Quote:
Originally Posted by MrForbes
Are you sure about the voltage thing? I don't really understand the whole voltage/current/percent power relationship with speed controllers...but my guess is you're providing it quite a bit more than 1.2 volts if you're commanding 10% power, using PWM.
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Yes, absolutely.
Model motor as a resistor, you can get it's resistance from the spec voltage of 12V and spec stall current of 133 Amps. V=IR gets you R = .09 Ohms.
Kt is torque/Amp, and therefore stall torque scales with voltage. So, at 10% of spec voltage, you have 10% current draw, which means 10% of spec stall torque.
Stalling a motor in no way leads to failure on it's own. Putting too high of a continuous current into a motor (whether that's at stall, or while rotating) is what fails it (due to heat building up wrecking insulation).
Quote:
Originally Posted by Thad House
Exactly. If you calculate it so you only need 10% of your stall torque to hold the object up, you only need 10% of the motors stall current to do so. Which you get if you input 10% of the nominal voltage. That becomes 13 amps at 1.2 volts, which is only 16 watts. A CIM can easily dissipate that much wattage, and will have no trouble doing that for the entire match.
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You can extrapolate this a tad further and do a BS calc where you say .2 or so of the motors mass is the armature weight in copper, and then do Q = m*C*deltaT, but change Q to Power input and deltaT to deltaT/sec. This lets you get an approximate temp rise per second that's pretty conservative as it neglects ALL heat transfer paths out of the motor.