Print

Please login or register.
1 Hour 1 Day 1 Week 1 Month Forever
Login with username, password and session length

[Jeremy McCreary]

Anyway, it's now proven by assuming the symmetry of the diameter and not the golden ratio. 

Could you elaborate on this symmetry? Not quite sure how it's defined or what it preserves.

[ta0]

Anyway, it's now proven by assuming the symmetry of the diameter and not the golden ratio. 

Could you elaborate on this symmetry? Not quite sure how it's defined or what it preserves.

Just saying that the centroid is in the middle of the segment between the point of contact and the opposite side of the figure. The golden ratio scaling then appears as a condition for the new centroid to be on the edge (i.e., on the end of the segment.)

Some observations from looking for golden ratio tops:
From a square you can create an infinite number of different tops: any position of the scaled square along the edge would work.
Only stars with an even number of points can be made golden.
Any triangle can be carved in 3 different ways, by placing the scaled triangle on each side in precise locations. But none of the resulting shapes has symmetry (non even for the equilateral triangle).

I tried to make a series of fish eating each other, but it was more difficult than I expected. I'm not very happy with what I have obtained so far:



The cool thing is that all the golden fish tops can be stacked to form a complete fish. Something like a Matryoshka doll.

[ortwin]

...
Only stars with an even number of points can be made golden.
..

Why is that? In any "normal" shape there is a line through the centroid  in which the centroid sits in the middle with respect to the edges. Do the points of the "hole star" cut through the the edges of the original star?

[ta0]

...
Only stars with an even number of points can be made golden.
..

Why is that? In any "normal" shape there is a line through the centroid  in which the centroid sits in the middle with respect to the edges. Do the points of the "hole star" cut through the the edges of the original star?

Yes, it seems to me that the some parts of the small star will always fall outside the large star, for an odd number of points. But I haven't proven it rigorously.

[ortwin]

Yes there seems to be a problem. In my five star example I just made, not by calculating but just by eye judgement and some functions of a drawing program, the large shape is  cut by the hole into two separate parts. That is probably why I felt like babbling something about "convex outline" some posts above.


Edit: looking at this, the colors remind me again of your avatar. Why are you not going for that? The shadow theme is also self referential, kind of ...

[ta0]

Ortwin: You need to move the star further along the segment with the centroid at the center, until it touches the other star. But then they will be crossing each other.

I'll look at the logo. But you give tough homework. 

[Jeremy McCreary]

Just saying that the centroid is in the middle of the segment between the point of contact and the opposite side of the figure. The golden ratio scaling then appears as a condition for the new centroid to be on the edge (i.e., on the end of the segment.)

So the golden ratio emerges when...
a. Three special points are colinear.
b. The final CM  falls on the inner edge of the hole.

Sorry I'm not getting this. Which 3 special points again? Where are they in the diagram from Reply #152 below...

On this diagram the original figure and the hole have the same shape and orientation and are in the Golden Ratio proportion (φ):



The hole also touches the edge of the original figure, in the direction the figure was displaced, but is otherwise completely enclosed.

C_H = centroid of hole
C_T = centroid of total figure
C_C = centroid of carved figure

[ta0]

Sorry I'm not getting this. Which 3 special points again? Where are they in the diagram from Reply #152 below...

Sorry, I should have drawn the complete segment of interest and indicated all the points on that diagram. Here it is again:



We are interested in making C_c coincide with Q'. The conclusion was that C_T has to be in the middle of QP (i.e. C_H in the middle of Q'P') and the scaling k has to be the Golden Ratio. From the levers you first find that y = k²-1. Then if you assume x = 2 b, you get k = φ, or alternatively, if you assume k = φ you get x = 2 b.

[Jeremy McCreary]

Thanks! Now I get it. Very nice!

To form the diagram in Reply #152, you might...
Step 1. Copy the outline of the blank (original shape, no hole)
Step 2. Scale down the copy by area so that A_copy = A_blank / k² , k > 1, while keeping the 2 centroids surperimposed.
Step 3. Translate the copy outward to kiss the blank's outline without rotation.
Step 4. Cut out the copy to form the hole.

Guessing that your derivation also needs the lack of rotation in Step 3. Is it also a requirement that corresponding points of the blank and copy kiss?

[ta0]

Yes, those are the steps. If you want the centroid to end up on the edge, the direction you select has to have the centroid in the center and should use k = φ.
Yes, no rotation. As you are moving the smaller one in a radial direction it will automatically kiss the other one in the corresponding place. This is critical for this construction and derivation.   This does not mean that it's impossible to find other cases that do not follow this construction.

[Bill Wells]

When not inverted, right? (Resting vertically on the body, not the stem.)

Yes, when not inverted. I have seen a tippe top invert and almost come to rest balanced upon the flat stem. That would be cool.

[ortwin]

Ortwin: You need to move the star further along the segment with the centroid at the center, until it touches the other star. But then they will be crossing each other.

I'll look at the logo. But you give tough homework. 

I don't give homework, only ideas you can work on- if you want- in your spare time.

[Jeremy McCreary]

When not inverted, right? (Resting vertically on the body, not the stem.)

Yes, when not inverted. I have seen a tippe top invert and almost come to rest balanced upon the flat stem. That would be cool.

I've found that it doesn't take much of a flat if the top's balanced well enough. The white/gray offset top at left (first up in the video) often comes to rest without falling. The less well-balanced one on the right never does.



It takes a 14x loupe to see the flats (tiny mold scars) on their identical tips! See Reply #28 for details.

[ortwin]

Ortwin: You need to move the star further along the segment with the centroid at the center, until it touches the other star. But then they will be crossing each other....

of course you are right again, I just stopped when the first thing  touched the outer edge. This should be a little bit closer to what you would get doing it the correct way:

 

[ortwin]

When not inverted, right? (Resting vertically on the body, not the stem.)

Yes, when not inverted. I have seen a tippe top invert and almost come to rest balanced upon the flat stem. That would be cool.

Maybe something like the balls wee discussed here would have a good chance to remain balanced on that hole when inverted? - In the not multiple flip version of course.
But then again an inverted ball is not as  impressive as a top with stem that reverses and stays balanced on that stem even without spinning. 

How about your flipping disks you mention here Jeremy? Any chance they stay balanced after inversion?