Chris Paveglio

iTrader: (2)

I do not think your idea about multiplying by 2 is correct; but I'd rather have someone with more engineering knowledge chime in on that with more authority.

wssix99

iTrader: (5)

One would take the force at the brake caliper piston and divide it by 2 in order to figure for the pad force on each side of the rotor, but... the forces would also have to be divide proportionally (first) between the front and rear because the pistons on those calipers are different sizes.

^ For the purposes of figuring out if the brake booster is sized for the calipers, this extra math isn't needed. One can simply look at the proportional difference in the surface areas of the pistons before and after. (This deals with total force applied, and being put out by the brake booster, and just doesn't divide it up between front/back and the sides of the rotor.)

eb110americana

iTrader: (4)

Since my rear brakes are the stock units, I don't think this enters into it, unless the fronts are confirmed to be of a different surface area, which would indicate an introduced imbalance front-to-rear. But going back to the fronts, I am not sure what you are saying exactly. In the first paragraph, it seems as if you are agreeing with what I stated, but then in the second I am not sure. Are you saying that you would then compare 3 pistons (half the 6-piston caliper) to 2 pistons on the floating caliper?

My thought is, in this following basic example, assume all pistons are 1" diameter for master and all calipers to keep numbers simple. You press the brake pedal to move the master 12" (1x12). Sent to the fixed 6-piston, each piston will move 2" (12/6 = 2"). If you send the same fluid to the floating caliper, assume the brake rotor is 3" away from the pads. The pads will push on one side until it hits the rotor, then the other side will begin to move, since the caliper can slide. The pads will then extend out another 3" from the other side from the same fluid continuing to flow into the same 2 cylinders. So each pad will move 3", or put another way, the piston will move 3", and the caliper will move in the opposite direction by 3" (12/2 = 6, and 6/2 = 3"). If it were a 2-piston fixed caliper, then each pad would move out by 6" (12/2 = 6").

wssix99

iTrader: (5)

The rear brakes are significant because they take proportional amount of force from the booster. Mathematically, they are almost like a "shock absorber" for this problem. The smaller they are in relation to the front caliper pistons, the greater you should feel the difference in the change of front calipers.

You can use hydraulic formulas to calculate piston movement, but that doesn't help what you want to know. You are interested in the situation where the pistons have completed their movement and the pads are all touching the rotors. At that point, your booster is developing pressure (PSI) and you just want to see how that pressure (PSI) is divided up among the total surface area of all the pistons the booster services. If the new total surface area of the pistons is greater than stock, then the pressure will be divided up across a wider area, resulting in less braking force. In order to make up for that, the booster will need to create more pressure - which will be delivered by your foot.

eb110americana

iTrader: (4)

Most of that makes sense to me. I was already thinking of the fact that movement may not equal or be proportional to pressure once the movement is stopped against the rotor. Let's try another simple example. Imagine a wheel cylinder with a piston at either end, pushing outwards in opposite directions. When you step on the master, the 2 pistons move outward from the center of the cylinder by the same distance and by the same amount. Let's pretend that they somehow push on a rotor, or two rotors to stop the car, the design isn't really important to the experiment. In this example, you would count the surface area of only one piston (a cross section of the cylinder)? Or would you add the two piston areas together?

PS: Sorry to OP for thread-jacking this brake problem request for help.

wssix99

iTrader: (5)

There's such a small gap of air between the pad and the rotor, that the movement of the pistons on the brakes is insignificant. The pad and the rotor are going to make contact very early in the pedal travel, so all the stuff that goes into how things feel at the brake pedal can be simplified to the static forces. (If we were looking at the hydraulics on a bulldozer, etc. then things would be different.) The force we really care about here is created by compression of the fluid inside the lines vs. the pistons moving and doing work.


You need to add all the pistons on all the brakes and put them into the equation. The rear pistons (even though they aren't changing) are still dillutive to the equation and buffer this type of thing. The smaller the rear pistons are in relation to the front, the larger the effect will be of putting bigger brakes up front. All things equal, below shows this effect. (Everything in the scenarios is the same, except I changed the size of the rear piston - this shows how it matters.)

Stock - Scenario A:
Fronts - 1.5 sq in.
Rears - 1 sq in.

Aftermarket - Scenario A:
Fronts - 2 sq in.
Rears - 1 sq in.

Stock Surface Area - Scenario A= 5
Aftermarket Surface Area - Scenario A= 6
Percentage increase = 20%


Stock - Scenario B:
Fronts - 1.5 sq in.
Rears - 1.25 sq in.

Aftermarket - Scenario B:
Fronts - 2 sq in.
Rears - 1.25 sq in.

Stock Surface Area - Scenario A= 5.5
Aftermarket Surface Area - Scenario A= 6.5
Percentage increase = 18%