Yeah true, but different Kv motors the same size should also have the same amount of copper. So the one with the lower Kv should have a higher resistance and vice versa. Therefore the copper loss shouldn’t be responsible for a large difference in efficiency.
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assume the motor amp limit is constant… therefore copper losses and torque are constant. torque in newton meters multiplied by the angular speed of the motor in radians per second is the mechanical power produced. as the motor spins faster (closer to no load), more and more mechanical power is produced for the same constant copper losses (constant motor amp limit), and more mechanical power for the same losses means greater efficiency.
with the same km, pack voltage, battery amps and motor amps, the 100kv motor is less efficient, even when assuming the 200kv motor has double the Io & iron loss.
suppose these 2 motors are geared such that for both the present ground speed is 20mph… both are the same KM… which is greater efficiency?
calculator source: bavaria-direct.co.za
^even with double the assumed iron losses & Io with the higher kv motor (io input doubled in this example) & the same km, the higher kv motor has greater efficiency, not less… 98% motor efficiency & 97.1% system efficiency for the 200kv compared to 95.5% motor efficiency & 94.6% system efficiency for the 100kv.
^in this example, for the same km but higher kv motor to draw the same current at the same effective voltage at the same ground speed, the higher kv motor must spin at more than double the angular speed of the lower kv motor (thanks to gearing), but the higher kv motor at same current has exactly half the torque as the lower kv motor, but half the torque & more than double the angular speed means the higher kv motor (same km) is actually generating more mechanical power with lower losses for the same input power, and therefore the higher kv motor is more efficient for the same km, when properly geared for the mechanical load.
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@professor_shartsis (professor of …)
the values you plugged in you have the winding resistance as a fourth of the value of the other not half which I think would make sense…double the turns double the resistance. and the Io should be the same I believe if it’s taken at the same rpm
http://powerditto.de/indexditto.html
heres a page I recently got that has a lot of links. pages and pages. it does explain how a higher kv motor will have more power similar to your explanation somewhere in one of the links. speed is a cheap addition to the motor with less losses even if the same copper volume in the motor.
don’t think we have to worry about skin effect at our frequencies though right?
this makes sense as a reason why greater torque is more inefficient following I2r
yes, half the turns half the resistance… double the cross section area of the wire (for same copper volume) and half the resistance again for the same km. 1/2 * 1/2 = 0.25
Io is measured at the maximum unloaded angular speed of the motor.
makes sense with the 1/4 resistance and halving twice thanks
ive seen the Io taken at different speeds but if it’s at the maximum speed of the motor it will be more than double for the double kv as the motor would be going double the speed and the eddy current being exponential based on speed
I currently run motormax 60A and battery 30A. While there be any benefit raising motormax to 80A? It’s a dual 6355 190kv setup (torque motors)
~%33.3 greater torque until you reach the 30a battery amps limit at higher rpms.
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As @professor_shartsis said more 
sauce in low rpms until you reach your battery amp limit. Basically better low end acceleration.