Page 71 - Cam Design Handbook

THB3 8/15/03 12:58 PM Page 59

MODIFIED CAM CURVES 59

Solution The time for a revolution = 60/300 = 0.2sec/rev. Let us divide the total action
into three parts, as shown in Fig. 3.3 with T and Q the tangent points of the curves, giving
the times
Ê 40 ˆ
00222 sec
t = . = .
02
1 Ë 360 ¯
Ê 30 ˆ
02
t = . = .
00167 sec
2 Ë 360 ¯
Ê 60 ˆ
t = . = .
02
00333 sec
3 Ë 360 ¯
From the last chapter, we have the constant acceleration displacement
1
y = V t + At in (2.24)
2
0
2
where V 0 = initial velocity, in./sec
t = time, sec
A = acceleration, in./sec 2
Therefore, the displacements and the velocity for the parabolic motion of parts 1 and
3 are
1
y = A t 2
1 1 1
2
1
y = v t + A t 2 (3.1)
3 Q 3 3 3
2
v = A t
T 11
v =- A t
Q
33
For constant velocity of part 2
v = v .
Q T
Combining yields
At =- A t .
11 3 3
Substituting yields
A 00333)
(
.
A 00222) =- ( .
3
1
1
or A =- 1 A .
1
2 3
Also, the displacement of part 2 is
y = v t .
2 T 2
We are given that the total displacement
y + y + y = 4.
1 2 3