Torque, gearboxs and chain gearing

[Trent B]

Hypothetical situation I can't seem to wrap my head around. (Not sure if this fits motors but I suppose it does given the relationship of gearings to motors)

Lets say I take a CIM motor and calibrate it to run at <40amps constantly (so it wont trip the 40amp fuse) and it is putting out some amount of ft-lbs of torque.

Banebots P80 gearboxes say to not use over 85ft-lbs of torque. So lets say that I find a gear ratio that perfectly puts of 85ft-lbs of torque.

Now lets say I run a 12 tooth on the p80 and a 24 tooth on the receiving end effectively doubling the torque to 170ft-lbs, would this mean the P80 now has an effective load of 85 ft-lbs on its output shaft or is it putting out 85ft-lbs of torque under the limit, and the gearing is causing 170ft-lbs on the receiving end.

I think the wording on banebots site sort of confused me into this scenario since when I think about that 150k:1 VEX Gearbox the beginning is effortless to turn and adding more and more reductions doesn't change torque required (correct?) thus the CIM should still operate at the same original level of torque and the gearbox at its 85ft-lbs or is there something I am missing.

Edit, now I think it may make sense to me, one point of a reduction gearbox is to get more torque without more work on the other end so I assume as long as the max torque of our CIM motor (controlled by current and CAN bus) stays below 85ft-lbs there would be no risk in gearing it up?

Trent

[James Tonthat]

Let's just say that you're using a 12:1 gearbox. That means that 12 turns of the input, gives you 1 turn of the output.

(Here's the USCS version)
Power [lbf*ft/s] = Torque [lbf*ft] * Angular Velocity [Rad/second]

Therefore (Power in is equal to power out, I know there are inefficients but for the sake of simplicity)
Torque [lbf*ft] * Angular Velocity [Rad/second] = Torque [lbf*ft] * Angular Velocity [Rad/second]
Torque of CIM * 12 = Torque output from gearbox * 1

So the output torque is 12 times the input torque. If you do the sprocket thing, that's an additional 2:1 (two rotations of the input, one rotation of the output), in this case you'd multiply the two.

Your overall ratio is 24:1 at this point.

What the Banebots spec is saying is that the OUTPUT of the gearbox cannot be more than 85 lbf * ft. This is probably because the final stage (piece of the gearbox) cannot handle stress greater than that spec (85 lbf*ft.) The input from the CIM would, of course, be much less than (85 lbf*ft), so yes there is no risk mechanically.

I probably didn't need to post this but just educational if you'd like. I'm going to guess you're going to be coming up with some sort of lift calculation. This will help you (remember keep track of your units.)

Power = Work / Time
Work = Force * Distance

[JesseK]

What to calculate and Why
The banebots warning isn't about the motor load; it's about the physical load put on the shaft. Planetary gearboxes can bind if too much load is put on a shaft that's off by even a few hundredths of a degree -- and most FRC bots are off by that tiny amount. Thus, at 85 ft-lbs of torque on the output shaft, the gears inside will have a tendency to break.

How to calculate
The motor will simply put out enough power to push that torque as fast as possible, regardless of what that torque is (provided it's not above stall torque). To see what speed the motor will turn at a certain torque, use a motor curve (search CD-Media for "calc*" and you'll find several spreadsheets). So ...

Conclusion
If you have 43lbs at the end of a 2ft arm that's powered by a shoulder joint that has a P80 gearbox, you are putting 86ft-lbs of torque onto the output shaft of the gearbox. The Banebots warning is trying to tell you that this is a bad idea, regardless of what the actual motor and overall gear ratio is.