(please excuse the top post, but with the HTML formatted original it's just easier than trying to add the tags to make my reply look right. Honestly I hate top posts. Just scroll down to see the rest of the converstation framed in html)

Nice job of the physics. The only thing you're missing is that on an angle the center of gravity will change, placing more weight on the rearmost wheels. On a car that was perfectly flat, say, a steel plate with tiny wheels, your math is perfect. On a car that was, say, 7 stories high, you can see that a small tilt would place *all* the weight on the rear wheels and the fronts would actually come up off the ground and it would tip over. Just prior to that the front wheel would have zero weight. At smaller angles, or a shorter vehicle, the shift would be someplace in between.

In a real car, much of the weight is low (drivetrain) and some of it is higher (greenhouse). The center of gravity is somewhere between the ground (steel plate) and 7 stories up (my tower car).

So a car on a slope will have some amount of increased weight distribution on the rearmost wheels, and therefore the /4 trick won't work even if the car is 50/50 on a level slope where /4 is correct.

I can't give you math for it, but I'd be the weight transfer on a moderate to steep driveway type hill would be on the order of 5% or so. Strictly a guess, but I suspect a generous one. BMW and other makers try to lower the center of gravity all the time for handling reasons, so it's probably not immense. And the number varies by the steepness of the hill of course. But I'm not up for a calculus function to describe the relationship, especially since I don't know what the center of gravity is to plug in.

So anyway most bimmers are close to 50/50 to start with. Let's go with that. Say going up the hill, it's now 45/55.

Front drivers are probably closer to 60/40 in general. Yes I know, I'm being very inexact here. But with the same transfer, you're now at 55/45. Amazing... the same weight on the drive axel in both cases.

My numbers are made up, poke at them all you want I don't mind. The concept is there though. Play with the numbers and get small variations.

But front drive still wins, no matter how you dice it up. I submit this. Back up you driveway in the FWD car. With math above, you have 65% of weight on the drive wheels, vs RWD's best of 55%. With your math, you have 60% vs 50%.

Now, your point about the direction of the vector of the force is valid. As the angle increases, the force decreases, until it reaches zero. Your car on the wall will indeed have zero force on the rear wheels, only because the vectors are straight down. At 89%, where there is still some amount of lateral force (not enough to produce enough friction to hold the car mind you) almost all the lateral force would be on the rear wheels, but the size of that horizontal vector has become very small. So yes, nearly 100% of the force is on the rear wheels and nearly zero on the fronts, but, the amount of this force in the useful direction is so little that it doesn't help the car to stay put and it slides down the hill. Consider a car on say a 70 degree angle. Add much more and the downward component of the vector will overcome the friction of the tires and the car will slide. Let's imagine that 70 degrees is very near this point. Now walk up to that car and lift the front bumper -- tah dah! You can. You're the Hulk! Because so much of the weight has shifted to the back wheels. Now go to the back and try to lift it. You can't. The weight is all there. Your angled vector is changing the actual size of the frictional force, yes. But, that portion of the force that remains to hold the car down is still much greater on the rear wheels. Imagine now that the front driver is trying to get up this very steep hill where it's very close to sliding back. (dry pavement and unlimited HP for this part of the discussion) The front wheels are barely staying down, all the weight has shifted to the back because of this insanely steep hill. The fronts are spinning like mad. Same car, rear drive. The rears are biting as best they can, they've got most of the available force on them -- but that total amount is now less, becuase much of it is vectored down the hill. On the tall car with the high center of gravity this effect is more pronounced. On my mythic sheet of steel car, the effect is almost zero. Real cars are somewhere in between.

So ok, we know that weight shift does occur on a hill. We don't really know how much, it vaires by the design of the car. The other factors that effect things are the coefficient of friction of the tires on a given surface, and the angle of the hill. At some point of angle with a given center of gravity, the weight over the drive wheels on a front drive going uphill equals that of a rear drive going uphill. On hills that are steeper yet the rear driver has more. At some later point of angle with a given center of gravity and given weight distribution the size of the force pulling down the hill exceeds that of the friction produced by the tires and the car slides down the hill. This would happen for a rear driver first, then a front driver (imagine both sitting on a lift bridge as it goes up)

And once final note: the AWD always wins. It has 100% of its weight on the drive axles, all the time. Plus, in a proper system, if one axle is deficient in traction (ice patch on the driveway) it shifts torque to the other one -- FWD or RWD just sit and spin. So, the AWD will always make the most of whatever force/weight/friction is available at any corner of the car which in the real world makes more difference than weight distribution.

-Russ.
1988 BMW 325iX (AWD)

(again, sorry for the top-post)





"Imad Al-Ghouleh" > wrote in message ...
how about we all part downhill and walk up!

Wolfgang Pawlinetz wrote:

"Fred W." <Fred.Wills@allspam myrealbox.com> wrote:

The rearwheel drive is fun and with all the electronic gimmicks it
will really do it's job. However at a certain climb angle or even
slipperyness of the road, the rearwheel drive gives in, then the FWD
and then the quattro.

Sorry, no. This is contrary to the laws of physics. If you assume equal
axle weights, as the car climbs it places more weight over the rear axle and
less over the front. So a rear wheel drive car would have an advantage over
a FWD in climbing. Obviously, an AWD car with the same weight and tires
would be better than either.

You almost got me there :-)

This is going to be a bit longer:

There's sort of a thinking error in your statement. It took me a while
to do the math (i.e. mechanics) but the outcome is, that the ratio
front/rear with regard to the friction force does _not_ change.

Let me elaborate:

The friction is depending on two parameters (yes, this is a
simplification for tires, but it's valid in all cases so bear with
with me): the friction coefficient µ and the force _orthogonal_ to the
surface. The formula for the friction force is Ff = Fn x µ.

The force pressing the car down onto the tarmac in this case is the
mass of the car x g (the earht acceleration 9,81) so you got Fn = mass
x 9,81

Now if you have the car on a level surface and assume a 50/50
distribution then the orthogonal force per tire is basically a quarter
of the Fn. So the result would be Fn/4 x µ.

The |
V indicates the direction of Fn


____
__/ | \__
|_ __V___ _|
____U______U_____

So far so good.

Now the worst case example:

Tilt the road and car 90° (don't sit in the car).

__
| | |
|C \
| | |
| | |
|C /
| |_|

In this case, the car would have to be held by something else, because
for Fr = µ x 0. I.e. there is no acceleration towards the tarmac and
so there is no resulting orthogonal force pressing the tires to the
tarmac and therefore no Friction. The car would slide.

So if you choose increasing angles between 0 and 90°, the orthogonal
force down on the tarmac slowly decreases on all four tires and is
gradually "converted" into a force wanting to push the car
"backwards".

But again, for all tires.

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.

plowing understeer than the RWD, which can be made to either under or over
steer with judicious input on the fun pedal.

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.