Lets calculate the horsepower of our brakes

[ciper]

We have enough rough data to figure it out, I just cant find the correct formula to do it.

While searching I did find a really good site with gems like
"If you multiply horsepower by the proper conversion factor, you discover that one horsepower generates 42.4 BTUs of heat per minute. If stopping a 4,000 lb. vehicle from 60 mph in roughly 150 feet requires 600 horsepower of force, it’s the equivalent of 25,440 BTUs of heat "

http://www.babcox.com/editorial/bf/bf10312.htm

I especially like how many aftermarket brake manufactures claim the stock brakes are too thin between the rotor surface and hub, when in fact its INTENTIONALLY as thin as possible! Look up "HEAT DAM" on that site.

[vrg3]

Well, let's try a very rough approximation using stock brakes as per the C&D preview of the 1990 LS AWD sedan:

http://www.legacycentral.org/articles/lsawd/lsawd_1.htm

The numbers I'm taking are 2982 lbs weight and 184 feet to stop from 70 mph to 0 mph.

I will make the rough approximation that velocity fell linearly from 70 to 0 (my gut says that's not too far off since the brakes get hot towards the end but their job gets easier), so:

average velocity = 70mph/2 = 35mph * 5280 ft/mi / 3600 s/hr = 51.3 ft/s
braking time = 184 ft / 51.3 ft/s = 3.6 s
2982 lbs * 2.2 lbs/kg = 6560.4 kg
70mph * 1.6 km/mi * 1000 m/km / 3600 s/hr = 31.1 m/s
KE = 0.5 * m * v^2 = 0.5 * 6560.4 kg * ( 31.1 m/s )^2 = 3172642.2 J
since KE at rest is 0, W = 3172642.2 J
P = W / T = 3172642.2 J / 3.6 s = 881289.5 W / 745 hp/W = 1182.9 hp

Hm. That number differs by an order of magnitude from the quote in the article... then again the article also says "600 horsepower of force" which is incorrect.

But this uses basic introductory kinematics that I learned back in 11th grade, so maybe there's a more intelligent way to do this analysis.

[rallysam]

VRG's calculation is very good. I think the assumption that the velocity falls linearly is a good one - because the limit of tire adhesion should be pretty constant.

Here's one part I take issue with. You have calculated the average power dissapation over the entire braking process. In reality, the power dissipation will be greater at higher speeds, and lower at lower speeds (even though the velocity changes linearly, the power dissipation does not!). Think about it this way, it takes much more work to slow the car from 70-60 than it does to slow the car from 10-0.

I think this is the easiest way to find the instantaneous power output:
like you said,
KE = 0.5*M*V^2
now differentiate this to find the rate at which KE is changing (because that is the rate at which power is being dissipated at that instant):
P = dKE / dT = 2*0.5 * M * V * dV/dT
P = M * V * dV/dT
P = M * V * A
but what is A - the deceleration? it's 31.1m/s divided by the 3.6 seconds it takes to stop: A = 8.6m/s^2
Substitute that in:
P = 6560.4 kg * 31.1m/s * 8.6m/s^2
P = 1,754,644 Watts = 2355 hp

It makes sense that that number I calculated is twice the number calculated by VRG, because the horsepower being dissipated will start at my number the instant you hit the brakes, then drops linearly to zero as you come to a stop. So VRG calculated the average over that time which would be half as much as the max.

BUT this raises the fundamental issue: you can generate even more brake power at top speed. In fact, you can generate just about any horsepower with any brake pad if you are going fast - just not for long. So although it is interesting, horsepower is not really any measure of performance and thus you never hear about it.

Ciper and VRG, I know you were just curious and you already know this so don't mind my lecturing. The real limit to braking ability is NOT how much horsepower it can grab, but how long can it do it without overheating and fading. So the relevent quesitons are how much power can it dissipate in the long run? This translates to: How much power can your rotors dissipate to keep temperatures low? How hot can the pad operate without fading?

If you are actually looking at how much power you could dissipate out of your rotors in the long run, I have a feeling it would actually be less than the 100 horsepower. The evidence of that is that you get brake fade on a road coarse with street brake pads. In other words, the rotors are actually unable to dissipate the amount of energy your engine is creating as you race around the coarse.

[vrg3]

Good point, Sam. Thanks for the explanation. I agree. The faster you go, the more power it will take to slow you down, but as you slow, the power required to continue slowing you decreases. Wow, that was a mouthful.

I think ciper was just trying to have some fun with the concept of horsepower in our cars coming from something other than the engine. :)

It kind of reminds me of a discussion my brother and I had about drag racing. He was telling me about a Supra that sheared its driveshaft through sheer torque when attempting to launch at a drag strip. He was saying how he couldn't believe that the 2JZ-GTE was making enough torque to do that. I pointed out that the instantaneous torque when the driver dumped the clutch was probably a lot bigger than the motor's peak torque, since the clutch was generating torque by accelerating the transmission shaft up to the engine's speed. That was my take on it, anyway.

[rallysam]

I pointed out that the instantaneous torque when the driver dumped the clutch was probably a lot bigger than the motor's peak torque, since the clutch was generating torque by accelerating the transmission shaft up to the engine's speed. That was my take on it, anyway.

I think you are right about that too. There is all the torque generated by the engine, PLUS additional torque generated as you dump all that momentum built up in the engine and flywheel into the rest of the car.

[evolutionmovement]

You're also working against the static rear drive of the car. All the weight and resistance of the diff and sticky tires that want to be moved in a very short span of time.

Steve

[vrg3]

Yeah, you're right, Steve. It's actually the torque coming through the transmission working against the torque coming through the differential that broke the driveshaft. Each is as responsible as the other.

[polychrome]

the preceeding calculations computed the weight in kg incorrectly.

kg = lb / 2.2

the weight should have been 1355.5 kg

the computed kinetic energy and power are overstated by a factor of 4.84

[ciper]

With similar brake setups I can notice the increase in time it takes to fade the brakes comparing the stock legacy NA setup with the replacement WRX components.

Yeah, I was just curious how much HP the brakes could generate. If your brakes are stock but in good working order I bet the stopping distance could be improved more with tires than pad compound.

[NemesisEJ22t]

Also, about the time to stop, 3.6 seconds assumes that the car is not decelerating during the 3.6 sec, it is just the time that the car would take to travel 184 feet at 51.3 ft/s. In order to figure out the time, you need to figure out the acc. This is done with the equation V^2 = Vo^2 + 2 a (x-xo). This gives an acceleration of -7.15 ft/s^2. Then use the equation: V=Vo + a t and find the time = 7.17 sec. This is the time that the car takes to stop from 70 - 0 mph in 184 feet.

[vrg3]

51.3 feet per second was the average velocity assuming a constant acceleration.

If you use the correct Vo (102.6 ft/s), your math comes out the same as mine.

[IronMonkeyL255]

Using the correct conversion for lb > kg and plugging that in to the rest of the calculations, I came up with 244.4089 HP.

[vrg3]

Oops, I multiplied there when I should have divided, didn't I?

[IronMonkeyL255]

Yup.

Don't worry. I didn't catch it until I read polychrome's post......

[NemesisEJ22t]

Oh, so THATS why i didn't do well in Physics 1. Oh well, now that i look at it again, i'm glad that it only takes 3.6s to stop.