# Stopping Abruptly to Save Break Pads

by babayevdavid
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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.
 Homework Sci Advisor HW Helper Thanks ∞ PF Gold P: 11,006 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.
 P: 3 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.
P: 108
 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
P: 17
 Quote by OCR pendantry? ...
Is that a funny word?
 Homework Sci Advisor HW Helper Thanks ∞ PF Gold P: 11,006 @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 ;) )
P: 17
 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
 Homework Sci Advisor HW Helper Thanks ∞ PF Gold P: 11,006 ... 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...
 P: 563 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!
P: 1
 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.
 P: 563 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.