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

Chebyshev's criterion — where it appears

The statement that the optimal conformal map of a region is the one whose scale factor is constant along the region's boundary. It converts a search over projections into a boundary-value problem, and it is one of the few results in the subject that characterises an optimum rather than approximating it.

Named by 13 essays across 4 fields — each of them below, with the objects they name alongside it.

The least distortion possible over a 30° region. Scale factor along a radius of the cap, each projection normalised to unit scale at the centre. Chebyshev's criterion names the projection whose scale is constant on the boundary as the conformal map of least scale variation, and for a cap that is the stereographic projection centred on it — reaching exactly sec²(ρ/2) = 1.0718 at the rim, marked. Every other conformal projection drawn here rises past that line before it gets there. It shows an optimum rather than a comparison.

Chebyshev's criterion

The only optimality theorem in the subject. Among all conformal projections of a region, the one with least scale variation is the one whose scale factor is constant on the boundary — a criterion with a proof, a unique answer, and a test anyone can run.

choosing · Choosing The scale factor across Europe, from the middle outwards. Each band is the range the larger principal scale factor takes on the ring at that distance from the centre of Europe: 0 is the middle, 1 the frontier. two of the three projections drawn are conformal, and every one of those reaches its maximum at the right-hand edge, because the logarithm of a conformal map's scale factor is subharmonic. None of them reaches a maximum inside the region.

Where the worst point is

The largest scale error on a conformal map of a country is always on the frontier, never inside it, whatever the country's shape and whichever conformal projection was chosen. It is a theorem rather than a tendency, and it is the reason Chebyshev's criterion works.

distortion · Tissot 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.

choosing · Condition The scale spread of every grid Britain could have adopted. A grid is conformal by requirement, so its angular deformation is zero everywhere and the whole design problem is the spread of its one remaining number: the largest scale factor over the region divided by the smallest, in parts per million. The grid's own scale factor does not enter — multiplying every scale by a constant leaves the ratio alone, which is why it is chosen last. The bottom bar is Chebyshev's optimum, the conformal map of this region whose scale is constant on its boundary, which no map of any family can beat; the adopted grid sits 1.98 times above it. Measured over a stated box rather than a coastline, because a coastline would put the vendor's generalisation into the answer.

The best grid a country could have had

A national grid is a conformal map chosen for one region, so its whole design problem is one number: the spread of its scale factor. That number has a theoretical floor, this site can now compute it, and the adopted grid turns out to be either exactly optimal or half as good again — depending entirely on which box the country is declared to be.

practice · Grid The bound was spherical, and the country is not. Three numbers per region, all in parts per million of scale spread. The first is the Chebyshev bound computed on the sphere. The second is that same optimal map used on the ellipsoid, which is what adopting it would actually deliver. The third is the bound with the ellipsoid-to-sphere factor put into the boundary condition, which is the real optimum. The penalty for using the spherical answer reaches 1.22 times — while a named candidate barely moves, because an optimal map has cancelled its own variation and has nothing left to hide a new one in.

The bound on the body the country is on

The best conformal grid a country could have had was computed against a bound solved on a sphere, with a note saying the flattening was second order and unquantified. It is second order for a named projection — every family's best candidate moves by under 0.7 per cent — and it is 22 per cent for the bound, because an optimal map has already cancelled its own variation and has nothing left to hide a new one in.

practice · Grid 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.

choosing · Condition 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.

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.

impossibility · Topology 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.

choosing · Condition 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.

choosing · Condition A sea with two islands in it, and the three circles it already is. A stated sea 60° across with two islands 12° across, their centres 12° from its own, drawn on the globe and drawn again under the stereographic projection centred on the sea. Stereographic projection carries every circle on the sphere to a circle in the plane, so the flat picture is bounded by three exact circles with no fitting and no iteration: the sea's edge at radius 0.5359, and the two islands at radius 0.1060 with their centres 0.2108 from the origin. Normalised, the region's three conformal invariants are 0.2361, 0.6848, 0.1221 — three numbers, which is what a region with two holes carries and no angle-keeping map can change.

Two islands are three numbers, and the proof does not survive them

A sea with one island at its centre has one conformal invariant and a forced best map. Put two islands in it and there are three invariants, readable in closed form because stereographic projection carries circles to circles — and the forcing is gone. The theorem that picked the map needed the region to be unchanged by rotation, not to have one hole, and a single harmonic term two hundredths in size beats the stereographic by 0.19 per cent the moment an island moves off centre.

impossibility · Topology The same two islands, and the cut that buys most is the central one's. A sea 60° across with a large island A, 12° across, and a small one B, 6° across, on opposite sides of its centre, drawn in the plane the stereographic makes of it. Each dashed line is one way of spending a single cut — from A to the open sea, from B to the open sea, or from A to B — and the solid line is the one that buys most, as a share of what two cuts buy. Left, A 18° from the centre and B 6°: A 9% · B 78% · between 21%. Right, A 6° and B 18°: A 81% · B 12% · between 11%. The better cut is the central island's in both, whichever island is large, and a cut joining the islands buys 21% and 11%.

One cut belongs to the island that would cost most alone

A sea with two islands pays a few per cent for being drawn whole by an angle-keeping map, and one cut buys back part of it. Which island to cut is a design choice that leaves the ground exactly where it was, and it is not decided by size: across twenty arrangements of a large island and a small one, the cut worth most goes to the island that would cost the sea most if it were alone in it — in eighteen of twenty, the other two near-ties — which is usually the more central one. A cut joining the two islands buys almost nothing, and the better cut is nearly always the longer.

impossibility · Topology The two islands dearest alone are the wrong pair to cut when they sit on the same side. One of forty three-island seas, 60° across, drawn in the plane the stereographic makes of it; beside each island, what it would cost the sea if it were alone in it — P 0.97%, Q 0.72%, R 1.68%. Left: the two dearest islands, P and R, cut to the open sea, which buys 75% of what three cuts buy. Right: the best pair, Q and R, which buys 97%. The dearest islands are 69° apart as seen from the centre, and their single cuts overlap by 40 points.

The second cut goes across the sea

With two islands and one cut, the island that would cost the sea most alone names the cut. With three islands and two cuts, the obvious rule — cut the two dearest — names the best pair in only 23 of 40 seas, and loses up to 23 points of what three cuts would buy. It fails when the two dearest sit on the same side of the sea, where their cuts do each other's work. The best pair keeps the dearest island in 37 of 40; its partner is the island across the centre from it, which is right in 32 of 40 and never costs more than 12 points. A cut joining two islands is never worth a second cut to the sea.

impossibility · Topology With five islands the two dearest are still the wrong pair when they share a bearing. One of 20 five-island seas, 60° across, drawn in the plane the stereographic makes of it; below, each island's price to the sea alone — P 0.73%, Q 1.30%, R 0.36%, S 1.96%, T 2.74%. Left: the two dearest, S, T, cut to the open sea, which buys 77% of what five cuts buy. Right: the best pair, Q, T, which buys 100%; the dearest island and then the one farthest from it in bearing — Q, T — buy 100%. One cut buys 77%, three 100%, four 100%.

A sea with five islands is seldom worth five cuts

With three islands in a sea, the second cut went across the sea from the first, and the number of cuts worth spending looked like the number of directions the islands point. With four and five islands that survives in part and breaks in part. Choosing the dearest island and then, each time, the island farthest in bearing from every island already cut names the best pair of cuts in 28 seas of 40, where the two dearest name it in 16 — and it fails at three cuts of five, because the best three cuts often do not contain the best two. The dearest island is no longer reliably among the best pair. And 39 seas of 40 are worth fewer cuts than they have islands: the count follows the widest gap in the islands' bearings more closely than it follows how many islands there are.

impossibility · Topology

Named alongside it

The objects these essays reach for when they reach for this one.

ConformalityScale factorVerificationMinimaxOptimal conformalInvariantPurposeSeamStereographicTopologyDegrees of freedomLeast-squares

All concepts