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

The residual is curvature, and it is cubic

Drawn here at the parameters it defaults to, with every essay that calls it.
The residual is curvature, and it is cubic. The worst trimetric residual inside a region, against the radius of that region, on logarithmic axes. The fitted slope is 2.99: doubling the region multiplies the residual by eight. That is the signature of a curvature effect on a distance — the relative error of flattening a patch grows as the square of its size, and a distance across the patch grows as its size, so the absolute error grows as the cube. At a radius of 2° the residual is 0.16 km; at 32° it is 636 km.

The worst trimetric residual inside a region, against the radius of that region, on logarithmic axes. The fitted slope is 2.99: doubling the region multiplies the residual by eight. That is the signature of a curvature effect on a distance — the relative error of flattening a patch grows as the square of its size, and a distance across the patch grows as its size, so the absolute error grows as the cube. At a radius of 2° the residual is 0.16 km; at 32° it is 636 km.

It is drawn by condition-figure with show: "condition-cost" — one member of a family of 3 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.

8 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.

Exact distances from two places, and from nowhere else. The two-point equidistant projection with London and Cape Town as its centres, 87.0° apart. The light circles are drawn in the map at radii of 30°, 60°, 90°, 120° about each centre; every one of them is a true distance circle on the sphere, to 1.8e-14 relative. Between two points that are not centres the drawn distance is wrong by up to 633 per cent. What each projection optimises

A projection written as a condition

Instead of a formula, a sentence: the distance from these two places must be exactly right. The map that satisfies it is found by intersecting two circles, it is exact to five parts in a hundred million million, and it exists over the whole sphere for a reason that belongs to the sphere rather than to the construction.

 Three distances, two degrees of freedom. The Chamberlin trimetric construction with its three centres marked. A point is placed by intersecting circles of its true distance from each centre — but three circles in a plane do not meet in a point, and the three candidate positions obtained by taking the conditions two at a time are 17.0 km apart at the sample point. They are drawn here magnified 120 times about their own centroid, which is where Chamberlin's rule puts the point. The residual is not an error in the arithmetic; it is the amount by which the sphere refuses to be a plane, and it grows as the cube of the size of the region. What each projection optimises

Three conditions are one too many

Two distances fix a point in a plane and a third has no freedom left to be satisfied with. Chamberlin's trimetric construction averages the three positions that satisfy two conditions each, and the spread between them — never zero anywhere, 22 km over North America, growing as the cube of the region — is the price of the extra clause.

 A map for pointing, not for locating. Craig's retroazimuthal projection, centred on Mecca. The straight line from any point to the centre makes an angle with the map's vertical equal to the true initial bearing from that place to the centre — checked here at 169 points, worst disagreement 5.7e-14 degrees. The bearings printed beside each line are computed on the sphere with no projection in them. What each projection optimises

A map that cannot be read backwards

Craig's projection answers one question exactly — lay a straight edge from any place to the centre and read the compass course, right to 6 × 10⁻¹⁴ of a degree. It pays by folding: 78°S and 48°S on the same meridian are drawn at the same point, so no inverse exists and nothing else can be read off it at all.

 Which scale fields a map could have, and which are only wishes. Liouville's equation — the Laplacian of log k equals 1/k² on the page — is the whole condition for a conformal map of a unit sphere to have a stated scale factor. The first two rows are the scale fields of real projections and they satisfy it to the differencing step. The rest are requests a designer might write, and every one of them fails — except one, which turns out to be a projection somebody already found. Asking for no distortion anywhere fails by exactly one, which is the curvature of the sphere. What each projection optimises

Not every distortion can be asked for

Six essays have written projections as conditions and asked how much freedom a condition leaves. The reverse question has never been put: a cartographer knows what distortion they want, so can they ask for it? For a conformal map the answer is a single equation, it is the Theorema Egregium in disguise, and asking for no distortion anywhere fails it by exactly the curvature of the sphere.

 The part of a request no conformal map can supply — scale falling with distance from the centre. The difference between the requested scale field and the nearest achievable one, over the patch, with the sign shown by colour. The RMS is 1.81e-1 in the logarithm of the scale and the worst single point is 4.91e-1. This is not an error: it is the part of the request that no conformal map of any kind can grant, and its SHAPE is the answer — it says where the request was impossible, which a single number cannot. What each projection optimises

The nearest map to an impossible request

Rung seven found that not every distortion can be asked for. It never asked what happens when one is asked for anyway — and the answer has a shape: the achievable fields are the solutions of an elliptic equation, a request is a point off that set, and the nearest point leaves a residual whose floor is the curvature rather than the size of the ask.

 The same four requests, put to the two conditions. Each request is a stated field over a square region, and the bar is what is left over after the nearest map satisfying the condition has been found. The conformal condition refuses: its achievable set is decided by boundary values, and the residual is the part of the request no conformal map of any kind can supply. The equal-area condition never refuses — every one of these is met to 3.0e-5, which is the quadrature's own noise — because a positive areal request is granted by a construction with no iteration in it and no boundary data. What each projection optimises

The nearest equal-area map to an impossible request

Every number in the previous rung is inside the conformal achievable set, because Liouville's is the conformal condition. The equal-area set is one equation on two functions and never refuses: the same four requests are met to nine parts in a billion, and charged for in angle instead.

 Where the condition holds, and where it was asked to. The boundary scale of a fit collocated at 20 points, drawn all the way round the boundary. The marked points are the ones the condition was imposed at, and the curve passes very near zero at every one of them; between them it does not. The largest departure on the samples is 3.92e-5 and the largest anywhere is 3.27e-4, and the second is the one the map has. What each projection optimises

A condition imposed at points is not a condition

Nine rungs state a condition and solve it, and every solve imposes the condition at a finite set of samples because that is what a linear system is. With barely more equations than unknowns the residual the solver reports is 8.3 times too good — and refining the collocation twentyfold does not improve the map at all, it only makes the report honest.

 One of these settles. The largest departure of the fitted map's boundary scale from constant — the quantity the previous rung showed the solver cannot see — against the number of collocation nodes, for the two placements. The clustered fit reaches 7.030e-5 at forty-eight nodes and returns exactly that at every count above it. The evenly spaced fit does not settle at all: it wanders by a factor of 1.43 across the same range, going up as often as down. Refining an evenly collocated fit is not convergence, and the previous rung's finding that more samples improve the report and not the map is this seen from one side. What each projection optimises

The nodes were evenly spaced

The previous rung showed that refining an evenly collocated fit improves the solver's report and leaves the map alone. Moving the same number of nodes to the Chebyshev positions — crowded toward the corners, where a conformal map of a polygon is singular — makes the fit settle: 7.030 × 10⁻⁵ at forty-eight nodes and exactly that at every count above it, against an even fit that wanders by 43 per cent and never converges at all.

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