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Articles about Performance Brake Upgrades

Balancing Brake Upgrades

Part 1 • Part 2 Part 3

The Numbers

So, let's start looking at some calculations. First, let's understand this front & rear balance thing with some numbers. Let's review the calculations to show the braking forces at the wheel. First, we need an input clamping force at the rotor. This is created with a certain pedal pressure, pedal leverage, master cylinder design, and caliper piston size. We're not trying to show the actual engineering here, just the principles with enough math to back it up, so we'll start with a set of assumptions that result in a certain caliper clamping force at the front & rear calipers. The components involved to this point will remain constant through our theoretical rotor and pad upgrades, so we'll use that same clamping force for all instances.

The clamping force is multiplied by the brake pad coefficient of friction. This results in a net clamping force on the moving rotor. Next, we have to convert the static force of lbs into a ft/lbs of torque (our rotor is rotating, and a rotating force is defined as torque). Next, to translate the rotor torque into force at the road surface, we have to account for the diameter of the tire. This ends up back to a simple linear force in lbs at the road / tire interface.

The table below shows the results of these calculations on four scenarios: the stock setup, front pad upgrade, front rotor and pad upgrade, and all four rotor and pad upgrade. The important net result is the balance of force between the front & rear, not the actual forces (which are variable based on driver's foot pedal pressure).

Stock Stock + Front Pads Front Rotor + Pads All Four Rotors / Pads
Front Rear Front Rear Front Rear Front Rear
Caliper Clamping Force (lbs) 2160 1440 2160 1440 2160 1440 2160 1440
Pad Cf 0.33 0.33 0.45 0.33 0.45 0.33 0.45 0.45
Net Clamping Force (lbs) 1425 950 1944 950 1944 950 1944 1296
Rotor Diameter 10.8"
Brake Torque (ft-lbs) 522 293 713 293 850 293 850 486
Tire Diameter (in) 24.9 24.9 24.9 24.9 24.9 24.9 24.9 24.9
Stopping Force (lbs) 503 282 687 282 819 282 819 468
Brake Force Balance 64% 36% 71% 29% 74% 26% 64% 36%


Caliper Clamping Force Fc (c = caliper)
Pad Cf Cf
Net Clamping Force Fr (rotor) = Fcal*Cf*2 (2 = 2 pads)
Rotor Diameter Dr (diameter of clamp point inches) (torque radius calculated 1" in from the rotor diameter)
Brake Torque Tb = Fr*(((Dr/2)-1")/12"/ft)
Tire Diameter Dt (t = tire)
Stopping Force Ft (tire) = Tb / ((Dt/2)/12"/ft)

OK, so now we see the relationship of how changing common bolt-on upgrades in the braking system affect the balance of front and rear braking forces. What does this do at the tire / road interface where the stopping action actually occurs?

Braking in racing is concerned primarily with performance during limit braking. With this type of braking, the brakes are applied with enough force to be just short of locking up the tires. One thing you must be very clear about, is that the frictional forces of the tire on the road is what slows a car down. Rotors, calipers, and pads do not stop the car. The brake rotor is used to apply a torque on the tire. That torque is translated into a linear force at the tire and road interface that resists the car's forward motion at the road surface. This is what stops the car. Therefore, when looking at brakes you must look at the impact of the forces applied at the road surface. All the other calculations in between are simply balancing the engineering for designing parts to get to that point.

A tire will generate a certain coefficient of friction based on the rubber compound, the size of the contact patch, the road conditions, etc. Another major factor in a tire's stopping power is the amount of downward force on the tire. Higher downward force increases a tire's traction. Because we're dealing with limit braking (the maximum end of the braking scale), we can simplify all the factors involved into two numbers. The downward force applied on the tire (which is a combination of a vehicle weight distribution and weight transfer under braking), and the tire's coefficient of friction (which reduces all the other factors into one number). If we multiply the downforce by the coefficient of friction, we get the maximum force the tire can create before it loses traction.

Let's look at some numbers and see how this actually affects braking.

Let's take a 3,000 car that has a static weight distribution of 55/45. That's 1,650 lbs on the front tires (825 each), and 1,350 lbs on the rear (675 each). Now, let's say under limit braking, we have an added weight tranfser shift towards the front of 265 lbs. That is now a total downforce of 1,915 lbs up front, and 1,085 lbs in the rear. Under braking, weight distribution is now 64/36.

Under our limit braking example, the front tires will have a total of 1,915 lbs of downforce. Multiply this times the coefficient of friction of a good sticky street tire of 1.3 and you have a total of 2,490 lbs of force (1,245 lbs for each tire). The rear tires contribute 1,085 lbs x 1.3 for a total of 1,410 lbs of force (705 lbs for each tire). For all four tires to be at or near their braking limit, this also means that the brake torque applied by each brake assembly must be the same as the tire forces above. So, from the arbitrary starting point in the table above, we actually need to increase pedal pressure to acheive 1,245 lbs of force at the front brakes, and 705 lbs at the rear brakes.

Now, be careful to remember that this amount of force could be generated by any size rotor, and any brake pad. The pedal pressure, master cylinder, and caliper could all be changed to account for various rotors & pads to generate this braking force. At this point, it really doesn't matter what the brake design is -- they can all lock up a tire. So, be careful not to get hung up on the fact that our sample table above shows different available brake torques. These are not maximum values -- they are sample values at a given brake pedal pressure (and other factors as mentioned).

If we take either the stock or the example four wheel modified brake system (both have 64/36 force ratios), and increase pedal pressure to generate 1,245 lbs of force up front, this will result in a force of 711 lbs in the rear (we used the four-wheel modifed brake system for the numbers). This is a simple ratio calculation using (1,245 / 819) * 468. That's less than 1% off -- almost perfect, and close enough for real world use. In a racing setup, this kind of small difference could be dialed out with a front / rear bias valve. Using either of these brake systems, the front and rear tires would be very close to their maximum potential at the same time under limit braking.

What happens when we use the front-only modified setup with a larger rotor and race pads? Again, the pedal pressure is increased to generate 1,245 lbs of front brake force. However, the rear brake force only generates 429 lbs of force at that pedal pressure ((1,245 / 819) * 282). For the rear tires to be at their braking limit, 705 lbs are needed. With only 429 lbs available, the rear tires are nowhere near their maximum braking potential. The rear tires are contributing far less to the overall braking performance than they could be. So with big front brakes, the rear brakes are not needed less, they are used less! Under racing limit braking, the big fancy brake upgrade may actually increase stopping distances! Brakes don't stop the car -- tires do. If you're not using 100% of all four tires, stopping performance suffers.

Available Brake Force, 705 lbs needed at Rear to maximize braking
Brake System Front tire force
at limit braking
Available rear tire
braking force
stopping force
Stock 1,245 lbs 698 lbs 1,943 lbs
Front Pads 1,245 lbs 511 lbs 1,756 lbs
Front Rotor and Pads 1,245 lbs 428 lbs 1,673 lbs
All Four Rotors and Pads 1,245 lbs 711 lbs 1,956 lbs

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