[VintageMA]

I think you guys are missing something in your calculations of high voltage = less amps. If you take the same motor with the same KV rating, higher volts will = higher amps.

If you take the equation V = R * I, if volts goes up for across the same amount of resistance then current has to go up proportionately.

To take an example, if you have a motor of 1000KV and a battery of 10V the max speed of the motor will be 10,000RPM. If you then double the source voltage the max speed of the motor will be 20,000RPM. Remember the concept that a BL motor will take as much power as it can to rev up to the RPMS it should hit for any given voltage. So if you hit the throttle to 50% on 10V, the motor will draw whatever amps it needs to get to 5,000 RPM. If you do the same with 20V it will try to rev up to 10,000RPM, doing this as fast as it can it will draw more current following along with Ohms law above.

I have seen this in action with a LMT 1950/7t connection up through a MGM 24160 in my LST2. I spec'ed out the system to run with 6S when I bought it, and it does so beautifully with very reasonable heat. I tried to put 8S in it (max S for the MGM controller) and in less than a few minutes the controller was up to 180 deg. and the motor hit around 140 deg.

In order to gain the benefit of higher voltage = less current, you need to take that into consideration when you are spec'ing out your project before you buy your motor. If you want to double the voltage of the source in order to get a lower current draw, you need to take this into account with the turns and KV of the motor you buy. For example, if you have spec'ed out a motor that is 2,000KV for a 4S setup, you can just go to 8S and say you'll get less amp draw because you're actually doubling your motor speed. If you want to go to 8S, you will have to lower your motor to a 1,000KV motor to stay in spec with your project and get the benefit of the lower amperage draw.