Hi. I’m a little confused on what a CIM motor’s maximum specs are within the context of FRC.
According to the Vex website, CIMs have a stall torque of 340 oz-in and a stall current of 133A. Because our CIMs are attached to 40A breakers, I believe that our motors won’t be able to reach the 340 oz-in limit.
How can I calculate what the maximum torque of a CIM is without tripping the 40A breaker?
From what I understand about the breakers, you can briefly exceed the 40A limit without tripping them. This has to be the case, because otherwise every time a team got in a pushing match where their CIMs experienced near-stall currents they would be tripping breakers instantly.
Edit: some quick research has illuminated the problem for me. The snap-action breakers are thermal breakers. This means that they trip when they reach a certain temperature, rather than the instant they hit a certain current. That’s why you can stall your drivetrain for brief moments without tripping them. They also include an electromagnetic component that will trip instantly when subject to major surges, like what you would see from a short circuit.
Don’t forget that you have a 120A breaker on the entire robot. That limits the current draw in addition to the 40A breakers, especially as the number of motors increase.
You can work the equations listed in the document for how much torque would be available for 20A (or another number, that’s just what my team has used), and use that torque and a gear ratio to find available torque at the wheels when only pulling 20A. If you can move at this current, use the gear ratio again to determine what torque at the wheels would stall the motors (this can be combined with the wheel size to get a max pushing for - ignoring weight of the robot).
I wrote a python script that can calculate the gear ratio to move a desired speed and the minimum number of motors to move at that gear ratio assuming a coefficient of static friction of 1 (static friction must be overcome to get the robot moving). I can post it if desired.