From: mshar@mentor.cc.purdue.edu (Mike Mshar)
Subject: Re: downforce and cornering g's (was which car handles the best?
Organization: Purdue University Computing Center
Date: Tue, 6 Oct 1992 04:29:22 GMT

> tsuit@gmuvax2.gmu.edu (Tom Suit) writes:

>  The significant thing you are ruling out is the contact area of the
>  tires. Wider tires give you more grip. (assuming dry pavement) I can't
>  BELIEVE you didn't consider that.

Actually, in the theoretical sense, this is not true. Believe it or not.
Take it from somebody that has taken some physics courses. The area of
contact for ideal surfaces does not affect the amount of friction force
that the surfaces produce. But, there is a catch, as I will explain
in a minute.

OK. Force due to friction is given by the simple equation FORCE=NORMAL FORCE
multiplied by the COEFFICIENT OF FRICTION. Notice that surface area does 
not appear anywhere in this equation. The normal force is the force that 
the road puts on the tires. This will always be equal to the weight of 
the car, when you neglect air foil usage. The coefficient of static friction
between rubber and concrete is about 1.0. Thus, using the above equation,
it would be impossible for a car to turn with a force greater than 1g. 
So how can cars corner at more than 1g???

Well, I took a long look at this, and there are several things that we
are not considering. First of all, not all rubber tires have a coefficient
of one. Drag slicks, for instance, can be very sticky, especially when 
you add that traction compound. So, drag cars can accelerate faster than
one `g`, partly for that reason. Secondly, we are all assuming that the 
coefficient of friction remains constant for varying normal forces. This
is not the case here. When a car is turning, there is a torque produced on
it because the center of gravity is not at road level. Whenever you apply
a force away from the center of gravity of an object, it causes the object
to rotate about its center of gravity. Now, there must be a force to repel
this torque, otherwise the car would roll along its center axis. What happens
is that the normal force on the outside tires increases; at the same time the
noraml force on the inside tires decreases. But through it all, the sum of
the 4 normal forces has to be equal to the weight of the car, otherwise
the car would rise of the pavement. But, rubber is a flexible substance. 
Since this is the case, its coefficient of friction can increase with an
increase in normal force. In simple terms, the tires `dig in` and mesh
with the pavement (like sandpaper on rough wood). The rubber gets hooked
up on the little cracks and crevices in the pavement. Also, the tires 
`squats` as this happens. This allows even more surface area to mesh with
the pavement. The result? The coefficient of friction on the outside tires
increases faster than the coefficient of friction on the inside tires 
decreases. Therefore, it can be possible for a car to corner at more than
1 `g`. That is why most performance cars use soft tires; they can more
easily mesh with a road surface than hard rubber tires can. 

Moral of the story...

Rubber and asphalt are not ideal surfaces. 

I would bet money, however, that a neutral downforce car could not corner
faster than one `g` on smooth, garage-like pavement. Why? Because the 
coefficient of friction would remain pretty much constant no matter how 
much normal force there was, because a super smooth concrete surface
has no cracks and crevices for the rubber to mesh with. 

I know this is long and tedious to read, but it should prove to be factual.
Anyway, back to that pesky homework....

P.S. This is not meant as a flame on anybody. I am just trying to clarify
some things that are obviously being disputed by many people. Don`t take
this personally.

-- 
|       Michael G. Mshar        | My Life-Long Goal: To keep common sense    |
|   mmshar@mn.ecn.purdue.edu    | alive and keep the idiots from taking over.|
|  mshar@mentor.cc.purdue.edu   |     \\President, Nice Guys Anonymous//     |
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