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Brake swept area? Wide tires better?


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Anyone that studied physics may have learned that the force required to slide an object over the surface of another object depends on the coefficient of friction between the two surfaces....keep reading this gets FUN.

 

F=uN

 

F-force required to move object

u-coeff. of friction between the materials

N-normal force of the objects (pressing down onto the sub-surface)

 

Car lingo:

F-handling/braking forces

u-sticky tires and grabby pads

N-weight of the car pressing the tires down, or the caliper piston force.

 

Notice anything missing?

 

AREA!!! So that tells you that no matter how much area of contact you have, the sliding force, F, depends only on the Mass of the object (contact force), and the coeff. of friction u.

 

 

When I first learned this, my hand shot up in class. My questions revealed where my mind was at the time. "So, why do wider tires make my car handle better?" "Why does everyone talk about more swept area when bragging about brakes?" AREA doesn't even get to play in the equation of force and friction!

 

If the AREA of contact does not even come into play when figuring out braking force or handling, based on the equation above, which is correct with assumptions, then why does everyone talk about MORE AREA for brake pads and tires?

 

 

CLASS, DISCUSS! :shock:

 

Quiz on Friday. I'll be on vacation until then at Hershey Park in PA!!!!

 

BTW, the Hershey Park Auto Museum has a 240Z on display. I can't wait to check it out. What size wrench do I need to pull the rear bumper? LOL.

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Umm... "U" can depend on the tire compound, width and tread pattern..

 

a set of 195 wide tires.. with a soft sticky compound and slick tires on rough asphault will probably give you the same coefficient of friction as a set of 275 wide tires with a hard compound and all season or rain tread pattern on smooth asphalt...

 

different areas.. roughly the same coefficient of friction

 

( the above statement is not an exact statement.. just an example )

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Wider tires can be made from softer compounds since there is less psi due to a larger contact area. Softer compounds have higher coefficient of friction and allow for more intimate contact with uneven road surfaces.

Edited by rossman
romoved statement. misread above statements
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All sounds right to me - AREA is only used for cooling in the braking surface after the work. After all a brakes job is thermal transfer of the work. Well this is my simple understanding of it all.

 

In a car = engine turns thermal energy ie the burning of fuel into motion via the power stroke in the ICE. The brakes take the motion and turns that into heat via the u (coeff. of friction between the materials and the force applied) the area comes into play in allowing these to cool down quicker ready for the next application of the brake and thus allowing better motion to heat transfer of energy.

 

As some might know after x laps on standard brakes they stop working as they are not cooling quick enough = less motion to heat (as the heat is already there) and the u can no long provide the correct balance of F and N which equals

 

F.......Far out that was one close almost a

N.......Nasty crash better get something the help with

U.......underpants replacement......= get better brakes so we can put the FuN back in the equation.

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Physics 101 is an oversimplification of everything to help everyone get the "big picture". Start taking too many variables at once, and things start to get confusing really fast.

There's a reason why it's called "fundamental physics" ;)

 

When the surfaces are conjoined, Coulomb friction becomes a very poor approximation (for example, adhesive tape resists sliding even when there is no normal force, or a negative normal force). In this case, the frictional force may depend strongly on the area of contact.
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It is all in the Mu.....or u. The table states that (Mu)u=coeff. of friction between the materials. As the two materials change, your Mu changes. So if you have a wider tire, you have a different Mu. Also, nobody stated how wide the road was...did they?

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Since road surfaces are not perfectly flat, and tires are not perfectly hard, tire material gets squished(technical term) into the voids, and the road surface peaks get forced into the tire. This means there is a shear force as well as a friction force created whenever there is a slip angle/ratio.

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There are so many variables that play a part in this subject that it will be ridiculouse to figure something out without positive field tests and more characteristic data. We will end up in thermodynamics, aerodynamics, and every other type of dynamics you could think of just to get an exact number. There are just too many bases to cover.

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the formula is mega simplified and doesn't really apply for elastic materials, and definitely not over rough surfaces. at the road level it's not just friction involved. there's mechanical engagement with the peaks in the asphalt, and more are better. so for tires, this discussion is fail.

 

for brake pads, well.. also fail, as someone else mentioned the surface area comes in when you're talking heat dissipation. friction is temperature dependent.

 

only reason you'll ever wanna cut back on tire and brake surface area are for other performance losses associated with big, unsprung, spinning parts.

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Here goes my input on the subject. I believe Dave and others are correct when using the F = uN formula when applying it to tires but post #14 points out some very good points about the road and tires not being perfect.

 

Where I disagree is on the braking.

 

When you apply a force to a solid object, the pressure is defined as the force applied divided by the area of application. The equation for pressure is:

 

P = F/A

 

where

 

P is the pressure

F is the applied force

A is the surface area where the force is applied

F/A is F divided by A

 

For example, if you push on an object with your hand with a force of 20 pounds, and the area of your hand is 10 square inches, then the pressure you are exerting is 20 / 10 = 2 pounds per square inch.

 

In car terms, if a brake pad pushes on a rotor with a force of 20 pounds, and the area of the pad is 10 square inches, then the pressure the pad is exerting is 20 / 10 = 2 pounds per square inch.

 

Granted as rotors increase, 80% of the time the pad does also. With that so does the MC and fluid properties do too (heat transfer) so better DOT fluid is needed. It is a complete system working together. But to say brake forces has nothing to do with area is wrong IMO.

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Let me start by saying I am by no means an engineer.

 

The way I'm lookig at this is the advantage you get by using more area whether it be in tires or brakes is the reduction of a failure point usually created by heat or the reduced ability to get rid of heat with the smaller area. I refer back to Joe Demers creation of a heat sink for the stock s30 caliper. Adding this to a atock setup made the stock brakes perform like a much larger caliper, pad and vented rotor.

Demonstrating this on a tire is much more difficult but can be explained by the larger tire patch sharing the total heat generated just before the point the tires begins to slide. The heat is generated by scrub. My thinking is the larger tire patch is able to handle more heat dissipation before creating so much heat the tire begins to slide.

 

Might not be completely accurate but that's why I say dissipation of heat is why bigger is better :)

 

Joe

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