Brakecalculations.

Why install bigger brakes on your car, you say. Isn't it enough that I install higher quality brakes, with the same size ?

No it isn't. Higher quality brakes will improve your braking some, with more vents (drilling), slotting and maybe more pistons. But the thing that will really improve braking performance is size. Size as in larger diameter discs and bigger, multi-piston calipers.

There are two main reasons why bigger brakes is the key to getting more stopping power:

• Bigger brakes (discs) have more surface area and can therefore get rid of more heat than smaller brakes, and heat is a brakes worst enemy. The heat can do three things: Melt the surface of your brakepads (glazing) which makes them smooth and reduce the friction between pad and rotor. Heat can make your calipers so hot that the brakefluid boils; boiling creates airbubbles and bubbles are compressible, which in turn allows you to press your brakepedal to the floor without any significant effect on your car's speed. And last but not least high temperatures is not good for the brakediscs: Discs will tolerate heat up to a certain temperature, but higher than that they will warp, or even crack if the temperature is high enough. That is unless you have carbon brakediscs like the ones used on F1 cars.
• Bigger brakes have more leverage on the tire so less braking force from the brake makes for more stopping power at the tire. The reason for this is showed in these equations:

Heat
One of the two reasons for bigger brakes is heat. The bigger discs will get rid of more heat than smaller discs because of it's increased surface area. It's fair to say that most of the heat will dissipate from the area of the brakedisk that is wented. I.e the surface the brakepad touches and the surface inside the vent channels in the disk. And of course the area of the holes in the disk if it's cross drilled.
If we assume that the standard disc of 278mm and a 340mm disc have the same height where the pad touch the disk (which is 55 mm for the standard disc) then the area dissipating heat will increase by almost 28%. Of course the new bigger disc will probably be drilled, and have a larger area inside of the braking-surface that also helps. All these factors combined will probably increase the amount of heat the disc can dissipate by about 40%.
To discover why the bigger discs are even more than 40% better; read on.

Forces
I calculate this exactly at the time when the wheel has stopped rotating and slides over the concrete. We have, as the wheel is stationary (stationary here means not rotating, but still sliding), that the sum of the torques acting on the wheel is zero (if not, it would rotate). In this equation:

• M_s = The torques acting on the wheel calculated from the center of the wheel.
• F_t = The force acting on the wheel (actually the tire) from the ground as it slides.
• r_w = The radius of the wheel inclusive tire.
• F_br = The force acting on the brakedisc from the brakepad in the caliper.
• r_br = The radius of the brakedisc.

I have solved the equation this way so we can look at the forces acting on the brakedisc, and therefore the wheel, from the brakepads (as a result of the pressure of the pistons on the pads inside the caliper). The "minus" in the lower equation just means that the force acting on the brakedisc from the brakepad acts in the opposite direction of the force acting on the tire from the ground, or else the wheel wouldn't stop. Direction here means rotating direction (all forces act perpendicular to the radius of the wheel).

Now we have 4 factors that have influence on the stopping power:

• The radius of the wheel which is dependent on wheelsize and tiresize, see wheels.
• The radius of the brakedisc.
• The force acting on the tire from the ground (only stickier tires will improve this).
• The force acting on the brakedisc from the brakepad, which will increase with bigger, higher quality brakes.

So when we look at the lower equation we get:

• If the radius of the wheel increases, as it will do when fitting 17 inch or 18 inch wheels with larger tires, then the force acting on the brakedisc from the brakepad must be incresed to provide the same stopping power (stopping power here means the force acting on the tire from the ground (or on the ground from the tire (Newton's third law: "For every action, there is an equal and opposite reaction."))).
• As the radius of the brakedisc increases the force acting on the brakedisc from the brakepad can be decreased while maintaining the same stopping power.
• If the pressure of the brakepads on the brakedisc increases this will increase stopping power as well (obviously).

So, how much does these things increase or decrease the stopping power of your car ? That depends on the sizes that are used, but more than you might think, as I here will show:

Before calculating anything we must assume a few things: As standard the car can brake at about 0,9 G, of which 85% is done by the front wheels, the car weights 1350 Kg with one person in it. If you think any of these assumptions are very wrong send me a mail and I'll fix it.
We have that the radius of the standard wheel is 304,44 mm. Brakedisc radius is 139 mm, effective radius (where the sum of the frictinal forces between the disc and the pad is) is: 120,5 mm.

Newton's second law F=ma: This gives us that the braking force, or stopping power is: 1350*0,9*9,81= 11919N. That is 11919*0,85*0,5=5066N for each front wheel.

Now we can calculate the braking force required from the brake. First for a standard wheel and brake: (5066 N * 304,44 mm)/120,5 mm = 12.799 N.
Then if one mounts 18 inch wheels with 235/40ZR18 tires (which have a radius of 322,59 mm) the braking force required will be: (5066 N * 322,59 mm)/120,5 mm = 13.562 N. An increase of 5,9%.
But then if you find out those standard brakes suck, and buy 340 mm AP-Racing brakes (which I have assumed have an effective radius of 150,75 mm), the braking force required of those brakes would be: (5066 N * 322,59 mm)/150,75 mm = 10.841 N. Which is 20% less than with the standard brakes.
Then if you use 16 inch wheels with standard tires and 315 mm brakes (effective radius of 138,25 mm). The braking force required would be: (5066 N * 304,44 mm)/138,25 mm = 11.156 N. Which is 12.8% less than with the standard brakes.
The trick to getting the most effective braking is bigger brakes without increasing the tire radius. If we go for my tire/wheel/brakes of choice: 8x18 inch Compomotive MO with 225/35ZR18 Bridgestone S02 PP tires. AP Racing 340mm 6-pot brakes. The required braking force will be: (5066 N * 307,34 mm)/150,75 mm = 10.328 N. Which is 19,3% less than with the standard brakes and wheel/tire combination.

The grand total
If you choose to go for 340mm brakediscs the increase in brake effectiveness will without doubt be very big. How big is difficult to say, but if we simply add the increases we've calculated here we will end up with brakes that are somewhere around 60% better than the original. Of course, that's not very scientific, but then we have not taken into consideration the increased effectiveness of a six-pot caliper, which is considerable. Anyway; it surely is a bit better.

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