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[ortwin]

Hm, if you don't care about any balancing and do very Q&D type documentation  one can be fast.



 

[ta0]

Nice, cuasi-Golden LEGO implementations!

I specially like that ortwin did his own homework! 

[Jeremy McCreary]

Hm, if you don't care about any balancing and do very Q&D type documentation  one can be fast.

Nice prototype!

The main thing I want out of a top is play value. And wobble mars play value for me. So if I were pursuing this double-diamond design, I'd have to see what I could do about the wobble. And that could take some time and effort.

[ortwin]

Yet another offset top that features  the Golden Ratio



This is basically the square version of the top in reply #81 .
Remaining question Nr.1: Can this way of making offset tops be generalized as the case with the wire frame that ta0 found to be possible for almost anything?
Remaining question Nr.2: What  would be an elegant prove for the fact that the position of the center of mass is where I put that red dot? I just did a straightforward calculation with the "lever rule" and was happy that the outcome was again the Golden Ratio, but a prove for that, that would be equally elegant as I think the resulting top is, would be different......

 


Not much couple unbalance to worry about. I think even reducing static unbalance could be done symmetrically  without introducing much couple unbalance.


Edit: This larger version can spin for over a minute on a mirror surface when started by hand:



The green piece is for balancing

[ta0]

Wow! Another apparition of the Golden Ratio! 
I checked your result, but it was not an elegant derivation. I need to think about your generalization question.

[Bill Wells]

I'll start by making a simple offset round top on my lathe. No golden ratio stuff for a while.

[ortwin]

I'll start by making a simple offset round top on my lathe. No golden ratio stuff for a while.

For which one do you intend to go?
The first in this topic?
The one with the hole from reply #81 ?
Ring in Ring from reply #90 ?
 I think I would start by making a circular hole close to the circumference of something like a large coin (it does not need to be a golden coin  ).
Hole to coin diameter ratio maybe 6/10 .
Then I would search for the center of mass by trying to balance that "holey coin" on some sort of needle.  CM should still be in the solid region of the coin since 6/10 makes a hole a bit smaller than the Golden Ratio.
Drill a little dimple in the coin at that CM spot, glue a ball bearing ball into the dimple.The circular stemless top with almost golden and optimum offset proportion is ready!The rim should be flat and wide enough so that it can stand on its edge. It will roll to a position on a flat table so that the hole is up.
If you give it a good spin with both hands as you would spin a regular coin,  the hole is moving downward during the spin - the reversing side of the top.

[ta0]

This is basically the square version of the top in reply #81 .
Remaining question Nr.1: Can this way of making offset tops be generalized as the case with the wire frame that ta0 found to be possible for almost anything?
Remaining question Nr.2: What  would be an elegant prove for the fact that the position of the center of mass is where I put that red dot? I just did a straightforward calculation with the "lever rule" and was happy that the outcome was again the Golden Ratio, but a prove for that, that would be equally elegant as I think the resulting top is, would be different......

[1] Maybe this can count. If you have a figure enclosed in a square, with it's centroid on the center of the square, and place stretched copies on two sides, the center of mass will fall on the corner of the square if the stretch factor is the Golden Ratio:



I don't even think the figure has to have circular symmetry: it could look different when stretched along the x and y axis. This is because what matters is that its center remains in the center of the rectangle and its area increases proportional to the stretch factor.

One problem for making a model is that the resulting center of mass can be outside the figure. You can fix this by replacing the original figure with one with the same area and also centered on the square, like I did on the right.

[2] I think the most direct derivation for the squares is to consider that the lever of the center of mass of the L shape and the one of the missing square should equate at the center of the big square. Mass small square = 1, Mass missing square = x², lever arm small square = (x-1), lever arm missing square = 1 (where I took the lever in the x direction to simplify). Equating everything you get the Golden Ratio equation.

[ortwin]

Quasi Golden Ratio LEGO top spinning for a minute plus

[Bill Wells]

>

For which one do you intend to go?

Ortwin, I am starting at step #1. It may seem simplistic, but I'm a simple person. And, I'm thinking of the manufacturing process, like densities of the wood species I have on hand; gluing; turning; etc. So I start with a round top with diameter 150mm and thickness about 20mm. Two semi-circles, one from a dense wood, the other from a lighter wood. Wood species yet undetermined, I will have to see what I can find in my shop.

Here is my in-elegant calculation for finding center of mass:

Find center of mass of composite disc, radius 75mm:
Moment about point “a”
(43.2 )(area A)(density A) + (106.8 )(area B)(density B) = ((area A)(density A) +  (area B)(density B))(X)
area A = area B

(43.2)(density A) + (106.8 )(density B) =(2) (density A +  density B)(X)
X=(43.2)(density A) + (106.8 )(density B))/ (2) (density A + density B)

NOTE: For some reason my .jpg image is not showing up on this reply. Any suggestions?

[the Earl of Whirl]

I do like those two offset tops by ortwin.  Very clever.

[ta0]

Quasi Golden Ratio LEGO top spinning for a minute plus

Worked very well. A keeper for the golden collection!

[ta0]

Ortwin, I found the generalization you were looking for. It covers the circular top with a circular hole of reply#80 and the square with a square hole of reply#123, and much more.

Golden Hole rule:
"If on a figure of arbitrary shape with area A you make a hole of arbitrary shape with area A/φ², the center of mass of the original figure will cut the segment between the new center of mass and the mass of the cutout by the Golden Ratio."

Proof
Equating levers of the two cutout figures at the center of mass of the original figure:
[A-A/x²] a =  A/x² ax => x²-x-1=0 => x=φ

[Jeremy McCreary]

Golden Hole rule:
"If on a figure of arbitrary shape with area A you make a hole of arbitrary shape with area A/φ², the center of mass of the original figure will cut the segment between the new center of mass and the mass of the cutout by the Golden Ratio."

Proof
Equating levers of the two cutout figures at the center of mass of the original figure:
[A-A/x²] a =  A/x² ax => x²-x-1=0 => x=φ

Trying to understand this very interesting conjecture and proof.

Q1: Could you elaborate on how a and x are defined, maybe with a diagram?

Q2: No symmetry requirements? Up till now, all figures and holes have had their CMs on a shared mirror plane.

[ortwin]

Ortwin, I found the generalization you were looking for. ....

EXCELLENT ta0 !!!
Not only a very broad generalization but also a most elegant proof that answers my second question at the same time!

I do not think I understand the proof 200% as would be necessary to explain a proof, but I try to say what I see:
 "[A-A/x²] a =  A/x² ax" The left side is the mass of the figure left after you cut a hole into it times the distance "a" that its CM is apart from the original CM. ("x" I would call the linear scaling factor.) 
The right side  is the mass of the figure you cut out, times the distance that its CM is apart from the original CM. Choosing that distance to be "ax" is the trick I believe. This basically secures for simple (convex?) geometries (and the case where the hole has the same shape and orientation as the original shape and the outside border of the original shape coincides with the border of the hole) that the new CM lies on the boundary where the cut is, or so I "feel".
Feelings are not good enough in math though, maybe it helped somehow anyway.