Brakes: Does size really matter?

D

Doc Bro

New member
I've thought myself into a connundrum which I'm sure has a really easy explanation that I'm overlooking.

Why does the size of the rotor matter so much for IT discussion and classification?

Aren't the brakes ability to quickly and efficiently stop the car a function of available traction of the tires. (ie threshold braking). If two cars are running the same Hoosier rs304's, and have similar weights, won't both lock (skid) at similar thresholds regardless of swept area size?

Sorry for being brain dead on this one!
R
 
Larger rotors equals more surface area. More surface area equals more heat transfer to the environment.

Assuming the brakes are cool enough to lock the wheel larger rotors don't help, but that's a pretty big assumption.
 
For qualifying you may have a case but I do not think there are many IT cars out there than can use the brakes full on every turn for 20 laps. Too much heat. That is where your big brakes excel.
 
Originally posted by dickita15@Jan 6 2006, 02:09 PM
...I do not think there are many IT cars out there than can use the brakes full on every turn for 20 laps.
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:figo:
 
I still don't get it totally.

Big brakes=more heat transfer. Agressive compounds=more friction=more heat. Yes? Why are agressive compounds so valuable then? If they only increase heat production quicker then wouldn't that result in quicker onset of fade?


But, is the weak link of the system really heat? or traction?

Another senario.

Big breaks with a firmer compound tire.

Is that better or worse than:

Smaller breaks with a stickier tire.

Which would offer better braking performance.

(For this discussion- tires are for stopping purposes only).

R
 
Think of it this way. Braking is all about dissipating energy. That means turning forward motion into heat. That heat has to be gotten rid of in some way or things bend/break/melt etc. Larger rotors shed heat faster therefore offer stopping power after many cycles. If you use a less aggresive pad there is less heat produced, but remember you are turning motion into heat, so less heat produced means less actual braking.

As far as your example above, the stickier tire theroetically offers more stopping power in a one time application no matter what size brakes are acting on it, assuming the brakes are big enough to lock the wheel. In practice things like brake modulation, caliper design and a variety of other factors come into play.

I hate to resort to a NASCAR analogy but think of this, at Daytona they run little tiny brakes because the unsprung weight is low and they only need to brake hard every 20-40 laps for pitstops or when creating multimillion dollar scrap heaps. The little brakes are enough to do the job when there is plenty of time to cool. But when they go to Watkins or Martinsville they put on brakes that are larger than some stock ITC wheels. Why? You need mass to shed heat.
 
The second part of the equation is that bigger rotors have more lever arm with which to stop the tire...therefore, they require less energy to do the same amount of work. Chuck
 
Makes you wonder why some guys don't run brake coolers.
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Because depending on the car, the track, how fast of a driver, it may not be necessary. For example, at LRP which is very easy on brakes it is not necessary for me to use brake ducts. Watkins Glen has that long straight and again is not necessary for my car. Summit Point, again not necessary. NHIS - this is the only reason I have them. Guess it can't hurt though; besides, I do like my aluminum soup cans.
 
Originally posted by chuck baader@Jan 6 2006, 04:54 PM
The second part of the equation is that bigger rotors have more lever arm with which to stop the tire...therefore, they require less energy to do the same amount of work. Chuck
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I think I have to take issue with this one Chuck. The physics definitions of work and energy in this context are the same. If you reduce a car from 80 mph to 40 mph you have dissipated the same amount of energy no matter what size brake or tire you used. Now, if you can brake more quickly because you have big brakes and sticky tires the instantaneous power dissipation will be higher, but the integrated energy dissipated will be the same. I think the advantage of the larger brakes' lever arm would be that the pads and rotors will be larger and therefore be both a larger heat reservoir and also have more surface area to dissipate the heat, thereby allowing a lower peak temperature and better cooling between corners, etc...
 
The tire limits the traction, but can't big brakes reach that traction limit faster than smaller brakes? If I'm going slow enough then the small and big brakes could lock the tires at the same time. If I'm going fast, though, the bigger brakes could dissipate energy faster and reach the traction limit faster than the small brakes thus reducing braking distance. Does this makes sense or am I missing something?

David
 
Sorry for the oversimplification. Yes, you are doing the same amount of work, therefore, you use the same amount of energy, however, let's just say a longer lever arm has to exert less force to do the same amount of work.
 
Originally posted by chuck baader@Jan 6 2006, 07:38 PM
Sorry for the oversimplification. Yes, you are doing the same amount of work, therefore, you use the same amount of energy, however, let's just say a longer lever arm has to exert less force to do the same amount of work.
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OK, now I completely agree with you. But unless you don't have enough power boost on your brakes or a strong enough leg then simply having more mechanical advantage in the brake system doesn't help you at all...
 
Originally posted by evanwebb@Jan 8 2006, 11:03 PM
OK, now I completely agree with you.  But unless you don't have enough power boost on your brakes or a strong enough leg then simply having more mechanical advantage in the brake system doesn't help you at all...
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Probably not applicable to IT, but to reinforce the concept of heat dissapation, take a look at the brake rotors on high end formula and sports racing cars. The outer edge is wavey, not round - equals more surface area on the edge of the rotor and I would emagine a different air flow over the surface.
 
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