BMWs (Finally) Get Brembo Brakes OEM...

[FredK]

You have an airstrip.

Car 1: 1000 kg, traveling at 20 m/s
Car 2: 2000 kg, traveling at 20 m/s

Both have identical braking systems--rotor diameter, thermal mass, pad area, compound, brake system hydraulic pressure, wheel geometry and cooling, everything.

They both hit the brakes at the same time.

Which brake rotor will have a higher temperature?

[bmw_e30]

I won't even get invloved in this discussion.

But many things are being said that are wrong.


But here is a general statement that is true in all cases:

Heavy cars are harder on brakes, tires and suspension points.

[brandondan1]

You have a bull doser traveling at 10mph.

You have a toy car traveling at 10mph.

Which one is going to take less force to stop? The toy car of course, because it weighs less.

How many times do you think you can bring the bull doser to a stop before you get worn out? 1, 2 times?
How many times can you bring the toy car to a halt? All day probably.

So, if there is no more heat being generated in a heavier car, why does braking performance deteriorate at a faster rate?

[Jesse30]

You have a bull doser traveling at 10mph.

You have a toy car traveling at 10mph.

Which one is going to take less force to stop? The toy car of course, because it weighs less.

How many times do you think you can bring the bull doser to a stop before you get worn out? 1, 2 times?
How many times can you bring the toy car to a halt? All day probably.

So, if there is no more heat being generated in a heavier car, why does braking performance deteriorate at a faster rate?

you obviously werent able to work with the equations provided.

mathematically you are wrong. not only that, the brakes on the bulldozer and the toy car are going to be of different size obviously..

there are a lot of factors to take into consideration. :)

[1991 318is]

I have worked on cars in two series that had V6 and V8 cars. ASA/All-Pro Stock cars and SCCA TransAM. In both series, the cubic inch to weight rule was adjusted. The V6 cars had additional advantages: better polar moment due to engine setback, the brakes performed better and lasted longer, and tire wear was better. End result - more weight added to the V6 cars.

[brandondan1]

Actually, you failed to see the simple point I was trying to make. I was just looking at the amount of force generated by each object. To stop a greater force you will need more energy..

Obviously the brakes on each object aren't going to be the same. But, hypothetically if they were, we know which one would have greater temp and therefore a higher rate of deterioration. I didn't bother to use any equations, because it just 'seems' logical.

The initial spark that ignited this debate was just a remark stating that the 135 would be a better performer if it had less weight (which is generally true with any car). On the track, being able to stop faster is a great advantage, and will lead to faster times. At the end of the day, isn't that what matters?

[brandondan1]

I am not talking about braking performance. Obviously a lighter car will perform better in all ways at a given power/brake level.

I'm talking about brake temperatures. All brakes do is turn one type of energy into another. Where does a heavier car get more energy to heat its brakes more?

If a lighter car is able to perform better (we do mean stop faster, right?) when slamming on the brakes, compared to a heavier car slamming on the brakes, then is it not true that the lighter car will be able to press on the brake pedal 'halfway' (reducing the amount of friction/heat) and be able to stop at the same rate as the heavier car that's still stomping on his brakes(more friction/heat)?

Sorry, I'm just trying to learn.

[matt]

You have an airstrip.

Car 1: 1000 kg, traveling at 20 m/s
Car 2: 2000 kg, traveling at 20 m/s

Both have identical braking systems--rotor diameter, thermal mass, pad area, compound, brake system hydraulic pressure, wheel geometry and cooling, everything.

They both hit the brakes at the same time.

Which brake rotor will have a higher temperature?

We don't have an airstrip... we have a racetrack.

In your example: Obviously car two would have more kinetic energy being heavier and having the save velocity... but where would it get that extra energy? It's motor!

If a lighter car is able to perform better (we do mean stop faster, right?) when slamming on the brakes, compared to a heavier car slamming on the brakes, then is it not true that the lighter car will be able to press on the brake pedal 'halfway' (reducing the amount of friction/heat) and be able to stop at the same rate as the heavier car that's still stomping on his brakes(more friction/heat)?

Sorry, I'm just trying to learn.

Yes, a lighter car will stop faster and in a shorter distance than a heavier car, all else the same (speed, brakes, driver, etc.)

But the amount of heat that brakes have to dissapate depends only on the amount of power the engine produces. To break it down once more: A car at rest with a full tank of gas has chemical energy. The motor converts that chemical energy into kinetic energy. The brakes convert that kinetic energy into heat.

The weight of the car provides no energy. :giggle:

[matt]

You have a bull doser traveling at 10mph.

You have a toy car traveling at 10mph.

Which one is going to take less force to stop? The toy car of course, because it weighs less.

How many times do you think you can bring the bull doser to a stop before you get worn out? 1, 2 times?
How many times can you bring the toy car to a halt? All day probably.

So, if there is no more heat being generated in a heavier car, why does braking performance deteriorate at a faster rate?

The bull dozer is going to take a shit ton more energy to get up to 10mph no? Then its the extra energy that makes it harder to stop, not the mass.

Another way to think about it... if you push equally hard on the bulldozer and on the toy car, then run in front to stop each of them, they will both stop just as easily.

Shit people, this is highschool physics.

The key thing that I think EVERYONE is missing here is that a track is essentially a closed system. No extra energy can be added, other than from the motor in the car. So, the heavier car WILL NOT be going the same speed as the lighter car given that they have the same motor.

[FredK]

In your example: Obviously car two would have more kinetic energy being heavier and having the same velocity... but where would it get that extra energy? It's motor!

Of course. By virtue of its power, it can enjoy the same velocity as a lighter car approaching a turn.

I'm talking about brake temperatures. All brakes do is turn one type of energy into another. Where does a heavier car get more energy to heat its brakes more?

.

[matt]

Oh, so you're on my side? LOL. My bad.

[FredK]

Let's update the example a little:

Car 1: 1000 kg, traveling at 20 m/s, 500 hp
Car 2: 2000 kg, traveling at 20 m/s, 1200 hp

Assume both engines have the same efficiency. Do both cars consume the same amount of fuel while racing?

[Emre]

In your example: Obviously car two would have more kinetic energy being heavier and having the save velocity... but where would it get that extra energy? It's motor!

I really don't understand the point you're trying so hard to make. Of course a car derives it's kinetic energy from its engine (or "motor" as you incorrectly call it). Where else would it come from?

Sure, a heavier car needs more power to reach the same speed as a lighter car. Duh. But that's not the question.

The question is: once both cars (heavier and lighter) are at a given speed, which one needs more force to stop?

Clearly, a heavier car going around the track at a higher speed will have greater demands on it's brakes. That's why they need to be bigger.

What exactly is your point?

Emre

[matt]

I was very precise with my terms:

On track, the weight of a car is irrelevant to how hard it is on brakes. High powered cars are hard on brakes.

A car on track, being a closed system, is a great example of the conservation of energy. No matter how heavy the car is, its brakes only need to dissapate as much energy as the engine can produce.

The question is: once both cars (heavier and lighter) are at a given speed, which one needs more force to stop?

That is obvious, and irrelevant to the statement I made.

Clearly, a heavier car going around the track at a higher speed will have greater demands on it's brakes. That's why they need to be bigger.

But a heavier car, with the same power (engine or motor) and the same driver: A) will not be going as fast B) will have the same demands on its brakes.

What exactly is your point?

My point is to educate the masses. Ignorance may be bliss, but knowlege is power.

[brandondan1]

That is exactly what I'm thinking, Emre. All he seemed to mention was that a bull doser takes more energy to move, and that's why it's harder to stop. That is a given, but why do you exclude mass when it is the exact thing that causes it to need excess energy in the first place? To me it just sounds like you're rewording it to fit your argument.

If a lighter car has more stopping power at its limit than a heavier car, then essentially, a lighter car could stop at the same rate as the heavier car using less braking force. Does stomping on the brakes not wear them out faster than pressing lighter?

I am looking from the same perspective as emre, but you seem to have added that the cars are no longer going the same speed which they wouldn't in a real world situation. I was basing everything I said earlier off of them going the same speed. The heavier car would have slower entry before the braking points, which would require less stress on the brakes to get it at optimal cornering speed...or not if the extra weight is in the wrong places and causes the car to need to go even slower. But this all equates to the car being slower anyway, so it's dumb to argue that heavier is better.

However, if the heavier car had a engine that allowed it to accelerate the same as the lighter car, the lighter car would still have a braking advantage, no? Are you saying that if you had a choice between at 2700lb 135 and 3000lb with a few more horsies, you would pick the latter? Basically, I think Erme's original point was that if you want a heavier car to be COMPETITIVE with a lighter one, it's going to require more stress on a lot of things, including the brakes.