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 51 
 on: July 12, 2020, 10:46:02 PM 
Started by paxl13 - Last post by Diz
Are you making sure the four spindles aren't rotating when the main shaft is twisting the strands together? I once forgot to do that and the results were similar to what you have. 

 52 
 on: July 12, 2020, 09:15:01 PM 
Started by ta0 - Last post by ta0
Thank for your input Mark. I think you should be the new ITSA Ambassador to Canada now that that position is open.

Here is a possible draft of rules so people have something concrete to discuss:

2020 World Spintop Online Contest
This contest is in lieu of the regular Spintop World Contest that cannot take place due to the Covid19 pandemic. The titles will be official world titles certified by ITSA.

A. Eligibility
1. Any player in the world can compete. Registration to compete is free but competitors need be members of ITSA or one of the affiliated national associations. Membership fee will be waived to those that request it because of economic hardship.

2. Players need to register as competitors at least 2 weeks before the contest deadline.

B. Deadline
Players will upload online their videos before the deadline of Sunday, September 27, 23:59 UTC/GMT.

C. Divisions
There will be the following divisions, provided that a minimum of 5 competitors enter each one:
1- Open freestyle (world title)
2- Traditional top freestyle (world title)
3- Sports ladder

D. Video Upload
1. The freestyle performances will be recorded from a single camera in landscape format in one continuous take and with live sound. The quality should be good enough to see the tricks clearly.
2. One performance per player (per division entered) will be uploaded to Youtube.com and published so it has a day stamp before the deadline. But it is recommended that it remains unlisted until the deadline.
3. Emails should be send to the Contest Organizer before the deadline with a link to the video. Several independent email addresses will be provided as backup.
4. The official logo should be printed and placed somewhere in view of the camera. If this is not possible it can be replaced by a sign saying 2020 World Spintop Online Contest.

E. Open Freestyle Rules
1. It consists of a 3-minute duration freestyle to music.
2. Any spinning top will be permitted, including fixed-tip, bearing-tip and one-way bearing tip, as well as over-sized tops. The number of spintops used or available for back-up is not limited, but the player cannot exit the view of the camera to retrieve them. All tops used must be wound by the player, no assistance will be permitted.

F. Traditional Top Freestyle Rules
1. It consists of a 3-minute duration freestyle to music.
2. A traditional top should be used, defined as one with the body made of solid wood and that has a fixed tip made of metal.

G. Sports ladder
1. The player will try to complete as many tricks as possible from the official ladder trick list.
2. Will perform the tricks in order and needs to complete a trick to pass to the next one.
3. Will have up to two misses forgiven. On the third miss the ladder is finished.
4. The players will not take more than 15 seconds between tricks.
5. Can change tops between tricks but has to use a fixed tip when the trick rules require it.

H. Freestyle Judging
1. A review panel from the organizers will first determine if the freestyles are of the minimum quality required for a World contest, and that they don’t break any rules, or for any reason cannot represent the contest (including inappropriate lyrics in the music). Competitors found at fault will be disqualified and not be part of the judging.
2. The freestyles that are not disqualified will be posted by ITSA on its website no later than September 29th. 
2. The freestyles will be judged from these videos by the fellow competitors.
3. Each competitor will provide a ranking of the other freestyles (but not his/hers) in his or her division: 1st, 2nd, 3rd, etc. No draws. 
4. The total score of a player is the addition of every ranking: e.g. 1 point for first, 4 points for fourth.
The winner will be the player with fewer points.
5. The performances should be ranked accordingly mainly to difficulty and quantity of tricks completed, originality and overall quality and interest of the performance. It is assumed that the competitors will be the best qualified to compare quality of the freestyles.
6. The competitors should send their rankings to the Organizers within 3 days after the freestyles are made public. Not providing rankings of the other players will disqualify the competitor.
7. Competitors should not share their ranking, including with other competitors, until the results are announced. The ranking provided by each competitor will at that time be public.

I. Sport ladder judging
1. The highest trick completed is the score of the player.
2. Of two players with the same highest trick completed, the one with the highest first miss will win.
3. If two players are still draw after 1 and 2, the one that completed the tricks in less time (from first throw to completion of last trick) wins.

J. Award Ceremony
The award ceremony will take place on Sunday October 4th live by .  . .
The prizes will be . . .

 53 
 on: July 12, 2020, 07:47:46 PM 
Started by Jeremy McCreary - Last post by ta0
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.

 54 
 on: July 12, 2020, 04:45:45 PM 
Started by Iacopo - Last post by Iacopo
PS: Make and model of your laser tach? Looks like a nice one.

It's a Nimex NI606 tachometer. It costs about € 40.
Yes, it's nice and works well, I think you will be happy about it.

 55 
 on: July 12, 2020, 01:49:07 PM 
Started by Jeremy McCreary - Last post by Jeremy McCreary
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.

 56 
 on: July 12, 2020, 01:37:15 PM 
Started by Jeremy McCreary - Last post by ta0
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

 57 
 on: July 12, 2020, 12:19:06 PM 
Started by Jeremy McCreary - Last post by Jeremy McCreary
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.

 58 
 on: July 12, 2020, 11:05:17 AM 
Started by Jeremy McCreary - Last post by ta0
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

 59 
 on: July 12, 2020, 04:23:58 AM 
Started by Jeremy McCreary - Last post by Jeremy McCreary
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.)

 60 
 on: July 11, 2020, 07:14:32 PM 
Started by paxl13 - Last post by ta0
After years of having a Neff string making machine I finally used it today  :-[ My excuse is that I got more string from others than I needed  ::)
I made several 4-strand (2 threads per strand) strings using two colors (both size 10, one color per strand). The string works well, but I'm getting a switch between alternating colors red-wihite-red-white at the ends and red-red-white-white at the center:



Not bad looking and didn't seem to affect play, but it was not my intention. Any advice on how to avoid it?

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