Cost of repair Audi BMW Saab...(crossposting)

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Momentum, forces... one should forbid phsyics students to ever read the internet and see the ways their science is abused for biased arguments... In my opinion, the most important driving aspect of FWD in the snow is that traction and steering are intimately connected, which makes the car very intuitive to drive. You can make either concept drive relaitvely well in the snow, and there are also examples for FWD that are undriveable in the snow. From anecdotical experience, while I lived in Germany, when a lot of snow fell I would never drive the BMW 320ci, it was very hard to drive, and mpossible to drive of summer tires. We also owned a cheap FWD Fiat Uno, and that car was a darling in the snow, you always felt what it was doing because the steering would feel connected to your hands, the BMW would regularly totally lose steering feel and you felt like you were just helpless. Downright scary, and twice when surprise by snow it was a miracle I made the journey from work to home (both in the city, 8 miles apart) in one piece.
There is no dount in my mindthe Saab is a very soothing bad weather car. It's my number one choice for being caught in a bad storm, and we also have a 4x4 SUV. The more it rains, the more it feels like it steers on tramlines. And don't be misled by the sunny California thing, we have some pretty awful storms here every once in a while that invariably hit when you're away from home and have no choice but heading back.
...pablo
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On Tue, 11 May 2004 23:57:11 GMT, Bill Bradley

The point exactly - and the reason FWD has an advantage over RWD (in general, you can always find exceptions). Most FWD cars have less than optimal weight distribution with a front bias. This is a feature in the snow (probably for acceleration too as it helps fight the FWD front end lift issue).
Give me a RWD car for *limit* handling. Give me a FWD car for snow.
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wrote:

equal.
Or a viscous AWD car for both. ;-)
-Russ. '88 iX
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-Bob- wrote:

    You have to be careful how you use the term "advantage." If you mean "doesn't get stuck in snow" then FWD _can_ have an advantage with their weight bias, BUT that's not the whole story. For one thing, how many FWD cars come with limited slip or a locking differential? Not a lot. A RWD with limited slip or locker will be at an advantage getting started in many low traction conditions (of course AWD with LSD or lockers trumps both, but "AWD" with open diffs often loses).     As a second point you've just stated that a RWD will *handle* better in the snow. How's that you may ask? The "limit" of handling applies no matter what the coefficient of friction. The same factors that make a RWD corner better at speed make it corner better in poor traction. The recommendation for where to put the "good" set of tires (if you have one better set) for winter is on the rear tires... to prevent skidding. That same front weight bias makes FWD more likely to spin out when cornering or braking. I won't call that an "exception" just a trade-off.     I play both sides of this issue myself, I have a Saab 900SPG and a E30 BMW 325 and I've been in situations where either one would have been the better choice (Saab loses goes up slippery hills with an open diff and FWD, BMW gets stuck when you need traction on the _front_ wheels to turn out of a tight space even when the rear tires are moving the car)
    Bill
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Bill Bradley wrote:

I'd say a RWD would spin out easier, because the amount of power you push to backwheels when they don't have grip. Car starts going sideways. Now that's a feature I just love with snow, ice and uphill. Our BMW (althought Compact) won't go anywhere, it's stuck. Tyres just spin, spin , spin and spin. Our MB with limited differential on the back, will also make tyres spin, then lock and then.. nothing. It's stuck also.
In same situation, our Toyota & Audi go forward, because they have grip in the snow/ice. Each of the cars have spiked wintertyres, yet they won't make miracles if there isn't enough weight on the back.
And it's also always nice to help taxis which use MB in the winter conditions, when it's been snowing a lot, they're stuck also. "c'mon passengers, help me a bit, push the car".
Nothing beats AWD, but FWD is a lot better in winter conditions, you don't get stuck. Whatever happens at the limit is usually pointless. When the weather is bad, you drive according to it. But there's no helping if the car won't move.
- Yak
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Bill Bradley schrieb:

Yep.
Must have been the late night yesterday :-)
In fact I did the torque equilibrium but got mislead by the gemotry I'd drawn up.
You are right, the CoG shifts back and the load on the rearwheel increases. That load is then split into a component orthogonal to the road (for friction) and one parallel to the road (pulling the car back).
So the total orthogonal force on the road for friction is still less then if the car would be on a horizontal plane, but it's higher than on the front wheels.
Regards
Wolfgang
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This is a multi-part message in MIME format. --------------070800080109060609090106 Content-Type: text/plain; charset=ISO-8859-1; format=flowed Content-Transfer-Encoding: 8bit
how about we all part downhill and walk up! ;)
Wolfgang Pawlinetz wrote:

--------------070800080109060609090106 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"> how about we all part downhill and walk up! ;)<br> <br> Wolfgang Pawlinetz wrote:<br> <blockquote type="cite" cite=" snipped-for-privacy@4ax.com"> <pre wrap="">"Fred W." <a class="moz-txt-link-rfc2396E" href="mailto: snipped-for-privacy@allspammyrealbox.com">&lt;Fred.Wills@allspam myrealbox.com&gt;</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 &micro; and the force _orthogonal_ to the surface. The formula for the friction force is Ff = Fn x &micro;.
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 &micro;.
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&deg; (don't sit in the car).
__ | | | |C \ | | | | | | |C / | |_|
In this case, the car would have to be held by something else, because for Fr = &micro; 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&deg;, 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&deg; 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> </body> </html>
--------------070800080109060609090106--
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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:

--------------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$ snipped-for-privacy@news20.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=" snipped-for-privacy@4ax.com"> <pre wrap="">"Fred W." <a class="moz-txt-link-rfc2396E" href="mailto: snipped-for-privacy@allspammyrealbox.com">&lt;Fred.Wills@allspam myrealbox.com&gt;</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 &micro; and the force _orthogonal_ to the surface. The formula for the friction force is Ff = Fn x &micro;.
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 &micro;.
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&deg; (don't sit in the car).
__ | | | |C \ | | | | | | |C / | |_|
In this case, the car would have to be held by something else, because for Fr = &micro; 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&deg;, 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&deg; 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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Imad Al-Ghouleh schrieb:

*LOL*
Yep, would do us good. :-)
Regards
Wolfgang
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* reply to wolfgang dot pawlinetz at chello dot at *
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(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)
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I don't, so there...
But maybe I hate formatted e-mails...
;-) DAS
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myrealbox.com> wrote:

You should be saying "AWD", not "Quattro", as that covers *only* Audi, and it is well-known that BMW and others *also* build AWD cars.

No; he *does* have you there.

Yeah; it does.

What you overlooked is the *practical* 'worst case example': 45 degrees. [This assumes that the tires can generate 1.0g of tractive force, otherwise the car slides down the slope.] Notice, at 45 degrees, where the CoG is. Depending on how high above the surface it lies, it could come to rest directly *over* the rear axle (even *behind it* in a tall or rear-heavy vehicle). At any rate, as long as it *is* above the surface, it will shift *toward* the rear axle as the angle increases. If you want a simple demonstration of this, think about moving a refrigerator. Lying on its side, the top part could be pretty heavy, but as you tilt it up, the upper end becomes lighter and lighter until you have shifted the CoG past the point where the bottom edge is on the floor. Then, the top side weight becomes *negative* and the thing falls over the other way. Therefore, a slope *does* influence the amount of weight (and traction) on the wheels on each end of the car, even if it's sitting still. -- C.R. Krieger (Been there; dropped that)
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C.R. Krieger wrote:

snipped-for-privacy@4ax.com
Regards
Wolfgang
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myrealbox.com> wrote:

and
over
<snipped a whole bunch of stuff>
No, the weight distribution (front/rear) will most certainly change as a) the car is tilted on the longitudinal axis and b) when accelerating.
-Fred W
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Say folks, do you read the thread?
I already said you were right.

Regards
Wolfgang
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Fred W. wrote:

This is true.

I think you're right. The best handling snow car I ever had or drove was my Alfa GTV6 -- w/ RWD. I liked it even better than the original Audi Quattro -- which had gobs of traction, but wasn't particularly nimble. The only problems the Alfa had in winter were low ground clearance, and a poor defroster.
My old 2002 never kept me from getting first tracks on a powder day, or home in time afterward. I drove right past plenty of 4WD cars stuck in snowbanks and ditches.
Around here we have ice storms, which are so bad it's dangerous to walk. Yet somehow the old farmers manage to get by in their old pickup trucks, without yuppie 4WD or highfalutin' Finnish winter tires. We're talking bargain basement 1982 Ford Rangers and Toyotas. Geez, how did people get around before Quattros and Xi-s?
Matt O.
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Ummm, slowly?
Seriously. Any FWD, RWD or AWD (*including* Quattro) can be made to go quite nicely in snow when the right tires are put on them. The rest is just varying dgrees of confidence at incremental speeds.
Some prefer the front weight biased FWD which is sort of a point and shoot dart approach. Others (myself included) prefer the rear driven, power-sliding cart before the horse. And yet a third category wouldn't be caught dead without their full time AWD pulling from both ends.
It's all preference and none are completely superior. As in all things in life, it is a matter of balancing trade-offs.
-Fred W
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Then there are the fourth category, who buy all-season tires which suck equally in all conditions, and probably don't know _which_ of their wheels are responsible for moving the car around. With the wrong tires, where the drive wheels are doesn't matter.

Yup. By the way, do I know you from another place, Fred?
Dave Hinz
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wrote:

    They bought snow tires, which shockingly few people do anymore; they assume all-weather tires are good enough, no matter what, then blame the car when it's the tires not gripping that's the problem.     I'm guilty of not buying snow tires in decades, but I'm in downtown Chicago and things are plowed pretty quickly here so I can get away with it. Also my winter beater is the 944 - easy to push out when it gets stuck.     Emanuel
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Just exactly the point I wanted to make!

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