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Author Topic: Interactive spin decay, critical speed, and spin time simulator  (Read 1616 times)

Jeremy McCreary

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For the record, ta0 unleashed this monster on the forum. Couldn't resist taking it for a ride.

Simple models can be dangerous when it comes to real top behavior, but sometimes they work quite well...

Interactive spin decay, critical speed, and spin time simulator

Iacopo, ta0, and Alan have been kind enough to post some useful (time, RPM) data sets or empirical spin decay curves (SDCs) over the years. I've plotted empirical SDCs from all but ta0's raw data (not posted) in Excel, and lo and behold: I can fit a purely exponential decay to each and every one with excellent closeness of fit -- all the way to the bitter end in some (like Iacopo's Nr. 25) and at all but the lowest recorded speeds in the rest. The low-speed SDC tails in the latter group become more linear than exponential as they approach the horizontal -- just what you'd expect as speed-dependent air resistance finally gives way to mostly weight-dependent tip resistance.

This interactive simulator graphs strictly exponential SDCs from key parameters adjusted with sliders. The good news: Many of the empirically linear SDC tails are still pretty close to exponential, with ta0's 2 peg tops and Iacopo's Nr. 14 finger top being the most notable exceptions. The bad news: Projected spin times can be rather sensitive to tail shape. So don't take them too seriously.

The graphs

1. Horizontal (x) axis = time t after release in seconds. Ignore everything below.
2. Vertical (y) axis = angular speed w in rad/s. Ignore everything to left. Remember, 1 rad/s = 9.54 RPM.
3. Solid curve w(t) = purely exponential SDC crossing the vertical (t = 0) axis at the specified release speed w0. The release speed slider maxes out at w0 = 500 rad/s = 4,775 RPM.
4. Dashed horizontal line = speed w(T) at one top "lifetime" T after release. The time spent above this line is T, which is akin to a half life. Lifetime grows with the top's absolute axial moment of inertia (AMI) and shrinks mainly with its air resistance. Importantly, a top with an exponential SDC will lose 63.2% of its speed over any time interval T, not just the one starting at release (t = 0).
5. Dotted horizontal line marks the critical speed wC for stability against gravity, a property of the top's mass distribution alone. The time spent above this line is the top's minimum spin time. Real tops generally begin their rapid death spirals shortly thereafter, but some manage to hang on longer now and then.

Be sure to fiddle with the w0 and T sliders controlling the solid SDC and the J3, J1, and H sliders controlling the dotted critical speed line (see below). The limits on these sliders cover posted data for Iacopo's tops, Alan's 3-inch disk top, and most of mine as well. Best to have a dotted line showing above the horizontal time axis before experimenting with w0 and T.

Things to try

A. Set release speed at w0 = 100 rad/s (954 RPM) and tap the solid SDC at any point of interest. The displayed y-value for that point will then be the percentage of w0 remaining.
B. Vary release speed w0 alone. Note how little that affects spin time when the lifetime T is very short. Exactly what happens in my highest-drag tops: Paltry percentage spin-time gains from huge percentage release speed bumps.
C. Vary lifetime T at constant starting and critical speeds (w0 and wC). Note the very strong effect on spin time. Streamlining pays!
D. Raise and lower the dotted critical speed line at constant w0 and T. Note how much spin time you can gain by shaving off a little critical speed.
E. Vary CM height H alone. Note how easy it is to push critical speed wC above release speed w0. A top in this situation never stays up. Critical speed is most sensitive to this parameter.
F. Vary central transverse moment of inertia per unit mass J1 alone. Critical speed is least sensitive to this parameter, but it's still pretty easy to push critical speed wC above release speed w0. Note that J1 is about the top's CM, not its tip.
G. Vary AMI per unit mass J3 alone. The effect on critical speed is much stronger than in (F).

The implications for top design are pretty clear.

NB: The simulator's main flaw is its failure to link lifetime T to AMI per unit mass J3. Purely exponential spin decay implies a total braking torque proportional to current speed. But the resulting deceleration is inversely proportional to the absolute AMI I3 = M J3, where M = total mass. And T is an inverse measure of deceleration rate.



Equations using notation above in SI units

i. w(t) = w0 exp(-t / T)

ii. T = (t2 - t1) / ln(w1 / w2) for any two time marks t2 > t1.

iii. w(t + T) = w(t) exp(-1) = 0.368 w(t) for every t.

iv. wC = 2 sqrt[9.81 (J1 + H²) H] / J3, where J3 = AMI per unit mass, J1 = central transverse moment of inertia per unit mass (about CM), H = CM height at 0° tilt, and 9.81 is the acceleration of gravity near sea level.

« Last Edit: July 09, 2020, 03:05:05 PM by Jeremy McCreary »
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Art is how we decorate space, music is how we decorate time ... and with spinning tops, we decorate both.
—after Jean-Michel Basquiat, 1960-1988

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ta0

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Re: Interactive spin decay, critical speed, and spin time simulator
« Reply #1 on: July 09, 2020, 09:22:28 AM »

I'm glad you are having fun with the monster  >:D
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ta0

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Re: Interactive spin decay, critical speed, and spin time simulator
« Reply #2 on: July 09, 2020, 09:35:02 AM »

For a rotational symmetric top (J1 = J2), the longitudinal moment of inertia J3 cannot be bigger than 2 times the transversal moment of inertia (the case of a flat disk). The graph lets you use as the maximum value on the J3 slider the variable 2J1. This software is quite flexible!
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Jeremy McCreary

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Re: Interactive spin decay, critical speed, and spin time simulator
« Reply #3 on: July 09, 2020, 01:03:38 PM »

For a rotational symmetric top (J1 = J2), the longitudinal moment of inertia J3 cannot be bigger than 2 times the transversal moment of inertia (the case of a flat disk). The graph lets you use as the maximum value on the J3 slider the variable 2J1. This software is quite flexible!

Good software tip! Really impressed with this graphing site.

That restriction on J3 definitely applies to all rotationally symmetric solids of orders 3 to infinity -- cylinders, spheroids, peg tops, regular polygonal plates, etc. But I don't think it extends to triaxial solids like triaxial ellipsoids, elliptical disks, and rectangular plates.

Can't afford to get on the bad side of my triaxial tops, you know. Some of them bite. :o







« Last Edit: July 09, 2020, 01:11:13 PM by Jeremy McCreary »
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Jeremy McCreary

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Re: Interactive spin decay, critical speed, and spin time simulator
« Reply #4 on: July 09, 2020, 01:23:10 PM »

I'm glad you are having fun with the monster  >:D

Say, do you still have the raw (time, RPM) data for the 2 peg tops you made SDCs for a few years ago?
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ta0

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Re: Interactive spin decay, critical speed, and spin time simulator
« Reply #5 on: July 09, 2020, 10:02:44 PM »

But I don't think it extends to triaxial solids like triaxial ellipsoids, elliptical disks, and rectangular plates.
Can't afford to get on the bad side of my triaxial tops, you know. Some of them bite. :o
If I1>I2>I3, it can obviously be I1>>I3 but I also think 2I2>I1>I2

That square angle top gives me the giggles.  :D

Say, do you still have the raw (time, RPM) data for the 2 peg tops you made SDCs for a few years ago?
I took those measurements well over 10 years ago. The excel file may be in some old hard drive, but I doubt I can find it.
By the way, I am not sure what SDC stands for (Spin Decay Curves?). Acronyms many times are not helpful, in my opinion.
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Jeremy McCreary

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Re: Interactive spin decay, critical speed, and spin time simulator
« Reply #6 on: July 09, 2020, 11:03:35 PM »

If I1>I2>I3, it can obviously be I1>>I3 but I also think 2I2>I1>I2

Should be simple enough to test with a cuboid and ellipsoid. We're talking about central moments, right? Stay tuned.

By the way, I am not sure what SDC stands for (Spin Decay Curves?). Acronyms many times are not helpful, in my opinion.

Yes, SDC = spin decay curve, defined on first use. And yes, we have different takes on acronyms. I've been up to my eyeballs in them for the last 50 years in 2 different scientific fields -- and generally found them pretty darned useful. Heck, I even think in acronyms now. Electronic engineering must've been full of them, too. Not helpful?

So realistically, my acronym habit isn't gonna change now. You know, CTODNT (can't teach an old dog new tricks)! >:D
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Jeremy McCreary

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Re: Interactive spin decay, critical speed, and spin time simulator
« Reply #7 on: July 12, 2020, 04:23:58 AM »

For a rotational symmetric top (J1 = J2), the longitudinal moment of inertia J3 cannot be bigger than 2 times the transversal moment of inertia (the case of a flat disk).

Let's see how this useful theorem for symmetric tops might extend to triaxial (aka asymmetric) tops. Unless otherwise noted, "axis" by itself will refer to one of the top's 3 principal axes, and "moment" by itself to one of the 3 central principal moments of inertia Ik about its center of mass (CM). As always, the axes are mutually orthogonal and pass through the CM. To connect these moments to the quote above, note that JkIk / M, where M is the top's mass.

In physics, a "symmetric top" has exactly 2 distinct moments. (All throwing tops and most simple finger tops are of this type.) Exactly 2 of the 3 moments must then be equal. In addition, the remaining moment must be about a principal axis of at least 3-fold rotational symmetry. We can then write

I3I2 = I1,

where subscript "3" refers only to the symmetry axis, and subscripts "1" and "2" mark the transverse axes. ta0's theorem then states that I3 / I1 ≤ 2 -- even when I3 > I1.

Question is, can we make any similar statements about the moment ratios in a triaxial top with I3I2I1? Turns out, we can!



Our starting point comes from Bächer et al. (2014), with emphasis added:
Quote
For an arbitrary rigid body, there exists an equivalent ellipsoid with the same inertial properties.

That means the same mass M, the same 3 principal axes, and the same 3 principal moments -- all referred to the same CM. Pretty powerful stuff, because in the absence of air and tip resistance, any 2 tops with the same equivalent ellipsoid and CM-contact distance H will behave exactly the same! No matter how different their shapes or sizes!

So instead of worrying about every conceivable triaxial top, we need only look for something like ta0's theorem in an arbitrary triaxial ellipsoid with principal radii (not diameters) a > b > c > 0. The moments are then

Ia = M (b² + c²) / 5
Ib = M (a² + c²) / 5
Ic = M (a² + b²) / 5 > Ib > Ia,

with Ic being the greatest moment, Ib the intermediate, and Ia the least. Since a triaxial ellipsoid has no symmetry axes of the kind mentioned above, we'll have to examine 3 different moment ratios...

Ic / Ia = (1 + β²) / (β² + γ²) > 1
Ic / Ib = (1 + β²) / (1 + γ²) > 1
Ib / Ia = (1 + γ²) / (β² + γ²) > 1

where the proportions β ≡ b / a and γ ≡ c / a always obey 1 > β > γ > 0 by definition. Otherwise, we're free to choose β and γ any way we please. So far.



To look for a triaxial version of ta0's theorem, we then ask...

Q1: Is Ic / Ia < 2 ever voilated? YES -- whenever γ² < MIN[β², ½(1- β²)]. One such set of violations: γ < β = ½ -- a perfectly reasonable family of ellipsoids.

Q2: Is Ic / Ib < 2 ever voilated? NEVER! Otherwise, we'd have γ < 0, which the setup clearly forbids.

Q3: Is Ib / Ia < 2 ever violated? YES -- whenever γ < β < 1/√2 ≈ 0.71. No shortage of offending ellipsoids there!

So what's the strongest statement we can make about the moment ratios in a triaxial top with central principal moments Ic > Ib > Ia?

Ic / Ib = Jc / Jb < 2

Who knew? (Maybe ta0 did.)
« Last Edit: July 12, 2020, 06:37:53 AM by Jeremy McCreary »
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ta0

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Re: Interactive spin decay, critical speed, and spin time simulator
« Reply #8 on: July 12, 2020, 11:05:17 AM »

I'm glad that you found the same result. The way I thought about it was by dividing the body into slices, so the body is many flat tops with the same stem. You can see that when you switch to a transversal axis, the moment will be minimum if the masses of all those tops are as close as possible to the height of that axis (as that minimizes all the radius of rotation for each mass). Therefore, the transversal moment is minimum if the body was a flat (finger) top to start with. From the perpendicular axis theorem the momentum around the stem is equal to the sum of the momentum along to perpendicular axis on the plane:  I1 = I2 + I3. Then, if I3 = 0, I1 = I2 and if I2 = I3, then I1 = 2 I2
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Jeremy McCreary

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Re: Interactive spin decay, critical speed, and spin time simulator
« Reply #9 on: July 12, 2020, 12:19:06 PM »

I'm glad that you found the same result. The way I thought about it was by dividing the body into slices, so the body is many flat tops with the same stem. You can see that when you switch to a transversal axis, the moment will be minimum if the masses of all those tops are as close as possible to the height of that axis (as that minimizes all the radius of rotation for each mass). Therefore, the transversal moment is minimum if the body was a flat (finger) top to start with. From the perpendicular axis theorem the momentum around the stem is equal to the sum of the momentum along to perpendicular axis on the plane:  I1 = I2 + I3. Then, if I3 = 0, I1 = I2 and if I2 = I3, then I1 = 2 I2

Interesting approach -- akin to the "washer" method for building up the moments of potentially hollow bodies of revolution by stacking many thin circular washers or solid disks and integrating their contributions, taking into account each one's radii and axial distance from the CM, the latter with the parallel axis theorm.

At some point, you get to an equation  I1 = I2 = ½ I3 + G, where G is a non-negative integral that vanishes only when the body of revolution was a thin non-hollow disc to begin with. Hence, I3  /  I1  ≤ 2 for tops of revolution. Other symmetric tops like dreidels might require a somewhat different approach.

Of course, in the purely triaxial case, no moment equalities are allowed, none of the usual symmety axes exist, and you have 2 different transverse central principal moments to wonder about. And stacking the right elliptical washers or discs would have given even Euler nightmares. So I had to come at it from a different angle.

But I don't get the minimization step in your process for symmetric tops. And I'm wondering about some typos in your last few equations.
« Last Edit: July 12, 2020, 12:58:01 PM by Jeremy McCreary »
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ta0

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Re: Interactive spin decay, critical speed, and spin time simulator
« Reply #10 on: July 12, 2020, 01:37:15 PM »

I think this explanation will be more clear. I just use two facts:

1 - If you squash a solid body into a flat figure, the transversal moments will decrease. This is because the distances to that axis for each mass will decrease (Pythagoras).
The distances to the longitudinal axis don't change.
2 - For a flat top, I1 = I2 + I3 (Perpendicular axis theorem). So if I2 > I3, then I1 < 2 I2

Combining them, I1original = I1flatened < 2 I2flatened < 2 I2original
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Jeremy McCreary

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Re: Interactive spin decay, critical speed, and spin time simulator
« Reply #11 on: July 12, 2020, 01:49:07 PM »

I think this explanation will be more clear. I just use two facts:

1 - If you squash a solid body into a flat figure, the transversal moments will decrease. This is because the distances to that axis for each mass will decrease (Pythagoras).
The distances to the longitudinal axis don't change.
2 - For a flat top, I1 = I2 + I3 (Perpendicular axis theorem). So if I2 > I3, then I1 < 2 I2

Combining them, I1original = I1flatened < 2 I2flatened < 2 I2original

(1) is the stretch theorem, so totally onboard there. (2) I'm still confused about. Which subscript points to the symmetry axis?

Got it now. Very clever! I'll have to remember that trick.
« Last Edit: July 12, 2020, 01:57:29 PM by Jeremy McCreary »
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ta0

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Re: Interactive spin decay, critical speed, and spin time simulator
« Reply #12 on: July 12, 2020, 07:47:46 PM »

Something interesting I did not mention is that the transversal principal axes of the squashed top can rotate with respect to the transversal principal axes of the original body. But this does not affect the reasoning as the perpendicular axis theorem works for any pair of perpendicular axes on the plane. That is, I2flatened does not have to be a principal axis of the flat top, just coincide with the principal axis of the original top.
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Jeremy McCreary

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Re: Interactive spin decay, critical speed, and spin time simulator
« Reply #13 on: July 31, 2020, 02:17:19 PM »

Iacopo, ta0, and Alan have been kind enough to post some useful (time, RPM) data sets or empirical spin decay curves (SDCs) over the years. I've plotted empirical SDCs from all but ta0's raw data (not posted) in Excel, and lo and behold: I can fit a purely exponential decay to each and every one with excellent closeness of fit -- all the way to the bitter end in some (like Iacopo's Nr. 25) and at all but the lowest recorded speeds in the rest.

So just how exponential are these measured SDCs? Here are 2 representative examples of the highest data quality...



Source: Raw data reported by Iacopo here.
Top: Simonelli Nr. 26b (see Iacopo's catalog photo below), consistently falling at 195 RPM.
Behavior: Quiet sleep.
Contact: Spiked tungsten carbide tip on polished tungsten carbide cup -- with and without oil lubrication.
Air pressure: Atmospheric.

To make these SDCs, I replotted Iacopo's raw (minute, RPM) data in Excel after converting to (s, rad/s) units, as the deceleration rate at any given moment is then just the slope of the SDC at that time. Excel's exponential fits (dashed curves) are least-squares regressions. In both data sets, Iacopo started the spin-down clock at w0 = 130.9 rad/s = 1,250 RPM. Hence, I forced both exponential fits to pass through (0, w0).

Nr. 26b is a smooth flat disk top, but posted peg top SDCs look much the same -- at least at higher speeds.



The R² values in the colored boxes measure goodness of fit over all the points plotted, with R² = 1 being a perfect fit. Granted, there are visible deviations -- especially in the dry SDC's low-speed tail. But at R² > 0.99 in both wet and dry cases, it's reasonable to ask...

Q1: What does an exponential SDC tell us about the underlying decay processes?
Q2: Are the deviations from pure exponential decay statistically significant? And if so, what are they trying to tell us about the relative contributions of air vs. ground resistance -- especially at the lowest speeds?

Since SDC shapes contain valuable information about the physical processes underlying spin decay, I plan to come back to Q1 in a subsequent post. The short answer to Q2: With so few measured SDCs available, too early to tell.



For now, note that lubrication significantly reduced Nr. 26b's deceleration rate (SDC slope) from the start. The result: A substantially longer "wet" spin time from w0 to fall. And FWIW, the wet SDC is also the more exponential of the two, though not by much.

Note also the vertical gap between the wet and dry SDCs. Not sure how much ground resistance the oil left. But there was a lot more in the dry case with the same air resistance. Dry ground resistance also became a significant player early on.
« Last Edit: July 31, 2020, 06:05:09 PM by Jeremy McCreary »
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