Page 98 - Cam Design Handbook

THB3 8/15/03 12:58 PM Page 86

86 CAM DESIGN HANDBOOK

which yields
4h 1 b 1
p h = b .
2 2
For the C1 curve the acceleration is
p h
y ¢¢ = 1
max
b 2
1
And for the H2 curve, the acceleration is
p 2 h
y ¢¢ = 2
max
4 b 2
2
It is given that y≤ max for C1 curve = 2 ¥ y≤ max for H2 curves.
Therefore
2
ph p h
1 =- 2
b 2 2 b 2
1 2
Now we have four equations to solve for h 1 , h 2 , b 1 , b 2 .
h + h = .
175 in
1 2
2094 rad
b 1 + b 2 = .
4 h b
1 = 1
p h b
2 2
p h p 2 h
2 =- 2
b 2 2b 2
1 2
This solves to yield
h = 04934 in
.
1
.
h = 1 2566 in
2
p
b = 40 deg ¥ = 06981 rad
.
1
180
p
b = 80 deg ¥ = 1 3963 rad
.
2
180
By following the above principles, the designer may complete the DRRD curve (utilizing
H3 and C4) and modify it to suit the design conditions of the machine.
Solution (b)
The 15 degrees falls in the cycloidal curve C1 region
15p
15 deg = = 0262 rad
.
180
The displacement
È 0262. 1 p (0262. ) ˘
.
y = 0493. - sin = 0400 in
Í Î 0698 p 0698 ˙ ˚
.