Is Rama stable?

[Dorfl]

Could a long rotating cylindrical space station keep rotating around its lengthwise axis without active stabilisation?

Last semester a teacher briefly mentioned a cylindrical satellite which unexpectedly began wobbling after a while, and eventually switched to rotating around an axis going through its sides, instead of the endpoints. When I read Rama slightly later, I didn't reflect over it, but this comic (http://www.schlockmercenary.com/d/20090911.html) reminded me again.

Of course, the Ramans probably wouldn't have any problems keeping Rama stable, if they wanted to.

[drkitten]

Could a long rotating cylindrical space station keep rotating around its lengthwise axis without active stabilisation?

Probably not. Butterfly effect and all that.

A cylinder rotating around its long axis is "metastable," in the same way that a cone resting on its vertex is. If it were balanced [i]perfectly there would be no particular reason why it would fall one way instead of another, and so would remain there until something broke the symmetry.

Something like, oh, say, a passing micrometeorite. Or movement inside the station as everyone goes to the north end for a baseball game. Or quantum-scale fluctuations.

[Doubt]

This problem was addressed in "Ringworld Engineers".

Larry Niven did not take the stability in the plane of rotation into account in his first Ringworld book. When fans pointed out the issue to him it became the source of the plot for the second book.

[Floyt]

Isn't Rama self-propelled? If so, there probably are attitude adjustor jets. Needn't be quite in the harnessed solar flare range either...

[dudalb]

This problem was addressed in "Ringworld Engineers".

Larry Niven did not take the stability in the plane of rotation into account in his first Ringworld book. When fans pointed out the issue to him it became the source of the plot for the second book.

I know several people who were there when people stood outside Niven;s balcony window at a hotel where a Sci Fi Con was being held chanting "the Ringworld is unstable!The Ringworld is unstable" in unison.......

[dasmiller]

Could a long rotating cylindrical space station keep rotating around its lengthwise axis without active stabilisation?

Now we're on my turf! No, assuming any vaguely smooth mass distribution, Rama wouldn't be passively stable about the spin axis. You could come up with mass distributions that would be (a very heavy hoop in the center, with the rest very lightweight), but I don't think Clarke gave us any indication of that.


Last semester a teacher briefly mentioned a cylindrical satellite which unexpectedly began wobbling after a while, and eventually switched to rotating around an axis going through its sides, instead of the endpoints. When I read Rama slightly later, I didn't reflect over it, but this comic (http://www.schlockmercenary.com/d/20090911.html) reminded me again.

We generally refer to that as "tumbling" or "going into a flat spin." Offhand, I know of only one spacecraft that was lost that way, though of course there have been a lot of spinning spacecraft that required active stabilization, and once the spacecraft is retired, the 'active' part stops so presumably there are a lot of tumbling spacecraft slightly above the GEO belt.

Of course, the Ramans probably wouldn't have any problems keeping Rama stable, if they wanted to.

There are a number of ways to stabilize a spinner, yes. Of course, we'd prefer not to use thrusters, but there are tricks with moving masses. If Rama had a big hunk of mass spinning at a different rate, then there are all sorts of games one can play.

[sol invictus]

[I believe that] A cylinder rotating around its long axis is "metastable," in the same way that a cone resting on its vertex is.

That's not metastable (unless the vertex is flattened or concave). Metastable means locally stable but globally unstable, like a ball in a bowl suspended over the floor.

Now we're on my turf! No, assuming any vaguely smooth mass distribution, Rama wouldn't be passively stable about the spin axis. You could come up with mass distributions that would be (a very heavy hoop in the center, with the rest very lightweight), but I don't think Clarke gave us any indication of that.

I thought rigid bodies were stable under rotations around the principal axes with the largest and smallest moments, and unstable around the intermediate one. The long axis (for a long cylinder) is the one with the smallest moment, so why isn't the rotation stable?

[BoogieWoogieWookie]

I thought rigid bodies were stable under rotations around the principal axes with the largest and smallest moments, and unstable around the intermediate one. The long axis (for a long cylinder) is the one with the smallest moment, so why isn't the rotation stable?


That is my understanding also. I have all four of the Rama books and I have read each of them three times. I have a solid mental picture of the spacecraft thanks to Clark's clear and vivid descriptions of them.

The spacecraft is described as a cylinder 54 km long and 20 km in diameter that rotates around its long axis to create artificial gravity along its inside curved walls. It has cities and even a sea along these curved walls which would be impossible, of course, if the craft rotated end-over-end style.

[Brian-M]

If the mass is distributed evenly... and compartmentalised so it can't become unevenly distributed over time, how could it possibly become unstable?

This isn't a rhetorical question... I genuinely want to know.

[nathan]

Could a long rotating cylindrical space station keep rotating around its lengthwise axis without active stabilisation?
no

Last semester a teacher briefly mentioned a cylindrical satellite which unexpectedly began wobbling after a while, and eventually switched to rotating around an axis going through its sides, instead of the endpoints. .

Right, I was taught the same lesson at university. The reason given was that real-world cylinders are not perfectly rigid. Internal flexing allows the axis of rotation to migrate. This happens with artificial satellites, and it apparently was a surprise when the first such cylindrical satellite did it.

[Dorfl]

Isn't Rama self-propelled? If so, there probably are attitude adjustor jets. Needn't be quite in the harnessed solar flare range either...

I'm not sure. I'd imagined it as being basically in stand-by, once it's bounced off of a star. Maybe it still has some small space-warpey adjusters active when travelling, though.

[Dorfl]

Right, I was taught the same lesson at university. The reason given was that real-world cylinders are not perfectly rigid. Internal flexing allows the axis of rotation to migrate. This happens with artificial satellites, and it apparently was a surprise when the first such cylindrical satellite did it.

Hmm... Would a perfectly rigid cylinder keep spinning around the right axis, even while being pelted by micrometeorites and stuff?

[Dorfl]

Now we're on my turf! No, assuming any vaguely smooth mass distribution, Rama wouldn't be passively stable about the spin axis. You could come up with mass distributions that would be (a very heavy hoop in the center, with the rest very lightweight), but I don't think Clarke gave us any indication of that.

Me neither. I guess a hoop could be fitted into either of the walls at the ends, but the interior of the ship is supposed to be empty.

We generally refer to that as "tumbling" or "going into a flat spin." Offhand, I know of only one spacecraft that was lost that way, though of course there have been a lot of spinning spacecraft that required active stabilization, and once the spacecraft is retired, the 'active' part stops so presumably there are a lot of tumbling spacecraft slightly above the GEO belt.

I guess people tend not to make the same mistake twice, when handling multi-million dollar spacecraft. Hmm... If parts break off of a satellite, is it more or less likely to produce dangerous space debris if that satellite is spinning?

There are a number of ways to stabilize a spinner, yes. Of course, we'd prefer not to use thrusters, but there are tricks with moving masses. If Rama had a big hunk of mass spinning at a different rate, then there are all sorts of games one can play.

So it could be made passively stable? Since it seems to be implied that Rama is pretty much inactive when not near a star.

[nathan]

Hmm... Would a perfectly rigid cylinder keep spinning around the right axis, even while being pelted by micrometeorites and stuff?

no. then there's an external source of torque. It'd depend on the distribution of the micrometeorites as to whether the axis wandered randomly or not though.

ETA:

I've realized a confusion here. With the external torque, the axis of rotation can now wander around the celestial sphere. Without that torque (as I'd read the original question), the axis remains fixed against the celestial sphere. With a non-rigid cylinder, the axis can wander relative to the cylinder's axes, with a rigid cylinder it cannot.

So, to be specific. With an isolated non-rigid cylinder, whatever star happens to be 'the pole star', will remain so, but the point on the surface of the cylinder directly below that pole star will change. Until the cylinder has reached the stable equilibrium of rotating around its highest inertial axis (at the lowest angular velocity).

At least that's my understanding, which is consistent with conservation of angular momentum.

[sol invictus]

With a non-rigid cylinder, the axis can wander relative to the cylinder's axes, with a rigid cylinder it cannot.

Rigid objects have two stable principal axes of rotation and one unstable. If you start with a rotation around the unstable axis, any tiny deviation (either in the initial condition or caused by some infinitesimal torque) will cause the rate of rotation about the other axes to grow exponentially (note that this does not mean the angular momentum is changing - it's actually a consequence of the fact that it's not changing).

On the contrary if you start with a rotation around either of the two stable axes, the perturbation doesn't grow. The long axis of a cylinder is one of the stable axes.

The non-rigid case is another matter, which sounds interesting.

[nathan]

Rigid objects have two stable principal axes of rotation and one unstable. If you start with a rotation around the unstable axis, any tiny deviation (either in the initial condition or caused by some infinitesimal torque) will cause the rate of rotation about the other axes to grow exponentially (note that this does not mean the angular momentum is changing - it's actually a consequence of the fact that it's not changing).

yeah, you're right. I'd got hung up about rotating the cylinder about its principle axes and forgot about other possibilities.

[Dorfl]

Rigid objects have two stable principal axes of rotation and one unstable. If you start with a rotation around the unstable axis, any tiny deviation (either in the initial condition or caused by some infinitesimal torque) will cause the rate of rotation about the other axes to grow exponentially (note that this does not mean the angular momentum is changing - it's actually a consequence of the fact that it's not changing).

Is this true even in the case of a cylinder?

[sol invictus]

Is this true even in the case of a cylinder?

Let me preface this by saying that I haven't thought about this kind of thing in a long time, so I might be wrong. But here's how I remember it: given an arbitrary rigid body you start by computing the moment of inertial tensor. It's real and symmetric, hence you can always diagonalize it. Do so. With the origin on the center of mass, the three basis vectors are the three principal axes, and their eigenvalues are their moments.

Then, as I remember it the one in the middle is unstable, and the two with max and min moments are stable. For a cylinder two are equal and one (the long one) is different; therefore the long one can never be the one in the middle; therefore it should be stable.

[Dorfl]

Let me preface this by saying that I haven't thought about this kind of thing in a long time, so I might be wrong. But here's how I remember it: given an arbitrary rigid body you start by computing the moment of inertial tensor. It's real and symmetric, hence you can always diagonalize it. With the origin on the center of mass, the three vectors in that basis are the three principal axes, and their eigenvalues are their moments.

Uh... I think we'll read about tensors next year, so for now I'll take your word for it.

Then, as I remember it the one in the middle is unstable, and the two with max and min moments are stable. For a cylinder two are equal and one (the long one) is different; therefore the long one can never be the one in the middle; therefore it should be stable.

That sounds reasonable. So a cylinder has to be non-rigid to spontaneously switch from the length-wise axis?

[sol invictus]

Uh... I think we'll read about tensors next year, so for now I'll take your word for it.

Just think of it as a 3X3 matrix.


That sounds reasonable. So a cylinder has to be non-rigid to spontaneously switch from the length-wise axis?

I think so, but I could be wrong.

[Dorfl]

Just think of it as a 3X3 matrix.
Then I think I follow you.

I think so, but I could be wrong.
Any idea on were I could read up on this? If people would demonstrate about Ringworld, presumably someone has written about Rama too.

[sol invictus]

Any idea on were I could read up on this? If people would demonstrate about Ringworld, presumably someone has written about Rama too.

I learned it from Goldstein, Classical Mechanics, but there must be hundreds of references.

As for the ringworld - a rigid hoop spinning around its circumference (like a wheel) should be stable by exactly the same argument (the moment for that rotation is either the max or the min, it can't be in the middle of the other two). But wasn't the ringworld supposed to be orbiting a star? That is unstable, but because of the gravity of the star, not because of the angular momentum of the ring.

[Dorfl]

As for the ringworld - a rigid hoop spinning around its circumference (like a wheel) should be stable by exactly the same argument (the moment for that rotation is either the max or the min, it can't be in the middle of the other two). But wasn't the ringworld supposed to be orbiting a star? That is unstable, but because of the gravity of the star, not because of the angular momentum of the ring.

Yes. I meant that if people will make fuzz about why the Ringworld is unstable, presumable they'd to that for Rama as well—not that they are unstable for similar reasons.

[nathan]

But wasn't the ringworld supposed to be orbiting a star? That is unstable, but because of the gravity of the star, not because of the angular momentum of the ring.

Yes, that's the problem with ringworld.

[dasmiller]

Let me preface this by saying that I haven't thought about this kind of thing in a long time, so I might be wrong. But here's how I remember it: given an arbitrary rigid body you start by computing the moment of inertial tensor. It's real and symmetric, hence you can always diagonalize it. Do so. With the origin on the center of mass, the three basis vectors are the three principal axes, and their eigenvalues are their moments.

Then, as I remember it the one in the middle is unstable, and the two with max and min moments are stable.

For a body rotating about the min axis, any tiny perturbation will cause it to diverge.

Imagine an even simpler case: 2 masses at opposite ends of a lightweight rigid rod. You can spin it along the long axis (the rod) without any problem. But if you perturb it just a bit, so the rod isn't quite along the spin axis anymore, then the two masses will be slightly off-center. Centrifugal force (yeah, I know, but we're in a rotating reference frame) will pull both masses farther from the spin axis. Of course, the forces on the two masses effectively create another torque on the system, so the actual motion is a bit more complicated. But that's the basic principle.

I'll try to post diagrams later.

[dasmiller]

Me neither. I guess a hoop could be fitted into either of the walls at the ends, but the interior of the ship is supposed to be empty.

I've pondered it a bit more. A more graceful passively-stable solution would be to make one of the end caps far more massive than the rest of the vessel. If I'm feeling inspired, I'll model it up (it just so happens that I have access to spacecraft mass properties modeling software).

I guess people tend not to make the same mistake twice, when handling multi-million dollar spacecraft.

You'd think that, wouldn't you?

Hmm... If parts break off of a satellite, is it more or less likely to produce dangerous space debris if that satellite is spinning?

For typical spacecraft, it really doesn't make much difference because orbital motion dominates.

If we take a spinning spacecraft such as a Hughes HS376 (it was a very popular spacecraft bus for many years), the outer edge of the spun section is moving at about 6 m/s due to the spacecraft's rotation. Meanwhile, orbital motion at GEO is about 3 km/s or about 500x as large.

So, imagine a piece of debris coming off of spacecraft A and hitting B, some time later. If the A and B are in exactly the same orbit (different positions, but same orbit), the debris would hit at 6 m/s. If A's orbit is inclined only 1 degree with respect to B, the impact would be 53 m/s or about 80X the impact.

If the orbits were at right angles to each other (which would be pretty weird at GEO), the closing speed would be over 4 km/s, for about 300,000X the impact.

(be warned that I just did that on a spreadsheet and haven't verified my math. But it looks about right to me).

[dasmiller]

Right, I was taught the same lesson at university. The reason given was that real-world cylinders are not perfectly rigid. Internal flexing allows the axis of rotation to migrate. This happens with artificial satellites, and it apparently was a surprise when the first such cylindrical satellite did it.

I'm puzzled by this one. I'm not aware of any structural dynamics effects playing a significant role in spacecraft stability. When I've seen "flex effects" in the context of stability, it's always refered to propellant motion.

(I've always felt that "flex effects" was a poor name for that phenomenon, by the way)

[dasmiller]

I've pondered it a bit more. A more graceful passively-stable solution would be to make one of the end caps far more massive than the rest of the vessel. If I'm feeling inspired, I'll model it up (it just so happens that I have access to spacecraft mass properties modeling software).

Whuf! To get the inertia ratio above one, one of the endcaps would have to have 98% of the total mass.

[Dorfl]

I've pondered it a bit more. A more graceful passively-stable solution would be to make one of the end caps far more massive than the rest of the vessel. If I'm feeling inspired, I'll model it up (it just so happens that I have access to spacecraft mass properties modeling software).
Whuf! To get the inertia ratio above one, one of the endcaps would have to have 98% of the total mass.

Oh. I suppose the Ramans would be able to overcome the technological difficulties that would create, but it still seems like a very inefficient solution.

Would a bunch of hoops on a string be better, do you think?

You'd think that, wouldn't you?

Heh. I guess I was a bit too optimistic there?

For typical spacecraft, it really doesn't make much difference because orbital motion dominates.

If we take a spinning spacecraft such as a Hughes HS376 (it was a very popular spacecraft bus for many years), the outer edge of the spun section is moving at about 6 m/s due to the spacecraft's rotation. Meanwhile, orbital motion at GEO is about 3 km/s or about 500x as large.

So, imagine a piece of debris coming off of spacecraft A and hitting B, some time later. If the A and B are in exactly the same orbit (different positions, but same orbit), the debris would hit at 6 m/s. If A's orbit is inclined only 1 degree with respect to B, the impact would be 53 m/s or about 80X the impact.

If the orbits were at right angles to each other (which would be pretty weird at GEO), the closing speed would be over 4 km/s, for about 300,000X the impact.

(be warned that I just did that on a spreadsheet and haven't verified my math. But it looks about right to me).
Sounds right to me. I hadn't considered how much bigger the orbital velocity is. Pity: it would've been nice if a broken satellite would just fling most debris to burn in the atmosphere or disappear into deep space.

[sol invictus]

For a body rotating about the min axis, any tiny perturbation will cause it to diverge.

I'm just about certain you're wrong: if the body is rigid and by "min axis" you mean the principal axis with the minimum moment, that rotation is stable. The unstable one is the intermediate axis.

Say you're rotating around axis 1. As I remember it, the equation governing small perturbations looks something like this.

w2''=-(m1-m2)(m1-m3)w2/D, w3''=-(m1-m2)(m1-m3)w3/D

where w2 is the angular velocity about axis 2, w2'' is its second time derivative, m1 is the moment of axis 1, and D is some positive denominator I've forgotten. The point is that (m1-m2)(m1-m3)>0 either if m1<m2 and m1<m3 or if m1>m2 and m1>m3. It's negative if m2>m1>m3 or if m3>m1>m2.

Therefore rotation around either the min or max principal axis is stable, but the rotation around the intermediate one is unstable.

[Darat]

Fascinating thread, not quite up the standard of the best Arthur C Clarke stories but really good. Thanks!

[nathan]

http://scienceworld.wolfram.com/physics/RotationalStability.html

I'm puzzled by what that is saying. For a long cylinder we have moments ordered A < B = C. A being the moment around the cylinder's axis and B & C being the other two orthogonal axes. Plugging those in gives

\Delta_C = 0

\Delta_A = {{B - A}\over{A}} > 0

but I thought rotation about C (or B) is more stable than that about A. Or perhaps I'm misunderstanding \Delta, and the larger it is the more unstable the rotation? (Hm, as A tends to zero, \Delta_A tends to \infty)

[Molinaro]

Now we're on my turf! No, assuming any vaguely smooth mass distribution, Rama wouldn't be passively stable about the spin axis. You could come up with mass distributions that would be (a very heavy hoop in the center, with the rest very lightweight), but I don't think Clarke gave us any indication of that.

Isn't the cylindrical sea just such a mass hoop? It was positioned halfway from either end of the cylinder.

[Dorfl]

Isn't the cylindrical sea just such a mass hoop? It was positioned halfway from either end of the cylinder.

It could be, provided the material the rest of Rama is made out of is really, really light, couldn't it? I don't remember if they said anything about the density of Raman building materials.

[sol invictus]

http://scienceworld.wolfram.com/physics/RotationalStability.html

I'm puzzled by what that is saying.

It agrees with what I was saying, except I'm not entirely sure how they've defined those \Deltas.

For a long cylinder we have moments ordered A < B = C. A being the moment around the cylinder's axis and B & C being the other two orthogonal axes. Plugging those in gives

\Delta_C = 0

\Delta_A = {{B - A}\over{A}} > 0

but I thought rotation about C (or B) is more stable than that about A.

No. As I keep saying, the min and the max are stable, while the mid is unstable. In this case the mid and max coincide, which makes both of them marginally stable (that's why Delta C=0). Physically that zero means those two axes can rotate into each other at a constant rate (rather than oscillating), which is just because the cylinder has a rotation invariance (so there's no unique way to decide which pair of axes are the principals).

[dasmiller]

Isn't the cylindrical sea just such a mass hoop? It was positioned halfway from either end of the cylinder.

Two complications.

First, as Dorfl points out, it's difficult (but perhaps not impossible) to imagine that the sea was massive enough to dominate Rama's mass properties. But it certainly is in the right place.

Second, the impact of partially-constrained liquids on the stability of spinning bodies is a bogglingly complex topic. (I have a theory that "mind-boggling" is redundant. What else would one boggle? Wait, don't answer that). Anyway, many smart people have devoted their careers to it and I don't think that any of them would characterize the generalized problem as 'well understood.' My intestines tell me that, in Rama's case, the ocean would tend to amplify perturbations, but my intestines aren't reliable on such topics.

It's conceivable that you could 'tune' the ocean to damp out perturbations. Now *there* would be a topic for a PhD thesis!

[Dorfl]

It's conceivable that you could 'tune' the ocean to damp out perturbations. Now *there* would be a topic for a PhD thesis!

Hmm... Wasn't the ocean sort of 'tuned'? I seem to remember there being a lot of little barriers underneath the surface to filter waves. I'm not sure how much that'd do for stability, though.

[dasmiller]

Therefore rotation around either the min or max principal axis is stable, but the rotation around the intermediate one is unstable.

I'll try to dig up the math, but I have a long, busy day ahead of me so it may be a while.

But until then - why constrain our spin axis to just the 3 principle axes (min, max, intermediate)? You can impart a spin about any of an infinite number of axes. It won't spin stably about most of them, but you can start with any.

So - to be certain that we're consistent on our point of disagreement

For a generalized body, with different inertias about the 3 principle axes, and a body given an initial spin about any arbitrary axis:

I believe that the body will eventually reorient so that its max axis is aligned with the spin axis. The amount of time required for the body to reorient depends on the principle inertias, the spin rate, and where the initial spin axis is with respect to the principle axes. If the initial spin axis is perfectly aligned with the min axis, then it will take infinitely long for the body to reorient, but even a tiny perturbation will speed things up dramatically.

You believe that for an initial axis of rotation sufficiently near the min axis, the body will reorient so that its min axis lines up with the axis of rotation.

Correct?

[dasmiller]

Hmm... Wasn't the ocean sort of 'tuned'? I seem to remember there being a lot of little barriers underneath the surface to filter waves. I'm not sure how much that'd do for stability, though.

Well, if I was to imagine how the ocean would be tuned, that sort of thing would definitely be part of it. So would the shape of the coastline, depth profile, etc etc - anything to manage how the water moved in response to Rama's motion.

[sol invictus]

You believe that for an initial axis of rotation sufficiently near the min axis, the body will reorient so that its min axis lines up with the axis of rotation.

Correct?

No. The angular velocity around the other two axes will oscillate with frequency (A-B)(A-C)/BC. Since they started small, they remain small. It will not reorient, as that would violate conservation of angular momentum.

This is all for a rigid body with no external torques acting.

[dasmiller]

No. The angular velocity around the other two axes will oscillate with frequency (A-B)(A-C)/BC. Since they started small, they remain small.

Then if any small perturbation will cause the spin axis to deviate slightly (but permanently) from the min axis, in what sense is it spinning stably about the mins axis?

It will not reorient, as that would violate conservation of angular momentum.

I was using 'reorient' because I didn't want to imply that the spin axis wasn't fixed in inertial space. The body changes orientation, but the axis of rotation doesn't. If this is a violation of conservation of angular momentum for the spin-axis-near-min-axis case, why is it not a violation of angular momentum for a body initially spinning about the intermediate axis?

[ben m]

Then if any small perturbation will cause the spin axis to deviate slightly (but permanently) from the min axis, in what sense is it spinning stably about the mins axis?


The deviation is only as big as the perturbation. Small perturbation, small deviation. That's how all stable systems work.

By contrast, for an object spinning around its "middle" axis, any arbitrarily small perturbation will make it tumble. The size of the deviation grows and becomes extremely large, even if the perturbation was small.

This is how all stable equilibria work. Think of, say, a mass sliding around the bottom of a frictionless bowl. If you give it a small push away from the bottom, it will undergo small oscillations about its rest position. A larger push will give larger oscillations. That's stable. By contrast, consider a mass sitting on the peak of a frictionless hill; a small push leads to a large excursion down the hill. That's unstable.

[sol invictus]

Then if any small perturbation will cause the spin axis to deviate slightly (but permanently) from the min axis, in what sense is it spinning stably about the mins axis?

As ben says, the definition of "stable" is that small perturbations do not grow but instead lead to small oscillations around the stable point.


I was using 'reorient' because I didn't want to imply that the spin axis wasn't fixed in inertial space. The body changes orientation, but the axis of rotation doesn't.

The body cannot reorient in the way you proposed unless either energy or angular momentum conservation is violated (such as if a torque is applied, or if the body can lose energy through some interactions).

If this is a violation of conservation of angular momentum for the spin-axis-near-min-axis case, why is it not a violation of angular momentum for a body initially spinning about the intermediate axis?

It is. A body initially spinning around the intermediate axis does not reorient to spin around one of the stable axes, at least not so long as its energy and angular momentum are constant. It just tumbles around a lot, going very far from the intermed axis and only occasionally coming back.

What you may be thinking of is that rotations around the min axis require the most energy. So if you have a process that reduces the energy which retaining the angular momentum (which is sometimes a good approximation for certain kinds of dissipation), the body will eventually re-orient itself to rotate around the max axis.

[Dorfl]

What you may be thinking of is that rotations around the min axis require the most energy. So if you have a process that reduces the energy which retaining the angular momentum (which is sometimes a good approximation for certain kinds of dissipation), the body will eventually re-orient itself to rotate around the max axis.

Doesn't every non-rigid body have the ability to reduce energy while retaining angular momentum?

[sol invictus]

Doesn't every non-rigid body have the ability to reduce energy while retaining angular momentum?

No, I wouldn't say every. But generally it is easier to lose energy than it is to lose angular momentum (that's why galaxies are flat and the planets orbit the sun in a plane).

I suppose a non-rigid body can lose energy by generating heat from internal motions and radiating it away. That process doesn't quite conserve the angular momentum of the body, but it should be close enough. However it could take an extremely long time for this to significantly affect the motion if the body is nearly rigid. Moreover not many bodies are totally isolated in space, so interactions with the surrounding environment are often going to be more important.

[dasmiller]

The deviation is only as big as the perturbation. Small perturbation, small deviation. That's how all stable systems work.

<snip>

This is how all stable equilibria work. Think of, say, a mass sliding around the bottom of a frictionless bowl. If you give it a small push away from the bottom, it will undergo small oscillations about its rest position. A larger push will give larger oscillations. That's stable. By contrast, consider a mass sitting on the peak of a frictionless hill; a small push leads to a large excursion down the hill. That's unstable.

Interestingly, we simply have a different use of terminology here. In aircraft dynamics, for example, if a small perturbation isn't damped, then the aircraft isn't considered "stable."

But that being said, I was operating under some confusion (see next post)

[dasmiller]

The body cannot reorient in the way you proposed unless either energy or angular momentum conservation is violated (such as if a torque is applied, or if the body can lose energy through some interactions).

So I woke up this morning troubled by the fact that in my view of rotating rigid bodies, the motions wouldn't necessarily be reversible. If one magically and perfectly reversed the rotation, the rigid body wouldn't eventually work its way back to the condition it was in when the perturbation was applied. That's a pretty serious flaw for a rigid body that isn't interacting with its environment.

After more pondering, I think I was led astray by way too many years working with satellites. Most real-world satellites don't act like rigid bodies because of fuel slosh. I believe that fuel motion provides the sort of energy loss (not necessarily angular momentum loss) I'd need to get the nutation to grow. Indeed, as I recall, the dedamping (the term "amplification" was reserved for other effects) constants were always associated with fuel motion.

But I was definitely confused on rigid body rotation. Now I'm not entirely sure that I'm unconfused on the rotation of a body with significant quantities of viscous fluids, but . . . that may be a bit ambitious for a Saturday morning.

[sol invictus]

Interestingly, we simply have a different use of terminology here. In aircraft dynamics, for example, if a small perturbation isn't damped, then the aircraft isn't considered "stable."

If there's dissipation, small oscillations will damp out so long as they don't grow due to some dynamical instability. So the two uses are consistent.

So I woke up this morning troubled by the fact that in my view of rotating rigid bodies, the motions wouldn't necessarily be reversible. If one magically and perfectly reversed the rotation, the rigid body wouldn't eventually work its way back to the condition it was in when the perturbation was applied.

Precisely. Damping is an irreversible process that involves an increase in entropy, so it can only happen when the "rigid" body is coupled to many other degrees of freedom (such as molecules of fuel).


But I was definitely confused on rigid body rotation. Now I'm not entirely sure that I'm unconfused on the rotation of a body with significant quantities of viscous fluids, but . . . that may be a bit ambitious for a Saturday morning.

There's a beautiful geometric way to visualize the motion which this thread reminded me of. Attach a coordinate frame to the rigid body, with the origin at the center of mass and the axes aligned with the principal axes of the body. Now use that coordinate system. It's non-inertial, which means the angular momentum vector in those coordinates doesn't remain constant. But because these coordinates are related to inertial coordinates by a rotation, the length of the angular momentum vector is constant. So is the total energy, of course. Therefore

${\rm total \, angular \, momentum}=\vec L^2 = L_x^2+L_y^2+L_z^2 ={\rm constant}$
${\rm kinetic \, energy}=T={1 \over 2}\left( L_x^2/A+L_y^2/B+L_z^2/C \right) = {\rm constant}$

The first equation is a sphere, the second is an ellipsoid. Therefore for some given angular momentum and energy, the motion will be along a line of intersection between those two shapes. That line of intersection is some closed curve on the sphere. It can be a small circle around the min axis or a small circle around the max axis. But it can't be a small circle around the mid axis, as you can see if you visualize the intersection - it must be a very large circle that only occasionally passes close to it.

So you can see the stability purely from that, which is pretty nice. No need to write any differential equations at all.

[Dorfl]

No, I wouldn't say every. But generally it is easier to lose energy than it is to lose angular momentum (that's why galaxies are flat and the planets orbit the sun in a plane).

I suppose a non-rigid body can lose energy by generating heat from internal motions and radiating it away. That process doesn't quite conserve the angular momentum of the body, but it should be close enough. However it could take an extremely long time for this to significantly affect the motion if the body is nearly rigid. Moreover not many bodies are totally isolated in space, so interactions with the surrounding environment are often going to be more important.

Sorry, I meant kinetic energy, not just energy. That can be decreased without changing angular momentum, or requiring transporting away, can't it?

Could Rama's atmosphere heating up then absorb some of the kinetic energy, leading to it eventually becoming unstable?

[sol invictus]

Sorry, I meant kinetic energy, not just energy. That can be decreased without changing angular momentum, or requiring transporting away, can't it?

Under some circumstances, yes.

Could Rama's atmosphere heating up then absorb some of the kinetic energy, leading to it eventually becoming unstable?

Any system will eventually evolve so as to maximize its free energy subject to whatever constraints are in place. In this case in the medium-long term that would probably correspond to a uniform rotation around the axis with the maximum moment while the excess energy goes into either heating the atmosphere/water or radiation. In the very long term I suspect it would correspond to zero rotation, even if Rama is completely isolated (because it can very slowly radiate away its angular momentum as well).

[dasmiller]

If there's dissipation, small oscillations will damp out so long as they don't grow due to some dynamical instability. So the two uses are consistent.

You'd think that I'd learn not to disagree, but the two uses don't sound consistent to me.

An aircraft is not considered to be stable if it simply maintains a perturbed orientation, or if it oscillates indefinitely due to a perturbation. The undamped rotating rigid body is considered to be stable even though the perturbation causes permanent oscillation.

Maybe we'll agree to disagree on that one.

Anyway, back to Rama and my earlier remark about 'tuning' the ocean. Barring any active measures (e.g. pumps), I suspect that the ocean can only dissipate energy. 'Tuning' would mean designing the ocean to get a particular dedamping constant, but since the ocean would always be dissipating energy, you'd never be able to get a negative dedamping constant. The ocean would always cause perturbations to grow.

[Myriad]

[obsessive sf fan]The Raman ocean is frozen except when it approaches a star system.[/obsessive sf fan]

Respectfully,
Myriad

[sol invictus]

You'd think that I'd learn not to disagree, but the two uses don't sound consistent to me.

An aircraft is not considered to be stable if it simply maintains a perturbed orientation, or if it oscillates indefinitely due to a perturbation. The undamped rotating rigid body is considered to be stable even though the perturbation causes permanent oscillation.

Normally, stable equilibria occur at the bottom of a potential well. Like ben said, think of a marble rolling around near the bottom of a bowl. If the marble is right at the bottom it stays there, but if you move it a little off it oscillates around the stable point. Give it a little friction and the oscillations damp, and eventually it's at rest at the stable point again.

This stable point around the min axis is very unusual, though, because it's not a minimum of the energy - it's a max (with fixed angular momentum at least). Nevertheless it's stable in the sense that small oscillations don't grow in the ideal case with no friction... but it does seem to me that damping actually makes them grow rather shrink. So in this case I think I agree with you that the two uses don't agree. And thanks for pointing it out, because I'm not sure I know another example like that.

The ocean would always cause perturbations to grow.

You mean, around the min axis? Around the max axis it should be stable in all senses of the word.

[dasmiller]

You mean, around the min axis? Around the max axis it should be stable in all senses of the word.

Yes, I was thinking specifically of Rama nominally spinning around the min axis.

[Dr Adequate]

Since we know that Rama does not obey Newton's Third Law, any analysis of its motion is going to be fraught with difficulties.

[Dorfl]

Since we know that Rama does not obey Newton's Third Law, any analysis of its motion is going to be fraught with difficulties.

Point. But I thought it behaved classically when in flight. It seemed like the non-Newtonian stuff only happened when it had just refuelled near a star.

[nathan]

Point. But I thought it behaved classically when in flight. It seemed like the non-Newtonian stuff only happened when it had just refuelled near a star.

Hey, this is like a detective story -- some participants have more information than others :)

[dasmiller]

Normally, stable equilibria occur at the bottom of a potential well. Like ben said, think of a marble rolling around near the bottom of a bowl. If the marble is right at the bottom it stays there, but if you move it a little off it oscillates around the stable point. Give it a little friction and the oscillations damp, and eventually it's at rest at the stable point again.

This stable point around the min axis is very unusual, though, because it's not a minimum of the energy - it's a max (with fixed angular momentum at least). Nevertheless it's stable in the sense that small oscillations don't grow in the ideal case with no friction... but it does seem to me that damping actually makes them grow rather shrink. So in this case I think I agree with you that the two uses don't agree. And thanks for pointing it out, because I'm not sure I know another example like that.


It seems to me that the marble-in-the-bowl might actually be another example. If I whip the marble around so that it's quickly circling just below the rim of the bowl, it that trajectory stable? (that's not a rhetorical question. I really don't know, but it seems like it could be). It's certainly not a minimum-energy case.

[sol invictus]

It seems to me that the marble-in-the-bowl might actually be another example. If I whip the marble around so that it's quickly circling just below the rim of the bowl, it that trajectory stable? (that's not a rhetorical question. I really don't know, but it seems like it could be). It's certainly not a minimum-energy case.

If it's just below the rim it's not, because it can fly out under a small perturbation. If it's farther down it might be, depending on the shape of the bowl.

But you make a good point. Orbits are examples of this, for example the earth's orbit around the sun. Even subject to no external perturbations, a two-body gravitational system will eventually spiral together, because that configuration has lower energy and the excess can be radiated away in various forms (including gravitational radiation) while increasing the entropy.

[ben m]

If it's just below the rim it's not, because it can fly out under a small perturbation. If it's farther down it might be, depending on the shape of the bowl.

But you make a good point. Orbits are examples of this, for example the earth's orbit around the sun. Even subject to no external perturbations, a two-body gravitational system will eventually spiral together, because that configuration has lower energy and the excess can be radiated away in various forms (including gravitational radiation) while increasing the entropy.

A Keplerian orbit isn't the minimum of a global energy function, but it is the minimum of an energy function constrained by angular momentum conservation. I wondered if their might not be a similar "effective energy" function for Rama, and a wobbly-Rama might be shown to be making small oscillations around the effective minimum. But on reflection I'm not sure this makes sense---I can't quite see how this breaks up into a dynamical variable and a constraint equation.

But I haven't done solid-body mechanics in ages.