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

The distances a construction holds, and the rest it decides

Drawn here at the parameters it defaults to, with every essay that calls it.
The distances a construction holds, and the rest it decides. The construction a surveyor would use, on London, New York, Tokyo, Sydney. The first place is put down anywhere, the second at its true distance along an axis, and every further place is trilaterated from those two — which holds 5 distances exactly, drawn heavy, and that is 2n − 3 for n = 4. Nothing is fitted. The remaining 1 distances, drawn thin, are then decided rather than chosen, and on places taken from a sphere they are decided wrongly: the worst is out by 6.87% and the closest to right by 6.867%.

The construction a surveyor would use, on London, New York, Tokyo, Sydney. The first place is put down anywhere, the second at its true distance along an axis, and every further place is trilaterated from those two — which holds 5 distances exactly, drawn heavy, and that is 2n − 3 for n = 4. Nothing is fitted. The remaining 1 distances, drawn thin, are then decided rather than chosen, and on places taken from a sphere they are decided wrongly: the worst is out by 6.87% and the closest to right by 6.867%.

It is drawn by curvature-figure with show: "distance-trilateration" — one member of a family of 29 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.

1 essay calls 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.

The share of its distances a flat picture can hold. A picture of n places on a plane has 2n coordinates and is unchanged by two translations and a rotation, so 2n − 3 numbers in it are genuinely free; each distance held exactly is one equation. So at most 2n − 3 of the n(n−1)/2 distances can be right, whatever the places are and however hard the map maker tries. The share falls as 4/n: five of six for four places, thirteen of twenty-eight for eight, and 197 of 4,950 — 4.0% — for a hundred. The dashed curve is 4/n, which the count approaches from below and never crosses. The impossibility

Five distances of six, and never more

A flat picture of n places has 2n − 3 free numbers and n(n−1)/2 distances, so it can be exactly right about 2n − 3 of them and no more — five of six for four places, twenty-nine of a hundred and twenty for sixteen, and 3.98 per cent for a hundred. The bound is reached by a construction with no fitting in it, and an azimuthal equidistant map spends only n − 1.

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