Concept

Estimator — where it appears

A rule for turning data into a number, as distinct from the number it is estimating. A maximum over a sample, a quadrature, a best fit and a finite difference are all estimators, and each has a bias and a spread of its own that the quantity it estimates does not.

Named by 20 essays across 6 fields — each of them below, with the objects they name alongside it.

The signal, and four instruments' noise. The spherical excess of an equilateral triangle at 45° north against its side, with the standard deviation of a measured excess — σ√3 — ruled for four instrument accuracies. A fifty-kilometre triangle, which is about the largest anybody routinely observed, has an excess of 5.49 seconds of arc. A theodolite reading to one second gives that excess a standard deviation of 1.73 seconds, so the measurement carries about three significant bits. Everything in this rung follows from that ratio.

How big a triangle it takes

Gauss proved that a surface-dweller can read the curvature off a triangle's angles. Doing it is another matter: a fifty-kilometre triangle has 5.49 seconds of excess, a one-second theodolite gives that excess a standard deviation of 1.73, and reading K to one per cent needs a side of 281 kilometres. Not one of the great surveys built a triangle within a factor of three of that.

impossibility · Curvature
The score does not settle at any resolution. The compactness of one stated boundary — a circle with cosine ripples at eight geometrically spaced wavenumbers, so it has structure at every scale — read at sixteen vertices up to two thousand and forty-eight. The ground score falls from 0.980 to 0.834, and it keeps falling: the boundary's length grows without bound as it is resolved while the area it encloses converges, so the quotient has no limit. The four page curves sit within a fraction of a per cent of the ground curve and of each other, which is the comparison this rung exists to make.

The score is not stable at any scale

One boundary, read at eight resolutions from sixteen points to two thousand and forty-eight: the compactness score falls from 0.980 to 0.834 and is still falling. Changing the projection instead moves it by 0.69 per cent. The two decisions are made by the same person on the same afternoon and only one of them is ever reported.

paths · Reach
The set a reach map shows, drawn from eight bearings. A geodesic disc of 4,000 km and the polygon a fan of eight bearings draws round it, on an equal-area azimuthal page centred on the disc so that the shaded ground is proportional to the ground it stands for. Every vertex of the polygon is on the true boundary and every edge between two of them is a chord, so the drawn set is inside the true one — always, at every count, for any convex reach set. The area it misses is 7.53% of 48,635,855 km², and it is not an error that care removes. It is what a finite fan is.

Every reach set ever drawn is too small

An isochrone is drawn by walking out along a finite number of bearings and joining the points, so its vertices are on the true boundary and its edges are chords — which puts the drawn set inside the true one, always, at every count, for any convex reach set. The deficit falls as the square of the count, and a spherical cap loses less than a circle by exactly cos t (1 + cos t)/2.

paths · Reach
The least error a flat picture can have, against how much sphere it spans. The same configuration of places, shrunk about its own centroid so that every bearing is kept and only the span changes, with the least worst-case relative error of the best flat picture at each size. Both axes are logarithmic. The fitted slope over the rows below ninety degrees is 2.0246: the error falls as the SQUARE of the diameter. That is Gauss's theorem arriving as a number for a finite set — curvature is a second derivative, so its first effect on a distance is quadratic in the separation — and it is why a county fits on a sheet and a hemisphere does not.

How wrong a flat picture has to be

The rung below proves no flat picture of four places is exact and leaves the size of the failure to a determinant nobody can read. Measured directly, the least error falls as the square of how much sphere the places span — fitted exponent 2.0088 — and the same exponent comes back from five different arrangements while the constant in front of it moves by a factor of seventeen.

impossibility · Embedding
The plan, the crossing that was flown, and the one a right forecast would have given. A crossing of NaN km planned against a jet forecast to sit on the fortieth parallel, flown through a jet that is actually 6 degrees further south, and re-planned six times on the way. Committing to the plan costs 167.81 hours. Re-planning costs 165.84. A vehicle that had known where the jet was would have taken 163.21. So the wrong forecast is worth 4.60 hours and re-planning recovers 1.97 of them.

A crossing is a chain of decisions

Eleven rungs hand back a curve and stop, and nothing anybody flies is decided once. Re-planning is worth exactly nothing when the forecast turns out right — a sub-path of an optimal path is optimal, and the chain is the plan. When it turns out wrong it recovers 19 per cent of the cost at two decisions and 83 at eighteen, and never all of it.

paths · Paths
Every projection against a picture that is not a map. The worst relative distance error over London, New York, Tokyo, Sydney, for each projection in the library and for the best flat arrangement of the same distances. Each azimuthal member is centred on the set's own centroid, which is the fair comparison. The free picture reaches 0.63% and the best projection, Azimuthal equidistant, reaches 8.28% — a ratio of 13.17. The free picture cannot lose, because every projection's own layout was handed to the search as a starting point; what the figure measures is how much the freedom is worth, and it is worth different amounts at different sizes.

The best flat picture is not a map

A set of dots whose separations are as nearly right as separations can be made beats every projection in this collection — by a factor of thirteen on four world cities, and by eight per cent on sixteen. The collapse between those two numbers is not about cartography. It is that a picture of n places has 2n − 3 free numbers and n(n − 1)/2 distances to spend them on.

impossibility · Embedding
One of these is a corner. The corner reported at the exact conformal seam, and the corner reported at a cube's gnomonic seam fifteen degrees along its edge, each measured over a shrinking arc. The gnomonic's is 21.4572° at every span from twenty-four degrees down to three — the same number to four decimals — because it is a corner. The conformal one halves whenever the span does, fitted exponent 0.990, because a tangent read from a finite chord of a curved image departs from the true tangent in proportion to the chord. It is not a corner; it is the instrument.

The exact map says the seam is smooth

The previous rung could only bound the conformal seam's corner at about two degrees, because the series it was fitted with holds its boundary condition to three parts in a thousand. An exact map exists — the stereographic projection composed with ∫dt/√(1 − t⁴) — and it settles it: the corner falls in exact proportion to the arc the tangent is read over, fitted exponent 0.99, while the gnomonic's 21.4572° is the same to four decimals at every span.

families · Polyhedral
Three face maps on a cube, over a shrinking span. The corner a feature gets crossing the seam of a cube, measured by reading the tangent over an arc and then shrinking the arc by a factor of sixteen. A chord differs from a tangent in proportion to the arc, so a perfectly smooth join reports a corner that HALVES when the span halves — a slope of one on these axes. None of the three lines has a slope of one. The fitted slopes are 0.000, 0.000, -0.018, which is a flat line in each case, and a flat line is a real corner. The three differ in size and not in kind: 20.1513°, 7.2772°, 0.7687°.

The span ladder, run on all five

A recorded shortfall said the exact map's span ladder was one loop away from settling the seam on the five Platonic solids. The loop was run and the answer is the other one: the series conformal face map has a corner that does not shrink with the measurement span on any of them — 4.10° on the tetrahedron, 0.77 on the cube, 0.004 on the icosahedron — and it belongs to the truncation rather than to conformality.

families · Polyhedral
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.

wrong · Identify
The threshold, and the two things it is a ratio of. The near-optimal set's fracture threshold on Robinson, against the size of the region, with the two quantities it is a ratio of drawn beside it. The threshold falls with fitted slope -1.293. The best score a region admits at all rises with slope 0.923 — a bigger region is harder to map — and the absolute score of the pass falls with slope -0.370. The first is the sum of the other two by construction, and the arithmetic says which of them is doing the work: the denominator carries 71 per cent of it.

The first break is mostly its denominator

Three rungs have fitted the near-optimal set's fracture threshold against region size and read the answer as a statement about the landscape. It is a ratio, and separating it takes one multiplication: the pass's own depth is constant to 12 per cent below twenty degrees of span, and the whole of the threshold's movement there is the denominator — the best score the region admits at all — rising with exponent 0.92.

choosing · Choosing
One quantity, one region, and the exponent left free. The scale departure of six projections over the world, aggregated as a p-norm, against p on a logarithmic axis. At p = 1 the best is Eckert IV; at p = 64 it is Winkel tripel. Nothing about the maps changed between the two ends of the axis — only how much of the region a bad point is allowed to spoil. Drawn in no projection: the axes are an exponent and a score.

The average was a choice of norm

Ten projections, one region, one measured quantity, and the only free decision left is how to turn a field into a number. Over the world's scale departure the ordering at the mean and the ordering at the worst case have a rank correlation of −0.04, all ten maps change position, and the exponent that produced each answer is stated nowhere.

wrong · Audit
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
The areal factor over 0° to 60° north, and the points it is measured at. Mercator's areal factor shaded over 0° to 60° north, from 1.0001 to 3.8473, with the 12 × 12 grid of cell centres a regional measurement uses drawn on top of it. The cross marks where the quantity is actually largest, found by a search that is allowed to leave the sample; the ring marks the largest value the sample contains. The grid's answer is 3.4639 and the real one is 4.0000, short by 13.40% — and the reason is visible in the picture, because no cell centre is ever on an edge.

The worst point is not on the grid

Every maximum distortion this collection has printed is a maximum over a sample, and a maximum over a sample is a lower bound. On Mercator over a sixty-degree band a twelve-by-twelve grid reports 3.464 where the answer is exactly 4, and the shortfall does not go away with refinement so much as decay at a rate that says where the extreme is hiding.

distortion · Sampling
One of these averages exists. Mercator's area-weighted mean areal factor, and its Kavrayskiy number, against how close the sampled band comes to the pole. The first is artanh(sin Φ)/sin Φ in closed form and has no limit: it passes 7.04 at a tenth of a degree from the pole and keeps going, gaining a fixed amount every time the remaining gap is halved. The second settles by 85° and does not move again. Both are published as summary distortion figures for the same map.

A mean that does not exist can still be printed

Mercator's area-weighted mean areal factor is artanh(sin Φ)/sin Φ, and it has no limit. A sampler asked for it returns the logarithm of its own sample count plus 1.512 — measured slope 1.001 against ln n — so the number is a property of the person who computed it. The Kavrayskiy number for the same map over the same sphere settles at 0.52124 and is a number.

distortion · Sampling
Three ways to cover a sphere with 900 points. The three samplers this collection's numbers are computed with, each with about 900 points, drawn on the Mollweide projection so that equal areas on the sphere are equal areas on the page and the crowding is the samplers' rather than the map's. The graticule piles points at the poles; the equal-area rings space them evenly by area and unevenly by distance; the Fibonacci lattice trades a little of each. None of them is equal-area, and no finite set is.

There is no equal-area lattice on a sphere

Every number in this collection is an average over a point set, and the three point sets available all fail to be equal-area in different ways. The equal-area ring sampler this site has used since its early essays is the one whose outermost ring sits half a step inside the rim — which is how a measured scale spread once came in 3.42 parts in a thousand below a proved bound.

distortion · Sampling

A refinement that stops moving

Doubling the sample and watching the answer settle is how every quadrature in every field is checked. On the Robinson projection the doubling ladder — 4, 8, 16, 32, 64, 128 — converges beautifully, with its increments halving at every step, on a limit that is wrong by a factor of twenty-seven. Whether a grid finds the answer is decided by whether n is a multiple of four.

distortion · Sampling

Which of these numbers are the sampler's

Four rungs have shown that a sampled maximum understates, a sampled mean can be a report on the sampler, no arrangement of points is neutral, and refining until the answer settles proves nothing. So the collection re-measured itself. The means move by at most 1.1 per cent, the worst points by up to 19, and the rankings — which is what the essays actually argue with — do not move at all.

distortion · Sampling

The area is unbiased and the perimeter is not

A boundary measured from noisy vertices comes out long, always, by σ²/d on every leg. The area enclosed by the same vertices comes out exactly right, because a shoelace is bilinear and the cross terms vanish. So densifying a boundary makes its area five times more precise and its perimeter three thousand times more wrong, and every compactness score computed from it falls short.

distortion · Precision

The sample was drawn on the page

Every mean in this collection integrates over the sphere, because that is where the ground is. A raster, a pixel loop and any figure that walks its own canvas integrate over the page instead, and the difference is exactly the covariance between the quantity being measured and the map's own area distortion — 7.2° of mean angular deformation on Miller becoming 18.0°.

distortion · Sampling

Named alongside it

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

VerificationToleranceClosed formConvergence ratePurposeSamplingRegional distortionConvergenceAreal factorDegeneracyQuadratureAngular deformation

All concepts