabel->{"t","xl"}]
gl=Plot[{xl=.l},{t,0,.02}]
Show[gl,j21]
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
TEAM LRN
390
Solutions
to the
Exercises
and
this energy
(a
remaining
sum of
kinetic
and
potential components
after
the
cutting
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:
TEAM LRN
Solutions
to the
Exercises

391
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
(h),
respectively,
thus,
Now
from
either Equation
(g) or
Equation (h),
we
derive
^4
as
follows:
Finally,
we

have
We
now
calculate
the
time
t
l
needed
by the
blade
to cut the
wire, which takes place
when
x =
0.096
m.
Therefore,
we may
write
or
In
the
MATHEMATICA
language,
we
solve
the
above-derived equation
as

follows:
fl=x"[t]+5000
x[t]+790.2
yl=DSolve[{fl==0,x[0]==0,x'[0]==-12.7},{x[t]},{t}]
jl=Plot[Evaluate[x[t]/.yl],{t,0,.0005},AxesLabel->{"t","x"}
TEAM LRN
392
Solutions
to the
Exercises
FIGURE
3E-2.2
Movement
of the
blade
x
z
[t]
during
the
cutting
process
(from
x
a
=
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
electromo-
tor is
calculated
from
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
TEAM LRN
Solutions
to the
Exercises
393
Thus,
From

the
initial conditions,
we find the
coefficient
A. For
time
t = 0, the
speed
CD
= 0.
Therefore,
and finally
From
here,
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
m has
travelled:
h =
<t>r
=
0.776 0.035
=
0.027m.
An
illustration
of the
solution
in
MATHEMATICA
language
follows.
fl=w'[t]+2.9w[t]-91
yl=DSolve[{fl==0,w[0]==0},{w},{t}]
jl=Plot[Evaluate[w[t]/.yl],{t,0,2},AxesLabel->{"t'V'w"}]
TEAM LRN
394
Solutions
to the
Exercises
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
0
+
T^co
- 4 -
0.1
CD.
TEAM LRN
Solutions
to the
Exercises
395
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
Thus,
From
the
initial conditions,
we find the
coefficient
A. For the
moment
t= 0, the
speed
CD
= 0.
Therefore,
and finally,
To
answer
the
question formulated
in the
problem,
we
substitute
t = 0.1 sec

into
Expression
(c):
Integrating Expression (c),
we find the
angle
of
rotation
<p(t)
of the
motor:
Thus,
Substituting
t-
0.1
sec
into (d),
we
obtain
TEAM LRN
396
Solutions
to the
Exercises
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
this
case
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.
fl
=0.001875
w'[t]-
2
15.7 (157-
w[t])/(24900-314
w[t]+w[t]*2)
jl=NDSolve[{fl==0,w[01==0},{w[tl},{t,0,l}]
yl=Plot[Evaluate[w[t]/.jl],{t,0,l},AxesLabel>{"t","w"}]
zl=Plot[2 15.7 (157-
w)/(24900-314w+wA2),{w,0,157}]

FIGURE
3E-3a).i
a)
Speed versus time
for AC
motor
drive;
b)
Characteristic
of
the
electromotor.
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
T
s
which, respec-
tively,
is

TEAM LRN
Solutions
to the
Exercises
397
Thus,
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
1
the
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\
+
T
0
=
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
=
(o
l
+
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
TEAM LRN
398
Solutions
to the
Exercises
Thus,
From
the
initial
conditions
we find the
coefficient
A For
time
t= 0, the
speed
a>
= 0.
Therefore,
and finally,
Integrating
Expression (c),
we find the
angle
of

rotation
<f)(t]
of the
motor:
Thus,
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
Equations
and
Formulas
(3.99),
3.100),
3.101),
and
(3.102).
Substituting
the
given parameters into
the
above-mentioned expression,
we
obtain
TEAM LRN
Solutions
to the
Exercises
399
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,
TEAM LRN
400
Solutions
to the
Exercises
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
m
2
is,
obviously,
To
solve this problem,
we use
Expression
(3.103),
which
is
derived
from
Equations
and
Formulas
(3.99),
3.100),
3.101),
and
(3.102).
Substituting
the
given parameters into
the
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"}]
TEAM LRN
Solutions
to the
Exercises
401
FIGURE
3E-4a).l
Displacement
of the
piston-rod
versus
time.

8
Solution
to
Exercise 3E-5
The
time sought
is a sum of
three
or
four
components.
Component
1
The first
component
t
0
is the
time needed
by the
pressure wave
to
travel
from
the
valve
to the
inlet
of the
cylinder:

r
0
=£/V
5
=
10m/340m/secE0.0294sec,
where
V
s
is the
speed
of
sound.
Component
2
The
second component
^
is the
time needed
to
reach
a
pressure
P
c
in
the
cylinder
volume

that
develops
a
force
equal
to the
resisting
force
acting
on the
piston-rod
(in
our
case this
is the
weight
mg of the
lifted
mass
m).
Obviously,
and for the
problem under consideration this
is
either
or
TEAM LRN