# Is it possible to create a vacuum ballon using concentric partially pressured spheres?

The old saying goes, nature abhors a vacuum, but that isn't exactly true. The universe is full of vacuum; what nature abhors large energy differentials.

So for example, a vacuum chamber has to be strong enough to withstand the ~101kPa of pressure on the surface, and typically the materials required to build such a structure are heavier than the volume that contain.

I've often heard references to Neal Stephenson's Diamond Age, and people saying that perhaps one day we'll be able to manufacture small strong nano structures that are lighter than air (indeed, it's quite plausible: http://www.euronews.com/2012/11/01/lighter-than-air-material-discovered/).

However, I'm curious why concentric partially pressured spheres wouldn't work as an approach.

If you consider a structure which is in cross-section:

Outside | P1 | P2 | P3 | ... PZ ... | P3 | P2 | P1 | Outside


Where the value of P(n) is something like, say, P(n-1) * 4 / 5 (or some other fractional amount), and PZ ~= 0, it seems to me there's no reason why it wouldn't be possible to build the structure out of relatively light, relatively weak materials with many concentric layers that wouldn't collapse.

Perhaps there would be some minimum gap distance required and there are also practical considerations like, 'what stops the spheres from falling due to gravity and touching; when two surfaces touch the pressure difference is effectively the delta over the two open spaces; which will eventually be external vs. internal vaccum and break the structure', 'how would you depressurize the spheres', etc.

...but lets say for an ideal situation in which the structure floats in a gas without a strong gravity source to mess things up, and the pressure has been magically set in advance in each pressure zone.

Does this work?

If not, why does the structure collapse?

Could it be possible to actually fabricate such a structure?

(for example, a single pipeline running from the edge to the center could maintain the position of the spheres and depressurize them one at a time as required; the amount of material to do this (despite being dense) would be related to the radius of the sphere, while the buoyancy would be related to the volume of the sphere. Plausibly a net positive lift could be generated, I imagine...)

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The "lighter than air" material you referred to is a network of carbon tubes so the air can obviously get through this "material" – most of the volume occupied by this material is the "air" – much like it can get through clothes (e.g. wool), so you can't use this "material" as the envelope for a balloon.

More generally, there obviously can't be a solid that would be lighter than the air at normal conditions. The reason is that the distance between the atoms or molecules of a solid has to be much shorter – comparable to the Bohr radius – than the average distance between an air molecule and its closest neighbor. Even if you place the lightest possible atoms at these distances, the density will already exceed the density of air by orders of magnitude. There simply can't be light-as-air impenetrable solids made out of any atoms and it's easy to see why.

Your "many layers, gradually changing pressure" construction doesn't solve anything. You may equally well send the room in between the shells to zero without changing anything substantial so having many layers only changes one thing, the total thickness of the balloon's envelope. And to beat one atmosphere which is really a big pressure, one really needs a thick material. Just to be sure, 1 atmosphere is 100,000 pascals or so. It's 100,000 newtons per squared meter – or, equivalently, 1 squared meter of the material has to keep its flat shape if you place 10 tons on it (in normal gravity).

If you want a balloon with those 3,000 cubic meters (of the balloon my dad just flew with, a gift we gave him to the birthday), note that the vacuum only saves 4,000 kilograms or so. A 9-meters-radius balloon has the surface of 1,000 squared meters or so. For the balloon to be lighter than the air, the envelope must have a surface density of less than 4 kilograms per squared meters. Even if you used a hypothetical material as light as water, 1,000 kg per cubic meter, it would have to be just 4 millimeters thin. And it's hard to imagine that such a thin material may support 10 tons per squared meter.

The usual solution – put a lighter gas inside – is clearly simpler and better. The hydrogen is approximately 15 times lighter than the air. Helium is 7 times lighter than the air. Hot air at doubled absolute temperature is 2 times lighter than the normal air and so on. With some lighter air inside, one may naturally balance the pressures.

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I'm not disputing what you've said, but I'm down voting this because it doesn't answer the question. –  Doug Jul 15 '13 at 13:55

I am afraid your structure does not look promising. First of all, the greatest challenge for vacuum balloons is a failure mode called buckling (loss of stability - see an illustration at http://www.youtube.com/watch?v=Zz95_VvTxZM ), rather than material failure due to insufficient compression strength. For example, a homogeneous spherical shell made of any existing material, even diamond, cannot be both strong enough to withstand atmospheric pressure and light enough to float in air due to buckling (please see the calculation in my US patent application 11/517915, Akhmeteli, Gavrilin, Layered Shell Vacuum Balloons, you can find it at USPTO site or at http://akhmeteli.org/wp-content/uploads/2011/08/vacuum_balloons_cip.pdf ). And your design seems worse than a homogeneous shell because the critical pressure for buckling of a thin spherical shell is proportional to (shell thickness/shell radius)^2 (http://traktoria.org/files/pressure_hull/spherical/buckling_of_spherical_shells.pdf ), so when you replace a homogeneous shell with several thinner shells of the same total mass you get less, not more stability.

On the other hand, sandwich structures made of existing materials can be both strong enough to withstand atmospheric pressure and light enough to float in air, according to our finite element analysis (please see our application). However, manufacturing such balloons is difficult and expensive, and it has not been done yet.

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This is irrelevant if the thinner shell is strong enough to withstand that pressure; which, if the force exerted on it is less due to the pressure in the gap between shells, is not a big deal. The question is why would a single shell with pressure P1 on one side and P2 on the other side, have a force exerted on it that is magically unrelated to the P1 and P2. –  Doug Jul 14 '13 at 5:57
@Doug: The force you are talking about is determined by P1-P2, so it is indeed related to both P1 and P2, and this difference must be less than the buckling critical pressure for the thin shell. So if you replace 1 homogeneous shell with 10 thinner shells with the same total mass, the pressure difference for each of them is roughly 10 times less than that for the initial shell, the thickness of each of them is roughly 10 times less than that of the initial shell, so the critical pressure for each of them is roughly 100 times less than that of the initial shell, so you have net loss, not gain. –  akhmeteli Jul 14 '13 at 6:22
I can see the basic point you're making; basically you're saying it's a dynamic equilibrium right? The pressure exerted on any shell is f(V,K) where V is the total volume of the structure and K is the number of shells, and the weight of the entire structure is f(V,K,M) where M is the material the shell is constructed of. You're positing that there's no M such that you can find values of V and K where the net volume of displaced air is > the total used mass. I'm not seeing this as a solid proof... I'm pretty sure you could find a very light, reasonably strong material that would do this. –  Doug Jul 15 '13 at 13:53
@Doug : I don't know what you mean by dynamic equilibrium in this case. You may read about buckling at en.wikipedia.org/wiki/Buckling . I am not saying I gave a solid proof. I gave some estimates that make me highly skeptical about your structure. I am afraid I don't have enough motivation to give an in-depth analysis of your structure. I hope you'll understand. –  akhmeteli Jul 16 '13 at 12:17

After reading quiet a few dozen articles on this subject, the math seems to fall into two catagories: 1. People who claim that the physics disallows vacuum spheres of ANY size: http://en.wikipedia.org/wiki/Vacuum_airship 2. Others who claim that a larger sphere's volume being greater in proportion to it's surface area respectively, affords buoyancy (levitation) for given materials, at a certain threshold radius (such as 100m @ 1/2 cm thick, cast iron...): http://aoi.com.au/Originals/VacuumBalloon.pdf The theory seems good, if the volume (proportion) is "ramped" high enough, then it should eventually "afford" the lift (displacement) required for said (required) thickness of material. This, of course; presumes that there is no difference in mechanical stress upon the decreased curvature of said surface area? Does anyone have a definitive answer? And, if this limit exists, what is the (lower) limit for titanium, aluminum, carbon this, that and the other? I'm curious to know?

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compressive yield strengths are not as high as tensile yields. In the case of concentric spheres, you get the wrong impression that the pressure load is equally distributed between each sphere, but this is not the case: the external pressure must be entirely sustained by the innermost sphere.

I'm not sure if the above argument applies if the structure is partly tensile, something like this:

http://blog.wolframalpha.com/2009/05/19/whats-in-the-logo-that-which-we-call-a-rhombic-hexecontahedron/

Where the cusps are sustained by trusses. The surface needs to be made of something strong and light, as monolayer grapheme

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Could you perhaps direct me to an explanation of why the pressure load is entirely sustained by the inner most sphere? I'm almost entirely certain this is not correct; for example, if you have a series of sheets of paper in parallel with an air gap between them, flicking the first sheet with enough force to break it doesn't 'magically' transfer this force to the other sheets. Pressure isn't special; it's just force. –  Doug Jul 14 '13 at 5:54
I am not getting it, either, lurscher. Why should the innermost sphere - which has similar pressures from both sides - sustain the whole pressure? –  Luboš Motl Jul 14 '13 at 8:32
Sorry, I didn't remember all the details: The actual problem with this is that a single shell of a given thickness will be always more stable than removing inner shell gaps and replacing them with air. The reason is that the material can be better to distribute the compressive load than air can be. It is a slightly different assertion than the one I made –  lurscher Jul 14 '13 at 9:01
@lurscher if you have any kind of link to something talking about that, or a name for that phenomenon that I can look up, I'd love to read about it. –  Doug Jul 15 '13 at 13:54