According to Wikipedia, a vacuum airship is a hypothetical airship that is evacuated instead of using a lighter gas. Really, once I first saw this, I thought, "Why did I never think about this before?!" Indeed, whenever I pictured an evacuated container, I thought it would just sit there.
But no. Instead here on Earth, a theoretical vacuum airship could lift 1.28 grams per liter of evacuated space, a 14% boost over helium, owing to the density of air.

Of course, the air is also the destroyer of this powerful capability; it exerts 14 psi, or ~0.1 MPa!! According to Wikipedia's calculations, a hemispherical shell would have to withstand $$450,000 kg^{-1} m^5 s^{-2}$$ which is over 3 times the buckling pressure of diamond.

Give up? Not yet. There is a couple of interesting ideas I thought about:

  1. Why hemispherical? Wouldn't other shapes do well also? For example, a sphere, or a cylinder (maybe a cylinder with radius=height). From what I know (not much..) about spherical items, I suppose it would distribute the pressure evenly. Wouldn't that mean that diamond could easily sustain 100 kPa (withstands up to 600 GPa)?

  2. Is there no other way around this obstacle? Could we make use of any exotic forces or structural methods?
    (One interesting suggestion: Gradually evacuate the helium-filled airship, slowly going up, so that when we reach a total vacuum the airship is basically in space? In that case there wouldn't be any powerful pressure from air...)

  • $\begingroup$ A sphere would provide the best rigidity and strength. Considering that the container would need to be made of metal (likely aluminum), it would need to be quite large to negate the weight of the metal.. $\endgroup$
    – LDC3
    Jul 21, 2015 at 1:39
  • $\begingroup$ There's no law that says it must be a large monolithic sphere like a hot air balloon. Larger sizes have increasing problems in compression due to buckling. You would, instead, use a large number of tiny vacuum spheres. This would get you the closest to the theoretical limits you've referenced here. At those sizes, you might be able to use electrical charge to balance the crushing pressure, and then you'd surpass the theoretical limits. How to build it is a different story. $\endgroup$ Jul 21, 2015 at 1:40
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    $\begingroup$ @AlanRominger I assume you mean a honeycomb-like structure that has been evacuated? Okay, now I'm interested.. :) $\endgroup$
    – CoilKid
    Jul 21, 2015 at 3:00
  • $\begingroup$ possible duplicate of Beryllium Vacuum Sphere Boat/Aircraft $\endgroup$
    – user10851
    Jul 21, 2015 at 6:25
  • $\begingroup$ Not much advantage over hydrogen (except for the flammability) :D $\endgroup$
    – orion
    Jul 21, 2015 at 8:06

1 Answer 1


Sometimes I feel Wikipedia is a funny place... In the article you quote they provide a calculation from our patent application (see, e.g., http://akhmeteli.org/wp-content/uploads/2011/08/vacuum_balloons_cip.pdf ) proving that a homogeneous shell made of any existing material cannot be both light enough to float in air and strong enough to withstand atmospheric pressure. They give the reference to our application, so I am not complaining:-) However, we show in the application, using the finite-element analysis, that a vacuum balloon can be made of existing materials if you use a non-homogeneous, sandwich-like spherical shell structure. It is, however, difficult to understand that from the Wikipedia article (as this question proves:-) ).

  • $\begingroup$ Have you (or someone else) ever tried to realize (i.e. actually construct) a vacuum balloon like the one in the patent? With buoyancy exceeding weight. $\endgroup$ Jul 21, 2015 at 7:06
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    $\begingroup$ Maybe it is time for yo to edit the Wikipedia page to make it clearer. $\endgroup$ Jul 21, 2015 at 9:23
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    $\begingroup$ @JeppeStigNielsen: I did some preliminary work, spent quite some time and quite a bit of money:-). The results were as expected (sorry, cannot be more specific right now). I need more time and money to go further and am currently lacking both:-) Hopefully, I'll be able to tell you more in the future:-) $\endgroup$
    – akhmeteli
    Jul 23, 2015 at 3:01
  • $\begingroup$ @IanRingrose: Maybe it is:-) $\endgroup$
    – akhmeteli
    Jul 23, 2015 at 3:01

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