Wolfgang Pawlinetz wrote:
> The core message is that the friction force is slowly reduced but
> equally on both front and rear tires as long as you don't change the
> center of gravity.
>
> Ok, now most likely I have made a complete fool out of myself, but if
> you are in doubt, then imagine a 90° sloped road. You'd need to
> support the car on the trunk because there is absolutely no way the
> tires would be able to hold the car in that position :-)
>
> In your theory, there would be a 100% load on the rear wheels and the
> car could still go.
>
> I'd be curious to learn if I am really wrong. Mathematically and
> physically I mean.

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. It's the same reason that your car will nosedive under braking
and lift the front end under acceleration.


> I agree. But getting away from a standstill is easier with the FWD
> because the RWD just slips sideways if it looses traction and you
> can't steer the direction vector.

If FWD slips you can't steer either, it's just that most FWD cars have
a front weight bias (due to having the engine, transmission, and other
such bits up front) so you have more traction all other things being equal.

Bill