Quote:
Originally Posted by Ether
Would you mind posting your calculation please
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Well, this is embarrassing: I can't find the scratch paper I did this on earlier, and I'm differing by a factor of two upon repeating the calculation. At any rate:
We'll omit the efficiency figure for the calculation; you can multiply by an estimate at the end if you wish.
Let's round up to 160lbs, so we have 40lbs per wheel. Assume all weight is distributed equally among the wheels. CoF is 1, so our maximum friction force from a single wheel is 40lb.
Let Wf be the free speed of a CIM, and Ts be the stall-torque of a CIM.
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).
If you plug in and calculate, you end up with ~25 feet per second, which is a factor of two off from what I got last time. For the life of me, I can't find where this calculation is wrong, though the result is much more surprising than what I had previously believed, and does not really mesh with my experience of wheel slippage while driving...