I just wanted to check my reasoning here, because it seems to me like kV rating has a lot less influence, in and of itself, on maximum possible torque a given motor can produce than the motor’s size.

Let’s start by assuming that the main limitation is how hot we can allow the motor itself to get, and assume that our driver MOSFETs are of relatively insiginificant “resistance” by comparison.

Let us also assume that the driver MOSFETs can handle whatever current we need.

Assume also, our driver circuit can always provide enough voltage, like in a scenario where you’re using the motor like a buck converter and the maximum time a phase of it is held high is only a fraction of the PWM period. Assume we’re never going so fast as to have the motor generating enough back EMF that the driving voltage cannot overcome it.

A motor’s torque is given by: `8.3*I/Kv_rating`

, with the torque in N*m, the current in amps and the kV rating provided in the standard rpm/volt units.

So now, considering the motor:

A high kV motor has fewer turns of thicker wire

A lower kV motor has more turns, of thinner wire

The resistance of the motor is determined by

R=(rho)LN/A

L being length of wire per turn, N being number of turns, A the area of a single strand of the wire

Within a given volume of motor we can fit a given amount of wire. To simplify it, by ignroing the space needed for the wire to turn around, these turning areas carrying current but not producing their B fields in the right directions to apply torques to the rotor, we can say we can fit N “turns” of wire in a motor, N=motor_volume/(AL).

The torque a motor produces is in proportion to:

I*N*L , the current flowing by the number of turns of wire it is flowing in, by the length of wire in each turn

The heat produced in a motor is R*I^2

One can calculate that to double the kV one must double the product of (length*number_of_turns) of wire, thereby halving its area to maintain the volume. The wire’s resistance increases 4 fold.

A motor with twice the kV needs twice the current to generate the same torque.

So I^2*R ends up the same for both motors, assuming the same resistivity of wire, which will be the case when both are copper (pretty much the best affordable metal for low resistances).

So if you want more torque from a motor, and the main limit on your system is the amount of heat generated in the motor, your only option is to increase motor volume. As well as improving turns*length, if you’ve fixed the area of the wire’s strands, but also gives more volume for the I^2R heat to be dissipated in such that the termpature doesn’t rise so fast.

Is it therefore the case that, for direct driven shafts, ignroing what gearboxes can do for you, there’s a fundamental rule that (for plausible resisitivity and a chosen set of thermal limits), at any given volume of motor there is a maximum torque it could ever hope to produce.

So if one can find a motor of lower kV, in the same size as a motor one already has, one won’t be able to get more torque from it whilst remaining subject to the same thermal constraints. And this is, effectively, a limitation from physics rather than the practicalities of engineering and manufacture.

Thanks