Maximum Brake Temperatures 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?
As this pops up #1 on &quot;mtb brake temperature&quot; 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 &quot;capability&quot; 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,

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 &quot;kill&quot; 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 &quot;braking strategies&quot; 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 &quot;pulsate&quot; 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,

taken from &quot;Thermal/Mechanical Measurement and Modeling of
Bicycle Disc Brakes&quot;, ISEA 2018, Ioan Feier and Robin Redfield.