# Torque and gear ratio help please?

Hi, I'm new to this forum but I've read over probably a dozen threads so I figured that someone should probably be able to help me; please forgive me if this is the wrong area to post this question as well.

Dilemma:
So the problem right now is that I'm trying to use a gear reduction system to get the most out of some small electrical motors (those of an NXT robot) and I'd like to be able to calculate the amount of weight that can be pulled upward if I know everything about the gears and motor(s) hooked to the gear reduction system. Any ideas as to how I could calculate this?

Thoughts:
I imagine that given W=FD and P=W/T that I could possibly use these two formulas as well as the rpms of the motor and the size of the gears to go backwards and mathematically calculate weight? Just a thought though. I've been thinking through a half dozen designs all day so as I write this I'm fried lol, please forgive this post if it comes out half thought through.
Oh p.s. heres a link showing a lot of thorough graphs for rpm(s), torque vs. mechanical power, etc. http://www.philohome.com/nxtmotor/nxtmotor.htm

Thank you in advance if you even look at this lol

## Answers and Replies

Well i those two equations are not that far off.lol
Just a quick idea, you have your output speed in rpm.
1) using the speed-torque graph, find the torque at that speed.
2) using the torque, check the power-torque graph and find the mech power the motor is producing in Watts
3) Convert the rpm speed into a linear speed in m/s = (rpm x pi/30) x (radius of the lifting pulley or gear..whatever)
4) divide the power by the linear speed to find the load on the gear/pulley at that speed.
i.e
Tangential load (weight) in Newtons = Power/linear speed
Hope this helps
Anyone correct me if I'm wrong somewhere.I'm in a rush

Filip Larsen
Gold Member
Welcome to PF!

Getting the most out of the motors depend a bit on what you mean exactly. If you want to be able to lift a mass as fast as possible you would want to go for maximum mechanical power, which the linked page mentions the motor will provide when loaded with a torque around 0.15 Nm. If you want to lift the mass as efficiently as possible (getting most mechanical work from your electrical source), diagrams on that page indicates that the torque should be a bit lower, around 0.10 Nm.

If your (frictionless) gear train has a reduction ratio of K (the output shaft rotates K times slower than the motor shaft, or equivalently, the output shaft has K times larger torque than the motor shaft), the output shaft is attached to a spool of radius R that winds the wire that lifts your mass M, and the motor shaft has torque N, then these relates as N*K = M*g*R or rearranged M = N*K / (g*R).

For example, if you attach an R = 0.05 m spool directly to the motor shaft (K=1) the maximum mass for N = 0.15 Nm motor torque is around 0.3 kg. If you instead put a K=10 gear train in, the mass increases 10-fold to around 3 kg. Likewise, if you put on a spool with half the radius you double the mass that can be lifted.

Furthermore, still assuming that the gear is frictionless (i.e. it transfers all power from the motor shaft to output shaft), you know that the lift power M*g*V must be equal to the mechanical power P provided by the motor, thus V = P/(M*g) = P*R / (N*K). So a change to K or R that allows you to lift double the mass will at the same time put the speed to half. Assuming P = 1.6 W, the last example above that gave M = 3 kg will give a speed of around 5 cm/s.

If the gear train has significant friction this will reduce the amount of torque on the output shaft that is available for lifting, thus reducing the maximum mass that can be lifted, but keeping the speed more or less the same.

If this did not help you feel free to elaborate your questions.