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[ta0]

I agree that proof was not clear. It was more a case of "if it's the Golden Ratio it works", than deriving the Golden Ratio.
Let me try again with an even more general result.

1. The center of mass (centroid) of the full figure has to be in the segment between the center of mass of the hollowed figure and the center of mass of the hole.
2. For the calculations you can consider the masses (i.e. areas) of the figures concentrated in their centroids, so they can be treated as point masses. Therefore, the shape and orientation of the figures doesn't matter.
3. Let the area of the hole be A/k², where A is the area of the full figure.
4. Let the distance between the centroid of the hollowed figure and the centroid of the full figure be: a
5. Let the distance between the centroid of the hole and the centroid of the full figure be: a x
Equating the levers at the centroid of the full figure:

(A-A/k²) a = A/k² a x => k² - 1 = x

So, for an arbitrary hole, the new centroid will be in line with the centroid of the original figure and the centroid of the hole, on the opposite side of the later, at a distance closer by a factor of 1/x = 1/(k²-1)

6. If k = φ (Golden Ration), x = φ² - 1 = φ

[Jeremy McCreary]

Well done! Off to find some LEGO-friendly figure and hole shapes that will really blow people's minds when the corresponding offset top spins without wobble.

[ta0]

Two clarifications:

1. This doesn't prove that the center of mass will fall on the edge of the new figure. I'm thinking about that condition.
2. Correction: this does not cover the case of reply #123 (the L shape) as in that case the hole area is φ²/(1+φ)² of the full area.

EDIT: After thinking more about it, although this is an interesting result, it is not the generalization that ortwin was looking for.   

[Jeremy McCreary]

2. Correction: this does not cover the case of reply #123 (the L shape) as in that case the hole area is φ²/(1+φ)² of the full area.

But 1+φ = φ², so A_hole / A_original = φ² / (1+φ)² = φ² / φ⁴ = 1 / φ².

Why then is Reply #123 is not covered?

[ta0]

2. Correction: this does not cover the case of reply #123 (the L shape) as in that case the hole area is φ²/(1+φ)² of the full area.

But 1+φ = φ², so A_hole / A_original = φ² / (1+φ)² = φ² / φ⁴ = 1 / φ².

Why then is Reply #123 is not covered?

Ha! You are right!    It's covered!  Good!

[ta0]

Well, I'm getting that if the figure has some symmetry and you scale it by 1/φ (without rotating) to make a hole touching on one edge, the center of mass will in fact be on the edge of the new figure. But cannot work more on this until tonight.

[Jeremy McCreary]

I love how you [ortwin] find all this elegant problems with nice mathematical solutions.

Just want to second that. For us geeks who like our tops with a large side of math and physics, this whole offset top thing has been great entertainment. The surprisingly pervasive golden ratio connection is just icing on the cake.

Also great show biz. Nothing like showing someone one of my tops with a totally unexpected feature at speed and seeing "Whoa, that's not right!" written all over their face. Color-mixing, blacklight, centrifugal, and puzzle tops are especially good for that.

Now I can add offset tops to the list. Everyone has an inner physicist. And when that physicist sees an offset top and expects a lot of wobble but sees none, that's a "Whoa, that's not right!" moment.

[Jeremy McCreary]

I can't add images to my posts anymore 
Have been trying to attach a photo to my last reply but to no avail.
Any suggestions? It started after I made some changes to my profile.

OK, I'm off to the shop to start making my first offset top.

Use the code tag to show us the code that failed.

[ta0]

I can see your image fine. 

[Jeremy McCreary]

Sorry, the code tag looks like this...

[X]paste your failed image code here[/X]

where "X" is the word "code". This tag just deactivates any code inside so readers can see the code text itself instead of what it does (or doesn't) do. Handy for troubleshooting.

[Bill Wells]

The code tag doesn't transmit. Tried [X]paste your failed image code here[/X]

So you are all seeing my photo attachments but I cannot?

[ortwin]

Well done! Off to find some LEGO-friendly figure and hole shapes that will really blow people's minds when the corresponding offset top spins without wobble.

I thought my stealth bomber from reply #124 does just that!   

[ta0]

The code tag doesn't transmit. Tried [X]paste your failed image code here[/X]

So you are all seeing my photo attachments but I cannot?

I see it both on the computer and on the phone. Can you see other people's images?

[ortwin]

Well, I'm getting that if the figure has some symmetry and you scale it by 1/φ (without rotating) to make a hole touching on one edge, the center of mass will in fact be on the edge of the new figure. But cannot work more on this until tonight.

Yes something along those lines I meant in reply #135 when I said :Choosing that distance to be "ax" is the trick I believe.
Thereby you put the condition in your formula that the lever lengths on the two sides of the original CM should have the same ratio as your scaling factor x.  Somehow you scale a shape by a factor x and push it to one edge and then you look into the edge of the old shape minus the new shape.
Somehow that reminds me strongly of 1/φ = φ - 1 
which is of course a defining formula for the Golden Ratio again. But I am not quite there yet......

[Bill Wells]

Now that my image attachment issue is resolved, I will revise and attach new diagram and centroid calculations. I have made the composite disc, half dense wood, the other half much lighter wood; waiting for glue to dry. Dimensions have changed. A work in progress.