Bill Bradley schrieb:

> Not a bad try, but you're missing the key factor: Torque. Since the
>center of gravity is NOT on the road, it has a torque arm to the point
>of contact of the tires. The SUM of the forces on the contact area is
>as you worked out, but it doesn't remain 50/50 front/rear since the rear
>axle is providing a counter-clockwise (if viewed as in your drawing)
>torque while the front axle can only provide a clockwise torque. To
>reach rotational equilibrium more of the weight force in on the rear
>axle.

Yep.

Must have been the late night yesterday :-)

In fact I did the torque equilibrium but got mislead by the gemotry
I'd drawn up.

You are right, the CoG shifts back and the load on the rearwheel
increases. That load is then split into a component orthogonal to the
road (for friction) and one parallel to the road (pulling the car
back).

So the total orthogonal force on the road for friction is still less
then if the car would be on a horizontal plane, but it's higher than
on the front wheels.

Regards

Wolfgang
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