This question of the worst descent speed for brake heating came up before so I'm cutting and pasting my answer from then:
"Let's calculate the speed at which the most power is being dissipated by the brakes. This is the difference between the drag power on the bike (assume all air drag for a fast descent) and the power exerted by gravity as the bike/rider descends the slope.

So let's take a bike/rider of mass m descending a slope of grade s at velocity v. The power exerted by gravity will be:
Pg = mgsv
and that of the air resistance will be (with 'c' a constant based on frontal area and shape):
Pa = cv^3
Terminal velocity (vT), or 'max. coasting speed' occurs when these two are equal, so
mgs(vT) = c(vT)^3, therefore c = mgs/(vT)^2
Therefore, the power dissipated by the brakes (Pb) will be:
Pb = Pg - Pa = mgsv - cv^3 = mgsv - mgsv^3/(vT)^2 = mgs[v - v^3/(vT)^2] = mgs(vT)[{(v/(vT) - v/(vT)**^3]
What we want to know is what value of v/(vT) gives the most heat dissipation by the brakes, so we can
substitute x = v/(vT):
Pb = mgs(vT)(x - x^3)
to find an extremum we differentiate and set equal to 0:
1 - 3x^2 = 0, so the maximum heat dissipation will be when x = sqrt(1/3) = v/(vT), and
the worst velocity v = vT/sqrt(3) or about 58% of the terminal velocity."

This isn't quite right since the brakes will be cooled by airflow over the pads and rims which will be more effective at higher speeds. So I'd expect the actual worst speed to be a little lower than calculated above.