Is it possible to construct an evacuated metal sphere which can float in the air without imploding?
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6 Answers
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I doubt if a vacuum metal sphere can be constructed that is light enough to have an overall density less than that of air. One of the lightest and strongest metals is beryllium ($d=1.85$) but since as the sphere has to be thin enough to save weight, it's unlikely to be able to withstand the outside pressure of $1\:\mathrm{bar}$.
But a large, thin beryllium sphere filled with hydrogen (at atmospheric pressure) could certainly be made to float in air, if you get the dimensions right.
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$\begingroup$ Gotcha, I think, graphene is full of holes.....? $\endgroup$– user140606Jan 18, 2017 at 15:22
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$\begingroup$ @Countto10-Hey, I see only now that you wrote this. I don't think I gotcha, I think, reading your comment... $\endgroup$ Jan 19, 2017 at 7:42
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The answer to "is it possible to build?" is "Not with current technology/materials".
However, in theory, this is of course possible.
Some years ago I did the math to reach an expression that relates thickness of a sphere shell and the density of its material, with the minimum diameter that would allow it to float in air. However, I tried some finite element method simulations with some well known materials, and the conclusion was clearly no (at least considering a spheric shell).
I can post the equation later when get home, but it's pretty straightforward to get there. At the time, I remember I tried considering metallic foams (such as aluminum foam) with a density of ~0.3 g/cm3, with a thickness of only a few centimeters, and the minimum diameter was something between 50 and 100 meters.
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1$\begingroup$ @cinico-But what happens if you make the radius, R, enormous? The surface area of the sphere varies with Rexp2, while the volume varies with Rexp3. Don't you reach, because of this, a point, while increasing R, where the uplifting force is big enough to let the sphere float and the thickness of the metal is big enough to withstand the outside air pressure? $\endgroup$ Jan 18, 2017 at 18:02
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It is not possible with a homogeneous spherical shell made of currently available materials (please see our US patent application 20070001053 at https://www.google.com/patents/US20070001053 ). The main failure mode is buckling. However, our finite element analysis (please see the same application) showed that it is possible using a spherical (inhomogeneous) sandwich structure, e.g., if the face sheets are made of beryllium and the core is made of commercially available aluminum honeycomb. To avoid the so called intracell buckling, the diameter of the shell made of these materials should be at least 3 m. This requirement can be relaxed using a different core.
answered Jan 18, 2017 at 14:37
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$\begingroup$ @descheleschilder: The scaling of the problem is quite different from that for helium balloons. If you look at the formulas for buckling (say, those in our application), you'll see that if you multiply all linear dimensions by the same number, the conditions for buckling will not change. You cannot increase all dimensions but the thickness of the shell, as the critical pressure for buckling is proportional to the squared ratio of the thickness and the radius. $\endgroup$ Jan 18, 2017 at 17:46
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There's a nice dimensional argument that says that you can't just consider very large spheres as a solution to this.
Consider an evacuated sphere of radius $R$, and consider cutting it in half. The force across this cut goes as $R^2$ obviously. The length of the cut goes only as $R$ though, so in order to keep the stress constant (and below the stress at which the material of the sphere fails), the thickness of the material must go like $R$ as the sphere becomes large.
The total volume of the material the sphere is made of therefore goes like $R^2\times R$, where the first term comes from the area and the second the thickness, as the sphere becomes large. So the mass of the sphere goes like $R^3$.
But the volume of the sphere, and thus the total lift, also goes like $R^3$. So, above a certain point if the sphere does not have enough lift then you can't fix this by making it larger.
Note: in practice buckling will be a much more serious problem as another reply said: I'm just putting this here because the existence of a bound is interesting, and it's too long to be a comment.
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$\begingroup$ @ttb-Nice one! In my answer, I arrived at the R-cubed dependence for the mass of the sphere in a much more difficult way! $\endgroup$ Jan 19, 2017 at 7:54
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It has been done, but with the help of helium inside the metal to aid in the displacement, when the Mythbusters flew a lead balloon. I know that the question is about an evacuated sphere, and not a balloon, but the difference is, from the point of view of physics, a question of engineering, not science.
answered Jan 19, 2017 at 6:49
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The weight of the air with a volume equal to the volume of the sphere, where the radius R reaches the inner side of the sphere is:
$\frac4 3\pi R^3\rho_{air}$
The weight of the metal sphere is:
$\frac {4} {3}\pi\rho_{metal} ((R+T)^3-R^3)=\frac 4 3\pi\rho_{metal}(3R^2T+3RT^2+T^3) $
Now
$ P_{crit}= (\frac T R)^2 $, so $T=\sqrt P_{crit}R$ (see the comment above, where T is the thickness of the sphere)
Substituting this in the previous expression gives the weight of the metal sphere:
$\frac4 3\pi\rho_{metal}R^3(3\sqrt(P_{crit})+3P_{crit}+P_{crit}\sqrt(P_{crit}))$
So if we can find a metal (which we haven't (yet)) for which or the density, or the critical pressure, or both are lower than is the case in present metals, we can let the sphere float in the air.
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$\begingroup$ This can be calculated with "Barlow's formula", which has been used for 150 years to calculate the strength of pressure vessels. $\endgroup$ Jan 19, 2017 at 12:58