Cartographic Projection
every map is wrong, and by exactly how much
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The thread: The trade-off is forced — page 4

Conformal means the principal scales are equal; equal-area means their product is one. Both at once forces them both to one, which is an isometry, which cannot exist. The trade-off is two lines of algebra. Essays 73 to 96 of 146.
The shortest crossing that never goes more than 745 km from an airfield. New York to London, drawn on a conformal conic, with a circle of 745 km round each of twenty airfields on both sides of the North Atlantic. The dashed line is the great circle, 5,540 km. The solid line is the shortest route that stays inside the circles, 5,698 km — 158 km or 2.85 per cent longer — and it bends only where two circles cross: between Narsarsuaq and Keflavík; Keflavík and Vágar. Nowhere else can it bend, because anywhere else a straight cut stays inside a circle. Paths and directions

The rule that keeps a route near land is pinned at both ends

A crossing that may never be more than a stated distance from an airfield is refused below one radius and free above another, and each wall is set by a single feature of the ocean. Between them the price falls continuously, with corners and no jumps — and at the lower wall it falls as the square root of the allowance, so the first kilometre is worth twelve of route.

 Two expansions of one integral. The worst error along the whole meridian, against the number of sine terms kept, for the series in the third flattening and the classical series in e². Both are checked against a Simpson's rule on the defining integral, which shares no algebra with either. At one and two terms they are the same number to four digits and the e² series is fractionally ahead; from the third term the n series pulls away, and at four it is 1175 times more accurate — 7.6e-8 metres against 9.0e-5. What is taught wrongly

Which small quantity the series is in

Every ellipsoidal formula in this collection is a truncated series in the third flattening, inherited from Krüger in 1912 and justified nowhere. Measured against the classical expansion in e², at four sine terms it is 1,175 times more accurate — and at one and two terms it is fractionally worse, which is not what the folklore implies.

 One field, one round trip between two cell schemes. Left: a stated field binned into an equal-angle grid of 36 by 18 cells. Right: the same field after being rebinned into an equal-area grid of 30 by 15 offset by six degrees of longitude, and rebinned back. Every step is exact area-weighted averaging, the total is preserved to 2 × 10⁻¹⁶, and the root-mean-square difference between the two pictures is 0.144 on a field whose own standard deviation is 0.370. What a machine does with it

The same data on two grids

Five essays have addressed, queried and ordered cells within one scheme and nobody has moved a number between two. Doing it exactly — area-weighted, both directions — preserves the total to 2 × 10⁻¹⁶ and loses 39 per cent of the field's own standard deviation in a single round trip; six round trips leave 23 per cent of its variance. The quantity that would reveal the damage is the one that never moves.

 Sixteen cities, each left out in turn. The least worst-case distance error any flat picture of these sixteen cities can have is 53.3%; the best projection in the library, Azimuthal equidistant, manages 57.6%. Each bar is that least error with one city left out. Leaving out Sydney brings it to 25.5%; leaving out Cape Town, the next most costly, to 46.7%; and leaving out any of five — New York, Tokyo, Anchorage, Singapore, Los Angeles — changes it not at all. The note on a row counts the pairs holding the picture at its worst that the city belongs to. The impossibility

The error belongs to a few of the places

The least error of any flat picture of sixteen cities is 53.3 per cent, and it is not spread over the sixteen. Six pairs hold it there and Sydney is in three of them: leave Sydney out and it falls to 25.5, while leaving out any of five other cities changes it not at all. Choosing which places to draw is worth more than any choice of how to draw them.

 One unit of a vector tile, in metres of ground. A vector tile's coordinates are integers on a lattice 4096 units across the tile, and the tile halves at every level, so one unit is a distance that halves too: 5.48 m at z10 and 0.086 m at z16, at 55°. It is also a different distance at every latitude, by cos φ, because the tile is in Web Mercator — the same factor that makes a grid metre a different quantity of ground at every latitude, arriving in the file format rather than in the projection. What a machine does with it

A vector tile has an integer grid

Six essays on this ladder treat a vector tile as the thing a raster tile is not: geometry, resolution-free, styled at draw time. Its coordinates are integers on a lattice 4,096 units across a tile, the tile halves at every level, and at 55° north one unit is 88 metres at zoom 6 and 21 millimetres at zoom 18.

 In a wind every circle moves upwind, and the route moves with them. New York to London with the sixty-minute allowance at a stated one-engine speed of 745 km/h. In still air every airfield's circle has a radius of 745 km, drawn faint and dashed. In a northerly of 150 km/h the places from which each airfield can be reached in sixty minutes form a circle of the same radius moved 150 km upwind, drawn solid. The dashed route is the shortest the still-air rule allows, 5,698 km; from one point of it the quickest diversion in this wind takes 68.7 minutes. The solid route stays inside the moved circles, 5,765 km, and its longest diversion takes 60.8 minutes — the sixty the moved circles stand for, to within what the Earth's curvature does to a wind that keeps one bearing. Paths and directions

A diversion allowance in a wind is the same circle, moved upwind

A diversion rule written in minutes is turned into a distance by assuming still air. In a uniform wind the places from which an airfield can be reached in sixty minutes are not a stretched region but the still-air circle moved upwind by the wind's run. A 150 km/h northerly makes the still-air route across the North Atlantic need a diversion of 68.7 minutes, a southerly makes it 54.0 — and the least radius at which the ocean can be crossed barely moves.

 The piece two schemes share, clipped rather than assumed. A cell of a gnomonic cube, whose four edges are great-circle arcs because a straight line on a gnomonic face is one, against a cell of a longitude–latitude grid, whose north and south edges are parallels and are not. Their overlap is neither a rectangle nor a spherical polygon of any standard kind, and it is 5965687 km² of the cube cell's 5965687 km² — 100.0 per cent. Computing it needs the arc of one boundary intersected with the plane of the other, which is three equations and two roots, and it is exact. What a machine does with it

Cells that are rectangles in no coordinate

What moving a field between two cell schemes costs was measured between two schemes whose cells are longitude–latitude rectangles — which is what made every overlap a rectangle with a closed-form area. The schemes anybody actually argues about have cells that are rectangles in no coordinate, and their overlaps have to be clipped.

 Walking the crossing all the way to a cube's vertex. The corner a feature gets crossing the seam, at crossings that approach the vertex geometrically — the last is 3.5e-4 degrees from it. Every face map is singular at a vertex, so the expectation is that the corner runs away. It does not. The gnomonic settles at 53.1295° and the equal-area map at 12.9656°, and both are finite. The dashed line is the solid's angle deficit, 90.0°, which is what the surface loses at that point and is not what either map's corner reaches. The families

The corner that is the curvature

Every face map is singular at a vertex, so a corner that grows as the crossing walks towards one should run away. It does not: the gnomonic settles at 53.1295° on a cube and the equal-area map at 12.9656, both finite, both reached like the first power of the remaining gap. What does not settle is the deficit beside them — 90 degrees, fixed by Descartes before any projection is chosen.

 Seven cuts of the same size, in different places on one body. The body's own colatitude and longitude, with the seven windows drawn on it. Each covers the same surface area to 0.24 per cent — the longitude extent is divided by sin θ and a scale is then solved per window — so their sizes are held and only their positions differ. Each is shaded by the areal spread of the conformal map solved on it, listed beside the grid and running from 1.03 to 1.36. The darker windows are the worse ones, and they are the ones over the body's lobes. What the numbers refer to

Cuts of the same size in different places

Where a body is cut decides how well it can be mapped, by a factor of ten — established with five windows of five different sizes, so *where* and *how much* were confounded and the factor could have been entirely about extent. Held to the same surface area to a quarter of a per cent, the answer survives at a factor of 1.32, and what predicts it is the curvature the window encloses.

 A curve built to have dimension 1.2619. The generator replaces every segment with four of equal length at headings 0, +60.0°, −60.0° and 0. Closing the displacement fixes the length ratio at 0.33335, and four copies at that ratio give a dimension of exactly log 4 / log(2 + 2 cos θ) = 1.2619. Nothing here is measured yet: this is the construction the measurement will be checked against. Drawn at depth 5, which is 1024 segments, with the second-level shape shown faint beneath it. What a machine does with it

A line has a length only at a scale

A curve given by a formula can be sampled as finely as the picture needs. A map is not that: the geometry that reaches the page has been through an algorithm whose job is to throw most of it away. The first thing that goes is the idea that the line had a length.

 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. What each projection optimises

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.

 Three named members of the family, and one the family has no name for. The equal-area pseudocylindricals are not a list of maps. They are the solutions of one equation — C(φ)·Y′(φ) = cos φ — in two unknown functions, so one function is free and the named members are points in a space of them. Writing Y as six numbers and searching over them finds a map at 25.03° of mean angular deformation against Eckert IV's 28.08, Mollweide's 31.83 and the sinusoidal's 38.86 — 12.2 per cent better than the best map the family has a name for, and every one of the four is equal-area to the same precision. The families

A family is a function, not a list

The equal-area pseudocylindricals are the solutions of one equation in two unknown functions, so one function is free and the named members are points in a space of them. Writing that function as six numbers and searching over them finds a map at 25.03° of mean angular deformation against Eckert IV's 28.08 — and two numbers are already enough to beat every map the family has a name for.

 50° S to 40° S, drawn conformally without a cut and with one. The band of latitude from 50° S to 40° S, drawn two ways that both keep every angle. On the left it is not cut: the only such map with least scale variation is the stereographic from the south pole, the ring closes up, and its scale varies by a factor of 1.0750 across the band. On the right it is cut along one meridian and opened into a conformal conic with cone constant 0.7080: the ring spans only 254.9° and leaves a gap of 105.1°, and its scale varies by 1.0038. Not cutting costs a factor of 1.071 in scale variation. The impossibility

A ring can be drawn whole, and only one way

A band of latitude has a hole in it, and a region with a hole has a number no angle-keeping map can change. Draw the band without cutting it and that number forces the map: the stereographic from one pole is the least-varying picture there is. One cut along a meridian frees the cone constant, and the scale variation the cut buys back runs from nothing on a polar cap to a factor of 2.13 between the tropics.

 Two regions, three answers. Britain and New Zealand, 166° apart, with the pole of the best oblique conic under each of three objectives. Pooling the samples and taking an area-weighted score puts the pole in one place; refusing to let either region be worse than the other puts it somewhere else. The regions are drawn on Mollweide so that equal ground areas are equal page areas. What each projection optimises

The pooled score abandons a region

Every criterion in this subject optimises for one region. An atlas is several, and pooling their samples into one area-weighted score is what everybody does — which on Britain and New Zealand serves Britain 1.2 times worse than it could be served alone and New Zealand 125 times worse. The worst-case objective makes them equal at 33 and 59, and the cost of sharing rises with separation from 1.4 to 59.

 Three boundaries, three tripoints. Three bilateral boundaries drawn through one nominal point, each described as the line between two monuments and each realised under a different convention — a geodesic, a rhumb line and a straight line on a Mercator sheet. The tripoint is defined three times, once by each pair of boundaries, and the three definitions are the three marked crossings. They are 21.33 kilometres apart at the widest and enclose 194.960 square kilometres. Under one convention throughout, the same construction puts all three crossings within 0.0 millimetres of each other — which is the refusal this figure carries, and the reason the triangle is a fact about the conventions rather than about the crossing arithmetic. Grids, and what a survey does

A tripoint defined three times

A tripoint is very often not a coordinate in any treaty. It is a description — the point where the boundary between A and B meets the boundary between B and C — and each of those boundaries is itself a description. So the point is defined three times, once by each pair, and under one convention throughout the three definitions agree to half a micrometre. Under three they enclose 6.69 square kilometres.

 Five bearings of four cities, held exactly. A flat picture of London, New York, Tokyo, Sydney holding five one-way bearings exactly, each drawn as an arrow from the city it is measured at: London to Tokyo, Sydney to London, New York to Tokyo, New York to Sydney, Tokyo to Sydney. Five is the most any flat picture of four places can hold, and of the 192 ways of choosing five pairs and a direction on each, this is one of eight that can be drawn with every arrow pointing forwards; all eight leave London–New York, dashed, unheld. The price is paid by the other seven bearings: the worst, London to New York, is 165.0° from the truth. The impossibility

Five bearings of twelve, and only eight ways to draw them

A flat picture of four places can hold five of their twelve one-way bearings exactly, the same count as distances, and the five are found by solving equations rather than searching. The equations cannot tell a line from its reverse. Of the 192 ways to choose five, eight can be drawn with every arrow pointing forwards, all eight leave London and New York unheld, and on sixteen cities no choice made at random can be drawn at all.

 Three lines of position, and a cocked hat that misses the ship. Three bodies whose azimuths from the ship are 20°, 140°, 250°, each observed with an independent error of about 2 km along its azimuth, give three lines that do not meet. They bound a triangle of 7.5 km², outlined, and the ship — the dot — lies outside it. With independent errors that is the usual outcome: three times in four, whatever the triangle's shape. Paths and directions

A cocked hat holds the ship one time in four

Three lines of position with any error in them bound a small triangle, and the instinct is that the ship is inside. With independent errors it is inside one time in four, whatever the triangle's shape. A common error in every sight makes that almost always when the stars surround the ship and almost never when they do not. And on a sheet that forgets the latitude, the triangle keeps its size and moves off the ship altogether.

 The picture is kept, at four tolerances. One closed curve of 3001 vertices, simplified at four tolerances. Douglas–Peucker's promise holds in every panel: no discarded vertex is further than ε from the line drawn in its place, measured at 0.1158 against 0.128 in the last. The picture survives. The enclosed area does not: it falls by 5.43 per cent, and it falls rather than wandering, because cutting a corner takes area off and never puts it back. What a machine does with it

A tolerance is a promise about the picture

Douglas–Peucker guarantees exactly one thing: no vertex it discarded is further than ε from the line drawn in its place. It says nothing about the enclosed area, nothing about which side of the boundary a point ends up on, and nothing about whether the curve still fails to cross itself — and all three are what the geometry is usually being asked.

 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%. What each projection optimises

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.

 One map made of two, and the latitude where they meet. Goode's homolosine drawn without its interruptions: sinusoidal between 40°44′11.98″ north and south and Mollweide beyond, joined at the latitude where a parallel has the same length in both. Dashed: the join. Heavy: four meridians, drawn from 60° south to 60° north, which cross it. The central meridian crosses smoothly; the others turn a corner there, reaching 13.30° at 110° of longitude, marked. The map is equal-area everywhere, and every parallel keeps its length across the join. The families

Joined where the parallels agree, and the meridians turn a corner

Goode's homolosine is two projections joined at 40°44′11.98″, and three quantities match there: the length of the parallel, the areal factor, and the spacing of the parallels on the central meridian. The fourth does not. Every meridian but the central one turns a corner at the join, up to 13.30° at 110° of longitude, and the angular deformation drops 26° across a line of no width. Moving the join decides where that corner is paid, not whether.

 Two defensible answers from the same four sights, and they are kilometres apart. Four bodies within sixty degrees of one another at azimuths of 60°, 80°, 100°, 120°, each sight carrying an independent error of about 2 km and the same common error of 3 km. The four lines are nearly parallel, so a step towards every body at once is almost exactly a step to the south. The two-unknown least-squares point takes that step and lands 3.66 km from the ship; the three-unknown fit, which solves for the common error as well, refuses it and lands 2.22 km away — its estimate of the common error being 1.87 km. Neither is wrong. They are the two ends of one exchange. Paths and directions

What the extra unknown costs where nothing can see it

Four lines of position can estimate a common error in every sight as well as the position. The estimate is honest — three kilometres put in comes back as 2.8 — and where the residual is blind, its scatter is plus or minus thirty-one, and the fix that carries it is scattered fourteen kilometres from the ship against the biased fix's two and a half. Letting the residual choose between the two is worse than either, because a test with no power is a one-in-twenty lottery on the worse answer.

 The same composite with a strip spliced in, and the meridians no longer turn a corner. Goode's composite redrawn with an equal-area strip of 8° of latitude spliced between the sinusoidal and the Mollweide cap, shaded. The strip's parallel half-length is the unique cubic matching the sinusoidal's value and slope at 36.74° and the Mollweide's at 44.74°, so every meridian arrives and leaves along one direction and the 11.2° corner the plain join turns at 60° of longitude is gone. Nothing else about the map is changed: the cap is the unmodified Mollweide, and the composite is equal-area throughout. The families

An equal-area strip removes the corner and charges nothing for it

Goode's seam matches three quantities and breaks the fourth: every meridian but the central one turns a corner there. A third map spliced between the two removes it — four conditions on one function, satisfied uniquely by a cubic — and the expected price does not arrive. The spliced composite carries less angular deformation across the seam than the plain join, not more: 34.7° against 37.7° at sixty degrees of longitude. What it does leave is the same defect one derivative up, a jump in curvature falling as the reciprocal of the strip's width.

 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. What each projection optimises

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.

 Two source geometries of 96 cells each, rebinned to the same three targets. Both curves start from a source of 96 cells and rebin to targets of 32, 128, 512 cells, so the count ratio is identical along them and the only difference is the shape of the source cells: gnomonic squares on a cube against rectangles in longitude and latitude. The ratio dominates — both curves fall by more than half across the range — and the shapes still separate by 25 points at the middle target. The cube loses less, because its cells are all much the same size and the lon/lat source's collapse towards the poles. What a machine does with it

The same number of cells, in two shapes

Moving a field between two cell schemes loses 18 per cent of it per cell in one geometry and 39 in another, and the earlier measurement could not say whether that was the shape of the cells or the ratio of their sizes, because changing the schemes changed both. Holding the counts settles it: the count ratio decides most of the loss, and the shape is still worth a quarter of the field.

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