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Heat dissipation from a brake pad

  1. Jul 29, 2005 #1
    Hi all,
    (this has originally been in the Materials and Chemical Section. I think it does not belong there).

    I'd like to add a question heat dissipation from a brake bad. Basically, let's assume a brake pad is pressed against a surface that quickly moves along (can be the bicycle brake pad on the wheel rim or, in the case of an inline skate with a heel brake, a brake pad and the road itself. Let's look at the latter example in the following). Consider a quasi-stationary case: rolling down a long decline with constant slope and speed. So we have to dissipate some constant power (in the order of a few hundred Watts [10% slope - 10m/s speed - 1 m/s vertical speed - 80 kg mass - 800 N weight - about maybe 50% power dissipated into air friction - makes 400 W]).

    In a first order approach I would say that the brake pad would not heat up higher than the temperature of the road. Because that road surface moves along quickly, it does not heat up significantly while the pad glides along. The temperature at the interface road/pad is thus constant. This is where the heat is generated. So the road basically acts like an infinite and "zero-resistive" heat sink.

    I think experience does not support this approximation. In particular, if you have an inline skate with a heel brake, the heel brake DOES heat up (does it?). How much? What's wrong with the first order approximation? What would be a more realistic model?

    Thanks, Hanno
  2. jcsd
  3. Jul 29, 2005 #2


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    This problem actually has applications in a wide number of areas, many of which have nothing to do with braking, so it's an interesting problem. I've not heard a whole lot of thought go into it though. I often wonder if there hasn't been some serious study on it. Here's how I've looked at this in the past. I'd be interested in other people's experience too.

    Obviously there's a lot of energy being changed from kinetic energy to heat energy. Conservation of energy tells you that this heat must go into the parts of the mechanism that are rubbing. One might conclude that the heat goes into each part in a proportion that results in a thermal equilibrium at the point of rubbing. In other words, the temperature at the point of contact is the highest temperature because that's where the energy is being generated (or converted if you like), and with the exception of the extremely small thermal resistance between the two parts, these two parts will have the same temperature.

    This is exactly the case you've suggested. It says that after some fairly slight initial heating of the pad, the temperature of the pad won't rise much more because any increase in temperature would result in heat transfer from the pad to the road. So rather than the pad continuing to increase in temperature, the thermal energy generated at the point of contact all goes into the road.

    It would seem like a reasonable argument, but I suspect the one assumption made is where we can find some significant discrepancy between this model we've created and reality. We've assumed that the thermal resistance between the two parts is small. I have my doubts that the contact resistance between these two parts is small. I suspect it's fairly high actually and that results in significantly higher temperatures at the brake pad.

    Note also that there's another assumption which underlies your original post, that the conduction of heat away from the point of contact in the road is very high. This might be true, but the energy is put into the road so quickly, the temperature at the surface of the road could still be fairly high as the brake pad goes over it. It might only take a second or two for the temperature of the road to drop due to thermal conductivity, but for the tiny fraction of a second the pad is in contact, the amount of road (ie: the mass of road) that is getting heated up is extremely small.

    I think there's another problem with the assumptions which I've not mentioned yet, and that is the proportion of energy going into each of the two parts may not be equal. We have to ask, "How does the rubbing of a material against another generate heat at the molecular level or at the surface of contact?" It could be that heat is not split uniformly between the two parts because of what happens at the surface of each part at the molecular level. One material could concievably have a larger percentage of the total heat going into it. If this is the case, the thermal resistance between the two parts plays an even bigger role, since this is yet another issue that results in a loss of the assumption made, that the two parts are in thermal equilibrium at the contact surface.
  4. Jul 29, 2005 #3
    Many thanks. A lot of good points!
    I'm now out for 3 weeks summer vacation. Will do some practical experiments! If anything useful comes out, I tell you. Hanno
  5. Jul 29, 2005 #4


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    ezskater, I'm not sure I've understood.

    In the case of a bicycle, the braking is done at the surface of the wheel rim. If you touch a wheel rim after braking down a long decent, you'll burn your finger. It does get hot. The friction between the tyre and the road here is irrelevant, as long as the wheel remains spinning whilst you're slowing down.
  6. Jul 29, 2005 #5


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    Brewnog, Note that the question ez asked about and the responce I gave him regards the brake pad on a roller skater where the pad is in contact with the road, or more ideally, a flat surface. I think the case of a brake pad on a disk or wheel rim is more interesting actually, but I didn't start this post! LOL
  7. Jul 29, 2005 #6
    Yes, sure the rim gets hot. But does the brake pad itself get HOTTER than the rim? And by how much? (In the case of a skate braked by a friction heel brake, rim ==> road. So the difference is that the rim comes back to your pad with every revolution of the wheel, while the road doesn't. You always get "fresh cool road" and don't care about the heat you leave behind!) Hanno
    Last edited: Jul 29, 2005
  8. Jul 31, 2005 #7

    after thinking about your answer, I guess it would be pretty naive to assume that intitially heat is distributed 50:50 between pad and road. I can easily imagine effects that transfer much more heat into the pad (as the softer part) than into the road. Like vibrations and erosion. So it would not be a too pessimistic approach to assume ALL heat is initally generated in the pad. Then it has to flow - via the thin air gap - into the road. Mmmmh ....

    (BTW, I son't see such a big difference between the "pad on a wheel rim" case and the "pad on a flat surfact" case. I'd even say, if you solved one, you have solved the other. Just let go wheel radius-> infinity to come from "wheel" to "flat". Or, the other way round, if you know how fast heat is dissipated from the rim, you get a steady state temperature of the rim -> replace rim by a "hot flat surface".)

    What about this semi-simplistic model?
    - ALL heat is generated in the pad, just above the interface
    - interface has some homogeneous and constant non-zero thermal resistance
    - road travels along under that interface at speed v
    - has some thermal conductivity and heat capacity.
    - I'd further assume that the pad is very wide (ignore edge effects) and
    - that vertical temperature gradients in the road are much larger than horizontal.
    - Lastly, we assume that everything has stabilized to a quasi-stationary solution ("quasi" because the road is moving)
    This should be enough to set up some modified heat conduction equation, is it? ("modified" because of the road moving along.) The only unknown we won't find in a book is the thermal resistance of the interface. This needed to be measured.

    What do you think?

    Regards, Hanno
  9. Jul 31, 2005 #8


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    The main difference between a brake pad on a road and a brake pad on a rim or disk or other rotating surface is the total thermal mass. The thermal mass of the road is essentially infinite so the heating is only the transient heating as the pad passes across it. In the case of the wheel rim, the wheel rim has limited thermal mass and can heat up significantly. It has to be cooled by the air flow over it.

    Regarding the 50:50 split, I don't really have any feel for that. I can't disagree with your conclusion, but can't agree with it either. I simply don't know. I have to believe this question has been researched and tested before and I'd be interested in finding out a bit about it because it has some application for me.

    Perhaps a subtle difference between the pad against the road and pad against a wheel rim will be the thermal contact resistance between the two. Brake pads on a rim or disk generally conform to the shape of the rim very nicely and eliminate most air gaps. A pad on a road such as on a skater's skate doesn't have the same surface going over it in a uniform way so there will be more of an air gap.

    Putting it all together, the heat transfer analysis gets pretty tricky. The rubbing creates heat instantly whereas the conduction of heat away from the point of contact is determined by the material's thermal conductivity. If it were me, I'd set up a computer program or spreadsheet that does an iterative calculation on it. You might be able to neglect lateral thermal conductivity as you suggest as a first go and just see how the temperature of the road warms up in responce to see if this is a reasonable assumption. And if you ever get some answers to this, I'd be interested in seeing what you come up with.
  10. Aug 5, 2005 #9

    I did some more investigation of how the heat distributes into the road. This turns out to be classical one-dimensional heat conduction theory. See http://www.ezskate.de/brake_pad.doc and http://www.ezskate.de/brake_pad.xlr. Temperature increase on the road surface is moderate (about 12 K in a typical case).

    What this does not answer is how much higher the temperature is in the brake pad. As you have said, due to the air gap and the interfaces heat transfer from the pad to the road may be pretty poor - means there will be a high temperature difference to drive the heat across the gap. Guess this needs to be measured somehow. Will certainly depend on roughness of the road and probably pressure.

    Any comments ... Hanno
    Last edited by a moderator: Apr 21, 2017
  11. Aug 16, 2005 #10
    Hi Q Goest

    missing your inputs! Have you read the memo? I added some lines yesterday.

    I start to see a basic difference between the bicycle rim and the inline skate heel brake. The heated layer (of rim or road) is pretty thin - in case of the road around 1/10 mm, in case of the rim it is about 1 mm because of the much better thermal conductivity of metal versus concrete. Now, if the roughness of the surface is small compared to this depth, it can more easily be assumed that the surface is flat. For a bicycle rim, I'd say this is the case. It is certainly very flat on a 1mm scale. A typical road, however, is certainly NOT flat on a 1/10 mm scale!

    Regards, Hanno
  12. Aug 17, 2005 #11


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    It's pretty simple. Brake pads need a high specific heat and toughness. Finding the right combination of both properties is what keeps us engineers employed.
  13. Aug 17, 2005 #12


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    Hey there ezskater. Sorry for not responding earlier. I looked over that 8 page letter you attached, and I'm thoroughly impressed. I'm glad you got something out of our discussion, it seems like you went quite a long way after that and developed a nice mathematical description of what happens.

    I honestly don't have the time and interest though to go through all your calculations and determine if they are correct or not. Usually when I have to go through something like this I find it best to review the calcs, but also to do them my own way and find out how close the two models come. To do that, I usually create a computer program because after I go through the calcs once, I generally don't want to have to do it again. The concepts (to me) don't benefit from doing the calculations over and over again, so if I do the math, I have to spend some time and do it right. I really don't have the time to do that though, and it sounds like your calculations are reasonable anyway. Perhaps someone else would be willing to review them.
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