My question is derived from Covering the surface of a sphere with circles with least overlap on Math SE.
In the referenced question, the problem of completely covering a sphere with the smallest possible spherical caps for a fixed number of congruent caps is addressed. For instance, if you are allowed five caps, you can minimize their size by placing the centers on the sphere at the corners of an inscribed $D_{3h}$ triangular bipyramid, in which case the required cap radius is the inverse tangent of $2$ and a portion of the sphere equal to $(3-\sqrt{5})/2$, about 38.2%, has pairs of overlapping caps. An alternative arrangement, such as a square pyramidal arrangement of the centers, requires a cap radius and overlap area greater than the quantities noted above.
As we go to many caps we expect the solution to tend towards the one in the plane with a hexagonal lattice arrangement, in which case the area of each circle divided by the hexagonal cell is $2\pi/3\sqrt3\approx1.209$, thus 20.9% of the area would be overlapped. But the table given in the answer seems to show the overlaps converging to about 26% instead. Why?
Is the convergence to the planar limit slower than expected?
Must some of the results referred to in the other question be nonoptimal, and only provisional table entries? For instance, spherical trigonometry calculations reveal that the table entry for $32$ caps is based on the icosahedrally symmetrical arrangement where the centers of the caps are on vertices and face centers of a regular icosahedron (calculations done with this configuration give the same cap radius as the tabulated one). This may "look" optimal but I suspect it isn't (the corresponding octahedrally symmetrical arrangement for $14$ caps is known to be nonoptimal; see * below).
Have there been more or improved solutions to shed light on the convergence to the limit defined above?
What, if anything, is wrong with my assumption about the limit? One feature I have noted is that on a sphere, some of the caps must have only five instead of six neighbors. The apparent limiting area ratio is about halfway between $1.209$ (for a circle covering a regular hexagon in the plane) and $1.321$ (for a circle covering a regular pentagon). Does the presence of even relatively few caps with five neighbors force this balance?
The table of cap radii and densities (100% plus the amount of pairwise overlap) are given below.
Source: Tarnal and Gaspar - Covering a sphere by equal circles, and the rigidity of its graph (DOI).
*Here is the argument that the octahedrally symmetrical arrangement of $14$ caps is nonoptimal. Place the cap centers at the vertices and centers of a regular octahedron, which is the symmetric arrangement. Connecting neighboring cap centers divides the sphere into 24 triangles that each have angles of $A=B=60°, C=90°$. Now the required cap radius is the circumradius of each triangle, given by the formula
$\cos R =\sqrt{\dfrac{\cot[\frac12(-A+B+C)]\cot[\frac12(A-B+C)]\cot[\frac12(A+B-C)]}{\cot[\frac12(-A+B+C)]+\cot[\frac12(A-B+C)]+\cot[\frac12(A+B-C)]}}.$
This works out to $R\approx36.20°$; but with $14$ caps the table identifies a better arrangement with a cap radius of $34.94°$. The conjecture here is that the corresponding icosahedral arrangement with $32$ caps, which is used for that count in the table, is in fact also nonoptimal.
