it certainly makes sense...but couldn't our smaller piston calipers exert a similiar force theoretically?
So maybe this is a better way to think about it...b/c at the end of the day we want to stop our rigs faster and a big part of that is how long the brake needs to travel in order to stop the damn thing.
If we push our brake pedal down 3" (just an example) will the smaller caliper exert the same force as the large caliper? Technically (someone correct me if i'm wrong) by pushing on the brake a hydraulic pump pushes fluid that pushes the piston within the caliper outward against the pad.
By pushing 3", are both the larger caliper and smaller caliper exerting the same force? My thought would be yes b/c the pump is exerting the same force on the liquid...but I have ZERO science to back up my idea...anyone much smarter than me that can chime in?
I'm not an expert at this either, but your correct. The pressure on the pistons regardless of size is the same for the 3" of pedal travel. Ok, thats the pressure in the equation. Now the surface area of the pistons is different, which would change the actual force measured in pounds exerted on the pads.
The actual pressure exerted on the pads is different because the size of the surface area of each piston is different even though the fluid pressure on each piston is the same.
That's how an air hoist works. It's still only 100psi in the hoist, but the surface area is much bigger, so bigger surface area * pressure = force, which has enough force to lift a vehicle.
I guess we would need to know how big round is each caliper piston, then we could use 800psi fluid pressure and then we could have some real numbers!
I guess I'll google the piston sizes.
EDIT: Dope! bpodskalny gave us the piston sizes in the first post! 75 and 79mm bore sizes.
Step1. Convert mm to inches. 75mm=2.952755905515 inches and 79mm=3.1102362204758003 inches.
Step2. Determine surface area in inches. Use formula A=pi*r2
reference found here
r=2.952755905515/2=1.4763779527575
r2=2.17969185938842690185380625
A=3.14159265*2.17969185938842690185380625=
6.8477039247195154499261890895241
Surface area of the smaller piston is 6.85 square inches.
r=3.1102362204758003/2=1.55511811023790015
r2=2.41839233678989776333002948137
A=3.14159265*2.41839233678989776333002948137=
7.5976035900754674075290601429554
Surface area of the larger piston is 7.60 square inches.
So now all we need is the psi pressure number on the piston due to pedal travel.
Lets say it's 800psi.
So let's do some math. 800*6.85 =
5480 pounds of force on the pads for the 1/2ton caliper, and 800*7.60=
6080 pounds of force on the pads for the 3/4ton caliper.
So the larger caliper exerts 600 more pounds of pressure which stops your truck better.
Whew! Hope that helps at least someone!