Cartographic Projection
every map is wrong, and by exactly how much
Concept

Covering — where it appears

A set of caps whose union is the whole sphere, whose density is the total cap area over the sphere's and is never less than one. The number of caps needed for a stated tolerance is what sets an atlas's sheet count, before any projection has been chosen for the sheets.

Named by 3 essays across 2 fields — each of them below, with the objects they name alongside it.

Sheets for a tolerance, from Chebyshev's bound and a covering. For each stated tolerance on the scale error, the cap radius at which the best possible conformal projection just meets it — sec²(ρ/2) − 1 = tolerance, which is Chebyshev's bound and has no fitting in it — and then the number of such caps needed to cover the sphere at the packing density a real arrangement achieves. One part in a thousand costs 1210 sheets of 403 kilometres radius. The slope is -0.989: a factor of ten in what the job will accept is a factor of ten in the atlas.

How many sheets an atlas needs

A tolerance on the scale error inverts, through Chebyshev's bound, into a sheet radius — and a covering problem turns the radius into a count. One part in a thousand costs 1,210 sheets of 403 kilometres radius, the count goes as the reciprocal of the tolerance exactly, and the projection multiplies it by anything from one to fifty-six.

choosing · Choosing The world after a map of degree 2. The graticule of the sphere, sent through the square of the stereographic coordinate and then drawn in Mollweide. Every parallel is still a closed curve and every meridian still runs pole to pole, and the whole world has been wrapped round the sphere 2 times: the 2 meridians that used to be 180° apart now lie on top of one another. Nothing has been torn. The degree measured by integrating the area the map sweeps is 2.0004.

How many times, not whether

Three rungs of this ladder answer yes or no and have no other kind of answer. The degree is the first quantity here that counts: a continuous map of the sphere to itself covers it a whole number of times, injectivity forces that number to ±1, and the number is recoverable three ways — by counting preimages, by integrating swept area, and from the rate at which the preimages coalesce.

impossibility · Topology How thinly a sphere can be covered by a few equal caps. The covering density of the best arrangement of n equal caps found for each n — the total area of the caps divided by the sphere's, so a value of one would be a perfect tiling with no overlap. The horizontal line is 2π/√27 = 1.2092, the thinnest covering density of the PLANE by equal discs, which this site has used for the sphere since its first atlas essay. It is wrong in both directions: at 2 caps the sphere is covered more thinly than any plane can be, because a cap may be a hemisphere, and at every count from 3 upwards more thickly — 1.5092 at 3, and 1.3377 at 14. The ringed points are the four counts whose optimum is proved: 2 at 90.00°, 4 at 70.53°, 6 at 54.74°, 12 at 37.38°. Everything else is an upper bound from a search, drawn as one, and the bound loosens as the count rises — the search reaches the proved optimum to 3.4 per cent at these counts and has no such check anywhere else.

The sphere is not the plane at small counts

The site's atlas arithmetic multiplies an ideal sheet count by 2π/√27, the thinnest covering density of the plane. At the four counts whose optimal covering of the sphere is a theorem the plane's number is 21 per cent high at two caps and 9, 5 and 2 per cent low at four, six and twelve — wrong in both directions, and the direction changes with the count.

choosing · Choosing

Named alongside it

The objects these essays reach for when they reach for this one.

Spherical capChebyshev's boundClosed formLower boundRepresentative fractionSheet layoutToleranceBranch pointConformalityContinuityDegreeEuler characteristic

All concepts