In this interview (part of the extra material from the Apollo 13 film), Tom Kelly of Grumman talks about the construction of the Apollo LEM. In particular he mentions that the "skin" of the LEM was made of 0.012" aluminum and that "You could easily, if you were careless, put your boot or your foot right through that wall".

Is he talking about the outer pressure hull, or could a careless astronaut really have put their foot through the side of the spaceship?

  • 5
    $\begingroup$ I'm not a materials engineer, but 0.012" is about 18 times the thickness of household aluminum foil. That seems to me difficult to puncture by boot-related accident, but not impossible, and easy to puncture with something like a screwdriver. $\endgroup$ Commented Dec 16, 2014 at 23:30
  • 4
    $\begingroup$ @RussellBorogove 0.012" is a third of a millimetre, which is pretty tiny and seems puncturable to me. A 70kg man wearing a 90kg spacesuit has a lot of momentum. $\endgroup$ Commented Dec 17, 2014 at 11:23
  • 3
    $\begingroup$ @RussellBorogove James McDivitt in an interview about Apollo 9 included in the series "When we left Earth" that the LEM had thick pads on the floor expressly to prevent the screwdriver drop piercing accident you describe. He describes being a little horrified on seeing the real LEM as opposed to a model (which looked sturdy to him) for the first time. $\endgroup$ Commented Dec 17, 2014 at 13:35
  • $\begingroup$ It is only a third as thick as the skin on a Boeing 747, but then astronauts tend to be a bit more careful than tourists. $\endgroup$ Commented Jun 27, 2021 at 12:05

2 Answers 2


0.012 inches (or approximately 0.305 mm) is actually fairly sturdy, under the right circumstances. See this related answer as well, but a soda can may have a wall thickness of 0.1 mm or even less and is capable of containing 30 pounds per square inch (psi) against a standard atmosphere's 14.7 psi pressure on the outside (a 15.3 psi positive pressure). (The ICAO definition of a standard ground-level atmosphere according to Wikipedia is 1013.25 hPa = 101325 Pa, which Google converts to just under 14.7 psi.) The LM was not tasked with such a pressure difference: according to Apollo Expeditions to the Moon, chapter 4.4 and Apollo Oxygen And Two-Gas Environment Problems, Apollo missions used a cabin pressure of approximately 4.8-5.0 psi in flight. (NASA states "5 psi" on the former page linked whereas the latter page linked states a more exact "4.8 psi"; whether it is more accurate, I don't know, but the difference between the two numbers is small enough to be insignificant for our purposes.)

Under a positive pressure, materials tend to expand. Provided that the strength of the material is adequate to hold in the positive pressure (if it wasn't, the LM would have ruptured violently most likely during ascent tucked away inside the Saturn V as the pressure difference between its interior and exterior grew), this also tends to make them rigid. As even a thought experiment, take an unopened soda can and deliberately try to puncture it. You may succeed in puncturing the can, but it's likely to take a fair amount of deliberate effort. The ratio of positive pressure inside the soda can to the thickness of its wall (about 15 psi to 0.1 mm, or 150 psi per mm of thickness) is significantly higher compared to that seen by the LM pressure vessel while in flight in space, which is the only thing it was designed to do (about 5 psi to 0.3 mm, or 15 psi per mm of thickness).

So while 0.3 mm might sound very thin, and at the face of it sound like it is at risk of being punctured, by the time you consider what it really needed to do and what conditions it faced, it turns out that it most likely was quite adequate for the job. Going with thinner materials also saved a lot of weight, which made quite a big difference in the difficulty in going to the Moon at all. If the people designing and building the LM had used something "obviously sturdy enough", it's entirely possible that there would have been no need for the LM at all because the tyranny of the rocket equation might have meant the additional mass made just the difference that killed the idea of the Trans-Lunar Injection (or even earlier). It certainly would have required the Saturn V to be even larger than it already was, to get the additional mass into a transfer orbit to the Moon.

And of course, if the hole was small enough, you could just patch it up. Even a "simple" space suit can handle a puncture. Only in movies do things violently blow up or decompress because of a small hole. Heck, people have survived being exposed to very low air pressure, including during Apollo equipment testing as well.

  • 3
    $\begingroup$ Two things to remember: 1) the pressure load reduces the walls ability to handle a "boot" load. 2) Most metals (including aluminum) actually fail due to shear. Pressurization will mostly induce a tensile load (think hoop stress in a balloon). Only a portion of a tensile load results in shear stress in a material. The rest creates a hydrostatic component of stress -- which doesn't contribute to yield/failure. So the tensile strength of a material is not a good measure of the struture's ability to withstand a "boot"/puncture load. $\endgroup$
    – Erik
    Commented Dec 20, 2014 at 1:20
  • $\begingroup$ @Erik I can only imagine the structural engineers and launch engineers going back and forth between calculations on wall thickness, strength, and mass calculations, and the Tsiolkovsky equations for both the Saturn V and the LEM, to make sure there was enough Delta-V to get everything done with Margin of Safety for both the mission profile and the astronauts's boots. $\endgroup$ Commented Jan 22 at 10:43

The shear strength of 6061-T6 AL is about 30,000 psi. So if we conservatively estimate the perimeter of a boot to be about 24", and assume a rigid boot applying its load uniformly, we get about 30,000 psi x 24" x 0.012" = 8,640 lbf of pressure to yield the wall. This is often called punching shear stress. More (much more) would be required to fracture the wall. Realistically, the wall would start to bend and the failure mode wouldn't be pure shear. This of course ignores other loads (pressure for example) that are being applied to the wall at the same time and/or stress concentrators (like attach fittings and holes) -- which would further complicate the failure mode. But 8K+ lbf seems like a lot of margin.

But this really depends on the radius of curvature of the boot edge and the boot's hardness. As Michael Kjörling pointed out, 12 mils is pretty strong under uniform pressure. However, if I subject this to a concentrated point load I can, like a soda can, easily "put my foot through it." You can see this by taking a reasonably sharp awl (small radius of curvature and high hardness) to a soda can. I can puncture it with very little effort. Boots are soft and compliant though compared to an awl. And I suspect that the astronauts weren't wearing spurs...

As a side note, this is a critical load case for the design of the panel that runs down the floor of the aisle in commercial jets. High heels are rough...

  • 3
    $\begingroup$ In all fairness, using an awl is one of the things I would consider "a fair amount of deliberate effort". $\endgroup$
    – user
    Commented Dec 19, 2014 at 22:00
  • 1
    $\begingroup$ I saw that 30kpsi figure while I was trying to learn enough to answer this question; does it really mean 30,000 pounds per inch of area per inch of thickness as your calculation implies? $\endgroup$ Commented Dec 19, 2014 at 22:05
  • $\begingroup$ I was wondering about the same thing as @RussellBorogove. Can you really go from psi to inches to lbf the way you do? I would also imagine that the interior pressure of the boot makes a difference in its rigidity compared to the wall of the LM containment vessel, and I think one of the pages linked in my answer stated suite pressure was about 1 psi lower. (Basically, it would seem like you're dealing with two containment vessels, each with its own pressure and pressure differential.) Either way, the numbers are interesting and certainly contribute another way of looking at the question. $\endgroup$
    – user
    Commented Dec 19, 2014 at 22:06
  • 2
    $\begingroup$ If you subject the wall to pure shear, you can use the 30K psi value. It is a force per unit area under shear. Of course, this means you would need to support the wall on the other side. The wall benefits from any deformation, so this is a conservative approach. Extremely conservative in my opinion. $\endgroup$
    – Erik
    Commented Dec 19, 2014 at 22:26
  • 1
    $\begingroup$ Agree on the deliberate effort Michael. $\endgroup$
    – Erik
    Commented Dec 19, 2014 at 22:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.