problems

Problem 8.1
Solutions to Chapter 8 Exercise Problems
A cam that is designed for cycloidal motion drives a flat-faced follower. During the rise, the
follower displaces 1 in for 180˚ of cam rotation. If the cam angular velocity is constant at 100 rpm,
determine the displacement, velocity, and acceleration of the follower at a cam angle of 60˚.
Solution:
The equation for cycloidal motion is:
y = L
1
sin
2
2
For L = 1, and =180˚= , then
y = L
1
sin
2
=1
2
1 sin
2
( 2 )= 1 2 sin2
( )
y ˙ = L
˙ y ˙ = 2L
1 cos
2
=
( 1 cos2)
2
sin 2
= 2 ( ) 2
sin2
The angular velocity is ˙ =100 rpm =100 2
=10.472 rad / s
60
When = 60˚= 3 ,
3
y =
1 ( sin2( 3)
2 ) = 1
1 ( sin(2 3)
3 2 )= 0.195 in
y ˙ = 1cos2 ( )= 10.472 ( 1 cos(2 3) )= 5.000
in.
sec.
˙ y ˙ = 2 ( ) 2
Problem 8.2
sin2 = 2 10.472 ( ) 2
sin(2 3) = 60.46 in.
sec2
A constant-velocity cam is designed for simple harmonic motion. If the flat-faced follower
displaces 2 in for 180˚ of cam rotation and the cam angular velocity is 100 rpm, determine the
displacement, velocity, and acceleration when the cam angle is 45˚.
Solution:
- 328 -

The equation for simple harmonic motion is:
y = L 2 1 cos
For L = 2, and =180˚= , then
y = L 2 1 cos
=
2 1 cos
2( )= ( 1 cos)
y ˙ = dy
dt = d ( 1 cos)=
dt ˙ sin
˙ y ˙ = d2y
dt2 = d dt ˙ ( sin)=
˙ 2cos
The angular velocity is ˙ =100 rpm =100 2
=10.472 rad / s
60
When = 45˚,
y = ( 1 cos)=
( 1 cos45˚)=10.707
= 0.292 in ,
y ˙ = ˙ sin =10.472 sin45˚=10.472 (0.707) = 7.405 in s
˙ y ˙ = ˙ 2 cos =10.4722 (0.707) = 77.531 in
s2
Problem 8.3
A cam drives a radial, knife-edged follower through a 1.5-in rise in 180˚ of cycloidal motion. Give
the displacement at 60˚ and 100˚. If this cam is rotating at 200 rpm, what are the velocity (ds/dt) and
the acceleration (d 2 s/dt 2 ) at = 60˚?
Solution:
The equation for cycloidal motion is:
y = L
1
sin
2
2
For L = 1.5, and =180˚= , then
y = L
1
sin
2
=1.5
2
1
sin
2
( 2 )=1.5 1 2 sin2
( )
y ˙ = L
1 cos
2
=
1.5 ( 1cos2 )
- 329 -

2
˙ y ˙ = 2L
sin 2
= 2(1.5) ( ) 2
sin2 = 3 ( ) 2
sin2
The angular velocity is ˙ = 200 rpm = 200 2
= 20.944 rad / s
60
When = 60˚= 3 ,
y =1.5
1
2 sin2
3
( )=1.5
1 2 sin 2 ( 3 ) = 0.293 in
y ˙ = 1.5 ( 1 cos2)=
1.5(20.944)
1cos
2 ( 3 )=15.00 in s
˙ y ˙ = 3 ( ) 2
=362.76
in
s2
When =100˚= 100
180 = 5 9 ,
sin2 = 3 20.944 ( ) 2
sin 2 3
( )
y =1.5
1
2 sin2
5 9
( )=1.5
1 2 sin10 = 0.915 in
9
Problem 8.4
Draw the displacement schedule for a follower that rises through a total displacement of 1.5 inches
with constant acceleration for 1/4th revolution, constant velocity for 1/8th revolution, and constant
deceleration for 1/4th revolution of the cam. The cam then dwells for 1/8th revolution, and returns
with simple harmonic motion in 1/4th revolution of the cam.
Solution:
The displacement profile can be easily computed using the equations in Chapter 8 using Matlab.
The curves are matched at the endpoints of each segment. The profile equations are:
For 0
2
y1 = a0 + a1 + a2 2
The boundary conditions at = 0 are y1 = 0 and y'1 = 0 . Therefore,
a0 = a1 = 0
So,
and
y1 = a2 2
y'1 = 2a2
where a2 is yet to be determined.
- 330 -

For 3
2 4
y2 = b0 + b1
( ) 2
The boundary conditions at =
2 are y1 = a2
2
and
Also
or
b0 + b1
2
( ) 2
= a2
2
0 = a2 ( 2 ) 2
+ b0 + b1
2
y'2 = b1 = a2
0 = a2 b1
For 3 5
4 4
y3 = c0 + c1 + c2 2
y'3 = c1 + 2c2
y"3 = 2c2
- 331 -
and y'1 = 2a2
2 = a2 . Then
The boundary conditions at = 3
4 are y2 = a2
4 + ( ) = a2 3
+
4 4
Also, at = 5
4 , y3 =1.5 and y'3 = 0 . Then, matching the conditions,
2
y3 = a2
2 = c0 +c1 3
4
y'3 = c1 + 2c2 3
= a2
4
1.5 = c0 + c1 5 5
+c2
4 4
( ) 2
3
+ c2( 4 ) 2
y'3 = c1 + 2c2 5
4 = 0
The boundary condition equations can be written as:
( )
= a2 2
2 and y'2 = a2 .

( ) 2
0 = a2
2
0 = a2 b1
2
0 a2
+ b0 + b1
2
2 + c0 +c1 3
4
0 = a2 + c1 + 2c2 3
4
1.5 = c0 + c1 5
4
+c2 5
4
( ) 2
+ c2 3
4
( ) 2
0 = c1 + 2c2 5
4
In matrix form,
0
0
0
0 =
0
1.5
( 2)
2
1
2
2
0
0
0
2
1 0
0 1
0
0
0
0
3
0
0
0
0
0
0 0
0 0
0 1
4
1
1
3
4
3
2
5
2
5
4
5
4
( ) 2
( ) 2
Solving for the constraints using Matlab,
a2
b0
b1
c0
c1
c2
0.2026
-0.5000
0.6366
= -1.6250
1.5915
-0.2026
The equations are then given in the following:
For 0
2
y1 = a2 2 = 0.2026 2
For 3
2 4
y2 = b0 + b1 = - 0.5000 + 0.6366
a2
b0
b1
c0
c1
c2
- 332 -

For 3
4
For 5
4
5
4
y3 = c0 + c1 + c2 2 = -1.6250 +1.5915 0.2026 2
3
2
y4 =1.5
For the return, 3
2 2 , and
y5 = L
2
3
1 +cos = ( 1+ cos2)
4
The displacement diagram is plotted in the following:
Normalized Follower Displacement, Velocity, and Acceleration
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
Position, max value is: 1.5
Velocity, max value is: 1.5 ( )
Acceleration, max value is: 3 ( 2 )
Follower Displacement, Velocity, and Acceleration Diagrams
TextEnd
0 50 100 150 200 250 300 350
Cam Angle
- 333 -

Problem 8.5
Draw the displacement schedule for a follower that rises through a total displacement of 20 mm
with constant acceleration for 1/8th revolution, constant velocity for 1/4th revolution, and constant
deceleration for 1/8th revolution of the cam. The cam then dwells for 1/4th revolution, and returns
with simple harmonic motion in 1/4th revolution of the cam.
Solution:
The displacement profile can be easily computed using the equations in Chapter 8 using Matlab.
The curves are matched at the endpoints of each segment. The profile equations are:
For 0
4
y1 = a0 + a1 + a2 2
The boundary conditions at = 0 are y1 = 0 and y'1 = 0 . Therefore,
a0 = a1 = 0
So,
and
y1 = a2 2
y'1 = 2a2
where a2 is yet to be determined.
For 3
4 8
y2 = b0 + b1
( ) 2
The bounary conditions at =
4 are y1 = a2
4
and
Also
or
b0 + b1
4
( ) 2
= a2
4
0 = a2 ( 4 ) 2
+ b0 + b1
4
y'2 = b1 = a2
2
0 = a2
b1
2
For 3
8
- 334 -
and y'1 = 2a2
4
= a2 . Then,
2

y3 = c0 + c1 + c2 2
y'3 = c1 + 2c2
y"3 = 2c2
The boundary conditions at = 3
8 are y2 = b0 + b1 3
8 and y'2 = b1. Also, at = , y3 = 20, and
y'3 = 0 . Then matching the conditions,
( ) 2
y3 = b0 +b1 3
8 = c0 + c1 3 3
+ c2
8 8
y'3 = c1 + 2c2 3
8 = b1
20 = c0 +c1 + c2 2
y'3 = c1 + 2c2 = 0
The boundary condition equations can be written as:
( ) 2
0 = a2
4
0 = a2
b1
2
+ b0 + b1
4
0 b0 b1 3
8 +c0 + c1 3
8
0 = b1 +c1 +c2 3
4
20 = c0 +c1 + c2 2
0 = c1 + 2c2
In matrix form,
( ) 2
3
+c2 ( 8 ) 2
0
0
0
0 =
0
20
1 0 0 0
4 4
0 1 0 0 0
2
0 1 3
1
8
3 3
8 ( 8 ) 2
3
0 0 1 0 1
4
0 0 0 0 1 2
0 0 0 1 2
Solving for the constraints using Matlab,
a2
b0
b1
c0
c1
c2
- 335 -

a2
b0
b1
c0
c1
c2
7.2051
-4.4444
11.3177
= -8.4444
18.1083
-2.8820
The equations are then given in the following:
For 0
4
y1 = a2 2 = 7.2051 2
For 3
4 8
For 3
8
y2 = b0 + b1 = - 4.4444 +11.3177
y3 = c0 + c1 + c2 2 = -8.4444 +18.1083 -2.8820 2
For 3
2
y4 = 20 mm
For the return, 3
2 2 , and
y5 = L
2
1 +cos
=10( 1+cos 2 )
The displacement diagram is plotted in the following:
- 336 -

Normalized Follower Displacement, Velocity, and Acceleration
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
Problem 8.6
Position, max value is: 20
Velocity, max value is: 20 ( )
Acceleration, max value is: 40 ( 2 )
Follower Displacement, Velocity, and Acceleration Diagrams
TextEnd
0 50 100 150 200 250 300 350
Cam Angle
Draw the displacement schedule for a follower that rises through a total displacement of 30 mm
with constant acceleration for 90˚ of rotation and constant deceleration for 45˚ of cam rotation. The
follower returns 15 mm with simple harmonic motion in 90˚ of cam rotation and dwells for 45˚ of
cam rotation. It then returns the remaining 15 mm with simple harmonic motion during the
remaining 90˚ of cam rotation.
Solution:
The displacement profile can be easily computed using the equations in Chapter 8 using Matlab.
The curves are matched at the endpoints of each segment. The profile equations are:
For 0
2
y1 = a0 + a1 + a2 2
The boundary conditions at = 0 are y1 = 0 and y'1 = 0 . Therefore,
a0 = a1 = 0
So,
and
y1 = a2 2
y'1 = 2a2
- 337 -

where a2 is yet to be determined.
For 3
2 4
y2 = b0 + b1 + b2 2
( ) 2
The boundary conditions at =
2 and y1 = a2
2
b0 + b1
+b2
2 ( 2)
2
= a2 ( 2 ) 2
and
Also
0 = a2 ( 2 ) 2
+ b0 + b1
2
y'2 = b1 + b2 = a2
( ) 2
+ b2
2
- 338 -
and y'1 = 2a2
2 = a2 . Then,
or
0 = a2 + b1 +b2
The boundary conditions at = 3
4 are y2 = 30 and y'2 = 0 . Then,
and
b0 + b1 3
4
( ) 2
+ b2 3
4
0 = b1 +2b2 3
4
= 30
( )= b1 + b2 3
( )
The four boundary condition equations can be summarized as:
( ) 2
0 = a2
2
0 = a2 + b1 +b2
30 = b0 + b1 3
4
( )
0 = b1 +b2 3
2
In matrix form,
( ) 2
+ b0 + b1
2
( ) 2
+b2 3
4
( ) 2
2
( ) 2
+ b2
2
0
0
=
30
0
1
2 2 2
0 1
0 1 3 3
4 ( 4 ) 2
0 0 1
3
2
a2
b0
b1
b2
7.5

Solving for the constraints using Matlab,
a2
b0
b1
b2
8.1057
-60.0000
=
76.3944
-16.2114
The equations are then given in the following:
For 0
2
y1 = a2 2 = 8.1057 2
For 3
2 4
y2 = b0 + b1 + b2 2 = -60.0000 + 76.3944 +-16.2114 2
For 3 5
4 4
=
2
and
y3 = L
1+ cos
2
= 15
2
( 1+ cos2)
This equation assumes that the curve is 15 mm high and that the curve falls to zero. However, the
actual curve begins at a height of 30 mm and returns to only 15 mm. Because of this, we need to
add 15 mm to the value for y3. Then,
y3 =15+ 7.5( 1+ cos2 )
For 5 6
4 4
y4 =15
For 6
2
4
=
2
and
y5 = L
2
1 +cos
= 15
2
( 1 +cos2)
The displacement diagram is plotted in the following:
- 339 -

Normalized Follower Displacement, Velocity, and Acceleration
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
Problem 8.7
Position, max value is: 30
Velocity, max value is: 25.4648 ( )
Acceleration, max value is: 32.4228 ( 2 )
Follower Displacement, Velocity, and Acceleration Diagrams
TextEnd
0 50 100 150 200 250 300 350
Cam Angle
Draw the displacement schedule for a follower that rises through a total displacement of 3 inches
with cycloidal motion in 120 degrees of cam rotation. The follower then dwells for 90˚ and returns
to zero with simple harmonic motion in 90˚ of cam rotation. The follower then dwells for 60˚
before repeating the cycle.
Solution:
The displacement profile can be easily computed using the equations in Chapter 8 using Matlab.
The curves are matched at the endpoints of each segment. The profile equations are:
For 0 2
3
and
= 2
3
y1 = L
( )
1 2 3 1
sin = 3
2 2 2 sin3
- 340 -

( )
y'1 = L 1 1 2 3 3
cos = 3 cos 3
2 2
y"1 = L 2
2 sin2
= 3 9
2 sin3 ( )
For 2 7
3 6
y2 = 3
For the return, 7
6
and
=
2
5
3 _
y3 = L
1+ cos
2
For the remainder of the cycle,
and
5
3
y4 = 0
2
=1.5( 1+ cos2)
The displacement diagram is plotted in the following:
- 341 -

Normalized Follower Displacement, Velocity, and Acceleration
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
Problem 8.8
Position, max value is: 3
Velocity, max value is: 3 ( )
Acceleration, max value is: 6 ( 2 )
Follower Displacement, Velocity, and Acceleration Diagrams
TextEnd
0 50 100 150 200 250 300 350
Cam Angle
A cam returns from a full lift of 1.2 in during its initial 60˚ rotation. The first 0.4 in of the return is
half-cycloidal. This is followed by a half-harmonic return. Determine 1 and 2 so that the motion
has continuous first and second derivatives. Draw a freehand sketch of y', y'', and y''' indicating
any possible mismatch in the third derivative.
Solution:
Follower Travel
1.2 in
0.4 in
- 342 -
Not to Scale
0˚ 60˚
β 1 β 2
The first part of the return is made up of a cycloidal curve and the second part is made up of a
harmonic curve. This is shown schematically in the figure below.

Follower Travel
1.2"
β1 β2
0.4"
The range for the cycloidal curve is given by
0 1 21
Cycloidal Curve
and the range for the harmonic curve is given by
Also,
1 2 2 1 + 2
2 =1 (1 2)
0˚ 60˚
β1
β2
- 343 -
2β
2
2 β1
Harmonic Curve
The general form for the cycloidal equation for a return is given in Section 7.8 as
y = L 1
+
1
sin
2
2
Half of the cycloidal return is 0.4 so return is 0.8. The range for 1 is 21. As
indicated in the figure above, the cycloidal curve is offset from the horizontal axis by 0.4".
Therefore, this much must be added to y. The cycloidal equation for the return is
y1 = L1 1 1 21 + 1
= 0.8 1.5 1
2 1
2 sin 21 21 +0.4 = 0.8 1 1 21 + 1 2 sin 1
1
+ 1 2 sin 1
1
+0.4
The harmonic curve is given by Eq. (812). Half of the harmonic return is (1.2"-0.4") = 0.8 so that
the whole return is 1.6". The range for 2 is 22 . Therefore, the equation for the harmonic part
of the return is:
y2 = L 2
2 1+ cos
2
= 0.8 1+ cos
2
2 2
2 2
We also know that 2 =1 (1 2) and 2 = 3
1
0.4"
(1)
(2)

Eqs. (1) and (2) can be reduced so that the only unknown is 1. To solve for the unknown, we can
equate the slopes at 1 = 1 . For the cycloidal equation,
y'1= 0.8
and at 1 = 1
2 1
y'1= 0.8
2 1
1 cos 1 1 1 cos 1 1 For the harmonic equation
y'2 = 0.4
sin 2
2
2 2
At 1 = 1, 2 = 2. Therefore,
y'2 = 0.4
2
sin 2 22 = 0.8
1
=
0.4
2
sin 2 22
= 0.4
2
Equation Eqs. (3) and (4) give the following equation
or
0.8
1
= 0.4
2
2
=
1 =
2 /3 1 This equation can be easily solved for 1. The result is: 1 =
- 344 -
2
= 0.4073436 radians.
3( +2)
We can now write y, y', and y" for each part of the curve. For 1 0.4073436
y = 0.8 1.5 1 + 1
y' = 0.8
2 1
y"= 0.8
2 1
1 cos 1 1 22 1 sin 1 1 2 sin 1
1
and for 0.4073436 /3
y = 0.8 1+cos
2
2 2
(3)
(4)
(4)

where
y' = 0.4
2
y"= 0.2 2
2 2
sin 2
22
cos 2
22
2 =1 (1 2) and 2 = 3
1
The results are plotted in the following
- 345 -

Problem 8.9
Assume that s is the cam-follower displacement and is the cam rotation. The rise is 1.0 cm after
1.0 radian of rotation, and the rise begins and ends at a dwell. The displacement equation for the
follower during the rise period is
n
i
s = h Ci
i =0
If the position, velocity, and acceleration are continuous at = 0, and the position and velocity are
continuous at = 1.0 rad, determine the value of n required in the equation, and find the coefficients
Ci if ˙ = 2 rad/s. Note: Use the minimum possible number of terms.
Solution:
Dwell
1.0
S
Β
Α 1.0
- 346 -
Dwell
First determine the number of terms required. There are a total of five conditions to match;
therefore, the number of terms is 5 making n = 4.
The conditions to match are:
At = 0 , s = ds
=
d2s
= 0
d d2
At = , s = h = 1.0
ds
= 0
d
Now,
s = Ci
n i
= C0 +C1
+ C2
2
+C3
3
+ C4
4
and
i=0
ds
d = C1 + 2C2
+
3C3
d2s
d 2 = 2C 2
2 + 6C3 2
2
+
12C4
2
+ 4C 4
2
3
θ, rad

Applying the conditions at = 0 ,
s = C0 = 0
ds
d = C 1
C1 = 0
d2s
d 2 = 2C 2
2 = 0 C2 = 0
Applying the conditions at = ,
or
s = C3
3
ds
d = 3C3
3C3 + 4C4 = 0
4
+C4
2
+ 4C 4
Solving for the constants,
and
C3 = 4
C4 = 3
Therefore,
s = 4
Problem 8.10
3
4
3
= C3 + C4 =1
3
= 3C3 + 4C4 = 0
Resolve Problem 8.9 if = 0.8 rad and ˙
= 200 rad/s.
Solution:
First determine the number of terms required. There are a total of five conditions to match;
therefore, the number of terms is 5 making n = 4.
The conditions to match are:
At = 0 , s = ds
=
d2s
= 0
d d2
At = , s = h = 1.0
- 347 -

Now,
and
ds
= 0
d
s = Ci
n i
= C0 +C1
i= 0
ds C1 2C2
= +
d
d2s 2C2 6C3
d 2 =
2 +
2
+ 3C3
+C2
2
+12C4
2
Applying the conditions at = 0 ,
s = C0 = 0
ds C1
=
d C1 = 0
d2s 2C2
d 2 =
2 = 0 C2 = 0
Applying the conditions at = ,
or
s = C3
3
+ C4
4
ds 3C3
=
d
3C3 + 4C4 = 0
2
+ 4C4
Solving for the constants,
and
C3 = 4
C4 = 3
Therefore,
s = 4
3
3
4
+ 4C4
2
= C3 +C4 =1
3
= 3C3
2
+C3
+ 4C4
3
= 0
- 348 -
3
+C4
Notice that this solution is EXACTLY the same as that for 8.9. The results are independent of both
and ˙ . This is one of the reasons for normalizing the problem with respect to .
4

Problem 8.11
For the cam displacement schedule given, h is the rise, is the angle through which the rise takes
place, and s is the displacement at any given angle . The displacement equation for the follower
during the rise period is
5
i
s = h ai
i =0
Determine the required values for a 0 ... a 5 such that the displacement, velocity, and acceleration
functions are continuous at the end points of the rise portion.
Solution:
Dwell
h
Α
S
Β
β
- 349 -
Dwell
θ, rad
There are a total of six conditions to match; therefore, the number of terms is 6 making n = 5.
The conditions to match are:
At = 0 , s = ds
=
d2s
= 0
d d2
At = , s = h
and
ds
=
d2s
= 0
d d 2
Now,
s = h Ci
n i
= hC0 +C1
+ C2
2
+C3
3
+ C4
4
+ C5
5
and
i=0
ds
d = h C1 + 2C2
+
3C3
d2s
d 2 = h 2C
2
2 + 6C3 2
2
+
12C4
2
+ 4C 4
2
+ 20C 5
2
3
+ 5C5
4
3

Applying the conditions at = 0 ,
s = C0 = 0
ds
d = h C 1
C1 = 0
d2s
d 2 = h 2C 2
2 = 0 C2 = 0
Applying the conditions at = ,
or
or
or
s = hC3
3
C3 + C4 +C5 =1
ds
d = h 3C3
+C4
3C3 + 4C4 +5C5 = 0
2
+ 4C 4
4
+ C5
d2s
d 2 = h 6C3
2
+
12C4
2
6C3 +12C4 + 20C5 = 0
Solving for the constants,
C3 =10; C4 = 15; C5 = 6
Therefore,
s = h 10
3
15
4
+ 6
5
Problem 8.12
5
= hC3 [ + C4 + C5]=h
3
= h
3C3 + 4C4 + 5C
5
= 0
2
+ 20C 5
2
3
= h
6C 3
Resolve Problem 8.11 if h = 20 mm and = 120˚.
Solution:
- 350 -
2 +12C4 2 + 20C5 2
= 0
There are a total of six conditions to match; therefore, the number of terms is 6 making n = 5.
The conditions to match are:
At = 0 , s = ds
d = d2s d2 = 0

At = , s = h
and
ds
d = d2s d 2 = 0
Now,
s = h Ci
n
and
i
= hC0 + C1
i= 0
2
+ C2 + C3
3
ds C1
= h
d
+ 2C2
d2s 2C2 6C3
d 2 = h
2 +
2
+ 3C3
Applying the conditions at = 0 ,
s = C0 = 0
ds C1
= h
d C1 = 0
d2s 2C2
d 2 = h
2 = 0 C2 = 0
Applying the conditions at = _
or
or
or
s = hC3
3
C3 +C4 +C5 = 1
ds 3C3
= h
d
+C4
3C3 + 4C4 +5C5 = 0
d2s 6C3
d 2 = h
2
2
+ 4C4
+ 12C4
2
2
+ 4C4
4
+ C5
5
3
+12C4
6C3 +12C4 + 20C5 = 0
Solving for the constants,
2
2
3C3
= h
2
+ 20C5
2
+ 20C5
2
3
- 351 -
+ 5C5
3
+C4
= hC3 [ +C4 + C5 ]= h
+ 4C4
3
+ 5C5
6C3
= h
4
= 0
2 +12C4
4
+ C5
5
20C5
2 +
2
= 0

C3 =10; C4 = 15; C5 = 6
Therefore,
s = h 10
3
15
If h=20 mm and = 120˚, then,
( ) 3
s = 20 10
120
4
15 ( 120)
4
+6
5
( ) 5
+6
120
Here, is assumed to be given in degrees. Note that the values for h and do not enter the
problem until the last step.
Problem 8.13
Resolve Problem 8.11 if h = 2 in and = 90˚.
Solution:
There are a total of six conditions to match; therefore, the number of terms is 6 making n = 5.
The conditions to match are:
At = 0 , s = ds
d = d2s d2 = 0
At = , s = h
and
ds
d = d2s d 2 = 0
Now,
s = h Ci
n
and
i
= hC0 + C1
i= 0
2
+ C2 + C3
3
ds C1
= h
d
+ 2C2
d2s 2C2 6C3
d 2 = h
2 +
2
+ 3C3
Applying the conditions at = 0 ,
s = C0 = 0
+ 12C4
2
2
+ 4C4
2
+ 20C5
2
3
- 352 -
+ 5C5
3
+C4
4
4
+ C5
5

ds C1
= h
d C1 = 0
d2s 2C2
d 2 = h
2 = 0 C2 = 0
Applying the conditions at = _
or
or
or
s = hC3
3
C3 +C4 +C5 = 1
ds 3C3
= h
d
+C4
3C3 + 4C4 +5C5 = 0
d2s 6C3
d 2 = h
2
2
+ 4C4
4
+ C5
5
3
+12C4
6C3 +12C4 + 20C5 = 0
Solving for the constants,
2
C3 =10; C4 = 15; C5 = 6
Therefore,
s = h 10
3
15
If h=2 in and = 90˚, then,
( ) 3
s = 210
90
4
15 ( 90)
4
3C3
= h
2
+6
+ 20C5
2
5
( ) 5
+6
90
= hC3 [ +C4 + C5 ]= h
+ 4C4
3
- 353 -
+ 5C5
6C3
= h
= 0
2 +12C4
20C5
2 +
2
= 0
Here, is assumed to be given in degrees. Note that the values for h and do not enter the
problem until the last step.

Problem 8.14
Assume that s is the cam-follower displacement and is the cam rotation. The rise is h after
degrees of rotation, and the rise begins at a dwell and ends with a constant velocity segment. The
displacement equation for the follower during the rise period is
n
i
s = h Ci
i =0
If the position, velocity, and acceleration are continuous at = 0 and the position and velocity are
continuous at = , determine the n required in the equation, and find the coefficients C i that will
satisfy the requirements if s = h = 1.0.
Solution:
Dwell
S
β
First determine the number of terms required. There are a total of five conditions to match;
therefore, the number of terms is 5 making n = 4.
The conditions to match are:
At = 0 , s = ds
=
d2s
= 0
d d2
At = , s = h = 1.0
ds
= tan45˚=1.0
d
Now,
s = Ci
n i
= C0 +C1
+ C2
2
+C3
3
+ C4
4
i=0
- 354 -
h
45˚
θ

and
ds
d = C1 + 2C2
+
3C3
d2s
d 2 = 2C 2
2 + 6C3 2
2
+
12C4
2
Applying the conditions at = 0 ,
s = C0 = 0
ds
d = C 1
C1 = 0
d2s
d 2 = 2C 2
2 = 0 C2 = 0
Applying the conditions at = ,
or
s = C3
3
ds
d = 3C3
3C3 + 4C4 =
4
+C4
2
+ 4C 4
Solving for the constants,
and
C3 = 4
C4 = 3
Therefore,
s =(4 )
Problem 8.15
3
+ 4C 4
2
= C3 + C4 =1
3
+( 3)
3
= 3C 3
+ 4C 4
=1
4
A follower moves with simple harmonic motion a distance of 20 mm in 45˚ of cam rotation. The
follower then moves 20 mm more with cycloidal motion to complete its rise. The follower then
dwells and returns 25 mm with cycloidal motion and then moves the remaining 15 mm with
harmonic motion in 45˚. Find the intervals of cam rotation for the cycloidal motions and dwell by
matching velocities and accelerations, then determine the equations for the displacement (S) as a
function of for the entire motion cycle.
Solution:
- 355 -

This is a curve matching problem. To begin the problem, consider the equations for harmonic and
cycloidal motions:
Harmonic:
y = L
2
Cycloidal:
1cos
y'= L
2 sin
y"= 2L
22 cos
y = L 1 2
sin
2
y'= L 1 1 2
cos
y"= L 2 2
2 sin
For both the harmonic and cycloidal motions, we must determine L and . There are a total of four
curves, so we need to determine four L’s and four ’s The geometry is shown in the following
figure.
Follower Travel
Cycloidal Curve
45˚
2
1
Harmonic Curve
20
20
25
- 356 -
15
Harmonic Curve
Cycloidal Curve
We can treat the rise and return separately, and then determine the dwell to ensure that there is a full
cycle of motion.
3
4
45˚

For the rise section, assume that the harmonic curve is half-harmonic, and the cycloidal curve is half
cycloidal. This will allow us to match the curves at their inflections points and will ensure curvature
continuity. For the harmonic curve,
and
2 =
2
L2 = 40
Therefore, the harmonic curve is
y = L 40
1cos = ( 1 cos2 )= 20( 1 cos2)
2 2
The slope equation is
y'= L
2 sin
1
= Lsin2 ( )
Or the cycloidal curve,
L1 = 40
and
y = L1 1 1 21 1 1 21
sin = 40 sin
2 2
1
The slope equation is
y'= L1 1 1 21
cos
1 1 1
To find 1, equate the slopes at the midpoint, then,
y'= L2 sin2 2 1 1 2 1
2
= L1 cos
1 1 1 2
or
or
40( sin2 )= 40 1 1
1 cos
sin ( 2)
1
1
1 1
=
1 cos
Then,
2
1 =1
or
1 = 2
The rise part of the curve is given by,
For <
4
1
1
- 357 -

y = 20( 1 cos2 )
The cycloidal curve starts at
= (2 1)/2 = ( 2
2 )/ 2 = 0.2146
Therefore,
1 = + 0.2146
< <
1
+
Then for
4 4 2
y = 40 1 1 21
[ + 0.2146]
sin = 40
2 2
1
sin[ +0.2146]
2
1
1
The angular distance to the dwell is
= 1
+
2
4 2
= ( +
4 2)=1+
4
For the return, use the same procedure. The harmonic part of the return is given by
y4 = L4
1+ cos4
2
4
Where
L4 = 2(15) = 30
and
4 =
2
Therefore,
y4 =15( 1+ cos24 )
For the cycloidal curve,
y3 = L3 L3 3 1 23
sin
2
Where
3
3
L3 = 2(25) = 50
To determine 3, equate the slopes for the two curves at their midpoints. Then,
or
y'4 = L4
sin
24 4
30 sin ( 2)
Solving for 3,
Or
2
3
= 3
5
3
4
2
= y'3 = L3 1
1 1
= 50
3 cos
3
1
3
- 358 -
cos 2
3
3
2

3 = 10
3
The cycloidal part of the curve will start at
= 2 ( 3 +4 )/ 2
Therefore, the dwell period will be
1
+
4 2 < < 2 3 ( +4 )/ 2
And the dwell distance will b,
y = 40
The cycloidal part of the curve occurs when
2 ( 3 +4 )/ 2 < < 2 3 / 2
And
y3 = L3 L3 3
3
1 23
sin
2
The curve will be shifted because L3 is 50. Therefore,
3
3
y = y3 10 = L3 L3 3 1 23
sin
2 10
The second harmonic part of the curve occurs when
And
2 3 / 2 < < 2
y4 = L4
2
1+ cos
4
The displacement diagram is shown in the following:
3
- 359 -

Normalized Follower Displacement, Velocity, and Acceleration
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
Problem 8.16
Position, max value is: 40
Velocity, max value is: 40 ( )
Acceleration, max value is: 80 ( 2 )
Follower Displacement, Velocity, and Acceleration Diagrams
TextEnd
0 50 100 150 200 250 300 350
Cam Angle
Construct the part of the profile of a disk cam that follows the displacement diagram shown below.
The cam has a 5-cm-diameter pitch circle and is rotating counterclockwise. The follower is a knifeedged,
radial, translating follower. Use 10-degree increments for the construction.
Solution:
Displacement, cm
1.5
1.0
0.5
0˚
45˚ 90˚
Cam Rotation Angle
Start by subdividing the follower diagram into 10˚ increments of cam rotation. Next draw the cam
pitch circle and lay off radial lines in the clockwise direction. The follower displacements can then
be taken directly from the displacement diagram. Use a smooth curve to draw the cam profile.
- 360 -

Problem 8.17
Displacement, cm
1.5
1.0
0.5
0 1 2 3 4 5 6 7 8 9 10
10
Cam Rotation Angle
Pitch
Circle
9
Construct the profile of a disk cam that follows the displacement diagram shown below. The
follower is a radial roller and has a diameter of 10 mm. The base circle diameter of the cam is to be
40 mm and the cam rotates clockwise.
Solution:
Follower Travel, mm
30
15
- 361 -
8
0 0˚ 60˚ 120˚ 180˚ 240˚ 300˚ 360˚
Cam Rotation
7
6
5
4
3
1
2
0
90˚

Use the follower diagram subdivisions of 20˚. Next draw the cam pitch circle and lay off radial
lines in the counterclockwise direction. The follower displacements can then be taken directly from
the displacement diagram. Draw the pitch curve. Draw the cam follower circles on the pitch curve.
Use a smooth curve to draw the cam profile tangent to the follower circles.
Problem 8.18
Prime Curve
9
8
10
7
6
Prime Circle
Base Circle
11
Cam
12
Accurately sketch one half of the cam profile (stations 0-6) for the cam follower, base circle, and
displacement diagram given below. The base circle diameter is 1.2 in.
Base Circle
- 362 -
Total Follow Travel
5
13
1.0
14
4
15
3
16
2
17
0 1 2 3 4 5 6 7
11 10 9 8
Station Point
Numbers
1
0

Solution:
First draw the displacement schedule to scale, and divide the cam into 12 equal sections. The cam is
rotating in the clockwise direction so that the cam is layed out in the counterclockwise direction.
Then draw the locations of the follower. The cam profile is then drawn such that the cam is tangent
to the follower in the different positions. The cam is drawn as follows:
Problem 8.19
5
4
6
7
8
- 363 -
3
2
1
12
11
9 10
Lay out a cam profile using a harmonic follower displacement (both rise and return). Assume that
the cam is to dwell at zero lift for the first 100˚ of the motion cycle and to dwell at a 1 in lift for cam
angles from 160˚ to 210˚. The cam is to have a translating, radial, roller follower with a 1-in roller
diameter, and the base circle radius is to be 1.5 in. The cam will rotate clockwise. Lay out the cam
profile using 20˚ plotting intervals.
Solution:
The displacement profile can be easily computed using the equations in Chapter 8 in a spreadsheet
or MATLAB program. The profile equations are:
For 0 100˚
y=0
For 100˚ 160˚
y = L
1 cos
2
where L = 1, = 100˚, and =160˚100˚= 60˚.
For 160˚ 210˚
y=1

For 210˚ 360˚
y = L 1+ cos
2
where L = 1, = 210˚, and = 360˚210˚=150˚
The displacement diagram is given below followed by a table of values for y at 20˚ increments of .
Theta Displacement
0.000 0.000
20.000 0.000
40.000 0.000
60.000 0.000
80.000 0.000
100.000 0.000
120.000 0.250
140.000 0.750
160.000 1.000
180.000 1.000
200.000 1.000
220.000 0.989
240.000 0.905
260.000 0.750
280.000 0.552
300.000 0.345
320.000 0.165
340.000 0.043
360.000 0.000
To lay out the cam, first draw the prime circle which has a radius of 1.5" + 0.5" = 2.0". Then lay
off radial lines at 20˚ increments and label the lines in the counterclockwise direction. Draw circle
arcs corresponding to the two dwells and lay off the distances for the other displacements along the
other radial lines. Draw 1" diameter circles through the endponts of the distances layed off along
the radial lines, and fit a smooth curve which is tangent to the circles corresponding to the roller
follower.
- 364 -

Problem 8.20
160˚
200˚
220˚
140˚
240˚
120˚
100˚
260˚ 280˚
- 365 -
300˚
320˚
0˚
340˚
1 inch
Prime circle
Lay out a cam profile using a cycloidal follower displacement (both rise and return) if the cam is to
dwell at zero lift for the first 80˚ of the motion cycle and to dwell at 2-in lift for cam angles from
120˚ to 190˚. The cam is to have a translating, radial, roller follower with a roller diameter of 0.8 in.
The cam will rotate counterclockwise, and the base circle diameter is 2 in. Lay out the cam profile
using 20˚ plotting intervals.
Solution:
The displacement profile can be easily computed using the equations in Chapter 8 in a spreadsheet
or MATLAB program. The profile equations are:
For 0 80˚
y=0
For 80˚ 120˚
y = L
1
sin
2
2
where L = 2, = 80˚, and =120˚80˚= 40˚.
For 120˚ 190˚
y=2
For 190˚ 360˚

y = L 1
+
1
sin
2
2
where L = 2, = 190˚, and = 360˚190˚=170˚
The displacement diagram is given below followed by a table of values for y at 20˚ increments of .
Theta Displacement
0.000 0.000
20.000 0.000
40.000 0.000
60.000 0.000
80.000 0.000
100.000 1.000
120.000 2.000
140.000 2.000
160.000 2.000
180.000 2.000
200.000 1.997
220.000 1.932
240.000 1.718
260.000 1.344
280.000 0.883
300.000 0.452
320.000 0.154
340.000 0.021
360.000 0.000
To lay out the cam, first draw the prime circle which has a radius of 1.0" + 0.4" = 1.4". Then lay
off radial lines at 20˚ increments and label the lines in the clockwise direction. Draw circle arcs
corresponding to the two dwells and lay off the distances for the other displacements along the
other radial lines. Draw 0.8" diameter circles through the endponts of the distances layed off along
the radial lines, and fit a smooth curve which is tangent to the circles corresponding to the roller
follower.
Notice the poor pressure angles in the range 80˚120˚. This would probably make the cam
unacceptable.
- 366 -

Problem 8.21
180˚
200˚
220˚
Prime circle
240˚
120˚
100˚
Lay out a cam profile assuming that an oscillating, roller follower starts from a dwell for 0˚ to 140˚
of cam rotation, and the cam rotates clockwise. The rise occurs with parabolic motion during the
cam rotation from 140˚ to 220˚. The follower then dwells for 40˚ of cam rotation, and the return
occurs with parabolic motion for the cam rotation from 260˚ to 360˚. The amplitude of the follower
rotation is 35˚, and the follower radius is 1 in. The base circle radius is 2 in, and the distance
between the cam axis and follower rotation axis is 4 in. Lay out the cam profile using 20˚ plotting
intervals such that the pressure angle is 0 when the follower is in the bottom dwell position.
Solution:
The displacement profile can be easily computed using the equations in Chapter 8 using a
spreadsheet or MATLAB program. Remember that the parabolic motion is represented by two
curves in each rise and return region. The curves are matched at the midpoints of the rise and
return. The profile equations are:
For 0 140˚
= 0
For the first part of the rise, 140˚ 180˚ , and
= 2L
2
where L = 35˚, = 140˚, and = 220˚140˚= 80˚.
For the second part of the rise, 180˚ 220˚ , and
260˚
- 367 -
80˚
280˚
300˚
320˚
340˚
0˚
1 inch

= L 12 1
2
where L = 35˚, = 140˚, and = 220˚140˚= 80˚.
For 220˚ 260˚
= 35˚
For the first part of the return, 260˚ 310˚, and
= L 12
2
where L = 35˚, = 260˚, and = 360˚260˚=100˚
For the second part of the return, 310˚ 360˚, and
= 2L 1
2
where L = 35˚, = 260˚, and = 360˚260˚=100˚
The displacement diagram is given below followed by a table of values for at 20˚ increments of
.
0.000 0.000
20.000 0.000
40.000 0.000
60.000 0.000
80.000 0.000
100.000 0.000
120.000 0.000
140.000 0.000
Theta Follower Angle
- 368 -
160.000 4.375
180.000 17.500
200.000 30.625
220.000 35.000
240.000 35.000
260.000 35.000
280.000 32.200
300.000 23.800

320.000 11.200
340.000 2.800
- 369 -
360.000 0.000
To lay out the cam, first draw the prime circle which has a radius of 2.0" + 1.0" = 3.0". Next draw
the pivot circle for the follower pivot. The radius of the pivot circle is 4". Draw the follower in the
initial position ( = 0˚ ) to determine the follower length (r3) and the position on the pivot circle
corresponding to = 0˚ . As indicated in Example 8.5, the length r3 is given by
r3 = r1 2 (rb +r0) = 4 2 (2 +1) 2 = 2.646"
Identify the point on the pivot circle corresponding to = 0˚ , lay off the radial lines at 20˚
increments from this point, and label the lines in the counterclockwise direction. Draw lines from
the intersections of the radal lines with the pivot circle tangent to the prime circle. Then lay off the
angular displacements from these tangent lines. Locate the center of the follower by the distance r3
from the pivot circle along these lines. Draw 1" radius circles through the endponts of the distances
layed off along these lines, and fit a smooth curve which is tangent to the circles corresponding to
the roller follower.
The cam profile is shown in the following figure.

220˚
240˚
Problem 8.22
200˚
260˚
180˚
280˚
160˚
300˚
140˚
- 370 -
320˚
120˚
340˚
0˚
100˚
80˚
20˚
1 inch
Lay out the rise portion of the cam profile if a flat-faced, translating, radial follower's motion is
uniform. The total rise is 1.5 in, and the rise occurs over 100˚ of can rotation. The follower dwells
for 90˚ of cam rotation prior to the beginning of the rise, and dwells for 80˚ of cam rotation at the
end of the rise. The cam will rotate counterclockwise, and the base circle radius is 3 in.
Solution:
The displacement profile can be easily computed using the equations in Chapter 8 in a spreadsheet
or MATLAB program. The profile equations are:
For 0 90˚
s = 0
60˚
40˚

For 90˚ 190˚ ,
s = L
where L = 1.5, = 90˚, and =190˚90˚=100˚.
For 190˚ 270˚ ,
s = 1.5
For 270˚ 360˚
s = L 1
where L = 1.5, = 270˚, and = 360˚270˚= 90˚
The displacement diagram is given below followed by a table of values for at 20˚ increments of
.
0.000 0.000
20.000 0.000
40.000 0.000
60.000 0.000
80.000 0.000
100.000 0.150
120.000 0.450
140.000 0.750
160.000 1.050
180.000 1.350
200.000 1.500
220.000 1.500
240.000 1.500
260.000 1.500
280.000 1.333
300.000 1.000
320.000 0.667
340.000 0.333
360.000 0.000
Theta Follower Angle
- 371 -

We will lay out only the rise portion of the cam. To lay out the cam, first draw the base circle which
has a radius of 3". Then lay off radial lines at 20˚ increments and label the lines in the clockwise
direction. Draw circle arcs corresponding to the two dwells and lay off the distances for the other
displacements along the other radial lines. Then lay off the distances from the base circle along the
radial lines. Draw a line perpendicular to the offset lines at the indicated distance from the base
circle. Finally fit a smooth curve tangent to the positions of the follower face indicated by the
perpendicular lines. The dwells and rise part of the resulting cam are shown below.
180˚
160˚
200˚
140˚
220˚
120˚
240˚
100˚
- 328 -
260˚ 280˚
80˚
60˚
1 inch
40˚
0˚
20˚

Problem 8.23
In the sketch shown, the disk cam is used to position the radial flat-faced follower in a computing
mechanism. The cam profile is to be designed to give a follower displacement S for a
counterclockwise cam rotation according to the function S = k 2 starting from dwell. For 60˚ of
cam rotation from the starting position, the lift of the follower is 1.0 cm. By analytical methods,
determine the distances R and L when the cam has been turned 45˚ from the starting position. Also
calculate whether cusps in the cam profile would occur in the total rotation of 60˚.
Solution:
Starting Position
Dwell
ω
2.5 cm radius
Spring
S
- 329 -
R
θ
L
Point of
Contact
Find k first given the values of S and at the given point. Note that must be converted to radians
to ensure consistent units.
S = k 2
Therefore,
k = S 2 =
1
2 = 0.9119
[ /3]
Now,
R = rb + S = rb + k 2 = 2.5+ 0.9119 2
Where rb is the radius of the base circle. Also,
At = 4 ,
and
L = dR
= 2k = 2(0.9119) = 1.8238
d
R = 2.5+ 0.91192 = 2.5 +0.9119 ( 4 ) 2
= 3.063 cm

L =1.8238 =1.8238 ( 4 ) =1.432 cm
Cusps will not occur if
or
rb + S + d2S
0
d2
rb + S +2k 0
Because,
(2.5+ S +1.18238 = 3.6824 + S) 0
and S is positive for all , there are no cusps for any value of .
Problem 8.24
Determine the cam profile assuming that the translating cylindrical-faced follower starts from a
dwell from 0˚ to 80˚, and the cam rotates clockwise. The rise occurs with cycloidal motion during
the cam rotation from 80˚ to 180˚. The follower then dwells for 60˚ of cam rotation, and the return
occurs with simple harmonic motion for the cam rotation from 240˚ to 360˚. The amplitude of the
follower translation is 3 cm, and the follower radius is 0.75 cm. The base circle radius is 5 cm, and
the offset is 0.5 cm.
Solution:
The displacement profile can be computed using the equations in Chapter 8 in a spreadsheet or
MATLAB program. The profile equations (displacement and first and second derivatives) are:
For 0 80˚
y=0
For 80˚ 180˚
y = L
1
sin
2
2
y' = L
1 cos2
y"= 2L
sin
2
2
where L = 3, = 80˚, and =180˚80˚=100˚.
For 180˚ 240˚
y=3
For 240˚ 360˚
- 330 -

y = L 1+ cos
2
y' = L
sin
2
2
y"= L 2
cos
where L = 3, = 240˚, and = 360˚240˚=120˚
The cam which will generate this follower displacement can be determined using the equations
given in Table 8.8. These equations have been programmed in the MATLAB program rf_cam.m.
The program calls an m-file called follower.m which contains the equations for the follower
displacement. This file is given as follows:
function [f]= follower(tt,rise)
%
% This function determines the follower displacement and derivatives
% for a full rotation cam. The routine is set up for the displacement
% schedule in Problem 8.24.
% The input values are:
%theta = cam angle (deg)
%rise = maximum follower displacement
% The results are returned in the variable f where f(1) is the
% displacement, f(2) is the derivative of the displacement with
% respect to theta, and f(3) is the second derivative with respect
% to theta. The derivatives are not used in this problem, but they are
% required by the program rf.cam.m.
% find the correct interval.
fact=pi/180;
theta=tt*fact;
if tt < 80
f(1)=0;
f(2)=0;
f(3)=0;
end
if tt >= 80 & tt < 180
beta=100*fact;
theta=(tt-80)*fact;
f(1)=rise*((theta/beta)-(1/(2*pi))*sin(2*pi*theta/beta));
f(2)=(rise/beta)*(1-cos(2*pi*theta/beta));
f(3)=(2*rise*pi/beta^2)*sin(2*pi*theta/beta);
end
if tt>=180 & tt< 240
f(1)=rise;
f(2)=0;
f(3)=0;
end
if tt >= 240 & tt

end
f(2)=-(pi*rise/(2*beta))*sin(pi*theta/beta);
f(3)=-(rise/2)*(pi/beta)^2*cos(pi*theta/beta);
The main program input is specified in the following:
Cam Synthesis for Axial Roller Follower
Enter 1 for file input and 2 for interactive input [1]: 2
Enter input file name (rf_camio.dat): prob8.24.dat
Enter base circle radius [2]: 5
Enter radius of cylindrical or roller follower [0.5]: 0.75
Enter follower offset [1]: 0.5
Enter follower rise [2]: 3
Enter cam rotation direction (CW(-), CCW(+)) [-]: -
Enter cam angle increment for design (deg) [10]: 5
The graphical results from the program are given in the following three plots. A file giving the cam
coordinates is also produced.
- 332 -

Problem 8.25
Resolve Problem 8.24 if the amplitude of the follower translation is 4 cm, and the follower radius is
1 cm. The base circle radius is 5 cm, and the offset is 1 cm.
Solution:
The displacement profile equations are the same as those in Problem 8.24. The cam profile can be
determined using the program rf_cam.m using the same follower routine as was used in Problem
8.24. The main program input is specified in the following:
Cam Synthesis for Axial Roller Follower
Enter 1 for file input and 2 for interactive input [1]: 2
Enter input file name (rf_camio.dat): prob8.25.dat
Enter base circle radius [2]: 5
Enter radius of cylindrical or roller follower [0.5]: 1
Enter follower offset [1]: 1
Enter follower rise [2]: 5
Enter cam rotation direction (CW(-), CCW(+)) [-]: -
Enter cam angle increment for design (deg) [10]: 5
The graphical results from the program are given in the following three plots. The displacement
diagram is the same as that in Prob. 8.24. Note that the cam shape is visually similar to that shown
with the solution in Prob. 8.24.
- 333 -

- 334 -

Problem 8.26
Solve Problem 8.24 if the cam rotates counterclockwise.
Solution:
Everything is the same except for the direction of motion of the cam. The main program input is
specified in the following:
Cam Synthesis for Axial Roller Follower
Enter 1 for file input and 2 for interactive input [1]: 2
Enter input file name (rf_camio.dat): prob8.26.dat
Enter base circle radius [2]: 5
Enter radius of cylindrical or roller follower [0.5]: 0.75
Enter follower offset [1]: 0.5
Enter follower rise [2]: 3
Enter cam rotation direction (CW(-), CCW(+)) [-]: +
Enter cam angle increment for design (deg) [10]: 5
The graphical results from the program are given in the following three plots.
- 335 -

Problem 8.27
Determine the cam profile assuming that the translating flat-faced follower starts from a dwell from
0˚ to 80˚ and rotates clockwise. The rise occurs with parabolic motion during the cam rotation from
80˚ to 180˚. The follower then dwells for 60˚ of cam rotation, and the return occurs with simple
harmonic motion for the cam rotation from 240˚ to 360˚. The amplitude of the follower translation
is 3 cm. First find the minimum base circle radius based on avoiding cusps, and use that base circle
to design the cam.
Solution:
The displacement profile (along with the first and second derivatives) can be computed using the
equations in Chapter 8 using a spreadsheet or MATLAB program. The profile equations are:
For 0 80˚
y=0
For the first part of the rise, 80˚ 130˚, and
- 336 -

2
y = 2L
y' = 4L
y"= 4L
2
where L = 3, = 80˚, and =180˚80˚=100˚.
For the second part of the rise, 130˚ 180˚ , and
y = L 1 21
2
y' = 4L
1
y' = 4L
2
where L = 3, = 80˚, and =180˚80˚=100˚.
For 180˚ 240˚
y=3
For 240˚ 360˚
y = L 1+ cos
2
y' = L
sin
2
2
y"= L 2
cos
where L = 3, = 240˚, and = 360˚240˚=120˚
The minimum base circle to avoid cusps for a flat-faced follower is given by
rb > y() y"()
and we must check both the rise and return regions. For the first part of the rise region,
- 337 -

2L
2
4L
2
For the second part of the rise region
rb > L 1 21
2
+
4L
2
and for the return region,
rb > L
1+ cos +
2
L 2
2
cos
- 338 -
(7.24.1)
(7.24.2)
(7.24.3)
To find the minimum value of the base circle for the rise and return, we must compute the value of
rb for all in the rise and return ranges. This is most easily done with a program such as
MATLAB.
When this is done, the minimum value in the rise region is 5.3755 cm and the minimum value in the
return region is 0.375 cm. Therefore, the minimum base radius to be used is 5.3755 cm.
The cam which will generate this follower displacement can be determined using the equations
given in Table 8.9, and these equations have also been programmed in the MATLAB program
ff_cam.m. The program calls an m-file called follower.m which contains the equations for the
follower displacement. This file is given as follows:
function [f]= follower(tt,rise)
% This function determines the follower displacement and derivatives
% for a full rotation cam. The routine is set up for the displacement
% schedule in Problem 8.27.
% The input values are:
%theta = cam angle (deg)
%rise = maximum follower displacement
% The results are returned in the variable f where f(1) is the
% displacement, f(2) is the derivative of the displacement with
% respect to theta, and f(3) is the second derivative with respect
% to theta.
% find the correct interval.
fact=pi/180;
theta=tt*fact;
if tt < 80
f(1)=0;
f(2)=0;
f(3)=0;
end
if tt >= 80 & tt < 130
beta=100*fact;
theta=(tt-80)*fact;
f(1)=rise*2*(theta/beta)^2;
f(2)=(4*rise/beta)*(theta/beta);
f(3)=4*rise/beta;

end
if tt >= 130 & tt < 180
beta=100*fact;
theta=(tt-80)*fact;
f(1)=rise*(1-2*(1-theta/beta)^2);
f(2)=(4*rise/beta)*(1-theta/beta);
f(3)=-4*rise/beta;
end
if tt>=180 & tt< 240
f(1)=rise;
f(2)=0;
f(3)=0;
end
if tt >=240 & tt

Problem 8.28
Solve Problem 8.27 if the cam rotates counterclockwise.
Solution:
Everything is the same except for the direction of motion of the cam. The main program input is
specified in the following:
Cam Synthesis for Axial Flat-Faced Follower
Enter 1 for file input and 2 for interactive input [1]: 2
Enter input file name (ff_camio.dat): prob8.28.dat
Enter base circle radius [3.2]: 5.3755
Enter follower rise [2]: 3
Enter distance from follower centerline to bottom of face [1.6]: 2.6
Enter distance from follower centerline to top of face [2.6]: 3.6
Enter cam rotation direction [CW(-), CCW(+)] [-]: +
- 340 -

Enter cam angle increment for design (deg) [10]: .5
The graphical results from the program are given in the following three plots.
- 341 -

Problem 8.29
Determine the cam profile assuming that an oscillating, cylindrical-faced follower dwells while the
cam rotates counterclockwise from 0˚ to 100˚. The rise occurs with 3-4-5 polynomial motion
during the cam rotation from 100˚ to 190˚. The follower then dwells for 80˚ of cam rotation, and the
return occurs with simple harmonic motion for the cam rotation from 270˚ to 360˚. The amplitude
of the follower oscillation is 25˚, and the follower radius is 0.75 in. The base circle radius is 2 in,
and the distance between pivots is 6 in. The length of the follower is to be determined such that the
pressure angle starts out at zero.
Solution:
The displacement profile (along with the first and second derivatives) can be computed using the
equations in Chapter 8 in a spreadsheet or MATLAB program. The profile equations are:
For 0 100˚
y=0
For 100˚ 190˚
y = L 10
3
4
15
+ 6
5
y' = 30L
2
2
3
+
4
y"= 60L
2
3
2
+ 2
3
where L = 25˚, = 100˚, and =190˚100˚= 90˚.
- 342 -

For 190˚ 270˚
y=25˚
For 270˚ 360˚
y = L 1+ cos
2
y' = L
sin
2
2
y"= L 2
cos
where L = 25˚, = 270˚, and = 360˚270˚= 90˚
The cam which will generate this follower displacement can be determined using the equations
given in Table 8.11. These equations have been programmed in the MATLAB program orf_cam.m.
The program calls an m-file called o_follower.m which contains the equations for the follower
displacement. This file is given as follows:
function [f]= o_follower(tt,rise)
% This function determines the follower displacement and derivatives
% for a full rotation cam. The routine is set up for the displacement
% schedule in Problem 8p29. The input values are:
%theta = cam angle (deg)
%rise = maximum follower rotation
% The results are returned in the variable f where f(1) is the
% displacement, f(2) is the derivative of the displacement with
% respect to theta, and f(3) is the second derivative with respect
% to theta.
% find the correct interval.
fact=pi/180;
theta=tt*fact;
if tt < 100
f(1)=0;
f(2)=0;
f(3)=0;
end
if tt >=100 & tt =190 & tt= 270
beta=90*fact;
- 343 -

end
theta=(tt-270)*fact;
ff=pi*theta/beta;
f(1)=(rise/2)*(1+cos(ff));
f(2)=-pi*rise/(2*beta)*sin(ff);
f(3)=-(rise/2)*(pi*beta)^2*cos(ff);
The main program input is specified in the following:
Cam Synthesis for Oscillating Cylindrical-Faced Follower
Enter 1 for file input and 2 for interactive input [1]: 2
Enter input file name (orf_camio.dat): prob8p29.dat
Enter base circle radius [2]: 2
Enter radius of cylindrical or roller follower [1]: 0.75
Enter distance between fixed pivots [3*rb]: 6
Enter follower length [sqrt(r1^2-(rb+r0)^2)]:
Enter follower rise (deg) [30]: 25
Enter cam rotation direction (CW(-), CCW(+)) [+]: +
Enter angle increment for design (deg) [10]: 1
The graphical results from the program are given in the following three plots.
- 344 -

Problem 8.30
Solve Problem 8.29 if the cam rotates clockwise.
Solution:
Everything is the same except for the direction of motion of the cam. The main program input is
specified in the following:
Cam Synthesis for Oscillating Cylindrical-Faced Follower
Enter 1 for file input and 2 for interactive input [1]: 2
Enter input file name (orf_camio.dat): prob8p30.dat
Enter base circle radius [2]: 2
Enter radius of cylindrical or roller follower [1]: 0.75
Enter distance between fixed pivots [3*rb]: 6
Enter follower length [sqrt(r1^2-(rb+r0)^2)]:
Enter follower rise (deg) [30]: 25
Enter cam rotation direction (CW(-), CCW(+)) [+]: -
Enter angle increment for design (deg) [10]: 1
The graphical results from the program are given in the following three plots.
- 345 -

- 346 -

Problem 8.31
Design a cam and oscillating roller follower assuming that the follower starts from a dwell for 0˚ to
80˚ of cam rotation and the cam rotates clockwise. The rise occurs with cycloidal motion during the
cam rotation from 80˚ to 200˚. The follower then dwells for 40˚ of cam rotation, and the return
occurs with cycloidal motion for the cam rotation from 240˚ to 360˚. The amplitude of the follower
rotation is 45˚. Determine the cam base circle radius, distance between cam and follower pivots, the
length of the follower, and the radius of the follower for acceptable performance.
Solution:
First establish the equations for the displacement profile. These can be determined (along with the
first and second derivatives) from the equations in Chapter 8 using a spreadsheet or MATLAB
program. The profile equations are:
For 0 80˚
y=0
For 80˚ 200˚
y = L
1
sin
2
2
y' = L
1 cos2
y"= 2L
sin
2
2
where L = 45˚, = 80˚, and = 200˚80˚=120˚.
For 200˚ 240˚
y=45˚
For 240˚ 360˚
y = L 1
+
1
sin
2
2
y' = L
1+cos
2
y"= 2L
sin
2
2
where L = 45˚, = 240˚, and = 360˚240˚=120˚
The cam which will generate this follower displacement can be determined using the equations
given in Table 8.11 once the cam base radius, distance between cam and follower pivots, the length
- 347 -

of the follower, and the radius of the follower are known. The equations have also been
programmed in the MATLAB program orf_cam.m. The program calls the m-file o_follower.m
which contains the equations for the follower displacement. This file is given as follows:
function [f]= o_follower(tt,rise)
% This function determines the follower displacement and derivatives
% for a full rotation cam. The routine is set up for the displacement
% schedule in Problem 8.31. The input values are:
%theta = cam angle (deg)
%rise = maximum follower rotation
% The results are returned in the variable f where f(1) is the
% displacement, f(2) is the derivative of the displacement with
% respect to theta, and f(3) is the second derivative with respect
% to theta.
% find the correct interval.
fact=pi/180;
theta=tt*fact;
if tt < 100
f(1)=0;
f(2)=0;
f(3)=0;
end
if tt >=100 & tt =190 & tt= 270
beta=90*fact;
theta=(tt-270)*fact;
ff=pi*theta/beta;
f(1)=(rise/2)*(1+cos(ff));
f(2)=-pi*rise/(2*beta)*sin(ff);
f(3)=-(rise/2)*(pi*beta)^2*cos(ff);
end
Before running the program, we need to establish values for base radius, follower radius, and a
distance
is to experiment with the program orf_cam.m and
determine the types of values which will work. It will be found that the base circle must be fairly
small and the radius of the follower fairly large if a straight stemmed follower is to be used without
interference with the cam. Similarly, the follower pivot must be fairly large to avoid interference
with the cam and follower pivot. One set of values which will work is a base radius of 1 in, a
follower radius of 1.5 in, and a distance between pivots of 7 in. These values and the others given
in the problem can be input into the program to determine the cam geometry. The basic program
input is as follows:
Cam Synthesis for Oscillating Cylindrical-Faced Follower
Enter 1 for file input and 2 for interactive input [1]: 2
Enter input file name (orf_camio.dat): prob8p31.dat
Enter base circle radius [2]: 1
Enter radius of cylindrical or roller follower [1]: 1.5
- 348 -

Enter distance between fixed pivots [3*rb]: 7
Enter follower length [sqrt(r1^2-(rb+r0)^2)]:
Enter follower rise (deg) [30]: 45
Enter cam rotation direction (CW(-), CCW(+)) [+]: -
Enter angle increment for design (deg) [10]: 2
The graphical results from the program are given in the following three plots.
- 349 -

Problem 8.32
Determine the cam profile assuming that an oscillating, flat-faced follower starts from a dwell from
0˚ to 45˚ and rotates counterclockwise. The rise occurs with simple harmonic motion during the
cam rotation from 45˚ to 180˚. The follower then dwells for 90˚ of cam rotation, and the return
occurs with simple harmonic motion for the cam rotation from 270˚ to 360˚. The amplitude of the
follower oscillation is 20˚, and the follower offset is 0.5 in. The base circle radius is 5 in, and the
distance between pivots is 8 in.
Solution:
The displacement profile can be computed using the equations in Chapter 8 using a spreadsheet or
MATLAB program. The profile equations are:
For 0 45˚
y=0
For 45˚ 180˚
y = L
1 cos
2
y' = L
sin
2
2
y"= L 2
cos
where L = 20˚, = 45˚, and =180˚45˚=135˚.
For 180˚ 270˚
y=20˚
- 350 -

For 270˚ 360˚
y = L 1+ cos
2
y' = L
sin
2
2
y"= L 2
cos
where L = 20˚, = 270˚, and = 360˚270˚= 90˚
The cam which will generate this follower displacement can be determined using the equations
given in Tables 8.12 and 8.13. These equations have also been programmed in the MATLAB
program off_cam.m. The program calls the m-file o_follower.m which contains the equations for
the follower displacement. This file is given as follows:
function [f]= o_follower(tt,rise)
% This function determines the follower displacement and derivatives
% for a full rotation cam. The routine is set up for the displacement
% schedule in Problem 8.32. The input values are:
%theta = cam angle (deg)
%rise = maximum follower rotation
% The results are returned in the variable f where f(1) is the
% displacement, f(2) is the derivative of the displacement with
% respect to theta, and f(3) is the second derivative with respect
% to theta.
% find the correct interval.
fact=pi/180;
theta=tt*fact;
if tt < 100
f(1)=0;
f(2)=0;
f(3)=0;
end
if tt >=100 & tt =190 & tt= 270
beta=90*fact;
theta=(tt-270)*fact;
ff=pi*theta/beta;
f(1)=(rise/2)*(1+cos(ff));
f(2)=-pi*rise/(2*beta)*sin(ff);
f(3)=-(rise/2)*(pi*beta)^2*cos(ff);
end
- 351 -

The basic program input is specified in the following. Default values are used for noncritical
values.
Cam Synthesis for Oscillating Flat-Faced Follower
Enter 1 for file input and 2 for interactive input [1]: 2
Enter input file name (off_camio.dat): prob8p32.dat
Enter base circle radius [2]: 5
Enter distance between fixed pivots [6]: 8
Length of follower face [1.5*r1]:
Enter follower offset [0.5]: 0.5
Enter follower rise (deg) [15]: 20
Enter cam rotation direction (CW(-), CCW(+)) [-]: +
Enter angle increment for design (deg) [10]: 1
The graphical results from the program are given in the following three plots.
- 352 -

Problem 8.33
Solve Problem 8.32 if the cam rotates clockwise.
Solution:
Everything is the same except for the direction of motion of the cam. The basic program input is
specified in the following. Default values are used for noncritical values.
Cam Synthesis for Oscillating Flat-Faced Follower
Enter 1 for file input and 2 for interactive input [1]: 2
Enter input file name (off_camio.dat): prob8p33.dat
Enter base circle radius [2]: 5
Enter distance between fixed pivots [6]: 8
Length of follower face [1.5*r1]:
Enter follower offset [0.5]: 0.5
Enter follower rise (deg) [15]: 20
Enter cam rotation direction (CW(-), CCW(+)) [-]: -
Enter angle increment for design (deg) [10]: 1
The graphical results from the program are given in the following three plots.
- 353 -

Problem 8.34
Design the cam system assuming that an oscillating, flat-faced follower starts from a dwell for 0˚ to
100˚ of cam rotation and the cam rotates counterclockwise. The rise occurs with uniform motion
during the cam rotation from 100˚ to 200˚. The follower then dwells for 40˚ of cam rotation, and the
return occurs with parabolic motion for the cam rotation from 240˚ to 360˚. The oscillation angle is
20˚.
Solution:
First establish the equations for the displacement profile. The displacement profile can be
computed using the equations in Chapter 8 in a spreadsheet or MATLAB program. The profile
equations are:
For 0 100˚
y=0
For 100˚ 200˚
- 354 -

y = L
y' = L
y"= 0
where L = 20˚, = 100˚, and = 200˚100˚=100˚.
For 200˚ 240˚
y=20˚
For the first part of the return, 240˚ 300˚, and
y = L 1 2
2
y' = 4L
y"= 4L
2
where L = 20˚, = 240˚, and = 360˚240˚=120˚
For the second part of the return, 300˚ 360˚, and
y = 2L 1
2
y' = 4L
1
y"= 4L
2
where L = 20˚, = 240˚, and = 360˚240˚=120˚
The cam which will generate this follower displacement can be determined using the equations
given in Tables 8.12 and 8.13 once the cam base radius, distance between cam and follower pivots,
the offset distance, and the length of the follower are known.. The program calls the m-file called
o_follower3.m which contains the equations for the follower displacement. This file is given as
follows:
function [f]= o_follower3(tt,rise)
% This function determines the follower displacement and derivatives
% for a full rotation cam. The routine is set up for the displacement
% schedule in Problem 8.34. The input values are:
- 355 -

%theta = cam angle (deg)
%rise = maximum follower rotation
% The results are returned in the variable f where f(1) is the
% displacement, f(2) is the derivative of the displacement with
% respect to theta, and f(3) is the second derivative with respect
% to theta.
% find the correct interval.
fact=pi/180;
theta=tt*fact;
if tt =100 & tt =200 & tt=240 & tt=300
beta=120*fact;
theta=(tt-240)*fact;
ff=theta/beta;
f(1)=2*rise*(1-ff)^2;
f(2)=-(4*rise/beta)*(1-ff);
f(3)=(4*rise/beta);
end
Before running the program, we need to establish values for the base radius and the distance
between pivots. The offset distance is arbitrarily taken as 0. One option is to experiment with the
program orf_cam and determine the types of values which will work. It will be found that the base
circle must be fairly large to avoid cusps. Similarly, the distance between pivots must be fairly large
to avoid interference with the cam and follower pivot. One set of values which will work is a base
radius of 3 inches and a distance between pivots of 5 inches. These values and the others given in
the problem can be input into the program to determine the cam geometry. The basic program input
is as follows. Default values are used for noncritical values.
Cam Synthesis for Oscillating Flat-Faced Follower
Enter 1 for file input and 2 for interactive input [1]: 2
Enter input file name (off_camio.dat): prob8p34.dat
Enter base circle radius [2]: 3
Enter distance between fixed pivots [6]: 5
Length of follower face [1.5*r1]:
- 356 -

Enter follower offset [0.5]: 0
Enter follower rise (deg) [15]: 20
Enter cam rotation direction (CW(-), CCW(+)) [-]: +
Enter angle increment for design (deg) [10]: 0.5
The graphical results from the program are given in the following three plots. The displacement
diagram does not show the acceleration "spikes" at the beginning and end of the dwell; however, the
discontinuities in the radius of curvature are indicated. This cam would perform poorly in a high
speed application.
- 357 -

Problem 8.35
Design the cam system assuming that an oscillating, flat-faced follower starts from a dwell for 0˚ to
50˚ of cam rotation and the cam rotates clockwise. The rise occurs with cycloidal motion during the
cam rotation from 50˚ to 200˚. The follower then dwells for 90˚ of cam rotation, and the return
occurs with harmonic motion for the cam rotation from 290˚ to 360˚. The oscillation angle is 25˚.
Solution:
First establish the equations for the displacement profile. The displacement profile (along with the
first and second derivatives) can be computed using the equations in Chapter 8 in a spreadsheet or
MATLAB program. The profile equations are:
For 0 50˚
y=0
For 50˚ 200˚
y = L
1
sin
2
2
y' = L
1 cos2
y"= 2L
sin
2
2
where L = 25, = 50˚, and = 200˚50˚=150˚.
For 200˚ 290˚
y=25
For 290˚ 360˚
- 358 -

y = L 1+ cos
2
y' = L
sin
2
2
y"= L 2
cos
where L = 25, = 290˚, and = 360˚290˚= 70˚
The cam which will generate this follower displacement can be determined using the equations
given in Tables 8.12 and 8.13 once the cam base radius, distance between cam and follower pivots,
and the offset distance. The length of the follower must be specified, but its value is not critical as
long as it is long enough for the follower to remain tangent to the cam. The default value in the
program can be used. The program calls an m-file called o_follower4.m which contains the
equations for the follower displacement. This file is given as follows:
function [f]= o_follower(tt,rise)
% This function determines the follower displacement and derivatives
% for a full rotation cam. The routine is set up for the displacement
% schedule in Problem 8.35. The input values are:
%theta = cam angle (deg)
%rise = maximum follower rotation
% The results are returned in the variable f where f(1) is the
% displacement, f(2) is the derivative of the displacement with
% respect to theta, and f(3) is the second derivative with respect
% to theta.
% find the correct interval.
fact=pi/180;
theta=tt*fact;
if tt < 50
f(1)=0;
f(2)=0;
f(3)=0;
end
if tt >= 50 & tt =200 & tt< 290
f(1)=rise;
f(2)=0;
f(3)=0;
end
if tt >= 290
beta=70*fact;
theta=(tt-290)*fact;
f(1)=(rise/2)*(1+cos(pi*theta/beta));
f(2)=-(pi*rise/(2*beta))*sin(pi*theta/beta);
f(3)=-(rise/2)*(pi/beta)^2*cos(pi*theta/beta);
end
- 359 -

Before running the program, we need to establish values for the base radius, the distance between
pivots, and the offset distance. One option is to experiment with the program off_cam and
determine the types of values which will work. It will be found that the base circle must be fairly
large to avoid cusps. Similarly, the distance between pivots must be fairly large to avoid
interference with the cam and follower pivot. One set of values which will work is a base radius of
2 inches and a distance between pivots of 4 inches. The offset distance is arbitrarily taken as 0.
These values and the others given in the problem can be input into the program to determine the
cam geometry. The program input is as follows. Default values are used for noncritical values.
Cam Synthesis for Oscillating Flat-Faced Follower
Enter 1 for file input and 2 for interactive input [1]: 2
Enter input file name (off_camio.dat): prob8p35.dat
Enter base circle radius [2]: 2
Enter distance between fixed pivots [6]: 4
Length of follower face [1.5*r1]:
Enter follower offset [0.5]: 0
Enter follower rise (deg) [15]: 25
Enter cam rotation direction (CW(-), CCW(+)) [-]: -
Enter angle increment for design (deg) [10]: 1
The graphical results from the program are given in the following three plots.
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Problem 8.36
Determine the cam profile assuming that the translating knife-edged follower starts from a dwell
from 0˚ to 80˚ and rotates clockwise. The rise occurs with cycloidal motion during the cam rotation
from 80˚ to 180˚. The follower then dwells for 60˚ of cam rotation, and the return occurs with
simple harmonic motion for the cam rotation from 240˚ to 360˚. The amplitude of the follower
translation is 4 cm. The base circle radius is 5 cm, and the offset is 0.5 cm.
Solution:
We can treat a knife-edged radial follower as a roller follower with a roller radius of zero. However,
we must first establish the equations for the displacement profile. The displacement profile (along
with the first and second derivatives) can be computed using the equations in Chapter 8 in a
spreadsheet or MATLAB program. The profile equations are:
For 0 80˚
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y=0
For 80˚180˚
y = L 1 2
sin
2
y' = L
2
1 cos
y"= 2L 2
2 sin
where L = 4, = 80˚, and = 180˚80˚=100˚.
For 180˚240˚
y=5
For 240˚360˚
y = L
2
1 + cos
y' = L
sin
2
y"= L
2
2
cos
where L = 5, = 240˚, and = 360˚240˚=120˚
The cam which will generate this follower displacement can be determined using the equations
given in Table 8.8 if we set r0 = 0. The equations are coded in the program called rf_cam. The
program calls an m-file called follower.m which contains the equations for the follower
displacement. This file is given as follows:
function [f]= follower(tt,rise)
% This function determines the follower displacement and derivatives
% for a full rotation cam. The routine is set up for the displacement
% schedule in Problem 8.36. The input values are:
%theta = cam angle (deg)
%rise = maximum follower rotation
% The results are returned in the variable f where f(1) is the
% displacement, f(2) is the derivative of the displacement with
% respect to theta, and f(3) is the second derivative with respect
% to theta.
% find the correct interval.
fact=pi/180;
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theta=tt*fact;
if tt < 80
f(1)=0;
f(2)=0;
f(3)=0;
end
if tt >= 80 & tt =180 & tt< 240
f(1)=rise;
f(2)=0;
f(3)=0;
end
if tt >= 240
beta=120*fact;
theta=(tt-240)*fact;
f(1)=(rise/2)*(1+cos(pi*theta/beta));
f(2)=-(pi*rise/(2*beta))*sin(pi*theta/beta);
f(3)=-(rise/2)*(pi/beta)^2*cos(pi*theta/beta);
end
Before running the program, we need to establish values for the base radius, the distance between
pivots, and the offset distance. One option is to experiment with the program orf_follower and
determine the types of values which will work. It will be found that the base circle must be fairly
large to avoid cusps. Similarly, the follower pivot must be fairly large to avoid interference with the
cam and follower pivot. One set of values which will work is a base radius of 2 inches and a
distance between pivots of 4 inches. The offset distance is arbitrarily taken as 0. These values and
the others given in the problem can be input into the program to determine the cam geometry. The
program input is as follows:
Cam Synthesis for Axial Roller Follower
Enter 1 for file input and 2 for interactive input [1]: 2
Enter input file name (rf_camio.dat): prob8p36.dat
Enter base circle radius [2]: 5
Enter radius of cylindrical or roller follower [0.5]: 0
Enter follower offset [1]: 0
Enter follower rise [2]: 5
Enter cam rotation direction (CW(-), CCW(+)) [-]: -
Enter cam angle increment for design (deg) [10]: 1
The graphical results from the program are given in the following three plots.
- 363 -

- 364 -

Problem 8.37
A radial flat-faced follower is to move through a total displacement of 20 mm with harmonic motion
with the cam rotates through 30˚. Find the minimum radius of the base circle that is necessary to
avoid cusps.
Solution:
The displacement profile can be computed using the equations in Chapter 8. The profile equation
for the rise is_
For 0 30˚
y = L
2
1cos
y'= L
2 sin
y"= L
2
cos
2
where L = 20 mm and = 30˚ . Cusps do not occur when
rb > f() f"( )
The minimum base circle radius is
rb = L
2
1cos
L
cos
2 2
= L
2
= L
2
+ L
2
+ L
2
1
2
( ) 2
1
/ 6
cos
cos
/ 6
= L
2
+
L
2 35 ( )cos6 = 18Lcos6
The value on the right is a maximum when = 30˚ . Then, the minimum base circle radius is
rb =18L =18(20) =360 mm
Problem 8.38
A radial flat-faced follower is to move through a total displacement of 3 in with cycloidal motion
with the cam rotates through 90˚. Find the minimum radius of the base circle that is necessary to
Solution:
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The displacement profile can be computed using the equations in Chapter 8. The profile equation
for the rise is:
For 0 90˚
y = L 1 2
sin
2
y'= L
2
1 cos
y"= 2L 2
2 sin
where L = 3 in and = 90˚ = / 2 . Cusps do not occur when
rb > f() f"( )
The minimum base circle radius is
rb = L
1 2 2L 2
sin
2 2 sin
= L L
+
2 sin2
2L 2
2 sin
= 2L L 8L 2 15
+ sin4 sin4 = L
2 2 sin4
[ ]
The value on the right is a maximum when drb
= 0 . Then,
d
or
or
[ ]
drb 2 60
= L
d 2 cos4
cos 4 = 2 / 30 =1/15.
L
= [ 2 30cos 4 ]= 0
= 1
acos(1/15) = 21.54˚, 66.54˚ = 0.376 rad, 1.161 rad
4
Substituting in the two values:
rb1 = L 2 15 2(0.376)
[ sin 4
2 ] = 3
15
2 sin86.16
= 6.42
rb2 = L 2 15
2 sin4
2(1.161)
[ ] = 3
15
sin 266.16
2
= 9.363 in
- 366 -

Therefore, the minimum base circle radius required is
rb = 9.363 in
Problem 8.39
A radial roller follower is to move through a total displacement of L=19 mm with harmonic motion
while the cam rotates 60˚. The roller radius is 5 mm. Use the program supplied with the book and
find the minimum radius necessary to avoid cusps during the interval indicated.
Solution:
The displacement profile can be computed using the equations in Chapter 8.
When 0 60˚
Now,
and
y = L
2
1cos
y'= L
2 sin
y"= L
2
cos
2
L =19 mm
=
3
To determine if there are cusps, run the program CAM2 under MAINMENU. Substitute in a base
circle radius of 0, the minimum base circle radius and run the analysis. From this it will be apparent
that there are no cusps for any base circle radius.
Problem 8.40
A radial roller follower is to move through a total displacement of L=45 mm with cycloidal motion.
The roller radius is 5 mm, and the cam rotates 90 degrees during the rise. Use the program
supplied with the book and find the minimum radius necessary to avoid cusps during the interval.
Solution:
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The displacement profile can be computed using the equations in Chapter 8.
When 0 90˚
y = L 1 2
sin
2
y'= L
2
1 cos
y"= 2L 2
2 sin
where L = 45 mm and = 90˚ = / 2 _
To determine if there are cusps, run the program CAM2 under MAINMENU. Substitute in a base
circle radius of 0, the minimum base circle radius and run the analysis. From this it will be apparent
that there are no cusps for any base circle radius.
Problem 8.41
Assume that a flat-faced translating follower is used with the displacement schedule in Problem
8.10. Determine if a cusp is present at = 60˚.
Solution:
When = 60˚ =
3 _
3
y =
1
2
( )
sin 2( 3)
1 1
= sin(2 3) = 0.195 in
3 2
y ˙ =
10.472
( 1 cos2 )= ( 1cos(2 3) )= 5.000
in.
sec .
˙ y ˙ = 2 ( ) 2
sin2 = 2 10.472 ( ) 2
sin(2 3) = 60.46 in.
sec2 To avoid a cusp,
Now
and
rb > f() f"( )
f( ) = 0.195
- 368 -

Therefore,
or
f"( ) = 2 1 ( )
2
rb > 0.195 0.551
rb > 0.746
sin2 = 2 ( ) sin(2 3) = 0.551
Therefore, any base radius will work and there will be no cusp.
Problem 8.42
Assume that a flat-faced translating follower is used with the displacement schedule in Problem
8.12. Determine if a cusp is present at = 90˚.
Solution:
When, = 90˚ =
2 _
f( ) = L 1 2
sin =1
2 1
sin 2( 2)
1 1
2 2
=
2 2 sin
y'= L
2 L
1 cos = ( 1 cos2)
f"() = 2L 2
sin2 = sin = 0
To avoid a cusp,
Now
and
Therefore,
or
rb > f() f"( )
f( ) = 0.5
f"( ) = 0
rb > 0.5 0
- 369 -
( )
= 0.5 in

0.5
Therefore, any base radius will work and there will be no cusp.
- 370 -