# Can Cloud Nine be built?

Cloud Nine is the name given by Buckminster Fuller to his proposed tensegrity sphere airborne habitats. The principle is simple and the physics seem to be sound.

For a sphere, as its radius is increased, the volume increase outpaces the surface area. The mass of the sphere depends primarily on surface area1. At some point the mass of the sphere is minimal compared to the mass of the air contained inside its volume. By slightly heating the air inside (and letting some escape), the density (and therefore total mass2) of the contained air can be significantly decreased with respect to the air outside. This will allow the sphere to float (via buoyancy) in the surrounding air.

None of this is unheard of, we're essentially talking about a hot air balloon with some thyroid problems.

So let's build one.

I want my sphere to be one kilometer in diameter and I want to keep the interior at 22 degrees Celsius (so I can live comfortably inside, of course). I'd like to be able to reach an altitude of one kilometer in a climate similar to the Rocky Mountains.

What materials, if any, can I use to build my Cloud Nine? Given the selected materials, how much mass can I lift in addition to the mass of the completed sphere?

Note: There seems to be an excellent design resource for tensegrity spheres here. The resource is full of equations, mathy stuff, which you'll need to answer this question. An answer of "the material needs to be, like, wicked strong and lightweight, dude" is not satisfactory. I want to know what materials will work (and calculations showing why), or if none exist, what materials would be required (and calculations showing why).

1: Of course the surface is not 2D, so the mass depends on the thickness too, which will need to grow in order to be self-supporting structure. However, I don't think the thickness will need to grow as fast as the volume.

2: See the ideal gas law, we want P and V to remain constant while T increases, thus reducing n in a constant volume. Fewer particles in the same volume means lower density and lower total mass.

Note: I am interested in the materials for construction of the sphere. The ability of a container to float due to temperature differential is not in question, hot air balloons perform such a feat daily. Assume a temperature differential can be achieved and will subsequently obey the ideal gas law as described.

Further Edit:
These aren't bubbles. A bubble has an increased internal air pressure to keep its surface inflated. I am fairly certain that a rigid structure is required to form the sphere. Otherwise the increased internal air pressure required to counteract the weight and surface tension of the enclosing material will negate the benefit of increasing the temperature. Again, see ideal gas law and footnote 2. A rigid sphere both can have the same air pressure as outside (no airlocks required) and maintain the spherical shape which optimizes the surface area to volume ratio (thus allowing maximal lift per unit mass of support structure). If you want to forego a rigid structure for your sphere, please use the correct volume for its final shape and/or use the final internal pressure value in your lift calculations.

• It can be built according to the resource, but I'm suspicious that the amount of heat to lift a village can be made, kept continuous, and not be unpleasant to those living there... I hope an answer includes maybe some gas expands the outer layers of fabric or something, to assist. Commented Feb 23, 2016 at 20:44
• @Mikey It shouldn't be too difficult to estimate heat loss through the entire surface area and energy absorption through the area exposed to the sun. I'm thinking the difference won't be so big it can't be engineered away. Commented Feb 23, 2016 at 20:51
• Must the material be rigid? Commented Feb 23, 2016 at 21:08
• @HDE226868 It certainly simplifies the living inside part. But, if you can find a way to achieve the broad goal without a rigid structure, I encourage it. Commented Feb 23, 2016 at 21:28
• If you're planning on an altitude of 1 km in the Rocky Mountains (or indeed, most of the US west of the Mississippi), your big problem is going to be digging a big enough hole. E.g. Colorado's lowest point is still a bit over 1 km altitude. Commented Feb 9, 2017 at 6:59

# Feasible Only with a Decreased Internal Pressure and Great Difficulty

This is a very complicated problem with many variables that affect the final outcome of the design. I made a few assumptions throughout and played with the numbers as much as I could in order to give a semi-complete answer. I feel like I could write an entire book on how this could or could not happen.

## Calculating Buoyancy with 1 Degree Temperature Difference

I'm going to use SI units in order to make this all simpler. I'm also going to focus primarily on the material requirements (since that is what the OP asked for) and am going to disregard the methods for heating up this behemoth. I am also going to use the assumption from Thucydides answer that the temperature within the sphere only needs to be raised 1 degree (I will use Celsius to make it easier and allow room for error).

In order to calculate the buoyant force, we must use the ideal gas law to find the internal density:

$$P = \rho T R_{specific}$$ $P$ is ambient air pressure, which is 101,325 Pa.

$$\rho = \frac{P}{T \cdot R_{specific}} = \frac{101,325\,\text{[Pa]}}{278\, \text{[J/kg K]}\cdot(295\text{[K]})} = 1.2355\,\text{kg/m}^3$$

so $\rho_i = \small 1.2355\,\text{kg/m}^3$

The density outside of the sphere is 1 degree cooler:

$$\rho_o = \ldots = \frac{101,325\,\text{[Pa]}}{278\, \text{[J/kg K]}\cdot(294\,\text{[K]})} = 1.2397\,\text{kg/m}^3$$

so $\rho_o = \small 1.2397\,\text{kg/m}^3$

The net buoyant force is equal to the density differential multiplied by volume and gravity:

\begin{align} F_b & = (\rho_o-\rho_i)V \cdot g \\ & = (1.2397\,\text{[kg/m^3]} - 1.2355\,\text{[kg/m^3]})\cdot\left(\frac{4\pi}{3}(500\,\text{[m]})^3\right)\cdot 9.81\,\text{[m/s^2]} \\ & = \text{21,585,879 N} \end{align}

$F_b$ = 21,585,879 N

You will need your sphere to weigh less than this in order for it to maintain lift, so the material required with need to be relatively light.

Most blimps use a Kevlar material for their crafts because it has low density in comparison to its tensile strength. Kevlar has a density of 1,440 kg/m^3 according to this table.

The surface area of our sphere is:

$A = 4\pi r^2 = 4\pi \cdot 500^2 = 3,141,590\,\text{m}^2$

Blimp Kevlar has a thickness between .5 and 4 mm according to this Goodyear Study. I will use .5 mm on the low end of the spectrum:

\begin{align} W & = A\cdot t_{hickness} \cdot \rho \cdot g \\ & = (3,141,590\,[m^2])(.0005\,[m])(1,440\,[kg/m^3])(9.81\,[m/s^2]) \\ & = \text{22,189,678.5 N}\end{align}

$\small F_b - W$ = 21,585,879 [N] - 22,189,678.5 [N] = NOT ENOUGH BUOYANCY TO LIFT JUST THE MATERIAL.

Allowed Density of Material

Okay, so what density would be allowed for such a material? Let's say you want to lift just the sphere and are disregarding lifting anything inside of it.

$\rho = \frac{F_b}{A\cdot t \cdot g} = 21,585,879\,[N]/{(3,141,590\,[m^2])(.0005\,[m])(9.81\,[m/s^2])}$

$\rho = 1,400.816\,kg/m^3$

You're so close! A slightly lighter material would work in this scenario, or you could create a greater temperature differential to increase buoyancy.

Greater Temperature Differential

If the temperature difference was increased to 10 degrees, the buoyancy would become 245,805,393.8 N, which would allow for the Kevlar material plus 223,615,715.3 N additional! A 20-degree difference would allow for an additional 464,201,618.3 N. You could carry any number of people and objects with this additional allowable weight.

## Accounting for Pressure Differential

Until now, I've been assuming atmospheric pressure inside and outside the Cloud 9, but according to this website the air pressure at an altitude of 1 Km is 89,908.62 Pa. The new density at this pressure and a temperature 1 degree cooler than the inside of 294 K would result in $\rho_o = \small 1.100\,042\,\text{kg/m}^3$. However, in order for the sphere to have any buoyancy, the outside density must be greater than the inside density. This could be achieved either by increasing the inside temperature or decreasing the outside temperature. Since you want the inside temperature to remain constant, let's look at the outside temperature.

The outside temperature would have to be a maximum of -12 degrees Celsius in order to provide any buoyancy for the sphere (Fb = 18,530,107.13 Pa) . However, you would have to be at -13 degrees to get even enough lift for the Kevlar that we investigated earlier. Each degree cooler will allow more buoyancy for the craft, so you might be okay on some cold winter nights in the Rockies, but don't expect your Cloud 9 to float in the summer or spring.

Reducing the Internal Pressure

You could try reducing the density inside the craft by decreasing the internal pressure. So let's say the people in the sphere are healthy and can live at .9 atmosphere. This would make the new internal density 1.1119 kg/m^3, so the temperature outside can now be a maximum of 289 K or 16 degrees Celsius to generate lift. This is a little more doable, but it does cause some issues if you have a particularly sunny day.

Decreasing the internal pressure any more than this will cause issues with our next problem: surface tension. In order for the material of our sphere to remain taut, the pressure inside Cloud 9 must be greater than the pressure outside. At .9 atm (or 90,179.25 Pa) our internal pressure is just greater than the outside pressure mentioned earlier of 89,908.62 Pa.

Therefore the pressure acting on our material is:

\begin{align} (P_i-P_o) & = 90,179.25\,[Pa] - 89,908.62\,[Pa] = 270.63\,[Pa] \end{align}

The tensile strength of Kevlar 29 is 2,860 MPa, so it would have no issue withstanding this pressure.

Alternative Materials

The next issue of course, would be how to construct anything within the sphere. You may want to go with a material stiffer than Kevlar in order to allow for buildings and structures to be built inside.

Any other material that you might investigate would most likely need to be thicker and denser to add any stiffness to the design. This causes all sorts of problems with your Buoyancy vs Weight difference. So let's assume you're living somewhere cold, like Antartica, where the average temperature is around -23.3 degrees Celsius. Your new buoyancy force is 604,753,584.9 Pa, which allows for a material with a thickness of 5 mm and a density less than 3,924 kg/m^3. There are any number of materials that would fit these specifications.

Let's say you make Cloud 9 out of Aluminum 6061 (see here) (which is well known for its low density). You would be able to lift the material, plus another 186,461,316.6 N. This number increases as the density of your selected material and the temperature outside decrease.

Geodesic Spheres

Geodesic spheres are spheres comprised of triangular elements which distribute the structural load throughout the system. It is possible that such a sphere, made of low density braces with high tensile and compression strengths would be structurally sound and light enough to act as a skeleton for Cloud 9. Stretching a strong and light material over this skeleton to hold in the heated air (perhaps something that harnesses the greenhouse effect), would allow for the buoyancy of the sphere.

The structural design and integrity of such spheres is a very advanced topic in itself and often requires specialized knowledge, tables, and/or software. Since I do not possess any of these things, I am not going to go into detail on this topic. Know that the weight of the sphere will cause there to be mostly compressive towards the bottom and most tensile at the top as the weight of the whole sphere pushes down on its self and tries to hold itself together at the top. The brace that experiences the most compressive force will most likely be your limiting factor, since metals undergo tensile stresses much more easily than compressive forces.

I could not calculate the weight vs buoyancy unless I knew the exact design of the skeleton, so I can not tell you whether this would work or not. However, it is a possibility.

# Conclusion

If you could maintain an internal atmosphere of .9 atm and an internal temperature of 22 degC, and you could ensure that there would be a maximum outside temperature of 16 degC, your sphere could be made of Kevlar and lift an additional 14,308,527.29 N (This is the equivalent weight of about 20,000 grown men).

If you want a stiffer material with a greater thickness, you could go somewhere colder, like Antarctica and use any material with a density lower than 3,924 kg/m^3. Aluminum for instance with a thickness of 5 mm, would allow an additional weight of 186,461,316.6 N (this is equivalent to the weight of about 261,400 grown men).

Finding a way to heat the inside and build any structures (a geodesic sphere could be the answer to this problem) are the next levels to this complicated problem.

• You should look at using MathJax in the posy. +1 for being the first to post hard-science! Commented Dec 30, 2016 at 17:54
• Would you kindly check/correct the URL for the Goodyear study? I'm getting a 403 error when I try to access it. Commented Dec 30, 2016 at 17:55
• Yea, the URL was abbreviated: I don’t think “...” is normally used as a path element. Commented Dec 30, 2016 at 17:59
• Why only 1 degree temperature difference? I would expect the ambient air 1 km above the rocky mountains to be rather colder than the 22°C interrior. Commented Dec 30, 2016 at 18:03
• made some edits, you may like to verify them Commented Dec 30, 2016 at 22:24

TLDR; Seems like it could be possible. I'm writing the TLDR at the end of writing the answer, and it honestly surprised me.

~~~~~~~~~~~~~

I think the big problem with it is actually generating the heat needed to cause the lift.

From the quote in Thucydides' answer, a sphere with radius 1320 feet would be able to float if the air inside were heated by one degree. Now, because science, I'm going to be assuming that that's one degree Celsius, because it makes the math oh so much easier. (Edit later: I noticed after the fact that the second quote specifies it as one degree Fahrenheit. That makes things better for us, as one degree Fahrenheit is less of a change than one degree Celsius, so we'd actually rise faster than expected by these calculations.)

The specific heat of a material is a measurement of how much energy is required to increase the temperature of one kilogram of the material by one degree Celsius. For air, the specific heat is about 1.007 Joules per kilogram per degree Celsius; for every kilogram of air, you need 1.007 Joules for every degree Celsius you want the temperature to increase.

A little bit of geometry and unit conversion thanks to WolframAlpha tells me that our 1320-foot radius sphere contains about 9.6 billion cubic feet of air (gotta love cubics), which comes out to about 300 billion liters (I believe WA computed it assuming NTP, but don't quote me on that). That means we'd need 20 trillion Joules of energy to heat the sphere by one degree. That's about 48 kilotons of TNT, so basically we'd need three of the bombs dropped on Hiroshima just to get off the ground.

Let's see if we can do it a bit less destructively. It was suggested in Thucydides' other quote that we could use the heat radiated by the human body to heat the air. Well, maybe not. Black-body radiation of a human is about 9 megajoules (as per Wikipedia); call it $10^7$ $J$ because we're just looking for a ballpark. We need $10^{14}$ $J$ to heat the air in our sphere. I don't feel like doing the math to figure out how many people we could fit on Cloud Nine, but I'm fairly certain it's lower than ten million - just try to imagine fitting everyone in New York City into a sphere less than half a mile across, and then cram in everyone from Philadelphia too. We wouldn't need quite that many in reality, as the black-body figure I quoted above is the human body at rest, and people would be working, moving around, etc, but it's a good enough estimate, as I don't think that would increase radiated heat by a factor of ten. Yes we would have different levels to fit as many people as we could, but there are two extra problems with that.

One, in order to fit more people you need more floor space. More floor space means more weight to lift off (in addition to more people meaning more weight, as people are denser than air [citation needed]), which means more heat we need to generate.

Two, more importantly, more people means less air. Every person that you add displaces a certain amount of air, and the ultimate volume of air that you have available to heat goes down. So in order to generate lift, you'd need to heat the (smaller) amount of air even hotter. Depending on how much air was displaced,

So basically it isn't going to work based off of human activity alone. The sun could help though. Our sphere has a cross-sectional area of about 509,000 square meters. According to a page I found from the University of Oregon (wouldn't you trust that formatting?) the Earth gets about 164 Watts per square meter, averaged over the course of the day. Therefore our sphere is (assuming perfect energy transfer blah blah blah) absorbing about $3.6*10^{12}$ $J$. So it is absorbing enough energy to heat the air, when coupled with human activity.

To finally get around to answering the core of OP's question, we could probably do it. We would need about two million people living inside it to generate the extra heat needed in addition to the sun. That is now looking like the big problem; how to fit enough people. I'm sure it would be possible to fit a lot of people by making various levels, but I've been writing this for a long time and don't really feel like doing the math to figure that part out. Leave that to an urban planner.

However: This does not take into consideration the fact that Cloud Nine will itself be radiating heat to the atmosphere. Especially at night, it would certainly cool down. Also, if we're having people living up there, we would need food, water, and material that wasn't included in the calculations.

My ultimate consensus is that it could very well be possible. It would depend on the size of the sphere, how many people are on it, how much stuff is on it, and (as was addressed in the comments) the strength/weight of the material used.

• Correct me if I'm wrong but the important parts wouldn't appear to be the total energy needed to heat the air up a couple of degrees, (you could do that quite slowly) but rather the energy radiated by the sphere while it's buoyant. If you can catch 3.6∗10^12 J worth of energy while avoiding radiating more than that much energy while keeping the temperature difference large enough then your sphere could float. Though another point is that the sphere would need to be structurally sound enough to withstand being caught in a large storm or hurricane without being shredded. Commented Feb 24, 2016 at 10:56
• I agree with Murphy here, you seem to be interchanging (or at least confusing) energy and power. It doesn't really matter how much energy you need to heat the air if there isn't a time limit to do it in. Also, you should be linking to your sources for hard-science questions :) Do you have any ideas on the materials required to build the sphere and its lifting ability? That was the focus of my question. Commented Feb 24, 2016 at 16:56
• I've clarified the question. Please check your answer to be sure it's actually answering the question asked. That is, this is a question about the structural material of the sphere. Additional details are welcome assuming the core question is answered in a hard-science fashion. Commented Jan 4, 2017 at 18:30
• Exclude 1,990,000 people and load some RTG generators and heat is not a problem. Human with cca 60W for ~80kg has lower power/mass coeficient than RTG, which can have values from 1 to 5 based on model. en.wikipedia.org/wiki/… Commented Jan 5, 2017 at 11:54
• What about heat leakage? Also, I'm confused about some of the numbers your giving. "9 megajoules per person" over what time period? How long does it take a person to omit 9 megajoules? Commented Feb 10, 2018 at 23:17

## Aluminum bars will work, with about six million lbs. left over.

I performed a simulation in an application called CADRE Pro for structural analysis of a 6V geodesic sphere. I chose this chord number so that the equator would be a horizontal line and for the relatively high approximation of a sphere.

I selected the members to be aluminum bars with a rectangular cross section of 10 inches by 8 inches. They are 94.08 lbs/ft. This brings the total structural weight to 33.3 million lbs. As calculated by kingledon the lifting force of the warmer air enclosed in a 1km diameter sphere is 44 million lbs. So, we have 11 million lbs. left for a covering and humans. That is, assuming this aluminum is strong enough...

And it appears to be. But only while airborne.

I added hydrostatic pressure (the force the sphere would experience via buoyancy) until the sphere cancelled its own weight and had a slightly positive resultant force (it's just lifted itself off the ground).

Numerous times before I got an error telling me a structural member had buckled and the simulation ended. But, with this particular size of aluminum bars, the structure held. The final shape is not a perfect sphere, but ends up stretched in the Z-axis by about 4% (too little to see in the resultant force diagram above).

By covering this structure in the material suggested by Faulkner, blimp kevlar, we use up another five million lbs. of our weight budget.

This leaves us with six million lbs. for fasteners, cabling, platforms, people & their stuff, and our old friend error.

• You could probably do much better than this by experimenting with different Aluminum alloys or using I-beams or pipes as opposed to bars. Commented Jul 12, 2022 at 18:00

## Lots and lots of insulation.

The previous answers have already delved into the difficulties faced by such a sphere in generating the necessary heat in order to float.

If the spheres are well insulated, it is possible to let the Sun do the majority of the heating work. After all, we are essentially dealing with a massive space heating issue here, and space heating costs fall when you have good insulation and double glazed windows.

The surface area of a sphere of 1320 feet in radius is approximately 2 million square meters, so assuming that Cloud Nine averages to have the insulative effectiveness of a single glazed window at $5.7 \text{W m}^{-2}\text{ K}^{-1}$, it will radiate around $0.55×5.7×2×10^6 = 6.3×10^6\text{W}$ of energy if it is one degree Fahrenheit above the ambient temperature.

Regarding heating the air itself, the hot air can simply be transported. By inflating Cloud Nine in the tropics and then moving it to the Rockies, the enormous heating costs can be avoided.

Six megawatts is a much easier energy output to sustain(simple solar heating alone of the $5×10^5\text{ m}^2$ of exposed surface area can provide 97MW of solar energy at $1700\text{kWh m}^{-2}\text{ year}^{-1}$ (ground level irradiance in the Rockies), and the thermal inertia of the massive amounts of air will simply do the rest.

• I would be very likely to be more concerned about over-heating when you consider common energy use for machines, electronics, etc. Say you had 100,000 occupants, consuming an average of 900 watts + 100 watts of body heat, i.e., 1000 watts total heating. You now have a 100 MW heat load, not considering waste heat from power generation or solar loading. It all depends upon number of occupants and how much energy they will use. Commented Feb 24, 2016 at 14:00
• Overheating isn't that big of a problem, just open the windows. Alternatively, air conditioners can be used to cool the air using solar energy. Evaporative cooling works too. Commented Feb 24, 2016 at 14:34
• Why the assumption of a 402 meter radius sphere? I asked for a one kilometer diameter. Do you have any ideas on the materials required to build the sphere and its lifting ability? That was the focus of my question. Commented Feb 24, 2016 at 17:01
• Actually the point was not that you could not prevent overheating, but that instead of worrying about heating, you are more likely to worry about cooling. Commented Feb 25, 2016 at 9:54
• I've clarified the question. Please check your answer to be sure it's actually answering the question asked. That is, this is a question about the structural material of the sphere. Additional details are welcome assuming the core question is answered in a hard-science fashion. Commented Jan 4, 2017 at 18:30

I will like to comment in the same post, but I need 50 rep..

I like SAMUEL calculation using Cadre Pro.. I like to hear that it does not need so much weight, even taking into account that 1 layer triangular shape is not the most efficient shape for a big geodesic dome, neither a v6 which is very low frequency for that size. Big geodesic dome are made with a 3d shape which use hexagons with light triangles connection in a second layer to maintain the structure shape. Using carbon struss like the Aeroscraft airship instead aluminum and layers of ETFE instead kevlar which provide a good insulation and high lifespan.

With that structure, I wonder how much pressure differential can support to increase the lift, lets take into account that spheres are very good to support uniform pressure. People can live without problem even at 0.5bar, also you can increase the oxygen to a 30 or 40% in any case.

About how to achieve the heat difference.. that is the most easy part.. Is a greenhouse good insulated and with huge volume/surface ratio that provide many days of thermal inertia, if inside the sphere, in the equator we add flat black panels able to rotate to receive the 100% of the sun or dodge all, then you have a way to control the heat inside to almost any temperature differential you want.

I made some calculation for my own before find this discussion, I have similar conclusion on the delta temperature needed.

One thing that nobody talks is what construction method to use to build such thing, and I guess I have the answer, but it will take another big comment.

• Angel welcome to the site. I can understand the frustration of not being able to comment (I recall doing the same thing at some point). That in mind please focus on asking questions and writing good answers until you are able to use the comments functionality! Trust me it doesn't take that long in the end :) Commented Feb 9, 2017 at 6:08
• Aside from a bit funny language structure, you can make this comment into a real answer if you flesh it out a bit. I, too, was frustrated over the time it felt that it takes to be allowed to comment - which turned out didn't take long time at all once I wrote an answer; all you need is five up-votes. I was told when I wanted to comment that one way is to, as example, build it into a real answer and starting it as "Expanding on the answer of [...]" and then add whatever calculations you have made and whatever extra you have discovered. Commented Feb 9, 2017 at 6:49

These 'balloons' need to be huge to lift enough weight, but seem feasible re physics and materials. (Though safety and sanity are different matters.)

We can learn from current blimp technology (about materials and rough weights) and extrapolate. IMHO, one needs truly vast volumes to loft the sorts of masses people would expect for a living or working environment. We can also make pretty good calculations for helium compared to barely-heated air for buoyancy. I suspect that will tell most of the tale.

Helium is monatomic and weighs about 0.18 Kg/m^3 at STP.
Whereas air weighs 1.3 Kg/m^3 at the same conditions.

So helium's buoyancy is 1.3 - 0.18 {call it 0.2} = 1.1 Kg/m^3 (almost as good as hydrogen and much safer!)

To make estimating easier, I'll assume a 3 degree (centigrade or Kelvin) increase. STP is close to 300 K, thus a 1% increase in absolute temperature. The density will go down correspondingly by 1%, and the lift is therefore the same 1%. So 1 our 3-degree heated air has a buoyancy of:

0.01 * 1.3 Kg = 0.013 -- thirteen miserable grams of lift (per cubic meter.) That's tiny compared to helium's lift.

Ratio (helium's lift vs 3K heated air): 0.013 / 1.1 = 0.0118 -- only 1 percent as much lift!

If we want to lift the same weight as with helium, we'll need something like 100x the volume as with helium! Assuming weight tracks volume, we'd need our thermo-balloon to be (cube root of 100 is ~5) five times larger (in linear dimension) than a corresponding helium blimp. And blimps are already huge compared to their payloads. Not impossible, but it's a big {huge actually!} engineering problem.

Here are URLs for two IMHO state-of-the-art blimp/ballon-like things; both use helium for most of their lift:

https://www.hybridairvehicles.com/aircraft/airlander-10

Note that these have payload capacities in the (few) tens of tons; the larger one is rated about 20 tons.

The OQ (original questioner) asked: [How] much mass can I lift in addition to the mass of the completed sphere?

If I make the following estimates:

Mass_avg_person = 100 Kg.

Mass ratio (person + all their stuff) / person's own mass: close order of 100. So each person + their stuff weighs 10 tonnes (metric ton, AKA megagram. How reasonable is this ratio? Big compared to that of nomadic people, but most nomads don't hang around in the sky, literally. This would have to include everything from structural mass to power sources to food and water.)

population onboard: 1000

we get a mass of 1000 * 100 * 100 = 1e7 Kg or 1e4 tonnes.
(IMHO low mass for realism, but plausible.) Cruise ships with similar numbers of people are much, much heavier, but we'll use low-density materials.

How big would our balloon need to be to loft this weight, with just a 3 K temperature differential?

Mass = 1e7 = (4/3)* pi * r^3 * 0.013

Solving for the radius, we get:

r = ((1e7 / ((4/3)* pi * 0.013))^(1/3) = 568.4 meters

This is in the plausible ballpark -- and close to Fuller's original size.

Faulkner wrote: "Blimp Kevlar has a thickness between .5 and 4 mm according to this Goodyear Study." Let's see what happens when we scale that up.

We may need a thicker skin to handle the larger structural and wind-related loads, but for now use the Goodyear data on thickness. If we take 2 mm as nominal for blimps and a specific gravity for kevlar of 1.4 (1400 Kg/m^3), how much would that weigh?

4 * pi * (1000^2) * .02 * 1400 = 3.5e8 Kg or 3,500 tonnes. Note that this nominal envelope mass is already 35% of what our 1 km radius sphere can lift

This is just the envelope, without any other structure. We may need carbon nanotubes. :-(

The key is that Bucky Fuller was right, the envelope mass can be small, compared with the airmass it encloses for a large enough balloon -- but that balloon is going to be enormous!)

If I use the same numbers to estimate the envelope-only mass of a 100 m diameter balloon at 2 mm thickness, I get 350 Kg., from which I surmise (compare to specs on blimps below) that the envelope is quite a small portion of those blimps' mass budgets, though I didn't find weight figures for those blimps :-(
(Please check me; that 350 Kg seems kinda low.) Structure, engines, fuel/batteries etc. look like they overwhelm the envelope mass, even at blimp scale.

We can already build big balloons and blimps. Those typically use materials with high strength to weight ratios, such as the Spectra family of polyethylene.

Wikipedia has some names and nominal values of high strength to weight materials: https://en.wikipedia.org/wiki/Specific_strength Suggest you look at Kevlar, Dyneema and Zylon as practical for use in envelopes, while we await economical, mass-produced carbon nanotubes.

• Ok, I'm having some trouble parsing this, but here is what I think you've got. 1) Helium's lift is much better (but not breathable and is 2.4 times the global annual use for a single sphere). 2) The radius of a sphere which describes the volume of heated air to lift 1k people + stuff is 568.4 meters, but that doesn't include anything to contain that air. 3) The mass of a thick skin covering is 350 tonnes. There are parts of the whole in there, but I'm not seeing A) What material for the sphere (structural and any covering)? and B) What's the remaining lifting capability of the empty sphere? Commented Dec 30, 2016 at 22:06
• @Samuel, these are estimates, not a detailed mass budget. We can see that the envelope weight (350 mT) is small (only 3.5%), compared to the total we could lift with a 3K temperature differential and 1 Km radius. That's what I was mainly driving at. Commented Dec 30, 2016 at 22:12
• @Catalyst This? Commented Dec 30, 2016 at 22:40
• If I use the same numbers to estimate the envelope-only mass of a 100 m diameter balloon at 2 mm thickness, I get 350 Kg. - no it not float, it have lifting force about 6500kg, and mass about 120 ton with 2mm wall thickness and 2000kg/m^3 density of the material, with 1% lifting hot/cold air difference. Commented Dec 31, 2016 at 0:15
• I've clarified the question. Please check your answer to be sure it's actually answering the question asked. That is, this is a question about the structural material of the sphere. Additional details are welcome assuming the core question is answered in a hard-science fashion. Commented Jan 4, 2017 at 18:31

First off, I am completely failing to comprehend the advantage of tensegrity structures in this context. So that may make my answer useless to you. My investigation seems to indicate that the primary stress that this structure will be subjected to is the tension between weight of gravity on the structures and inhabitants (which are presumably at or near the bottom of the sphere) and the lift force which is applied to the top half of the sphere. Since this is a simple one-dimensional force (forces in two direction on one axis), the obvious solution, to me, is a guy cable. More on this in the second section below.

So I'm going to investigate some material properties needed for this design, but it is possible (perhaps probable?) that there is some tensegrity-based structure which will have superior performance to what I am proposing.

# Calculating lift force

As mentioned above, the primary tension is going to be between lift force pulling up and wight pulling down. We need to calculate the magnitudes of these forces.

Using the ideal gas law to calculate lift force, and with pressure and volume kept constant we have an equality between the interior (designated 'int') and the displaced air (designated 'air'): $$n_{int}RT_{int} = n_{air}RT_{air}\tag{1}.$$ We want to use this to calculate lift force, which I do do by $$\text{lift} = \text{weight}_{air} - \text{weight}_{int} = gA\left(n_{air} - n_{int}\right)\tag{2}$$ where $g$ is force of gravity, $A$ is the atomic weight of air (.029 kg/mol).

The mols of air inside an 0.5km radius sphere can be solved from the ideal gas law. At a height of 1km, I will use 89kPa air pressure and 10 C. $$n_{air} = \frac{PV}{RT} = \frac{89000 \text{ Pa}\cdot 5.2\times10^{8} \text{ m}^3}{8.314 \text{J/mol-K}\cdot 283 \text{ K}} = 1.98\times10^{10} \text{mol}.$$ Solving (1) we can get $n_{int}$ in terms of $n_{air}$ as $$n_{int} = n_{air}\frac{T_{air}}{T_{int}}.\tag{3}$$

Plugging (3) into (2) we get $$\text{lift} = gAn_{air}\left(1-\frac{T_{air}}{T_{int}}\right) = 9.8\cdot.029\cdot1.98\times10^{10}\left(1-\frac{283}{295}\right) = 2.29\times10^{8} \text{ N}$$ or about 23,000 tons; about two AEGIS cruisers (or a quarter of an aircraft carrier).

Coming soon

# Spherical cow estimate of a self-supporting balloon envelope.

Let us say that all of the lift force is applied to the top half of the sphere while all the weight of the contents is applied to the bottom half of the sphere. Half of the weight of the balloon will also be applied to the bottom half; the other half will be subtracted from the lift force on the top half.

Now lets consider the tensile stress on the 'equator' of the sphere. The total (one-dimensional) stress equals upwards force of lift plus the downwards weight of the contents and half the balloon.

To restrict ourselves to real-life materials, we will make the balloon out of the strongest fabric that I can find. HEXCEL woven carbon fiber fabrics (specifically the IM10-12K) have an advertised tensile strength of around 6000 MPa and density of 1800 kg/m$^3$. The thinnest sold carbon fiber fabrics are about 0.11mm thick at 0.20 kg/m$^2$.

The circumference of our sphere at the equator is 3141 meters. The total cross sectional area of the envelope, on which force is being applied is $3141 \text{ m} \cdot t$ where t is the thickness of the envelope in meters.

The mass of the either half of the sphere is half the surface area of the sphere times the thickness times the density: $1570000 \text{ m}^2 \cdot t \cdot 1790 \text{ kg/m}^3$.

The tensile stress applied to the cross-section at the equator is equal to twice the lift force minus the weight of half the envelope divided by the cross sectional area ($A$). Given the ultimate tensile strength of 6000 MPa, lets say we do not want this stress to exceed 2000 MPa for safety ( I don't know what an appropriate safety margin is). In equation form, this is $$\frac{2\cdot\left(\text{lift}-\text{weight}\right)}{A} = \frac{2\cdot\left(2.29\times10^{8}-9.8\cdot2.81\times10^{9}\cdot t\right)}{3141\cdot t}\lt 200000000000$$ We want to find the minimum thickness that meets this inequality. I solve that to be $t = 0.07 \text{mm}$, which is conveniently close to our 0.1mm minimum available fabric thickness.

If we can get 0.07mm thickness fabric, then the total mass of the envelope is about 393 tons, leaving about 22500 tons of lift capacity remaining for the occupants of said sphere.

Conclusion: I'm as surprised as you, but until I find a math error (or realize that my strength tolerance of 3:1 is ludicrous), it looks like you can lift a 1km diameter bubble and support ~23 thousand tons of cargo with just a carbon fiber fabric envelope and no further internal structural support.

# Spherical cow estimate with internal guys for support

Coming soon!!

Also coming soon, testing Kevlar, Nylon, and Dacron to see if we can get better performance.

Also also coming soon, accounting for wind shear

• I think the tensile on top and compression on bottom assumption is an error. Each unit sized latitude ring (henceforth lat-ring) will both be pushing up the lat-ring above (compression) and pulling up the lat-ring below (tension) (except the tipy-top and bity-bottom, technically speaking). This is the whole idea of tensegrity, structural units contain both a compression and tension member. Fuller's spheres are spheres rather than bubbles. A bubble (no structure) won't work, because the surface tension is equalized by increased internal pressure. More pressure negates the lifting force. Commented Dec 30, 2016 at 22:20
• @Samuel What force from a lat ring is pushing upwards on the lat-ring above it? Commented Dec 30, 2016 at 22:28
• It's part of the force required to hold that upper lat-ring up. The lift can't be simplified/abstracted as a cable attached to the top pole of the structure, right? Because it's generated by the entire volume of the sphere. It just looks to me like you might be thinking of it as a bubble, with just surface tension required, but the inside of bubbles have increased air pressure to hold the shape. I don't think the air pressure required to inflate this bubble would be habitable. I could be wrong in my thinking here, so take all this as just my thought process, not fact. Commented Dec 30, 2016 at 22:36
• But in any case, the entire point of the tensegrity sphere was to distribute the forces across the entire surface of the sphere (geodesic dome style, and incidentally, bubble style). Actually they're kind of like a rigid bubble. A non-rigid bubble could possibly work, but to get lift the bubble needs to expand while always having an increased internal air pressure to keep the shape. Programmable tension would be useful there, but at the cost of significantly increased complexity. Sorry if this just got a lot less fun... Review is usually more fun than learning something complex and new. Commented Dec 30, 2016 at 23:07
• @Samuel You bring up excellent points. Basically, my proposal involves a 'hoop' around the equator that provides outward radial pressure, while the lift force of the interior gas provides upward radial pressure to maintain the shape. I didn't mention that, so I should demonstrate that it works (if it does). I will add a section demonstrating that the shape can maintain itself. This is why you start on day 1, so there are 6 more days before the bounty expires :) Commented Dec 31, 2016 at 0:47

I started doing some research on this question and it get's complicated pretty quickly and I wasn't up for a full hard science answer today, but I found a few interesting building material possibilities not already mentioned that I thought should be added.

So for this structure you want to maximize material strength while minimizing density which presents a few good materials (a simple useful list I found https://en.wikipedia.org/wiki/Ultimate_tensile_strength and more detailed material properties can be found on http://www.matweb.com/). I ended up making a strange comparison unit of tensile strength/density (MPa/[g/cm^3]) which is what units all the following number are in.

Nylon (800) and Kevlar (2600) look pretty good but tensegrity structures require elements in both tension and compression and they don't do well in compression, (like pushing a rope) they may however be used in conjunction with other material elements in compression.

Metals generally weren't comparing well due to their higher densities; Steels (70-200), Aluminum Alloys (50-150), Titanium (50-80); most of these are used in aerospace for not just strength/lightness but also higher temperature application which isn't an issue in this case.

Carbon Fiber laminates (900) looks pretty good and one material that I found doing unexpected well was bamboo (1250).

Another material that I think would be very useful not for structural purposes but for a liner material would be silica aerogels, They are incredibly light around 0.001-0.002 g/cc (air is 0.0012), they are great insulators and are optically clear, which would maximize the solar heating and minimize heat loss to keep the structure afloat.

• You at least get a +1 for being the first one to mention that this structure requires more than tensile strength. Commented Jan 5, 2017 at 22:05

This is in response to a comment by Samuel, but is going to be an answer due to the extra length.

Buckmaster Fuller's insight into the properties of a geodesic dome came from a realization that increasing the size of the dome increased the volume enclosed in a square/cubed relationship. The larger the dome, the more quickly the volume of entrapped air increased, so eventually only a small change in temperature would allow a dome to lift off.

[The following extract from a paper posted to GEODESIC by Robert T. Bowers explains the idea.] When considering a geodesic sphere, the weight of the sphere is a function of the surface of the sphere. The amount the sphere is lifted by warm air is a function of the volume of the sphere. In mathematical terms, weight is a function of the radius squared, while volume is a function of the radius cubed. This is very significant. Even as the radius of a sphere increases, thus increasing the sphere's weight, the lift of the sphere increases more. If you image a sphere that could grow larger, as the sphere gained a little weight, it would gain much lift.

Buckminster Fuller proposed that as spheres of great size are considered, the amount of air enclosed grows huge compared to the weight of the sphere. Of a sphere with a radius of 1320 feet, the weight of the enclosed air is 1000 times greater than the weight of the sphere's structure. If that volume of air was heated only one degree, the sphere would begin to float!

http://www.geniusstuff.com/blog/flying-cities-buckminster-fuller/

I know it sounds like science-fiction, but here’s how Bucky proposed a Cloud Nine would work. A half mile (0.8 kilometer) diameter geodesic sphere would weigh only one-thousandth of the weight of the air inside of it. If the internal air were heated by either solar energy or even just the average human activity inside, it would only take a 1 degree shift in Fahrenheit over the external temperature to make the sphere float. Since the internal air would get denser when it cooled, Bucky imagined using polyethylene curtains to slow the rate that air entered the sphere

Evidently Fuller considered this more of a thought experiment than a serious proposal, and aside from a bit of discussion about anchoring Cloud Nine's to mountain tops or using them as free flying city-states, he evidently never explored the idea much further.

Fuller himself:

$$$$Of course, domes of even greater sizes would be required if that sphere were to carry any additional weight. But it is not inconceivable that floating geodesic spheres could carry aloft entire communities. Perhaps the concept of a floating dome of one half a mile diameter is too much for most people to seriously consider. Regardless, it does demonstrate the scope of projects that are made possible with geodesic domes.'' -Robert T. Bowers Fuller quote from I Seem To Be A Verb

• Hmm, I'm not seeing anything new here, just a slight expansion of background details I've already included in the question. This doesn't appear to answer any part of my question, so, all of my original questions remain. What materials, if any, can I use to build my Cloud Nine? Given the selected materials, how much mass can I lift in addition to the mass of the completed sphere? I want to know what materials will work (and calculations showing why), or if none exist, what materials would be required (and calculations showing why). Any ideas on the specific details I'm asking about? Commented Feb 24, 2016 at 0:40
• I've clarified the question. Please check your answer to be sure it's actually answering the question asked. That is, this is a question about the structural material of the sphere. Additional details are welcome assuming the core question is answered in a hard-science fashion. Commented Jan 4, 2017 at 18:31

Ok let me revise this...

First off we need to know how much would a geodesic sphere weigh... before anything else...

For that, to get anywhere close I need to know the information at this website: http://www.desertdomes.com/dome6calc.html
And here's a better calculator, but not the one i used unfortunately - http://acidome.ru/lab/calc/#1/1_Inscribed_Fullerene_on_Piped_D108_3V_R500_beams_150x50

60 lengths of 130.0m = 7,800m
60 lengths of 152.3m = 9,138m
120 lengths of 145.5m = 17,460m
180 lengths of 162.2m = 29,196m
60 lengths of 149.9m = 8,994m
120 lengths of 158.4m = 19,008m
240 lengths of 164.7m = 39,528m
120 lengths of 172.2m = 20,664m
120 lenths of 173.3m = 20,796m

giving us a total material length of 172,584m
let's say 10 centimeter width and thickness
That would give us 1,725.84$m^3$

That would give us a total of 3,624.264kg if we use: https://en.wikipedia.org/wiki/Metallic_microlattice

It will likely have to be heavier still though cuz you're probably not going to get 100+ meter segments like that, but who knows.

The next is Covering or what goes between those geodesic struts...

The covering would likely be made out of Carbon Nanotubes while if you just fill in between the struts you're going to use the same metal.

In the latter case you just have to find area of the geodesic and then multiply by the weight of the material... I can't find a way to get the area so no clue.

The Former case is - The surface area will be 2,010,000 $m^2$.

Miralon® Sheet and Tape

These products are available with standard areal densities of 12 and 25 grams per square meter and can be post-processed for specific applications. Miralon sheets can be prepregged with a variety of resin systems or infiltrated with various polymer systems to better suit your application, all with industry standard equipment.

Let say half of mass will be proper for the task resin, so we get 50 gram per square meter and the whole sphere will weigh 100,480,000 grams or 100 tons.

The lifting force of such sphere depends on the difference of temperatures, which is given by ideal gas law:

$$PV=nRT$$

at 10C and a difference of temperatures about 15 degrees - lifting force will be about 5% of the atmosphere density at that altitude - with air it is about 68.9 gram per cubic meter or less(with higher altitude).

So potentially the sphere may lift 17,638,400 kg.

The next piece of weight is going to come from creating a solid floor and roof. I assume you can use the metal latice to do that, but it depends on where you put it for how much weight you are going to add on. If you put it at the center it obviously will give you more room, but add more weight. The max weight would be something like 105,557.508kg a the top or bottom part of the habitat area which you'd need 2 of for 211,115.016kg

You also definitely want to segment the habitat like this because then you can heat the air as you like, keeping the lower portion just hot enough to keep your inhabitants at comfortable temperature and the upper portion being hot enough to do the primary lifting of the structure. This would however need two heaters, reducing how much you can carry, but "probably" increasing the overall lifting capacity.

An average medieval city population would weigh about 3,200,000kg
To feed this population you need 3 acres of top soil which comes to about: 14,568,720kg.

Reduce this by half and you could probably get everything you need in one of these with about 20,000 people, probably a few less if you're not looking to be vegan, You'd also need to walk everywhere, need fairly thin walls, etc, much like medieval society really. However, if you did set this up like this you'd also want to set up medical facilities and again have to reduce your population due to heavy machines like catscanners, x-ray machines and such.

So I was wrong initially and this is possible, but doesn't seem like a practical use of space and you'd always have to be heating which i have no numbers for since you'd needs 2 pretty big heaters and lots of fuels stored somewhere. Not to mention water supply and insulation from more radiation from being at least 4 meters up... I suppose you could wrap the habitat area with a water insulator to solve two problems at once. The bottom line is that it is "possible", but a seemingly engineering nightmare to get everything all balanced and stabilized.

Here's a quick layout of how i'd set this up. Fuel and heater sizes are random, water sheath is bigger than scale due to it being too small to see at scale.

As mentioned above, if you do this you'd be able to heat the upper area a lot more and get more lift while the bottom could be heated, but kept cooler to provide lift as well. Other people would have to figure out the optimals for the locations and heat ranges for those though...

• Comments are not for extended discussion; this conversation has been moved to chat. Commented Jan 5, 2017 at 19:45
• I was one of the downvoters, and I've removed my downvote. I haven't checked your numbers yet, but this is a huge improvement over the original post. Thank you. Commented Jan 5, 2017 at 20:37

# Feasibility of Cloud Nine

I did some exploration into this idea and thought I'd share my results here. TL;DR at the bottom.
In this answer, I attempt to explore the feasibility of a Cloud Nine style airborne habitat.
I’ll look at:

• modeling atmospheric conditions for different altitudes;
• calculating buoyancy & lift;
• estimating the mass of the geodesic tensegrity superstructure;
• estimating the mass of the the envelope/membrane;
• optimal operating altitudes;
• building a model Cloud Nine;
• energy requirements, solar-thermal input, and losses to waste heat;
• and interior architecture for buildings and living structures.

First, I’ll describe some of the equations and methods I’ll use, then I’ll apply them to build a model Cloud Nine and inspect its feasibility.

## Atmosphere Model

From NASA’s document on U.S. Standard Atmosphere, I get the following formula for modeling barometric pressure at various height regimes:

$$P=P_{b}\exp\left(\frac{-g_{0}M_{air}\left(h-h_{b}\right)}{RT_{b}}\right)$$

• $$P_b$$ = reference pressure [Pa]
• $$T_b$$ = reference temperature [K]
• $$h$$ = height at which pressure & density are calculated [m]
• $$h_b$$ = height of reference level (meters; hb = 11,000 m)
• $$R$$ = universal gas constant: 8.3144598 J/(mol·K)
• $$g_0$$ = gravitational acceleration: 9.80665 m/s^2
• $$M_{air}$$ = molar mass of Earth’s air: 0.0289644 kg/mol

Next, from the same source, I get the the following formula for modeling atmospheric density:

$$d=d_{b}\exp\left(\frac{-g_{0}M_{air}\left(h-h_{b}\right)}{RT_{b}}\right)$$

• $$d_b$$ = reference density [kg/m^3]

From the table supplied in the paper, I use the pressure & temperature reference values calculated for heights less than 11,000 m:

• $$h_b$$ = 11,000 m
• $$P_b$$ = 22,632.10 Pa
• $$T_b$$ = 216.65 K
• $$d_b$$ = 0.36391 kg/m^3

To model temperature, I use this approximation from NASA for temperatures at heights less than 11,000 m, adjusted for units in Kelvin:

$$T=288.19-0.00649h$$

(It's surprisingly accurate, compared against empirical data I was using before it.)

## Buoyancy & Lift Equations

The buoyant force of an airship, regardless what mixture is used within the envelope, is the difference in densities between interior and exterior, multiplied by the volume of the balloon, multiplied by gravitational acceleration:

$$F_{b}=\left(d-d_{i}\right)\cdot V_{sphere}\cdot g_{0}$$

I calculate the external density, temperature, and pressure using my model atmosphere, and the internal density using the Ideal Gas Law:

$$P=\frac{d_i}{M_{air}}RT$$

Rearranging, we get:

$$d_{i}=\frac{PM_{air}}{RT}$$

where:

• $$d_i$$ = internal density [kg/m^3]
• $$P$$ = atmospheric pressure [Pa]
• $$T$$ = desired internal temperature [K]
• $$M_{air}$$ = mean molar mass of Earth’s air: 0.0289644 kg/mol
• $$R$$ = universal gas constant: 8.3144598 J/(mol·K)

From the equations, we can see that our options for increasing the density differential (thereby increasing buoyancy) are increasing either the pressure or temperature differentials. Building a multi-kilometer-wide pressure vessel sounds annoying compared to just heating the interior up, so I’ll go with the latter. This means I'll use the outside pressure from the atmosphere model for the internal pressure of the balloon.

## Geodesic structural mass

The formula to calculate the various strut lengths of a geodesic sphere is actually pretty simple. You just need the chord factors, the number of strut members for each given factor, and the sphere radius. Chord factors are constants that pop out in unit space and remain identical no matter how you resize the structure or what your units are, like angles. Each chord factor is resized according to the sphere/dome radius, and multiplied by the number of struts per strut type (A-I). You can find those figures and also a handy calculator here. I’ll be using the values for a 6V geodesic sphere.

total strut length [m] = sphere radius [m] * chord factor * NOF strut members
total combined length [m] = sum( total strut length [A - I] [m] )

Alternatively, using math from this paper, you can calculate the total combined strut length of a geodesic icosahedron, $$L_{total}$$, using the following equations:

$$S_{PCF}=2\sum_{a=1}^{f}\sum_{b=1}^{a}f_{PCF}\left(a,b\right)$$

$$f_{PCF}\left(a,b\right)=2\sin\left(\frac{1}{2}\operatorname{arccot}\left(t\left(a\right)+\left(\frac{a}{2}-b\right)\left(\frac{a}{2}-b+1\right)t\left(a\right)^{-1}\right)\right)$$

$$t\left(a\right)=\frac{1}{2}\sqrt{\left(f-a\right)\left(f-3a\right)+f^{2}\cot^{2}\left(\frac{\arctan\left(2\right)}{2}\right)}$$

$$L_{total}=20S_{PCF}\cdot r_{sphere}$$

where:

• $$f$$ = chord frequency (1, 2, 3,...)
• $$r_{sphere}$$ = radius of the sphere

To estimate the mass of the total structure, take the total combined length of all the strut types (A-I) and, treating them as cylindrical spars of a certain radius, compute their volume and multiply by the chosen material’s density.

volume [m^3] = pi * strut radius^2 [m^2] * combined length [m]
mass [kg] = volume [m^3] * density [kg/m^3]

Cloud Nine is a tensegrity structure with “floating compression” members. The calculation I’ve just shown is for the mass of the compression members. What I’m missing is the mass of the tension members, e.g., cables. I’m not sure how to find the lengths of those mathematically, but, as a rough approximation, we can just double the result of the mass of the compression members. It’s likely way off because, for certain ultimate stresses, tension members are usually less massive than compression members (think cables versus I-beams). It usually takes more to pull something apart than to crumple it up. The way I see it, doubling the compression member mass is a very conservative estimate. As we’ll see later, we have a lot of lift-mass to play with, anyway.

## Envelope/Membrane Mass

Finding envelope mass is as simple as multiplying the surface area by the thickness of the membrane, getting an approximate volume, then by material density.

envelope mass [kg] = surface area [m^2] * thickness [m] * material density [kg/m^3]

I could calculate the volume of a spherical shell instead, which would be more accurate, but for large-scale objects like Cloud Nine with paper-thin skins, the difference is beyond negligible.

## Optimal Altitude for Flight

For populated, non-pressurized Cloud Nines, I figure the maximum altitude will be around the highest elevation humans tolerate living at. The highest-elevation urban settlement in the world is La Rinconada, Peru, with 30,000 residents at 16,700 ft (5,100 m), living at half standard pressure. Cloud Nine appears to function even better at higher altitudes than this, but for the sake of comfort, I’ll stop there. The optimal altitude for maximum buoyancy is probably where the ratio of air density to ambient temperature is greatest (when we are constrained to a steady-state internal temperature).

Basically, there comes a point along increasing altitude where density begins falling faster than temperature, and we start losing buoyancy. At best guess, I'd say that point could be as high as 40 km.

## Building a model Cloud Nine

Using the equations above, let’s construct a Cloud Nine tensegrity sphere with the following properties:

• Sphere diameter: 2,000 m (1.24 mi)
• Height above sea level: 5.1 km, elevation of La Rinconada, Peru
• Internal temperature: 300 K (80 F)

The atmospheric conditions at 5,100 m are:

• Pressure: 57,380 Pa
• Temperature: 255.1 K
• Density: 0.923 kg/m^3

Our 2 km wide Cloud Nine has an internal density of 0.67 kg/m^3, creates a density differential of 0.26 kg/m^3, and exhibits a buoyant force of over 10.5B N, or a “lift-mass” of over 1B kg or 1M metric tonnes. The mass of the supporting superstructure needs to be less than that to maintain buoyancy at 5.1 km altitude. Indeed, it needs to be a lot less than that to support the mass of a whole “village”. Let’s look at that.

Running through the geodesic strut calculations, we arrive at a total material length of 215,780 m of struts. If we use Aluminum alloy 6061, with density 2,700 kg/m^3, as our construction material (known for its low density and being a common “aircraft aluminum”, see here), and if each strut is 20 cm in diameter, then we calculate a total mass of ~18,300 metric tonnes for the compression members.
Assuming the tension members weigh just as much (which they likely don’t, but let’s be generous), we double that figure to get ~36,600 metric tonnes for the geodesic tensegrity skeleton. (There are many options to explore for the compression and tension member materials--way too many.)

Most blimps use Kevlar as a membrane material for its properties of low density and high tensile strength. According to this table, Kevlar has a density of 1,440 kg/m^3. According to this Goodyear study, most blimps have a Kevlar membrane thickness between 0.5 mm and 4 mm. I’ll use the lowest value, choosing 0.5 mm. Doing the math, we get an envelope mass of ~9,000 metric tonnes. Choosing 2 mm would bring the mass up to around the mass of the tensegrity superstructure itself at ~36,000 tonnes.

The total mass of the geodesic tensegrity sphere + Kevlar envelope is around ~47,000 metric tonnes. A tiny divet into our lift-mass budget of over 1M tonnes. Our Cloud Nine can lift over 22.5 times its own structure’s mass, or over 23 Lexington-class battlecruisers. If each inhabitant on average requires 10 tonnes of material (ballpark figure), then the sphere can carry over 100,000 people. Definitely a village’s worth.

To find the power requirements of the internal heat engines I find the energy lost via transmittance through the Kevlar membrane, which represents the energy that needs to be added back into the system to maintain a steady-state temperature condition. For that, I look at the thermal transmittance of the entire Cloud Nine. The paper-thin Kevlar envelope has a thermal conductivity of 0.04 W/(m•K) (source). Dividing by our chosen membrane thickness of 0.5 mm gives us its thermal transmittance U-factor of 80 W/(m^2•K). Cloud Nine will leak heat like nobody’s business.
I figure the envelope could be painted with a dark material, further insulated with a foam layer, or both to bring that value down to that of at least single-glazed glass at 5.7 W/(m^2•K).

thermal transmittance [W] = surface area [m^2] * U-factor [W/(m^2*K)] * (internal temp - external temp [K])

Crunching the numbers, we get a transmittance over the entire sphere’s surface of 3.22 GW. Yup, gigawatts. That’s how much power needs to be pumped back in to maintain our 300 K internal temperature. It shouldn’t be all that surprising considering the sheer quantity of air inside the sphere. Our insulation could be better than that of darkened glass. A well-insulated roof has a thermal transmittance of only 0.15 W/(m^2•K), which would drop the energy requirement into the megawatt range.

## Gains from Solar Flux

The original Cloud Nine concept was made buoyant entirely by heat energy from the Sun. Let’s see how close I can get. According to this paper, Kevlar has a thermal absorptivity coefficient of 0.47. To find the solar energy absorbed, we multiply the solar irradiance times the surface area normal to the light rays times the (dimensionless) absorptivity coefficient.

solar input [W] = solar irradiance [W/m^2] * cross-sectional area [m^2] * absorptivity

Solar irradiance at sea level on a clear day is usually about 1,000 W/m^2, and that value only gets larger as you go higher due to there being less air between you and the top of the atmosphere to attenuate the Sun’s rays. I’ll just use 1,000 W/m^2 as a constant. Cranking the analytical engine, we get an input energy of ~1.5 GW when the Sun is out. Close to half of the 3.22 GW requirement for maintaining a 300 K internal temperature. We get some irradiance on the bottom of the sphere, too, reflected up from Earth’s surface, but for simplicity's sake I’ll ignore that.

We could improve the absorptivity of the envelope with a coat of paint, a different membrane material, or an additional layer of some high-absorption substance. With an improvement in insulation, the gains in efficiency could drastically reduce the power requirements and the mass of the heat engines.
Out of curiosity, I rearranged the thermal transmittance equation to find the steady-state temperature with solar irradiance alone. I calculate the Sun will heat our Cloud Nine sphere up to ~275.7 K. Plugging that into the buoyancy equations as our new internal temperature, we can find the new free lift-mass after subtracting the structure mass: ~780,000 metric tonnes. Cloud Nine can hover at 5.1 km altitude while carrying almost 18 Lexington-class battlecruisers without any energy input from onboard heat sources.

## Power Station Requirements

Our power requirement during the day, when the Sun lifts some of our energy burdens, is 3.22 - 1.5 GW = ~1.72 GW. At night, depending on how well-insulated the structure is, the requirement may be the full 3.22 GW -- or it may be a fraction of that. Over 3 GW is the power output of a good-sized nuclear power station, so it’s probably worth the extra mass to cake on extra insulation, especially since it’s such a small proportion of our total lift-mass budget.

If we increase the insulation to that of a well-insulated roof, our power requirements drop to ~85 MW, far smaller than what we receive from solar energy. With that kind of insulation, we could warm up during the day and cool down at night while losing very little internal temperature and altitude, day by day, ad infinitum. 85 MW is about 40 Hoover dams, close to the output of some of the smallest nuclear reactors which usually bottom-out at a shameful 50 MW. If the nuclear reactor mass can be made as little 5-10 kg per kW, then the entire powerplant could weigh as little as 423-847 metric tonnes. If instead it is as high as 500 kg per kW, then the powerplant would weigh ~42,000 metric tons, about half an aircraft carrier or about as much as the geodesic tensegrity structure itself. We would still have much of the original 1M tonnes to work with for other structures.

With current battery technology as heavy as it is, storing excess power to any meaningful extent seems out of the question.

There may be the possibility of cladding the envelope with a thin solar array to more efficiently draw solar energy. Using figures from Space Future, “current best” aerospace solar cell arrays can reach 4,600 W/kg with a density of 30 g/m^2. Assuming protection against the environment reduces that by an order of magnitude, 460 W/kg, and assuming half the sphere is clad in the solar cell film, I get a total mass of ~200,000 kg and an input of ~87 MW. Not tremendous, but nothing to sneeze at either.
Another potential option is outrigging large mirrors or reflective foils to focus more sunlight onto collectors for internal heating, or outrigging wind turbines to collect wind power. With such a large cross-sectional area, space-based microwave beam power seems like an enticing option, and if fusion power could be realized and built small enough, that could potentially work too.

## Interior Architecture

This section is a little less rigorous and is mainly just me guessing. We have approximately 1M tonnes to work with, mass enough for basically any number of people and structures. I assume we want to take advantage of how the tensegrity structure spreads tension forces over the surface of the sphere and hang everything inside on guy wires. There are plenty of material options for the necessary tensile strength. Structures and levels might be suspended like suspension bridges, with additional supports guarding against bad oscillations/resonances, shearing, and torques.

## Conclusion

Cloud Nine seems possible.

I've calculated that a 2,000 m diameter tensegrity sphere, made of 20 cm diameter Aluminum 6061 spars and tension cables, enveloped in a 0.5 mm Kevlar membrane, masses ~47,000 metric tonnes. With an internal temperature of 300 K (80 F), at an altitude of 5.1 km, the sphere displaces over 1 million metric tonnes of air.

The sphere absorbs 1.5 GW of solar energy and naturally heats to an internal temperature of 275.7 K. With no energy input, the sphere can hover at 5.1 km altitude while carrying 780,000 metric tonnes of cargo, structures, and people. When heated to an internal temperature of 300 K, a well-insulated sphere radiates as little as 85 MW of power, requiring that much power during the nighttime to stay aloft. Conceivably, it wouldn't need any power input if the sphere were allowed to drop in altitude by a kilometer or two at night; however, a small nuclear reactor massing between 420 and 42,000 metric tons could supply the power, depending on W/kg assumptions. Conversely, a poorly-insulated sphere radiates as much as 3.22 GW and would require a major power station to stay aloft.

What I'm missing from my analysis is an estimate of the stresses across the structure. Samuel, another user, calculated the forces of a buoyant geodesic sphere in their answer, but not a tensegrity sphere. Reading through the other answers and comments, I’m not entirely sure where the major loads would be or how to estimate their magnitude. The general consensus that I've seen is that high-tensile strength materials would be enough. Fuller believed so, and he made his calculations in the 1950s with the materials available at the time.
From my geodesic strut calculations, the longest struts would be over 400 m in length, nearly 4x the size of the largest wind turbine blades. A larger geodesic chord number with a greater number of, yet on average smaller, struts might be needed, increasing the mass of the geodesic tensegrity skeleton, but offering less load on the compression members.