12.75:1 enough?

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?

The maximum frictional force is not 150lbs + 40lbs. The trailer wheels are not powered and so don’t add to your maximum tractive force. The tongue weight, however, will help you - let’s call it 10 lbs (probably an overestimate). So your downforce is 160lbs.

This means that the maximum tractive force (with 0.2 CoF) is .2*160lbs = 32 lbf.

Max torque = 32lbf * 3in = 96 lbf*in

Torque needed/torque provided = 96/8.5 = 11.3

And realistically the CoF is 0.1 or below, so I think direct drive off the ToughBoxes provides more than enough torque (with 1 motor per side).

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

Retired Starman built a competition type trailer and measured the weight (35 lbs), and the tongue weight (5 lbs).

I was thinking more of a 8-9:1 Ratio with one motor per side. But thatz just me.