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stopping time for a BLDC motor
- Thread starter jag27
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time = ((wk2 in ft-lb)*the change in rpm)/(308 * Tq in ft-lb)
here wk2 and rpm are at the motor. I assume you actually want to stop a load so you would have to reflect the wk2 of the load through the gear to the motor. So you actually have the gear wk2 and the load wk2. To reflect the wk2 of load back through the gear, divide the load wk2 by the square of the gear ratio. Load wk2=18 lbft^2 gear equal 3:1 therefore wk2 at motor is 18/9=2.
here wk2 and rpm are at the motor. I assume you actually want to stop a load so you would have to reflect the wk2 of the load through the gear to the motor. So you actually have the gear wk2 and the load wk2. To reflect the wk2 of load back through the gear, divide the load wk2 by the square of the gear ratio. Load wk2=18 lbft^2 gear equal 3:1 therefore wk2 at motor is 18/9=2.
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