# Stopping Abruptly to Save Break Pads

1. Jan 5, 2013

### babayevdavid

Hi,

I am curious, would stopping a car more quickly, or more abruptly, allow its break pads to last even a little longer?
I understand that stopping short may have adverse effects on other parts of the breaking system or on the car or person riding it. I do not intend to stop short in an effort to save my break pads.

It seems to me that the shorter the time in which the break pads are in contact with the spinning inside of the wheel, the less kinetic friction there will be between the two, and the less wear the breaks will suffer from.

Am I missing something? Is there a tradeoff of some sort that will make this action useless? I assume it would be the same for a bicycle

2. Jan 6, 2013

### Simon Bridge

The breaks stop the car by converting it's kinetic energy into heat.
Stopping more quickly from the same speed just dissipates the same energy faster... you can fill in the rest.

Of course, if you hit the brakes really hard (and you don't have ABS) then the wheels will lock and you'll trade break wear for tire wear.

Certainly the less energy dissipated in the brakes the longer the pads will last. You can do this by moderating your speed.

3. Jan 6, 2013

### Redbelly98

Staff Emeritus
But it is not the heat that wears out the brake pad, is it? There is also mechanical wear from the brake pads rubbing against the wheel's metal disk. It is the mechanical wear -- i.e. loss of brake pad material -- which eventually causes the pads to need replacing, not their heating up.

So the basic question is, will pressing the pad against the wheel harder (thereby increasing the frictional force) over a shorter distance create more wear, less wear, or the same?

Yes. I think we need to assume the braking is not so sudden that the tires start skidding.

Last edited: Jan 6, 2013
4. Jan 6, 2013

### Simon Bridge

How else does the heat get into the break pad?
Hot break pads wear out faster.

Break pads are cooled - sometimes special arrangements are made for this.
If you heat the pads faster than they are cooled you can melt them.

You should be able to get your pads to last longer by breaking only gently, and intermittently ... so they can cool between application.
Of course, you'd slow down slower...

Presumably not all the energy goes into wearing out the pad...
I don't think there is a straight-forward, all-situations, answer to the question.

Last edited: Jan 6, 2013
5. Jan 6, 2013

### Staff: Mentor

Re: Stopping Abruptly to Save Brake Pads

Yep... You have to spend some time talking brake pads with people who design, maintain, and drive race cars to really appreciate just how many variables there are here. Probably the best practical answer is that smooth non-jerky driving is fastest and easiest on all the mechanical components of a car.

6. Jan 6, 2013

### Simon Bridge

I kinda have ;) notice that in post #2 I kinda left the details a bit open ... huge great manuals of specifications have been written on this subject. A good starting point for a physics student is to think in terms of the energy and where it goes and how it gets there. Part of what we try to do here is get the questioner to do their own thinking right?

It is a nice example of how apparently simple every-day actions turn out to be annoyingly messy on closer inspection.
I think it's time to hear from OP :)

7. Jan 6, 2013

### CatastrophicF

Re: Stopping Abruptly to Save Brake Pads

A for instance... I really enjoy driving video games. The pinacle of realistic driving simulators has always been Gran Turismo. I was baffled when I first started playing Gran Turismo 5 because there were no brake upgrades available for any of the cars. So I wondered why is that? Well I have a degree in automotive technology and I will tell you why. In regards to brake performance the measure is NOT how fast they can stop the car or with how much force the calipers can squeeze the pads into the rotors... the measure of brake performance is measured by longevity of the pads and heat dissipation. If the brakes can operate at higher temperatures without fade then they can yield more performance. Gran Turismo left this simulation of brake performance/wear/heat dissipation out of the game because of the MONUMENTAL amount of energy needed to make these calculations. There are so many variables and so many variables that do not change in a linear way you would have to play the game on a supercomputer. So Gran Turismo just made the brakes a constant, they never wear, never fade and work just as well at 200F or 1800F.

8. Jan 6, 2013

### babayevdavid

Re: Stopping Abruptly to Save Brake Pads

So, whether you ease on the break and your car takes longer to stop, or you break faster, in the end, the break pads will dissipate the same amount of the wheel's kinetic energy? I suppose that not only makes sense, but has to make sense, otherwise the car wouldn't stop for both cases.

What is "ABS"? Also, is it the break pads that are locking the wheels? If so, this is the answer that I was looking for, for lack of a better phrase. I wanted to know specifically if break pads could be saved even slightly by breaking hard, regardless of consequence to other parts of the car.

I should have specified. I'm talking about your casual everyday car on the road, if that makes a difference.

9. Jan 6, 2013

### SteamKing

Staff Emeritus
ABS stands for anti-lock braking system. It is installed in some cars to prevent the brakes from locking up in sudden stops.

Your approach to jamming on the brakes constantly might save the brake pads, but they will probably slowly destroy the rest of your car. Also, others driving around you may not be prepared for all of your sudden stops, and an accident may result.

10. Jan 6, 2013

### Studiot

It is interesting to note the change in driving technique as taught today.

I learned in the days when it was considered poor driving to need to use the brakes much at all. To quote my instructor "The brake should only be used to stop the car eg at a traffic light. Anything else is poor anticipation and a waste of energy"

Today drivers are taught to slow the car down for junctions, roundabouts etc to and then select an appropriate gear.

At my last service I had to have new front disks - because I don't use my brakes enough according to the mechanic!

11. Jan 6, 2013

### babayevdavid

Ah I see. And yes, as I said, I'm aware of the consequences and dangers. That "might" is what I want to clear up. This was more a thought experiment than anything else

12. Jan 6, 2013

### Q_Goest

I think what's missing from the discussion here is how wear rate is affected by those factors required to make a vehicle stop faster.

Wear rate is generally measured by pressing a sample of material against a moving surface. (See ASTM D3702 for example) The rate at which material comes off of the sample can then be measured as a function of the contact pressure (P) and the velocity (V). So rate of wear is equal to pressure times velocity:
K = P V
where K = rate of wear
P = Contact pressure
V = Velocity

The total amount of material removed from the sample can then be calculated by multiplying K times the time.

For typical materials, and for relatively low values of PV, the rate of wear (K) is linear. Double the contact pressure and you double the wear rate. But all materials will exhibit increased wear rate as PV increases. This increase in wear rate may be minor for low values of PV such that you can assume the relationship is linear, but will increase at an increasing rate as PV gets closer to some limit. Below is a good example of wear rate versus PV. This graph shows 4 different materials. Note that each material seems to have a wear rate that is fairly linear with a line that would trace back to a wear rate of zero at a PV of zero. Note also that above some point, wear rate suddenly increases. This graph is only intended to be representative of materials in general:

The rest of the calculations to determine if you can reduce the wear of brake pads on a car should be relatively straightforward. As an example, consider doubling the pressure on the brake pad and assume all other variables such as coefficient of kinetic friction stay constant. If you double the pressure, how does that affect the deceleration of the car and how does wear rate increase? Since wear rate also tends to be increased at higher temperature, how might that affect the results?

I think you should find that braking harder will not reduce the rate of wear on your brake pads and may even increase it.

13. Jan 6, 2013

### babayevdavid

Wow, thank you for the in-depth explanation. It seems to me though, that when stopping more quickly, even if the rate of wear were to increased, the duration of wear would decrease, so the amount of material lost it seems would not necessarily be greater. And that's aside from the possibility of slamming the breaks, which would cut the time of wear even shorter.

14. Jan 6, 2013

### pantaz

Re: Stopping Abruptly to Save Brake Pads

You need a new mechanic. Seriously, I can't imagine any possible reason for that claim.

Sorry, but after seeing "break" used incorrectly so often in this thread, I have to go pedant...

BREAK
1. to smash, split, or divide into parts violently; reduce to pieces or fragments
BRAKE
1. a device for slowing or stopping a vehicle or other moving mechanism by the absorption or transfer of the energy of momentum, usually by means of friction.
2. brakes, the drums, shoes, tubes, levers, etc., making up such a device on a vehicle.
3. anything that has a slowing or stopping effect.
- Source: http://dictionary.reference.com

15. Jan 6, 2013

### babayevdavid

Re: Stopping Abruptly to Save Brake Pads

Oh yes, I always forget to make the distinction. Thanks for pointing it out. I can definitely appreciate pendantry, as I generally am the same :tongue:

16. Jan 6, 2013

### Q_Goest

Try breaking down the problem into small, managable pieces. For the sake of this example, let's assume disc brakes are being used. If the contact pressure of the brake pads against the disc doubles, that will double the frictional force, right? If you double the frictional force of the brake pads on the disc, that doubles the torque on the wheel. So how fast does the car decelerate? Consider the rate of deceleration, then determine the difference in the time it takes to stop. Try to figure out how the two cases compare quantitatively.

17. Jan 6, 2013

### Simon Bridge

Erm... So for varying speed, instantaneous wear would be: K(t)=Pv(t) ... K(t) would be the instantaneous rate that material comes off the er material ... brake-pad ... which would be proportional to dm/dt where m is the mass removed?

(On the face of it, K does not have dimensions of a mass rate of change though.)

Looks like the total wear should be proportional to the area under the v-t curve while the brakes are applied?

@pantaz: time to take a break from brakes and have some brekkie?

18. Jan 7, 2013

### bahamagreen

Rolling friction is still happening during braking, right?

If a portion of slowing down is contributed by rolling friction, and that portion increases as speed decreases, then I'm thinking that braking easy allowing a slow deceleration provides more exposure to low speed rolling friction.

That the gross kinetic energy transfered to heat is basically the same for slow and fast stopping is a good start; but wouldn't more of that transfer be diverted through rolling friction in the slow stopping scenario than in the quick stop, so the pads themselves transfer less net energy with slow stops?

19. Jan 7, 2013

### Q_Goest

Hi Simon, good catch...
I made a mistake on my previous post (#12). The graph I posted was the one that shows depth wear rate as a function of PV. Graph repeated below for convenience:

http://ars.els-cdn.com/content/image/1-s2.0-S0043164811005916-gr2.jpg

This shows the “depth wear rate” for various materials given the PV. The amount of material lost to wear is the depth wear rate (units nm/s) on the Y axis shown in the graph times time. So if you know PV and the material, you go along the X axis of the graph and up to the material to find the depth wear rate. You can then multiply by time to get how much material is shed. What I was trying to point out is that as you double the PV (by say doubling P), the depth wear rate also doubles.

Also, I shouldn’t have shown K = PV as an equation, I meant to only point out that the rate of wear is proportional to PV. What I should have said is that there is a "specific wear rate" which is generally thought of as being a constant for a material up to a given point as shown in the graph below. If you multiply the specific wear rate times PV you should get the depth wear rate. These graphs are both taken from a paper I found online here.

It should be noted that the wear rate varies depending on the material (there are 4 different materials shown in that graph) but also due to surface roughness and hardness, temperature, any gasses or fluids in contact, etc… so it is an emperically determined value. Note also that the graphs are for various polymers and they're being used only as an example, but brake pads will have similar wear characteristics.

To get back to the OP, the point I was trying to make was that
- The rate at which material is shed from a brake pad is linearly proportional to PV. If you double the PV, the rate at which material is shed is doubled.
- The rate at which a car decelerates is also linearly proportional to the PV of the brake pad.
- If you increase PV above some point, wear rate can be greater than linearly proportional which can happen for example if the temperature goes up. (consider for example, the point you (Simon) raised earlier (post #2) about the amount of energy that has to be absorbed by the brakes in order to stop the vehicle)

Given those factors, it should be fairly clear whether or not brake pad wear will increase, decrease or stay the same under various conditions.

Sorry for the confusion.

20. Jan 7, 2013

### Simon Bridge

OK - so $\dot{w} = kPv(t)$ where w is the linear depth of material removed and k is a constant of proportionality. That still makes w proportional to the area under the v-t graph - the stopping distance (in this case) - as well as the applied pressure.

But since the stopping distance, d, is inversely proportional to the retarding torque (i.e. the applied pressure) ... that would make P=c/d where c is another constant of proportionality ... so w=kc i.e. a constant, independent of the amount of time the brakes are applied for.

I know/knew that was what you were saying - it's just that sometimes it helps to spell things out for people.

However, as you (and others) observed, IRL there are other factors affecting brake wear than just the speed and the pressure. This is reflected in the graphs. I suspect that the sharp upwards curl at higher speeds is the result of the material heating up for example. The graphs seem to show that we'd expect very high PV braking for short duration to potentially produce more wear (ceteris paribus) than low PV braking for a longer duration.

This introduction of real-World data appears to have strongly answered the original question.