Page 79 - Cam Design Handbook

THB3 8/15/03 12:58 PM Page 67

MODIFIED CAM CURVES 67

7 q
for £ £ 1,
8 b
Ê q Ê q ˆˆ
.
y = hÁ06110155 03889845 - 00309544sin Á4p ˜˜
+ .
.
Ë b Ë b ¯¯
h Ê Ê q ˆˆ
.
y¢ = Á03889845 - 03889845cos Á4p ˜˜
.
b Ë Ë b ¯¯
h Ê Ê q ˆˆ
y¢¢ = Á -4 888124sin Á4p ˜˜
.
b 2 Ë Ë b ¯¯
h Ê q ˆ
y¢¢¢ = 61 425975 cos Á4p ˜.
.
b 3 Ë b ¯
A computer solution is employed to establish the incremental displacement value and the
characteristic curves of the action.
The modified trapezoidal curve has the following peak values
h
y ¢ = 2
b
h
y ¢¢ = 4 888
.
b 2
h
.
y ¢¢¢ = 61 43 . (3.9)
b 3
The nondimensional factors of the displacement, the velocity, and the acceleration of this
curve are given in App. B.
EXAMPLE A cam rotates at 300rpm. A symmetrical modified trapezoidal acceleration
curve (parabolic motion combined with the cycloidal curve) is to be drawn with the ratio
1
b = / 8 . The total rise is 4 inches in 160 degrees of cam rotation. Find pertinent values of
all the characteristics and plot the curves without the use of Eqs. (3.7) through (3.9).
Solution In Fig. 3.6 we see the basic cycloidal curve from which the combination
curve is developed. This figure also shows the modified trapezoidal acceleration curve.
The variables pertaining to the cycloidal sector will be denoted by the primed symbols
(Fig. 3.6a). In Fig. 3.6b, let us divide one-half of the rise into its three component
1
parts. Since b = / 8 and the angle b/2 is 80 degrees, we see that the cam angle for parts 1
and 3 is q 1 = q 3 = 20 degrees = b¢/4. This gives q 2 = 40 degrees. The angular velocity of
the cam is
w = 300 60 ¥ 2 p = 31 4 rad sec
.
The characteristics of the cycloidal curve from Eqs. (2.58), (2.59), and (2.60) are

Ê q 1 2 pq ˆ
Displacement y = hÁ - sin ˜ in
Ë b 2 p b ¯
w h Ê 2 pq ˆ
Velocity y = w y¢ = Á1 - cos ˜ ips
˙
b Ë b ¯
2
Ê w 2 hp 2 pq ˆ
y = Á
Acceleration ˙˙ y = w 21 sin ˜ in sec 2
Ë b 2 b ¯