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.
>
>
>
>>-Fred W
>>
>>
>>
>
>Regards
>
>Wolfgang
>
>
>