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

Two square worlds, either of which would carry the same quadtree

Drawn here at the parameters it defaults to, with every essay that calls it.
Two square worlds, either of which would carry the same quadtree. Both projections map the sphere to a square, so either could be the root tile of a pyramid of square images. Mercator reaches its square by being cut at 85.05°, which discards 1,901,487 square kilometres; the cylindrical equal-area with standard parallels at ±55.65° — the solution of π cos²φ₀ = 1 — is square with no cut at all, and carries the poles as lines. At 60° north the Mercator panel inflates area by 4.00 and deforms no angle; the equal-area panel keeps area to 1.000000000 and deforms angles by 13.8°. The choice between them is not about the square.

Both projections map the sphere to a square, so either could be the root tile of a pyramid of square images. Mercator reaches its square by being cut at 85.05°, which discards 1,901,487 square kilometres; the cylindrical equal-area with standard parallels at ±55.65° — the solution of π cos²φ₀ = 1 — is square with no cut at all, and carries the poles as lines. At 60° north the Mercator panel inflates area by 4.00 and deforms no angle; the equal-area panel keeps area to 1.000000000 and deforms angles by 13.8°. The choice between them is not about the square.

It is one of 12 figures drawn by the same construction, and the drawing above is this one with nothing changed.

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.

Two square worlds: Mercator's areas against an equal-area scheme's angles. Both projections give a world exactly as wide as it is tall, so either could carry a quadtree of square tiles. Mercator keeps every angle and inflates area by sec²φ — 132-fold at 85°. The cylindrical equal-area scheme whose world is square has standard parallels at ±55.65° — the solution of π cos²φ₀ = 1 — and keeps every area exactly, at a cost of nothing there and 145° of angular deformation at the edges. At the standard parallel itself Mercator's areal factor is 3.142, which is π, because the square-world condition and sec²φ are the same equation. What a machine does with it

The pyramid did not have to be Mercator

The usual defence is that a quadtree needs a square world and Mercator supplies one. So does the cylindrical equal-area with standard parallels at ±55.654° — the solution of π cos²φ₀ = 1 — and it needs no polar cut at all. What Mercator actually buys is conformality, and the price of giving it up is 13.8° of shear at 60° north.

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