93 teal terror
Founding Member
Your right except that theres more load on the drilled ones.......Mustang5L5 said:Ok lets try this....
F = pressure x area
F = 1000psi x 12 in^2 (rough estimate of max brake pressure and stock rotor surface area)
F = 1000 psi x 11in^2 (area of crossdrilled rotor)
Fsolid = 12000 lbs
Fdrilled = 11000 lbs
F is our normal force N
Now we use this equation Fk = Muk * N (Mu is a symbol that my computer can't generate but you engineering guys should know what i am talking about.
Fk is the frictional force
Muk is the coefficient of friction
N is the normal force from above
I have the Coeff of friction of leather on metal is 0.50. I have no idea what the actually MuK is so we will use that number since it's the closest i could find to brake pad on metal
.5 X 12000 = 6000 lbs-F (Frictional Force on solid face rotor)
.5 X 11000 = 5500 lbs-F (frictional force on drilled rotor)
I may not be 100% right on units, but that's what i have to prove that a larger contact area of a pad will give you more stopping force.
Does that check out with any of your other engineering/physics guys?
The brake pad delivers whatever force the piston produces no matter what. With the drilled rotors you have less contact area with the rotor so lets say the Distributed load on the solid one is 50N/mm and the drilled one is 61N/mm
multiply that time surface area to get equivelant loading. So lets say the Suface area of the drilled is 984mm^2 and the SA. of the nondrilled rotor is 1200mm^2. When you multiply 50N/mm^2*1200mm^2 you get 60,000N. When you multiply 984mm^2*61mm^2 you still get 60,000N. Conservation of engery. When you calculate friction, you have to use the equivelant loading. So its the same Ffr no matter what. All less SA. is doing is increasing the pressure on the rotors. Does that make sense Mustang5L5?