# Thought Experiment: If you fell out of a plane at 36,000 ft (11km), would the additional air pressure from falling allow you to breathe?

Imagine you fell out of a plane at 36,000 ft (11km), roughly cruising altitude, with a parachute. If you deployed your parachute immediately, you would be stuck at high altitude with no oxygen, and considering oxygen at 36,000 ft is roughly 25% of oxygen at sea level, you'd quickly lose consciousness and suffocate.

Now if you waited to deploy your parachute, you'd rapidly reach terminal velocity for that air pressure, and if you fell with your mouth open, basically, would the additional air pressure from all that air you're falling into compress the air into your lungs sufficiently to reach or exceed normal breathing pressure? So in theory could you just spend the first few miles/km falling and breathing, and then deploy the parachute once you pass into breathable air?

• Not sure, but if you fell face down with your body spread out, at terminal velocity the pressure on your face would be your weight divided by your cross-sectional area exposed to the wind. For a female of 70kg and area of .8 square meters, that is about 9gm/sq cm. I'm not sure how that compares to sea level pressure. – Mike Dunlavey Nov 18 '16 at 18:50
• @MikeDunlavey That would come out to 900 Pa pressure force, which is about 0.01 atm. – Chet Miller Nov 18 '16 at 19:09
• @MikeDunlavey Clearly, body spread out is not the way to go, but that math is a very bad estimator of actual falling because of the fluid mechanics involved, so it could be off by orders of magnitude. That's why I included the fluid dynamics tag – TheEnvironmentalist Nov 18 '16 at 19:13
• I'm too lazy to do this. But here is what I would do. Calculate the velocity vs altitude assuming free fall with no air resistance. Then calculate the stagnation pressure at each altitude. See how this compares with 1 atm and how long it takes to reach the calculated stagnation pressure. – Chet Miller Nov 18 '16 at 19:33
• @ChesterMiller: Wouldn't you come to terminal velocity pretty quickly, and then drag = weight? Seems to me, at that point, speed doesn't matter much, because drag = pressure * area. If you're right about 0.01 atm, the diver is out of luck. – Mike Dunlavey Nov 18 '16 at 21:53

Here is a very rough calculation of the stagnation pressure. At standard conditions, the density of air is roughly 29/22.4 gm/liter, or 1.3 kg/m^3. At 11 km, the air density is on the order of about 0.2 times this, or 0.26 kg/m^3. Assuming that you are falling at a pretty high velocity of about 100 m/sec (328 ft/sec = 224 mph), the stagnation pressure at this velocity is $\rho \frac{v^2}{2}=0.26\frac{100^2}{2}=1300\ Pa$. This is roughly 0.01 atm above the ambient pressure of roughly 0.2 atm at 11 km. This is consistent with what we calculated in the comments using Mike Dunlavey's approach. So the conclusion is that the higher pressure resulting from falling through the atmosphere at 11 km (i.e., the added pressure at the entrance to your nostrils) would represent only a tiny fraction of what is needed to give ground-level pressure.