## Stopping Abruptly to Save Break Pads

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?

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Hi Simon, good catch...
 Quote by 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?
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:

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.
 Recognitions: Homework Help 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.
 No, i've learned in auto shop classes and common sense that when you apply your brakes quickly that they wear out much quicker. I don't have any math or calculations or anything of the sort, but when you brake quickly, as said before, a lot of heat energy is created, and when two metal objects rub up against each other, even if they didn't get hot, which they would, then material would get lost. The hotter your brake pads the more wear and tear they go through, hence braking quickly and sporadically is not only unsafe, but your vehicles brake pads wear out much faster.

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 Quote by Simon 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.
 Quote by TheAnalogKid2 No. ...
No?!
 Quote by TheAnalogKid2 i've learned in auto shop classes and common sense that when you apply your brakes quickly that they wear out much quicker.
This lesson/observation is supported by the graphs - the graphs also give you an idea of how quick "quickly" is to get higher brake wear.

 Quote by babayevdavid I can definitely appreciate pendantry, as I generally am the same
pendantry? ...

 Quote by Simon Bridge I know/knew that was what you were saying - it's just that sometimes it helps to spell things out for people.
 Quote by Simon Bridge independent of the amount of time the brakes are applied for.
Only because, I really am a pedantic person...
 independent of the amount of time the brakes are applied for.

OCR... lol

 Quote by OCR pendantry? ...
Is that a funny word?
 Recognitions: Gold Member Science Advisor Whoa up there fellows..no pun intended. Granted pads wear due to pressure and velocity but...remember, long repetitive braking, such as one which occurs during a mountain descent, will result in a brake fluid temperature rise and may cause brake fluid vaporization. You can boil the brake fluid! This may be a concern particularly for passenger cars equipped with aluminum calipers and with a limited air flow to the brakes. Brake performance of a car or truck for that matter, can be significantly affected by the temperature rise in the brake components. High temperature during braking may cause brake fade, premature wear, brake fluid vaporization, bearing failure, thermal cracks in the brake pads, and vibration. Disk brakes are exposed to large thermal stresses during routine braking and extraordinary thermal stresses during hard braking. High-g decelerations typical of passenger vehicles are known to generate temperatures as high as 900C in a fraction of a second. These large temperature excursions have two possible outcomes: thermal shock that generates surface cracks in the brake pad material and/or large amounts of plastic deformation in the brake rotor ( warped rotors). In fact you can even cause macroscopic cracks in rotors with enough High G panic stops. Bottom line is - Heat generation due to friction in the sliding contact of rotor and brake pad influences the friction and wear characteristics of entire brake system. The more aggressive you drive and apply the whoa pedal, the more thermal shock and chances of causing excessive heat , thus wear. Having worked on 100’s of cars over the years. about 25% of the time brake pads are replaced, it is due to the metal to metal contact of the wear indicator on the brake pads. This little tab makes contact with the brake rotor when the brake pad is worn to the recommended replacement point. The contact during braking makes an unpleasant noise alerting the driver it is time to do maintenance. The other 754% of the time I replaced pads was because the thermal cracks were evident during routine inspection. I would not want to drive on cracked brake pads,,,,would you?
 Recognitions: Homework Help @babayevdavid: well, a pantry is a closet reserved for storing trousers and an inventory is where you invent things... like having to replace 779% of faulty brake pads (which can be done if you are replacing good brake pads and telling the owner they were faulty ;) )

 Quote by Simon Bridge @babayevdavid: well, a pantry is a closet reserved for storing trousers and an inventory is where you invent things... like having to replace 779% of faulty brake pads (which can be done if you are replacing good brake pads and telling the owner they were faulty ;) )
Uh, why am I being given the definition of a pantry again? lol
 Recognitions: Homework Help ... well, by extension, a "pendantry" (you asked the question) is clearly a small closet for storing pendants, the place where pendants are made or the art and craft of pendant-making. ;) Presumably a pedantry is an institution for the care and treatment of pedants. Which leaves me to look for a punnery...
 Brake pad and rotor wear are tied pretty closely to peak temperatures reached during braking, all else being equal. Lowering temps by 200F makes a rather large difference in brake life, an especially critical parameter in endurance racing. "Get thee to a punnery!" indeed!

 Quote by 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.
You may double the frictional force at first, but as the pad heats up, the friction is reduced and at the same time the wear rate increases due to the heat so you don't have a linear progression.
 Stopping abruptly will cause your front brake pads to wear down quicker due to the weight of the car shifting forward onto them. I think, 99% of the equation cancels itself out. You are driving your car, and it has X amount of kinetic energy that, when you apply the brakes, turns into thermal energy. Applying the brakes faster just transfers the thermal energy from, say, a 60/40 split front/rear brakes to an 80/20 split. Braking wears down the pads. When you brake hard, you wear them down faster (think of scraping chalk against the ground) yet you also stop faster. The heat is not really an issue here, since you will be applying the same amount of heat no matter how fast you stop if it is from the same speed. Sure you can overheat your brakes, but they are designed to work better at a certain range (high performance brakes actually don't work their best until they are in the 1200+ range, and they are generally ceramic) In the end, if you really want to save your brakes, lighten your car. less kinetic energy = easier to stop. I wish I had a picture of the brake pads I just changed out. The front's were down to about 30% and the rears were around 70% left. I brake hard, and corner fast, so the front's took a beating. p.s. If you drive a manual, downshift, that will save your pads!
 Temperature is the issue. Same heat over a shorter period of time = higher temperature. High temp = brake pads wearing out quicker. Using the engine to slow down instead of the brakes = more \$ spent (brakes are much cheaper than engines, clutches and transmissions). Don't downshift to save the brakes.

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