"Haqsau" > wrote in message >...
> Way too complicated. With no suspension deflection and two orthogonal
> forces (longitudinal and lateral) wouldn't it be possible to solve each
> plane individually and then add the results? For example:
>
> cg height = Zcg
> longitudinal force = Fx
> lateral force = Fy
> wheelbase = WB
> track = T
>
> longitudinal weight transfer = (Fx * Zcg) / WB
> lateral weight transfer = (Fy * Zcg) / T
>
> Now just add the weight transfers with the correct sign at each wheel. I
> believe this will give you the correct solution given the conditions of the
> problem as you state it.

I don't think I understand how you arrive at this. Doesn't it matter
where the centre of mass is located - apart from its height?

Also, this is assuming the wheels are in a rectangle, which may not be
the case. If you are being accurate about steering, then the wheels
rotate about a point which is external to the wheel. Also, the front
and rear track may be different.

Ruben