I was lucky enough to have only one exam the week of the 12 Hours of Sebring. I took the exam, got some sleep, and drove almost 12 hours myself just to get there. I had an excellent experience, and wanted to be sure to write a few articles detailing some basic Engineering involving the cars there. Add that to my series I’ve wanted to make using the Machinery’s Handbook, and you get this.

One nice thing about being there was seeing the corners, and the distance markers, and being able to see and hear when the cars begin braking. Seeing the brakes light up red at Sebring’s hairpin at night got me thinking about how much force actually goes into those brakes. Well, why not take a stab at it, then hopefully google an answer to see how close I can get?

This will probably involve just a few simple equations. I’ll jumble them right here before I try tackling this specific example and list assumptions.

$F = m * a$

$F = \mu * N$

$M_o = F * r$

EXAMPLE

Oreca 03 LMP car with Brembo brakes

Let’s do this for the Oreca 03 (LMP) I got some pictures of at Sebring 2012. I want to estimate how much normal force the brakes must produce to stop the car. Note, I am trying to totally guess how to do this before looking up if there are common equations used.

First, lets list some assumptions:

Tire Dia = .656m (I found the size of the tires)
Dia of Brake Rotor = .38m (Partial guess)
Coefficient of Friction = .16 (Assuming this is kinetic coef [Machinery Handbook])
No tire slip/deformation
Assume 3 g’s in braking (guess)
Assume all 4 tires produce equal braking force (Won’t happen in real life, I may account for this later)
ASSUME CONSTANT ACCELERATION. Obviously at high speed braking is better than low speed.
Vehicle Mass = 900Kg
Ignore angular momentum of tires/wheels
Ignore air drag acting to decelerate the vehicle

$F_{tires} = m * a = 900[Kg] * 3*9.81[\frac{m}{s^2}] = 26,487 [N] (6,622/wheel)$

$M_o = F_{tire} * r_{tire} = 6,622[N] * \frac{.656}{2}[m] = 2,172[Nm]$

$2,172 = F_{Brake} * \frac{.38}{2}[m]$

$F_{Brake} = 11,432 = \mu * N = .16 * N_{Caliper}$

$N_{Caliper} = 71,450[N]$

To me, that sounds too high. Let’s check the values by finding comparable products online with good data.

StopTech says that a high brake line pressure is 2000psi which is about 13.8MPa. According to Brembo the P/N XA8.Z4.01/04 which is recommended for single seaters, has a Piston Area of 55.35 square centimeters. The normal force, which is aparently also known as the clamping force, should now be found by:

$N_{Caliper} = P * A = 13.8[MPa] * .005535[m^2] = 76,383[N]$

This shows my guesses were within a reasonable range. However, I found a mistake I made. It is unreasonable to think that the clamping force of the brake pad is exerted right on the outside diameter of the brake rotor. Let’s assume the force is exerted 25mm inboard of the outside diameter of the brake rotor.

$2,172 = F_{Brake} * (\frac{.38}{2}-.025)[m]$

$F_{Brake} = 13,164 = \mu * N = .16 * N_{Caliper}$

$N_{Caliper} = 82,275[N]$

This demonstrates why it is desireable to have the largest diameter brake rotors that can fit within the wheel. Larger diameters can exert more braking force per unit of clamping force. Also note, by assuming 25mm offset from the outside diameter of the rotor, I was assuming a 50mm brake pad from the Brembo Catalog linked above. A smaller brake pad would have the benefit of exerting the distributed load farther out on the rotor diameter. However, this would concentrate the heat generated in a smaller area. So a tradeoff may have been found. In tinkering with very basic assumptions and equations we “discovered” some things that are likely common knowledge to auto mechanics.

Unfortunately after googling I have been unable to find a good Clamping Force value.

Values of interest to help reduce assumptions and increase accuracy of my calculations would be:
Braking G Force. If that can’t be found, the velocity at the 300m and 200m marker would give the acceleration (assuming it is constant), 400m to 300m would be even better since the acceleration will be greater due to the higher downforce of the car.
Documented brake line pressure for high braking LMPs.
Not to mention, eliminating my assumption that all 4 tires would definately change the results. In that case, the front brake calipers would have to provide a HIGHER clamping force in order to stop the vehicle at 3 G’s.

Sources Machinerys Handbook Pg. 158
Hard Carbon on Carbon Coeff Friction Unlubircated .16

In the near future I’d like to compare the way I guessed how to do this to the website StopTech White Papers that had the brake line pressure I used. In addition I ‘d like to remove the assumption that all 4 tires are exerting the same braking force.

The main purpose of this was to show how quickly practical answers can be found. When I compare this to established methods, we will see how practical, or unpractical, the results of all my guesses were. Another improvement would be to integrate the moment created over the surface of the brake pad assuming a uniformly distributed load.

Hope this was interesting, and I really suggest you check out the Stop Tech site. At first glance it seems pretty good.

Machinery’s Handbook Part 3

Sebring 2012 Brakes Part 2