



Question
Efficiency, motor constant and copper loss....
Does efficiency affect torque constant? I.e the kv formula solves X torque per amp, but would that be solved X torque per amp*efficiency?
Also, concerning efficiency, I hear the usual "the motor is 80% efficient, so 20% of the output power is wasted as heat" but isn't energy wasted as heat I^2/R? An example 2 motors identical torque constant, driving the same load and both 80% efficient but motor a has a higher phase rm but with the formula above I'm supposed to believe a motor with higher resistance is going to make as much waste heat as a motor with less? 






Kv and Kt are a physical characteristic. Efficiency is just power out divided by power in. For instance, with no shaft load, the motor's operating efficiency is zero, but Kt is still Kt. Efficiency is no set parameter  it varies greatly depending on the motor's load and operating conditions.
Winding losses are defined by (I^2)*R. That's not motor efficiency, though  just copper losses. You still have iron losses and friction to account for. Note that you multiply by resistance; division would yield higher losses at lower resistances. As for your example, no. If both motors have identical Kt, then the only (sensible) way for Rm to drop is by using thinner windings, which will amplify the (I^2)*R losses. You can't just peg efficiency at 80%  losses will be higher if you're doing the same work (i.e. torque @ RPM) with higher phase resistance, because torque is proportional to current, and winding losses are equal to (I^2)*R. This is why denser windings and stronger magnets are so valuable  anything that allows a reduction in Rm with no corresponding drop in Kt results in a motor capable of higher peak efficiency and greater total power throughput. 
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You send power into the motor and the motor does its thing. Some power gets turned into winding heat, some becomes stator heat, some becomes bearing heat, and the rest goes out at the shaft.
Efficiency is just somebody's evaluation of those outputs relative to the total input. Generally, the motor's discreet losses all get lumped together as "heat", while the useful remainder is what exits through the shaft. Can you define the conflict more clearly? 
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"I^2*R really is total energy. The heat dissippated will be a fraction of that. You would be absolutely correct if were were looking at a heater, which converts all of it's energy into heat, but motor systems have to convert at least some of their energy into mechanical work. You can check this by substitution. A Sidewinder (~100 Amps) on 3S (12V) has total resistance of ~ 120mOhm. 100 Amps^2* 120mOhms = 1,200W, which is the same as 12V*100Amps total power. Assuming 90% efficiency means ~120W of waste heat.
Total_Power = P_out + P_heat_loss, so P_Heat_Loss = Total_Power  P_out Efficiency = P_out / Total_Power" This was the explanation I got 


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