- #1
JJ1989
- 5
- 0
I would like to run 8 pulleys off of one motor. Each "pulley" would be 19" in diameter with 20lb of weight at 30RPM. I would like to know how to calculate the inch-lb that would be required of the motor. I read this on masterdrives.com
Torque(inch-lb) = (.03908 x N x W x R^2)/t
Where N = difference between final and initial rpm, W = weight of rim in lb, R = radius in ft, and t = time required to affect speed change in seconds.
So if I wanted it to come up to seed in 5 seconds I calculate with one
Torque = (0.03908 x 30 x 20 x 2.51)/5 = ~11.77 inch-lb of torque.
If I were to run the motor at 90 rpm with a pulley a third of the diameter I would be using a third of the torque or ~4 inch-lb.
Then since there's 8 do I need a motor capable of 90rpm and ~32 inch-lb of torque?
Torque(inch-lb) = (.03908 x N x W x R^2)/t
Where N = difference between final and initial rpm, W = weight of rim in lb, R = radius in ft, and t = time required to affect speed change in seconds.
So if I wanted it to come up to seed in 5 seconds I calculate with one
Torque = (0.03908 x 30 x 20 x 2.51)/5 = ~11.77 inch-lb of torque.
If I were to run the motor at 90 rpm with a pulley a third of the diameter I would be using a third of the torque or ~4 inch-lb.
Then since there's 8 do I need a motor capable of 90rpm and ~32 inch-lb of torque?