Stanford Torus

Artist's conception of a Stanford Torus. [Source]

Consider a Stanford Torus of radius $ 4km $, with the inner radius of the torus ring being $1km$. Disregard structural stability issues. To simulate $1g$ of centripetal acceleration, it would rotate at around $0.5 rpm$ generating a rotational speed of ~$200m/s$ at the edge of the torus. This structure is then filled with air to maintain $1 atm$ of pressure, with a mixture of gases matching Earth's own atmosphere.

What would happen if the hull is punctured? How rapidly would air escape from the structure? How would this variable change depending on the size and location of the puncture? How long would it take for the entire structure to fully decompress?

  • $\begingroup$ Homework problem? $\endgroup$ Commented Nov 4, 2015 at 18:14
  • $\begingroup$ @OrganicMarble This is a problem far more complicated than what my school would give as homework ;) No, it's a project I'm working on. I personally don't know the mathematics that go into calculating air venting through a hole of any certain size, so I came here for help. $\endgroup$ Commented Nov 4, 2015 at 18:23
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    $\begingroup$ Try googling for compressible flow equations and/or choked flow. The wikipedia article on choked flow has the equation you seek, then you can calculate the rest for yourself. $\endgroup$ Commented Nov 4, 2015 at 18:33
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    $\begingroup$ "How rapid would air loss be in the event that a supermassive torus is punctured?" For a small puncture, in a structure of that size? Minimal. At least, that is what the engineering students who worked on the initial designs determined. $\endgroup$ Commented Nov 4, 2015 at 22:14
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    $\begingroup$ First off, the loss would be quite proportional to the puncture size. You can't expect the same results from a millimeter micrometeorite and from a space shuttle crashing into it and creating a 300-meter long breach as the rotation drags it through the coating. $\endgroup$
    – SF.
    Commented Nov 6, 2015 at 10:06

1 Answer 1


Plugging my values into the Wolfram Alpha Venturi flow equation calculator, I got my result:

320,000 litres per second


$$Q = (\pi (D_1^2 \sqrt{\frac{2 (P_1-P_2)}{\rho}}\over(4 \sqrt{\frac{D_1^4}{D_2^4}-1))}$$

Q | flow rate
rho | fluid density
P_1 | upstream pressure
P_2 | downstream pressure
D_1 | upstream pipe diameter
D_2 | downstream pipe diameter
(Q is the flow rate as measured by a Venturi flow meter)

I cannot verify this answer as correct.

  • $\begingroup$ Got a link? Can you show the formula? $\endgroup$
    – kim holder
    Commented Nov 5, 2015 at 19:49
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    $\begingroup$ What value did you use for the size of the puncture? $\endgroup$
    – called2voyage
    Commented Nov 5, 2015 at 20:02
  • $\begingroup$ @called2voyage I used the larger diameter as 1000 metres, and the smaller diameter as 1 meter. $\endgroup$ Commented Nov 5, 2015 at 20:45
  • $\begingroup$ 1 meter does not sound like a small puncture, and 1000 meters is huge. $\endgroup$
    – called2voyage
    Commented Nov 5, 2015 at 20:48
  • $\begingroup$ Are you using this: wolframalpha.com/input/?i=Venturi+flowmeter&lk=3 ? $\endgroup$
    – called2voyage
    Commented Nov 5, 2015 at 21:23

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