i mean PARK!

Imad Al-Ghouleh wrote:

> 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
>>
>>
>>