The band names the exponent for a compact region, and not for a thin one

One exponent names one map, and at tight bands a better one replaced a specification that had forty-two answers with one that has one. Asked for the conformal map of a region that keeps the most area inside a band about true scale, a surveyor gets a criterion that is not convex and a search that stops on whichever ledge it reaches first. Asked instead for the map that minimises the mean of |ln k| raised to a stated power p, the surveyor gets a convex criterion and a single map for every p from one upward.

The price of the substitution was a second question. The power that keeps the most area inside a given band depended on the band. On a 25° spherical square it was p = 1 — the mean departure — at every band up to about one per cent, and the least worst case, the limit of large p, once the band was wide enough to hold that case whole. Between them was a short stretch, from about 1.25 to 1.4 per cent, where an intermediate power was best.

That makes the exponent a function of the band and of the region, and the obvious hope is a rule that names it from the band alone, so that a surveyor is not asked for both a tolerance and an exponent. The earlier measurement pointed at the one number such a rule would use: the region’s least worst departure, the smallest worst-case |ln k| any conformal map of it can reach. The square’s was 1.476 per cent, and its switch sat at nine tenths of that. Whether nine tenths is a property of squares or of bands is a question one square cannot answer.

Ten regions, read against their own worst case #

The regions are chosen to vary one thing at a time. The ellipses run from round to thin at a fixed long radius: a cap, then axes of 1 : 0.8, 0.6, 0.45, 0.3, 0.2 and 0.12. The polygons vary the number of corners at a fixed radius — three, four and six — with every edge a great circle. All ten are solved exactly as the square was: a conformal map built from a series of twelve complex terms, 6,400 samples in forty rings weighted by their area, and for each of twelve exponents from 1 to 32 the one map that minimises the mean of |ln k|^p, found by reweighted least squares continued from p = 2 in both directions.

Their least worst departures span a factor of thirty-five, from 2.38 per cent on the cap to 0.068 on the thinnest ellipse. A band of half a per cent is therefore a tight band on the cap and a very loose one on the thin ellipse, which keeps its whole area inside any band wider than 0.07 per cent. Reading every band in per cent would compare ten different questions. So every band below is stated as a fraction of its own region’s least worst departure, r = τ / w*, and the question is whether the exponent that serves best is a function of r alone.

The region’s size drops out #

The first thing r buys is that size stops mattering. The square’s least worst departure is 0.225 per cent at a long radius of 10°, 1.476 at 25° and 4.134 at 40°, a factor of eighteen; the thin ellipse’s runs from 0.029 to 0.467. Read against those, the area each map keeps inside the band is the same at all three sizes to within three points at every ratio, and for the square to within about one.

That is what the scale field of a small conformal map should do. Its departure from true scale grows as the square of the region’s size, as total curvature and the scale rule found, and to first order the whole field scales by that factor without changing its shape. Dividing the band by w* divides out exactly that factor. What is left over at 40°, where the square’s worst departure is over four per cent, is the second-order part, and it is small. A rule for the exponent, if there is one, can be written in r and in the region’s shape, and never in its size.

Compact regions switch at nine tenths #

On the square the rule the earlier essay suggested holds almost exactly. The mean-departure map is the best of the twelve at every ratio up to 0.85 — it keeps 66.8 per cent at half the least worst case, 81.1 at seven tenths and 88.6 at 0.85, against 48.7, 69.1 and 83.1 for the least worst case. At 0.9 the least worst case draws level and above it wins outright, because by 1.0 it keeps everything and the mean-departure map, whose corners run to more than twice w*, keeps under 91 per cent. The intermediate exponents are best only in the last tenth, and never by more than a point or two.

The triangle and the hexagon do the same, and so, roughly, do the round and the moderate ellipses. What the rule gives up — the largest gap, over bands from 0.3 to 1.0, between the area the best exponent keeps and the area the rule’s exponent keeps — is 1.2 points on the square and on the hexagon, 1.9 on the triangle, 2.0 on the ellipse of 1 : 0.8 and 2.3 on the cap. For a surveyor asked to choose between a map that minimises the mean departure and one that minimises the worst, “the mean below nine tenths of the region’s least worst case, the worst case above” is as good as knowing the best exponent, to within two points of area, on every compact region tried.

The cap is compact in a degenerate way. Every exponent draws the same map on it — the stereographic projection centred on the cap, which is the conformal map with circular scale contours — and the exponents differ only in the overall scale they choose, which sets how the region’s departure is split between too large and too small. So the cap’s best exponent wanders between one and 1.5 without ever mattering by more than two points, and the rule’s loss on it is a loss of scale setting rather than of map.

A thin region passes through every exponent between #

The thin ellipses do not switch. On the ellipse of 1 : 0.2, p = 1.25 is already best at six tenths of the least worst departure; p = 2 at seven tenths; 3 at eight; 4 at 0.85; 8 at nine tenths. The best exponent climbs steadily through every power between the two ends, and over most of that climb it is well ahead of both. At eight tenths of the least worst case the best member of the family keeps 93.4 per cent of the ellipse, the mean-departure map 88.0 and the least worst case 85.5. On the thinnest ellipse, at the same ratio, 94.7 against 88.1 and 87.1.

In that stretch the two ends of the family are the worst choices available rather than the best, and the rule’s loss is correspondingly large: 5.6 points on the ellipse of 1 : 0.3, 5.7 on 1 : 0.2, and 6.6 on 1 : 0.12. It sits between the compact regions’ two points and the thin ellipses’ six for the ellipses between — 3.1 at 1 : 0.6 and 3.8 at 1 : 0.45, which already pass through powers from 1.5 to 6 in the last fifth. So the break is not sharp. Elongation opens the stretch of intermediate exponents gradually, from the last tenth on a square to the last four tenths on the thinnest ellipse.

No switch point rescues a two-exponent rule on a thin region #

The obvious repair is to move the switch. If the thin ellipses leave p = 1 at six tenths rather than nine, a rule that switches earlier for thin regions might recover what the fixed rule loses. It does not. With the switch put at whatever ratio suits each region best, a two-exponent rule gives up 5.6 points on the ellipse of 1 : 0.3, 5.4 on 1 : 0.2 and 6.6 on 1 : 0.12 — within a third of a point of what the fixed switch at 0.9 gives up. On the compact regions the best switch is 0.9 or near it, and it changes nothing there either, except on the cap, where 1.3 replaces 2.3.

So the loss on a thin region is not the wrong switch point. It is the missing middle of the family. Wherever the rule changes from one end to the other, it spends some stretch of bands on an end that is several points short of an intermediate map, because on a thin region both ends are several points short of the intermediate maps over a stretch four tenths wide.

Smooth rules fare worse, which is worth recording because they are the natural next attempt. A rule raising the exponent as 1/(1 − r) reaches p = 2 at a band of half the least worst case and costs every region between three and ten points, because on the compact ones p = 1 is still clearly best at half. Any rule that serves the thin regions by raising p early costs the compact ones by raising it too early for them.

The least worst map spreads its departure differently #

What separates the two kinds of region can be read off the least worst map itself, without solving the rest of the family. On the square, the least worst map’s departure is spread almost evenly over the region’s area: the share it keeps inside a band is very nearly the band’s fraction of its worst departure, 29.3 per cent at three tenths and 48.7 at half. On the triangle and the hexagon the same holds to a point or two. On the thin ellipses it does not. The least worst map of the ellipse of 1 : 0.12 keeps 36.6 per cent at three tenths and 57.0 at half, because much of a thin region lies along its long axis, where the map’s departure is small and nearly constant.

That changes what the two ends of the family trade. On a compact region the least worst map is well behind the mean-departure map at every band below nine tenths — twelve points at seven tenths on the square — and it catches up only when the band is nearly wide enough to hold everything. On a thin region it is closer behind throughout, eleven points at seven tenths on the ellipse of 1 : 0.2 and seven on the thinnest, while the mean-departure map stalls: it keeps 82 per cent of the thinnest ellipse at six tenths and only 91 at nine. Neither end is good in the stretch between, and a map that gives the tail a little more weight than the mean and a good deal less than the worst case beats both.

So the practical rule has two numbers in it after all. The band, divided by the region’s least worst departure, says where on the family the answer lies. How evenly the least worst map spreads its departure — readable as the share it keeps at half its own worst case, 47 to 50 per cent for every compact region here and 51 to 57 for the thin ones — says whether that answer is one of the two ends or somewhere between.

Why the size of a region was never the variable #

It is worth being clear about why r works as well as it does for size and fails for shape, since both are properties of the region.

A conformal map’s scale factor satisfies Liouville’s equation: the Laplacian of ln k is fixed by the curvature of the sphere. Not every distortion can be asked for found that equation as the one condition a wished-for conformal distortion has to meet, and the trade-off is two lines is why a conformal map has only this one field to spend. Over a small region the curvature is nearly constant, so ln k is a harmonic function plus a fixed quadratic, and everything about a conformal map of the region — which harmonic function, which level — is chosen to fit that quadratic to the region’s boundary. Enlarging the region by a factor multiplies the quadratic by the factor squared and the harmonic part along with it, so the whole field is the same field, larger. Dividing by w* removes exactly that factor, and the second-order terms — the curvature not quite constant across 40° — are the three points left.

Shape does not scale out, because it decides which harmonic function fits. On a disc the quadratic is already constant on the boundary and nothing is added; on a square, the fit leaves the corners out; on a thin ellipse, it leaves the tips out and makes the long axis nearly flat. The spread of departure over the area is that fit’s shape, and no rescaling of the band can make a thin ellipse’s spread look like a square’s. Chebyshev’s map is the best at its worst and not on average found the worst-case map spending everything on its extremes; the distribution figure shows how much of the region each shape leaves near those extremes, and that is the second number.

Where the named criteria came from #

Airy’s criterion of 1861, a mean square, is p = 2; Chebyshev’s of 1856, the least worst case, is the limit of large p. Both were proposed as general principles — a best map for a region, full stop — and the argument between them, carried on through Kavrayskiy’s work in the twentieth century, has usually been conducted as an argument about purpose: a map for measurement wants a bounded error, a map for display a small typical one. Which projection is best sets out that framing, and every projection minimises something the observation under it: a criterion is a statement of what the map is for.

The tolerance band gives the argument a number to settle it with. For a compact region the band either lies below nine tenths of the region’s least worst departure, where Airy’s end is nearly right and the mean is better still, or above it, where Chebyshev’s is. For a thin one — a coastal strip, a long valley, a country the shape of Chile — neither of the classical criteria is the answer to a band in the middle of the range, and the answer that is has no name.

What each number was held to #

Size must drop out. The square and the ellipse of 1 : 0.2, solved at 10° and at 40°, must keep the same area at every ratio to within three and a half points, against a spread of about twenty between shapes. They agree to within 3.0.

The mean departure must be best, or within half a point of best, from three tenths to one half of the least worst case, on every region but the cap, whose family is one map at different scales. It is.

The large-power end must be the least worst case. On every region the map at p = 32 must come within half a per cent of the least worst departure any member reaches. The first run of the iteration failed this on the ellipse of 1 : 0.2: at large powers only a handful of samples carry weight, the reweighted fit lost its conditioning, and a step that made the map worse was taken anyway, sending the worst departure from 0.00185 to 2.76. With the weights normalised each round, the step allowed to shorten much further, and a step that improves nothing ending the iteration instead, the same ellipse reaches 0.00183 — Chebyshev’s value for it — and the square’s numbers are unchanged.

The finding must be there to be found. The thinnest ellipse must lose more than three times what the square loses under the two-exponent rule. It loses 6.6 points against 1.2.

Where the ten regions stop #

Ten shapes, one family each. Ellipses and regular polygons are the simplest shapes with a controllable elongation and a controllable corner count. A real territory has both at once, and a concave coast adds something neither has — a boundary that bends inward, where the map that keeps the most ground inside a tolerance would put the band’s edge somewhere else again.

Twelve exponents. The best of twelve is not the best of all, and between neighbouring powers on the doubling scale a finer search would gain a fraction of a point. The rule’s losses are measured against the best of these twelve and would be slightly larger against a continuum.

Area weighted evenly. Every share counts a square kilometre of the region the same, which the area weighting was a readership all along found is an assumption about who reads the map. A readership concentrated along a thin region’s long axis would favour the maps that are good there, which are the least worst map’s strength.

The band criterion itself is not solved here. Every comparison is within the exponent family. The earlier essay found the band criterion’s own search could beat the family at mid bands on the square, by as much as eleven points at half a per cent, and nothing here says whether that gap is larger or smaller on a thin region.

Still open: whether the spread of departure predicts the best exponent #

The distribution figure suggests a stronger rule than the one found. If the thin regions leave the ends of the family because their least worst map holds so much area at small departures, then the exponent that serves a band might be predictable from that map’s distribution alone — from the whole curve, not just its value at one half. The mean-departure map’s own distribution would be the other input, and the best exponent would be the member whose distribution crosses the band highest.

That would turn the exponent from a choice into a reading: solve the two ends, which a surveyor needs anyway to know the least worst departure, and read off where between them the band’s answer lies. Whether the members of the family interpolate their distributions smoothly enough for two ends to predict the middle, and whether the prediction holds on a region with a concave coast where neither end’s distribution is simple, are questions ten convex regions and twelve exponents cannot settle.