Back driving a CIM

[matthewdenny]

Does anyone know how much torque it takes to back drive an unpowered cim motor? Does it mater if it is connected to a speed controller or not if the motor is programmed to 0?

[asid61]

Look up the cim specs on Andymark. All torques etc. are listed there.
Look for "torque at stall current".

[matthewdenny]

I don't want the torque at stall current. I'm trying to find the torque at zero current. That number does not appear to be listed.

[Jared]

Originally Posted by matthewdenny

Does anyone know how much torque it takes to back drive an unpowered cim motor? Does it mater if it is connected to a speed controller or not if the motor is programmed to 0?

It doesn't take very much to backdrive an unpowered CIM at low speed. To get the motor going quickly takes some more torque.

This information isn't on the datasheet.

If the speed controller is in brake mode and it receives a command of zero, it will short together the motor output. The current generated by the motor works to oppose the rotation of the shaft making it harder to backdrive. If the speed controller isn't in brake mode, it won't affect the CIM at all when it is command to zero power.

What are you looking to do with a CIM motor?

[Nyle]

I think he's asking how much torque it takes to spin a CIM that is not receiving any voltage, which would not be the stall torque. I don't see this anywhere on the datasheet for the CIM, although it might be possible to calculate it from data in there. Unless someone on here has already measured it and has the value available, you'll probably have to just measure it yourself. I'm not sure if connecting it to a speed controller would change the value or not, I believe that would depend on if/how the motor leads are electrically connected when the controller is set to zero. I'd guess that they are not, but I don't know exactly how they work. What do you need this for? For most cases I can think of where you might need it, approximating it as zero would probably not make any significant difference.

[Richard Wallace]

Free speed drag torque for a CIM can be calculated based on data sheet parameters as follows:

Free power at 12V = Free current x 12V = 2.5A x 12V = 30 Watts

Effective resistance = 12V / Stall current = 12V/130A = 0.092 Ohm
Winding loss at free current = 0.092 Ohm x (2.5A)^2 = 0.57 Watt

Drag power at free speed = Free power - Winding loss = 29.4 Watt

Drag torque at free speed = 29.4 Watt / (5300 RPM x pi/30) = 0.053 Newton-meter

I can confirm from measurements in my motor lab that this theoretical figure is pretty close to what we really get, after the motor warms up a bit. Initially the drag (and the free current) are a bit higher because the grease and the brushes are a little stiff when cold.

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Back to the OP: yes it matters a lot if the motor is connected to a speed controller. The calculations above give drag at zero current; when the CIM is connected to a speed controller its induced voltage (back-EMF) will excite the controller and that excitation will cause current to flow, charging up the controller's capacitors and creating additional drag torque. Anyone who has pushed an FRC robot and seen the little LEDs come on knows what I mean.

[cad321]

Relating to the op's question, would disabling 1 motor in a 3 cim gearbox and continuing to power the other 2 cause any damage to either the speed controller on the unpowered cim, or the unpowered cim itself? An example of a practical application for this would be in a 6 cim drive, if the current draw exceeds a certain amount, disable one motor on both gearboxes to lower the overall current draw.

[matthewdenny]

We are thinking about the possibility of a drive motor being only occasionally used, and freely spun by the gear box when not used for acceleration/pushing.

We also have a swerve design where to rotate the wheel 90 degrees it would have to back drive a cim a rotation or two.

We would like to know how much torque it takes to do this. Also is the torque required the same regardless of the speed of rotation?

In both cases the coma would be hooked up speed controllers, but not receiving voltage from them.

[nathannfm]

Originally Posted by matthewdenny

We are thinking about the possibility of a drive motor being only occasionally used, and freely spun by the gear box when not used for acceleration/pushing.

As long as you don't set it to brake mode this should not be a problem but acceleration and pushing make up almost all motion in an FRC match because by definition to change direction you must accelerate.

Originally Posted by matthewdenny

We also have a swerve design where to rotate the wheel 90 degrees it would have to back drive a cim a rotation or two.

In most (if not all) swerve drives that use bevel gears when the pod rotates instead of backdriving the cim the wheel turns slightly.

[brndn]

Originally Posted by matthewdenny

We also have a swerve design where to rotate the wheel 90 degrees it would have to back drive a cim a rotation or two.

We would like to know how much torque it takes to do this.

In this case, aren't you powering the motor? I'm assuming the swerve module causes the wheels to rotate a little and you want to counteract that by powering them in the opposite direction.