-Bob- wrote:
> On Tue, 11 May 2004 23:57:11 GMT, Bill Bradley
> <senator2@NOSPAMearthlink.net> wrote:
>
>> 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.
>
> 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 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

 

dizzy wrote:
> On Tue, 11 May 2004 15:14:15 +0200, Peter Bozz
> <fake-user@fake.email.address> wrote:
>
>
>>The top selling cars in Holland in 2003 were:
>
>
>> 2. Peugeot (52.412 exemplaren)
>> 3. Renault (47.159 nieuwe auto's)
>
>
> You mean someone outside of France buys those Frenchy cars? Wow.
>
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.

 

Dori A Schmetterling wrote:
> At least in Germany seeing lots of BMWs and Mercs is no illusion! They are
> top sellers:
> http://www.kfz-auskunft.de/kfz/pkw_n...ngen_2003.html
>
> In the Netherlands I seem to see Mercs on every corner...
>
> DAS

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:

> 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.

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
--
* Audi A6 Avant TDI *
* reply to wolfgang dot pawlinetz at chello dot at *

 

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
>
>
>

 

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
>>
>>
>>

 

Imad Al-Ghouleh schrieb:

>how about we all part downhill and walk up!

*LOL*

Yep, would do us good.

Regards

Wolfgang
--
* Audi A6 Avant TDI *
* reply to wolfgang dot pawlinetz at chello dot at *

 

(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)





"Imad Al-Ghouleh" <ighoul@po-box.mcgill.ca> wrote in message news:r_ooc.44884$FH5.1028635@news20.bellglobal.com ...
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.

 

Bill Bradley wrote:
> 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'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

 

Wolfgang Pawlinetz <mille@afm.at> wrote in message news:<bnc2a0lgtrf9deedu74ocpuv2d7u0trrm3@4ax.com>. ..
> "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.

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.

> >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.
>
> You almost got me there

No; he *does* have you there.

> 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.

Yeah; it does.

>
> 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 /
> | |_|

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)