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

Similarity transformation — where it appears

A rotation, a uniform scaling and a translation of the plane, which preserves every shape and every angle. It is the four-parameter map a projection identification removes to account for how a map was drawn or scanned, and it also absorbs almost all of a datum shift.

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

OSGB36 to WGS84, one parameter at a time. Each bar is how far the mark at 2.0° west, 54.5° north moves under one of the seven parameters with the other six set to zero. seven of the seven are not zero for OSGB36. The units hide the comparison: one arcsecond of rotation moves this point 30.8 metres and one part per million of scale moves it 6.36 metres, so OSGB36's scale term contributes 130 metres — more than one of its three translations.

The seven parameters, and what each one does

A datum transformation is published as three metres, three arcseconds and one part per million, in three different units — which hides the fact that all seven are the same size of thing. Converted to ground displacement, the scale term beats two of the translations.

datums · Datum What 4 parts per million of network strain leaves behind. Each arrow is where the best-fitting seven-parameter transformation leaves a marker, over 36 markers laid out as a grid across OSGB36's ground. The RMS residual is 1.64 metres and the worst is 3.00 metres. Seven parameters span the constant and linear parts of a displacement field; this one is quadratic, so no choice of the seven can reach it. The arrows are drawn 27× life size.

Where a fit leaves residuals

Seven parameters can carry a rigid motion and a size exactly. A triangulation network is neither, so the best possible transformation between two datums leaves metres on the table — in a pattern, not as noise — and which seven parameters come out depends on where the markers were.

datums · Datum 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.

wrong · Identify Fitting a polynomial to Mollweide, and what each order buys. An affine, a quadratic and a cubic transformation fitted by least squares between the sphere and Mollweide over patches from 8° down to 0.5° radius, centred at 20°E 40°N. Each is a straight line on these axes and its slope is one more than its own degree: 1 → 2.00, 2 → 3.00, 3 → 4.00. That is not a coincidence and it is this ladder's subject: the first thing a model of degree d cannot represent is the term of degree d+1, so the affine model's error is governed by the second derivative — the flexion and skewness measured everywhere else here.

A local model has an order

Every georeferencing tool fits a polynomial between two coordinate systems and the choice of degree is usually made by counting control points. What it buys is an order of convergence — 2, 3 and 4, measured — and the first term an affine model cannot hold is the second derivative this ladder has spent five essays on.

distortion · Flexion 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.

wrong · Identify The same map on four differently prepared pages. Mollweide at 30°E 20°N, then the same projection with a rotation and a magnification applied to the page, then with one axis stretched by 1.6, then with a shear of 0.5. The similarity changes nothing: flexion, skewness, ω and the anisotropy are identical to every printed figure, and only the last column — the same turning measured per unit of page arc rather than per unit of ground arc — moves, by exactly the magnification. The stretch and the shear change all of them, and flexion by 35 per cent.

The second derivative is not an invariant

The first-order ladder established which quantities survive a change of coordinates and which are artefacts of the parameterisation. Asked of the second order, the answer is that flexion survives a rotation and a magnification of the page exactly, and survives nothing else: a stretch of 1.6 in one axis moves it by eleven per cent, a shear of 0.5 by thirty-five, and a shear of 3 leaves a ranking of eight world maps with a rank correlation of −0.07 to the one it started with.

distortion · Flexion Two parameter sets 100 metres apart, over the region they were fitted to. Each marker is drawn at a size proportional to how far the two transformations put it apart. The second set differs from the first by 100 metres of translation along the direction this network can least see, with the rotations and the scale re-fitted to absorb it — which is what a second agency's adjustment does when it chooses a different constraint. The worst disagreement anywhere in the region is 5.64 metres and the mean is 3.61. Applied at south-eastern Australia the same two sets differ by 193 metres, because the rotation that absorbed the translation here is a rotation of the whole Earth.

Two parameter sets, one transformation

Agencies publish seven-parameter datum transformations that differ by hundreds of metres in translation, and the usual reading is that one of them is better. Over the region either was fitted to they are the same transformation: a hundred metres of translation, re-absorbed by the rotations and the scale, moves a British coordinate by 5.6 metres and an Australian one by 193.

datums · Datum What a 20 mm closure tolerance lets each station hide. A closed traverse of seven stations in plan, each labelled with the angle blunder that would leave the closure inside a 20 millimetre tolerance. An angle error at a station rotates everything downstream of it about that station, so the closing point moves by the distance from the station to the close — and the last station before the close stands 72 metres from it and can hide 57 arcseconds, against 2.1 at the worst-placed station. The check is not insensitive; it is unevenly sensitive, and nothing in the specification says so.

What a closed figure cannot see

A closed traverse imposes exactly two conditions on its observations, so everything else is free — and the freedom is not spread evenly. At a twenty-millimetre tolerance the worst-placed station in a seven-station loop hides an angle blunder of 2.1 arcseconds and the station standing seventy-two metres from the close hides 57, because a rotation about a point near the finish moves the finish hardly at all.

practice · Reduction 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.

wrong · Identify 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.

wrong · Identify What a sheet does to the points before anybody measures them. The graticule crossings of a map drawn on Conformal conic, with an arrow at each one showing where the same crossing has moved to after the sheet dried — 0.1 per cent along the grain and 0.4 across it, with the grain at 23° to the map's axis, and the displacement magnified 60 times so it can be seen at all. The pattern is a stretch along one direction and a squeeze along the perpendicular, which is what an anisotropic scaling looks like. It is a property of the paper and has nothing to do with the map printed on it. A similarity fit to these points leaves 3.66e-4 of the map's own width unexplained, against 1.22e-10 on the unshrunk sheet.

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.

wrong · Identify 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.

wrong · Identify Millimetres of horizontal move per kilometre of assumed height. How far the transformed latitude and longitude move when the height fed into a datum transformation changes by one kilometre, for four published transformations at five places. It runs from 7.7 to 121 millimetres per kilometre. A latitude and a longitude do not carry a height, so a two-dimensional coordinate cannot be transformed until somebody supplies one that is not in it.

The height a coordinate does not carry

Ten rungs treat a datum shift as a map from one pair of angles to another. It is not one: the transformation runs through Cartesian coordinates, so the answer depends on the height — by 7.7 to 121 millimetres per kilometre depending on the datum and the place, which puts Lhasa 432 millimetres from where a height of zero would have said.

datums · Datum

Named alongside it

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

ResidualToleranceVerificationLeast-squaresDegeneracyDatumHelmert transformationControl pointsIdentificationProjection identificationRealisationReproduction

All concepts