I have a 4 wheel skid steer robot. My motor can deliver up to 6 Nm with a gearbox efficiency of 70% up to 170 rpm. I want to find out whether I will be able to make a pivot turn and move.

So I assumed that pivot turn force = force to move the blocked wheel.

The force required to move a 100 kg weight is F = c * m*a = 0.02 (friction, not a rolling friction as I am taking worst case scenario) * 100 * 9.81 = 20 N.

The wheel will be placed 10 cm below the weight so the torque required is 0.1 * 20 = 2 Nm. Is that correct?

  • $\begingroup$ Your friction factor seems too low. The expected range is from 30% to 70%. In rubber it could be more. $\endgroup$ – kamran Jun 16 '19 at 16:10

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