More or less amps?

[SpEEdyBL]

Sorry, it's actually 1352/kv in (oz.in/amp). Here are some equations to get you to that number.

Power = Volts x Amps = Velocity (m/s) x Force (Newtons)

Lets say you have a motor with X kv (rpm/volt). Volts = V, Amps = I. For now, the efficiency of this motor will be 100%

With V voltage the motor is spinning at VX rpm = 2(pi)(V)(X)/60 radians per second (r/s).

If radius = 1 meter, r/s = meters per second (m/s) away from the center of the shaft.

Efficiency = 100%
Power in = VI = Power out = (m/s)(Newtons) = 2(pi)(V)(X)(Newtons)/60

If Power out (VI) = 2(pi)(V)(X)(Newtons)/60, then
The torque in Newtons = 60VI/(2(pi)(V)(X)) newtons a meter away from the center of the shaft.

To convert to oz.in/amp,
39.37 inches/meter and 3.6 oz/newton

oz.in = 39.37(3.6)(60)(V)(I)/(2(pi)(V)(X))
oz.in = 1353(I)/(X) the V's cancel out.
oz.in/amp = 1353/X = 1353/kv

Put in any value for kv, amps, voltage and you will get that number every time. The reason why I used 100% efficiency is so I don't need to factor in resistance of the motor. 100% efficiency assumes the motor has a constant kv at any amp rate and has 0 ohms resitance, but you can find efficiency just by finding the rpm drop constant. With the I = V/r equation, you can find the max amp rate where the motor is spinnin at 0 rpm and is 0% efficient. Basically the rpm drops linearly until it reaches that point. rpm drop = kv(I)(r). Rpm = (kv)(volts) - (kv)(amps)(resistance). Efficiency = (V-rI)/V.