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Let $\gamma=\gamma(L)$ be a simple (non-self-intersecting) closed curve of length $L$ on the unit-radius sphere $S$. So if $L=2\pi$, $\gamma$ could be a great circle.

I am seeking the most equitably distributed $\gamma(L)$, distributed in the sense that the length of $\gamma$ within any disk is minimized. This is something like placing repelling electrons on a sphere, but here the curve self-repels. So there should be no "clots" of $\gamma$ anywhere on $S$. I am especially interested in large $L$. A possible $\gamma$ is shown below, surely not optimal for its length:


           


Here is an attempt to capture more formally "equitably distributed." I find this an awkward definition, and perhaps there is a more natural definition.

Around a point $c \in S$, measure the $r$-density of $\gamma$ as the total length within an $r$-disk: $$d_\gamma(c,r) = | \gamma \cap D(c,r)|$$ where $D(c,r)$ is the disk of geodesic radius $r$ centered on $c$. Then define $d_\gamma(r)$ as the maximum of $d_\gamma(c,r)$ over all $c \in S$.

Finally, we can say that, for two curves $\gamma_1$ and $\gamma_2$ of the same length $L$, that $\gamma_1 \le \gamma_2$ if $d_{\gamma_1}(r) \le d_{\gamma_2}(r)$ for all $r \in (0,\pi)$, i.e., $\gamma_1$ is less concentrated than $\gamma_2$ for all $r$ up to a hemisphere.

This definition provides a partial order on curves of a given length $L$. One version of my question is:

Q. What do the minimal elements of this poset look like, especially as $L$ gets large?

These minimal curves are in some sense nowhere densely clotted.


Update. Acknowledging Gerhard Paseman's remark, I thought I would include this attractive image of a space-filling curve on a sphere:
  HilbertCurve
  (Image from this website).
But notice it is certainly not equidistributed in any sense, crowding near the northpole.

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    $\begingroup$ I imagine something like a hamiltonian circuit on a dodecahedron, extended to hamiltonians on a planar graph with metric borrowed from a stereographic projection. Perhaps what you want for large L are the inverse images ( under s. p. ) of approximations to space filling curves inside large disks. Gerhard "Such Thoughts Fill My Mind" Paseman, 2014.11.26 $\endgroup$ Nov 26, 2014 at 21:11
  • $\begingroup$ Why must you have a great circle if $L = 2 \pi$? $\endgroup$ Nov 26, 2014 at 21:43
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    $\begingroup$ Also, this appears to be in a similar (though also clearly distinct) vein to mathoverflow.net/questions/26212 $\endgroup$ Nov 26, 2014 at 21:46
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    $\begingroup$ Erm... Since your minimizer should satisfy continuum comparisons with each other single curve, why are you so sure it exists at all? $\endgroup$
    – fedja
    Nov 26, 2014 at 23:00
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    $\begingroup$ about existence, not worried, see en.wikipedia.org/wiki/Knot_energy We can replace the ambient distance by geodesic distance. Likely the winner for your picture length is the laces on a baseball; i would expect to be able to prove smoothness of the optimizer. As usual, Rob Kusner has written about closely related things. $\endgroup$
    – Will Jagy
    Nov 27, 2014 at 1:38

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I suppose one heuristic would be to find a shortest tour through uniformly distributed points on a sphere. The following image applies Mathematica's FindShortestTour command to $10000$ points generated by your own sphere command from Computational Geometry in C.

enter image description here

We could also use more regularly distributed points to obtain a path that (I suspect) has better local properties. This image uses 5000 points generated by the algorithm described here.

enter image description here

I don't know that I quite follow your $\gamma(L)$ function, but it might be something like an inverse of the function obtained by plotting the length $L$ as a function of the number of points input to this procedure.

enter image description here

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    $\begingroup$ I love that image of Markus Deserno's algorithm---Thanks! $\endgroup$ Dec 10, 2014 at 15:16
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This is from "Population coding under normalization":

Numerical solutions of a necklace of electrons

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  • $\begingroup$ Ooops... The solution to $\alpha = 1.6$ corresponds to $\alpha = 2.0$ and vice-versa. This was done by starting with a circle with $\alpha = 0.8$, incrementing the length a little bit, and letting the solution evolve, and repeating the cycle. $\endgroup$ Mar 13, 2021 at 15:38
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to me the best possible solution seems to be the sequence of closed Hilbert curves on the Cubed Sphere;
that curve consists of six ordinary Hilbert Curves, one for each face of a cube, which, when appropriately connected, yield a space-filling Jordan Curve for the cube's surface.
Then centrally projecting those curves onto a cocentric sphere yields a solution without crowding, albeit a slightly higher density around the projections of the cube's corners.

A further improvement may result from rounding away the corners in a similar fashion as is used for tennis balls, which is the rounding applied to your example.

Yet another, maybe even better, bet would be Hamilton Cycles of Hamiltonian Fullerene Graphs (cf e.g. http://arxiv.org/pdf/0801.3854.pdf for a discussion); that bet is based on the honeycomb conjecture (cf e.g. http://en.wikipedia.org/wiki/Honeycomb_conjecture) and on the fact, that Fullerene graphs are the closest one can get to a hexagonal tiling of the sphere (there are twelve inevitable pentagons).
Examples of Hamiltonian Fullerene graphs are depicted here: http://www.academia.edu/4269519/On_the_Hamiltonicity_of_Fullerenes.

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  • $\begingroup$ I especially like your Fullerine graphs idea---Thanks! $\endgroup$ Nov 27, 2014 at 12:20
  • $\begingroup$ The curve obtained by glueing together and then deforming 6 Hilbert curves is also algorithmically described and pictured at s2geometry.io/devguide/s2cell_hierarchy. $\endgroup$
    – Alex M.
    Jun 27, 2021 at 12:01

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