BMWs (Finally) Get Brembo Brakes OEM...

[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?

[matt]

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

[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]

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.

[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:

[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.

[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?

[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.

[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. :)

[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?

[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.

[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?

[matt]

Bingo. Even though they weigh different amounts, the kinetic energy of the car is always equal to the amount of energy put in minus the amount of energy taken out.

[FredK]

neither

[matt]

Now do the same amount of work to both of them... which has more kinetic energy then?