Quote:
Originally Posted by DMetalKong
Okay, I was working with some numbers and this is what I came up with for a 2 CIM drive (1 driving each side of the robot).
Torque supplied by 1 CIM at peak efficiency = 0.48 N m = 4.25 lbf in.
Torque supplied by 2 CIMs = 4.25 * 2 = 8.5 lbf in.
Max. Frictional Force = [150 lbf (robot) + 40 lbf (trailer)] * 0.2 (rounding the COF up quite a bit) = 38 lbf.
Max. Torque = 38 lbf * 3 in (wheel radius) = 114 lbf in.
Torque needed/Torque provided = 114/8.5 = 13.4.
A toughbox provides a 12.75:1 reduction. Does anyone think that this will be sufficient for powering a robot?
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You should take a look at the motor curves, you'll almost never be running your CIMs at peak efficiency. It's better to look at 50% stall (peak power) and 25% stall, which is where you'd like your motors running most of the time.
50% Stall torque = 2 x 21.5 lbf-in x 50% x 12.75 = 275 lbf-in.
25% Stall torque = 2 x 21.5 lbf-in x 25% x 12.75 = 137 lbf-in.
So I think a 2 CIM drive with the stock transmissions and a 1:1 sprocket ratio will work out just fine. Especially as you won't be getting the full 40 lbs out of the trailer, and the CoF isn't .2