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.
It is drawn by screen-figure with
show: "pyramid-tradeoff" — one member of a family of
12 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.
3 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.
A screen map is a pyramid of tiles
The scheme every slippy map runs on is a coordinate system with three integers and one projection, and almost all of it is forced. A square world is what makes the quadtree work, the levels are exact powers of two, and the published resolution — 156,543 metres per pixel at zoom zero — is a distance on the ground at exactly one latitude.
The square costs the poles
The cut at 85.0511287798° is the most-quoted number in web mapping and is almost never derived. It is where Mercator's northing equals half the world's width — the condition for a square — and it drops 1,901,487 square kilometres. A two-tile root would have reached 89.786° and dropped 3,558.
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.