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

Control points — where it appears

A place on a map whose ground position is known, used to recover what the map was drawn with. The identification methods here are handed graticule crossings; a map with no graticule supplies none, and the correspondence between ink and ground has to be found rather than given.

Named by 5 essays across one field — each of them below, with the objects they name alongside it.

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 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.

wrong · Identify 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.

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

Named alongside it

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

IdentificationResidualConfoundingDatumSimilarity transformationAffineDegeneracyEllipsoidEstimatorGeoreferencingIdentifiabilityMargin

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