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

Solving the conformal condition, and finding Mercator

Drawn here at the parameters it defaults to, with every essay that calls it.
Solving the conformal condition, and finding Mercator. The cylindrical family is x = λ and y = Y(φ), one unknown function. Requiring the map to be conformal makes the condition a first-order differential equation in Y — Y′ = sec φ — and this curve is the Runge–Kutta integration of it, drawn against Mercator's own closed form as a dashed line. They agree to 1.4e-12 of the ordinate, so the named projection is not quoted anywhere in the calculation: it is what the condition returns.

The cylindrical family is x = λ and y = Y(φ), one unknown function. Requiring the map to be conformal makes the condition a first-order differential equation in Y — Y′ = sec φ — and this curve is the Runge–Kutta integration of it, drawn against Mercator's own closed form as a dashed line. They agree to 1.4e-12 of the ordinate, so the named projection is not quoted anywhere in the calculation: it is what the condition returns.

It is one of 7 figures drawn by the same construction, and the drawing above is this one with nothing changed.

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.

Three solutions of one condition, all exactly equal-area. The pseudocylindrical ansatz is x = λ·C(φ) and y = Y(φ) — two unknown functions — and the equal-area condition is one equation, C·Y′ = cos φ. So Y may be chosen freely and C follows, and these three choices give three maps that are equal-area to 1.8e-11 and look nothing like one another: their pole lines are 10%, 100%, 19% of their own equators. That is why the cylindrical family has one equal-area member and this family has as many as anybody cares to name. The families

The condition does not always decide the map

Write a family as a shape with an unknown function in it and every classical property becomes a differential equation. In three families the equation has one solution and the named projection is what comes back. In the fourth it has a whole function of solutions, which is why that family has forty members and the others have three.

 The same patch, pinned six ways. Two vertices have to be held or the conformal energy has a similarity's worth of null space. Which two turns out to decide two of the three numbers reported. The median angular deformation is the same to 8 per cent across all six — that is the map. The areal spread runs from 2.68 to 7.44, so the 3.07 reported for this body was a statement about its corners. And pinning two adjacent vertices, which fixes the similarity through a very short lever, ruins the worst point without touching the median: a badly conditioned constraint pays for its scale in one corner. What the numbers refer to

The map depends on where it was cut

Solving the discrete conformal equations on a triangulated body gave an areal spread of 3.07, with the number possibly belonging to the patch, the boundary and the two pinned vertices rather than to the surface. It belongs to the pins: hold a different pair and it runs from 2.68 to 7.44, while the typical angular deformation does not move at all.

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