if one were to calculate the line pressures generated by a master cylinder and then the clamping force at the rotor - how can you calculate the effect a power booster has on clamping of the caliper. the power booster would influence the (brake pedal lever ratio x the force applied by your leg), but by how much? i guess the amount of vacuum generated and the dia of the booster are several variables. anybody know the formula?

i may have answered my own question - take the (vacuum (inHG) x .4912 = psi) x the in^2 of the booster to obtain a force (lbs). this force would be added to the force applied to the MC via your leg and the pedal ratio??

Yes, but use some reasonable vaccum like 18" or so. You don't get the whole booster effect unless you are pressing the pedal hard enough to open the valve inside which allows atmospheric pressure to enter the booster.

I'm no math whiz but it goes something like this:

Camaro pedal ratio for power brakes is 3.8 to 1, manual brake ratio is 6.25 to 1.

so, to calculate, foot pressure in lbs, X pedal ratio plus booster pressure in lbs = total force on master cyl piston in lbs.

Piston force lbs X piston sq inches = line pressure (PSI).
Line pressure X caliper bore in sq inches = clamping lbs. (Use one side of a four piston caliper piston area or the whole area of a single piston caliper.)

caliper clamp lbs times pad friction coeficient = caliper drag. (pad size doesn't matter here)
caliper drag X rotor (effective) radius to pad center, (not rotor OD) = rotor torque in inch/lbs.
Rotor inch/lbs torque divided by tire radius in inches = lbs of braking force at tire tread.

If you want to do more estimate forward weight transfer and add that to the front wheel load.

Tire loaded weight X tire coeficient of friction = tire limit of adhesion, if brake lbs exceed this, the wheel locks and skids.
David

excellent, thanks david!

Originally posted by davidpozzi:
Yes, but use some reasonable vaccum like 18" or so. You don't get the whole booster effect unless you are pressing the pedal hard enough to open the valve inside which allows atmospheric pressure to enter the booster.

I'm no math whiz but it goes something like this:

Camaro pedal ratio for power brakes is 3.8 to 1, manual brake ratio is 6.25 to 1.

so, to calculate, foot pressure in lbs, X pedal ratio plus booster pressure in lbs = total force on master cyl piston in lbs.

Piston force lbs X piston sq inches = line pressure (PSI).
Line pressure X caliper bore in sq inches = clamping lbs. (Use one side of a four piston caliper piston area or the whole area of a single piston caliper.)

caliper clamp lbs times pad friction coeficient = caliper drag. (pad size doesn't matter here)
caliper drag X rotor (effective) radius to pad center, (not rotor OD) = rotor torque in inch/lbs.
Rotor inch/lbs torque divided by tire radius in inches = lbs of braking force at tire tread.

If you want to do more estimate forward weight transfer and add that to the front wheel load.

Tire loaded weight X tire coeficient of friction = tire limit of adhesion, if brake lbs exceed this, the wheel locks and skids.
David

Sorry to revive an ancient thread, but I wanted to point out a little problem to possibly save people some confusion.


Piston force lbs X piston sq inches = line pressure (PSI)

This should read, Piston force lbs / piston sq inches = line pressure (PSI).

I should admit I've been lurking, reading, and enjoying this forum for quite some time and now finally have found a good excuse to post. :)

Sorry to revive an ancient thread, but I wanted to point out a little problem to possibly save people some confusion.



This should read, Piston force lbs / piston sq inches = line pressure (PSI).

I should admit I've been lurking, reading, and enjoying this forum for quite some time and now finally have found a good excuse to post. :)no, not divided by but times the piston sqare inches. The larger the caliper piston the more piston force at the same line pressure.

Welcome, Speed. You are correct.
Rounding off figures
50 lbs on the pedal times 6:1 ratio = 300 lbs of force divided by 0.785 sq in piston (1 inch dia) = 382 PSI.

no, not divided by but times the piston sqare inches. The larger the caliper piston the more piston force at the same line pressure.

That's true, but the equation in question refers to the master cylinder piston and line pressure, not the caliper piston and clamping force.

Welcome, Speed. You are correct.
Rounding off figures
50 lbs on the pedal times 6:1 ratio = 300 lbs of force divided by 0.785 sq in piston (1 inch dia) = 382 PSI.

Thanks for your welcome and for the confirmation. :thumbsup:

Thanks for the correction.