# Vacuum balloon / airship with balloons as shell segments

What are the problems with the following idea for building a vacuum balloon / airship? The question is targeted at the "making it float" part, ignoring other problems with super-light airships (steering, weather, etc).

Build the shell of the vacuum chamber out of smaller gas-filled balloons, like this:

The blue parts are gas-filled chambers, either using air, or for an extra lift, helium. The red part is vacuum. It is drawn as a circle, but I'm obviously thinking of a sphere.

The blue chambers would obviously not stay concave on their inwards-facing surface. That is just my poor drawing skills.

The material needed for the blue chambers would have to withstand the gas pressure vs. vacuum. A cheap toy balloon won't, but as far as a quick internet search goes, making a shell that can hold e.g. 1 atm of air vs. vacuum is certainly possible.

Fixing the individual segments is probably the interesting part. The "wedge" shape is intended to make them lock against each other from the air pressing them together, but being non-rigid, that alone will not be sufficient. The segments are glued together to fix that problem.

• Why do you need the blue parts? Wouldn't it be more effective just having the vacuum part? Commented Aug 31, 2019 at 16:29
• you cannot do this without compressive stress Commented Aug 31, 2019 at 19:34

So in your proposal, the balloons themselves provide only tensile strength. The resistance to compression comes from the pressure of the air inside the chambers. Let $$R$$ be the inside radius of the balloon shell, and let $$a$$ be the thickness of the shell. (So the outside radius is $$R+a$$.)

Now consider what happens when the radius shrinks to $$R-dR$$. The outside radius shrinks to $$R+a-dR$$, meaning that the device now displaces $$4\pi(R+a)^2dR$$ less air. This is energetically favourable, leading to a reduction of energy of $$4\pi(R+a)^2dR \cdot 1$$atm. On the other hand, the air inside the chambers now occupies $$4\pi((R+a)^2-R^2)dR$$ less volume, which is energetically unfavourable. If $$P$$ is the pressure inside the chambers, then the energy increases by $$4\pi((R+a)^2-R^2)dR \cdot P$$. Thus, the overall change in energy is:

$$\Delta E = 4\pi dR \left( P((R+a)^2-R^2) - 1\text{atm}(R+a)^2 \right)$$

If $$\Delta E < 0$$, the device will be crushed down to a smaller size by atmospheric pressure. Thus, in order to have the device be stable, the following inequality must be satisfied.

$$\Delta E = 4\pi dR \left( P((R+a)^2-R^2) - 1\text{atm}(R+a)^2 \right) \geq 0$$

Therefore:

$$P \geq 1\text{atm} \frac{(R+a)^2}{(R+a)^2-R^2}$$

Now let's compute the mass of this thing. Suppose that the density of air at atmospheric pressure is $$\rho$$. The total mass of air displaced by your balloon will be:

$$\frac{4\pi}{3}\rho(R+a)^3$$

On the other hand, the mass of the balloon, ignoring the mass of the balloon fabric, will be:

$$\rho \frac{P}{1\text{atm}} \frac{4\pi}{3} ((R+a)^3-R^3)$$

(Since density is proportional to pressure.) This is equal to:

$$\frac{4\pi}{3} \rho ((R+a)^3-R^3) \frac{(R+a)^2}{(R+a)^2-R^2}$$

$$=\frac{4\pi}{3} \rho (R+a)^3 \frac{(R+a)^3-R^3}{(R+a)^3} \frac{(R+a)^2}{(R+a)^2-R^2}$$

$$=\frac{4\pi}{3} \rho (R+a)^3 \frac{(R+a)^3-R^3}{(R+a)^3-R^2(R+a)}$$

Since $$R$$ and $$a$$ are both positive, we have:

$$\frac{(R+a)^3-R^3}{(R+a)^3-R^2(R+a)} > 1$$

Thus, the balloon must be more massive than the air that it displaces. Preventing atmospheric pressure from simply crushing the balloon requires a very high $$P$$, which means that the air in the shell has a high density. This high density adds enough mass to prevent the balloon from feeling any lift.

If you use helium instead of ordinary air, then you can use a smaller value for $$\rho$$ in the equation for the mass of the balloon, so you might get lift. However it is clear from the equations that you are best off making $$R=0$$. i.e. creating an ordinary helium balloon with no vacuum inside.

This structure was proposed and analyzed in https://arxiv.org/abs/physics/0610222 (although for a cylindrical geometry). The author's conclusion: "pressurizing with air can never lead to a structure that is lighter-than-air". I analyzed a similar structure for a spherical geometry and came to the same conclusion.

Together with my co-author, I proposed a viable design of a vacuum balloon using currently available materials (https://arxiv.org/abs/1903.05171 and references there).

• What would happen if you used a multi-layered approach where the innermost chamber is a vacuum, the next layer is 1/n atm, the next 2/n atm, ... and the last is (n-1)/n atm, where one layer of the structure would be able to support a difference of 1/n atm. Given today's 3D printing technology, though it might be difficult to evacuate the spacer layers of support material (a non-closed cell separation layer would have to be used), one could conceivably make this material very strong and lightweight and might be more technically feasible than the design you gave, given a large enough printer. Commented Jan 26, 2020 at 18:55
• @Adrian : I have not analyzed such a design, but I am highly skeptical about it, as it seems the layers would have to be very lightweight. Again, I cannot be positive about that. Commented Jan 26, 2020 at 19:03
• Of course, the number of layers would have to be minimised. Currently, SLM can do layers around 20 micrometers and have been used in creating rocket engines and other items used in aerospace. Commented Jan 26, 2020 at 19:46
• @Adrian : So I don't quite understand two things. Why this approach is more technically feasible than that of our preprint and why the 3D printing technology cannot be applied to the design of our preprint (please note that we refer to a 3D printing technology in the preprint). Commented Jan 26, 2020 at 20:24
• Sorry, where did you refer to 3D printing in that paper? I did a search on it just now and didn't find the word 3D or print in it anywhere. What is the "preprint" document that you are referring to? You cannot use 3D printing as it it stands now because you would require supports to keep the voids in the honeycomb. After the print is complete, how do you expect to remove the support material? They are closed cell, meaning that the support material has nowhere to go. Commented Jan 27, 2020 at 2:41