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Unread 20-05-2011, 17:02
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Ether Ether is offline
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Re: 4" vs. 6" + 8" Diameter Wheels

Quote:
Originally Posted by James Critchley View Post
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.

Code:
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.


Attached Files
File Type: pdf compare wheel accel.pdf (11.3 KB, 45 views)

Last edited by Ether : 20-05-2011 at 20:54.