Although I have seen a lot of literature online about how internal pressure resistance is calculated (i.e the pressure in a gas tank) I haven't found anything about the opposite scenario (i.e the pressures on a deep sea submarine).

How would one calculate the thickness of material needed and stresses involved at high pressures externally? Please assume a very basic level of physics knowledge (I have GCSE knowledge for sure and can probably work out and search up some of the more basic stuff, but I'm certainly no engineer).

  • 1
    $\begingroup$ It would depend significantly on the desired shape and the expected pressures. It also depends on what you mean by "needed". I'm not sure what constitutes a success and a failure in terms of your vessel, as I do not know it's purpose. $\endgroup$
    – JMac
    Jun 6, 2017 at 12:00
  • $\begingroup$ So, first start with the knowing pressure at the depth you're working at, this should help: pressure = density * gravity * height (depth in this case) + pressure due to the atmosphere. An example is here : grc.nasa.gov/www/k-12/WindTunnel/Activities/fluid_pressure.html $\endgroup$
    – Solar Mike
    Jun 6, 2017 at 14:35
  • $\begingroup$ Related: engineering.stackexchange.com/questions/3668/… $\endgroup$
    – Mark
    Jun 6, 2017 at 17:01

2 Answers 2


The mathematically determined formula for a sphere with a thin wall under external pressure (which is very inaccurate - keep reading!) is simply:

$$ \frac{2Et^2}{r^2\sqrt{3(1-\nu^2)}} $$

Where $E$ is the modulus of elasticity, $t$ is the thickness, $\nu$ is the Poisson's Ratio and $r$ is the radius of the sphere (to the midplane). This is a result of Euler's method of buckling when applied to sphere's. Ignoring how we derive this formula for now, it was tested in various laboratories before use.

When tested, however, this formula was found to be completely inaccurate. Spheres would collapse way before this value was ever reached. It turns out that any imperfections - any flat spots, or any wrinkles would get vastly exacerbated during the collapse and cause premature failure. There was no rhyme or reason to the methods.

Eventually, instead of the profound mathematical $\frac{2}{\sqrt{3(1-\nu^2)}}$, the empirically determined factor for most metals was determined to be 0.365. (Only 32% of the maximum possible). Other materials (or even different manufacturing methods for the same material!) have different empirical values, and various regulatory agencies and scientific bodies have dedicated hundreds of people and millions of testing dollars to uncovering these factors.

Different shapes, with different formulas also began to creep in, with similar problems - if they were perfectly curved they could reach the theoretical values, but they would cave within about 1/3 of the maximum theoretical value! To uncover the proper values for various shapes and materials you'd have to discuss with ASME or ASTM or other engineering organizations to find out what they've discovered. Mil Handbook 17F has a useful formula for most composites to reach close values, but it requires taking 900 computations and finding the smallest one!


You can find the formulas in petroleum industry ,like API, for cylinders = tubulars. This is a routine calculation done in all oil/gas well design. It will be listed under "collapse". It is in API bulletin 5C3, unfortunately they now charge $ for this.


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