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

Condition — the ladder

11 distinct arguments against one idea, from the one that introduces it to the one that assumes the rest.
  1. 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.

    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.

    rung 1 · choosing
  2. 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.

    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.

    rung 2 · choosing
  3. 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.

    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.

    rung 3 · choosing
  4. The minimum-distortion conformal map of an elongated region, 30° by 10°. The conformal projection of an elongated region, 30° by 10° whose scale is constant on the boundary, which is Chebyshev's criterion, obtained by fitting eight terms of a series rather than by choosing a named projection. The scale factor runs from 0.98640 to 1.00000, a spread of 1.01379; on the boundary itself the largest departure from constancy is 2.13e-7 in the log, which is what the fit achieved and not what it was told. Each dot is an interior sample shaded by its own departure from the boundary's scale.

    Solving for the map instead of choosing it

    Chebyshev's criterion has sat on this site since its second phase with one case it could be applied to: the spherical cap, whose answer is the stereographic projection. For any other region the site stated the criterion and stopped. It is a linear least-squares fit, and the fitted map beats every named projection over the region it was fitted to.

    rung 4 · choosing
  5. What a solved map is, as a list of numbers. A map with no formula is a list of coefficients, and this is the list. The Chebyshev map of an elongated region, 30° by 10° has its coefficients falling by a factor of 1.5e+13 from the first to the fourteenth, so a table of a dozen numbers carries the whole projection; the conformal cube face's coefficients, marked separately, fall far more slowly because the map has a singularity at each corner. How fast this line falls is exactly how portable the map is — and neither map has a name, an inverse in closed form, or a formula anybody could quote.

    A map with no formula

    The solved projection has no name, no formula and no closed-form inverse. It is fourteen numbers — and the rate at which those numbers fall away decides whether a map can be shipped at all: geometrically for a smooth region, and like a power for one with corners.

    rung 5 · choosing
  6. An equal-area map nobody would publish. Mollweide, with a row-dependent horizontal displacement applied to the page afterwards. That plane map has Jacobian determinant 1 everywhere, so every areal scale factor is untouched: the worst departure from 1 anywhere sampled here is 6.0e-12, which is the arithmetic's own floor. It satisfies the equal-area condition exactly and completely, and it is a ruin. The angular deformation at 30°E 20°N has gone from 11.0° to 60.5°.

    Every equal-area map is every other one

    Take Mollweide and slide every row of the page sideways by an amount that depends on the row. The result satisfies the equal-area condition to 6 × 10⁻¹², exactly as well as Mollweide does, and it is a ruin — the angular deformation at one ordinary point has gone from 11° to 60°. Equal-area is one equation, and one equation leaves a whole function free.

    rung 6 · choosing
  7. 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.

    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.

    rung 7 · choosing
  8. 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.

    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.

    rung 8 · choosing
  9. 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.

    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.

    rung 9 · choosing
  10. 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.

    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.

    rung 10 · choosing
  11. 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.

    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.

    rung 11 · choosing

All ladders