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Date: Wed, 12 May 2004 09:08:52 -0400
From: Imad Al-Ghouleh <ighoulnospamox.mcgill.ca>
Subject: Re: Cost of repair Audi BMW Saab...(still crossposting)


This is a multi-part message in MIME format. --------------050405080401020904090405 Content-Type: text/plain; charset=ISO-8859-1; format=flowed Content-Transfer-Encoding: 8bit i mean PARK! :) Imad Al-Ghouleh wrote: > how about we all part downhill and walk up! ;) > > Wolfgang Pawlinetz wrote: > >>"Fred W." <Fred.Willsnospampam 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 90ƒ (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 90ƒ, 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 90ƒ 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 >> >> >> --------------050405080401020904090405 Content-Type: text/html; charset=us-ascii Content-Transfer-Encoding: 7bit <!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"> <html> <head> <meta http-equiv="Content-Type" content="text/html;charset=ISO-8859-1"> <title></title> </head> <body text="#000000" bgcolor="#ffffff"> i mean PARK! :)<br> <br> Imad Al-Ghouleh wrote:<br> <blockquote type="cite" cite="midr_ooc.44884$FH5.1028635nospam20.bellglobal.com"> <meta http-equiv="Content-Type" content="text/html;"> <title></title> how about we all part downhill and walk up! ;)<br> <br> Wolfgang Pawlinetz wrote:<br> <blockquote type="cite" cite="midbnc2a0lgtrf9deedu74ocpuv2d7u0trrm3nospamcom"> <pre wrap="">"Fred W." <a class="moz-txt-link-rfc2396E" href="mailto:Fred.Willsnospampammyrealbox.com"><Fred.Willsnospampam myrealbox.com></a> wrote: </pre> <blockquote type="cite"> <blockquote type="cite"> <pre wrap="">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. </pre> </blockquote> <pre wrap="">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. </pre> </blockquote> <pre wrap=""><!----> 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. </pre> <blockquote type="cite"> <pre wrap="">plowing understeer than the RWD, which can be made to either under or over steer with judicious input on the fun pedal. </pre> </blockquote> <pre wrap=""><!----> 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. </pre> <blockquote type="cite"> <pre wrap="">-Fred W </pre> </blockquote> <pre wrap=""><!----> Regards Wolfgang </pre> </blockquote> </blockquote> </body> </html> --------------050405080401020904090405--

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