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Tucson_Tim
2007-Oct-18, 09:07 PM

Can an object, say like an orbiting planet, have multiple axes of rotation? I can easily see where an object is spinning on one axis, like the Earth, and then could be spinning pole-over-pole also. But is this a transient thing and eventually stopped by interaction with the Sun? I'm not talking about precession here (at least I don't think I am). It would seem to me like a solid object traveling in inter-stellar space could have many axes of rotation. To take it one step farther, couldn't a body have an infinite number of axes of rotation?

Swift
2007-Oct-18, 09:40 PM
My understanding is that asteroids can tumble. Here (http://www-personal.umich.edu/~scheeres/conferences/AIAA-2004_1446.pdf) and here (http://wise-obs.tau.ac.il/~david/OnSpinEvolution.htm) are some papers about it. I don't know about larger objects.

IsaacKuo
2007-Oct-18, 09:50 PM
Something that is roughly spherical will more or less have only one axis of rotation. However, this axis can wobble due to precession effects and other effects.

A non-spherical shape can rotate around multiple axes (i.e. tumbling). However, internal flexing will tend to eventually dampen out the tumbling into rotation about a single axis.

So tumbling is possible for small asteroids, but not for objects large enough to be roughly spherical.

hhEb09'1
2007-Oct-18, 10:19 PM
If you're not talking about precession (by which I assume you mean, influenced by "outside" forces) and you mean axes fixed to the object, then probably there can only be one rotational axis. An object rotating rapidly about a long axis and tumbling over that axis is not really tumbling about an axis fixed to the object, since the object is rotating.

Ken G
2007-Oct-19, 12:44 AM
Or another way to say all this, a sphere has a moment of inertia that is a single number, which means that its angular momentum completely determines its angular velocity. In the absence of important torques, as is typically the case (i.e., precession is slow), the angular momentum is a constant, so the angular velocity is too. One axis.

hhEb09'1
2007-Oct-19, 12:57 AM
Of course, we're talking about Toutatis (http://www.solarviews.com/eng/toutatis.htm) :)

Toutatis is one of the strangest objects in the solar system, with a highly irregular shape and an extraordinarily complex "tumbling" rotation. Both its shape and rotation are thought to be the outcome of a history of violent collisions. "The vast majority of asteroids, and all the planets, spin about a single axis, like a football thrown in a perfect spiral, but Toutatis tumbles like a flubbed pass," said Dr. Scott Hudson of Washington State University. One consequence of this strange rotation is that Toutatis does not have a fixed north pole like the Earth. Instead, its north pole wanders along a curve on the asteroid about every 5.4 days. "The stars viewed from Toutatis wouldn't repeatedly follow circular paths, but would crisscross the sky, never following the same path twice," Hudson said.

publius
2007-Oct-19, 01:20 AM
General rotation is a very complex problem. *Very* complex. See a mechanics text for a taste of it. In general, the moment of inertia is a tensor, not a scalar. We can write:

L = I * w. Where, L and w are vectors, and I is the inertia tensor. That inertia tensor is complex thing, and it has coordinate dependencies that make things even more complex.

Now, if I is just a scalar time the identity matrix, which is the case for a sphere, things are simple. L and w lie along the same axis, they always point in the same direction and the one is just a scalar multiple of the other.

Now, torque is the time rate of change of angular momentum, so you can write:

T = dL/dt = d/dt(I * w). In the simple case of scalar times identity, that's very simple, T = k*dw/dt.

When I is not so simple, things get complex. That derivative operation gets involved, but it can be shown that:

T = I*dw/dt - w x L.

If no external torque is applied, T = 0, and

I*dw/dt = w x L. (I forget, there may be minus sign I'm missing)

If w and L are parallel, as in the simple case, the axis of rotation doesn't move. However, if they are not for a non-trivial I, then there must be a dw/dt in the absence of torque.

That is precession. Note that L remains constant. Angular momentum is constant, that's the defintion of torque. However, that does not mean that w, the axis of rotation, must be constant.

L is what is constant, not w, in the general case. And that's just weird. :)

The translation momentum case is p = m*v. m is a simple scalar, and so a change in p means a change in v (well, the rocket equation looks at something with a variable mass, but that's a reaction thing, throwing it off, and using conservation of total momentum to simplify things. The variation of the I tensor is something different).

-Richard

hhEb09'1
2007-Oct-19, 02:35 AM
If w and L are parallel, as in the simple case, the axis of rotation doesn't move. However, if they are not for a non-trivial I, then there must be a dw/dt in the absence of torque. In the case of Toutatis, it's a loose lump of a couple loose lumps, so it's bound to be complicated.

But, we still have the question of the OP. You show the axis changing, but what does it mean to have rotation around two axes.

Tucson_Tim
2007-Oct-19, 02:46 AM
Here's a hypothetical scenario, a thought experiment:

Say you have an unihabited planet, size of the Earth (we'll call it Earth-2), that has been ejected into interstellar space (never mind how) but the result is that Earth-2 is traveling between the stars where the gravitational influence of other stars and large bodies is negligible, and it is still spinning on its axis. Along comes another large body and collides with Earth-2, a glancing blow, that imparts a rotation of pole-over-pole. Since this is a thought experiment, let's not get involved with how this glancing blow can occur without ripping both bodies apart - let's just assume it happened. Will Earth-2 have a rotation around two axes? If so, will it last long?

Tucson_Tim
2007-Oct-19, 02:49 AM
Ken G,

I think you're saying that if Earth-2 (in my above example) is perfectly spherical, it can only have one axis of rotation, even after the "collision".

tdvance
2007-Oct-19, 03:11 AM
There is only one axis of rotation--it's sort of like vectors in that if you combine two rotations, you get a single "resultant rotation". This is true whether the object is spherical or not. A collision will have the effect of moving the axis. Saying there are two axes is like saying that 7 is two numbers, because it is 4 combined with 3. You could claim that an axis is the combination of any number of axes in an infinite number of ways.

On the other hand, various forces can cause the axis of rotation to move--e.g. gravity from the sun and moon acting on the equatorial bulge causing precession and nutation of the Earth (relative to "the fixed stars", not relative to the Earth itself).

a1call
2007-Oct-19, 03:15 AM
There seems to be observational evidence that in a planetary system such as ours the influence of the neighbors will favor the stability of a single axis of rotation. I remember (I think in Cosmos the series) mention of actual existence of an expected very slow rotation of the moon in accordance with the impact which has caused the whitish apron-ed crater. I assume this is an axis different from the main Lunar rotational axis.

From strictly physical point of view I fail to see why 2 or more axis can't act in conjunction.

Here is how I think it could look (http://perfext.com/virtual/axis.html)

Click on the two buttons to stop and start rotation.
You can also left click and drag to rotate the globe or right click and drag to zoom in and out.

Tucson_Tim
2007-Oct-19, 03:21 AM
There seems to be observational evidence that in a planetary system such as ours the influence of the neighbors will favor the stability of a single axis of rotation. I remember (I think in Cosmos the series) mention of actual existence of an expected very slow rotation of the moon in accordance with the impact which has caused the whitish apron-ed crater. I assume this is an axis different from the main Lunar rotational axis.

From strictly physical point of view I fail to see why 2 or more axis can't act in conjunction.

Here is how I think it could look (http://perfext.com/virtual/axis.html)

Click on the two buttons to stop and start rotation.
You can also left click and drag to rotate the globe or right click and drag to zoom in and out.

That's good! It was what I was thinking. But is it possible?

Tucson_Tim
2007-Oct-19, 03:24 AM
On the other hand, various forces can cause the axis of rotation to move--e.g. gravity from the sun and moon acting on the equatorial bulge causing precession and nutation of the Earth (relative to "the fixed stars", not relative to the Earth itself).

I understand the effects of precession (I used to understand the physics behind it but it was too many years ago) and I think I want to ignore it for this thought experiment. Assume no outside forces on the sphere.

Tucson_Tim
2007-Oct-19, 03:56 AM
General rotation is a very complex problem. *Very* complex. See a mechanics text for a taste of it. In general, the moment of inertia is a tensor, not a scalar. We can write:

L = I * w. Where, L and w are vectors, and I is the inertia tensor. That inertia tensor is complex thing, and it has coordinate dependencies that make things even more complex.

Now, if I is just a scalar time the identity matrix, which is the case for a sphere, things are simple. L and w lie along the same axis, they always point in the same direction and the one is just a scalar multiple of the other.

Now, torque is the time rate of change of angular momentum, so you can write:

T = dL/dt = d/dt(I * w). In the simple case of scalar times identity, that's very simple, T = k*dw/dt.

When I is not so simple, things get complex. That derivative operation gets involved, but it can be shown that:

T = I*dw/dt - w x L.

If no external torque is applied, T = 0, and

I*dw/dt = w x L. (I forget, there may be minus sign I'm missing)

If w and L are parallel, as in the simple case, the axis of rotation doesn't move. However, if they are not for a non-trivial I, then there must be a dw/dt in the absence of torque.

That is precession. Note that L remains constant. Angular momentum is constant, that's the defintion of torque. However, that does not mean that w, the axis of rotation, must be constant.

L is what is constant, not w, in the general case. And that's just weird. :)

The translation momentum case is p = m*v. m is a simple scalar, and so a change in p means a change in v (well, the rocket equation looks at something with a variable mass, but that's a reaction thing, throwing it off, and using conservation of total momentum to simplify things. The variation of the I tensor is something different).

-Richard

Richard,

To say you lost me would be an understatement. Years ago I might have followed you but now. . . Sorry.

Tim

publius
2007-Oct-19, 03:58 AM
But, we still have the question of the OP. You show the axis changing, but what does it mean to have rotation around two axes.

Consult a good mechanics text to see this -- general rotation always made my head hurt. :lol:

I think that two or more different axes of rotation will always sum to some resultant axis. For example, you can think of this relation as the defintion of the w vector:

v = r x w.

You can let w = w1 + w2 +... and that distributes out as

v = r x w1 + r x w2 + ....

-Richard

George
2007-Oct-19, 04:09 AM
I would have guessed three axis would define the worst case. This can be found in a spaceship's pitch, roll, and yaw.

Tucson_Tim
2007-Oct-19, 02:48 PM
Here is how I think it could look (http://perfext.com/virtual/axis.html)

Click on the two buttons to stop and start rotation.
You can also left click and drag to rotate the globe or right click and drag to zoom in and out.

All right, I need some input on this link that a1call provided. This is exactly the type of movement I was thinking about when I started this thread. Now, is this multi-axis rotation possible?

Thanks again for the link a1call.

a1call
2007-Oct-19, 02:57 PM
My pleasure Tucson_Tim. :)

Ken G
2007-Oct-19, 03:13 PM
Ken G,

I think you're saying that if Earth-2 (in my above example) is perfectly spherical, it can only have one axis of rotation, even after the "collision".

Right, if there's no torque, and it's a sphere, then there's only one axis.

Tucson_Tim
2007-Oct-19, 03:15 PM
Right, if there's no torque, and it's a sphere, then there's only one axis.

Ken, then where would the axis be in the animation of a1call's link? Or is that motion not possible without an outside force?

thecolorofash
2007-Oct-19, 03:23 PM
Consult a good mechanics text to see this -- general rotation always made my head hurt. :lol:

I think that two or more different axes of rotation will always sum to some resultant axis. For example, you can think of this relation as the defintion of the w vector:

v = r x w.

You can let w = w1 + w2 +... and that distributes out as

v = r x w1 + r x w2 + ....

-Richard

That would be true if all the axes had the same rotation period which is generally not the case. Think of a gyroscope in space.

Jerry
2007-Oct-19, 05:57 PM
There is a very good explanation of space-axial orientation here:

http://einstein.stanford.edu/highlights/hl_polhode_story.html

Tucson_Tim
2007-Oct-19, 06:10 PM
There is a very good explanation of space-axial orientation here:

http://einstein.stanford.edu/highlights/hl_polhode_story.html

Thanks. That is a very good article - good videos and animation also. Polhode motion, eh?

a1call
2007-Oct-19, 07:24 PM
Highly enlightening Jerry,
Many thanks.

tdvance
2007-Oct-19, 09:06 PM
There seems to be observational evidence that in a planetary system such as ours the influence of the neighbors will favor the stability of a single axis of rotation. I remember (I think in Cosmos the series) mention of actual existence of an expected very slow rotation of the moon in accordance with the impact which has caused the whitish apron-ed crater. I assume this is an axis different from the main Lunar rotational axis.

From strictly physical point of view I fail to see why 2 or more axis can't act in conjunction.

Here is how I think it could look (http://perfext.com/virtual/axis.html)

Click on the two buttons to stop and start rotation.
You can also left click and drag to rotate the globe or right click and drag to zoom in and out.

Look carefully--there is only one axis of rotation at one time. It keeps switching between one and the other--something prohibited by conservation of angular momentum--unless a force keeps causing it to switch between one and the other.

Todd

Tucson_Tim
2007-Oct-19, 09:21 PM
Look carefully--there is only one axis of rotation at one time. It keeps switching between one and the other--something prohibited by conservation of angular momentum--unless a force keeps causing it to switch between one and the other.

Todd

Not sure about that. It looks like it is exhibiting polhode motion that was described in the link supplied by Jerry:

http://einstein.stanford.edu/highlig...ode_story.html

publius
2007-Oct-19, 10:38 PM
That would be true if all the axes had the same rotation period which is generally not the case. Think of a gyroscope in space.

Again, I'll just say consult a good mechanics text for a discussion of how solid body rotation is modelled. The 'w' vector is a vector we can right as w*n, where w is the scalar rotational velocity (radians per time), and n is a unit vector along the spatial axis. That can be resolved into whatever components you wish.

For example, an w vector of sqrt(2) units pointing at a 45 degree angle in the X-Y plane can be resolved into two components, w1 and w2 of 1 unit along the x and y axis.

IOW, a sphere centered at the origin rotating at 1414 RPM aligned as above can be seen as the sum of two rotations of 1000RPM about the x and y axes.

Now, a gryoscope is something different. The rotor is a solid body rotation, but it is in a cage of gimbals that require *differential* movement. Just forgetting about the rotor itself, and imagining spinning the gimbals, you have a differential rotation -- different solid parts of the object are moving relative to each other and constrained to a particular axis.

The above is about rigid, solid body rotation. Take a blob and get it rotating -- not different pieces moving relative to each other.

-Richard

publius
2007-Oct-19, 10:47 PM
Look carefully--there is only one axis of rotation at one time. It keeps switching between one and the other--something prohibited by conservation of angular momentum--unless a force keeps causing it to switch between one and the other.

Todd

This was the point I was trying to make above. In the general case, the angular momentum vector, L, and the axis of rotation *do not point in the same direction*. They are not parallel. What is conserved is L, not w. In the simple case of rotation about a principle axis, L and w will be parallel (and the principle axes have to do with eigenvectors of the inertia tensor). However, for other rotations, they are not parallel.

So, in order to conserve L, w must be vary, and that's torque-free precession. Like I mentioned, it's complicated. :)

There are also some principles about stability that I forget just how they go. In general, the inertia tensor will have off diagonal components, but you can diagonalize it by finding the principle axes and using coordinates aligned with them. That operation is finding the "natural axes" of rotation so to speak. (For a sphere, basically you're diagonal using any coordinate system). The math is then simpler to resolve a general axis of rotation about those principle axes.

In general, the diagonal components will not be equal. If you start something spinning about any one of those axes, L will align with w, and it will not precess. But what if you perturb it slightly? There is a theorem about stable rotation that relates to the relative magnitudes of the diagonal components, I1, I2, and I3. Depending on that relation, a slight perturbation can cause the object to start "tumbling" (precessing) and it gets worse and worse. That's an unstable principle axis. With a stable axis, a slight bump does not get worse.

-Richard

Ken G
2007-Oct-20, 02:53 AM
In general, the diagonal components will not be equal. If you start something spinning about any one of those axes, L will align with w, and it will not precess. But what if you perturb it slightly? There is a theorem about stable rotation that relates to the relative magnitudes of the diagonal components, I1, I2, and I3. Depending on that relation, a slight perturbation can cause the object to start "tumbling" (precessing) and it gets worse and worse.

The unstable axis is the middle I diagonal element, the largest and smallest are stable. A tennis racket is a good example-- you can spin it around its handle (smallest diagonal I), or around an axis through its face (so hold it normally and flip it, catching the handle), but if you hold it by the handle with its face up, and try to flip the handle back to your hand, it will tend to spin the racket head. I think the extreme I tend to be stable because the angular momentum would have to go only "up" or "down" in I, which it won't do, but it can spread out both ways.

publius
2007-Oct-20, 05:30 AM
Ken,

Ah, so. Thanks.

I knew it was something. :lol: That comes from an old character here, long since passed away, I knew as a kid. He would sit around and listen to various mechanical or whatever problems being discussed-- say a truck wouldn't run -- and offer vague and general advice from time to time on what could be wrong that was usually well off the mark, due mostly to age. If the last thing you worked on was a T-model, well.....

When you found out what the problem was, he'd say, "I knew it was something".

-Richard

publius
2007-Oct-21, 12:14 AM
When the polhode rolls on the herpolhode
invariably in the plane,
the forces tend to be free.

When the polhode slips on the herpolhode,
the forces are not mundane,
they're nonholonomic, you see!

David N. Williams

That's what the general solid body rotation leads to. :lol:

-Richard

Ken G
2007-Oct-21, 04:30 PM
Crystal clear. :)

Tucson_Tim
2007-Oct-21, 04:50 PM

End-of-line.

mugaliens
2007-Oct-21, 04:50 PM

Can an object, say like an orbiting planet, have multiple axes of rotation? I can easily see where an object is spinning on one axis, like the Earth, and then could be spinning pole-over-pole also. But is this a transient thing and eventually stopped by interaction with the Sun? I'm not talking about precession here (at least I don't think I am). It would seem to me like a solid object traveling in inter-stellar space could have many axes of rotation. To take it one step farther, couldn't a body have an infinite number of axes of rotation?

At any given moment in time, all objects have only one axis of rotation. It may change over time due to either external forces or internal friction which favors rotation around the largest modulus.

George
2007-Oct-21, 08:16 PM
True, but I have fun just watching them have fun. :) Herpoholde and other terms here are new to me, but interesting. I prefer tadpoles, myself. Since I will probably end up croaking, let's stick with them.

Let's say we have a trained tadpole, Herpie. We teach Herpie to swim forward. Then, we teach him to roll to his right as he swims forward. Then we teach him, as he swims forward, to change course to the left, while maintaining his depth. He now can roll and yaw, which is movement about two separate axis.

But Herpie isn't just any trained tadpole, he decides to swim forward and do his two motions, roll and yaw, at the same time, obviously to impress us, his teachers. Interestingly, he can and by doing so he can be seen to do this by rotating in only one axis that would be located between the forward line of motion axis (for rolling motion) and the vertical axis (for yawing motion). This works the same as combing two vectors, which is what Publius originally was showing, though he appropriately included the moment of inertia term.

Without dusting-off my static book, moment of inertia is simply a way to express the resistance an object has to rotation. Take a 10 lb. bowling ball and spin it and note how hard it is to spin. Now take a 10 lb. bowling ball -- you can have mine since my wife has much more success with hers -- and cut it in half. Place each half on a long, lightweight pole. Now spin this 10 lb. weight [about an axis perpendicular to the connecting pole] and note how much more difficult it is to spin. There are handbooks that give this moment of inertia info for many different homogenous shapes. For odd shapes and varying densities, it takes a fair amount of effort to calculate the moment of inertia for that object.

I could be all wet, as it has been 30 years since I've messed with this stuff. But if the above is mature, than I can say I toad y'all. [I hope no one objects to the gross amount of fun I have around here. :)]

[BTW. I vaguelly recall that when a jet is out-of-control, the pilot will correct one of the three motions at a time, rather than try to determine what the rotation problem is around the net rotational axis. For instance, he, or she, might eliminate the roll motion first, then tackle pitch and yaw separately.]

publius
2007-Oct-21, 09:17 PM
Let's see if I can still show how the general inertia tensor comes about. From elementary physics, you learn about the scalar moment of inertia, and that's sort of a mr^2 summation/integration, but where 'r' is the perpendicular distance to the axis. But, it turns out that in general this notion of "moment of inertia" has to be a (rank-2, 3D) tensor.

How is angular momentum defined? Well, it's

L = r x p;

Torque is defined as

T = r x F = r x dp/dt {Note that T = dL/dt; the above would suggest two terms, a dr/dt and dp/dt, but work things out, with the help of a mechanics test, and you'll see how things shake out}

Note the (r x) form; r in front.

Now, what is r? Well, in the general case, that can be any convienent coordinate position vector (center of mass of a body helps).

Let's look at the L expression.

L = m*(r x v). Now what is v? Well one can think of the w vector as being defined by this relation:

v = w x r {Note this has to (x r), opposite of the (r x) form. This form is the most convienent way to say it. You can think of an inverse relation, and it does like that w = r x v/|r^2| of sorts, but that form just doesn't work out so nicely. There is some stuff you can do with quaternions that can make this "division by vectors" sort of make sense. }

Anyway, plug that in for L:

L = m*(r x w x r).

That's a triple cross product, which can be expanded by the BAC-CAB rule:

L = m *[ w(r dot r) - r(r dot w) ]

Now, time to play with the dyadic notion of tensors. At any rate, a defintion of the dyad (or outer product) of two vectors A and B can be this defintion:

(AB) dot C = A*(B dot C)

The term on the right is well defined for regular vectors. So the above expression can be written:

L = m *[ r^2 * w - (rr) dot w]

Matric algebra allows a way to get a component representation of this dyad thing. Consider the 3 component r vector. You can think of the dot product as a matrix multiplication of a 1 x 3 vector times a 3 x 1, which results in a 1 x 1 product, or a scalar.

IOW, a row vector times a column vector (and you've got to decide conventions, such as if a regular vector you're going to work with is going to normally be a row or column, then use the transpose to flip it -- that is the beginning of "index gymnastics" -- how you define these things is important ).

Now, the "outer product" of a vector is then the reverse of that, a
3 x 1 times a 1 x 3 which results in a 3 x 3 product. That is then a column vector times a row vector.

So the above expression is a type of curious commutator. It's the difference between the inner and outer products, or r^T*r - r*r^T where the T notation means "transpose". Making use of the identity matric I, 1s down the diagonal, and 0 otherwise, we can write our L thus:

L = m [ r^2 I - rr] dot w

The matric in brackets is our inertia tensor. You can see the diagonal elements are (x^2 + y^2 + z^2) - r_i^2, or

Now calling I the inertia tensor and not the identity matrix :lol:,

I11 = y^2 + z^2, I22 = x^2 + z^2, and I33 = x^2 + y^2. The cross terms are then various xy, yz, and zx terms.

The diagonal terms are sometimes called moments of inertia, and the cross terms "products of inertia".

For a sphere, all the cross terms are zero, and the diagonal terms are equal, and that whole mess above just collapses to a simple scalar -- you don't "notice" the tensor form.

Because of that tensor form, as I stressed before, L and w do not have to point in the same direction. We might ask for what values of w, what spin axes, does L and w point in the same direction. That is, I dot w is simply a constant times w. That question is then:

I dot w = k*w

And that is a classic eignevalue problem. The eigenvalues 'k' are then scalar moments of inertia, and the eigenvectors are the principle axes.

You can also look at kinetic energy, and relate to that to simple scalar notions of moment of inertia. The general kinetic energy expression is another cute thing:

E = 1/2 (w dot I dot w ), which is the familiar form of an "operator" sandwiched between two vectors that students of quantum theory get burned in their heads. :)

-Richard

publius
2007-Oct-21, 09:43 PM
I fear many of you may see the above as some obtuse mathematical arcana and didn't get anything much out it. Well, that's sort of the physicist's way of looking at it, and the final commutator, eigenvalue and "operator sandwich" form is just slick as the proverbial owl droppings. When you think like a physicist, that is.

To me, that form is just insightful and boils things down to the fundamentals. But if you don't think that way, it probably doesn't.

But any rate, the fundamentals I want to stress are that L and w do not have to point in the same direction. What is conserved is L, not w, and that's the root of precession and tumbling.

Because of the complex form of the I tensor, the equation of solid body rotation, dL/dt = T becomes very complex.

But they can be simplified in specific ways, and symmetries and constaints can help, and all sorts of interesting, but non-intuitive stuff can fall out of those equations.

One thing is the notion of "balance". Spin something about a non principle axis, but mechanically constrain it so that axis can't move. And that's just about thing where you stick an axle through and spin it and keep it fixed to terra firma.

If you're not along a principle axis, then you're going to have to exert a torque to keep that axis fixed. What you're doing is holding w fixed, and making L move around. That's generally called "wobbling". If something is out of balance, spinning it cause the bearings to have to take "crazy sideways" forces to exert that torque.

A complication with wobbling is a translational component. For example, spin a disc about a non-central axis, and you have to apply a translational force that rotates around. If you let it go, it would fly off translationally, and spin about a central axis, but no precess. And that is something different from this non-principle axis thing. With the non-principle axis thing, you have to exert a torque, but the net translational component of the forces is zero.

So there's sort of two "wobbles" there, and that can get complex.

But in the big picture, you need some symmetry and balance to make something rotate "nicely" about an axis.

-Richard

a1call
2007-Oct-21, 10:30 PM
This is how I would recap:

Take a bicycle wheel and turn it as fast as you can. Assuming the gyro's axis as x, try rotating it around the y axis. Assuming the wheel is turning fast enough to prevent you from giving it a complete rotation the resulting spin will be complicated by a wobble. This wobble may be circular which can be considered the 2nd axis of rotation or elliptical which will be whatever it is.
In case of a circular wobble/polhode which is extreme enough to have an axis perpendicular to the 1st axis you end up with a similar system as the java applet that I have posted.
From the link provided by Jerry I would conclude that the wobbling rotation is inherently unstable and because perfect symmetry is unachievable will shift towards the "shortest axis of symmetry", eventually. In case of the Earth the wobble is reportedly present now.

I find this topic very interesting in that the most "perfect" spheres manufactured by mankind was not perfect enough to maintain a constant wobble and would shift axis so dramatically in matter of hours.

I would also conclude that a a wobbling rotation would subject it's mass to a constantly varying acceleration vector which results in energy dissipation in form of elastic movements of solids.

publius
2007-Oct-21, 10:35 PM
Now, maybe we can shed some light on polhodes and herpolhodes. The former means simply "path of the pole". The latter means "serpentine path" (or so it is said), and some say Poinsot (of Poinsot's construction fame) actually misused the term.

Like I said above, because L and w don't point in the same direction necessarily, the torgue free, dL/dt = 0 motion of a rigid rotator can get very complex. But, Euler and Poinsot figured out some general properties of the motion that are very useful even though a full analytic solution is impossible.

Now, one can show that w obeys a couple of contraints. One is that is must lie on an ellipsoid related to I1, I2, and I3 in turn along the principle axes. That is the endpoint of the w vector must lie on that ellipsoid. It can move all around it, but it must lie on it. You can imagine an ellipsoid as being the most symmetric of bodies that can have 3 moments of inertia. So, in effect, all solid bodies have an equivalent ellipsoid in some sense.

The curve it traces on that ellipsoid is the polhode motion.

Now, L is of course constant, and the plane normal to L make a natural reference and that's called the invariable plane. And that is used for "orbital" and well as spin L. The solar system has such a plane (but it depends on whether you're considering just orbital L or total L -- tidal interactions can exhange the two. It is only the total L whose plane is truly invariable).

Now, consider the curve traced out by the projection of the w vector in that invariable plane. That is the herpolhode motion.

Now, various relations between those two can be divined, and that business about the polhode rolling on the herpolhode can be seen as "something" slip-free rolling along the herpolhode. If it doesn't slip, things are nice. If it does slip, things are messy.

publius
2007-Oct-22, 01:41 AM
All this is bringing back memories from school. You may recall I mentioned getting -1's a lot. Well, it was my mechanics professor who was bad to give out those -1's. Even ff you got the right answer, but got a little sloppy or did something not exactly kosher, he'd give you a -1. He was persnickety about things.

He was a wondeful man, and quite funny, and taught well. He was Indian, and had this sing-song voice. It was hard to understand him at first, because of that sing-song cadence, but once you got the knack of it, well, it was just fun to listen.

He prounounced "moment of inertia", as sort of "muhMENTuvinerSHA" and one other student misheard that as "momentum inertia".

We got into solid body rotation, not too deep but deep enough. He had to give himself some -1's there. I remember he botched something on the blackboard, one student caught it, and we all shouted "Minus one!" He put on some Broadway production in jest, throwing the chalk up the air and storming out of the room.

So, there you go. General rotation "spun" even him for a loop sometimes.

-Richard

George
2007-Oct-22, 03:49 PM
My Indian professor, from India, would say "wectors" for vectors. He was one of the best ME profs. I had, nevertheless.

Unfortunately, I am still unclear on all this. It seems to me, that a single net axis of rotation will not work. This may just be because I can't visualize a rolling and yawing motion in an object reduced to a simple rotating object about another unique axis. This, of course, is contrary to my last post that bought into the one axis idea. Or, could Herpie really be seen as having one rotation about one axis when he does both of his rolling and yawing demonstration?

Are y'all perhaps saying that the net final rotational motion will be around one axis (ie the one with the greatest I value), assuming no, or very little, external torque? [Meaning rotation about 2 or 3 axis, unresolvable in one, are only temporary.]

hhEb09'1
2007-Oct-22, 05:11 PM
Are y'all perhaps saying that the net final rotational motion will be around one axis (ie the one with the greatest I value), assuming no, or very little, external torque? [Meaning rotation about 2 or 3 axis, unresolvable in one, are only temporary.]I think what is being said that the rotation is resolvable but it may be moving.

In a1call's animation (http://www.bautforum.com/questions-answers/66030-can-object-space-have-many-axes-rotation.html#post1092674), the object rotates around the y-axis of the object, but about the x-axis of space. It's a kind of mix of two different concepts.

DyerWolf
2007-Oct-22, 06:35 PM
Of course, we're talking about Toutatis (http://www.solarviews.com/eng/toutatis.htm)

One consequence of this strange rotation is that Toutatis does not have a fixed north pole like the Earth. Instead, its north pole wanders along a curve on the asteroid about every 5.4 days. "The stars viewed from Toutatis wouldn't repeatedly follow circular paths, but would crisscross the sky, never following the same path twice," Hudson said. :)

For some reason reading this made me a bit seasick - as I imagined watching those stars wheel overhead.

"Weeeee! ugh-mmmph- RETCH," said the intrepid explorer.

publius
2007-Oct-22, 08:25 PM
Unfortunately, I am still unclear on all this. It seems to me, that a single net axis of rotation will not work.

George,

Visualizing general rotation does indeed get complex. That's sort of what I've been lamenting the whole time. But, at any point in time, the motion can be seen as a rotation about a single axis. That axis can just change with time.

Consider what is meant by a "rigid body". Imagine two particles joined by a rigid connection. We can see the constraint is the distance between the two (a scalar) must be constant, but yet you can see the thing can "rotate" in some fashion. Expressing that gets a bit complex.

But I think we can intuitively see that this rigid body constraint means that there is some coordinate system, the "body frame", where nothing moves. In the non-rigid case, there would not be such a frame. So in this body frame system, all parts of the body are stationary.

In general, this body frame will be both accelerating, both translational and rotational. Showing how this works, and figuring out the best way to do it is not trivial. That's what a good mechanics text is for. :)

But, it turns out the principle axes can define that body frame (the most natural ones to do so). So imagine the center of mass of the body with its little principle axes attached.

You can resolve things into two pieces: the motion of the center of mass, and then the motion of the body frame about that center of mass. Imagine the axes of the body frame attached to a point on the COM trajectory, being sort of transported.

The question is then, in the presence of forces, what is the trajectory of the COM, and what is the motion of those body frame basis vectors as it moves. Not a trivial problem at all.

It turns out that is possible to resolve the body frame vector motion as a single axis of rotation at any given time. IOW, wherever the 3 little orthogonal body vectors are, the instantaneous motion about the COM can be seen as single w vector riding with the center of mass.

So imagine a little 'w' vector riding along the COM trajectory, and moving all around as it does so. Imagine the COM curve in space; a little w vector is attached to that and moves all around (with a variable length) as it slides along the COM trajectory.

The thing to appreciate here, is *w is not fixed relative to the body frame*. Imagine you're riding along with that body, and are at a fixed location relative to the body. You see your axis of rotation, w, moving around your own self. This is important to appreciate.

The guy in inertial space watching that show sees both w and the body frame moving around. But you, riding along that non-inertial body frame see w moving as well.

What you see of the w motion is the "polhode" motion. What your own pole is doing relative to you. The herpolhode motion is roughly what the inertial guy sees of you w vector relative a fixed set of axes moving with your COM.

I think you can see why this makes my head hurt. :) But the motion is specified by knowing, R(t), the trajectory of the COM, and the w(t), the axis of rotation.

If we know those functions, we know the motion of the body frame relative to inertial space.

-Richard

publius
2007-Oct-22, 08:58 PM
Now that we know what polhode motion is, the earth itself has a little bit of that, and it's called the Chandler Wobble. Our axis of rotation moves around relative to the earth itself. While that motion is irregular, it has a rough 14 month period.

Euler and Poinsot figured out the treatment of solid bodies I was rambling about above, and, making certain estimates, came up a polhode period of about 9-10 months for the earth. When Chandler first announced the 14 month period, no one much believed him, since his result was outside Poinsot's prediction. :)

But Chandler was right as was demonstrated soon thereafter. The problem was the earth is not truly rigid, and Poinsot's derivation didn't apply. If the body can move (and with internal dissipation as it moves), the motion gets even more complex, and it was shown the ~14 month was consistent with that.

Note the Chandler wobble is this "wobbling" effect due to L and w not pointing in the same direction. The larger ~26KY precession of the equinoxes is another thing, due to "real torque" on the earth caused by complex tidal interactions.

Over a short period, you can figure the earth's L is constant, and the Chandler wobble is the variation of w with constant L. The precession of the equinoxes is due to changing L over much longer time periods.

-Richard

publius
2007-Oct-22, 10:53 PM
There's actually more to the Chandler wobble than this. The Euler-Poinsot treatment (for want of a better term) assumes perfect rigidity and therefore conservation of rotational kinetic energy in the absence of external forces and torques.

But bodies aren't completely rigid. So the notion of a body frame as I used it above is complicated by this. But still you can imagine a "semi-rigid" body, perhaps with elastic forces where the notion of body can still be roughly defined, but there is some differential movement, some distortion in that frame.

You can see that centrifugal forces (and yes, I use that term refering to the body frame, and I'm not going to bother with "romper room" pedantic sillyness about that -- when one has reached the level of Euler and Poinsot, all that mess needs to be behind you! If it's not, you're not prepared anyway :lol: ).

Anyway, you can see that centrifugal forces will tend to stretch the body radially from the w axis. Well, what happens when w is moving relative to the body? That stretching changes. When there is frictional dissipation involved with that stretching motion, you're going to lose kinetic energy, and that loss is going to have to effect the motion.

What happens is this. Remember Ken reminding us that given
I1 < I2 < I3, rotation about the min and the max is stable, but that above the middle, I2 is not. When the body is perfectly rigid, that holds exactly.

However, where there is frictional deformation forces that loose energy, that gets modified. Perturbations about I1, the lowest moment axis (this would be the "longest" axis of a general body), become unstable over the long haul.

As it looses energy due to the flexing, the precession gets worse and worse until it gets large enough to kick in the "tumbling" mode about I2, and then it goes crazy. It will eventually stablize about I3, the highest moment axis, or the shortest body axis.

Rotation there gives you the least energy per L. IOW, the scalar E/L ratio is a minimum for a given L.

So, allow a body to dissipate like that and it wants to fall into the minimum about its I3 axis. Once it does that, w stablizes, and dissipation ends and there she sits. Rotation about I3 is then completely stable in the general case.

The earth's pole is about the I3 axis, because that's what the deformation does to a sphere anyway.

So, actually, because of this non-rigid deformation, the Chandler Wobble should damp out. But it does not. Something keeps it going. And that something has to be active.

You can imagine being on a rigid rotating body, and then taking a stick of dynamite or something and moving pieces around. Do that and you're going to make it wobble. However, it will stabilize around whatever the new configuration is eventually.

To keep it going, you've got to have something driving it, and keeping it from reaching that equilibrium.

And I stress this only applies to non-rigid bodies where kinetic energy can vary. The "Chandler Wobble" would not dissipate for a perfectly rigid lumpy sphere.

-Richard

George
2007-Oct-22, 10:54 PM
It turns out that is possible to resolve the body frame vector motion as a single axis of rotation at any given time. IOW, wherever the 3 little orthogonal body vectors are, the instantaneous motion about the COM can be seen as single w vector riding with the center of mass. Ah, that explains your earlier work. You are saying that at any given instant of time the body is rotating about an axis that is on the move relative to us, the outside observers.

So imagine a little 'w' vector riding along the COM trajectory, and moving all around as it does so. Imagine the COM curve in space; a little w vector is attached to that and moves all around (with a variable length) as it slides along the COM trajectory. So, for instance, if I were to "take the com", and then looked along the translational line of motion, plus used a laser beam to shine along the instaneous net rotational axis, the beam would be going all over the place, assuming motions appropriate for addressing the OP. But, if done properly, only one laser beam would be required. If so, that is the key point I think y'all are making.

The thing to appreciate here, is *w is not fixed relative to the body frame*. Imagine you're riding along with that body, and are at a fixed location relative to the body. You see your axis of rotation, w, moving around your own self. This is important to appreciate.

The guy in inertial space watching that show sees both w and the body frame moving around. But you, riding along that non-inertial body frame see w moving as well. I suppose I should have read your whole post first, and that was my intent, but... :)

What you see of the w motion is the "polhode" motion. What your own pole is doing relative to you. The herpolhode motion is roughly what the inertial guy sees of you w vector relative a fixed set of axes moving with your COM. I think above I was considering myself the inertial kind of guy, though not completely inert, but close. If I understand this, though I doubt it, both guys would agree on where the laser points siderally, but the dw/dt would be different for each. Is this what you are saying?

George
2007-Oct-22, 11:21 PM
So, allow a body to dissipate like that and it wants to fall into the minimum about its I3 axis. Once it does that, w stablizes, and dissipation ends and there she sits. Rotation about I3 is then completely stable in the general case. Yes, excluding external factors, no doubt. Jerry's link to a video of astronaut Owen Garriott demonstrating this in space was excellent.

You can imagine being on a rigid rotating body, and then taking a stick of dynamite or something and moving pieces around. Do that and you're going to make it wobble. However, it will stabilize around whatever the new configuration is eventually. Yikes, I assume these are unlit sticks, or are we on a boat engaged in serious fishing. ;)

The Earth's fluid interior and atmospheric mass variations should keep our little dw/dt on the move, no doubt. [Our air mass movements cause rotation rates to vary by about 100 ms or so, IIRC. They are pretty significant.] Then there is the Sun and Moon to consider, too.

And I stress this only applies to non-rigid bodies where kinetic energy can vary. The "Chandler Wobble" would not dissipate for a perfectly rigid lumpy sphere. For asteroids, however, you would have varying grav fields along its translation, as well as, thermal stresses as it rotates relative to the Sun. Then there is Solar wind and microimpacts to consider, too, I suppose.

publius
2007-Oct-22, 11:23 PM
I think above I was considering myself the inertial kind of guy, though not completely inert, but close. If I understand this, though I doubt it, both guys would agree on where the laser points siderally, but the dw/dt would be different for each. Is this what you are saying?

That's it. Imagine we have three laser beams for the body axes, say red, green, and blue, and another for the w axis, say yellow.

An inertial observer will see all four of those (in the general case) moving all around, and will see the yellow w beam moving relative to the RGB beams.

In the body frame, we see RGB as fixed by defintion, but see w moving. If we imagine the inertial observer has him 3 laser beams, XYZ colors, the body frame observer sees those moving, and w moving relative to them.

But, if we align yet another laser beam with the L vector, the inertial observer sees that as constant, not moving at all (without external torques). The body frame observer will see that moving, but it will remain constant relative to the XYZ axes. So just align say Z with L, and there you go.

Polhode motion is basically what w does relative to the RGB axes. Herpolhode is what w does relative to L/XYZ.

Add what the inertial observer sees RGB doing, and you've got that polhode rolling across the herpolhode business. :)

-Richard

-Richard

George
2007-Oct-23, 12:45 AM
That's it. Imagine we have three laser beams for the body axes, say red, green, and blue, and another for the w axis, say yellow.

An inertial observer will see all four of those (in the general case) moving all around, and will see the yellow w beam moving relative to the RGB beams.

In the body frame, we see RGB as fixed by defintion, but see w moving. If we imagine the inertial observer has him 3 laser beams, XYZ colors, the body frame observer sees those moving, and w moving relative to them. Maybe I am close with all this. Since it's nearly haloween, if we add an orange laser to be on the nose of a flying orange-dressed witch's broom stick, with her on it, and her broom always points along the net rotational axis, the motioin of our object will always appear to rotate around her and her orange beam.

Placing a purple witch at the COM, also, but having her steady with your RGB lasers will cause her to see the orange witch move around the COM and point in varying directions over time. The object, however, will not appear to be rotating about any fixed axis for the purple witch.

Our orange witch will see the RGB lasers moving all around, though orthogonal to each other, of course. [I don't quite see where the w vector would be located, and am reluctant to ask.]

Can I go lie down now? :shifty: I think we've exceeded the number of actual laser colors. :) Oh, there's more, uh oh...

But, if we align yet another laser beam with the L vector, the inertial observer sees that as constant, not moving at all (without external torques). The body frame observer will see that moving, but it will remain constant relative to the XYZ axes. So just align say Z with L, and there you go. Is the L-vector moving along with our our net single axis (and not the w vector); where angular momentum is determined at that point in time about this axis, though w is varying? Is the L-vector found always steming from the orange laser spot, even though its length is constant, I think?

Polhode motion is basically what w does relative to the RGB axes. Herpolhode is what w does relative to L/XYZ. So the difference is simply knowing which witch is which. ;) Thanks a lot. :)

publius
2007-Oct-23, 01:25 AM
I think that's it, although I'm getting confused on which witch is which, as you noted.

Where is w located and L for that matter. Well, those are axial vectors (psuedovectors, actually -- and don't ask about the meaning of that, because I'd botch it. It has to do with coordinate systems, transformations, and the "handedness" of them, and other stuff). It doesn't matter where they are located, just what their direction and length is.

Take a position vector. That is referenced to an origin. However, "where is the velocity vector"? Well, you can sort of think of the end of the velocity vector being attached to the location of the particle, but that doesn't matter. L and w are like that, but even more so.

At any rate, just think of them as sort of riding along with the COM, their little tails sitting right on that point as it moves along. It's not really important, just that they play a mathematical role.

L is the angular momentum vector. It is just the sum of all the r x p of the little pieces that make up the rigid body (and the r there can be any position vector, at any time). That is conserved, and is constant with no external torque being applied.

So imagine a rotating body with no external forces or torques. The COM just moves in a staight line inertially. L is constant as well. So we see a little L vector of constant length pointing in the same direction just sliding along a straight line path.

But, because of that complex tensor relation, w, the actual axis of rotation does not have to point in the same direction as L. And so things go wobbly in the maddeningly complex way I've been rambling about above.

w, the instantaneous axis of rotation, wobbles all around as it slides along, and the body axes themselves (RGB) wobble all around as well.

-Richard

publius
2007-Oct-23, 01:48 AM
Another thing I want to stress is all of the above can be rigorously defined mathematically. My rambling about it ain't so rigorous, but the math itself is. The motion of a rigid body can be completely described as the motion of the COM plus a rotation. That axis of rotation may vary, but any instant, it has a single w(t). It's just like an accelerating point particle has an instantaneous velocity which can vary with time.

v is just dr/dt, v and r being vectors.

Well, w can be seen as a type of derivative itself, and rigorously defined as such. Only problem is that defintion get into more high powered 'rithmetic than we're used to, involving tensors (to see things in their full general glory) and coordinate transforms and all that good stuff. I'd botch that big time trying to write it out, heck, I'd have to pull textbooks out and go over it all again. And probably still botch it.

But imagine this. We have a set of three little orthogonal basis vectors, a triad as it is sometimes called (in relativity, you have a tetrad, which includes the local proper time "direction" as well -- the tetrad "field" gives us the local ruler and clock of observers moving through space-time).

We put the origina of out little triad at the COM and let it slide it along. Now imagine letting that little triad "rotate" around however it may. Only constraint is they remain always unit and always orthogonal. Orthomormal, IOW. But we just do whatever we wish with them otherwise, spin it this way, stop, then that way, etc, etc.

The definition of our 'w' vector is in the time derivatives of that triad. You can sort of see you've got three little vectors, and three time derivatives of each, for nine components. That suggests a tensor. But the constraints collapse that down to only three independent components. And those make up the 'w' vector. And it has the direct interpretation as an instantaneous axis of rotation. Its magnitude is the instantaneous angular speed, and it is directed along the physical axis.

-Richard

George
2007-Oct-23, 02:26 AM
At any rate, just think of them as sort of riding along with the COM, their little tails sitting right on that point as it moves along. It's not really important, just that they play a mathematical role. That helps a little, as I was trying to draw a vector from the point where the axis extended to an imaginary exterior globe. That shouldn't be wrong, just a little harder to deal with.

L is the angular momentum vector. It is just the sum of all the r x p of the little pieces that make up the rigid body (and the r there can be any position vector, at any time). That is conserved, and is constant with no external torque being applied. Yes. For our object, our single axis of rotation, or net axis (the orange one), for any given instant should be the reference axis for r. Is this right?

So imagine a rotating body with no external forces or torques. The COM just moves in a staight line inertially. L is constant as well. So we see a little L vector of constant length pointing in the same direction just sliding along a straight line path. Wouldn't this vector point in the direction of the rotational axis and not the translational direction? That is how I've been seeing it. The angular momentum would be constant, but not about a fixed axis relative to our RGB lasers, since angular velocity is a vector. If so, then another vector could point in the direction to where the axis is headed next. It is this rascal that is getting me, perhaps, since it will be tangential to your polholde and will describe it.

But, because of that complex tensor relation, w, the actual axis of rotation does not have to point in the same direction as L. And so things go wobbly in the maddeningly complex way I've been rambling about above. I think I am far too rusty with vectors to get this. L = Iw, so why is their direction different?

w, the instantaneous axis of rotation, wobbles all around as it slides along, and the body axes themselves (RGB) wobble all around as well. So w is along our orange axis, not L? :doh:

George
2007-Oct-23, 02:30 AM
Now imagine letting that little triad "rotate" around however it may. Only constraint is they remain always unit and always orthogonal. Orthomormal, IOW. But we just do whatever we wish with them otherwise, spin it this way, stop, then that way, etc, etc.

The definition of our 'w' vector is in the time derivatives of that triad. You can sort of see you've got three little vectors, and three time derivatives of each, for nine components. That suggests a tensor. But the constraints collapse that down to only three independent components. And those make up the 'w' vector. And it has the direct interpretation as an instantaneous axis of rotation. Its magnitude is the instantaneous angular speed, and it is directed along the physical axis. That helps, but I see three component derivative vectors, dw/dt, only. I suppose we could add acceleration vectors, too, just for fun. ;)

publius
2007-Oct-23, 02:46 AM
George,

Ah. Okay, L is whatever it is. Whatever the angular momentum happens to be. We "spin" our general object however we wish. But no matter what we do, we get some L vector. It can be of any length and point in any direction. But it is constant (in the external torque free case). So, imagine this arbitrary L riding along with the COM. It can point anywhere, but it doesnt' change.

Why do L and w not have to point in the same direction? Well, read over a couple of my rambling posts above, the one where I showed the triple cross, L = m *r x w x r form and the expansion.

And it's not really a vector thing. It's a *tensor* thing. So you're not rusty on vectors, you've just not pulled the tensors out of the shrink wrap yet. :lol:

THe expression L = I*w is actually a tensor expression. I is a tensor, not a scalar (although it collpases to a scalar equivalent in the simple cases you learn in Physics 101).

Think of this I tensor as 2D matrix, and think of L and w and 1D column vectors. I*w is then a matrix mulitplication of a 3x3 times a 3 x 1 which yields a 3 x 1, which is another column vector. But that matrix can do things to that w vector that a simple scalar cannot. It can make it change direction.

Take the simple case where I is just I1, I2, and I3 down the diagnonal (and that's the principle axis, body frame representation of the I tensor, we define the body frame so this tensor takes the simplest form). You can see that if those 3 are not all equal, it will change the direction of the product.

And so when L is conserved, w itself does not have to remain constant. That's what makes it so complicated. You can show that
dw/dt = w x L for the rigid, zero torque case. IOW, if L and w are not parallel to begin with, w *must* move in order to conserve L.

Tensors do stuff that vectors alone can't do. Now you can see why the statement "gravity must be a tensor field" as opposed to a vector field means that gravity gets a wee bit complicted. :lol: And you can probably see why they don't teach them in Physics 101 -- they leave stuff out like L and w not pointing in the same direction and making things very complicated.

-Richard

publius
2007-Oct-23, 02:50 AM
That helps, but I see three component derivative vectors, dw/dt, only. I suppose we could add acceleration vectors, too, just for fun. ;)

The components of the derivatives of the triad vectors themselves. Three vectors, three components each, three time derivatives. The constraints on the triad, allow us only 3 independent components in the end, and those make up w. This is used to show the part about a single, albeit time varying axis can specify the motion about the COM.

dw/dt is of course the derivative of that. It involves the second time derivatives of our triad.

-Richard

George
2007-Oct-23, 04:38 AM
Okay, L is whatever it is. Whatever the angular momentum happens to be. We "spin" our general object however we wish. But no matter what we do, we get some L vector. It can be of any length and point in any direction. But it is constant (in the external torque free case). So, imagine this arbitrary L riding along with the COM. It can point anywhere, but it doesnt' change.Hmmmm. After looking at the angular momentum page in Wiki (illustrations only, of course, no text), the L vector is along the axis and follows a right hand rule convention, apparently. But since our object is dancing all over the place, I expect L to move along our orange laser axis (net axis), though with a constant magnitude. Is this what you meant?

Why do L and w not have to point in the same direction? Well, read over a couple of my rambling posts above, the one where I showed the triple cross, L = m *r x w x r form and the expansion. I've gotten soft with general explanations, apparently. That looks like a -w result, but I would have to review my vectors of old.

And it's not really a vector thing. It's a *tensor* thing. So you're not rusty on vectors, you've just not pulled the tensors out of the shrink wrap yet. :lol: Yes. The very sound of the word expresses stress and I consider it a fair and deliberate warning, so I've eschewed it. Evidence has always supported this initial assumption, too.

THe expression L = I*w is actually a tensor expression. I is a tensor, not a scalar (although it collpases to a scalar equivalent in the simple cases you learn in Physics 101). And the evidence continues to mount up. :) For a rigid body, in what way does I vary along any given axis? [I already regret this question.] Or, is it because we are varying the net axis which results in a variance in I, along with a matching variance in w?

George
2007-Oct-23, 04:45 AM
The components of the derivatives of the triad vectors themselves. Three vectors, three components each, three time derivatives. I picture the w vector to have just three componets, namely wx, wy, wz. Are you saying, for one axis, the three would be wx, dwx/dt, d2wx/dt2?

publius
2007-Oct-23, 05:18 AM
I picture the w vector to have just three componets, namely wx, wy, wz. Are you saying, for one axis, the three would be wx, dwx/dt, d2wx/dt2?

George,

I'm afraid I may have just confused you with this. The notion of 'w' can be sort of vague, and what I was driving was there is a very rigorous way to define it. But that defintion gets mathematically involved, which is why it is usually introduced in a more "hand-wavy" sort of fashion.

But let's try this. Take the your thumb, forefinger and middle birdy finger and make a triad, 3 orthogonal unit vectors, i, j, and k. Now, keep those mutually orthogonal, but twist and rotate your hand all around in any fashion you like.

That is an arbitrarily rotating triad. Now, you've got three basis vectors. i, j, k. They are not constant with time, since you're moving them all around.

We have di/dt, dj/dt, and dk/dt. Each of those has three components. So we have a total of 9 components. Say i is your thumb. di/dt is how your thumb is changing. Say j is your forefinger. dj/dt is how it is changing, and dk/dt is how your birdy finger is changing.

But, we have constraints. i, j and k must remain of constant length, and they must remain orthogonal. Those contrainsts collapse those 9 components down to only 3 indepedent components.

So, we have only three degrees of freedom to rotate our triad. From those three come the components of our w vector. We can then define our w vector in terms of the triad derivatives, which obey those constraints.

Note I'm not telling you how to get w from those triad derivatives. That is complex expression, and I'd have to look it up. :) But I was trying to get the concept across that even though we have three basis vectors which can rotate around, we have only 3 independent ways we can do that. And that is just an instantaneous axis of rotation. That axis can vary with time, but at any point in time, it's just a single vector.

-Richard

publius
2007-Oct-23, 06:40 AM
Hmmmm. After looking at the angular momentum page in Wiki (illustrations only, of course, no text), the L vector is along the axis and follows a right hand rule convention, apparently...

George,

That page doesn't go into this mess here, just the elementary basics. Wiki has stuff on this mess, it's scattered about and written by various authors with slightly different styles. It would make for a very bad textbook. Here's some that get into what I'm trying to ramble about:

http://en.wikipedia.org/wiki/Euler%27s_equations

http://en.wikipedia.org/wiki/Poinsot%27s_construction

http://en.wikipedia.org/wiki/Moment_of_inertia_tensor

None of these are really all that great. Nothing *wrong* that I can see, just not anything too good to learn from.

A good intermediate mechanics text is where to start. A complete treatment is the province of advanced (very) mechanics texts.

-Richard

George
2007-Oct-23, 02:19 PM
...That is an arbitrarily rotating triad. Now, you've got three basis vectors. i, j, k. They are not constant with time, since you're moving them all around.

We have di/dt, dj/dt, and dk/dt. Each of those has three components. And these three for each are used to define the direction and magnitude of motion for each, right? Since all three axis are fixed relative to each other, then any of the three components are necessary to define all three axis.

I'm still trying to jump across the math to get to the physical picture of our objects rotational motioin. [Now it is more of a pole vault, though my pole hold is weak. ;) (sorry)] See if this sounds right... The OP object, with its complex motion, will have a single axis of rotation that is constantly on the move and not fixed to any spacial coordinate system. The motion of this axis and the rotation rates of the object is defined as you have shown.

publius
2007-Oct-23, 07:53 PM
And these three for each are used to define the direction and magnitude of motion for each, right? Since all three axis are fixed relative to each other, then any of the three components are necessary to define all three axis.

Those three define the body frame. We can resolve 'w' according to them, or we can resolve from the unchanging triad of an interial observer watching it.

It turns out that the equations of motion become easier if you do things from the body frame, the one that is moving willy-nilly. And that's because of the complex form of the inertia tensor. The components will be constant only in the body frame.

Euler's equations are for 'w' in the body frame. That is, what the observer fixed on the body sees of his rotational axis in terms of his own (rotating) triad. That is the polhode motion.

Now, knowing that, you can transform back to the intertial frame and find out what the body frame itself does. That would be the solution you're after, how the body moves relative to inertial space. The w vector will have a resolution in those inertial coordinates as well, which will be different from the polhode solution. That (well it is related) is the herpolhode motion.

In the abscence of external torques or forces, the inertial observers sees the angular momentum vector, L, as absolutely constant. He sees the body wobbling and bobbing all around, and the instantaneous axis of rotation moving all around as well. But because of the complicated tensor form of L:

L = (InertiaTensor) dot w ;

w itself can vary even though L does not. Think of it this way. As the body moves relative to inertial space(the body triad flips all around as it does), the components of the inertia tensor change around, w then *is required to change* itself in order to keep L constant.

In the case of w being along a principle axis exactly, L and w then point in the same direction, and I becomes an effective constant. Then w will be constant in order to conserve L.

But perturb it a bit about the intermediate axis, and it goes wild. Perturb it about the other two, and it just precesses a bit.

But in the big picture, nothing is going on but conservation of L. It's just that, with the complexities of the inertia tensor, that conservation of L requires some fancy footwork by w and the body triad.

-Richard

publius
2007-Oct-23, 08:02 PM
Oh, and another thing that may cause confusion. I know just enough about tensors to be dangerous, and would drive off in the ditch on my own quickly. I need the high priests to hold my hand, still. But, I throw some expression out without realizing the meaning may be unknown. For that, I need my butt kicked, I'm certain.

For example, L = I dot w.

Those familiar with vectors will see that as a scalar. A dot product returns a scalar. Well, that's true with both are vectors. Here, "the dot product" of a tensor with a vector returns another vector. You can see that in components as a matrix multiplication, a 3 x 3 matrix front multiplying a
3 x 1 column vector. The result is a 3 x 1 column vector. L in this case.

And there are similiar cute things with vector calculus like operations. You remember a surface integral of a vector over a closed surface is equal to the volume integral of the divergence. Stokes Theorem.

Well, you can think of integrating a tensor against that surface, giving a vector. And Stokes Theorem holds there as well (just a more specific expression of much more general principle). The "divergence" of a tensor is a vector. And the dot product of a tensor with the surface normal is a vector. Same thing holds, just increasing the rank.

-Richard

publius
2007-Oct-23, 08:22 PM
Let me try to summarize all this. To specify the motion of a rigid body, we need to know two vector functions of time. The first is the motion of the COM and the second is the w vector. Call those R(t) and w(t).

Forgetting about the COM (we're moving with it) all we need to know to deterimine all the ways that body frame, that triad, can wobble around is the
w(t) vector.

That is three independent functions of time. Know that, and we know exactly how the body is going to move.

Those three functions can be anything in the general case, but at any instant in time, we have some wx, wy, and wz, and at that instant, we can say the body is rotating about an axis defined by a vector of those three components. That vector may vary with time, but at any instant, it is what it is.

If it turned out we had four or more independent components, then we couldn't say there was a single axis of rotation. But in 3D space, we have only three components, and we can see it as a single axis at all times. That axis just moves over time.

-Richard

publius
2007-Oct-24, 02:13 AM
On a lighter note about all this, while it does make my head hurt, I've always been fascinated by rotation and conservation of angular momentum. Try to move the axis of a spinning wheel (make it point in a different direction), and you get all sorts of weird feeling torques. If you did that in free space, you would get yourself rotating around such that the total L was conserved.

And to that end, something I would love to do -- well watch -- is someone trying to use a big right angle grinder floating in orbit. :) The motor is along one axis that turns the wheel through a right angle gearbox. Starting that up in free space would be interesting. Dangerous, actually. But it would be something to watch. And you'd better use a cordless version, cause I imagine a drop (float here) cord would tend to get tangled fairly quickly.

-Richard

publius
2007-Oct-24, 02:27 AM
And now I imagine a little simple mechanics homework problem for us, since we all have this downpat, now.

You find yourself tumbling "out of control" in free space, perhaps due to trying to use a right angle grinder and letting the thing go. Now, we can change our inertia tensor, simply by moving our limbs out and in, curling ourselves up in a ball, etc. Note this is another thing rotational systems can do that linear systems can't. You can't change your mass without throwing some away. Because of the simple relation between p and v, you can't change v without changing p, and you can't change m. So unless you throw something away, you can't change your velocity.

But, you can change your inertia tensor in various ways, and you can make your w do all sorts of things while L remains constant.

So you find yourself tumbling out of control. What you can do to try to stabilize your motion? Note you could easily make it much worse by flailing around.

-Richard

publius
2007-Oct-24, 03:50 AM
For obvious reasons, NASA has been very interested in this tumbling man in free space problem. :) This is their current system:

http://en.wikipedia.org/wiki/Simplified_Aid_for_EVA_Rescue

"SAFER". This is a system that can "detumble" and stabilize a hapless astronaut. They decided the system had to be automatic and local, worn by the astronaut.

Imagine trying to detumble yourself manually using some thruster system. That would be hard enough for a simple rotation, but imagine the tumbling, wild willy-nilly 'w' we've been discussing. You'd just loose all orientation.

So you use an automatic system that can detect its rotation and stop it. Rotation (proper at least, remember good ol GR) is something that can be determined locally. Your brain just isn't good at it. But the proper sensors can, and then use that to determine how to fire thrusters to stop it.

-Richard

matthewota
2007-Oct-24, 04:55 AM
The Saturnian moon Hyperion tumbles, due to gravitational interaction with Titan. It is an interesting satellite to study, because it is the largest non-spherical satellite in the Solar System.

Matthew Ota

George
2007-Oct-24, 01:59 PM
Yes. It is all interesting to me. I am impressed especially with the fact an object must, with energy loss, rotate around the shortest axis, which is the longest moment of inertia.

What would happen to the rotation of a large object that was clobbered by another large object if both were orbiting in the same plane? If the impact was along the equatorial axis that had been perpendicular to the ecliptic, would the loss of mass along the equator cause the new moment of inertia to be along the original axis, roughly? If this object must eventually rotate around an axis that is about 90 deg. from its original axis then it would appear much like Uranus, which is probably how the impact theory for it is supported, right?

John Mendenhall
2007-Oct-24, 02:30 PM
Imagine trying to detumble yourself manually using some thruster system. That would be hard enough for a simple rotation, but imagine the tumbling, wild willy-nilly 'w' we've been discussing. You'd just loose all orientation.

-Richard

Richard, your posts here are great. Thank you for your time. A wicked (Wikied?) little tensor problem in ordinary Newtonian space! Who would have guessed?

George
2007-Oct-24, 09:51 PM
I wonder if it is just a simple mercury switch on each axis kind-of-thing? That would be too simple, I suppose.

publius
2007-Oct-24, 11:56 PM
Richard, your posts here are great. Thank you for your time. A wicked (Wikied?) little tensor problem in ordinary Newtonian space! Who would have guessed?

Thanks -- that made my day. While this certainly isn't trivial to say the least, it's still nothing compared to how complex mechanics problem can become, even if they don't require tensors.

The Lagrangian formulation allows you to tackle problems that would be intractable using F = ma. To give a taste, I'll mention one problem I had to solve at the last of my mechanics courses, and I actually think "solving" was just finding the equations of motion. You didn't worry about solving those. Indeed, the biggest hurdle can be finding the equations of motion. It may be some non-linear humdinger, but now that's just a matter of cranking it numerically.

Anyway, imagine a weight on turntable that can move along a track radially. It cannot move tangentially, just slide radially. Now attack a string to that and run it to the center and thread it down through a hole in that center. Attach a second weight on the end of that and allow it to oscillate as a pendulum in a plane.

Initial conditons: Turntable is spinning at w. First weight is at some
r0, and the second weight is at some angle with the vertical. What is the motion? There were two versions. One was the turntable had a motor that maintained consant speed. Second was the turntable was just a flywheel. As the weights moved, the moment of inertia would change. And also, kinetic energy of the flywheel could be extracted to raise the second weight against gravity. The pendulum motion complicated that. Just a little bit. :)

It took two semesters of mechanics learnin' to be able to tackle that problem.

F = ma, while it holds, just becomes intractable. Main thing is constaints, and odd configurations. The Lagrangian formalism tackes it from the "Least Action" perspective, and is particularly well suited for arbitrary, odd constraints. Ie these two parts are connected in this way, but can otherwise move as they will. That means the force maintaining that constraint is simply whatever it has to be to prevent it from breaking it.
ma = Whatever_it_has_be is problematic.

The other beauty is coordinates. Rather than imposing some external coordinate system on the problem, you let the problem itself determine the coordiantes. In the above turntable pendulum, we have just two free variables if the turntable speed is held constant, r(t) of the first weight, and the angle of the second weight. If the turntable is free wheeling, you have a third, its angular speed.

-Richard

publius
2007-Oct-25, 12:20 AM
I wonder if it is just a simple mercury switch on each axis kind-of-thing? That would be too simple, I suppose.

Yes, it would. :) We need to know our w vector, which way it points and what its magnitude is, so we can apply the right torque to kill it.

Can you do that with accelerometers? Yes, you can in principle. How accurate they would have to be and all that good stuff, I don't have a clue.

I think I'll leave this is an exercise ('cause I'm lazy). But some guiding principles. First, we won't worry about possible variation in w. We are riding on something with a fixed axis of rotation that is otherwise unknown. We are at the center of mass with three little accelerometers fixed on the end of a triad. Those accelerometers measure acceleration parallel to their axis.

Now, centrifugal force in that rotating frame is given by this expression:

F = m*(w x w x r). That's another triple vector product (and note the cross product is NOT associative. A x B x C associates as A x (B x C) )

Note the vector form of that. If we pull the magnitudes out, we get
w^2r, which looks familiar, but we've then got that mess with three unit vectors. If w and r are perpendicular, as it would be on a rotating flat disc, the scalar versions of that work.

This is another example of how the simple scalar forms you learn in Physics 101 are actually more complex in the general case. :)

Now, what are the components of acceleration that each accelerometer feels? Can they determine w completely?

Now, suppose you were accelerating translationally as well as rotating. We've got three more possible components to worry about there. So I think it's apparent that three measurements won't be able to determine 6 components there.

So, unless there is some other constraint, if our COM can accelerate as well, we're going to need to take 6 measurements.

Think about all that. How would you do those six. Where you would put them?

-Richard

publius
2007-Oct-25, 06:44 AM
Yes. It is all interesting to me. I am impressed especially with the fact an object must, with energy loss, rotate around the shortest axis, which is the longest moment of inertia.

What would happen to the rotation of a large object that was clobbered by another large object if both were orbiting in the same plane? If the impact was along the equatorial axis that had been perpendicular to the ecliptic, would the loss of mass along the equator cause the new moment of inertia to be along the original axis, roughly? If this object must eventually rotate around an axis that is about 90 deg. from its original axis then it would appear much like Uranus, which is probably how the impact theory for it is supported, right?

George,

Yes that sounds about right in generally, but there are always possible complications. :) Simulations of planets colliding in ways that change the spin axis have been done, and those would be the ones to tell us the details. There might be some surprises there, too.

-Richard

ASEI
2007-Oct-25, 10:54 AM
If you want to express the inertia tensor in multiple frames of reference, you need the time varying rotation matrix between those frames of reference.

I_frameA = inv(R(t))*I_frameB*R(t) where R is some series of rotations between them.

Just my two cents. It occurs to me that if you're solving for the time dependence of the angles involved in R, R would be R(at least 3 angles). I'd have to dig out my spacecraft attitude dynamics book to remember it all.

publius
2007-Oct-25, 01:10 PM
You know, there is a way to do rotations with quaternions that is computationally superior to the rotation matrix. And I'll bet you might do the intertia tensor with them as well in some way, but I've never seen it.

The problems with the rotation matrix are related to a physical problem with gyroscopes known as the infamous "gimbal lock". The best way to see gimbal lock is to imagine you are tracking the motion of an airplane in the sky with a transit, which has two gimbals that give you the two angles of a spherical coordinate system. Suppose the airplane is coming straight at you.

Azimuth is constant, and elevation is increasing. You get pointed vertical, and suddenly the airplane makes a hard right angle turn. Your gimbals are parallel at that point and you can't make the turn. You need to quickly rotate the azimuth 90 degrees to allow you to move the sight back down.

The 3-angle way of resolving rotations can "mess up" computationally, terms blow up or become otherwise poorly conditioned, due to that same effect. The trouble is it isn't unique. The physical way to solve the gimbal lock problem is to add another gimbal, and drive it around so you always have an axis to move.

And there is an equivalent thing to do computationally. However, quaternions do not suffer from that at all. I'm not sure of the details beyond the above, but you can find them if interested.

-Richard

George
2007-Oct-25, 02:30 PM
Regarding the viability of using a simple mercury switch on each axis for sensing...

Yes, it would. :) We need to know our w vector, which way it points and what its magnitude is, so we can apply the right torque to kill it.

Can you do that with accelerometers? Yes, you can in principle. How accurate they would have to be and all that good stuff, I don't have a clue.
Accelerometers would be superior, no doubt. Solenoid control from simple mercury switches might work in this EVA application since the counter thrusts should be significant to the amount of minor spin for our astronaut, though I am not speaking with much experience with this design. The delay time, however, from the switch to the thruster would be the big issue, I suspect. A wild and wooly motion of a tumbling body could be worsened by such a design due to too much thruster dealy; reinforcing thrust might occur.

Automatic controls for several types of construction machines (eg curb machines) use either servos or solenoids, depending on mfg. In both cases, however, they use potentiometers to counteract the changes much more smoothly.

F = m*(w x w x r). That's another triple vector product (and note the cross product is NOT associative. A x B x C associates as A x (B x C) ) Ah, I had forgoten that.

Note the vector form of that. If we pull the magnitudes out, we get w^2r, which looks familiar... Yep, I new the acceleration vector would be critical.

Now, what are the components of acceleration that each accelerometer feels? Can they determine w completely? Interesting question. The question engineers will want to know is if they can respond to w accurately enough for the range of likely conditions. For EVA work, I would not expect the wild and wooly event to happen.

Now, suppose you were accelerating translationally as well as rotating. Jerks are always problematic. :)

publius
2007-Oct-25, 11:45 PM
George,

Actually they use angular rate sensors, which are sometimes called "gryos on a chip". :lol: It is something that can detect the angular rate of rotation of the device itself about a single axis.

One type actually uses uses the coriolis force (not centrifugal!). It causes a vibration in some internal mechanism that is oscillating. This is small enough to easily fit on printed circuit board. Just a tiny little component.

And the Sagnac effect is used to advantage as well. You would think the difference would be too small for small w's, but they can be very sensitive, apparently. And very expensive too. :)

The SAFER system uses three linear accelerometers and three gyros on a chip to determine what it's doing.

Nowadays, with the revolution in fancy and cheap electronics, you rest assured that someone has figured out to measure just about anything using some small, cheap eletronic component. Now, if life was going to depend on it, you want reliability and redunancy and that gets expensive.

However, I think you can roll rate sensors for around \$10 a pop, now (that may be the price if you're going to buy a 1000 or so....)

-Richard

George
2007-Oct-26, 12:29 AM
The chip gyros sound very interesting and I think I understand what they are doing by what you've stated. I am surprised that they would have such sensitivity to be useful in such low w, as should be found in SAFER.

The accelerometers make sense, along with some simple Boolean algorithm.