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Im making a robot that must sense objects and pick them up. I need to order some motors, so i wanted to work out the torque needed just to push one wheel (as it may pivot, so one wheel will need to push the full load of the robot). I want an rpm of around 150-200. I'm using wheels that have a diameter of 125mm (but we can change size to get optimized torque/speed ratio later). The weight of the robot will be no more than 10kg's and the surface that the wheels will grip is similar to lino (say a friction coefficient less than 1, which is pretty high). '

I drew a picture of what i thought would work but not sure if the workings are correct. The big assumption was that the friction coefficeint was 1, as im not sure what it should be.

I am ordering the motors through maxon, and had a look at what they had and their motors put out around 5-20mNm for their motors then a gearbox of say 50:1 would be added to the motor, then multiply by the motor efficiency (60%) give an output of 600mNm. The RPM of the motors are around 5k-8k, so reduce to 100-150RPM.

I'm guessing its suitable for my application but not sure.If any one has some info on this stuff that would be great.



If anyone has some proper formula's to work this out it would be great.

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sorry here's the link to the drawing – Cameron May 15 '12 at 8:16
Possible duplicates: and links therein. – Qmechanic Apr 16 '14 at 4:35

What matters is the friction in the drive train i.e. in the gearboxes and axles in your robot. The friction between the wheel and the floor will control whether your robot skids or not, but it won't affect the robots ability to move.

I would guess it's going to be very difficult to calculate the friction in your drive train. I think the best you can do is measure it. Presumably it will be load dependant, so you probably want to measure it when your robot is carrying various different loads.

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+1 concise and complete – Pygmalion May 15 '12 at 9:51
Minor nit-pick: if friction with the floor is too low, or your torque is too high, you could skid out and have trouble moving because of that. Other than that, good explanation. – Colin Fredericks May 15 '12 at 14:21

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