15
$\begingroup$

The Kalpana One presents itself as a practical modern design for a space station with artificial gravity. One of the tweaks that it made was to limit the length to diameter ratio. This leaves it looking somewhat stubby compared to the old O'Neil cylinder type or other fictional stations. Their justification:

The radius is the minimum necessary to provide 1g at the hull when rotating at no more than 2rpm. The length is the longest possible while ensuring rotational stability.

Kalpana One Station

I don't expect that this would absolutely rule out the slender designs forever, but there is some concern about disturbances that longer stations would be subject to, and I can't quite understand what. You'll have some wobble control either way, but what is it exactly that would make longer designs more challenging to manage?

$\endgroup$
  • $\begingroup$ What's the difference in rate of rotation? My first thought was it could be related to smaller gyroscopic effect stabilisation on the axis or rotation? Also, could you please quote those concerns that you mention? $\endgroup$ – TildalWave Nov 7 '13 at 19:57
  • $\begingroup$ @TildalWave Well I'm kind of asking what are those concerns, because it wasn't clear to me from reading the NSS piece. O'Neil's Island Three threw around numbers of 8km (diameter) x 32 km (long). Everything can be assumed to be tuned to Earth gravity, so this tells you the difference in rotation rates. Sci-fi authors propose larger diameters, but that proposal was serious, just long-term. $\endgroup$ – AlanSE Nov 7 '13 at 20:09
  • $\begingroup$ Well there is some mention of rotational stability in the link you provide: The new design fixes a rotational stability problem, which shrinks the settlement so the new population target is 3,000 residents. And there's a list of other problems with earlier designs, mentioning lack of wobble control for O'Neill Cylinders, so I'm thinking this could go hand in hand, and create a resonance chamber effect on the whole toroid, if it's too long. Together with minimal clearance to non-rotating shield mass, that could spell disaster. $\endgroup$ – TildalWave Nov 7 '13 at 21:09
15
$\begingroup$

If you can have dissipation of energy by any means, i.e. the body is not completely rigid (and there will be plenty of stuff moving around inside a space station, making it not rigid), then the angular energy will be minimized by shifting the angular momentum axis to the principal axis that has the greatest moment of inertia. The axis of a long cylinder has the least moment of inertia, so it is not stable under energy dissipation.

In fact, exactly this happened to the first US satellite, Explorer 1, where the energy dissipation was via its flexible antennas. From Wikipedia:

The elongated body of the spacecraft had been designed to spin about its long (least-inertia) axis but refused to do so, and instead started precessing due to energy dissipation from flexible structural elements. Later it was understood that on general grounds, the body ends up in the spin state that minimizes the kinetic rotational energy for a fixed angular momentum (this being the maximal-inertia axis).

Explorer 1

| improve this answer | |
$\endgroup$
4
$\begingroup$

The biggest point about the instability of the longer cylinders is that it's inherent to an object spinning around a long axis moment of inertia, and dynamically increasingly unstable. Unless constantly corrected against the slightest movements causing it to wobble further off axis, it must get worse. The Kalpana chooses to limit the spin axis to keep it inherently stable, like a gyroscope -which inherently resists being pushed off axis. Hence coming out more like a torus with the massive spinning rim.

Reducing the spin rate will make things slower, taking longer, but the inherent instability remains.

| improve this answer | |
$\endgroup$
3
$\begingroup$

Probably you have seen this answer already on physics.SE, but I'm going to quote the relevant part:

Another potential problem is that if the mass distribution of the ship is not symmetric, the ship would wobble. Suppose that the ship is balanced at some time, but then an object of mass $m$ is moved a distance $d$ along the side of the cylinder towards the end. Then the frequency of the wobble should be on the order of $\omega md/MR$ and the amplitude should be on the order of $md/M$. The jerk you feel would then be on the order of $\omega^3 m^2d^2/M^2 R$, which should be small as long as it's a fairly small mass moving a small distance.

How small can these effects be? That relies mostly on the size of the ship, as we saw earlier. If the ship is going to simulate one Earth gravity, it can't be too big. Eventually, the tension in the sides of the ship would be so great the ship would tear itself apart. If your ship were a loop, that critical radius would be $T/\lambda g$, with $T$ the tension and $\lambda$ the linear mass density of the wall. For carbon nanotubes this gets you a ship size of up to $10^7 {\rm > m}$, a figure so large (bigger than Earth) that all effects could be made inconsequential. For steel it's about $5*10^3 {\rm m}$, meaning that the effects on to human-sized might be in general $.001$ or $.0001$ of gravity

| improve this answer | |
$\endgroup$
3
$\begingroup$

What is it exactly that would make longer designs more challenging to manage?

I was looking for the paper that an acquaintance (Al Globus, the author of the Kalpana proposal) quoted when he defended the "stubbyness" of Kalpana in personal conversation.

Google Search found your question instead of that paper.

The tumbling of Explorer1 that Mark Adler mentioned was also mentioned by "Veritasium" in the YouTube video The Bizarre Behavior of Rotating Bodies, Explained.

Note that "Veritasium" used the clip Dancing T-handle in zero-g, HD

posted by "Plasma Ben".

The Dzhanibekov Effect or The Tennis Raket Theorem or The Intermediate Axis Theorem

The instability - or rather the tendency to reorient for the minimum kinetic rotational energy for a fixed angular momentum (this being the maximal-inertia axis) - becomes a problem if the material inside the rotating cylinder can move around. And if there are people living inside the habitat, they will move around and they will move things around, too. Al told me that the height to width ratio of the rotating cylindrical space habitat should be 2.45 or less.

The T-handle and the wing nut in the videos are rotating on the axis that has intermediate kinetic energy and intermediate angular momentum so their rotation is unstable.

A "stubby" cylinder should be "stable". A sphere might have a similar problem but the dumbbell and torus designs should be much more stable.

| improve this answer | |
$\endgroup$
  • $\begingroup$ That 2nd video is really nice. $\endgroup$ – Organic Marble Feb 12 at 17:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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