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[Russpin]

This experiment seems to demonstrate that an higher transversal moment of inertia does not slow down precession speed.

For high spin rates, that is what the slow precession equation for tops predicts.
Nice experiment !

[Russpin]

I wrote a script to calculate the mass properties of a simple gyroscope. The following parameters were used:

Rotor:
radius = 1.
width = .25
density = 11.3 (Lead)

Inner gimbal:
inner radius = 1.1875
outer radius = 1.25
width = .25
density = 2.7 (Aluminium)

Outer gimbal:
inner radius = 1.4375
outer radius = 1.5
width = .25
density = 2.7 (Aluminium)

The script predicts a precession of 1.5 w

I have checked the numbers three times and cannot find an error.

All the intermediate calculations from script are shown here.

[Jeremy McCreary]

Elegant as predicted, Iacopo! Within experimental error, your results are completely consistent with both of the surprises in the approximate slow precession rate formula

p ~ M g H / I₃ s,

where p is the precession rate, M is the top's mass, g is the acceleration of gravity, H is tip-CM distance, I₃ is the AMI, and s is the pure spin rate. The surprise Russpin noted is that I₁, the TMI about the tip, is missing in action. The other is that the tilt angle is missing, too.

NB: To separate s and p this cleanly, you must assume that s >> p. Fortunately, that's usually a pretty safe bet in the slow precession regime.

mmmmmmmmmmmmmmmmmmmmmm

Couldn't resist calculating how TMI grows with disk separation in your experiment...

Setup:
1. Subscript "t" points to top disk, "b" to bottom disk, and "c" to either disk CM.
2. Subscript "1" points to transverse axis through tip (line of nodes), "3" to spin and symmetry axis.
3. The identical disks each have mass m, radius R, and axial length L.
4. Axle mass is negligible, so total top mass is M = 2 m, and top AMI and TMI are due solely to the disks.
5. Parameters m, M, R, L, and H are all held constant.
6. The distance 2 D between disk CMs remains centered on the top's CM at all times as required by constant H.

Distances from tip to disk CMs are respectively H_b = H - D and H_t = H + D.

Disk AMIs are I_3b = I_3t = ½ m R², so top AMI is I₃ = I_3b + I_3t = m R² = ½ M R² regardless of disk separation.

Individual disk TMIs about their own CMs are respectively I_1bc = I_1tc = m (3 R² + L²) / 12, so disk TMIs about the tip are respectively I_1b = I_1bc + m H_b² and I_1t = I_1tc + m H_t². Top TMI about the tip is then I₁ = I_1b + I_1t = m (3 R² + L²) / 6 + m [ (H - D)² + (H + D)²].

After some algebra, the final result:

I₁ = m (3 R² + L² + 12 H² + 12 D²) / 6 = M [3 R² + L² + 12 (H² + D²)] / 12

So I₁ varies with the square of D, as one might guess from the parallel axis theorem. Note that H and D have more influence on TMI than m, R and L do!

This TMI is the same one obtained by moving the CM of a single disk of mass M and length L to a new height H' = sqrt(H² + D²) above the tip. And if L / H << 1, you'll get almost the same TMI by pushing your 2 disks together and raising the joint between them to H'. This gives you a way to do the reverse experiment with variable overall tip-CM distance and TMI approximately constant!

Since you kept m, M, R, L, and H constant, the rate of change of the TMI with disk half-separation D is

dI₁ / dD = 4 m D = 2 M D

So the TMI grows with D at a rate proportional to D. If instead you kept D constant and varied H, you'd have

dI₁ / dH = 4 m H = 2 M H

[ta0]

Russpin: I seriously doubt that they use lead and aluminum on the gyros of missiles  . We may be civilians, but we also want to play with the good stuff: tungsten for the flywheel and beryllium for the gimbals. Keeping the same dimensions will increase the moment of inertia of the flywheel by 19.1/11.3 = 1.69 and decrease the one of the gimbals by 1.84/2.7 = 0.68 or the ratio will increase by 2.49. Beryllium has much better mechanical properties than aluminum (4.37 times the Young modulus and 2.66 times the tensile strength), so it might be possible to make them smaller, even with a bigger flywheel.
I think this may take us where I wanted, without the need to use platinum flywheels 

[Iacopo]

For high spin rates, that is what the slow precession equation for tops predicts.
Nice experiment !

Thanks Russpin. I am thinking about the formula.

[Iacopo]

The surprise Russpin noted is that I₁, the TMI about the tip, is missing in action. The other is that the tilt angle is missing, too.

They are in fact the two things I was thinking about.

The lack of influence of the TMI in slow precession I am not sure what tha cause could be.

But for the lack of influence of the angle of tilting I think I have found the solution:
There is no doubt that a stronger torque makes for a faster precession.
Also, no doubt that a more tilted position of the top creates a stronger torque.

But the stronger torque of a more tilted top cancels out with a required faster transverse speed of the flywheel, at parity of angular speed of the precession.  I explain:   In a top very tilted down (let's say 90 degrees, the rotation axis of the top horizontal), 1 degree of tilting of the top axis along the precession trajectory corresponds, of course, to one degree of precession. 
But if the top is nearly vertical, one degree of motion of the axis of the top corresponds to a much larger angle of precession.
For example, if the axis of the top is 0.5 degrees far from the vertical axis, a shift of just one degree of the top axis would make this axis to be in the other side of the vertical axis, which corresponds to 180 degrees of precession.
One degree of tilting of the top axis corresponds now to 180 degrees of precession.
So, when the top is nearly vertical, a slower motion of the top axis, which is a slower transversal speed of the flywheel, (that is proportional to torque), is needed for having the same angular speed of the precession.
It is something related to geometry.

[Russpin]

Keeping the same dimensions will increase the moment of inertia of the flywheel by 19.1/11.3 = 1.69 and decrease the one of the gimbals by 1.84/2.7 = 0.68 or the ratio will increase by 2.49.

Increasing the density of the rotor increases both I1 and I3.
using I₁ = 2.265, I₃ = 4.437 and Ig = 0.664 from the lead and aluminum case. Then
using tungsten and beryllium:
p = 1₃*1.7/(I₁*1.7 + Ig*0.685) = 1.75 w

using platinum and beryllium:
p = I₃*1.9/(I₁*1.9 + Ig*0.685) =  1.77 w

The script gives the same results.

[Russpin]

The lack of influence of the TMI in slow precession I am not sure what tha cause could be.

Check out this derivation of the slow precession formula. It's the simplest one I have seen.
http://hyperphysics.phy-astr.gsu.edu/hbase/top.html

[ta0]

Increasing the density of the rotor increases both I1 and I3.
using I₁ = 2.265, I₃ = 4.437 and Ig = 0.664 from the lead and aluminum case. Then
using tungsten and beryllium:
p = 1₃*1.7/(I₁*1.7 + Ig*0.685) = 1.75 w

As the rotor here is a cylinder, not a disk, the "free space precession" is a little less than 2:
p_fs = 4.437/2.265 = 1.96
So, for tungsten and beryllium we have 1.75/1.96 = 89.3% of the ideal free space value. With the superior mechanical properties of the beryllium we can likely tweak the gimbals and get even closer to the ideal.
I think this gyro would be useful in demonstrating inertial "floating" behavior without having to go to space. Of course, for the precision needed on a navigation gryo you would have to take into account the effect of the gimbals.

[Iacopo]

The lack of influence of the TMI in slow precession I am not sure what tha cause could be.

Check out this derivation of the slow precession formula. It's the simplest one I have seen.
http://hyperphysics.phy-astr.gsu.edu/hbase/top.html

Thank you, anyway I am not sure yet about the answer. 
It isn't that I doubt about the role of the TMI; after the experiment, and what you and Jeremy have said, I do believe that the TMI does not influence the precession rate. 
I tend very much to think in geometrical terms when I try to understand these things.
Even not knowing the terminology and the math language, often I understand the logicality of these forces and motions, at an elementary level of course, (and at times I am wrong even for simple things).

Given the direction of the torque in a top, the first idea could be that both the moments of inertia partecipate, as an inertial resistance, to that torque. 
Now i know that this is wrong.
 
So I am come to think that, (today I started to have this idea), the transversal accelerations on the flywheel happen solely in a direction parallel to the axis of the top: this would make the TMI of no influence on precession speed.

At first glance this could seem strange; during precession, (the normal slow one), for each spin of the top, its mass points are subjected to two accelerations and two decelerations in a direction parallel or approximately parallel to the axis of the top.
Anyway the spin axis has not lateral movements synchronous with these accelerations;  it only traces a conical trajectory.
Since the top is not made of rubber and doesn't bend while spinning, how can it be that the mass points of the flywheel go up and down and the stem instead go straight along the precession trajectory ?

The answer is that the top doesn't bend and that the precession movement is the visible result of these vertical accelerations.
The stem movement unaffected by the vertical accelerations in the flywheel is a clue that those accelerations happen only parallel to the axis of the top, so that they have no influence on the trajectory of the stem.

If the accelerations on the flywheel happened about the tip, or the CM, (which I can't imagine), so that the trajectory of these accelerations would be curves, and the kind of acceleration angular, not through straight lines, then the stem should move in some way accordingly to these accelerations in the flywheel.
It is difficult to conceive clearly these vertical accelerations in the flywheel, but the motion of the stem (axis of the top) is easier;
it has no lateral movements, (apart from precessing, assuming that there is not nutation nor unbalance), so it should mean that the shape of the trajectories of the mass points in the flywheel during the vertical accelerations can only be straight lines parallel to the axis of the top.
This maybe would explain why the transverse moment of inertia does not influence precession speed.

[Jeremy McCreary]

Check out this derivation of the slow precession formula. It's the simplest one I have seen.
http://hyperphysics.phy-astr.gsu.edu/hbase/top.html

Yes, a nice little derivation similar to the one Feynman used in this Lectures in Physics. His essential insight about precession due to a gravitation torque: "You have to go down a little, to go around."

[Russpin]

Yes, a nice little derivation similar to the one Feynman used in this Lectures in Physics.

Off Topic but I was in a bookstore years ago and came very close to buying them. I'm glad I didn't because they are online for free.

[Jeremy McCreary]

I tend very much to think in geometrical terms when I try to understand these things.
Even not knowing the terminology and the math language, often I understand the logicality of these forces and motions, at an elementary level of course...

I'm often guilty of the opposite. When dealing with real-world tops, it's usually better to have the gut-level geometric/mechanical understanding in the end. You're very good at that.

As I've said before, I find formulas most useful, not to calculate things, but to remind me of the players directly involved and how they influence the outcome in a qualitative way. The slow precession formula reminds us that the steady precession rate grows with tip-CM distance and slows with increasing AMI per unit mass and spin rate, while absolute mass, TMI, and tilt have no direct effect. That's pretty valuable even if it doesn't really explain the phenomenon.

But there's a caveat: One must always apply simple formulas to the real world with eyes wide open regarding the underlying simplifications, which can often turn out to be deal-killers. Since it often takes some real digging to bring all the assumptions to light, people often skip this step without saying so and are then led astray. This is a serious problem even among professional scientists.

...(and at times I am wrong even for simple things).

Welcome to the club, my friend!

I'll have to think about the rest of your post.

[Iacopo]

gyro would be useful in demonstrating inertial "floating" behavior without having to go to space.

A poor's man gyro of that kind could be a rotor suspended to thin strings, it would oscillate with a ratio very near to 2:1...

[Russpin]

A poor's man gyro of that kind could be a rotor suspended to thin strings, it would oscillate with a ratio very near to 2:1...

Time for another experiment.
I'm guessing that twisting modes will develop causing the height of rotor CM to go up and down changing the dynamics from that of free precession.