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Warmed brakes

  1. Jan 18, 2013 #1
    I have got one interesting question for you.
    Imagine that you are driving a car down the hill. What is the speed to the most warmed brakes? Today I thought about it and I wonder how it somehow calculate numerically but I don't know how. Have you got any ideas?
  2. jcsd
  3. Jan 18, 2013 #2


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    What? What does "the speed to the most warmed brakes" mean?
  4. Jan 18, 2013 #3

    Simon Bridge

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    Interpreting "speed of most warmed brakes" as requiring information about how brakes perform as they are heated...
    You need to know how the brakes warm up ... it will depend on how long you have applied the brakes for, how fast you are going, how hard the brakes are being applied, what the brakes are made out of, how they are cooled, how long since you last applied them, etc etc etc.
  5. Jan 18, 2013 #4


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    Put the car in first gear, drive down the hill with your foot flat to the floor on the acclerator pedal, and use the brakes to keep the engine at the RPM for maximum torque.

    That should get close to the maximum possible brake temperature :biggrin:
  6. Jan 18, 2013 #5


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    Find out the mass of your car - if you do not know assume 1000 kg for a small car.
    Pick a speed say 100 kph --> convert to m/s ( 28 m/s )
    calculate the kinetic energy of the car at that speed. (400 kJ)
    to stop the car the brakes convert KE into heat energy. ( same value ) (400 kJ)
    how much that KE could warm up 1 litre of water from 0C to a temperature T ( 95C )
    ( I always thought it was more than that !!! , did I miss a factor )
    decide if you can now have a nice warm drink or two :)
  7. Jan 19, 2013 #6


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    Front brakes usually dissipate the most energy and are usually disc brakes. Rear brakes still tend to be drums and are designed not to get very hot - to avoid distortion and brake fade. Assume the discs are about 5kg each and that they are cast iron / steel. The answer to your sums could give a value well in excess of 100C temperature rise. Racing car discs can be seen to glow red hot after a lot of braking - or a long hill with the brakes applied frequently, I guess.
  8. Jan 19, 2013 #7
    As others have noted, getting a numerical answer will involve details about the car, including its braking system.

    However, one can observe that, for constant velocity, and unit time, loss of potential energy has to equal the energy dissipated by the brakes plus the energy dissipated by drag. The latter will be almost all aerodynamic drag.

    Since the loss of potential energy is proportional to velocity, but energy dissipated as aerodynamic drag is proportional to the cube of the velocity, it's clear that there will be two velocities at which no energy is dissipated in the brakes - zero, and some higher value.

    Plugging in some appropriate constants that would have to be measured for the car in question, and you can get an equation describing the energy dissipated by the brakes as a function of velocity. Find the maximum of that by differentiation, and you'll obtain the the required velocity, on the assumption that the temperature of the brakes is a rising function of energy dissipation.

    The result will quite likely imply an energy dissipation in the brakes that they are incapable of sustaining, which is something I urge you to consider before testing the result experimentally.

    Last edited: Jan 19, 2013
  9. Jan 19, 2013 #8
    Simon Bridge: Thank you so much for translating others.

    256bits: I can consider fuel as water and solve it by simple calorimetric equation without losses? Thermal energy is certainly larger than the kinetic energy. Why 0 ° C? It can be even higher. I think this is too simplistic, but thanks very much.

    Sylvia: Numerical answer will involve details about the car... Yes, i know it. When we come to a conclusion, I will search some details about average car. You wrote it very nicely, but I'm not sure if I properly translated and understood it, so I'll ask you about it again.
    so, for constant velocity:
    loss of potential energy (loss is given by moving downhill?) = energy dissipated by the brakes (it is the loss of energy (intended efficiency)?) + energy dissipated by drag (energy required to move?)
    It is valid for a car going from downhill?
    I will express speed of this total energy?

    Seek maximum ... I have to derive? it is not for me yet, but I'll try it another way.
  10. Jan 19, 2013 #9
    Numeriprimi I think you have the general idea, though if you're not able to differentiate the equation, I'd have to feel that you're trying to run before you can walk.

    In any case, the result will depend not just on the car, but on the slope of the hill, and the density of the air, which varies with the weather. Indeed, the density of the air increases as the car descends. This means that the velocity for maximum brake heating varies not just from car to car, hill to hill, or day to day, but continuously during the descent.
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