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[ta0]

I love how you find all this elegant problems with nice mathematical solutions.
Inspired by you, keeping the center of mass on the small circle, I put it's center on the big circle. Guess what? The ratio of the two radiuses (radii) is the Golden Ratio! 

[ortwin]

So you say it is something like this what you get?




Could also make some nice piece of jewelry! Since the golden ratio is in there it would make sense to take golden rings.
As a tip for the stemless top I would choose one of those ruby balls that you see in EDC tops.

[ta0]

Yes, that is it.
I think this can be generalized to pairs of similar polygons with a parallel edge.

[ortwin]

Wow,  that seems like quite a generalization!
It reminds me also of that site with the many constructions of the golden ratio.
Did we mention here that the Golden Ratio is the most irrational number there is? In the sense that it can be least well be approximated by rationals.

[Jeremy McCreary]

Haven't done the math, but something like this should also be possible.



Dynamic balance could then be had by making the triangles coplanar.

[ta0]

Yes, that's one of the cases that I was referring to. If the barycenter of the smaller triangle is on the largest triangle (at the vertex), the ratio of dimensions between the triangles is the golden ratio.

For regular polygons with odd number of sides you need to mirror one and there are two ways of doing it.
In all these cases, if x is the ratio of sizes and 1 is the distance between the smaller polygon barycenter (centroid) and the overall center of mass, then you get the equation:
(x-1) x = 1

[ta0]

If I'm thinking well, it works with any figure that is symmetric with respect to the line that connects the two centroids. You have to mirror one of them, scale it by the golden ratio and place the smaller one so its centroid is on the larger figure. The centroid of the combined figure is on the corresponding point of the small figure.



Edit: I now realize this can be generalized to an arbitrary line figure. Also the smaller shape does not have to have the center of mass on top of the line of the larger figure. It suffices that the figures have a size ratio equal to the Golden Ratio and that they are mirrored (not rotated or shifted sideways). Then, the overall center of mass will be in the same position with respect to the smaller figure as the center of mass of the smaller figure is with respect to the larger figure.

[ortwin]

Hm, I have some doubts that we are talking about the same thing now:
In reply #90 and onward I was  not considering the area in side the circle, I only assigned some arbitrary weight per length of the circumference.That makes no difference for the position of the center of mass in this case, but the weight is different: it scales only with R instead of with R² if you consider the area.
I did not check any of your calculations ta0, but somehow I have the feeling that when you are thinking about polygons or other figures you think of them as if you would cut that shape from a metal sheet, not as formed from  a wire. Do you want to force me to think myself, or do you want to tell me of which type of polygon and shapes you were talking?

[ta0]

I'm thinking wire frames too. It doesn't otherwise.

[ortwin]

Thank you for not forcing me to think.

[Bill Wells]

... keeping the center of mass on the small circle, I put it's center on the big circle. Guess what? The ratio of the two radiuses (radii) is the Golden Ratio!

Since I just got to the party, could you post the equation for the above? I tried to prove this solution on AutoCad. Gave up.

Bill

[Bill Wells]

Now I see this equation from earlier reply:   "...if x is the ratio of sizes and 1 is the distance between the smaller polygon barycenter (centroid) and the overall center of mass, then you get the equation:
(x-1) x = 1"

I think centroid and center of mass are synonymous for a flat uniform plate? I am used to "barycenter" as used in astronomy, the center of mass of two distant celestial objects.

 

[ortwin]

...
Since I just got to the party,...

Good to see you at this party Bill!
Hopefully ta0 is sending you that equation you are looking for (maybe a snapshot from his notes?).
I was thinking since you have a lathe and all, maybe you could make a top like the one in reply #81 of this topic from wood or aluminum? I would go stemless, just a ball bearing ball as a tip. -----and the outer diameter two inches at the most, thickness maybe 3 mm. What do you think, doable?
I don't mean that you should make it for me, just for yourself and show it here on the forum. You wanted to practice posting a video anyways, right?

[ta0]

I think centroid and center of mass are synonymous for a flat uniform plate? I am used to "barycenter" as used in astronomy, the center of mass of two distant celestial objects.

Yes, for a uniform object they are the same thing. Bary-center literally means (from Greek) weight-center. When I was in school (in Spanish) we called the centroid of a triangle the barycenter.

Hopefully ta0 is sending you that equation you are looking for (maybe a snapshot from his notes?).

I added a couple of dimensions to the drawing above.
If you equate the lever of both figures, you get:
1 m = (x -1) m x
where m is the mass of the small line figure, and the ratio of dimensions is x. From there the equation x² - x - 1 = 0, which gives x equal to the Golden Ratio.

If you were attaching two planar figures, the equation would be:
1 m = (x -1) m x²
as ortwin pointed out. The solution would be x = 1.46557

[Jeremy McCreary]

An interesting twist arises when the polygon under consideration is a non-equilateral isoceles triangle. Then the CM of the wire frame never coincides with that of the filled triangle.

Consider an isoceles triangle of base length b, leg length a, and height h above the base. Let α = a / b > 1/2 by the triangle inequality.

For the filled triangle, the CM height is H = h / 3 above the base for all valid α. But for the wire frame, the CM height is S = 3 H α / (2 α +1).

Note that S > H when α > 1, and S < H when α < 1. Only in the excluded equilateral case (α = 1) does S = H.