[Behind the Lines] Ep. 3 with Ken Stafford on Motors and Gearboxes!

[Jared]

For the calculation:

First, we must find the motor's current peak performance.
This is 41% efficient (given by datasheet). Assuming all losses are due to resistance of the motor (this isn't true, but I'm just trying to see if I can get anywhere close to the measured 30% difference), we can calculate the resistance R = v/I = 12/68 = 0.17647.

Copper has a resistivity constant of around 0.0039 per *C. So, the new resistance at a temperature 100 *C higher would be 0.17647*(1+0.0039*100) = 0.245 ohms. Our loss of power is proportional to resistance, and the ratio of the two resistances is 1.38, meaning there is a 38% loss of power at the higher temperature. This is higher than the actual value, because my approximation of the original resistance is too high. It could be more accurately measured if I had a motor in front of me.