# 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? – nicoguaro Aug 31 at 16:29
• you cannot do this without compressive stress – lurscher Aug 31 at 19:34

## 2 Answers

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).