I should note, before I begin, I'm a programmer, not a mathematician, so if my question seems stupid, there's a reason for that. :-j

I'm trying to create a 3d model of a sphere with 6 rings around it. Where the rings intersect are important; they need to be equidistant from one another, and there can't be any more than 2 rings intersecting at any point. This is for a game, where the points of intersection are areas of maximum flexibility.

As a degenerate case, consider the case for 3 rings;


As the image shows, each point where 2 rings intersect has at least 90 degrees of arc between itself and any other point of intersection. What I want is a model with 6 rings, where the distance between any arbitrary point of intersection and it's closest neighbor is maximized.

Here's the openSCAD code I used to create this image. I expect people on this forum will mostly be interested in the rotation matrix I'm using.

orbit_ring_rotations = [
    [[    0,   0,   0]],
    [[    0,  90,   0]],
    [[    0,  90,  90]]
    for (i = [0:1:3 -1])
            rotate_extrude(angle = 360, $fn=quality)
                translate([3, 0, 0])
                    circle(r = .1); 

I really don't have any idea how to attack this problem from here, except to write a program that tries all of the combinations (for integer values between 0 and 359 in the rotation matrix, anyway). But seeing as that's kinda complicated, if there's a math-y solution, I'd love to hear it.


Icosidodecahedron may be a suitable choice. Each face is either a regular pentagon or equilateral triangle with total spherical symmetry. The model can be made out either of six rings or six ( 10 sided) polygons.It was possible to make a model using discarded Venitian blind strips. All rings are great circles. You can google other sites as well where rings were assembled.



It depend's on whether the rings are restricted as great circles.

Analogous to Thomson Problem distributing charges on a sphere. Say you have one ring at the equator, there're 5 rings left. It's most likely have 5 meridian rings. But there'll be two kinds of distances of intersections.

If not, try the rings on every faces of a cube.

  • $\begingroup$ +1 If the OP needs great circles and equidistant intersections, 2 rings at a time, I'm fairly sure this will be impossible. Their example with 3 rings configured like an octahedron's edges is nice, but a fluke. $\endgroup$ – pjs36 Mar 4 '16 at 13:34
  • $\begingroup$ Interesting. Taking a look at this: en.wikipedia.org/wiki/Thomson_problem#Known_solutions The several of the known solutions are just polyhedra. So they'll be an intersection at each vertex of an X vertex polyhedra, where X is the number of intersections (for 6 rings, X = 30). $\endgroup$ – Jason Trout Mar 4 '16 at 15:24

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