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If I am given a set of N congruent circles with radii R, where the number of circles allows them to be placed on a sphere so the distance between the centers of every pair of different circles in the set is separated by the same distance, can I figure out if the sphere's covered?

Hopefully without requiring access to a supercomputer... (Didn't think that'd be a possibility, but I'm working with 'large' numbers...)

I previously asked: "What is the minimum radius of N congruent circles that are placed on a sphere equidistant from each other, so that the sphere is covered by circles?"

That was intended to produce a general answer to a generalization of the question: "How wide a radius of a field do I need 10,000 field emitters to emit, so the total field covers the Earth", but since the previous question can't produce the answer I wanted, I'm making this question.

Discussion about the differences between the questions, go here please...

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Malady
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    What does equidistant mean here? – Hagen von Eitzen Aug 06 '15 at 20:35
  • @HagenvonEitzen - Where every circle has the same distance to every other circle, measured from their centers... Should I have said "over a sphere" or "ontop of a sphere", I suppose? Or perhaps the "equidistant from every other circle"? If you can think of a better phrasing, please, edit it in! – Malady Aug 06 '15 at 20:40
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    That condition implies $N\le 4$, doesn't it? – Hagen von Eitzen Aug 06 '15 at 20:41
  • @HagenvonEitzen - Well 2 circles and 1 circle can both generate the the equidistant property, I'm not sure about 3 circles, and the other question has pdfs and diagrams, that relate to circle placement, but I'm not quite sure what they're saying about it... – Malady Aug 06 '15 at 20:48
  • What sort of minimum and maximum values do you anticipate for the $N$ values? I ask this because it may be easier to get an approximate answer for large $N$. And then how close would you need an answer to be, e.g. within $\pm1$%? All this bearing in mind that in a practical situation placement at certain locations may be cheaper. – Marconius Aug 06 '15 at 21:33
  • @Marconius - Well, within... 10% on either side, might have worked for my purposes, but now I'm just leaving the question here for math-based educational reasons or something, 'cause that hexagon answer by Aretino seems good enough for my purposes... – Malady Aug 06 '15 at 21:38
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    The main issue of this question is that "equidistantly spaced upon a sphere" means nothing, in general. – Jack D'Aurizio Aug 06 '15 at 23:36
  • For example, you could put all the centres of the circles equally spaced around the equator. Then they don't cover the whole sphere unless each is at least a hemisphere. – Robert Israel Aug 07 '15 at 07:37
  • Equidistant spacing could be taken to mean equal minimal distance to neighbors with 2D polar symmetry. – Narasimham Aug 07 '15 at 10:04
  • @RobertIsrael - Mmm... As was discussed here, I was trying for: "a set of congruent circles, where the distance between every pair of different circles in the set is the same distance.", but I don't know to concisely say that... And apparently, its impossible... or something like that... – Malady Aug 07 '15 at 11:55
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    You can't have 5 points equidistant from each other on the sphere, or in three-dimensional space for that matter. – Robert Israel Aug 07 '15 at 14:28

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I presume that by "equidistant from each other" you mean "equidistant from their neighbours". The problem is you can't in general put $N$ points equidistant on a sphere (you can do that only for some small values of $N$).

But if $N$ is large you can safely approximate each circle with a regular hexagon and treat the sphere as if it were a plane covered with hexagonal tiles. Since the area of a hexagon of radius $R$ is $(3/2)\sqrt3 R^2$ and the area of a sphere or radius $R_{sph}$ is $4\pi R_{sph}^2$, you need approximately $$ N={8\pi \over 3\sqrt3}{R_{sph}^2\over R^2} $$ circles to cover your sphere. By inverting this you can find $R$ if $N$ is given.

Intelligenti pauca
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Yes, there is an algorithm to answer this question for any given configuration:

Let $x_1,\ldots, x_N$ be the centres of the circles. For each face $x_ix_jx_k$ of their convex hull find the centre of the circumscribed circle, project it (i.e., prolong it) to the sphere and check if the point found this way is covered by one of the circles around $x_i,x_j,x_k$.

(Actually, I'm not sure if centre of circumscribed circle is exactly the right point to check, but I cannot see what could possibly go wrong)

Hagen von Eitzen
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  • ... Well, I want to request something that I don't need a supercomputer to do... But that's more a personal problem than a math one... I suppose ... Yeah! TvTropes! – Malady Aug 06 '15 at 20:50
  • @Malandy I guess this is $O(n^2)$, no supercomputer needed – Hagen von Eitzen Aug 06 '15 at 21:49
  • How do you algorithmically determine the faces of the convex hull? I assume it can be done, but I just wonder about the complexity of that task :-) – Jyrki Lahtonen Aug 06 '15 at 21:54
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Any R that makes circles that have a surface area = 2*(sphere surface area / N), would always produce a covering set of circles... So there's a loose upper bound to the radii that won't work at least...

And any R that makes circles that have a surface area = (sphere surface area / N) is a loose lower bound to the radii that won't work...

Malady
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  • Those hypothetical radii are also bounding the radii that would work... but phrasing that way is proving problematic because it seems to result in strange, and possibly phrasings involving the possibilities that those radii have "guaranteed success" or "guaranteed failure"... – Malady Aug 06 '15 at 21:01