Sizing DC Motor

Hi Everyone, I'm a bit stuck on a prototype I've been working on where I've designed a linkage system driven by a worm drive gear box. For cost purposes, the gear box I'm using is this one from Tamiya: and I've swapped out the built in tiny DC motor for a 6V 130-size DC motor: Now, speed and torque are both pretty critical in this prototype. Ideally, I need the output shaft to turn at roughly 25-30rpm, but it also needs to be able to lift about 5lb/ft (roughly 900 oz/in). According to the pololu page, using a 120:1 gear box with that DC motor produces 75 oz/in. I chose the 441:1 gear ratio in the Tamiya gear box because it gives me about 26rpm (motor turns at 11,500rpm)... however once I extrapolate the torque using the 441:1 ratio - the amount of output torque is only 1.43lb/ft (275 oz/in). I can choose the next gear ratio up in the chain, which is 1543:1 which would likely put me at about 5lb/ft of torque... but then my speed is too slow (only about 8rpm). So, my question is... Is there a stronger motor (of the same size as the 130-type DC motor) which would allow me to use the 441:1 ratio but generate more torque? Or, is there a faster motor (again of the same size) which would allow me to use the 1543:1 ratio but still maintain roughly 25-30 rpm? I've actually already built my prototype... so trying to use the same configuration (or something similar would be pretty key). Do you have any suggestions? Thanks for your help. -Andy

If you're stuck right now, you might consider alternatives like altering your built prototype (isn't that what a prototype is for ?) while waiting for someone to hand you a solution. You'll see, once you've changed your current setup, someone will pop up with that solution for you.

but it also needs to be able to lift about 5lb/ft (roughly 900 oz/in).

What are pounds per foot and ounces per inch? Not torque I'm afraid, I think you mean "foot pounds-force" (ft-lbf) and "inch ounce-force" (in-ozf) :)

Use SI units of newton metres (Nm) to make things easy to calculate. 5 ft-lbf is 6.7 Nm

power = torque x angular velocity (no conversion factors needed - torque in Nm, angular velocity in radians/second and power in watts.

30rpm is pi radians/s, so power needed is = 6.7 x 3.14 = 21W. Of course we need a safety margin and to allow for gear losses, so call it 30 or 40W. If your motor / power-supply cannot handle that sort of power level you'll need to find ones that can (is this continuous or intermittant duty BTW?)

Smallish DC motors are not greatly efficient, so you'll have to allow for perhaps 10 to 15W heat dissipation in the motor as well, and make sure there is enough air-flow (these motors will have built-in fans).