6 rings with equidistant intersection points

Question

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;

http://s15.postimg.org/e5k18koyj/ring_problem.png

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(orbit_ring_rotations[i][0])
            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.

Answer 1

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 Venetian blind strips. All rings are great circles. You can google other sites as well where rings were assembled.

EDIT1:

I found the symmetric assembly of six great circles so pleasing.. to 3d print an icosidodecahedron later.

IcosidodecaHedron

Answer 2

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.