# Would a Hot Air Balloon work in an O'Neill Cylinder?

We had a long discussion in my office about this - if we were in a smallish 500m diameter O'Neil cylinder would hot air balloons work?

• I assume that, like a conventional O'Neil design, it's spun to generate gravity? – Cadence Sep 27 '19 at 6:46
• @Cadence Yes correct. – flox Sep 27 '19 at 6:46
• Well, yes, obviously. Why wouldn't they? The apparent gravitational force is due to the rotating frame of reference. With respect to the rotating frame of reference the ballon would appear to be rising in the air. To make the question interesting it would be helpful to indicate why you think that they might not work. – AlexP Sep 27 '19 at 8:02
• @AlexP I can understand your response. It's just sometimes that things that seem obvious, may not be so obvious. The nature of the physical world is quite often non-intuitive. Personally I'm glad to see the question. Nothing like seeing an issue clarified. – a4android Sep 28 '19 at 6:02
• Thanks for asking the question. I tend not to take for granted that things work the same with centrifugal gravity compared to normal gravitation. Hot air balloons in O'Neil cylinders is one of them. This puts my mind at rest. – a4android Sep 28 '19 at 6:06

Yes, certainly.

The rotation creates a gravity-like effect for everything that follows the rotation, which includes the air. If the hot-air balloon is lighter than the air, it will rise. Even in so small a cylinder, there will be a small density gradient between the air at the rim and that nearer to the center, since the inner air will not carry as much weight of air as the outer air. This should make it possible to stabilize the hot-air balloon at a certain altitude-

There are some coriolis effects that create wind patterns, which will affect any hot-air balloons: As warm air rises, it moves to regions where the rotational velocity is smaller, creating wind in the direction of rotation in this region. As cool air falls, it enters regions where the rotational velocity is greater, creating wind counter to the rotation. This is likely to result in rolling vortices of air where air rises, moves forward, cools and falls, and moves backwards. The smaller the diameter of the cylinder, the more pronounced this effect will be, though the air will never move faster than the speed of rotation. Around the center of the cylinder, there will be a vortex rotating in the opposite direction as the outer vortices.

• But wouldn't the density difference be marginal and most of the air pressure come from the cylinder being pressurized? – Infrisios Sep 27 '19 at 8:48
• @Infrisios. Certainly, but you wouldn't need much of a gradient. Hot-air ballons are generally heated until they lift off the ground and quickly find a stable height at a few hundred meters. At (say) 250 meters on Earth, the air pressure is 97% of the surface pressure, not that much of a difference. – Klaus Æ. Mogensen Sep 27 '19 at 8:54

Simple answer: yes. Hot air balloons rise due to their low density compared to the air around them. Gas centrifuges exist which use centrifugal forces to cause density stratification in a gas (in the case of the link, separating lighter 235UF6 from heavier 238UF6. Uranium gas centrifuges are low pressure, but the technique works for gases more amenable to high pressure work (see this paper on natrual gas/CO2 separation, with a 5 bar fill pressure). Thus, in a spun habitat with an earth-like atmosphere (at least at the outer edge of the volume), your hot air balloon can rise, regardless of whether or not the air pressure is caused by pumping lots of air into a small volume, or by centrifugal effects.

You may get some odd coriolis effects (as you move up, you'll appear to be accelerated forwards, and as you move forwards you'll appear to be accelerated down, etc etc) but as the speed of a balloon is usually quite sedate this shoudl have negligible effect.

As the radius increases though, things start chaning a bit. For example, the flight profile won't be quite the same as they would on earth for various reasons.

From McKendree's paper on rotating habitats, section 5.2.3, you get this (not entirely legible) formula for atmospheric density:

suggesting that the scale height of the atmosphere is not dissimilar to Earth., so operating altitudes for a balloon in a large habitat would be similar to Earth. Conversely, in this Space Exploration SE answer (Radial variation of atmospheric pressure in rotating O'Neill cylinder-like ship?, answer by "Atmospheric Prison Escape" who sometimes shows up here) suggests that the scale height is very large compared to the radius of the station, meaning you could fly up Quite High, which will get you interesting gravitational effects.

$$P(r) = P_0 \exp(-\frac{\Omega_0^2}{c^2_s}\frac{r_0^2-r^2}{2})$$

I couldn't tell you who is right here (though I wonder if APE is closer to the truth due to more sensible handling of the effects of a rotating reference frame), but it seems that modelling of pressure and turbulence in a rotating station hasn't been done, and it isn't clear how hard the problem has been studied, if at all. You could, for example, have quite strong rotary winds that would just swing you round and round in a slightly unpleasant way and at mildly alarming speeds. Not like the sort of balloon flights you might get on earth at all.