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.
It is drawn by region-figure with
show: "sheet-ladder" — one member of a family of
7 figures
that share a generator, so the drawing above is what that generator returns when it is asked
for this one and given nothing else.
2 essays call it. Every call below passes it something, because a placement that passes nothing draws whichever member of the family the generator happens to default to rather than the one its essay argues about.
Where it is called
Changing this changes every one of these figures.
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.
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.