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[ta0]

I should have replied that you described the case of the unbalance wobble mixed with the torque free precession.
It is possible to have pure unbalance wobble too, in which case the stem trajectory is a circle; a bigger unbalance will produce a larger circle, at parity of spin speed.

Ok. I agree, if the top is spinning around the actual principal axis of the unbalanced top, there will be no torque free precession, just a circular movement by the stem. I guess you can correctly call it pure unbalance, although the top is spinning perfectly stable: it's just the stem that it's in the wrong place! 

[Iacopo]

It took me some time to understand why the cones were drawn in that way, flanked at the left and one inside the other at the right.
They both produce the same kind of motion.
Then I realized that in the first case the spin motion and the inertial precession motion have opposite the same direction in the first case, and the same opposite direction in the second case.

It means that oblate tops which spin clockwise, have inertial precession clockwise.
Prolate tops instead, when spinning clockwise, have inertial precession counterclockwise.

[Jeremy McCreary]

Still trying to understand the theoretical points being made. Meanwhile...

Data point: In the sleeping  LEGO test tops I made, the stem trace of pure static unbalace appears to be a fairly clean circle centered on the vertical. This is what Iacopo sees, but in a mass property domain quite removed from his.

When the unbalanced top tilts into what averages out to a steady precession at mean angle A., the stem trace around the mean axis precessing about the vertical still appears to be a small circle.  These are just the antics you'd expect from whirl.

[Iacopo]

I guess you can correctly call it pure unbalance, although the top is spinning perfectly stable: it's just the stem that it's in the wrong place! 

If the top spins like a balanced top with just a bended stem, I think that I could call it "false unbalance".

I see the real unbalanced top as the top with the tip and the center of mass vertically misaligned.
The behaviour is different, the real unbalanced top may appear nearly well balanced at high speed, but then it makes a circular trajectory with the stem which becomes wider and wider by the time, until toppling down. The spin time is reduced.

The "false unbalanced top" instead, (just the stem bended), would have the stem wobbling since from the start, and its circular trajectory would not become wider and wider by the time, at least until towards the end of the spin; it would move like a balanced top, (hence "false unbalance").   

[Jeremy McCreary]

I guess you can correctly call it pure unbalance, although the top is spinning perfectly stable: it's just the stem that it's in the wrong place! 

If the top spins like a balanced top with just a bended stem, I think that I could call it "false unbalance".

There's already a name for that -- "couple unbalance". Or maybe "dynamic unbalance" if the bend also causes some static unbalance. In any case, no false unbalace here.

Pure couple unbalance occurs when the top is constrained to spin about an axis passing through the CM but not one of the top's principal axes. More than one way to end up there.

[ta0]

It took me some time to understand why the cones were drawn in that way, flanked at the left and one inside the other at the right.
They both produce the same kind of motion.
Then I realized that in the first case the spin motion and the inertial precession motion have opposite direction in the first case, and the same direction in the second case.

It means that oblate tops which spin clockwise, have inertial precession clockwise.
Prolate tops instead, when spinning clockwise, have inertial precession counterclockwise.

You got them mixed up. In the left side (prolate case) the precession and spin are in the same direction while on the right side (oblate case) they are in opposite direction.
The oblate case is hard to visualize.

[ta0]

If the top spins like a balanced top with just a bended stem, I think that I could call it "false unbalance".

I see the real unbalanced top as the top with the tip and the center of mass vertically misaligned.
The behaviour is different, the real unbalanced top may appear nearly well balanced at high speed, but then it makes a circular trajectory with the stem which becomes wider and wider by the time, until toppling down. The spin time is reduced.

The "false unbalanced top" instead, (just the stem bended), would have the stem wobbling since from the start, and its circular trajectory would not become wider and wider by the time, at least until towards the end of the spin; it would move like a balanced top, (hence "false unbalance").   

I was thinking only about unbalanced tops supported at the center of mass, so the only source of imbalance was the direction of the principal moment axis aligned in a direction different than the stem. In that case the only pure imbalance would be the false imbalance. But I can see that if the center of mass is away from the support the top, the top cannot have a stable spin. At this time I cannot tell if there can be "pure imbalance" with no (or much smaller) torque-free precession. 

[Iacopo]

You got them mixed up. In the left side (prolate case) the precession and spin are in the same direction while on the right side (oblate case) they are in opposite direction.

You are right... but there must be an error somewhere..
My oblate tops have the spin and the inertial precession in the same direction, not opposite.

[Iacopo]

Pure couple unbalance occurs when the top is constrained to spin about an axis passing through the CM but not one of the top's principal axes.

I was thinking only about unbalanced tops supported at the center of mass, so the only source of imbalance was the direction of the principal moment axis aligned in a direction different than the stem. In that case the only pure imbalance would be the false imbalance. But I can see that if the center of mass is away from the support the top, the top cannot have a stable spin. At this time I cannot tell if there can be "pure imbalance" with no (or much smaller) torque-free precession. 

To say it in different words, if a top can spin in sleeping position, with the principal moment axis staying vertical, and passing through both the center of mass and the tip, I would say that the top is balanced; if such a top had the stem bended, the stem would wobble, but the imbalance would be apparent, not real.

The imbalance is real when the center of mass is away from the support of the top. 
I saw high CM tops to spin unbalanced and with some torque-free precession, I am not sure whether these tops can spin unbalanced without the torque-free precession.
But I can tell that my tops with low CM, when unbalanced, (I mean real imbalance), can spin without torque-free precession, in which case they trace a simple circular trajectory. I call this "pure imbalance wobble".

[ta0]

You got them mixed up. In the left side (prolate case) the precession and spin are in the same direction while on the right side (oblate case) they are in opposite direction.

You are right... but there must be an error somewhere..
My oblate tops have the spin and the inertial precession in the same direction, not opposite.

Ok, I think I found where the error is. There are two ways of defining the spin of the top. One is absolute, with respect to fixed coordinates. In these coordinates, the torque-free precession and the spin, s, for an oblate top, are in the same direction, as you measured. But, w_o on Butikov diagram is taken on a coordinate system that rotates with the precession. The total spin s is the sum of w_o and the contribution of the precession to the spin. For a plate rotating at a small tilt, p is about 2 w_o but opposite, so s = - w_o.

Looking carefully at the oblate diagram, the cone does spin in the same direction as it rolls, although it spins backwards with respect to a radius that follows the rolling.

[Iacopo]

I observed those cones and thought some more, I think now I understand them:

I made some mess before; the fixed cones are the ones with the letter "z", not the letter "n", (I thought the opposite but it's my fault because you explained it).

I observed cylinders spinning with torque-free precession in mid air: they always precess in the same direction of the spin, never opposite, the only difference is that the oblate cylinders precess faster and the prolate ones precess slower.

So the way the cones are drawn is not for a change of direction of motion, which doesn't exist;
in both the pairs of the drawn cones in fact, spin and precession have the same direction.

The difference which now I see between the two drawings, is that in the first case, at the left, the precession cannot be faster than the spin, while, in the second case instead, the precession cannot be slower than the spin.
The more narrow the fixed cone, the littler the difference between the two speeds.
 
If, instead of the fixed cone, there was a simple vertical line, and the cone of the top rolled without slipping on this vertical line, the speed of the spin and that of the precession would be the same, (the case of the three axes of inertia being equal). 
 
All of this if I stay in the frame of reference of the observer.  It seems to me now that it works even staying in this frame, as for what I explained here.  Looking from the frame of reference of the precession confuses me a bit.
     

[ta0]

I agree with everything you said above.

I can see a case of pure imbalance precession (or nutation if you prefer): if the top just goes around like a stone whirled at the end of a string, where the center of mass is the stone and the support is the hand. Because both the support point and the CM are fixed with respect to the top, it will have to spin once per rotation: precession = spin (or in the rotating frame w_o = 0) The line between the support and the center of mass will be perpendicular to the axis of the precession.

[Jeremy McCreary]

I can see a case of pure imbalance precession (or nutation if you prefer): if the top just goes around like a stone whirled at the end of a string, where the center of mass is the stone and the support is the hand. Because both the support point and the CM are fixed with respect to the top, it will have to spin once per rotation: precession = spin (or in the rotating frame w_o = 0) The line between the support and the center of mass will be perpendicular to the axis of the precession.

Nice description of whirl (in the engineering sense) due to pure static unbalance.

The vast engineering literature on this topic (usually under the banner of "rotor dynamics") has much to teach us about real top behavior.

Several very useful free intros online in PDF form. I've posted links to them several times. Wikipedia is also a good start.

[Iacopo]

I can see a case of pure imbalance precession (or nutation if you prefer): if the top just goes around like a stone whirled at the end of a string, where the center of mass is the stone and the support is the hand. Because both the support point and the CM are fixed with respect to the top, it will have to spin once per rotation: precession = spin (or in the rotating frame w_o = 0) The line between the support and the center of mass will be perpendicular to the axis of the precession.

Now that I know that "precession" is a kinematic term, and is already used for both the torque free and the torque induced wobbles, at this point I suppose that it could be used for imbalance too. It sounds better than "imbalance wobble".

I agree that imbalance precession and spin are synchronized. Not sure about your last sentence.
I think that the imbalance precession axis, (if the motion is pure, with no other wobbles superposed), is always vertical.
The motion happens about only one axis in pure unbalance precession.

This is different from the other two pure precessions, where the top moves always about two axes, the spin axis in the stem, and the vertical precession axis.   

[ta0]

Now that I know that "precession" is a kinematic term, and is already used for both the torque free and the torque induced wobbles, at this point I suppose that it could be used for imbalance too. It sounds better than "imbalance wobble".

I agree that imbalance precession and spin are synchronized. Not sure about your last sentence.
I think that the imbalance precession axis, (if the motion is pure, with no other wobbles superposed), is always vertical.
The motion happens about only one axis in pure unbalance precession.

This is different from the other two pure precessions, where the top moves always about two axes, the spin axis in the stem, and the vertical precession axis.   

I actually kind of like using wobble with imbalance, as it's undesirable. 

I was neglecting gravity, but that is true only for a very fast precession nutation wobble whirl. If not, the whirl axis of a pure imbalance has to be vertical and the line between the support and the CM will be slanted downwards (what will make the stem describe a cone).