After reading the following question about anti-gravity, I wondered what the largest hollow structure you can make in space is. None of the questions about hollow earth seem to answer this. To put it clearly into a question:

What is the largest structure that can be created in space?

The structure I imagine isn't completely hollow. It is filled with a supporting structure. This can be a honeycomb structure for example. The idea is to make it as large as possible, without it caving in structurally thanks to the gravity it generates. I wonder if it might even be possible to make the hollow structure large enough that you can continue near indefinitely to enlarge the structure, because of the low mass and thus low gravity.


  • The structure itself must be solid enough to have hollow spaces.
  • existing or theoretically existing materials are allowed. So no Unobtanium, but if large superstructures of carbon nano-tubes are theoretically possible, they are fair game.
  • There are no limitations in amount of material in existence.
  • There are no technological limitation on how the structure is created. It is basically as if a god placed the structure there in space in an instant. I'm only interested in the materials used to make a large as possible structure.
  • Although all forms of structures are allowed, a sphere is probably the strongest.
  • It just needs to be a structure floating in space. No habitats, computers, or even other celestial bodies in the neighbourhood are required, unless you think it somehow adds to the strength of the structure (like electrifying some part of the structure and it becomes stronger due to the electricity).
  • The structure doesn't have to be rigid. It just needs to be able to exist and be as large as possible.
  • Hollow in this case means that a large part can be filled with oxygen. This is regardless if it'll just immediately be blown away into the vacuum of space.
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    $\begingroup$ It depends on what you want to do with the structure. It very much depends on what you want to do with the structure. Since the question does not specify any design goals, then the answer is "infinitely large". For example, have a cosmonaut take a can of compressed air (that the "oxygen filling") and press the button. The structure consists of the ever expanding cloud of gas. $\endgroup$
    – AlexP
    Commented Aug 17, 2020 at 11:36
  • $\begingroup$ @AlexP I'll make an adjustment that the (supporting) structure must be solid. But it is just a structure without a purpose but existing. On the other hand, it can't be called a hollow structure if it is an expanding cloud of gas. $\endgroup$
    – Trioxidane
    Commented Aug 17, 2020 at 11:51
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    $\begingroup$ What does "solid" mean? It is crucial to indicate the expected load. Otherwise you can have six very thin and very long wires forming the edges of a tetrahedron and call it a structure. Yes, its load bearing strength will be as close to zero as to be irrelevant; but the question does not indicate any design requirement. $\endgroup$
    – AlexP
    Commented Aug 17, 2020 at 11:53
  • $\begingroup$ Does anyone know if you could simply add gas to the inside of the sphere to offset any gravitational effects? I'm envisioning the equivalent of an ultra-low pressure mylar balloon the size of several galaxies. It would collapse the instant something hit it, but by the question rules, that doesn't matter. Maybe something enclosing the entire known universe so no solid material will contact it? Or could the pressure of radiation keep it "inflated"? $\endgroup$
    – DWKraus
    Commented Aug 17, 2020 at 18:25
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    $\begingroup$ @DWKraus Adding gas to the inside of a structure just adds more mass to it, and makes it more likely to collapse if it was close to that before. $\endgroup$
    – NomadMaker
    Commented Aug 18, 2020 at 3:44

9 Answers 9


Using a lattice structure and steel we arrive at ~ 2800 km across, a few hundred km smaller than earths moon. only about 2% of the volume are actually filled.

The problem with thin shells

You can't make a shell with an arbitrarily thin shell, as thin shells will buckle, long before the stress reaches the compressive strength. Approximately half of the other answers don't take this into account and are wrong.

I interpret the spirit of the question: Can there be a porous, honeycomb structure with lots of internal volume?

Luckily, lattices (3D-honeycombs) have been investigated and can be approximated as a bulk material. In a built structure - unlike a planet, which is by definition in hydrostatic equilibrium - the upper layers can carry their own weight and need not exert pressure on the lower layers. This is why thin shells are so popular in the answers here.

Approach 1
We can approximate a sphere as series of layers, each designed to carry their own weight. Analytically, we treat them as infinitely thin, however as we assume lower layers we need not worry about buckling. It might make sense structurally to transfer some load downward, but then the math gets more hairy. The upside of this approach is that the interior of the structure is less filled with stuff.

What we need to do is take the formula for the stress in a thin shell under it's own weight, and modify for the lower density and strength of the lattice.

The Force $F$ acting on piece of size $A$ of the outermost shell with thickness $t$ depens on the mass of the whole structure and is given with:

$$F = A \rho t G \rho \frac{4}{3} \pi$$

Note how $\rho$ goes into the left part - weight of the shell-element - and right part - total gravity - of the right hand side. If we move $A$ to the left by division we arrive at a sort of pressure acting on the shell. The hoop stress in pressure vessel is given by $\sigma = \frac{Pr}{2t}$, this relationship holds here too, its just compressive stress (not tensile stess). For the stress in our outtermost shell we arrive at:

$$\sigma = \frac{2}{3}G \rho^2 r^2 \pi$$

The paper mentioned above gives the following link between density and yield strength: enter image description here

We see that by picking a pyramidical lattice and 0.02 denstiy - meaning 2% of the available volume is filled by the material - we get about 1% of the yield strength. Presumably the pyramidical lattice looks something like this:

enter image description here

Now it's just entering numbers for your favorite material, with my favorite material (concrete), these are a compressive strength of 20-80 Mpa and a denstiy of about 2600 kg/m³. We'll assume 20 Mpa to account for a safety factor and arrive at 727 km radius, and 84 million tons. This is almost twice as large as Ceres, but far lighter.

Now, how about mild steel? Values for compressive strength for steel are hard to find, as metal rods under compression usually fail in shear or buckling. However the strength is higher than the tensile strength. So we assume a high strength alloy with a yield strength of 690 MPa and a density of 7.8 g/cm³. For funs sake, no safety factor is assumed. With these values I arrive 1426 km radius and 1.8 billion tons. As above, the surface gravity is on order of magnitude of 10^-5 m/s² - not enough to hold an atmosphere. The radius is only 300 km less than earths moon!

Why are these so small? Remember, the outtermost layer has to carry it's own weight. This means that at any circular hoop the weight of one hemisphere presses against the other, causing compressive stress. The weight scales (assuming constant gravity) with the square of the radius, the are only linearily. The same reason pressure vessels and pipe become weaker against internal pressure with larger size and constant wall thickness.

Note that my approach rests on assuming thin shells and in practice the thinnest thinkable shell in a lattice structure is at least as strong as a truss is long, this may introduce major errors - I simply don't know & don't know how to solve without doing a finite element analysis (which I also don't know how to do).

A view from the inside

From the image above we see that one cell of our lattice has 24 outer trusses and 12 inner trusses, but half of the outer trusses "belong" to other cubes so for the following we assume a total of 24 trusses. With the length of a truss $l$, the cube has an edge length of $l_c = \sqrt{2}l$. The filled part of this cube is $V_f=lr^2\pi n$, with $n$ the number of trusses. From all of this we see that:

$$r = l \sqrt{\frac{\sqrt{2}^3 \rho_{rel}}{24}}$$

$\rho_{rel}$ is the relative density, our 2% from above. If wes assume trusses 10 km long to allow some flying in out structure, each truss will be about 960m thick (diameter). For a proper analysis we would need to calculate the load on an individual truss and prove that it does not exceed the critical load causing buckling and I won't do that. However, thiscritical load on a slender column scales with $\frac{r^4}{l^2}$. Since, to keep our $\rho_{rel}$ constant, $r$ scales with $l$ we can just make the column longer and thicker to make it stronger. If you want to fly around in your structure, it surely could be engineered that way!

You could also have hollow trusses for the same total mass, say 1.4 km outer diameter and 960 m inner diameter, with the inside of the trusses filled with an atmosphere.

Approach 2
This maybe something another user want to play with: a sequence of thin shells, but each shell is a geodesic structure with minimal support between shells. Don' know enough about geodesics to try it myself, the beauty is that it would give a better visual feel for the finished structure.

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    $\begingroup$ I'll wait a bit longer before I choose my answer then. I do like that you're tackling the question with internal structure, as that is what I was curious about. Do note that honeycomb structure was a suggestion. Any internal structure is applicable. $\endgroup$
    – Trioxidane
    Commented Aug 19, 2020 at 10:38
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    $\begingroup$ I've been thinking about a megastructure like this, filled with water and gasses, as a setting for a space opera or planetary romance or so myself for a while. $\endgroup$
    – mart
    Commented Aug 19, 2020 at 10:57

TLDR: Almost As Large As You Like (but not due to the Shell Theorem)

To really understand this, let's first think about what causes a mega-structure to collapse- Gravity.

Since this structure is filling 3-D space, any point on the outer surface of the structure is going to feel a gravitational force due to all the other components. On the surface of your structure, that will result in a gravitational force pointing to the center-of-mass. Wherever that center-of-mass is.

(Yes, this is even true if the structure is a hollow shell. In that case, anything bouncing around inside doesn't feel gravity, but the shell itself does! This can be found considering the gravity acting on something on the surface, or something that is the surface.)

So how can we overcome gravity? Put it to work by spinning.

The reason why spinning objects (or curving objects...) can go in a circle is because some force is acting to push the object toward the center of curvature. It could be a rope, friction from a car's wheels, or ... the gravity of a megastructure!

So, choose your rotation correctly, choose a nice shape (like a cylinder, series of cylinders approximating a sphere, etc.), and build to your heart's content. A "nice" shape lets you use the force of gravity to keep the structure together, so you'll want something with more-or-less uniform distance from the center of rotation.

What about Longitudinal Forces?

There is the question of forces which exist along the axis of rotation. This is a question of ingenuity and creativity- there are likely many solutions which let you go up in scale.

An fun solution for this is for the structure to not be rigid, but dynamic. Imagine a series of rings which approximate a sphere. They're spinning at the right speeds so they don't feel any radial stress. Remove half of the rings such that they can flatten down into a disk. Arrange them so the along-the-axis-of-rotation forces make the overall sphere collapse into a disc, then back out into a sphere, and you've just solved this issue. It'll oscillate forever until you do something, like put air in between the rings. There are likely other solutions to this problem, but this doesn't impose any material limitation.

A less fun solution is to just make a thin hollow rod which forms a really large ring, and keep increasing R until you've realized that this lets you fill in infinite amount of gas, fulfilling the "hollow" criteria. After a certain point, the difference in gravitational force between the top and equator won't matter anymore, so no bending issues!

Speed of Light Issues

Okay, so the speed of light is a limit on how fast you can go. If you take Newton's law of gravity and set it equal to the formula for centripetal force, you can get a rought theoretical limit for any structure. I happened to do this for a radially symmetric structure, which gives a general equation of

$$\int_{0}^{2\pi} \frac{m-dm}{r} = \frac{c^{2}}{G} $$

with c being the speed of light, m being your total mass (also depends on radius), and G being the universal gravitational constant.

Some important items to consider:

  • For newton's law, examine a small slice of your structure as your second mass, the first mass being the 'total' mass (that's an approximation there)
  • You can express the small slice of mass (dm) in terms of a small angle (a dtheta) times r, the cross-sectional area, and density. Integrate over the whole structure to this down to simple algebra. (This substitution changes for each choice of structure: rings will do differently than cylinders...)
  • Once integrated, you solve for r to get your maximum size limit.

I don't know what the ideal shape is for this, though. I do know this gives you the limit. I may come back and solve for some suggestions later...

The Real Challenge Is Building It

Building that structure is another question entirely. Your ideal balance between speed and structure only works once built. Getting there involves applying a force and having incomplete geometry, which means the structure will need to be load-bearing.

  • $\begingroup$ I think instead of being a cylinder or a sphere, it could become a larger structure by being a slowly spinning torus, anyway great answer $\endgroup$
    – user35577
    Commented Aug 17, 2020 at 20:55
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    $\begingroup$ I mentioned torus because 3d filling works along "pseudo centrifugal force counteracting gravity" (equatorial). If considering a sphere, 3d filling between poles has to withstand monumental compression forces. $\endgroup$
    – user35577
    Commented Aug 17, 2020 at 21:05
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    $\begingroup$ You can't spin it on all axes. The poles get no support, once it's big enough it collapses.. $\endgroup$ Commented Aug 17, 2020 at 22:45
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    $\begingroup$ This does not work: Gravitation falls off with r^-2, centrifugal force with r^-1. I.e. the forces will cancel out just at a specific distance from the center, but everywhere else you have a force difference that, if large enough, will simply rip the structure. $\endgroup$
    – toolforger
    Commented Aug 18, 2020 at 8:00
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    $\begingroup$ This answer only works for a ring where there are no poles to collapse. Spheres have an upper structural bound. And, rotation also has an upper bound at speed of light. So “unlimited” is not quite valid. $\endgroup$
    – SRM
    Commented Aug 18, 2020 at 14:21

TLDR: when considering materials and manufacturing methods available to modern humans (bar the scale of course), possibly a few thousand light-years, or the size or a small galaxy.

I'm basing this heavily on an answer to a similar question (How large could a decoy planet made of extruded polystyrene be?): https://worldbuilding.stackexchange.com/a/138280/29103 The conclusion that answer comes to, for a thin spherical shell, is something like this:

$$R < \frac{P}{\pi T \rho^2 G},$$

where $R$ is the radius of the shell, $T$ its thickness, $\rho$ its density, $P$ its tensile strength, $\pi=\tau/2$ and $G\approx6.6\cdot 10^{-11}\ \mathrm{m^3\ kg^{-1}\ s^{-2}}$ is the gravitational constant.

For EPS and a thickness of 1 m, this comes to $4.71\cdot 10^{11} \ \mathrm{m^2}$, or around 3 AU. It can be easily computed for different materials (e.g. more than 10 times the size for 3D graphene); somebody into hi-tech materials could provide the numbers for something extremely strong and lightweight, to give some more numbers.

This formula shows the problem with the "shell theorem" discussion. While you can increase the size by making the shell thinner (e.g. over 0.5 light-year for 1 mm 3D graphene), and in theory you can have as large shell as you want if you keep it infinitely thin, with real materials (made of solid matter) you cannot get under the thickness of 1 molecule. So the particles on the OUTSIDE of the shell WILL be pulled towards the center by the gravity of the rest of the shell. You could "lighten" your shell by making it sparser, or introducing "holes", but the more you do that, the less of a perfect sphere it is, and you'll hit the limits soon.

  • $\begingroup$ An important note: this answer also assumes a sphere, and will not hold for any complicated structures or schemes. Important to note in the question: it is asking for "largest", not "contains the most volume with the least surface-area-to-volume ratio". $\endgroup$
    – PipperChip
    Commented Aug 18, 2020 at 20:30
  • $\begingroup$ This is very good! Although the sphere doesn't have an internal support structure, this answer already supplies me with enough handholds to do a lot on my own. Thanks! $\endgroup$
    – Trioxidane
    Commented Aug 18, 2020 at 20:58
  • $\begingroup$ This answer does not take buckling into account. $\endgroup$
    – mart
    Commented Aug 19, 2020 at 7:13

The one constraint that is always with you no matter which materials you are using is the gravitational field. Its magnitude is determent by the Gauss' flux theorem for gravity. Basically it says that the flux through the closed surface is proportional to mass inside this closed surface. Then from this flux considering that the shape of the construction is spherical, one can calculate the gravitational acceleration on the surface(or inside) the construction: $$\begin{gather} g(r) = -\frac{GM}{r^2},\ M = \rho V = \rho \frac{4\pi r^3}{3} \\ g(r) = -G\rho \frac{4\pi r}{3} \end{gather}$$

  • $M$: structure's mass
  • $\rho$: structure's density
  • $V$: structure's volume
  • $r$: structure's radius
  • $g$: gravitational acceleration.

Knowing the acceleration it is possible to calculate the weight of objects on the surface of the construction. Then it is possible to calculate the pressure of the "upper" structures to "lower" structures: $$P = \frac{m_u \cdot g(r_c)}{S}$$ where

  • $P$: pressure from upper structures to lower structures
  • $m_u$: the mass of upper structures
  • $r_c$: the mass center of the upper structure
  • $S$: the surface of contact

Then to determent the critical radius of the structure, one should solve the equation with respect to $r$ - structure radius while on the left side is the critical pressure when the weakest construction will collapse. In the spherically symmetric case, the weakest is somewhere low, because it has the most pressure on it.

This approach will provide the upper boundary for the radius depending on the critical pressure of the weakest point of the structure.

For the more precise answer(especially in the form of the number) one needs to find the data for the honeycomb structure and solve the equation.

  • $\begingroup$ Yes, but the point is that the question does not specify any load or effort acting on the structure; so that if the structure is floating in empty space you can make a geodesic dome with very thin beams and the self gravitation will be vanishingly small. $\endgroup$
    – AlexP
    Commented Aug 17, 2020 at 14:38
  • $\begingroup$ That's true and such structure will make g(r) to be almost zero in every point of it, so nothing inconsistent with such way of calculating it $\endgroup$
    – FrogOfJuly
    Commented Aug 17, 2020 at 14:54
  • $\begingroup$ I wonder if @AlexP's point isn't, "you didn't answer the question." In fact, off the top of my head, I wonder why this is applicable at all? Since there is no gravity inside a sphere (Shell Theorem), what does it matter how much gravity is experienced outside the sphere? So, how big can the shell be? $\endgroup$
    – JBH
    Commented Aug 17, 2020 at 14:57
  • $\begingroup$ If it is hollow g(r) is just a constant which is proportional to delta r instead of just r. From first sight, it means that this such structure might have an arbitrary size because g(r) is independent of the radius of the sphere itself. $\endgroup$
    – FrogOfJuly
    Commented Aug 17, 2020 at 15:07
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    $\begingroup$ It can't have zero thickness, because it must be a boundary. It doesn't matter if there is a load, because it will attract itself, just like the planets pulled themselves into spheres because of gravitational attraction with their own mass. $\endgroup$
    – NomadMaker
    Commented Aug 18, 2020 at 3:48


This answer is for more or less habitable hollow structures in space that are at least partially habitable. Uninhabitable structures which are just monuments or artworks floating in space might possibly be larger.

Short Answer:

The place to start researching this question is "Bigger Than Worlds", Larry Niven, Analog Science Fiction/Science Fact, March, 1974, which has been reprinted many times.

Long Answer:

One type of hollow structure in space that is often discussed is a hollow cylinder that rotates to simulate gravity in the inner surface.

There are structural limitations to how many miles wide such a structure might be but possibly no structural limitations on how long it could be, or limitations which show up only after it gets very long.

Thus a hollow rotating cylinder in space could be 1 kilometer or mile wide,

or 10 kilometers or miles wide,

or 100 kilometers or miles wide,

or possibly 1,000 kilometers or miles wide.

And that hollow rotating cylinder could be

1 kilometer or mile long,

or 10 kilometers or miles long,

or 100 kilometers or miles long,

or 1,000 kilometers or miles long,

or 10,000 kilometers or miles long,

or 100,000 kilometers or miles long,

or 1,000,000 kilometers or miles long,

or 10,000,000 kilometers or miles long,

or 100,000,000 kilometers or miles long,

or 1,000,000,000 kilometers or miles long,

And so on and so on.

See here:

https://en.wikipedia.org/wiki/Topopolis 1

And there have been discussions of other types of very large structures in space.

Many far out ideas for vast structures in outer space were discussed by Larry Niven in "Bigger Than Worlds", Analog Science Fiction/Science Fact, March, 1974, which has been reprinted many times.

http://www.isfdb.org/cgi-bin/title.cgi?133302 2

https://en.wikipedia.org/wiki/Bigger_Than_Worlds 3

And of course that article was published 46 years ago, and there could have been many ideas about mega structures in space and their structural limitations since them.

See also:

https://tvtropes.org/pmwiki/pmwiki.php/Main/DysonSphere 4

https://tvtropes.org/pmwiki/pmwiki.php/Literature/Ringworld 5


This answer is for more or less habitable hollow structures in space that are at least partially habitable. Uninhabitable structures which are just monuments or artworks floating in space might possibly be larger.

  • 4
    $\begingroup$ It's not protected from self-gravity along it's length. Eventually it collapses. $\endgroup$ Commented Aug 17, 2020 at 22:46
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    $\begingroup$ Your links don't work right. I get to a blank page. $\endgroup$
    – NomadMaker
    Commented Aug 17, 2020 at 23:46
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    $\begingroup$ Even spinning cylinders will have gravitational problems if they are large enough. $\endgroup$
    – NomadMaker
    Commented Aug 17, 2020 at 23:47
  • $\begingroup$ Many links are broken, remove the [n] part at the end. $\endgroup$ Commented Aug 18, 2020 at 7:31
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    $\begingroup$ Links should work now. $\endgroup$
    – SRM
    Commented Aug 18, 2020 at 14:23

How many angels can dance on the head of a pin? Answer: As many as wanting.

Assumption: The structure is built in intergalactic space.

Now, granted, we don't necessarily know a lot about intergalactic space. For all we know there are swarms of giant space-bees out there. But, insofar as we understand everything right now, the space between galaxies is very empty and the interfering gravitational forces are very small.

Which means we can build a structure out of aluminum beams and tarps from your local hardware store — and it can be at least as large as half the distance to the nearest galaxy.

  • The Shell Theorem teaches us that gravity is zero inside the shell of an object. That means there's no gravity at all inside a hollow shell. So long as we don't have anything inside like the marble or ball bearing inside a can of spray paint that could bounce around, gain momentum, and eventually rip the thing apart, nothing on the inside can hurt the structure.

  • The things on the outside would include moving rocks (asteroids, meteors, rogue planets, streams of intergalactic dust...), gravity (which is intentionally very light and, more or less, statistically balanced in this scenario), and light (which has pressure, but at those distances it ain't very much). I'm banking that none of those have a significant influence.

It is important to realize that external gravity can cause problems. This structure is obviously humongous and will have considerable gravity — heck, it might have enough gravity to affect all those nearby galaxies. I'm ignoring that because I'm not prepared to calculate the actual mass of the object I'm describing. THAT would be a fair and legitimate limiting factor in the ultimate size of the any such object. (It isn't that outside gravitational influences would be great enough to hurt it, it's that it would draw galaxies into itself, which would be bad.) Let's assume for the sake of argument that the gravitational attraction of our structure must be equal to or less than 1% of the mass of the closest nearby galaxy. That limitation, based on what building materials and techniques are available, would limit the size of the structure and probably (in fact, certainly) force it to be less than the size I've proposed. My thanks to @BThompson for pointing out this deficiency in my answer.

  • I am ruling out alien interference. I'm not being sarcastic, something that big sitting between galaxies is bound to attract attention and there must be somebody else out there when we're talking about surrounding galaxies....

I don't have the time to figure out an arbitrary point of intragalactic space and then calculate half the distance to the closest galaxy to provide a precise estimate. I'm not sure it's relevant. The structure could be a whole lot larger than that (if the gravitational forces are small enough), I'm just assuming that at the 50% point that closest galaxy might have enough gravitic influence to start deforming (and eventually destroying) the shell.

But my point is, for all intent and purposes, it's so ginormous that it might as well be considered infinitely large. It's a space that could surround multiple galaxies and yet, due to its flimsy nature, would exert so little gravity itself that it wouldn't change (I believe) anything in the universe.

Might be a good place to put the bees, though. :-)

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    $\begingroup$ The answer considers gravitational effects on the structure from objects inside and outside, but not those of the structure itself. Which will indeed collapse under its own weight, as all planets do (that's why they are roughly spherical). $\endgroup$
    – toolforger
    Commented Aug 18, 2020 at 7:56
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    $\begingroup$ @Trioxidane The tarp is for humor. The ridiculousness of both the answer and the question were not as obvious as I expected. $\endgroup$
    – JBH
    Commented Aug 19, 2020 at 1:00
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    $\begingroup$ @JBH even if it's hollow, the sheer amount of mass, even if it's thinly distributed and most of it is very far away, will exert the same gravitational force as if it were concentrated at the center. I don't have the formulae ready, but just just posit some hull thickness (1 millimeter would be fine), and calculate the mass at 1 AE. I'd expect that it's going to be more mass than the Sun (I don't have the formulae ready though), and since the thing doesn't have centrifugal force, it will simply contract even if the mass is much less than the Sun's. $\endgroup$
    – toolforger
    Commented Aug 19, 2020 at 9:45
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    $\begingroup$ @JBH A Dyson swarm would not collapse because each item would be orbiting around a common center. If the items were somehow just all sitting "motionless" in a spherical shell, then they would all start moving inward to a common center, regardless of the size of this structure. $\endgroup$ Commented Aug 19, 2020 at 14:59
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    $\begingroup$ @JBH Given a material that can withstand the stresses of such motion, the 'equatorial' region of your object could, indeed, support itself from collapse by the outward force generated by this spin. The farther from the equatorial region, the less support the material receives, until the gravitational force again dominates, and the structure collapses inward. As to you 'either/or' I'm unsure of your knowledge base. There is no minimum amount of mass (other than actually zero) where an object becomes immune to gravity. A Dyson swarm is literally a swarm of satellites. $\endgroup$ Commented Aug 19, 2020 at 18:26

Assuming you will permit an active structure, as big as you have mass for.

Inside the shell you have spinning bands, they exert an outward force. This balances the inward pressure of the self-gravity of the shell. You can drive the overall force to zero, the only strength required is between supports and if you have enough bands you can drive this as low as you want. Other than the maglev coupling between the bands and the shell you could build it out of tissue paper. (Although it would no doubt be cheaper to use something stronger.)

  • $\begingroup$ How do you transfer the outwards force of the spinning bands to the hull? In particular since the bands will be on intersecting trajectories. $\endgroup$
    – toolforger
    Commented Aug 18, 2020 at 8:13
  • $\begingroup$ @toolforger The spinning portion can be contained within non-rotating torus shell. Each ring is slightly offset so that they do not intersect, but can be connected to neighboring shells. $\endgroup$ Commented Aug 18, 2020 at 18:36
  • $\begingroup$ @toolforger Maglev does it fine--after all, it holds trains up. $\endgroup$ Commented Aug 19, 2020 at 2:29
  • $\begingroup$ @LorenPechtel Maglev has no interference from intersecting tracks. There are various complications involved: Distance from band to surface (you need space for an intersecting band), centimeter-level precision for orbit-sized structures, band-to-band and band-to-band-to-surface interaction between the magnetic fields, and energy distribution unless you have a star at the center and use photo cells. I guess it's solvable with enough detail engineering and power sources (but how to you get power into an intergalactic structure?) $\endgroup$
    – toolforger
    Commented Aug 19, 2020 at 9:37
  • $\begingroup$ @toolforger As you say, everything you're talking about is engineering details, not theoretical limits. Note, also, that you can avoid a lot of the interaction problem by bending one track a little bit more before an intersection, thus causing it to pass above the other track in a straight line. This means it doesn't need support at that point. $\endgroup$ Commented Aug 19, 2020 at 14:50

Other answers have correctly concluded that rings may be as large as you want in Newtonian gravity. And the speed of light has been mentioned as a limit in relativity. But the true relativistic limit is cosmological: if your ring is bigger than a cluster of galaxies, the expansion of space due to dark energy will stretch it until it breaks. Maybe 10 megaparsecs in size, depending on your local matter density. However, at this scale things happen very slowly: your ring could last billions of years.


Provided the structure is built as a relatively thin shell it can be arbitrarily large. In a large enough void in space far away from any large gravitational fields, ignoring construction “difficulties” and assuming sufficient mass was available, the structure could be many light years in diameter.

  • $\begingroup$ In the limit, the shell can have zero thickness... $\endgroup$
    – AlexP
    Commented Aug 17, 2020 at 15:22
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    $\begingroup$ Could you provide some explanation for how you came to your conclusion? $\endgroup$ Commented Aug 18, 2020 at 0:48

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