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Is it possible to make a solid rigid evacuated "balloon" out of Beryllium or other elements or alloys?

The critical buckling pressure at which an evacuated sphere is given as $$ P_1=\frac{2E\cdot\left(h/R\right)^2}{\sqrt{3\left(1-\nu^2\right)}} $$ where $E$ is Young's modulus, $h$ the shell's thickness, $R$ the radius to the midsurface shell, and $\nu$ is Poisson's ratio.

Of course this is theoretical and experimental and literature values for submersibles hover around 1/4 of the theoretical pressure. Beryllium has the following properties:

  • Density=1.85 g/cm$^3$
  • Young's Modulus=287*10$^9$ Pa
  • Poisson Ratio=0.032

Calculating the upward force, buoyancy at 101.3 kPa atmospheric pressure and Radius of 10m gives 4.24*10$^8$ N. The weight of this sphere with shell thickness 1mm has 2.32*10$^7$ N downward force of Beryllium material, so it will theoretically float at ground level.

Calculating the critical buckling stress of this beryllium sphere is 331.57 kPa, which is 3.27 times the atmospheric pressure of 101.3 kPa, so theoretically it won't buckle. But in experimentation done in earlier experiments as stated before, due to construction/material non-ideal situations, the critical buckling formula predicts about 4 times stronger than actual, so it would need about 400 kPa, not 331 kPa as Beryllium sphere with this specification would hold. Increasing the strength capability also increases weight and therefore lowers it's ability to float, but Beryllium is the closest element with this situation I have calculated that could work if ideal (I have only been using periodic table elements and household materials)

Is it possible with an alloy of elements? Using the equations for buckling and buoyancy, the following is needed. What alloys/materials come to your mind?

Material considerations:

  1. Material with larger Poisson ratio
  2. Material with larger Young's Modulus
  3. Less dense material
  4. Construction (is it possible)

Material considerations affect on lift

  1. Thickness can be larger, increases strength by the squared. But also increases weight. If a material with lower density with similar structural properties as Beryllium is used, it would help.
  2. Radius can be lower for same reason, but also lowers buoyancy (volume)

My other designs (don't worry about these just yet, just if you are curious is a vacuum sphere with a pressurized "outer tube" with spokes in tension (carbon fiber nanotubes maybe) to keep the 2 spheres in position. Another is a spherical truss out of carbon fiber nanotubes with a light membrane around. The truss prevents the outer membrane from collapsing due to the atmospheric pressure

The "point" of this if you ask, as to why not use a traditional balloon is it can be used as a boat by filling the sphere with air, and then used as a aircraft at will by evacuating the sphere. An alternative mode of transportation for sea vessels etc.

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    $\begingroup$ In air, the weight of air excluded from a given volume decreases with altitude, so a balloon of fixed volume has an altitude limit. High altitude balloons must be capable of expanding. $\endgroup$ – Mike Dunlavey Apr 6 '14 at 16:48
  • $\begingroup$ Just want to comment here that Beryllium is a seriously difficult material to handle because it so very very toxic/poisonous. Single exposure can lead to fatality. $\endgroup$ – tom Nov 19 '14 at 21:17
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Your figures seem problematic. For example, the volume of a 10m radius sphere is below 10^4m3, air density at normal conditions is, very roughly, 1kg/m3, so how do you get more than 10^8 N buoyancy force?

According to calculations in 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 ), no homogeneous spherical shell made of existing materials can be both strong enough to withstand atmospheric pressure and light enough to float in air. The critical failure mode is indeed buckling.

However, according to finite element analysis, sandwich structures made of existing materials can be both strong enough and light enough to float in air (please see the same patent application).

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  • $\begingroup$ This answer has not had enough attention. "Yes it can be done and I tried to patent it.". That. Is. Awesome. $\endgroup$ – Floris Jul 24 '15 at 20:39
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    $\begingroup$ @Floris There are actually two un-issued patent applications from 2006 (App 11/127613) which was abandoned 10/2006 and 2007 (App 11/517915) which was abandoned 6/2013. There's prior art in Armstrong, US patent 1,390,745 going back to 1919. For those of you who'd like to explore this technology: patents.google.com/patent/US1390745 $\endgroup$ – Bob Jacobsen May 16 '18 at 4:01
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To see if this is possible we can use a formal material selection approach of the kind presented in "Material Selection in Mechanical Design", Elsevier, M.F.Ashby. Our objective is to maximize buoyancy while keeping within the no-buckling constraint. To make things easier I suggest some simplifications.

First, Poisson's Ratio does not vary much for most metals. I took the constant parts of the buckling equation: $$ k={2 \over \sqrt{3(1-\nu^2)}} $$ and calculated it over a database of almost 3700 materials including all common aerospace composites, metals and alloys. This k varies from just 1.15 to 1.33. So I will treat it as a constant.

Next, assuming we don't really care too much how big our balloon is we define $\gamma=\frac{h}{R}$.

So now the bucking criterion is just: $$P<kE\gamma^2$$

For the buoyancy we have a force of $F_B = \frac{4}{3}\pi R^3 g \rho_A$ against a shell weight of $F_W = 4\pi R^2 h \rho_M g$ where $\rho_A$ and $\rho_M$ are the densities of the air and shell material respectively. So we define a lift factor Q: $$Q=\frac{F_B}{F_W}={R\rho_A \over 3h\rho_M}={\rho_A \over 3\gamma\rho_M}$$

As our quantity to maximize (or at least get bigger than 1!) Since we do not really care about the size of the balloon at this stage, we treat $\gamma$ as a free variable and combine to get: $$Q<\frac{\rho_A}{3}\sqrt{\frac{k}{P}}\left(\frac{E^{1/2}}{\rho_M}\right)$$ This defines the maximum achievable lift factor for any given material under the constraint that the sphere can withstand buckling. Since density and pressure of air are related by: $P=\rho_A R_s T$ where $R_s$ is the specific gas constant for air and T is the absolute temperature, we can rewrite this as: $$Q<{\frac{\sqrt{Pk}}{3R_s T}}\left(\frac{E^{1/2}}{\rho_M}\right)$$. So in general, to maximize the buoyancy we must find a material with a maximal value for $E^{1/2} / \rho_M$. Using the database I have access to, the best material I could find was Cyanate ester + high modulus carbon fiber composite which gives a value of 378,5 $N^{1/2}m^2 / kg$. As you rightly suggest Beryllium is close behind. Several alloys have similar values but the best I could find was for I-250 grade hot isostatically pressed Beryllium which gives a value of 305 $N^{1/2}m^2 / kg$.

In practice however, since we require $Q>1$ to achieve lift and P and T are defined for our planet then we can find a specific lower limit for this term. Taking T=300 K, P=101325 Pa and $R_s$=287 J/(kg·K). We obtain a minimum value of 730 $N^{1/2}m^2 / kg$ to achieve lift.

So in short, the best materials fall short of achieving our aim by a factor of just over 2. It is conceivable we might push up the stiffness / density ratio by perhaps developing some kind of Beryllium foam. For example, a closed cell foam of this Beryllium with a relative density of 0,041 would give a value of about 920 at the expense of lowering the Young's modulus to about 600MPa - however, I have no idea if such a foam is even possible. Alternatively it might be possible to come up with some clever engineering of the envelope geometry to overcome the buckling constraint. However, I suspect the effort is unlikely to pay off with a better boyancy ratio than is already achievable by conventional balloons.

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    $\begingroup$ A honeycomb structure might overcome the buckling concern. Otherwise I really like your analysis - especially coming up with a "minimum strength parameter" required. $\endgroup$ – Floris Jul 24 '15 at 20:38