Is Rama stable?

[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

[latex]\Delta_C = 0[/latex]

[latex]\Delta_A = {{B - A}\over{A}} > 0[/latex]

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

 

[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

[latex]\Delta_C = 0[/latex]

[latex]\Delta_A = {{B - A}\over{A}} > 0[/latex]

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.