CIM Motor RPM

[Free Willy]

What is the rpm of a CIM motor with a 150 lb load?

[John]

150 lbs how far from axis of rotation. You need to know the torque on the motor to calculate the rpm. For example, 150 lbs one inch away from the axis of rotation is 150 inch pounds of torque, but 150 lbs two inches away from the axis of rotation would be 300 inch pounds of torque. The motor will spin faster in the first case.

[theprgramerdude]

You might want to look into how DC motors work, as you've given vastly insufficient info to give an answer.

FYI, loads on DC motors on given in terms of Torque, which are linear forces acting on a given radius. Your 150 Lb load could mean any amount of torque, depending on the radius.

Second, the speed of the motor is linearly dependent on the applied input voltage; for the FRC, this is typically variable from -12 to 12 VDC.

Last, the typical free speed for the CIM motor at 12V DC is 5310 RPM, +/- a few percent, with a normal operating speed of 4320 RPM.

Andymark's documentation can be found here: http://files.andymark.com/CIM-motor-curve.pdf

You're welcome to recalculate the load and do the math yourself, or repost your torque measurements here if you need some additional help or resources.

[Free Willy]

This would be for a drivetrain assuming that the total weight of the robot (battery and bumpers included) was around 150 lbs. We know the no load rpm of the CIM motors, but we are wondering if you were to essentially do a direct drive system what the rpm on the CIM motors would be with the weight of the robot on top of it.

[theprgramerdude]

Are you talking about directly driving the wheels from the output on the CIM's? If so, then theoretically the max speed for the wheels at 12V would technically be 5310 RPM; the actual max speed is a statics problem.

The static max speed of the robot occurs when the rotational force of the motors happens to be equal to the frictional rotational force that's impeding the acceleration. The frictional force is experimentally determined, but tends to be rather proportional to the quality of your bearings and such, and the robot weight.

IMHO, though, you should never direct drive the CIMs due to the issue with actually accelerating up to the top speed in the first place. Acceleration is going to be proportional to the applied torque to the wheels: gearboxes are used to amplify this torque so max speed is reached relatively quickly and the amount of current flowing through the motors during acceleration is limited. Acceleration is actually the hardest part on the robots during the match, and because the output torque from the CIM's is relatively low to geared CIM's, they're going to accelerate slowly and probably throw breakers/destroy themselves if you try this for any extended period.

Edit: If you want an in-depth explanation of how the DC Motors and gear trains work, MIT's article explains it in a complicated but complete way.
http://ocw.mit.edu/courses/mechanica...lecture-notes/

[Free Willy]

We are not actually planning on using direct. We are trying to figure out what kind of gear ratio we want for our robot to go a certain speed so we want to know what the rpm of would be without any gears. We would like two know what the estimated rpm would be given we have one CIM motor driving each side of a 150 lb robot.

[John]

If you don't gear down the CIM motors at all, and you used six inch wheels, you would be geared for:
5310 rpm * (6inch * pi /revolution) * (1foot/12inch) * (1minute/60seconds) =
139 feet per second

Typically, a ~150 lb FIRST robot is geared for no more than about 16 feet per second, and usually less. Anything much higher draws too much current, and heats up the motors too fast.

[EricH]

If you're driving directly from the CIMs, with 150 lbf of robot on top of them...

Let's assume 6" wheels (3" moment arm), and 4 wheels, each with its own CIM. 150/4=37.5 lbf per wheel supported (which we'll call N). At 3", assuming a coefficient of friction (mu) of 1 (F=N*mu), that's 112.5 lbf-in of torque exerted on the wheel (F*d). That converts into 1800 oz-in to get moving at all.

Stall torque of a CIM is 343 oz-in, and it draws 133 amps. So, not only do you not move, you spend all match tripping breakers trying.

Now, I've given you some of the calculations for a no-gearing situation. Rework it to include gearing to find your minimum gearing. Or, rework it using a different coefficient of friction. Or a different number of powered wheels.

I'd also comment on the wisdom of running one CIM per side, but I think you can work that out as well using similar calculations. It might help to post the calculations when you're done, so that any mistakes can be caught early.

[Ether]

Originally Posted by EricH

Let's assume 6" wheels (3" moment arm), and 4 wheels, each with its own CIM. 150/4=37.5 lbf per wheel supported (which we'll call N). At 3", assuming a coefficient of friction (mu) of 1 (F=N*mu), that's 112.5 lbf-in of torque exerted on the wheel (F*d). That converts into 1800 oz-in to get moving at all.

The 1800oz-in number is the max wheel torque that can be supported before you lose traction... not the torque required to start moving the vehicle.

[jwfoss]

For practical/safe purposes in a drivetrain you can assume that loaded RPM will be around 80% of the free speed of the motor. I strongly suggest downloading JVNs Mechanical Design Calculator and running your gearing numbers to see approximate output speed and current draw.

[Clem1640]

Use the CIM motor curves. Available from FRC and attached.