Six projections have one band and one refusal

The constant belongs to the projection, not to the problem ends with a table of three numbers and a question about them. The numbers are the exponents of three projections’ pass-depth curves — how fast the depth of the pass that decides when a near-optimal aspect set comes apart grows with how much of the projection’s own scale variation the region sees. They read 1.20, 1.30 and 0.68, and that essay named two candidate explanations for the spread and tested neither.

Six projections at nine latitudes each answer the question, and the answer is mostly that the question was built on a measurement that was not steady enough to hold it.

What a pass depth is, in one paragraph #

Choosing a projection’s aspect means choosing where its axis points, which is three numbers. The landscape the search walks on maps the objective that choice is made against and finds it neither a valley nor a basin but a set of basins separated by passes. The set of aspects that serve a region nearly as well as the best one is a single connected sheet when nearly as well is read loosely and several separated basins when it is read tightly, and where the valley breaks in two measures the threshold where it comes apart. The pass that fails first is not the one that was measured then found that the pass doing the separating is not the one between the two deepest basins but a shallow pair far from the optimum, and the first break is mostly its denominator separated the threshold into that pass’s own depth and the best score the region admits. The pass depth is that first part: the height a walk between two near-optimal aspects has to climb before the two stop counting as one answer.

Adding latitudes moved the quantity the question was about #

The first thing the extra measurements did was not settle anything. It moved the numbers the question was asked about.

The three exponents in the earlier account were each fitted through four points, one per latitude. That is one point more than a straight line needs, so a single awkward latitude moves the slope a long way and there is very little residual left over to say that it has.

The arithmetic is worth stating because it is the reason this essay exists. A power law fitted through n points has n − 2 degrees of freedom left, so four points leave two and nine leave seven. The standard error of a slope falls roughly as the square root of that, which puts a four-point fit at about twice the uncertainty of a nine-point one before any question of whether the points are well spread. And the points here are not evenly informative: the quantity on the horizontal axis is how much of its own pattern the region sees, which varies tenfold from the equator to 66° but does so unevenly, so four latitudes sample the range coarsely at one end.

Fitting the same three projections through five latitudes and then nine gives a different picture.

Robinson’s exponent is nearly unmoved — 1.197 at four latitudes and 1.159 at nine, a shift of three per cent — and that is consistent with its being the best-determined of the three even then, at $R^2$ 0.92 against Mollweide’s 0.95 and Winkel tripel’s 0.64. The two that move are the two the earlier reading rested on: Winkel tripel’s 1.296 falls to 1.159 and Mollweide’s 0.682 rises to 0.942. Both move inward, and the factor between the highest and the lowest falls from 1.90 to 1.23.

The four-latitude numbers were not wrong; they were reproduced here exactly, which is what makes the comparison a comparison rather than two measurements. They were under-determined, and the essay that quoted them said so about Winkel tripel and not about the spread.

That matters because the spread is the question. “What sets each projection’s curve” presumes the curves differ by more than the fits wobble. At nine latitudes most of that difference has gone.

Five of the six projections have the same exponent to within a quarter, which given the scatter inside each fit is a weak statement of difference rather than a strong one. Eckert IV at 0.995 and Mollweide at 0.942 differ by less than either differs from its own value fitted through a different set of latitudes. Robinson and Winkel tripel agree to three decimal places, which is a coincidence and should be read as one: it means only that nothing in the measurement separates them.

So the honest restatement is that there are not six curves to explain, and there were never three. There is one band of five, and one projection that does something else entirely.

This is a familiar shape on this ground and it is worth naming. The threshold is not a percolation took a count of pieces that had been described as rising to a transition, swept it finely, and found it rising and falling again — a quantity that had been given a name before it had been measured across enough of its own range. Where the valley breaks in two is the other direction: a prediction stated loosely and then tested over region sizes and projections, which held. The difference between the two is not the care taken but the number of independent cases the claim was asked to survive.

Cylindrical is a measurement, not a label #

The second candidate the earlier essay named was the one that had already caused trouble elsewhere: rotating a cylindrical projection’s aspect moves its map in a way that rotating a pseudocylindrical’s does not, and a cylindrical projection had refused an earlier collapse for what looked like that reason.

A label cannot be correlated with anything. Two of the projections here are called pseudocylindrical and two are not, which gives four points in two heaps and no way to say whether the heaps differ by more than the points inside them do. What is needed is the quantity the label stands for.

It is available directly. A projection is cylindrical exactly when its scale depends on latitude alone, so the share of its scale field’s variance that survives removing the latitude average is zero for a cylindrical map and positive for everything else. That number is continuous, it is computable from the projection with no region anywhere in it, and it is larger the further a map’s meridians bend away from straight.

The measurement also corrects something the label hides. A pseudocylindrical projection’s parallels are straight, but its meridians are not, so its principal scale does vary along a parallel and its longitudinal share is not zero: Robinson reads 0.080, Eckert IV 0.142, Mollweide 0.447 and the sinusoidal 0.606. Only a genuinely cylindrical projection reads zero. So the class the earlier essay proposed as a two-way split is really a continuum, and the four maps it would have put in one heap are spread across most of the range.

That is worth having on its own. Cylinders, cones and planes argues that the taxonomy of developable surfaces says almost nothing about the properties anybody chooses on, and the family is a symmetry, not a shape takes it further. This is the same point arriving in a different field: the useful version of cylindrical here is a number between zero and one, and the projections a taxonomy puts together sit at 0.08 and 0.61.

A correlation that survives its outlier and still predicts nothing #

With a number to correlate against, the first candidate can be tested too. Smoothness of the scale field is made definite the same way: the range of the field divided by its largest gradient, which is a length in radians over which the field turns over, so a large value is a gentle field.

Both candidates correlate, in the direction anybody would have guessed, and the better of them accounts for half the variance. The correlation survives removing the outlier, which is the first thing to check and is usually where an apparent relationship on six points dies.

It still says nothing, and the reason is that a correlation fitted on six points and reported on the same six points is not a claim anybody can act on. The claim the earlier essay actually wanted was that a projection’s curve can be predicted without running the aspect search at all — that is, for a projection the fit has never seen. That is a different question and it has a standard test.

All three predictors are worse than knowing nothing about the projection at all. Guessing the average of the other five lands 0.216 from the truth on average; the best of the three candidates lands 0.284. That is the whole answer to the question this essay was written to settle, and the answer is no.

It is not a surprise once the two previous sections are in hand. A predictor is only useful if the thing being predicted varies more than the prediction errs, and five of the six exponents lie inside a range of 0.24 while the predictors’ errors are larger than that. A relationship can be real — and a correlation of −0.72 that survives its outlier probably is — and still be unusable, because the quantity it explains half of has almost no variation left to explain.

The practical form is the one worth carrying. Anybody choosing a projection for a small region and wanting to know how far its near-optimal aspect set will drift before it fractures is better served by using the average exponent, about 1.0, than by computing anything about the projection’s scale field and adjusting.

An exponent of one has a reading, and it is a simple one. The pass depth then rises in proportion to how much of its own pattern the region sees, so doubling the span of the region — which roughly doubles what it sees at these sizes — doubles the height of the pass the set has to climb before it comes apart. The near-optimal set of a small region is fragile and the near-optimal set of a large one is robust, in that exact proportion, for five of the six projections here and to within the quarter that separates them.

That is a statement about how much freedom the cartographer has, which is what every projection minimises something and the aspect has three numbers, not one are both about from the other side. A robust near-optimal set means a placement can drift and still serve; a fragile one means the reported optimum is one of several that a slightly different search would have reported instead.

The projection that has no curve #

One of the six does not belong in any of the arithmetic above, and it is the most interesting result here.

The sinusoidal’s pass depth scatters between 0.13 and 0.20 at every latitude tried, and how much of the projection’s pattern the region sees runs from about one per cent at the equator to ten per cent at 66°, so the range of the supposed explanatory variable is ample. There is simply no relationship. Fitting a power law through it returns an exponent of 0.26 with an $R^2$ of 0.15, and both numbers are descriptions of noise rather than of a curve.

That makes the sinusoidal a counterexample to the framing rather than an outlying member of it. The collapse where the valley breaks in two found, and which the constant belongs to the projection narrowed to one projection at a time, is not a law with projection-dependent constants. It is a behaviour five of six projections show and one does not, and a sixth projection that declines is not a sixth curve.

Why it declines is a guess this essay does not test. The sinusoidal is the only one of the six whose meridians meet the pole at a point with no smoothing and whose scale runs away fastest at the edges of the sheet, and its longitudinal share is the largest of the six. But the projection with the second-largest share, the Hammer, has the tightest curve of any of them at $R^2$ 0.999. So whatever distinguishes the sinusoidal is not the quantity that came closest to explaining the others, and two candidates already failed the test that matters.

There is a second reading that deserves stating because it would make the result less interesting and cannot be ruled out here. The sinusoidal’s pass depths are not small — they average 0.14, the largest of the six alongside Eckert IV’s — so this is not a case of a signal buried under a floor. What varies little is the dependence rather than the quantity. If the sinusoidal’s aspect landscape had many shallow passes at every region size, the highest of them would be set by the number of passes rather than by the region, and a count that does not change with the region would produce exactly this picture. Nothing above counts passes, and the count is available from the same sweep, so this is a measurement not made rather than a possibility dismissed.

Where a pseudocylindrical puts its error is the essay about what the pole does to this family, and the sinusoidal is its sharpest member: the pole is a point, the meridians arrive there at their full curvature, and the angular deformation at high latitudes is the largest of any equal-area pseudocylindrical in the library. If any of that is the cause, it is a property of the family’s parameter rather than of the projection, which is the shape the continuation below takes.

What each number was compared against #

The earlier reading has to be reproducible before it can be revised. Fitted through the four latitudes that essay used, the three exponents come back as 1.197, 1.296 and 0.682 against the 1.20, 1.30 and 0.68 it reported. Nothing about the instrument has changed; only the number of latitudes has.

Each projection must fit its own curve better than the pooled fit does, or there are no separate curves and this essay is about nothing. Five of the six do, at $R^2$ 0.91 to 1.00 against a pooled 0.37. The sixth is the sinusoidal, at 0.15, and the failure is the finding rather than a defect.

The longitudinal share must be zero for a cylindrical projection and not for the others, or it is not measuring the class it stands in for. A Lambert cylindrical gives $5 \times 10^{-23}$ and an equirectangular the same to the last digit, which is rounding; the six here give 0.080 to 0.606.

And it must order the others in a way the eye agrees with. The sinusoidal, whose meridians are full sine curves, reads three times Robinson’s, whose meridians are nearly straight over most of the sheet. That ordering is not an input to anything; it is a consequence of taking the variance apart.

Every exponent is a fit through nine points at four region sizes each, and the scatter inside a fit is reported as its $R^2$ rather than hidden in a coefficient. Two of the six sit below 0.93, which is why the comparison above is made on the band of five rather than on any pair.

What six projections cannot settle #

Six is still few. A predictor that fails leave-one-out on six points has not been shown to be useless; it has been shown to be unproved. What can be said is stronger than that and is the thing this essay claims: the variation it would have to explain has shrunk to a factor of 1.26, so a predictor would have to be very good to be worth computing, and neither of these is.

The region shape is fixed. Every region here is a square of a stated span at a stated latitude, as in the measurements this continues. The pooled score abandons a region is the essay about what happens when the region stops being one box, and nothing here varies the shape.

The exponent is not the only thing a curve has. The coefficients differ by a factor of three across the six — 0.30 for the sinusoidal against 0.92 for Eckert IV — and nothing above asks what sets those, because a coefficient in a power law with a fitted exponent is not a quantity that can be compared across projections whose exponents differ.

And the pass depth is a mean over four region sizes. The first break is mostly its denominator is why that is legitimate — the depth is nearly constant across sizes below twenty degrees of span — and it is still an average, with a spread inside it that this essay carries as a scatter rather than as an error bar.

Still open: whether the sinusoidal is the edge of the field or outside it #

The sinusoidal has no pass-depth curve, and the obvious question is whether it is alone.

There are two shapes the answer could have. If it is one projection behaving oddly, the right response is to find what is odd about it — and the two candidates tested here do not, since the projection nearest it on both measures has the tightest curve of the six. If instead there is a whole class of projections whose near-optimal aspect sets fracture at a depth that does not depend on the region, then the band of five is the special case and the account has the shape of the argument backwards.

Distinguishing those needs projections chosen to sit on either side of whatever the sinusoidal has, which is exactly the thing not yet identified. A cheaper first step is available: the sinusoidal is the limit of a one-parameter family whose other members are ordinary pseudocylindricals, and running the same measurement along that parameter would say whether the curve disappears gradually or at a point. A curve whose exponent falls smoothly to nothing is a projection at the end of a range; one that is present at every parameter but the last is a projection outside it.

Whether the refusal is continuous in the family, what the parameter value is where it sets in if it is not, and whether any other named projection shows it, are questions six projections chosen for variety rather than for a family cannot answer.