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

You are now forgetting the most important point here, i.e., as it is impossible to keep perfectly straight steering - for one thing nothing is perfect, for another lateral slopes will spoil the rest of any good attempt at this. This, in practice, in the real world, zillion light years from where you live, means that when slippery enough and from certain climbing angles on the RWD will start swinging its butt so badly that no amount of wheel work will be able to compensate in order to keep the car on the right path. If you don't acknowledge this, it is only because you've never experienced that before. Actually, it does happen even with Quattro, given a sufficient amount of torque and pressure on the accelerator, and that's precisely because of your explanation above.

When climbing under those conditions Quattro has no FWD rivals, RWD is totally out of the question, but FWD is the easiest to handle.

The reason FWD is way more effective on slippery ground than RWD can be read in my previous point. Any attempt at countering this would suggest a tremendous lack of hands-on experience.

Well, Floyd, my argument does not hold true for your 330xi, but the point is that RWD will render their rear tyres unusable much faster than Quattro will render either front or rear or both, that's simply because any burst of acceleration is evenly distributed. You know the worst for tyre life is drift spinning, and that's the only thing I envy from those driving BMW RWDs. Of course I must concede RWD on the dry is way better fun than either FWD or Quattro.

He certainly wouldn't if he'd been driving an RWD.

JP Roberts

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.

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.

Regards

Wolfgang

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.

Not a bad try, but you're missing the key factor: Torque. Since the center of gravity is NOT on the road, it has a torque arm to the point of contact of the tires. The SUM of the forces on the contact area is as you worked out, but it doesn't remain 50/50 front/rear since the rear axle is providing a counter-clockwise (if viewed as in your drawing) torque while the front axle can only provide a clockwise torque. To reach rotational equilibrium more of the weight force in on the rear axle. It's the same reason that your car will nosedive under braking and lift the front end under acceleration.

If FWD slips you can't steer either, it's just that most FWD cars have a front weight bias (due to having the engine, transmission, and other such bits up front) so you have more traction all other things being equal. Bill

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

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.

You mean someone outside of France buys those Frenchy cars? Wow.

Or a viscous AWD car for both. ;-)

-Russ. '88 iX

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

I know, it's shocking. I guess buyers are lured by the abundant "standard" gadgetry the French are so fond of, the pseudo-futuristic looks (you seen the new Megane?), and who knows, maybe they're fun to drive the 50,000 miles you can manage in them before you throw them away.

A short drive around where I live revealed 8 Mercs. Now, it's no Beverly Hills, just a fairly affluent neighborhood of a big Dutch city (actually, I crossed over into the affluent neighborhood from the not-so-affluent part of town I live in). There were a couple of Saabs and a few of the ubiquitous V40/V70 Volvos. I counted 6 BMWs and at least one A6, two A4s, a Jaguar XJ and a Lexus LS400. Most of the Mercs were of course C series. I didn't count the CLK Cabrio my local drug dealer drives: he's hardly representative of the general population and might skew my empirical data. I guestimate that I must have seen about

500 cars.

Most people here seem to have a, shall we say, predilection for spacious MPV type of cars, mostly uninspiring brands I coudn't even tell apart. Think Ford, Opel, Peugeot, Fiat, some Japanese and Korean brands, whatever.

I wonder, what city was it that you say you saw Mercs on every corner?

Peter

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

Imad Al-Ghouleh schrieb:

*LOL*

Yep, would do us good. :-)

Regards

Wolfgang

(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

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

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)