Apex Arc8 wheels to e46 m3

[325ix]

lol I don't think that's how it works...

arc's look sweetest.

no it's not. I learned about traction and such in physics and area of contact isn't even in the formula.

[Be30mer]

no it's not. I learned about traction and such in physics and area of contact isn't even in the formula.

Than explain. Because I calculate, the tires have about 6-8 (ill say 7) inches of contact in y direction and 8-10 inches in x depending on width of tire.
7x8x4= 224 inches squared of contact area for all tires.
7x10x4= 280 inches
280-224= 56
64 was a close estimate.
Of course, I am basing this assumption off the idea that
More area = more traction. << Is that not true?
But yes I want the arcs because they look awesome as well.

[325ix]

Fr = μN

where:

•Fr is the resistive force of friction
•μ is the coefficient of friction for the two surfaces (Greek letter "mu")
•N is the normal or perpendicular force pushing the two objects together
•μN is μ times N
Fr and N are measured in units of force, which are pounds or newtons. μ is a number between 0 (zero) and ∞ (infinity).
Taken from http://www.school-for-champions.com/...n_equation.htm

I'll explain a little better when I can get to my school notes. As you can see though, there is no varible for surface area or contact patch. It's the extra weight of the wider wheel/tire combo that gets you the extra traction.

[Be30mer]

Fr = μN

where:

•Fr is the resistive force of friction
•μ is the coefficient of friction for the two surfaces (Greek letter "mu")
•N is the normal or perpendicular force pushing the two objects together
•μN is μ times N
Fr and N are measured in units of force, which are pounds or newtons. μ is a number between 0 (zero) and ∞ (infinity).
Taken from http://www.school-for-champions.com/...n_equation.htm

I'll explain a little better when I can get to my school notes. As you can see though, there is no varible for surface area or contact patch. It's the extra weight of the wider wheel/tire combo that gets you the extra traction.

^^ thats for static friction. I think area plays a role in kinetic friction, but im a little dusty in the physics department. I do remember that the coefficient of friction M= force of friction/ normal force, where the normal force is just mass in kilograms x acceleration from gravity (9.8m/s/s). I cant remember if there was any specific formula other than that ^^ to solve for the force of friction, and cant remember much about kinetic friction.

[325ix]

^oops wrong one. I'll grab my physics stuff at school tomorrow and post up the formulas.

[E30_Pare]

I like how this turned from a pchop to a physics lecture.

[Nic01101011]

Fr = μN

where:

•Fr is the resistive force of friction
•μ is the coefficient of friction for the two surfaces (Greek letter "mu")
•N is the normal or perpendicular force pushing the two objects together
•μN is μ times N
Fr and N are measured in units of force, which are pounds or newtons. μ is a number between 0 (zero) and ∞ (infinity).
Taken from http://www.school-for-champions.com/...n_equation.htm

I'll explain a little better when I can get to my school notes. As you can see though, there is no varible for surface area or contact patch. It's the extra weight of the wider wheel/tire combo that gets you the extra traction.

high school / GE lower div college level physics dont explain much in the real world.
Even grad level dynamics courses only explain the behavior of nothing more than simplified models..

Long story short, your coefficient of friction isn't simply a static number, and does depends on overall contact patch area.