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Discussion Starter #1
Hey folks, I got another one for you. THis one relates to Friction, mu, and the normal force.

Given the function: F=mu*R

Where,
R = the downward force on an object (i.e. a tire)
mu = the static coefficent of friction
F = the resistence to slipage

Does an increase in area (i.e. contact patch) increase the grip, all else being equal? If so, what is the equation?


Here is my current thinking:

We know from "the Physics of Racing" that a tire gains grip as it is loaded up, then at some point, begins to become overloaded with weight and begins to lose grip. We also know that a tire gains grip a slip angle increases, but at some point begins to lose grip and enter a slide.

I think it is a safe assumption that as a contact patch increases in size, the wieght on the tire does not change significantly. The load per inch does, but the total load does not.

Why am I asking? I am building a model that will take the increase in contact patch, change in the stickiness of the tire, changes in downforce, etc, into account and I want it to be as accurate as possible.

I would also add, that for my purposes, it is not important to point out the effect of slope on the equation as it relates to the normal force. As there is nothing in my model to take that into account and it is constantly changing on the track/street anyway.
 

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I'd imagine the total weight on the contact patch has a lot to do with it....but I'm not a physics major.
 

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I choose B
 

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kinda what qball said. for this friction / force equation contact area doesn't matter. If the contact area got bigger, the same force is spread out more, but the coef. of friction would remain the same, so area is negligible.
 

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As surface area grows, the overall traction grows. It would be a constant multiplied against your (mu) friction. Your equation F=mu*R should be F=mu*R*surfacearea. Are you writing a computer simulation? If so are you simulating the suspension? You need to.
 

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As the contact patch grows, force per unit area decreases, so theoretically your grip wouldn't change. In real life mu is not a constant with respect to force, temperature, surface, or slip angle but has a complex relationship with each... it will actually begin to decrease with higher loads/unit area, which is one of the reasons why a larger contact patch does lead to more grip in many real life cases.

Note that in wet weather a smaller contact patch is sometimes desirable, and the higher loading per unit area helps "pump" water out from under the treads.
 

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traction =/= friction

friction is one part of it but not the whole story... wider tires do increase traction but not friction (except when turning and in a bad way)
 

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Does an increase in area (i.e. contact patch) increase the grip, all else being equal? If so, what is the equation?
Yes, it does. I don't think there's a simple equation. I've seen references to a book that deals entirely with tires, and is supposed to be excellent. It might have models/approximations. I can't remember the exact title, I could probably find it if necessary.

My understanding is that the classical equations you quote only apply to perfect, smooth surfaces. A tire is not a smooth surface. And pavement is far from smooth. Essentially, a (soft) tire deforms, and interlocks with the pavement surface to some degree. So the simple equations dealing with smooth surfaces are a poor approximation of what is really going on.

Just multiplying by the size of the contact patch, like Hello Kitty suggested, sounds wrong. If that were true, you could get enormous amounts of grip with flat tires... It's much more complex than that.
 

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Discussion Starter #11
Cone, I agree.

Here is what I think at this point. Given what I mentioned about the two traction graphs as a function of slip angle and normal force (the opposite of weight), I think that an increase in area, allows more space for the friction to act. To say that in terms, perhaps more clear, if you slip at 10 degree with 10 square inches, perhaps you slip at 9 degrees with 11 square inches. Subsequently, if you are over loaded at 1000lbs (100 lbs per sq inch) with 10 sq inches, then perhaps, with 11 sq inches, you are not over loaded until 1100lbs (100lbs per square inch).

This should account for the true statement that mu does not change, but grip does as surface area is increased.

In the end, I am trying to find a simple, but "reasonably accurate" equation that demonstrates this relationship. if it is true at all.
 

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FWIW, tires are very non-linear in their behavior, so your simple equation wouldn't work. Once to temperature, tires are soft enough that at the detailed level, they can mold around the roughness in the road. For sticky tires, this effect is even more pronounced. (I see Cone Fusion has already said this.)

In other words, mu is not a constant.

If you're interested in modeling their behavior, suggest you look at Carroll Smith's book "Drive to Win". It describes (and graphs) tire behavior at the limit.

ed
 

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Cone, I agree.

Here is what I think at this point. Given what I mentioned about the two traction graphs as a function of slip angle and normal force (the opposite of weight), I think that an increase in area, allows more space for the friction to act. To say that in terms, perhaps more clear, if you slip at 10 degree with 10 square inches, perhaps you slip at 9 degrees with 11 square inches. Subsequently, if you are over loaded at 1000lbs (100 lbs per sq inch) with 10 sq inches, then perhaps, with 11 sq inches, you are not over loaded until 1100lbs (100lbs per square inch).

This should account for the true statement that mu does not change, but grip does as surface area is increased.

In the end, I am trying to find a simple, but "reasonably accurate" equation that demonstrates this relationship. if it is true at all.
Are you saying that you're getting the same normal force at 10 degree slip and 10 sq in. as you'd get at 9 degree slip at 11 sq in., given the same car weight? <strike>If so, that implies a different mu for different slip angles (which I believe is correct), not a constant mu, since for the two cases F1=F2 and R1!=R2.</strike>

Oops, posted too fast. It implies that in that case the change in mu due to loading (decrease) perfectly offsets the change in mu due to slip angle (increase), since F1=F1 and R1=R2...
 

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