Let´s do some little math:
weigt of the disc brake: 0,2kg
specific heat absorption capacity of metal: 460 Joul / kg.
> spec. heat absorpt. capacity of 0,2 kg disc brake: =92 Joul
Speed: 45km/h (ok, let´s say 60kmh to be realistic): 16,6 m/s.
The deceleration in an emergency brake (half of the speed in meters) => 30 meters: 4,2 m/s²
weigt of bike + rider + gear of the rider: 110kg
But how to go on?
Nobody seems to have addressed your math/physics question for some reason, even though you came sooo close. What you were missing is E = 1/2 m v^2
You actually don't need to worry too much about how long your emergency stop is. If you are cruising at 60 kph on a 110kg gross weight vehicle, your kinetic energy at that point is
E = 1/2 m v^2 = 1/2 110kg (16.6m/s)^2 = 15266 Joules
If your disk brake rotor has a heat capacity 92 joules / oC, and you do all your braking just with the one disk brake so all this heat is dumped into the rotor, then the rotor temperature would increase by:
15266 Joules / 92 Joule/oC = 165oC.
That's hot, you'd burn your finger touching the brake rotor, but it's not a temperature that would compromise the rotor or the brake pads themselves. In practice, some of your energy will be lost to rolling drag, air drag, to heat on the caliper pads, and some of the generated heat will be shed to passing airflow, so you wouldn't see it _all_ get dumped into the rotor like this, but it gives a good worst case number.
It's not coming to a quick stop that pushes disk rotors to their limit, but long steady descents where the heat just keeps building up. Calculate your Joules from atop a mountain pass and that 15266 number becomes quite small. At this point, the heat capacity of the disk rotor isn't as much a factor as how effectively the rotor can shed heat to the passing airflow.
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