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

Condition — the series

20 essays on 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.

    part 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.

    part 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.

    part 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.

    part 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.

    part 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.

    part 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.

    part 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.

    part 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 measurement before this one 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.

    part 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

    Every projection written as a condition is solved by imposing that 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.

    part 10 · choosing
  11. One of these settles. The largest departure of the fitted map's boundary scale from constant — the quantity 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 finding that more samples improve the report and not the map is this seen from one side.

    The nodes were evenly spaced

    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.

    part 11 · choosing
  12. Two conformal maps of one square: the worst case held, and the mean square held. Lines of equal scale inside a spherical square of 25° circumradius, every quarter of a per cent about each map's own mean; one ink above the mean and another below. Left, Chebyshev's map: the boundary is itself a line of equal scale, 1.44% above the mean, and the scale varies by 2.98% in all. Right, the map whose scale is closest to constant in the mean square over the area: the lines cross the boundary, the corners rise to 2.05% above its mean, and the scale varies by 3.47% — but its root-mean-square departure is 0.804% against Chebyshev's 0.871%.

    Chebyshev's map is the best at its worst, and not on average

    Chebyshev's criterion makes a conformal map's scale constant on the boundary of a region, and no conformal map has a smaller range of scale inside it. The same series, fitted instead to make the scale as nearly constant as possible over the area, is a different map on every region but a cap. On a spherical square Chebyshev's map is 8.3 per cent worse in root-mean-square, the other map's range is 16.5 per cent wider, and every map between them trades one for the other.

    part 12 · choosing
  13. Three criteria, three maps, and the tolerance band each one encloses. A spherical square of 25° circumradius under three conformal maps, with the two contours at ±0.5% of true scale drawn heavy and the region between them the area each map keeps inside that tolerance. Chebyshev's map holds 37.0 per cent of the area there, the mean-square map 45.4, and the map fitted to the tolerance itself 64.1. The faint lines are further multiples of the tolerance, and they say where each map spends what it is not counting: the third map's outermost contours are much further out than the other two's.

    The map that keeps the most ground inside a tolerance

    A surveyor does not want the smallest worst case or the smallest mean square. They want as much of the territory as possible within a stated tolerance of true scale and are indifferent to how far outside the rest goes. That is a third criterion, it is answered by a third map, and on a 25° square at half a per cent it keeps 64.1 per cent of the area where the mean-square map keeps 45.4 and Chebyshev's 37.0 — bought by sending a tail out to 5.6 per cent where Chebyshev's worst is 1.5.

    part 13 · choosing
  14. Forty-two starts, forty-two answers. Every local optimum coordinate ascent reaches from 42 starts on the same region and the same criterion — a 25° spherical square at a tolerance of ±0.5% — plotted at the share of area it keeps. They run from 36.98 per cent to 67.50, a spread of 30.5 points, and no two of them are the same map: clustered on their scale fields rather than their coefficients, 42 of the 42 are distinct. The two marked starts are the ones anybody would actually use — Chebyshev's map reaches 36.98 per cent and the mean-square map 64.05, and neither is the best this search found.

    A criterion worth using is one whose answer is not unique

    Forty-two starts on one region and one tolerance give forty-two different maps, keeping from 36.98 per cent of the area to 67.50. The two starts anybody would actually use — Chebyshev's map and the mean-square map — end 27.1 points apart, which is more than the criterion buys over either of them. And the spread is widest at exactly the tolerance where the criterion is worth using at all, because both facts are the same fact: an objective with a flat top and a sharp edge has somewhere to hide answers.

    part 14 · choosing
  15. The second clause moves the map without moving what the first clause measures. The plateau the tie-break changes most, drawn before and after. The contours are where the scale departs from true by the stated half a per cent, so the ground between them is the ground the specification asks to be kept. The rule leaves that ground alone — 52.23 per cent of the square before and 52.54 after — and spends its freedom outside it, where the worst departure falls from 27.16 per cent to 23.77. Of the 42 plateaux this is the largest change any of them shows.

    A second clause decides the half that was already decided

    A specification written as a band admits forty-two maps keeping between 37.0 and 67.5 per cent of a region. Adding a tie-break — among those, take the one whose worst departure is least — is a linear programme, and it works: it makes each of the forty-two definite. It also merges three neighbours at the bottom into one, leaves 29.1 of the 30.5 points of spread in area kept, and names a single map only at the tolerance where every map keeps everything and the first clause has stopped asking for anything at all.

    part 15 · choosing
  16. Three exponents, three maps, and each is the only answer its specification has. The conformal maps of a 25° spherical square that minimise the mean of |ln k| raised to three exponents. The heavy contours are where the scale departs from true by half a per cent either way, so the ground between them is what a surveyor asking for that band would keep. At p = 1, the mean departure, the map keeps 52.8 per cent of the square inside the band; at p = 2, the mean square, 44.2; at p = 32, which is within a per cent of the least worst case, 32.7. Its worst departure falls the other way, from 3.25 per cent to 1.94 and 1.48. For every exponent from one upward the specification is convex and has one answer.

    One exponent names one map, and at tight bands a better one than the band usually finds

    A specification written as a band keeps a count of area and throws the rest of the scale field away, which is why it had forty-two answers and why a tie-break could not name one. Written instead as the mean of |ln k| raised to a stated power, it is convex for every power from one up, and has one answer. At a band of a tenth of a per cent that single answer keeps more of the region than forty-one of the band's forty-two, and at half a per cent fewer than nine of them: the band's best answer leads it by 14.7 points there, and the band's own search started from it reaches within two points of that best. Only as the band approaches the least worst case does a large power take over.

    part 16 · choosing
  17. A square switches from p = 1 to the worst case at one band; a thin ellipse passes through every exponent between. Area kept inside a band, per cent, against the band as a fraction of each region's least worst departure. Solid: the map minimising the mean of |ln k|, p = 1. Dotted: the least worst case, p = 32. Thin line: the best of twelve exponents from 1 to 32. Shaded: what a rule taking p = 1 below 0.9 and the worst case above it gives up against the best exponent. On the 25° square the best exponent is p = 1 up to 0.85 and the rule gives up at most 0.9 points. On an ellipse with axes 1 : 0.2 an intermediate exponent is best from 0.60 and the rule gives up 5.9.

    The band names the exponent for a compact region, and not for a thin one

    A tolerance band is best served by the mean departure when it is tight and by the least worst case when it is wide. Read as a fraction of the region's own least worst departure, the band decides which, and the region's size drops out: a square, a triangle or a hexagon at 10°, 25° or 40° switches from one to the other at nine tenths, and a surveyor told only that loses at most two points of area on every compact region but the degenerate cap. A thin ellipse does not switch. From six tenths upward it is served best by exponents in between, and any rule with two exponents in it gives up five to seven points there, wherever its switch is put.

    part 17 · choosing
  18. Two solved maps and the line between them keep what the whole family of exponents keeps. Area inside a band, per cent, against the band as a fraction of each region's least worst departure, from 0.3 to 1. Heavy: the best map on the line between the p = 1 map and the least worst map, found by a search along the line alone. Thin: the best of twelve exponents from 1 to 32, each solved separately. Dashed: the rule taking p = 1 below 0.9 and the least worst case above it; shaded, what that rule gives up. On the square the line trails the best exponent by at most 0.00 points and beats it by up to 1.3; the rule gives up 0.9. On the ellipse the line trails by at most 1.1 and the rule gives up 5.9.

    Two maps and the line between them do the work of every exponent

    A conformal map's log scale is linear in its coefficients, so every blend of the map that minimises the mean departure and the least worst map is itself a conformal map, available the moment the two are solved. Searched along that one line for each band, the blend keeps within 0.17 points of the best of twelve exponents on every compact region and beats all twelve by up to 1.35 on the square; it trails by 1.5 at most on the thinnest ellipse and 0.74 on four regions with a bay, where the rule of switching between the two ends gave up as much as 7.1. The family's middle members lie up to two thirds of their own departure off the line. The line does not reproduce them; it does their work.

    part 18 · choosing
  19. Three solved maps span a plane of maps, and the band's best map lies well outside their triangle. Every conformal map of a 25° spherical square whose ln k is the p = 1 map's plus s times its difference from the least worst map plus u times its difference from the mean-square map, for s and u from −4 to 4, contoured by the share of the square it keeps inside a band of ±0.74 per cent — 0.5 of the least worst departure. The three maps are the corners of the shaded triangle at (0, 0), (1, 0) and (0, 1); the line between the first two is its lower edge, and the exponents from 1 to 32, dashed, run from one corner to the other through the mean-square map and out past it. The best map of the twelve exponents keeps 66.4 per cent, the best on the line 66.5, the best inside the triangle 66.5. The best in the plane sits at s = 0.72, u = −2.81, outside every one of them, and keeps 75.8.

    Three maps make a plane, and the band's best map lies outside their triangle

    Every conformal map whose ln k is a fixed blend of three solved maps — the p = 1 map, the least worst map and the mean-square map — is itself a conformal map, so three solved maps are a plane of them. Inside their triangle the third map recovers most of what the line between the first two gave up on thin regions: 0.63 points short of the best exponent on the thinnest ellipse, where the line was 1.48 short. But the band's best map in the plane is not between the three. On 172 of 182 regions and bands it lies outside the triangle, and it keeps up to twenty points more than any exponent; started from it, a search over all twenty-five coefficients adds about a point on four regions of five and seven on a hexagon. Its price is a worst departure up to nearly five times the least worst.

    part 19 · choosing
  20. Each cap on the worst departure is a long convex band of maps, and the best map under it rides the band's edge. The plane of conformal maps of a 25° spherical square whose ln k is the p = 1 map's plus s times its difference from the least worst map plus u times its difference from the mean-square map, for s and u from −4 to 4. Dashed: the edges of the maps whose worst departure is at most 1.2, 1.5, 2, 3 times the least worst — convex, nested, long and narrow, each running diagonally through the least worst corner. Thin contour: the maps keeping as much of the region inside a band of ±0.74 per cent as the best exponent's 66.4 per cent; heavy contour: three points under the uncapped best, 72.8 per cent. Dots: the map keeping the most inside the band under each cap — 1.2: 59.8 per cent at (−0.30, 1.56); 1.5: 61.3 per cent at (−0.17, 1.03); 2: 65.8 per cent at (0.50, −0.36); 2.5: 69.1 per cent at (0.51, −1.00); 3: 72.1 per cent at (0.55, −1.63); 4: 75.8 per cent at (0.72, −2.81); the large dot is the uncapped best, 75.8 per cent at a worst of 3.82 times the least worst.

    A cap on the worst departure keeps the thin regions' gain and gives back the square's

    The conformal map that keeps the most of a region inside a band can sit far out in the plane of three solved maps and let its edges go to nearly five times the least worst departure. Capped at twice the least worst, the plane still matches or beats the best exponent on 69 region-and-band pairs of 91: on the thin ellipses it keeps the whole of its gain, because that gain sits in a small pocket just past the exponents' own worst departure and arrives all at once; on the square and the bays it gives most of it back, because there the gain is a ramp that runs out to four times the least worst. Under any one cap the plane still beats the exponents that cap admits, by up to eighteen points — and it can name maps that no weighting of area against worst departure would ever choose.

    part 20 · choosing

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