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?

  • 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
  • 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

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.


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).


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:

  • 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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