Page 474 - Cam Design Handbook

THB14 9/19/03 7:58 PM Page 462

462 CAM DESIGN HANDBOOK

of the cam-follower mechanism. Since we are concerned with the maximum pressure
angle

E
tana = tant = . (14.8)
m m
M
EXAMPLE An eccentric circle cam drives a translating roller follower. This circle has an
eccentricity of 0.500in. and a diameter of 2.500 in. and rotates at 180rpm. The roller
follower has a diameter of 0.750in. Find (1) the characteristics of the follower
motion after the cam has rotated 30° from its lowest position and (2) the maximum
pressure angle.
Solution The equivalent mechanism has a crank length E equal to 0.500in. The
connecting rod length equals the eccentric circle radius plus the roller radius,
giving
+
=
.
.
M =1 2500 375 1 625 in.
.
The cam angular velocity
2 p
w =180 ¥ =18 85 rad sec.
.
60
From Eq. (14.4)
E
sint = sine
M
Ê . 0500 ˆ
=
t = sin -1 sin308 .85 . ∞
Ë . 1 625 ¯
The displacement from Eq. (14.2)
-
y = E Ecose + M cost - M
- .625 0500
= . 0500 1 - . cos30 + . 1 625 cos . 8 85
= . 00475 in.
The velocity from Eq. (14.6)
e
˙ y = w sin - cos tan ) t
e
E (
. )(
.
= (0500 )(18 85 sin30 - cos30 tan . )
8 85
= . 344 lb in 2 .
The acceleration, Eq. (14.7)
2
Ê E cos e ˆ
˙˙ y = Ew 2 cos - + sin tant
e
e
Ë M cos t ¯
3
2
. )
.
tan .85
= (0500 )(18 85 2 Ê cos30 - . 0500 cos 30 + sin308 ˆ
Ë . 1 625 cos . 8 85 ¯
3
2
=125 .2 in sec .
The maximum pressure angle from Eq. (14.8)