In a message dated 2/14/2000 6:38:02 PM Eastern Standard Time,
James.Nazarian@colorado.edu writes:
> What I am down to now is I have a coefficient of
> friction of a set of tires but can't relate that to a tire size. What I
> have is 500lbs on a tire with coefficient of friction of .8 so 400lbs of
> traction force (round numbers chosen and ignoring weight transfer etc for
> simplicity). So I have this but I don't have a tire size. Obviously if I
> put bicycle tires on the car they would loose traction, but if I put drag
> tires on they wouldn't even though both might have Cof of .8
>
James,
I'm an electrical engineer, not a mechanical, but I'll take a stab at this.
First of all, your statement that a bicycle tire and a drag tire, both with
the same CoF, won't give the same traction is true, but not for the reason
you may think. With 500 pounds on the tire, and tire pressure of 25psi, both
tires will "try" to put 20 square inches of rubber on the ground. To do this,
a 20 inch wide tire will only need to have a one inch long contact patch,
whereas a one inch wide tire will need to put down a 20 inch long patch.
Obviously, a reasonable diameter bicycle tire can't do this.
A more reasonable comparison would be to compare a 10 inch wide tire to a 5
inch wide tire. With 500 pounds and 25 psi, the 10 inch tire would give a 2
inch long patch, while the 5 inch tire would give a 4 inch long patch. From a
pure frictional standpoint, both tires (assuming made of the same material)
would give the same traction effort.
However, pure friction is only part of the story. A 10 in wide by 1 inch long
patch will corner much better than a 5 inch wide by 4 inch long patch. When a
tire is cornering, the tire material must "walk" as the turn is made. That
is, the material at the front of the tire will come down on the road surface
just a little bit towards the outside of the turn compared to the material
already on the road. The longer and narrower the tire contact patch, the
further outboard the displacement will be. The further outboard the
displacement, the further the tire strays from the desired path. That's the
whole rationale behind super wide tires for racing.
Another thing to consider is that a tire can have a CoF greater than 1. Due
to the irregularities in both the tire material and the road surface, the
tire actually "grips" the road, rather than depending on friction alone. It's
like turning one egg carton upside down and dragging it across another. The
bumps in one "hook" into the valleys in the other. That is why race tires are
very soft, and why they don't last very long - the road literally rips pieces
of the tire off.
> So how do I relate Cof to traction force and to tire contact patch area?
As far as I know, you can't. There are so many variable involved, that the
only way to tell is to test. I'm sure tire engineers have dozens of
complicated formulas, including such esoteric items as angle of the sides of
the tread blocks, number of threads per inch in the side wall material, etc.
but in the final analysis, they spend hours and hours in testing. So do the
race teams, at least the successful teams. Even if you had the formulas, you
would still not be able to do the calculations, as you wouldn't be able to
get the data to plug into the equations.
Nevertheless, I'm sure there are "rule of thumb" formulas to get you into the
ball park. Hopefully, someone on the list will know at least some of them.
Dan
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