First, let’s dispel the notion of “friction” heating. At speeds where the air gets heated up enough to matter, the problem is *compression* heating. Air heats up when you compress it; this is true whether you capture the air in a diesel engine and squeeze it with a piston, or slam it into a flat surface so fast that it’s crushed by its own momentum. The latter phenomenon is the issue here: the hottest surfaces on a supersonic object are the surfaces facing forward, where the air is moving the slowest and has been compressed the most. On an aircraft, this is primarily the nose and the leading edges of the wings, but the air still isn’t back up to slipstream-speed as it passes along the rest of the aircraft, so there’s still plenty of warmth there as well.

Bernoulli’s equation governs the relationship between pressure, density, and velocity in a moving fluid:

P[sub]1[/sub]+0.5*rho[sub]1[/sub]*V[sub]1[/sub][sup]2[/sup] = P[sub]2[/sub] + 0.5*rho[sub]1[/sub]*V[sub]2[/sub][sup]2[/sup]

The subscripts 1 and 2 denote two different states of the fluid. In your case, state 1 has P=101 kPa, T1 = 0C, and V1 = the velocity of your aircraft. state 2 is the condition of the fluid after it’s slammed into the nose or wing of the aircraft and reached its lowest relative velocity (call it V[sub]2[/sub] = 0). For liquids, the density is constant, and so you can solve this equation by itself directly for P2. For air, the density changes as you go from state 1 to state 2, and so you need to add another equation for adiabatic compression:

P[sub]1[/sub]*(1/rho[sub]1[/sub])[sup]gamma[/sup]=P[sub]2[/sub]*(1/rho[sub]2[/sub])[sup]gamma[/sup]

“gamma” is the ratio of specific heats; for air, this is 1.4.

Now you have two equations, and two unknowns (P2 and rho2), so you can solve for your unknowns.

Once you’re done with that, you know P[sub]1[/sub], T[sub]1[/sub] and P[sub]2[/sub], so now you can figure out what your final temperature T[sub]2[/sub] is using another equation related to adiabatic compression:

P[sub]1[/sub]sup[/sup]*T[sub]1[/sub][sup]gamma[/sup]=P[sub]2[/sub]sup[/sup]*T[sub]2[/sub][sup]gamma[/sup]

The math gets a little bit easier when you have access to software that solves systems of polynomial equations while simultaneously calling up thermodynamic properties of substances. Assuming a vehicle moving through air at sea level (so P[sub]1[/sub]=101 kPa) and T=0C, I came up with this:

```
V[sub]1[/sub] (km/hr) T[sub]2[/sub] (C)
0 0
100 0.5366
200 2.137
300 4.773
400 8.403
500 12.97
600 18.4
700 24.64
800 31.6
900 39.2
1000 47.39
1100 56.09
1200 65.24
1300 74.77
1400 84.65
1500 94.83
1600 105.3
1700 115.9
1800 126.7
1900 137.7
2000 148.9
```

The point at which “heating overcomes cooling” will depend on what temperature you’re trying to maintain on your moving object. From the table, you can see that 600 km/hr results in a stagnation temperature pretty close to “room temperature” - although if you’ve ever ridden a motorcycle in a T-shirt when it’s 18C out, then you know it still feels pretty damn cold. OTOH, if you’re inside a capsule so that the 600-km/hr breeze isn’t ruffling your shirt, then 18.4 C just feels slightly chilly.

Passenger jets travel at something close to 900-1000 km/hr, but they do this at an altitude where the ambient pressure, temperature, and density are much lower; ambient temperature at cruising altitude is somewhere around -50C, and the result is that the stagnation temperature ends up being something close to 0C.