[Ether]

Originally Posted by James Critchley

Sorry Ether, I'm just not being clear.

No, you were quite clear.

The torque available "at the motor" and "by the motor" is the same for all wheel sizes.

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keeping the same top speed and low end torque requires a different gear box

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So the torque applied TO THE WHEEL must go through a different gearbox, and will then be a different torque.

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Having so designed all gearbox-wheel combinations, at stall the force at the exterior rim of any wheel will actually be the same (no losses). So the applied torque as indicated is actually F_stall * R which was used correctly.

If a wheel of mass M_w and moment I and radius R is sitting on the floor and is free to accelerate, then the force F that it exerts on the floor when a torque tau is applied is not equal to tau/R.

It is equal to tau*M_w*R / (M_w*R² + I). That approximately equals tau/R only if I is negligible compared to M_w*R².

That wasn't made clear in your post.


If four of the above wheels are on a vehicle of mass M_v which is free to accelerate, then the force F with which each wheel pushes against the floor is given by

F = tau*R*(4*M_w+M_v) / (R²*(4*M_w+M_v)+4*I)

The above approximately equals tau/R only if I is negligible compared to R²*(M_w+M_v/4).


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The acceleration is given by a = 4*tau*R / (R²*(4*M_w+M_v) + 4*I)

Letting I=K*M*R² this becomes:

a = 4*(tau/R) / (M_v+4*M_w*(K+1))

Compare the acceleration of two vehicles, one with wheels of mass M_w1, K= K₁, radius R₁, applied torque tau₁, and vehicle weight M_v1, and the other with wheels of mass M_w2, K= K₂, radius R₂=2*R₁, applied torque tau₂=2*tau₁, and vehicle weight M_v2. Then

a1/a2 = (M_v2+4*M_w2*(K₂+1)) / (M_v1+4*M_w1*(K₁+1))

... and R does not appear in the ratio, as you said.