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We know that regular hexagons can tile the plane but not in a self-similar fashion. However we can construct a fractal known as a Gosper island, that has the same area as the hexagon but has the property that when surrounded by 6 identical copies produces a similar shape, but with dimensions scaled by a factor of $\sqrt{7}$.

What is the distance between two of the centers? Is it the same as the distance between hexagons of the same area? ie. If I start with a hexagon of area A, then construct a Gosper island and place it next to an identical copy, would the distance still be the same as if they were hexagons? Or does the scaling factor come into play somewhere? Right now I think the answer is $\sqrt{3}/2$, as for the hexagon.

The reason I ask is that I'm trying to calculate the Gosper island's moment of inertia through an axis through its centre of mass and perpendicular to the plane of the island.

If we assume that the moment of inertia is always proportional to the mass, and proportional to the square of a characteristic length scale, then $$ I = \gamma Ml^2, $$ where $\gamma$ is a constant, $l$ is the 'diameter' of the island, in a hexagon this would be the distance between two opposite vertices. Shrink the Gosper island by the scaling factor and surround it by six others. This self-similarity technique is super cute, and can be used to calculate the moment of inertia of an equilateral triangle, and can be extended to a square/rectangle quite easily. Fractals, having a high degree of self-similarity, seem amenable to this technique - here I calculate the moment of inertia for a Koch snowflake.

$\hspace{1.3cm}$ Scaling and tesselating

Using the principle of superposition, $$ I = I_{\text{centre}} + 6I_{\text{edge}}, $$ where $$ I_{\text{centre}} =\gamma \frac{M}{7}\left(\frac{l}{\sqrt{7}}\right)^2 = \gamma \frac{Ml^2}{49} = \frac{I}{49}. $$

Now, by the parallel axis theorem $\displaystyle I_{\text{edge}} = I_{\text{COM}} + Md^2$ where $$ \displaystyle I_{\text{COM}} = \frac{I}{49} $$ and $\displaystyle d= \frac{\sqrt{3} l}{2} $ (this was one source of error), so $\displaystyle I_{\text{edge}} = \frac{I}{49} + \frac{3Ml^2}{4},$ and \begin{align*} I &= \frac{I}{49} + 6\left(\frac{I}{49}+ \frac{3Ml^2}{4}\right), \\ I & = \frac{I}{7} + \frac{9Ml^2}{2}, \\ \frac{6I}{7} & = \frac{9Ml^2}{2}, \\ I & = \frac{21Ml^2}{4}. \end{align*}

This seems incorrect? It feels wrong, comparing to a disk of radius $l/2$ which has moment of inertia $Ml^2/4$ it seems far too large.

It would also be nice if we could verify our answer numerically or otherwise. Any references are also appreciated.

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    $\begingroup$ The center of mass of the Gosper island is indeed the same as the center of the initial hexagon. This, as well as the moment of inertia can be computed using the technique of self-similar integration as in my answer to this question. I could provide a few more details but I get the feeling this might be an exercise? $\endgroup$ – Mark McClure May 17 '16 at 13:18
  • $\begingroup$ Nope, just something I thought up after teaching the self-similarity method for finding the MOI of a triangle/square. Thankyou so much for the link! I suppose this can be done for an fractal with enough rotational symmetry to tile the plane? Please provide as many details as you wish, in an answer if you like. $\endgroup$ – Bennett Gardiner May 18 '16 at 2:05
  • $\begingroup$ By the way, the polar inertia of a disk is not $ml^2/4$ (That is the x or y inertia) but $Ml^2/2$. $\endgroup$ – Urukann May 20 '16 at 4:40
  • $\begingroup$ Isn't that only if $l$ is the radius though? In my formulation I was trying to use the diagonal of the hexagon, so it made more sense to compare to a disk with $l$ as the diameter. $\endgroup$ – Bennett Gardiner May 20 '16 at 6:19
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    $\begingroup$ @BennettGardiner: IMO this is a source of confusion (especially if you don't explain it). The standard formula is $I=MR^2$ where $R$ is an equivalent radius (AKA gyration). I found it more effective to work with unit area shapes, so that the mass can be simplified, and $R$ conveys all shape information. $\endgroup$ – Yves Daoust May 20 '16 at 14:40
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I think that your $\frac{\sqrt{3}}{2}$ for the distance between hexagons center is not right... The distance from an hexagon center to an edge is $\frac{r\sqrt{3}}{2}$, thus the distance between two hexagons of radius $\frac{l}{2\sqrt{7}}$ that touch by an edge should be: $$d = l\sqrt{\frac{3}{28}}$$

Now, if we take back your computation: $I = \frac{I}{49} + 6 \left(\frac{I}{49} + \frac{3Ml^2}{28}\right)$

This gives us : $I = \frac{3}{4} M l^2$, which may still not be the right answer.

It would be nice to compare this value to numerical methods, I'll look into it.

e/ I ran some numerical simulation, unfortunately it does not seem to confirm the above result.

The code.

How I did it:

  • Starting from the code linked in this page, I generated the all points composing the outer shape of the gosper island.
  • Then, I trimmed the generating points to remove duplicates
  • I approximate the Gosper island as being star-shaped (This is not perfectly true, at least for $n=5$). Then its inertia is the sum of the inertia of the triangles composed of the origin and two consecutive points. Note that to get the actual inertia, one has to divide by the shape's area (Because when evaluating the triangle's inertia, we consider it has a surfacic mass of 1.). All triangle related formulas are available on Wolfram Alpha.

The results show that indeed the inertia is proportional to $l^2$, but its ratio to the value conjectured above is not 1, but a constant close to $\frac{4}{7}$: $$ I_{gosper} \approx 0.568582263418 \frac{3}{4} M l^2$$

Unfortunately, it seems that this error is not related to the star-shaped approximation: I ran another experiment, this time using the Seidel program from the University of North Carolina at Chapel Hill. It allowed me to find a triangulation of the inner area of the Gosper island. Using another (similar) code I could check that the computation for a radius of 1. does yield the same ration between expected inertia (0.75) and the actual inertia (0.42856647032), with a similar ratio of 0.571421960426. Note that this inertia is very close to $\frac{3}{7}$, (best fractional approximation with a denominator inferior to fifteen thousand).

Actually, I had forgotten that the characteristic dimension is not the diameter, but the radius, thus the ratio is 0.142855490107, very close to $\frac{1}{7}$.

Using this method for a Koch snowflake yields pretty correct results: for a snowflake of radius $r \approx 1.44$ (I forgot to scale the step size properly) I get an inertia of $I_{koch} \approx 0.736036125705$ while the one given by $Ml^2/11 \approx 0.7435999999999999$

e/ I found the error: The mass of a "small" Gosper island is not $M$, but $\frac{M}{7}$, thus the missing factor 7. This is due to the fact that we make the assumption of a uniform density Gosper Island, thus its mass is proportional to its area.

We can rewrite our original equation: $$ I_{edge} = I_{center} + \frac{M}{7}d^2\\ d = l \sqrt{\frac{3}{28}}$$

Which gives us: $$ I = \frac{I}{49} + 6 \left( \frac{I}{49} + \frac{3Ml^2}{7\cdot 28}\right)$$

And finally:

$$I = \frac{3Ml^2}{28}$$

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  • $\begingroup$ Bah, forgot to scale the distance $d$. Thanks, this sounds much more reasonable. If you could confirm using numerical methods with some explanation of how you did it, it would put it to rest and I'd definitely accept/award the bounty. $\endgroup$ – Bennett Gardiner May 19 '16 at 12:42
  • $\begingroup$ I just added some numerical simulation. However, it does not confirm my above conjecture. It may be because my code is quite dirty, but it may also be an error in the reasoning... Cheers $\endgroup$ – Urukann May 20 '16 at 4:37
  • $\begingroup$ Interesting, thanks for your efforts, looks like we still have some way to go. Can you use the code to check the Koch snowflake case I linked? It code provide a check on the code For the snowflake, $I = Ml^2/11$ where $l$ is the largest 'diameter' of the snowflake (tip to tip) I really felt like this method should work, I wonder where the error lies! $\endgroup$ – Bennett Gardiner May 20 '16 at 6:22
  • $\begingroup$ Hum, I checked with the Koch snowflake (my code was easier to modify than expected !): I get "correct" results. I noticed that I forgot to use the diameter instead of the radius, so we are off by another factor 4 I'm guessing... $\endgroup$ – Urukann May 20 '16 at 6:44
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    $\begingroup$ Sorry, this is getting confusing, but no: I calculated the MOI for a radius 1 Gosper Island to be 0.42856647032, that is to say, $I = \frac{3Mr^2}{7} = \frac{3Md^2}{7 \cdot 4}$, meaning that we are missing a factor $\frac{1}{7}$. $\endgroup$ – Urukann May 20 '16 at 6:58
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Yes, the distance between those hexagon-like shapes in a hexagonal lattice is the same as the distance between centres of hexagons of the same area in a hexagonal lattice.

The reason is that "in the long run" the area per mesh must be the same. That is, a sufficiently large circular disk contains both Gosper islands and hexagons in a number approximately equal to area of the disk dividied by area of the shape (with an error in the order of magnitude proportional to the cirumference of the disk).

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The area inside the hexagon is made of one whole island and exactly six thirds, by symmetry.

enter image description here

Hence for an island of unit area, the distance between centers is such that

$$3\frac{\sqrt3}2d^2=3,$$

$$d=\sqrt{\frac2{\sqrt3}}.$$


To compute the moment of inertia, we will work with the radius of gyration and for an island of unit area.

$$I=MR^2$$ where $M$ is the mass.

For the assembly in the figure, enlarging by the factor $\sqrt7$ and using the axis theorem,

$$7M7R^2=49I=I+6(I+Md^2),$$

giving

$$R=\frac d{\sqrt7}=\sqrt{\frac{2}{7\sqrt3}}=0.406149258\cdots$$

This compares to the radius of gyration of a unit area disk,

$$R'=\frac1{\sqrt{2\pi}}=0.398942280\cdots$$

and that of a unit area hexagon,

$$R''=\sqrt{\frac{10}{36\sqrt3}}=0.400468569\cdots$$

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  • $\begingroup$ Interesting, so to scale this to an island with diameter $l$, would we simply scale this distance by $l$? $\endgroup$ – Bennett Gardiner May 20 '16 at 6:57
  • $\begingroup$ Or $\sqrt{3/28}l$, like in the other poster's answer? I'm afraid I've confused myself quite a bit with the length scales. $\endgroup$ – Bennett Gardiner May 20 '16 at 7:00
  • $\begingroup$ @BennettGardiner: I have no idea of the value of the diameter $l$, but you don't need it for the computation of the moment. $\endgroup$ – Yves Daoust May 20 '16 at 7:03
  • $\begingroup$ I'm still trying to figure out your distance $d$. If that is the radius of gyration, what would the moment of inertia be? $\endgroup$ – Bennett Gardiner May 20 '16 at 7:32
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    $\begingroup$ My confusion is part of why I asked the question. I am inclined to believe your answer more now that you have posted a comparison with the hexagon of unit area, in my mind they should have been very similar. $\endgroup$ – Bennett Gardiner May 20 '16 at 14:18

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