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[ta0]

I love that you could make such clean examples with LEGO. Even one very close to being a maximal offset for the ratio of densities!  Certainly keepers!

I also liked that you could insert the golden ration there!
That got me thinking. To have a square divided by the golden ratio φ (so ratio of larger to smaller square = ratio of full square to large square) and the stem at the boundary, the ratio of densities should be φ² = 1 + φ ~ 2.6. It should be possible to 3-D print by playing with the filling ratio.

[Jeremy McCreary]

I love that you could make such clean examples with LEGO. Even one very close to being a maximal offset for the ratio of densities!  Certainly keepers!
I also liked that you could insert the golden ration there!
That got me thinking. To have a square divided by the golden ratio φ (so ratio of larger to smaller square = ratio of full square to large square) and the stem at the boundary, the ratio of densities should be φ² = 1 + φ ~ 2.6. It should be possible to 3-D print by playing with the filling ratio.

Thanks! The golden ratio connection was my favorite part.

Not sure how you're dividing the square here, though. The usual golden ratio diagram divides a golden rectangle into a square and a smaller golden rectangle, but that wouldn't work here. The only "golden" way I could think of to divide a square rotor was to assign a fraction β = 2 - φ = 0.382 to the small rectangle and the rest, 1 - β = φ - 1 = 0.618, to the large one. Then I crossed my fingers and plugged β = 2 - φ into my formula for q. To my great surprise and delight, the q that popped out corresponded to δ =  β!

[Jeremy McCreary]

Q: Given a square offset top with its CM pinned to its boundary, what happens to its AMI I₃ as its areal density ratio δ increases?

This is a question best addressed with a spreadsheet and answered with a plot, as the AMI equations turned out to be way too complicated to give a feel for what's going on.

Recall that δ = α₂ / α₁, where α is areal density, and the subscripts "1" and "2" point to the small and large rectangles, resp. To make the graph below, I fixed the rotor's side length at S = 16 (for LEGO purposes) and fixed the large rectangle's areal density at α₂ = 1. Growing δ from 0.05 to 0.60 then amounted to reducing the small rectangle's areal density α₁ from 20 to 1.7. At every increment, this effectively removed mass from each rectangle and the rotor as a whole.



The AMI units here are arbitrary but consistent. The rectangle AMIs (red and blue) are about the top's CM, not their own. Ditto for the complete rotor (green). The downward trends in all AMI plots are mainly due to the mass losses noted above, but the large rectangle and complete rotor also saw small progressive losses in specific AMI. The small rectangle's specific AMI grew slowly with δ but not fast enough to offset its loss of mass.

ortwin will be happy to see that reducing δ to increase the astonishment factor also increases play value by increasing spin time by increasing AMI. Life is good.

Two things I didn't see coming:
1. At every step, the rotor's AMI would have been greater had its total mass been evenly distributed throughout the square (purple line.)
2. At δ = 0.25, the large rectangle has twice the area but nearly the same AMI (red) as the small rectangle (blue). At higher δ values, area continues to shift from the large to small rectangle, but their AMIs only get closer.

[ortwin]

..... Astonishment value for the general public probably 9/10 for both shapes. ....
Remember, most people aren't used to seeing square tops. As long as the top has a conspicuous asymmetry but spins smoothly anyway, I foresee a decent surprise factor even when the stem and tip aren't on the border but still way off center....

That is precisely why I think that the circular version has the better "astonishment value" for the right reasons. If you had the square version at your booth, a conversation with a visitor could go like this:
Visitor: "Oh you have a square top over there, that surely does not spin well. Everybody knows that wheels have to be round!"
Jeremy: "Let me show you." (and J. gives it a gentle start after which the square offset top spins happily and wobble free for 32 seconds )
      Edit: and can you show us also Jeremy? A simple video or even a gif with the cellphone would suffice
Jeremy: "And look, the stem and tip of the the top are not even in the center."
Visitor:  "Yeah, sure, I told you a square would not spin nicely, you had to make up for that, no surprise to me!"
Jeremy: "Um,... sorry I need to run, I need something to drink, urgently,..."
 
What I am trying to say, I think it is generally better not to present two surprises at once. They might even cancel each other out.
Somehow that would be like giving a single equations with two unknowns. Since it is not solvable unambiguously, people loose interest or come up with silly solutions.
I am all for a circular version: silver and acrylic is a combination Iacopo tested color wise at least. The density ratio is around 10. 
 

Golden ratio: I think I will put a little problem on that in the NSTR section since the two of you seem to like that number.

[Jeremy McCreary]

@ortwin: After 9 years of experience with the crowds at our LEGO displays, pretty sure a square offset top will be popular. And since I have neither the means nor the desire to make a round one, that will have to do.

As they say, perfect is the enemy of good.

[ta0]

Not sure how you're dividing the square here, though. The usual golden ratio diagram divides a golden rectangle into a square and a smaller golden rectangle, but that wouldn't work here.

What you are talking about are rectangles in which the width and length have the golden ratio.
But the fundamental geometric definition of the golden ratio is for a segment: you divide it into two segments so that the ratio of the big part to the small part is the same as the whole to the long part. Of course, if you do that with one side of a rectangle, the areas of the rectangles will have the same property.

EDIT: I just realized that as this would be true for any rectangle, it's also true for golden rectangles. And in fact, it's the reason that you get a square and a new golden rectangle when you cut it like that.

[Jeremy McCreary]

Not sure how you're dividing the square here, though. The usual golden ratio diagram divides a golden rectangle into a square and a smaller golden rectangle, but that wouldn't work here.

What you are talking about are rectangles in which the width and length have the golden ratio.
But the fundamental geometric definition of the golden ratio is for a segment: you divide it into two segments so that the ratio of the big part to the small part is the same as the whole to the long part. Of course, if you do that with one side of a rectangle, the areas of the rectangles will have the same property.

OK, then we're both dividing one side of the square with the segment approach. After normalizing that side to unit length, I ended up with β = 2 - φ = 0.382 for the shorter segment.

[ta0]

OK, then we're both dividing one side of the square with the segment approach. After normalizing that side to unit length, I ended up with β = 2 - φ = 0.382 for the shorter segment.

2 - φ = 1 - (φ -1) = 1 - 1 / φ = (φ - 1) /  φ = 1 / φ² 

If you divide a segment by the golden ratio, the ratio between the whole and the small part is φ² what can be seen directly from c/a = c/b b/a = φ φ

But what I'm saying is that if you divide any rectangle into the golden ratio, the ratio of densities should be φ² to have the stem at the borderline.
Of course, if you are doing that, better to use a golden rectangle and the larger side will be a square.

[Jeremy McCreary]

OK, then we're both dividing one side of the square with the segment approach. After normalizing that side to unit length, I ended up with β = 2 - φ = 0.382 for the shorter segment.

2 - φ = 1 - (φ -1) = 1 - 1 / φ = (φ - 1) /  φ = 1 / φ² 

If you divide a segment by the golden ratio, the ratio between the whole and the small part is φ² what can be seen directly from b/a c/b = φ φ

But what I'm saying is that if you divide any rectangle into the golden ratio, the ratio of densities should be φ² to have the stem at the borderline.
Of course, if your are doing that, better to use a golden rectangle and the larger side will be a square.

So are we agreeing?

[ta0]

So are we agreeing?

 

[Jeremy McCreary]

Maybe I've confused things a bit by defining my areal density ratio as δ = (areal density of larger, less dense rectangle) / (areal density of smaller, denser rectangle) so that δ < 1. The result  δ  =  β = 0.382 in the golden case then makes sense, as the smaller rectangle will then be 1 / δ =  φ² = φ + 1 = 2.618 times denser than the larger one.

[ortwin]

Again I had to ask my son for help and his LEGO.
We present a round LEGO offset top!

This should be incentive enough for Jeremy to show us that it can be done a lot better!
Some people even make metal LEGO parts. That is asking for some Iacopo Jeremy cooperation: molded lead LEGO parts from Italy!

[ta0]

@Jeremy: yes, as the golden ratio is defined as a ratio greater than one, it became a little confusing there. But your equations are all fine.

Nice circular offset LEGO top ortwin! 

[Jeremy McCreary]

Two LEGO-friendly square offset tops we'll call "White" and "Black" after the colors of their smaller, areally denser rectangles. You saw White above. Black is new.



Both tops are based on square LEGO plates S = 16 "studs" on a side. (1 stud = 8.0 mm.) 



White: The short side of the white rectangle is B = 4 studs, so β = B / S = 4/16 = 0.250. Since it's uniformly 3 plates thick, the top's areal density ratio (larger rectangle/smaller) is exactly δ = 1/3 = 0.333. The resulting CM location is exactly y_CM = 6 studs, well away from the boundary.

What's so remarkable about White is how perfectly balanced it is. So perfectly, in fact, that sometimes it forgets to fall. (No visible flat on small ball tip under high magnification, but surely there must be one.) White then takes about 120 s to come to a stop. But when it does fall, it lasts about 70 s. And no visible wobble whatsoever till the bitter end!

Black: This one was trickier. The short side of the black rectangle is B = 6 studs, so β = 6/16 = 0.375. Problem is, getting the CM exactly on the boundary (so that y_CM = 6 studs as well) called for a density ratio of δ = 0.360. Not so LEGO-friendly.



So I had to get creative. Despite the small couple unbalance introduced by this solution, Black spins very smoothly at all but the lowest speeds.

Black's probably the closest I'll ever get to a golden square offset top (δ = β = 0.382) in LEGO.

Note on White's tip: Switching White's red tip holder from the taller to shorter version below lowered the top's CM by 6 mm. This alone bumped spin time from 50 to 70 s (when it remembered to fall).



Tip design is one of the harder parts of LEGO topmaking. Any wiggle in the tip assembly will show up as wobble in the top, and suitable contact surfaces are few and far between.

[ta0]

It's amazing that these offset square tops made with the constraints to discrete steps of LEGO, can spin so smoothly.

There is something I don't understand. The golden ratio black top needs a density radio of 3.6. But on the photos it looks like the heavy side has three layers, with the bottom layer carved out, so I would expect it to be more like 2.6