Vacuum airship

 ( Airship Configurations ) 

The density of air at standard temperature and pressure is 1.28 g/L, so 1 liter of displaced air has sufficient buoyant force to lift 1.28 g. Airships use a bag to displace a large volume of air; the bag is usually filled with a lightweight gas such as helium or hydrogen. The total lift generated by an airship is equal to the weight of the air it displaces, minus the weight of the materials used in its construction, including the gas used to fill the bag.

Vacuum airships would replace the lifting gas with a near-vacuum environment. Having no mass, the density of this body would be near to 0.00 g/L, which would theoretically be able to provide the full lift potential of displaced air, so every liter of vacuum could lift 1.28 g. Using the molar volume, the mass of 1 liter of helium (at 1 atmospheres of pressure) is found to be 0.178 g. If helium is used instead of vacuum, the lifting power of every litre is reduced by 0.178 g, so the effective lift is reduced by 13.90625%. A 1-litre volume of hydrogen has a mass of 0.090 g, reducing the effective lift by 7.03125%.

The main problem with the concept of vacuum airships is that, with a near-vacuum inside the airbag, the exterior atmospheric pressure is not balanced by any internal pressure. This enormous imbalance of forces would cause the airbag to collapse unless it were extremely strong (in an ordinary airship, the force is balanced by the pressure of the lifting gas, making this unnecessary). Thus the difficulty is in constructing an airbag with the additional strength to resist this extreme net force, without weighing the structure down so much that the greater lifting power of the vacuum is negated.

The total force on a hemi-spherical shell of radius $R$ by an external pressure $P$ is $\backslash pi\; R^2\; P$. Since the force on each hemisphere has to balance along the equator, assuming

$\backslash sigma\; =\; \backslash pi\; R^2\; P\; /\; 2\; \backslash pi\; R\; h\; =\; R\; P\; /\; 2\; h$

Neutral buoyancy occurs when the shell has the same mass as the displaced air, which occurs when $h/R\; =\; \backslash rho\_a/(3\; \backslash rho\_s)$, where $\backslash rho\_a$ is the air density and $\backslash rho\_s$ is the shell density, assumed to be homogeneous. Combining with the stress equation gives

$\backslash sigma\; =\; (3/2)(\backslash rho\_s/\backslash rho\_a)P$.

$P\_\{cr\}\; =\; \backslash frac\{2Eh^2\}\{\backslash sqrt\{3(1-\backslash mu^2)\}\}\backslash frac\{1\}\{R^2\}$

$E/\backslash rho\_s^2\; =\; \backslash frac\{9P\_\{cr\}\backslash sqrt\{3(1-\backslash mu^2)\}\}\{2\backslash rho\_a^2\}$

Akhmeteli and Gavrilin assert that this cannot even be achieved using diamond ($E/\backslash rho\_s^2\; \backslash approx\; 1\backslash cdot\; 10^5$), and propose that dropping the assumption that the shell is a homogeneous material may allow lighter and stiffer structures (e.g. a honeycomb structure).

$k\_\{\backslash rm\; L\}\; =\; 2.79\backslash cdot\; \backslash frac\{\backslash rho\_s\}\{\backslash rho\_\{\backslash rm\; atm\}\}\; \backslash cdot\; \backslash sqrt\{\backslash frac\{P\_\{\backslash rm\; atm\}\}\{E\}\}\; \backslash cdot\; (1-\backslash mu^2)^\{0.25\}$ (or, when $\backslash mu$ is unknown, $k\_\{\backslash rm\; L\}\; \backslash approx\; 2.71\backslash cdot\; \backslash frac\{\backslash rho\_s\}\{\backslash rho\_\{\backslash rm\; atm\}\}\; \backslash cdot\; \backslash sqrt\{\backslash frac\{P\_\{\backslash rm\; atm\}\}\{E\}\}$ with an error of order of 3% or less);

$L\_\{\backslash rm\; a\}\; =\; \backslash frac\{\backslash rho\_a\}\{\backslash rho\_\{\backslash rm\; atm\}\}\; \backslash cdot\; \backslash sqrt\{\backslash frac\{P\_\{\backslash rm\; atm\}\}\{P\}\}$ (or, when $\backslash rho\_a$ is unknown, $L\_\{\backslash rm\; a\}\; =\; 10\; \backslash cdot\; \backslash sqrt\{\backslash frac\{P\_\{\backslash rm\; atm\}\}\{P\}\}\; \backslash cdot\; \backslash frac\{M\_a\}\{T\_a\}$),