Quote:
Originally Posted by Oblarg
Say our drive is geared to a top linear speed S, with effective stalled-torque-at-wheel T. Let our wheel radius be denoted r (we need not specify a value, as it is divided out later). Then (Wf * 2 * pi * r)/S = T/Ts. Let T/r = 40lb, i.e. the force to stall our wheel is precisely equal to the available friction force. Then we have (Wf * 2 * pi * r)/S = (40lb * r)/Ts, which yields S = (Wf * 2 * pi * r)*Ts/(40lb * r). Our wheel radius term cancels, leaving S = (Wf*2*pi*Ts)/(40lb).
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I think you may need to take a look at your derivation there, it's difficult to see where you're going.
It would probably be simpler to create a formula for the required gear ratio to slip the wheels at stall torque. (Gear ratio is lacking in your above formula). The formula would be as follows,
Assuming 4 wheels and 4 CIMs;
Desired drive force = (Gear Ratio * Stall torque)/Radius
Desired drive force/(stall torque * radius) = gear ratio
Once you have the required gear ratio you can then calculate the speed you would achieve.
Using the numbers you provided, this would give a gear ratio of about 4:1, with a top speed of ~23FPS, assuming 0 inefficiencies or losses. However, in the real world, with losses and FRC batteries, the maximum speed to slip the wheels is going to be substantially lower.