In my previouse post on Sebring Race Car LMP Brembo Brake Mathematical Analysis I tried to guess how to find the clamping force necessary to make an Oreca 03 LMP car stop at 3G’s. My only googling method I could find at the time was somewhat sketchy at best. I would now like to throw out another guess, and see if this method will also work. In this case, I will try an energy method.

A quick back story:

I am also designing a hovering airplane, similar to an Ekranoplan (YouTube this, Example). I invented an equation that looked completely wrong, but no matter how I checked it, the equation checked out. Finally I brought it to a professor. I had made the equation using a system similar to my last post. What he did was he used an energy method to tackle the same problem, and see if the results came out the same. When it did, he assured me that despite how unorthodox the equation I made was, it was indeed correct.

Let’s try applying the energy method to the braking example from my previous post.

First, I’ll guess what equations I’ll need and see if that works out.

$E = \frac{1}{2}*m*v^2$

$W = F*d$

$W = M_o * \theta$

EXAMPLE

Oreca 03 LMP car with Brembo brakes

Based on the first equation, I need to know the velocities if I want to find the reduction in kinetic energy necessary. If you don’t know how to derive this, let me know and I can detail it. I want the vehicle to stop at 3 G’s for 100m with a starting velocity of 80m/s which I am guessing is a reasonable LMP speed entering the Sebring hairpin. For a more complete list of assumptions see my previouse post.

$100 = \frac{1}{2}*-3*9.81*t^2 + 80t$

$V_2 = -3*9.81*t + 80$

$t = 1.95[s]$

$V_2 = 22.7[\frac{m}{s}]$

$\Delta E = \frac{1}{2}*900*80^2 - \frac{1}{2}*900*22.7^2 = 2.65[MJ]$

$E = F*d = 2.65[MJ] = 100[m] * F$

$F = 26,481 / 4 = 6,620[N] PerWheel$

That is the first quick check. Before, using forces instead of energy, I found the stopping force of each tire needed to be 6,622[N]. Looks like we are on the right track so far.

Now I want to find how much kinetic energy the brakes must transform into heat (assuming 100% efficiency). To do this, I’d prefer to find the moment the brakes must produce, and to find that I need to know how many times the tires will rotate so I know how many degrees that moment acts on to compare it to the kinetic energy that must be dissipated.

$d = r * \theta = 100[m] = \frac{.656}{2} * \theta$

$\theta = 304.9[radians]$

Another way:

$1 Tire Rotation = 2 * \pi * r = 2 * \pi * \frac{.656}{2} = 2.061[m]$

$100[m]/2.061[m]= 48.52$

$48.52 rotations * 2 \pi = 304.9[radians]$

Looks like that checks out. Not lets find how stong a moment the brakes most produce.

$E = M_o*\theta = 2.65[MJ] = M_o * 304.9$

$M_o = 8685 / 4 = 2,171[Nm] per wheel$

Compare this to 2,172[Nm] per wheel on the previouse post.

The rest of this will be the same as before.

$M = F * d = 2,172 = F * (\frac{.38}{2} - .025)$

$F = 13,159[N] = \mu * N = .16 * N$

$N = ClampingForce = 82,244[N]$

Compare this to 82,275[N] from the other method and it can be seen that they produce the same result.

Why are the numbers not exactly the same you may ask? Throughout this example I have been using at most 4 significant figures, which I’m sure you remember from high school chemistry. Had I carried every decimal place, which I don’t care to type in here, the results would have been the same.

This shows that my previous answer stands a good chance of being pretty good. Why then, is my answer above that predicted by the high brake line pressure times the piston area?

Well there are several reasons.First, I assumed the car used that caliper, and that the cars max brake line pressure was 2000[psi]. Also, I made many simplifying assumptions. First, that the tire diameter when on the road is actually .656m, and that the brake rotor dia was .38m. More importantly however, I assumed that the coefficient of friction was actually .16 as the Machinery’s Handbook said. This does not take in to account the fact that racing brakes perform better as they heat up over time. Likely the coefficient of friction INCREASES above the .16 value used, which would reduce the clamping force necessary to stop the 900[kg] car at 3G’s.

Sources Machinerys Handbook Pg. 158
Hard Carbon on Carbon Coeff Friction Unlubircated .16
Acceleration, Energy, and all that can be found starting at page 167.

So ways to improve beyond this article would be to find the actual equations others use, to estimate the efficiency of the brakes, to account for weight transfer during braking, and to have better data instead of having to guess for so many values. Hopefully I can do at least some of these in the near future.