# Radial variation of atmospheric pressure in rotating O'Neill cylinder-like ship? (Rendezvous with Rama)

Sir Arthur C. Clarke was a science writer as well as a prolific writer of science fiction (including hard SF*), and his stories usually had a substantial footing in science.

His book Rendezvous with Rama describes a perfect hollow cylinder, 20 kilometers in diameter and 54 kilometers long, rotating at 0.25 rpm (versus the O'Neill cylinder at 8km diameter and ~0.5 rpm) to produce artificial gravity on the inside walls.

For a small rotating spacecraft the atmospheric pressure would be uniform, but in this case the diameter is of the order of a scale height on Earth!

Question: If the (atmospheric) pressure at the "surface" (the inner wall) were 1 standard atmosphere, how would the pressure vary moving towards the axis, and what would be the minimum pressure? What atmospheric pressure would be reasonable, and would the air rotate en masse at 0.25 rpm, or would rotational forces create shear or other effects, producing turbulent wind at the surface?

* Hard science fiction is a category of science fiction characterized by an emphasis on scientific accuracy or technical detail or both. The term was first used in print in 1957 by P. Schuyler Miller in a review of John W. Campbell, Jr.'s Islands of Space in Astounding Science Fiction Wikipedia

• I think this is more of a physics questions and thus may be more suitable to Physics.SE. The actual question here seems to be: given a cylinder with a diameter of 20km (minus something for the walls) rotating at 0.25rpm, which forces are experienced at the inner wall? Oct 19, 2018 at 20:33
• "One of the aspects of its propulsion system was a slow continuous acceleration rate that over a long time frame took the vessel to .99 speed of light." Where are you getting that? Its speed was 100 000 km/hr. Oct 19, 2018 at 21:39
• These are different questions but somewhat related, and have some interesting answers and comments: Can birds fly inside an O'Neill cylinder?, and What is the gravity inside a rotating cylinder?, and What would determine the interior temperature of a large space station?, and Can humans play basketball in simulated gravity?
– uhoh
Oct 20, 2018 at 0:26
• Close vote retracted - but I still think this pretty speculative, but could have an instructive answer. Oct 20, 2018 at 1:23
• @OrganicMarble there are several well-received questions here about propulsive farting and Santa Claus. The answer to this question will be a solution to a differential equation, so I think it should stay.
– uhoh
Oct 20, 2018 at 1:58

The full problem of an O'Neill cylinder is pretty tough to solve.

1. A simple approach

Analytically, one would assume to be in hydrostatic equilibrium, without fluid motion on the surface of the O'Neill cylinder. Then one can use the equations of planetary atmospheres in cylindrical coordinates to derive an approximation to the pressure profile.
The radial velocity equation on the surface $$z=z_0=r_0$$ with boundary condition $$v_{\rm cylinder} = \Omega_0 r_0$$ becomes $$\Omega^2_0 r = -\frac{1}{\rho}\partial_rP$$

Which in the simple, isothermal case with Temperature $$T$$ and corresponding sound speed $$c^2_s = k_B T /\mu$$, with boundary condition $$P(r) = P_0$$ has the solution

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

where we keep in mind that the radial coordinate and the vertical one are connected via $$r=r_0-z$$.
Thus, one gets a gaussian pressure profile with a modified scale-height $$H=c_s/\Omega_0$$ in this, the simplest case.
It is instructive rewrite this scale-height as ratio of velocities, $$H=\frac{c_s}{\Omega_0} = \frac{c_s}{v_{cylinder}} r_0$$ because then we can relate scale-height to cylinder radius. I take $$c_s = 350m/s$$ like in an Earth-like atmosphere and $$v_{cyl}$$ becomes with your data $$v_{cyl}=83m/s$$, so we see that the scale height is very large compared to the cylinder radius, thus there wouldn't be any strong density stratification in the cylinder.

Then, dynamical effects can have a huge impact over the dynamics in the cylinder. Through friction, the rotating surface will inject lateral momentum into the atmosphere, forcing it to co-rotate. At the center of the cylinder nothing can co-rotate. Thus clearly the situation is unstable, and there will be some adjustment of the pressure profile.

2. The full approach and the rotation profile

One would have to solve the set of vector equations

$$(\vec v \cdot \vec \nabla)\vec v + 2 \vec \Omega \times \vec v + \Omega^2 \vec r_{\perp} = -\frac{\vec \nabla P}{\rho} + \frac{\nu}{\rho} \Delta \vec v$$

where only gravity is missing now.

The question you've had whether the air would rotate en masse as a solid body, or whether there would be shear, depends strongly on how effective turbulent viscosity would is in transporting momentum from the wall into the atmosphere. I don't know the answer to that, I think one would need to simulate this.

My intuition tells me that upwards momentum transport could be much more effective than on Earth, due to the weak stratification. But even then one needs some type of instability for the turbulent vortices to pump momentum into the systematic rotational velocity of the gas. So it might as well be that the gas will simply destabilize and breakup into large vortices in the radial-rotational direction.

3. In the book
Rama starts out much colder, so what I'm speculating is that the author has done the same little solution as we have above and said "oh I'll do a transition from strong to weak stratification by letting $$c_s initially and then $$c_s>v_{cylinder}$$ which should then create a storm".

• I tried to solve this but quickly stopped because of the pie-shaped section of atmosphere. In the flat case, each slice with thickness $dz$ has the same area $dA$, but in the cylinder it's not so; there's going to be another $r$ or $r^{1/2}$ or $ln(r)$ in there somewhere. Is that taken into account in your solution?
• I only solved the local case in cylindrical coordinates, I didn't take the global equations in the symmetric assumption. But this is not necessary if you take only the two terms into account that I did: Once averaged, the equation looks identical, and the differential element $\nabla_r = \partial / \partial r$ in cylindrical coordinates. Oct 30, 2018 at 19:10
• @uhoh: Why the whole solution may look confusingly like the plane-parallel atmosphere is because the rotational term goes linear with r. So we might rewrite the whole thing into a cartesian like-DEQ by simply exchanging r and z and suddenly it's like on Earth. I've thought quite long if $\Omega$ should be a function of r, which would change the whole solution, but I don't see why. It is only the rotation of the coordinate system after all. Oct 30, 2018 at 19:15