Quote:
Originally Posted by ksafin
Gotcha, thanks so much!
Just a quick question.
Right now we have it set up such that the motor has an 80 tooth gear, turning a 40 tooth gear, which is on the same axle as a rack and pinion gear (32 tooth I believe).
I feel like the calculation is a bit different in this case.
My assumption is that I can say the motor torque "doubles" from the gear ratio, lending to a 283 * 2 oz-in torque on the axle w/ the 40 tooth and pinion gear.
At this point, do I use the radius of the pinion gear to determine linear force on the rack and pinion?
|
Correct. Don't forget that free speed will be halved by that ratio as well.
For a Given gear ratio N, To = N*Ti and wo = wi/N.
To = output Torque
Ti = input Torque
wo = output speed
wi = input speed
I'd wager you already knew that, but I figured others reading might not.
Also, for multiple stages of gear reduction, lets say N1... N4. The total reduction Ntot = N1*N1*N2*N4.
EDIT: Magnets is correct that each stage of gearing will cause some energy loss. However I do disagree that a significant safety factor is always necessary.