CIM Motor RPM

[EricH]

If you're driving directly from the CIMs, with 150 lbf of robot on top of them...

Let's assume 6" wheels (3" moment arm), and 4 wheels, each with its own CIM. 150/4=37.5 lbf per wheel supported (which we'll call N). At 3", assuming a coefficient of friction (mu) of 1 (F=N*mu), that's 112.5 lbf-in of torque exerted on the wheel (F*d). That converts into 1800 oz-in to get moving at all.

Stall torque of a CIM is 343 oz-in, and it draws 133 amps. So, not only do you not move, you spend all match tripping breakers trying.

Now, I've given you some of the calculations for a no-gearing situation. Rework it to include gearing to find your minimum gearing. Or, rework it using a different coefficient of friction. Or a different number of powered wheels.

I'd also comment on the wisdom of running one CIM per side, but I think you can work that out as well using similar calculations. It might help to post the calculations when you're done, so that any mistakes can be caught early.