I follow aviation accidents pretty closely. For as long as I can remember, I have been fascinated with air and space travel and the associated risks. I even took a flying lesson at Teterboro airport 15 years ago or so, but I never got a chance to get certified.
The latest Boeing 737 Max 9 accident1 resulted in no loss of life, but the idea of a door, even a fake door, blowing out mid-flight warrants a little analysis. Some questions come to mind: How could this happen, why it happened, under what conditions (the proximal cause), and what could have happened under slightly different conditions? First, a warning: this is not an analysis of how the defect was introduced during manufacturing and assembly; I know nothing about that. This is an analysis of the conditions under which a loose door (I will call that, even though it was technically a door plug) could separate from the fuselage.
The Facts
On January 5, 2024, at 5:00 p.m. local time, Alaska Airlines Flight 1282 took off from Portland International Airport en route to Ontario, California. The sunset was at 5:25 p.m., so it was already fairly dark. When the aircraft reached approximately 16,300 feet (about 5,000 meters), the passengers heard a loud boom as the door blew out from the side of the aircraft.
Everyone on board survived, and the aircraft landed in Portland shortly after, but I am sure most passengers will have a different experience with air travel from that point forward. Many years ago, I was in a much less dangerous but scary landing that did not go as planned (no, not the runway miss with a fly-around, which I experienced twice and is not a big deal), but something more extreme. Since then, I have never been entirely comfortable during unexpected turbulence.
Conditions at 16,300 Feet and Above
The air temperature at this altitude is approximately 0 F (-17 C), and outside air pressure is about 54 kPa (kilo-Paskals) or about 8 PSI (pounds per square inch). For reference, sea-level air pressure is about 101 kPa or 15 PSI. This pressure is the column of air pushing on you as you stand on the ground, and by convention, this gives us yet another unit of pressure — 1 atm (atmosphere). Pressure is related to the number of air molecules available for each breath since the pressure is directly proportional to the density (this comes from the ideal gas law, which I will touch on later). At the cruising altitude of 33,000 feet (about 10,000 meters), the air temperature is about -50 (here, C and F units are close to each other), and air pressure is about 19.3 kPA.
At 8,000 feet and higher, there are not enough air molecules for most people to breathe, so modern aircraft are pressurized (sealed with cabin pressure controlled by the Environmental Control System).
Ideal gas law and what happens during the change in altitude
The fact the door blew out during the ascent was no accident, pardon the pun, and can be explained by the ideal gas law.
\[
pV = nRT
\]
The key quantities are Pressure \(p\), Volume \(V\), and Temperature \(T\). The other quantities are constants, so we can write:
\[
pV \propto T
\]
From the ideal gas law, pressure is inversely proportional to volume—as pressure decreases, the volume of gas increases, and vice versa. This makes some intuitive sense, as you can imagine what happens when you reduce the volume of a sealed container; the pressure inside goes up and vice versa.
When the pressure outside the sealed container rises, say during a submerging process, the walls experience an inward pressure due to a pressure differential, and when the outside pressure falls, say during an ascent, the wall experiences an outward pressure.
This is why, during a scuba lesson, you are told to keep your mouth open on the ascent so that expanding air doesn’t damage your lungs. For the same reason, when a flight attendant hands you a bag of chips at 30,000 feet, the bag appears inflated2.
This outward pressure on the fuselage caused the blowout, and we will now try to estimate how much pressure it took for the door to separate.
Computing the pressure on the fuselage
As a statistician, it always blows my mind how much we can learn from n=1 “experiments” if we are willing to bring some background knowledge, the ideal gas law in this case, to bear on the problem. Physicists would not be impressed, as to them, this is par for the course.
Anyway, back to our problem. To make the calculations, we need to make a couple of key assumptions, namely the pressure outside the aircraft as it climbs and the pressure inside the cabin. The drop in atmospheric pressure with altitude is well-known and follows an exponential decay according to the Barometric formula. This is the blue line in the following diagram.
The second assumption we need is the pressure inside the cabin. First, I assume that the target inside pressure is equivalent to about 6,000 feet (1,800 meters), which is at the lower end of the reported range. This type of pressurization balances the passengers’ comfort with the force that the fuselage has to withstand.
Second, I (erroneously) assumed that the pressure is gradually adjusted to reach the target at the cruising altitude of 33,000 feet, the green curve on the above diagram. I later learned that the target pressure is typically reached relatively quickly after takeoff and that the cabin pressure was most likely at its target at the time of the accident. This means that my green curve would drop much more quickly and remain flat for the rest of the flight. The green and red vertical bars at the height of the accident represent this error.
To compute the outward pressure on the fuselage (red), we take the difference between the two curves at 16,300 feet, which gives us 26 kPa, assuming the aircraft was pressurized to about 1,800 meters at the time of the accident. (If we incorrectly assume gradual pressurization, the pressure would be about 36 kPa.)
To make this more interpretable, we can compute how much force (in lb or kg-equivalent units) was applied to the door. We will assume the door is nearly rectangular with 72 x 34 inches (183 x 86 cm) dimensions, taken from a 737 manual. To compute the weight in pounds:
\[
W = P \cdot A \cdot 2.2/g
\]
P is the pressure in kPa, A is the area in square meters, and g is the gravitation constant. This turns out to be about 9,400 lb (4,300 kg), which I rounded to 9,000 lb in the diagram. At the cruising altitude of 10,000 meters, the weight on the door would be approximately 19,000 lb (~ 8,600 kg).
Conclusions
When the aircraft finally landed, the passengers were treated to the following view of the airfield.
The calculations can make us appreciate how much force every 1.6 square meters of fuselage must withstand, so constructing a well-pressurized cabin is an impressive engineering achievement, while forgetting to put a few screws into the door plug during a final assembly is a massive management oversight.
One can imagine a situation where the door was a bit more secure than it was (but still not fully installed) and that the separation would have happened at 33,000 feet instead of at 16,000. What then? I am not sure what the immediate depressurization at that altitude would do to a human body (if your mouth and nose are closed, your lungs may rupture; you may also suffer from decompression sickness; if you don’t put on the oxygen mask, you will suffocate), much less if the pilots could descend fast enough to avoid hypothermia and other pleasantries.
Up or Down
In June 2023, a submersible operated by OceanGate imploded near the remains of the Titanic. The Titanic is located at 12,500 feet (3,800 meters) under the sea, and the pressure at that depth is about 38,000 kPa, 380 times more than at the surface (~ 380 atmospheres). When the implosion was originally detected, the Coast Guard undertook a massive search and rescue operation. The rescue part was a fool’s errand — at this pressure, the passengers were instantly disintegrated.
- In 2019 and 2020, the 737 Max was involved in two accidents relating to the angle of attack sensors. ↩︎
- Why should the bag inflate inside the plane? That’s a good question, and this should be the clue that the cabin’s pressure is not kept at sea level but adjusted downward as the plane climbs. ↩︎