Airplanes can fly because the pressure on the top of the wing is lower than on the bottom. But the difference in pressure must be huge in order to lift an enire airplane. Also, the difference in air pressure between, let's say, Mount Everest and sea level is not neglectable, thus, air pressure is decreasing the heigher you go.

This is where I asked myself:
Can we theoretically build a material which is light enough or high enough (or both), that can levitate just due to the difference in pressure on the top vs. the bottom like an on an airplane wing.

For example imagine you're holding a piece of paper horizontally, then air pressure on the top is slightly lower due to the decreasing air pressure and now just make the piece of paper as light as it needs to be to stay in the air like a wing on a flying airplane

Is this theoretically possible or are there other effects or simplifications I overlooked?

  • 16
    $\begingroup$ Re, "Airplanes can fly because the pressure on the top of the wing is lower than on the bottom." That explanation has been making the rounds for almost as long as airplanes have been flying. Unfortunately, it is not true: grc.nasa.gov/www/k-12/airplane/wrong1.html $\endgroup$ Mar 10, 2020 at 1:27
  • 26
    $\begingroup$ @SolomonSlow and at the same time, it is true. The pressure on the bottom of the wing is higher than on the top of the wing as a necessary condition for developing lift. It just has less to do with Bernoulli and more to do the with the wing, basically, pushing air out of the way. $\endgroup$
    – hobbs
    Mar 10, 2020 at 6:36
  • 4
    $\begingroup$ @hobbs, Touché. You are absolutely right. I just assumed that when somebody starts by talking about the pressure difference, they must be informed by the old trope. Whereas, if somebody starts by talking about the mass of the air and the wings beating it down, then they must be better informed. $\endgroup$ Mar 10, 2020 at 12:09
  • 4
    $\begingroup$ Last month's Scientific American had an article about how airplane wings work. There are multiple processes involved, and scientists aren't totally sure of the physics. $\endgroup$
    – Barmar
    Mar 10, 2020 at 16:45
  • 4
    $\begingroup$ @JMac Bernoulli itself is not the wrong answer, at least at speeds where compressibility is not an issue. Where Bernoulli-based explanations usually go wrong is in fallacious attempts to explain the origin of the differences in airspeed in the vicinity of an airfoil, such as the equal transit-time fallacy, which is simply wrong, and would not be sufficient even if it were correct. $\endgroup$
    – sdenham
    Mar 10, 2020 at 23:14

6 Answers 6


A hot air balloon, or a helium-filled balloon floats in air, so either might meet your criteria.

If you're looking for a solid material, perhaps a sphere of very sparse aerogel, with its outside surface sealed with a thin layer of plastic then evacuated, could come close to what you have in mind. But whatever the "material" is, it would need to have a lower mass density than the ambient air.

  • 109
    $\begingroup$ @J.Doe This is not another effect - buoyancy is precisely the force that you're talking about (a net imbalance in pressure above and below the object) and being less dense than air is precisely the criterion required such that the buoyant force overcomes gravity. $\endgroup$
    – J. Murray
    Mar 9, 2020 at 22:05
  • 8
    $\begingroup$ I feel the answer is incomplete without mentioning that this evacuated aerogel exists and is lighter than air. Wikipedia: "The lowest-density silica nanofoam weighs 1,000 g/m3,[19] which is the evacuated version of the record-aerogel of 1,900 g/m3.[20] The density of air is 1,200 g/m3 (at 20 °C and 1 atm).[21] As of 2013, aerographene had a lower density at 160 g/m3, or 13% the density of air at room temperature.[22]" $\endgroup$
    – JollyJoker
    Mar 10, 2020 at 9:18
  • 8
    $\begingroup$ @JollyJoker Ok, so the aerogel is less dense than air. But I couldn't find anything about the mechanical strength of aerogel. Would the wrapped and evacuated version be strong enough to withstand atmospheric pressure, or would it just collapse under the pressure of the surrounding air? $\endgroup$ Mar 10, 2020 at 13:48
  • 5
    $\begingroup$ @J.Doe "Also the material does not necessarily need to be lighter than air because anything that is lighter than air is levitating anyways." Being lighter than air is the only way in which something can levitate (based on its atmospheric environment and not some additional outside force such as a magnet or wind cannon - which would similarly invalidate the question). If you define levitation as hanging still in the air instead of ascending, then the object needs the exact same density as the air surrounding it. $\endgroup$
    – Flater
    Mar 10, 2020 at 15:01
  • 3
    $\begingroup$ @FabianRöling there is (we can assume) already fluid on all sides of the object. It doesn't need to fall down to fill in an "empty" space. So what creates an upward force on the object? The fact that there is fluid pressure on all sides of the object, but the pressure on the bottom of the object is greater than the pressure on the top (because of the pressure gradient in the fluid). $\endgroup$
    – hobbs
    Mar 10, 2020 at 22:23

Is this theoretically possible with only one single material? Yes. Any object occupying a volume of gas (or liquid, or solid) has a slightly higher pressure underneath it than the pressure on top of it (while sitting still) due to the pressure decreasing with height. If it is less dense than the air it occupies (combined weight of both its outside envelope and the mass inside), then the slightly differential air pressure will cause it to rise. It will continue to rise until the differential air pressure matches the weight of the object.

A balloon's weight, combined with lighter than air gas inside achieves this. The helium in a balloon doesn't push on the top inner side of the balloon to make it go up. It is the stronger outside pressure on the bottom of the balloon than is on the top that makes it rise. The helium has enough pressure to maintain the envelope of the balloon, but low enough mass so that the entire weight of the balloon is less than the amount of air that occupies the same volume of the balloon. Same idea for things that float on a liquid (the differential pressure is larger, allowing large ships made from metal).

It is not just like an airplane wing. Airplane wings accelerate the airflow downward (with a bit of Bernoulli as well), like a water skier, and the wing needs to be moving through the air.

Is this theoretically possible or are there other effects or simplifications I overlooked?

Theoretically, two configurations I can think of:

  • A single material heavier than air, but with a vacuum inside, light enough to be less dense overall, but strong enough to not only prevent air from leaking inside, but prevent the air from crushing it as well; or
  • As you alluded to, a solid whose density is less than that of air.
  • 7
    $\begingroup$ It is not just theoretically possible, it is actually how buoyancy works. You don't seem to understand that. Everything that floats works in that way. $\endgroup$
    – rghome
    Mar 10, 2020 at 11:34
  • $\begingroup$ "It is the stronger external pressure on the bottom of the balloon than is on the top that makes it rise." $\endgroup$
    – CJ Dennis
    Mar 11, 2020 at 2:06
  • $\begingroup$ Again, the question isn't asking whether lighter than air flight is possible. It is asking whether it is possible purely because air density varies with height which is not how normal buoyancy works. $\endgroup$
    – matt_black
    Mar 11, 2020 at 21:50
  • $\begingroup$ @matt_black The question does not mention density at all! And buoyancy (normal buoyancy - the only sort there is) works the same way regardless of whether the fluid is constant-density or not, as a consequence of a difference in pressure. $\endgroup$
    – sdenham
    Mar 11, 2020 at 22:16
  • $\begingroup$ @sdenham I was, perhaps falsely, assuming the OP wasn't talking about the explanation for normal buoyancy but was asking whether the difference in pressure with height would be enough to cause buoyancy. Perhaps I overinterpreted it. $\endgroup$
    – matt_black
    Mar 11, 2020 at 22:46

Congratulations! You have figured out the basis of Archimedes' principle. The article linked here gives a satisfactory explanation, though the main point can be made simply:

Consider a vertically-oriented cylinder within the fluid. The difference in pressure between the top and bottom of that cylinder is due to the weight of the fluid within it, and is just that weight divided by its cross-sectional area. Substitute a solid body of the same dimensions, and the difference in the force exerted by the fluid on the top and the bottom of that body -- the upthrust on it -- is the difference in pressure multiplied by the cross-sectional area. But that is just the weight of the fluid that formerly occupied the space taken by that body, and has now been displaced by it.

Considering any irregularly-shaped object as a bundle of thin cylinders, each buoyed by the fluid it displaces, it is clear that shape does not matter, only volume. The upthrust is greater than the body's weight if the body weighs less than the fluid it displaces, and as they are the same volumes, this is just when the body is less dense than the fluid.

The answer to your question, therefore, is that anything of a density equal to or less than than air will float in air, and balloons containing hydrogen or helium are the most common examples. Other answers have suggested vacuum aerogels, but an aerogel containing hydrogen or helium at atmospheric pressure is a more straightforward candidate, as it does not require the gel to have any great strength. For example, the lightest evacuated aerogel achieved so far has a density of 1000 $g/m^3$, and room temperature and pressure hydrogen has a density of 83.2 $g/m^3$, giving a density for the gel, when infused with hydrogen, of no more than 1083.2 $g/m^3$, less that the density of air in the same conditions - 1200 $g/m^3$ (the surface of the aerogel would have to be sealed with a membrane, but its contribution to the density would decrease with increasing volume, by the familiar surface area / volume scaling.)

More radically, this aerographene has an evacuated density of only 160 $g/m^3$.

  • $\begingroup$ Not answering the question! Buoyancy in a constant density fluid doesn't depend on the different density of the fluid at different heights, just on whether or not the suspended object is of a lower density. $\endgroup$
    – matt_black
    Mar 11, 2020 at 21:53
  • $\begingroup$ @matt_black Firstly, the last paragraph answers the question. Secondly, where do you see any reference to different densities of the fluid at different heights? $\endgroup$
    – sdenham
    Mar 11, 2020 at 22:07

You're asking about air, I think, but if you mean gasses in general, then aerogels are a kind of substance that can float on xenon. Wikipedia and YouTube have lots of information.

  • $\begingroup$ Could it float in the middle of a gravitationally caused density gradient of Xenon (or another gas)? $\endgroup$ Mar 10, 2020 at 7:44
  • $\begingroup$ Simply put, "yes". $\endgroup$
    – S. McGrew
    Mar 10, 2020 at 23:39

An alternative to baloons is a spheric shell, with vacuum inside, with a thickness and material strength designed to resist to atmospheric pressure. The force upwards is the equivalent to the weight of air. And its weight is proportional to the surface and thickness. For a thin shell:

$$F_\text{up} = \mu_\text{air} g (4/3)\pi r^3$$
$$\text{weight} = \mu_\text{shell} g 4\pi r^2 \cdot \text{thickness}$$

In order to fluctuate $F_\text{up} \ge \text{weight} \Rightarrow \mu_\text{air}/\mu_\text{shell} \ge 3\cdot\text{thickness}/r$

For a shell made from steel ($\mu = 7850\ \mathrm{kg/m^3}$) with $10\ \mathrm{mm}$ of thickness, and for $\mu_\text{air} = 1.2\ \mathrm{kg/m^3}$

$$r_\text{min}= \frac{7850\ \mathrm{kg/m^3}}{1.2\ \mathrm{kg/m^3}} \times 3 \times 0.010\ \mathrm m = 196\ \mathrm m $$

If it could resist to the pressure of $1\ \mathrm{kg/cm^2}$ without collapsing is another question.

  • $\begingroup$ Please note that descriptive terms or names of quantities shall not be arranged in the form of an equation. $\endgroup$
    – user59991
    Mar 10, 2020 at 19:42
  • 1
    $\begingroup$ @Loong, Your way is more clear, but I've never heard of any such law. The symbols we use, like F for force or a for acceleration are just shorthand for the names. $\endgroup$
    – The Photon
    Mar 11, 2020 at 2:27
  • $\begingroup$ @ThePhoton No. Quantity symbols such as $F$ and unit symbols such as $\mathrm N$ are mathematical entities and not abbreviations. Quantity names and unit names are not mathematical entities. That’s why such names shall not be arranged in the form of an equation (e.g. write “$\mathrm{kg\cdot m^2}$” or “kilogram metre squared”, not “$\text{kilogram}\cdot\text{metre}^2$”). Furthermore, symbols and names shall not be mixed (e.g. write “$\mathrm{g/mol}$” or “gram per mole”, neither “$\mathrm{g}/\text{mole}$”, “g per mole”, “$\text{gram}/\mathrm{mol}$”, nor “gram per mol”). $\endgroup$
    – user59991
    Mar 11, 2020 at 7:16
  • 2
    $\begingroup$ @Loong, when you say "shall not" it sounds like you think there's some legal authority or word of God that created this rule. What authority is making this absolute rule that you're expressing here? $\endgroup$
    – The Photon
    Mar 11, 2020 at 14:55

It's a vacuum balloon. A sphere of implausibly strong material, with a vacuum inside. It will rise until the weight of the air it is displacing equals the weight of that implausibly strong sphere. And in Science Fiction, it can stay aloft indefinitely and adjust its altitude by means of a small vacuum pump, batteries, solar panels and a leak valve ....

For a proof of concept (were one needed) that might actually work up to a few thousand metres altitude, I'd look for the strongest aerogel I could find, wrap it in a thin tough plastic membrane, and suck the air out using a vacuum pump. If the aerogel did not collapse under the pressure, you may have your vacuum balloon.

In the real world we fill a membrane with helium or hydrogen, and this has no need for implausibly strong materials. They can rise until they go pop as the gas inside expands. Google has a "loon" project for balloons which regulate their altitude so they don't pop, creating high-altitude non-satellites to bring broadband and phone services to truly remote areas.

  • $\begingroup$ Have you done basic calculations as to pressure, thickness of gel, etc.? I have a feeling it would work fine but I'm too tired $\endgroup$ Mar 13, 2020 at 4:06
  • $\begingroup$ No. Not even sure how I would. It would mean modelling a huge sparsely connected set of semi-random nodes in three dimensions, and knowing a lot about the strength distribution of the connections. You'd need a big computer. My instincts tell me that local collapses would at least initially lead to stress relief and that the network would be self-strengthening, up to the point of catastrophic failure. An experimental test would be very much easier! $\endgroup$
    – nigel222
    Mar 17, 2020 at 12:11
  • $\begingroup$ I'm not talking modelling at the atom level. There are videos showing 2g of aerogel holding up a 2kg brick. Sea level air pressure is 1.03kg per sq cm, so would need 1g aerogel per sq cm. I don't know what thickness of gel would give you 1g at 1cmx1cm, but knowing that just set weight(x) = buoyancy(x), where x is the centimeters, and solve for x. weight would be a sphere of x radius times aerogel density, minus sphere of x-thickness radius times aerogel density for the hollow center. You might end up with basically a solid balloon that would crumple on pretty much any impact... $\endgroup$ Mar 20, 2020 at 5:57

Not the answer you're looking for? Browse other questions tagged or ask your own question.