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Over in the thread "Evenly distributing n points on a sphere" this topic is touched upon: https://stackoverflow.com/questions/9600801/evenly-distributing-n-points-on-a-sphere.

But what I would like to know is: "Is the Fibonacci lattice the very best way to evenly distribute N points on a sphere? So far it seems that it is the best. Does anyone know of a better method?"

I have a Ph.D. in physics and may have an application for some of this research in physics.

I came across this wonderful paper:

http://arxiv.org/pdf/0912.4540.pdf "Measurement of areas on a sphere using Fibonacci and latitude–longitude lattices"

The paper states, "The Fibonacci lattice is a particularly appealing alternative [15, 16, 17, 23, 65, 42, 66, 67, 68, 76, 52, 28, 56, 55]. Being easy to construct, it can have any odd number of points [68], and these are evenly distributed (Fig. 1) with each point representing almost the same area. For the numerical integration of continuous functions on a sphere, it has distinct advantages over other lattices [28, 56]."

It the Fibonacci lattice the very best way to distribute N points on a sphere so that they are evenly distributed? Is there any way that is better?

As seen above, the paper states, "with each point representing almost the same area. "

Is it impossible, in principle (except for special rare cases of N such as 4, etc.), to exactly evenly distribute N points on a sphere so that each point/region has the exact same are?

So far it seems to me that the Fibonacci lattice the very best way to distribute N points on a sphere so that they are evenly distributed. Do you feel this to be correct?

Thanks so much!

Community
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Physics Ph.D.
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  • Technically speaking, it is impossible to equidistribute points on a sphere unless they are the corners of a Platonic solid, so N=4,6,8,12,20. So, for any N's other than this, what is the best way to equidistribute N points on a sphere? So far it seems that the Fibonacci lattice the very best way to distribute N points on a sphere so that they are evenly distributed. Does anyone know of a better method to distribute points equally on a sphere? Thanks! – Physics Ph.D. Jul 12 '15 at 15:48
  • I should note that I originally asked the question here: http://stackoverflow.com/questions/31363178/is-the-fibonacci-lattice-the-very-best-way-to-evenly-distribute-n-points-on-a-sp Also, I have emailed some experts, and I will let everyone know what I find out! :) – Physics Ph.D. Jul 12 '15 at 17:45
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    More cool information found!

    I came across an epic page here comparing some methods visually: http://bendwavy.org/pack/pack.htm

    The middle column, representing "a golden section of the circle," seems to be the most symmetric under rotation?

    For instance Rusin's and Saff & Kuijlaars methods seem to have poles, so one would be able to note the rotation of the spheres.

    Having noted that, would it be logical to say that the center method utilizing the golden section (the Fibonacci lattice) provides the best way to symmetrically distribute N points on a sphere in an equidistant manner?

     – Physics Ph.D. Jul 12 '15 at 18:49
  • (bendwavy.org c'est moi.) The golden sector spiral also has poles, though they're perhaps less obvious to the eye. – Anton Sherwood Nov 03 '15 at 09:20
  • @PhysicsPh.D. "Technically speaking, it is impossible to equidistribute points on a sphere unless they are the corners of a Platonic solid, so N=4,6,8,12,20." -- I have two counterexamples: N=2 (antipodes) and N=3 (equilateral triangle on a plane passing through the sphere's center). Whether a triangular bipyramid counts as a solution for N=5 or not will depend on how you define "equidistributed". – Travis Reed May 20 '19 at 17:58
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    The N=2,3 are correct, but obvious ones. The reasons he forgot to mention them, is that they don't form platonic solids. They don't form solids at all. The Triangular bipyramid lacks the symmetry we seek, since the three middle points are 120 degrees apart but the pole ones are 90 degrees apart from nearest neighbors. Hence the poles are not the same as the middle ones. – K. Sadri Jul 19 '19 at 08:19

1 Answers1

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The Fibonacci lattice is not the best way to evenly distribute points on a sphere. The problem of distributing N points evenly on a unit sphere is only known for specific N.

Moreover, the vertices of Platonic solids are not always optimal. This is succinctly described on the Wolfram Mathworld site:

“For two points, the points should be at opposite ends of a diameter. For four points, they should be placed at the polyhedron vertices of an inscribed regular tetrahedron. There is no unique best solution for five points since the distance cannot be reduced below that for six points. For six points, they should be placed at the polyhedron vertices of an inscribed regular octahedron. For seven points, the best solution is four equilateral spherical triangles with angles of 80 degrees. For eight points, the best dispersal is not the polyhedron vertices of the inscribed cube, but of a square antiprism with equal polyhedron edges. The solution for nine points is eight equilateral spherical triangles with angles of arcos(1/4). For 12 points, the solution is an inscribed regular icosahedron.”

There are many approximate solvers for this ( SpherePoints[] and Offset Lattice ). The Fibonacci Spiral is easy to program, but is not optimal.

Aaron T. Becker
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    "For eight points, the best dispersal is not the polyhedron vertices of the inscribed cube" This is mind blowing, how could an inscribed cube fail to sample the sphere uniformly. – Gappy Hilmore May 20 '21 at 20:11
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    The cube does sample the sphere uniformly, but there are huge gaps in all those square faces. A square antiprism is just as uniform (all vertices symmetrically equivalent), and has fewer gaps with only two square faces instead of six; and the square antiprism turns out to be the true eight-point optimum. – Oscar Lanzi Aug 13 '22 at 00:45
  • This is one of those things (referring to the antiprism being a better distribution by some metrics than the cube) that I imagine would blow the minds of any hypothetical Flatland mathematicians. – Brian Tung Nov 21 '24 at 23:52