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I have a robot that has a load of 60kg and it has to move using 4 stepper motors. I saw from previous searches that the torque required by the motors to move is Force times the wheel radius.

COF = 0.55

Torque = [(60 x 9.8) x 0.55] x 0.05 = 16.17 Nm

I would like to verify if the motor I have available will do the job. I assumed that aerodynamic drag is negligible since it runs 0.5 m/s at most. There is also no slope on the floor.

And since I would be using omnidirectional wheels, are there additional torque needed when it moves sideways? Thank you

Motor: Wheels:

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Lester Narciso
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  • Hint: you can make your question nicer by not posting the full URLs, but by hyperlinking - e.g. [description text](the url). Try editing to see how it works. Your question is otherwise fine. I think you don't take into account that you have four motors in your calculations? – anrieff Nov 14 '17 at 02:57
  • 0.55 COF seems high for a ball bearing wheel. Also your equation does not factor in how fast you want your mass to accelerate. – Trevor_G Nov 14 '17 at 03:02
  • Do you mean the COF in the bearing wheel or the COF between rubber and tiles? – Lester Narciso Nov 14 '17 at 03:38
  • to clarify,if I used 4 motors, will the total torque be divided equally into 4? – Lester Narciso Nov 14 '17 at 03:40
  • As long as all wheels are on the floor and nothing is slipping yes. – Trevor_G Nov 14 '17 at 03:47
  • COF between rubber and tires dictates how hard you can accelerate before the wheels slip and is not a load. COF in the bearing is the one you need to overcome plus the force you need to accelerate the mass horizontally. – Trevor_G Nov 14 '17 at 03:48
  • Correction.. the wheels that are in-line will deliver the same torque. The thing will go round in big circles if you don't match the speeds on either side. Once you do that the torques will match. – Trevor_G Nov 14 '17 at 03:54
  • So for the torque: T = [F(friction) + F(acceleration)] x R(wheel) ? – Lester Narciso Nov 14 '17 at 04:45
  • @LesterNarciso yes, assuming that is bearing/rolling friction. – Trevor_G Nov 14 '17 at 19:29

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