Cancelling effects on a larger brake rotor

[Tachyon son]

Hello all,

I am struggling to find out what exactly makes to have very little advantage in terms of cold stopping distance when you fit larger brakes on a car.

Important note, the discussion is only regarding the braking power in the very first stop, before the heat starts to become a problem for the smaller brakes. For the biggers brakes, the advantage of coping better with thermal stress is very direct and obvious when hard driving.

Part of my job is teaching how to drive fast around tracks for the past 20 years, and thanks to this I have the direct experience about the behavior of many types of brakes fitted in many types of cars.
Besides brake pedal feeling, heat resistance and overall performance, the empirical truth is that when you arrange braking tests focused only in raw stopping distance that starts with cold brakes, the difference in stopping distance between stock size brakes and big brakes is very similar. (Same car, same weight, same tyres, same pads)
When I say "very similar" I am referring to an advantage as good as of 1 or 2 meters for the bigger brakes in test up to 100 km/h.
At higher speeds, more than 200 km/h for example, we have not measured the distance in detail like in the low speed tests, but thanks to several track visual references we know that the advantage is still not a big deal.

The experience is the same for others racing drivers I work with.

More data that supports this behaviour: Zeperfs.com
This web is an excellent database that gathers real results of many telemetry-guided car tests.
You will find, for example, this results between a small cheap car (Seat Ibiza 1.0 115hp) and a supercar (Ferrari F430 490hp):
From 100 km/h to 0: 34,2 metres for the Seat VS 34,1 metres for the Ferrari
From 130 km/h to 0: 58,1m VS 57,6m

I know tyre grip capacity plays one main role here. Because once you reach ABS kicking point there almost no gain in cold stopping distance.
And modern brake pumps, even the ones fitted on small cheap cars, are extremely strong when generating clamping force.

Another important key factor is that the friction force is not dependant on brakes surface area. (F = μN)

From this point my question arise, because there must be some physical property on bigger rotors that somehow cancels the advantage of applying the force farther away from the axis point. (Center of the wheel in this case)

Two cons of bigger rotors I consider for the cancelling explanation:
- The bigger the rotor, the heavier. (Generally speaking)
- Increased rotational speed on the outer part of the rotor.

Is this correct? What am I missing in the whole picture?

Is also surprising when you realize that a fixed caliper is not performing better than a floating one in this topic.
Many thanks!

 

[Baluncore]

I know tyre grip capacity plays one main role here. Because once you reach ABS kicking point there almost no gain in cold stopping distance.

I would think that tyre grip explains it all. Why do you think it is otherwise ?

 

[Tachyon son]

I would think that tyre grip explains it all. Why do you think it is otherwise ?

The heavier mass of the bigger rotor requires more energy to be stopped.
I suppose the higher rotational speed of the bigger rotor requires more energy to be stopped.

 

[Baluncore]

The linear kinetic energy of the vehicle, including the brakes, must be decelerated to zero.
No more brake force is needed than that limited by tyre grip.

The rotational energy of a heavy rotor must also be decelerated to zero. The additional braking force needed to stop rotation of the heavy rotors is provided by the bigger rotors. But that does not require any tyre grip since the rotation is relative to the vehicle frame.

 

[Spinnor]

As long as the smaller rotor system can generate enough torque to lock up the wheels then minimum stopping distance will be similar to the car with a larger rotor system, other factors being similar, as long as the smaller rotor system stays below the temperature where the brakes begin to fade? There are advantages to larger rotor systems but that was not your question.

 

[Tachyon son]

As long as the smaller rotor system can generate enough torque to lock up the wheels then minimum stopping distance will be similar to the car with a larger rotor system, other factors being similar, as long as the smaller rotor system stays below the temperature where the brakes begin to fade? There are advantages to larger rotor systems but that was not your question.

Ok, very good recap.

So, if at the end stopping distance is all about how fast can brakes generate enough torque to lock up the wheels, then is it correct to derive that the higher the speed the more advantage to bigger rotors? (The threshold before the ABS kick in is bigger at higher speeds, you notice this very well when race driving a car without ABS)

Do actually bigger rotors really have more clamping force to lock up the wheels?

 

[Lnewqban]

Hello all,

I am struggling to find out what exactly makes to have very little advantage in terms of cold stopping distance when you fit larger brakes on a car.

Please, see:
https://ebcbrakes.com/how-to-choose-the-best-big-brake-kit/

https://www.brembo.com/en/car/sporting-use/discs

https://www.brembo.com/en/car/formula-1/f1-infographics

 

[jack action]

The force that decelerates a vehicle is the friction force at the tire-road contact patch, $F_b$. The maximum force $F_{b\ max}$ a tire can do cannot be surpassed, no matter how the brake system works. The corresponding maximum wheel braking torque $T_{b\ max}$ will be $F_{b\ max} r$, where $r$ is the tire radius. Any braking system that can produce that torque will be sufficient to create the maximum deceleration possible. With friction brake systems, the question is always: how long can the brake system maintain that force?

With a disk brake system, the torque produced is $F_c r_d$, where $F_c$ is the caliper friction force and $r_d$ is the disk radius (where the force is applied, that is).

What does it means to have larger rotors? You need a smaller caliper force, hence a smaller caliper should do the job, i.e. less force means less stress on the part. But this smaller caliper must be able to handle the same amount of braking energy converted to heat. So the 'smaller' caliper may no be as small as one can think. You couldn't use a bicycle brake system on a semi; even if it could create the necessary force, it wouldn't last long before the heat destroyed it.

So, if at the end stopping distance is all about how fast can brakes generate enough torque to lock up the wheels, then is it correct to derive that the higher the speed the more advantage to bigger rotors?

It is not about how fast it generates torque, it is about how fast it can remove the heat coming from the kinetic energy. The heat build up destroys the friction performance.

Do actually bigger rotors really have more clamping force to lock up the wheels?

No. But high performance calipers usually have a better force distribution along the brake pads. It doesn't create a higher force, but it disperses the heat better, thus less chance of overheating.

 

[Tachyon son]

No. But high performance calipers usually have a better force distribution along the brake pads. It doesn't create a higher force, but it disperses the heat better, thus less chance of overheating.

Why not? Because the force is applied further away from the axis on bigger rotors. With your hand is much easier to stop a spinning bicycle wheel from the outer side (tyre) than from near the hub.

 

[russ_watters]

When I say "very similar" I am referring to an advantage as good as of 1 or 2 meters for the bigger brakes in test up to 100 km/h.
At higher speeds, more than 200 km/h for example, we have not measured the distance in detail like in the low speed tests, but thanks to several track visual references we know that the advantage is still not a big deal.

It is not about how fast it generates torque, it is about how fast it can remove the heat coming from the kinetic energy.

It's worth noting that the kinetic energy dissipation is a square function of speed. A set of brakes decelerating from 200 km/hr dissipates 4x as much heat (power) as one decelerating from 100 km/hr, at the same braking force.

 

[jack action]

Why not? Because the force is applied further away from the axis on bigger rotors. With your hand is much easier to stop a spinning bicycle wheel from the outer side (tyre) than from near the hub.

The force you apply is not greater; the torque is, because the lever arm is greater. We already established that there is a maximum torque $T_{b\ max}$ that can be applied. The limiting factor is friction at the tire-road contact patch.

If you have a caliper acting on a disc that can produce that maximum torque, increasing the disc diameter only means that you will need a smaller friction force from your caliper to reach that same maximum torque. Again, you gain nothing by going over the maximum torque allowed by the tire-road friction.

Having a smaller caliper friction force would mean less heat, but it is canceled out by the larger velocity of the larger disc (same work done, same power). But having a greater disc diameter still gives you better heat dissipation overall due to cooling.

 

[Tachyon son]

Please, see:
https://ebcbrakes.com/how-to-choose-the-best-big-brake-kit/

https://www.brembo.com/en/car/sporting-use/discs

https://www.brembo.com/en/car/formula-1/f1-infographics

Nice links, specially the first one, very comprehensive.

I have read EBC one many times and there is something I am not able to fully grasp yet:

- It is said that Braking torque = 2µFr (Where F is clamping force applied by caliper)
- Later, the author highlights that F = Pressure x Area
- But then I remember that friction force is not dependant on brakes surface area. (F = μN)

How can I get rid of the apparent conflict between second and third point?
Are pistons area involved in the caliper clamping force (F) but not pads area?

If you have a caliper acting on a disc that can produce that maximum torque, increasing the disc diameter only means that you will need a smaller friction force from your caliper to reach that same maximum torque. Again, you gain nothing by going over the maximum torque allowed by the tire-road friction.

This asseveration clearly fits my racing experience.
Then the point is that, for a given tyre, the faster you achieve the "maximum torque allowed by the tire-road friction" the shorter stopping distance. And this effect is harder to accomplish the faster you travel.
So at the end is all about tyre grip, basically, because is the ultimate limitation factor in cold stoppages.
In racing cars with no servo on brakes nor ABS the maximum power you are able to apply on brake pedal until reaching wheel locking point changes dramatically depending on tyre selection, specially comparing rain ones and dry soft compound. Tyre-adapted regressive braking is used to achieve max deceleration each time, as hard as enjoyable...

 

[jack action]

I have read EBC one many times and there is something I am not able to fully grasp yet:

- It is said that Braking torque = 2µFr (Where F is clamping force applied by caliper)
- Later, the author highlights that F = Pressure x Area
- But then I remember that friction force is not dependant on brakes surface area. (F = μN)

How can I get rid of the apparent conflict between second and third point?
Are pistons area involved in the caliper clamping force (F) but not pads area?

When applying those equations to brake calipers, it should be written like so:

• Braking torque = 2µFr
• N = Pressure x Area
• F = μN

The second equation depicts the relationship between a pressure P and a force N, both acting in the same direction;
The third equation depicts the relationship between two forces perpendicular to each other.

 

[Lnewqban]

Nice links, specially the first one, very comprehensive.

I have read EBC one many times and there is something I am not able to fully grasp yet:

- It is said that Braking torque = 2µFr (Where F is clamping force applied by caliper)
- Later, the author highlights that F = Pressure x Area
- But then I remember that friction force is not dependant on brakes surface area. (F = μN)

How can I get rid of the apparent conflict between second and third point?
Are pistons area involved in the caliper clamping force (F) but not pads area?

The author describes two principles simultaneously used in any hydraulic brake system, but that are very different: friction and Pascal's.

Please, note that the number 2 in the braking torque means that two forces of opposite directions and similar magnitude F are acting on the disc-pads interphase, regardless the magnitude of the area of that interphase.
That total force acting on the surfaces in close contact induces a resistive force which direction is aligned with the relative sliding and which magnitude is certain percentage of that total force (which we call μ after finding its approximate value via experimentation).

Please, see:
https://en.m.wikipedia.org/wiki/Tribology

If the author was talking about mechanically actuated brakes, like for an old motorcycle, he had no mentioned area at all, but the force with which the pads pressed on the metal surface of the drum.

He talks about area and pressure because he is discussing the hydraulic system that transfers the force and movement of the driver's foot onto the pads.
That combination of force and movement of the brake pedal is mechanical work, which is limited to certain normal amount.

That limited mechanical work becomes pressure and movement of the brake fluid and it comes out at the opposite ends of the system: the pistons of the calipers.
For each caliper, the forces that those pistons are able to apply upon the brake pads, as well as their displacement or linear movement in the caliper, depend on their total area, which depends on individual diameter (respect to the diameter of the master cylinder's plunge) and number of pistons.

Please, see:
https://en.m.wikipedia.org/wiki/Pascal's_law

 

[Tachyon son]

The author describes two principles simultaneously used in any hydraulic brake system, but that are very different: friction and Pascal's.

Please, note that the number 2 in the braking torque means that two forces of opposite directions and similar magnitude F are acting on the disc-pads interphase, regardless the magnitude of the area of that interphase.
That total force acting on the surfaces in close contact induces a resistive force which direction is aligned with the relative sliding and which magnitude is certain percentage of that total force (which we call μ after finding its approximate value via experimentation).

Please, see:
https://en.m.wikipedia.org/wiki/Tribology

If the author was talking about mechanically actuated brakes, like for an old motorcycle, he had no mentioned area at all, but the force with which the pads pressed on the metal surface of the drum.

He talks about area and pressure because he is discussing the hydraulic system that transfers the force and movement of the driver's foot onto the pads.
That combination of force and movement of the brake pedal is mechanical work, which is limited to certain normal amount.

That limited mechanical work becomes pressure and movement of the brake fluid and it comes out at the opposite ends of the system: the pistons of the calipers.
For each caliper, the forces that those pistons are able to apply upon the brake pads, as well as their displacement or linear movement in the caliper, depend on their total area, which depends on individual diameter (respect to the diameter of the master cylinder's plunge) and number of pistons.

Please, see:
https://en.m.wikipedia.org/wiki/Pascal's_law

View attachment 273079

Brilliant! Fully understood now.