Other way round!!!
Voltage loss across the coil is the power that drives the voltage loss across the resistor is lost heating power,
Btw, in the motors we use typical Ri is around 0,015Ohm
That’s what I also started thinking when I wrote the post as any power in a resistor is just converted to heat. So the voltage loss across the coil actually corresponds to the created magnetic field’s strength. Correct?
That’s why I also thought the simplified circuit was a bit odd, because if the motor was represented as a resistor all the power in the multiple hundreds of watts would be pure heat loss 
. When in fact only a small part of the power loss happens in the winding resistance and most of it goes to creating the magnetic field.
correct.
Except that I don’t like the words “voltage loss” for the coils, it should actually be “voltage gain” … 
The simplified model has to include the coil (C and L), I just put it as back-EMF voltage. Otherwise the model is useless.
Then I think the plus / minus for the e are somewhat misleading as this is AC voltage. What does the source of your picture say?
The power output of the motor is the product of torque (T) and angular velocity (w). w is proportional to RPM Rotational energy: E = 1/2 * M * w^2 ; (M = moment of inertia)
P = dE/dt = 1/2 * M * 2 * w * dw/dt M * dw/dt = T (definition of torque)
=> P = T * w torque T is proportional to the current.
Thanks Mathias, makes sense
I’ve pointed out in several threads that torque of a motor is proportional to the current in the motor. In fact every motor has a so called torque constant Kt. It’s related to the Kv: Kt=1/Kv
=> P = Kt * I * w = I * w / Kv (careful with units!)
Regarding to this:
It’s all about matching the load to the motor. Wouldn’t in this example a 50 kV hub motor have the same loading when you account the 12:36 reduction with the 150 kV motor. The real world problem being that there really aren’t same size hub motors that were so high wound to get the 50 kV rating? So therefore you have to use higher kV hub motors which causes bad load matching (impedance matching) and tun in the less efficient (right side in the graph) area.
But you would get higher efficiency with a same size 50 kV hub motor, because it wouldn’t have the belt friction losses? Problem being that the motors don’t really exist in the market.
No ramping, no FOC, no further limitations due to the electronics? 100% I would guess because this is what you need to ramp up the motor current at maximum speed. You’ll hit the limit fast though. You can watch this video to get an idea. In this case there is a motor current limit of 50 A, and ramping as limitation of the duty cycle change:
Very simple: if you go full throttle, your duty cycle is 100%
I suppose this is not the answer you expected, but it’s true.
Your correct question should be: "If a motor’s winding resistance is 0.007 ohm, battery voltage is 50V, and you set a Motor Amp Limit of 750 amps… at 0 rpm, full throttle:
What will the voltage and current in the motor be? "
And the correct answer to this question is: there is not enough information to answer this question. In order to calculate it we would have to know 1) C and L of the motor and it will and 2) it is also a function of time or RPM because the motor spins up and then the values change.
No. At 0 RPM, 0 current the things are unrelated.
The voltage is proportional to the change of the current. You want full throttle, so you want 100% until you reach 750 A.
I don’t disagree with @PB1. He is correct. I only told you what you should do if you want to increase the current as fast as possible. But I already said that. What your ESC should do is to give you the torque that you want. For that it should send the right current through the motor. To ramp the current up, increase the duty cycle and vice versa.
@devin, don’t quote me out of context!!! I will not play that game!
I said if you pull full throttle, duty cycle is 100%, @Mathias says the same.
Then I said we can’t calculate the current going through the motor.
No games please!
The inductance at first, then the motor current limit. Also the voltage over the motor drops as the (negative) voltage produced by the motor is proportional to the RPM.