4" vs. 6" + 8" Diameter Wheels

[Ether]

Originally Posted by James Critchley

now I'm just waiting to be surprised by the results Ether's calculations

Well I wouldn't want to disappoint :-)

Looking at the wheel without the vehicle (as you did):

I*alpha is equal to the net torque on the wheel. If "tau" is the driving torque on the wheel and F is the floor reaction force responsible for the linear acceleration of the wheel, then a free-body analysis of the torques and forces on the wheel gives:

tau - F*R = I*alpha

tau - (M*A)*R = (K*M*R*R)*(A/R)

tau = = M*A*R*K + M*A*R

A = tau/(M*R*(K+1))



For the analysis of wheel plus vehicle, see attached PDF.

a = tau/(I/r +r*(Mw+Mv/4))

a is vehicle acceleration;
tau is driving torque on wheel from gearbox;
I is wheel moment;
r is wheel radius;
Mw is wheel mass;
Mv is vehicle mass.