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Old 07-26-2009, 06:33 AM
strictlypersonl strictlypersonl is offline
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Oh boy! Another opportunity to do the math...

Brake pad coefficient of friction, .4-.6. Let's assume .5 for this setup.

Radius of the center of pressure for the brake pad on an 11" rotor: ~5" (.42 ft)

Diameter of Grenada caliper piston - a pure guess: 2.3", which yields a piston area of 4.15 sq.in. Because this is a floating piston, the effective area is double that or 8.3 sq.in. (The piston effort is exerted on both sides of the rotor.)

Weight (under heavy braking) on each front wheel: ~900 lbs. > 900 lbs force at the contact patch for 1G braking

Radius of front wheel: ~1.1 ft.

Torque at the wheel required to generate 1G: 900 x 1.1 = 990 lbft. Let's round that to 1000 lbft. (We're doing a lot of back-of-envelope calculations here. )

So, the caliper must generate that much torque on the rotor/hub.

1000=.5 x .42' x 8.3 sq.in. x line pressure (psi)
or
1000/1.743 =574 psi line pressure required for a 1 G stop.

Ta Da!

(The big assumptions here are the area of the caliper piston and the pad coefficient of friction, which might change under heavy braking. If my guesses are off, it could change the answer significantly.)
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