joshuabardwell
Elite Member
I was pulling a load of firewood in my garden cart today, and I thought to myself, "Wouldn't it be neat if I put an electric motor on this?" I have a zillion projects in the queue at this point, so who knows if I'll ever get to it, but as I was thinking about it, I realized that I didn't have much idea how to select the motor, and I'm a sucker for a good physics problem.
Like, one factor is the motor's available torque. Torque is measured in foot-pounds and the force required to move the cart is measured in pounds. So if I just work the units, then it seems like I would divide the motor's torque by the radius of the wheel to get the pounds of motive force delivered. Is that right? How does that equation change if, for example, there is a drive gear, a gear on the axle, and then the diameter of the wheel? The torque is affected by the relative size of all of those things, right?
The other factor would be the motor's RPM and how that translates to the speed of motion. With no gearing, the formula should be rotations-per-minute * wheel circumference. With gearing, then it should be rotations-per-minute * (drive-gear / driven-gear) * wheel-circumference. Is that right?
Like, one factor is the motor's available torque. Torque is measured in foot-pounds and the force required to move the cart is measured in pounds. So if I just work the units, then it seems like I would divide the motor's torque by the radius of the wheel to get the pounds of motive force delivered. Is that right? How does that equation change if, for example, there is a drive gear, a gear on the axle, and then the diameter of the wheel? The torque is affected by the relative size of all of those things, right?
The other factor would be the motor's RPM and how that translates to the speed of motion. With no gearing, the formula should be rotations-per-minute * wheel circumference. With gearing, then it should be rotations-per-minute * (drive-gear / driven-gear) * wheel-circumference. Is that right?