Quote:
Originally Posted by Lunabot
Hello,
I would like urgent help regarding the procedure to calculate/determine the torque of a CIM Gearmotor (specifications mentioned below) for the Gear ratios of 27:1, 36:1, 48:1, and 64:1.
Link for the CIM Gearmotor Specsheet: http://content.vexrobotics.com/docs/...otor-specs.pdf
Regards,
Lunabot |
It is difficult to answer your question outright, without more information.
I can try to describe some of the principles involved...
The motor will exert as much torque as it can to overcome whatever load you put on it, until that load exceeds its stall torque. The more load you put on it the slower it goes. Once you hit the stall torque... it stalls, and stops moving.
The stall torque of the gearbox is equal to the motor stall torque multiplied by the gear ratio.
The motor motor load, the higher the motor's current draw.
Typically in FRC... max loading for a CIM motor is calculated based on an approximately 40-amp load, since we run these motors through 40 amp circuit breakers (which can actually run at 45 or 50 amps for limited periods of time).
Torque is proportional to current draw. At stall, the motor draws the stall current. The motor draws 40 amps when it has a torque applied on it equivalent to (Stall Torque) x (40 amps / Stall Current).
Does this make sense? If you provide more details on your application I'm sure someone can help you more.
-John