Identify — the ladder · Cartographic Projection

Every candidate fitted to one map's graticule, ranked by what is left over. The map is drawn in Conformal conic over a region 40° tall centred at 45° north, and the projection is not told to the fit. Each candidate is evaluated at the same 121 graticule crossings, its own parameters are searched, the best plane similarity between its output and the picture is removed, and what remains is drawn as a proportion of the map's own width on a logarithmic scale. Conformal conic fits to 1.3e-10, which is the arithmetic's floor; the next candidate is 3.7e+7 times worse.

A map does not say what it is

Every map on this site is one the site drew, from a projection it chose. Every map a reader has ever used is the other kind — a picture whose projection is a sentence in a corner, a legend, or nothing at all. The graticule is enough to recover it: fit every candidate to the crossings and rank what is left over.

rung 1 · wrong

How large a map has to be before a wrong projection stops fitting it. Three rivals fitted to a Mercator graticule centred at 45° north, over regions from 1° to 70° of half-extent. The residual is the shape difference alone, with the best scale, rotation and offset removed, and both axes are logarithmic. The horizontal rule is a fifth of a per cent of the map's width — about the width of a drawn line on a printed sheet — and where a curve is below it, no measurement of that map can tell the two projections apart, however carefully it is made.

Two projections that cannot be told apart

Over a small enough region every projection is the same picture, so the question is how small. The answer is not one number: two conformal projections need twelve degrees of extent before their graticules can be separated, and two projections with different anisotropy separate below one.

rung 2 · wrong

The half-extent at which each rival stops fitting a Gall–Peters graticule. For each candidate, the size of region at which its best fit to a Gall–Peters map first leaves a residual of a fifth of a per cent of the map's width. Below that size the two are the same picture. The numbers are half-extents in degrees of latitude, at 20° north, with the plane affine transformation removed; a bar at 90° is a rival that never separates at all within the range searched.

What a careless copy hides

A photocopier that stretches one axis is a nuisance to remove before a projection can be identified. Removing it costs more than it looks: an affine fit absorbs the entire difference between the cylindrical equal-area projections, so Gall-Peters and Behrmann become the same picture at any size, to sixteen decimal places.

rung 3 · wrong

The same ranking, with the right answer removed from the library. A map drawn in Mercator, fitted by every candidate except Mercator. Something still wins: Conformal conic, by a factor of 1.64 over the runner-up, leaving 0.4 per cent of the map's width unexplained. With Mercator in the library the winner's margin is 1.2e+13. The ranking always produces a name; what tells the two situations apart is how far ahead the name is.

When the answer is not in the library

Fitting twenty candidates to a map and ranking the residuals always produces a winner, which makes it a ceremony unless it can also produce a refusal. Held out of its own library, a Mercator map is named as a conformal conic, leaving 0.4 per cent of the map's width unexplained — and the quantity that tells the two situations apart is not the residual but the margin, which is 1.6 when the truth is absent and 10¹³ when it is present.

rung 4 · wrong

Two explanations for one residual, against the size of the region. A map in the right projection whose control coordinates are on the wrong datum, and a map fitted with the wrong projection and the right datum, both measured as a residual after the reproduction's scale, rotation and offset have been removed. The datum shift's residual is flat: it is the same 1.3e-6 at every size, because a similarity fit is very nearly what a datum shift is and it absorbs the rest. The wrong projection's grows by three orders of magnitude with the region. They are the same number below about a degree, which is where a residual stops saying anything about either of them.

A residual has more than one explanation

The method names a projection by fitting every candidate to a set of control points and taking the smallest residual. It has never been asked what else a small residual could be. A map drawn in the right projection from coordinates on the wrong datum leaves a residual of one part in a million — indistinguishable from noise, at every region size, because a similarity fit absorbs a datum shift almost exactly.

rung 5 · wrong

What each configuration can see, and what it cannot. The smallest eigenvalue of the fit's own normal matrix — how much the residual changes for a unit move in the worst direction of parameter space — for three parameterised candidates against six configurations of the same size. A zero is not a hard fit: it is a direction the control points cannot see at all, so every value of the parameter along it gives an identical residual. 2 of 18 are at the floor of double precision, and they are not the ones a reader would guess. Where none is, the spread between the best and worst arrangement is still Infinity at a fixed point count.

Where the control points are

Five rungs fit a library to a map and ask how much to trust the winner. None asks whether the parameters are recoverable at all. On control points along one parallel an equirectangular's standard parallel is not merely hard to find — it is invisible, exactly, and five hundred and twelve points on the same parallel are as blind as eight.

rung 6 · wrong

A plane fit swallows a datum shift, and keeps swallowing it. A map of OSGB36's ground drawn as though it were on WGS84, at seven sheet sizes. The upper curve is how far the drawing moves — a real 99-metre error on the ground — and the lower one is what survives the best similarity between the two, which is what a fit for the projection removes for free. The absorption runs from 99.68 per cent on a 1° sheet to 85.16 on a 60° one, and what is left is 3.9 parts per million of the map's own size even then — two microns on a sheet half a metre across.

The datum hides inside the projection's parameters

Six essays recover a projection from control points with the body it was drawn on taken as known. It is not known, and the fit cannot find it: a 99-metre datum shift is absorbed to 99.7 per cent on a two-degree sheet and to 85 per cent on a hemisphere, leaving eleven parts per million of the map behind.

rung 7 · wrong

Which rotation a projection cannot see is decided by the projection. The same rotation — 2.455 arcseconds, DHDN's polar one — applied about each of the three axes in turn, with the residual after the best plane fit drawn on a logarithmic scale. A cylindrical and a conic in their normal aspects hide the polar rotation to arithmetic noise, 3e+5 times better than either equatorial one, because a change of longitude is a symmetry of both. A pseudocylindrical hides none of them — its horizontal coordinate carries a factor in latitude, so a longitude shift is a shear rather than a translation. And an azimuthal centred on the equator hides the equatorial rotation instead.

A rotation is not absorbed the way a shift is

The previous rung expected a datum's rotations to be the part a plane fit could not swallow. They are the part it swallows best — 93 times better than the translations over a hemisphere, and 24 times better per metre moved — and the reason is that the rotation which matters is a change of longitude, which is a symmetry of the map.

rung 8 · wrong

How many projections the map could be in. The number of candidates whose residual sits below the measurement noise, against the size of the region, at four noise levels. At one per cent of the map's width — a hand-digitised graticule — a four-degree region admits ten of the twenty candidates and a forty-degree one admits exactly one. Every curve falls, none of them crosses another, and all four end at one: identification works, and what it needs is extent rather than precision.

The answer is a set

Eight rungs have produced a best fit — one projection, ranked first, with a margin. A best fit without a spread is not a measurement, and the spread is free: a control point has a width, and every candidate whose residual is inside that width has not been ruled out. At one per cent noise a four-degree region admits ten of twenty candidates and a forty-degree one admits exactly one.

rung 9 · wrong

The sheet moved before it was measured

Nine rungs take control points off a map and assume the sheet they came from is the sheet the cartographer drew. Paper shrinks across its grain three times as fast as along it, and on a map whose grain runs along its own axis that shrinkage is EXACTLY a change of standard parallel — one per cent moves the recovered parallel by 0.57 degrees with the residual sitting at the solver's floor.

rung 10 · wrong

The whole of what an unlabelled map gives you. The outline of Japan as drawn on Conformal conic, delivered as an ordered list of page positions with nothing attached to any of them. No latitude, no longitude, no scale, no north. The rung's question is whether a projection can be recovered from that, and it can: the correspondence between the ink and the ground is found by sweeping the starting point round the curve and both directions, and the true candidate comes back with a residual of 2.88e-14 against the runner-up's 4.57e-4.

A map with no graticule

Ten rungs are handed control points, and a great many maps have none. Handed an outline with no labels on it at all, the method still works — and works better: the correspondence between ink and ground is recoverable exactly, because a similarity preserves ratios of arc length, and the margin on clean observations is 1.6 × 10¹⁰ against a graticule's 9.9 × 10⁶. What breaks it is noise, at three parts in a thousand.

rung 11 · wrong