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How hot does a bicycle disk get? I’ve been working on heat rejection for auto/truck disk brakes with some success. But the disk temperature needs to reach 700 degrees to be effective. This seems a bit toasty for a bicycle disk. Any thoughts?

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  • Dunno - I've experience brake fade on three descents (2x road bike and 1x MTB) and remember checking the front MTB rim - it was hot to hold through full fingered bike gloves. I've also had a road descent with a lot of braking where the front tube punctured - possibly heat stressing a weak point by the valve stem. Comment because no values given. – Criggie Jun 13 '17 at 20:18
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    It's hot enough to cause the disk rotors to blue and for hydraulic fluid to boil..... – RoboKaren Jun 14 '17 at 8:23
  • Dark blue is usually 550 to 600 degrees C. for openers. It doesn't seem that brakes survive at this temp. – TomO Jun 14 '17 at 15:49
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    Hot enough to leave a disk brake tattoo on your arm that lasts for several weeks and itches like crazy .... however hot that is. – SteveJ Jun 17 '17 at 7:01
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    I've bought an IR thermometer with the intention of taking it to bits and mounting the sensor on my fork. So far I haven't had time to do this, but will post something when I do. Even just a short flat test ride with lots of braking, then getting off and measuring got me to 70°C, a decent hill would put a lot more energy in – Chris H Aug 10 '17 at 14:59
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As this pops up #1 on "mtb brake temperature" search, let us continue the answer of @Emyr. On descents the potential energy will be converted into the heat of the braking system, and then dissipated into the air. There are two main physical processes going on:

  • accumulation of (thermal) energy in the brakes, governed by P * t = C * M * DT, where P=power, t=time, C=heat capacity (of steel/aluminium, or whatever), M=mass of the disc, DT = change of the disc temperature from the ambient temperature, and

  • dissipation of heat into the air, governed by P = k * S * DT, where S is the area of contact, k is the convective heat transfer coefficient; so for given DT, there is a given power that will be radiated away,

Now the heat transfer coefficient is actually a "capability" depending on many things (like geometry), but also - and mainly, mostly linearly - on the airspeed. There are different values you can find, but I include some typical ones below, enter image description here so let's say it is 35 W/(K * m^2) for 16kmh and 70 W/(K * m^2) for 32kmh (and let's ignore the dependence of these coefficients on the temperature).

Now these processes play as follows. On a long descent the heat capacity will quickly be saturated, and all the potential energy lost will have to be radiated away, i.e. the temperature of the disc will go up until the point where the disc radiates away as much as is generated from the descent. Assuming 800W generated (typical biker on ~16kmh descent on somewhat steep slope, and both brakes working), and a disc brake of (2-side) surface (both sides together) 5 dm^2 = 0.05 m^2 disc area, we get to

DT = 400 / (35 * 0.05) = 228 °C

Of course, should the slope be less steep, and allow for same potential energy generated at twice the speed, we would get 114°C. That's why it is the steep slopes that "kill" the brakes.

Now before the steady state is achieved, the disc will store the energy as heat. Assuming the disc is 2mm in diameter, so 0.05dm^3 in volume, so 400g of steel [it is usually much less due to perforations], with heat capacity of 500 J / (kg * °K) we get roughly

DT = 2°C * t [s],

so without the convective losses the disc would warm up by 2°C per second. (Incidently, if your disc is worn out, say to 1mm, it will warm up twice as fast.)

One can now play with different "braking strategies" etc. but almost for sure the following will hold:

  • increasing the radiating area will help; S → 2S leads to DT → DT/2 (use both brakes, and/or brakes that have somewhat greater area from which heat is dissipated),
  • you may increase the radiated power if you allow for a greater temperature (even for limited period of time), but only if the airspeed is the same; it is almost certainly a bad idea to get to "pulsate" by getting to greater speeds and brake to much lower ones (as at the point after the heat is generated, the airspeed will become low, and hence the transfer will be lower).
  • also notice that for a given slope, the generated power (potential energy lost), and the radiative heat transfer are both proportional to the velocity; hence the steady state temperature of the brakes will be independent of the (steady) velocity.

Let us finish with a view of braking/working disc, and the source of these simulations, enter image description here taken from "Thermal/Mechanical Measurement and Modeling of Bicycle Disc Brakes", ISEA 2018, Ioan Feier and Robin Redfield.

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    Great answer! I’m surprised those cooling fins actually do anything - I always thought they were more of a placebo and cash grab. – MaplePanda Apr 4 at 18:26
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    @MaplePanda I suspect that pad fins are useful mainly because they are closely coupled to the heat source, and the other thermal paths from the pads are poor and heat the fluid. That means they don't have to be great to be good – Chris H Apr 4 at 20:45
  • On the "pulsed braking" problem, the frequency would probably be ~1Hz, faster would be hard to do consistently and slower would mean a slower average speed while maintaining the same peak speed. The benefit of pulsing is to reduce the time when the pad is in contact with the hot rotor, so less of the total heat produced ends up in the brake fluid. – Emyr Apr 15 at 13:20
  • @Emyr The pads are heated directly by the braking action, and they have a much harder time to get rid of that heat. As such, I would always assume that it's the disk that is in contact with the pads which helps to dissipate the heat generated at the pad's surface, not the other way round. – cmaster - reinstate monica Apr 15 at 15:34
  • Truly a great answer. Nevertheless, your example number are a bit low for riding on public roads: Typical mountain descends on such roads ask for speeds in excess of 50 km/h. True, most riders would roll without using their brakes for the largest part of such descends (using atmospheric braking), but if the road is curvy due to serpentines, you can easily get an average heat generation of several kilowatts. – cmaster - reinstate monica Apr 15 at 15:42
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To simplify this a bit, I'm going to ignore rolling resistance and aerodynamic drag, so all the work is done by braking. Also, when you brake hard, the front brake does almost all the work, so ignoring the rear brake to simplify the scenario.

Stopping

Mass of bike + rider = 100kg

Speed = 10 m/s (36km/h, about 22.4mph)

kinetic energy = 0.5 m v^2 = 5000J

A basic 180mm steel brake rotor weighs about 150g and the specific heat of steel is about 0.5 J/gK (Joules per gram-degree), so converting 5kJ of kinetic energy into heat in the rotor should increase the rotor temperature by 67 degrees (C), assuming none of the energy is absorbed by the brake pad, caliper body or brake fluid (not realistic, but useful to identify an upper bound for the rotor).

Descents

To maintain constant speed on an even slope, the change in gravitational potential energy must equal the heat energy extracted by braking (plus other sources of friction and drag that we're ignoring).

Descending a 1 in 10 slope for a 100m distance gives a height change of 9.95m (not 10m, because the distance is along the hypotenuse).

Gravity on earth is about 9.8 m/s^2.

Energy change = m g h = 100 * 9.8 * 9.95 = 9751J

This would increase the rotor temperature by 130 degrees.

Heat Dissipation and the limits of my knowledge

The two calculations above don't have a duration because I haven't allowed for heat loss during the event; here are some of the complicating factors:

  • The friction creates a film of pad material on the rotor. This film insulates the rotor, but also is necessary for cohesive friction, which generates heat without consuming the pad material, but gives way to destructive friction as pressure increases.
  • The heat is generated at two patches on either side of the rotor, not uniformly throughout the rotor material, so as the rotor turns through the contact patch, heat penetrates from the surface to the core of the rotor, then as the rotor leaves the contact patch, the surface is cooled proportionally to the surface-air temperature difference, and once the surface is colder than the core, heat moves from core to surface. Heat is also conducted parallel to the surface, from hotter to colder parts of the core.
  • The airflow path length across the rotor is not consistent.
  • The airflow around the rotor is turbulent due to the disruption caused by the leading part of the wheel and, for the upper aft part of the rotor, the fork leg and brake caliper.

Modelling heat in rotors is sufficiently complicated that it is a topic used in Engineering project assignments for university students, and the software typically used to perform the calculations is very expensive.

"Experimental and Numerical Thermal Analysis of Formula Student Racing Car Disc Brake Design", Manthan Vidiya1 and Balbir Singh, Manipal Institute of Technology. Published in Journal of Engineering Science and Technology Review 10(1)(2017)138-147

Example model "Heat Generation in a Disc Brake" for COMSOL Multiphysics:

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  • Given the other assumptions I reckon you can probably treat the rotor as all being at the same temperature as the surface measured opposite the caliper, at least for the purposes of losing heat to the air. The error in this would be much smaller than the error in the airflow model. As in the linked question we do have to consider loss through the caliper/pad as well (and for fade that's where the heat matters). The heat loss will soon become significant in descents -- which is just as well, as 1kJ/m (for 100kg) would build up pretty quickly -- you'd melt the steel; after 100m vertical – Chris H Aug 10 '17 at 15:09
  • Anyway +1 and greetings from your alma mater – Chris H Aug 10 '17 at 15:09
  • Hi Chris! I reached a dead-end on calculating a steady-state temperature for the rotor given a constant ~1KW flux at 10m/s airspeed. The most useful property Thermal Resistance (unit K/W) used for heatsinks in electronics is readily available for copper and aluminium alloys, but not steel. I probably haven't found the right formulas or input values... – Emyr Aug 10 '17 at 16:20
  • There are so many different steels it would be hard to know where to start -- and their thermal performance varies, I'm sure. But I suspect the dominant effect would be due to the airflow. With the airflow over rotor being affected the fork and caliper assembly this would be hard to assess. I'm an experimentalist (though I've done some thermal modelling -- just conduction) hence my intention to measure it. – Chris H Aug 10 '17 at 16:46
  • Engineeringtoolbox.com gives 12--45 W/mK (the units I'm used to for modelling and more suitable if you want to define the shape). Carbon steel is a little higher but aluminium is 204 W/mK. – Chris H Aug 10 '17 at 16:49

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