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

Enumeration — where it appears

Settling a question by listing every case rather than by an argument that covers them all. It answers what an argument may not reach — which of 18,564 choices of six constraints can be held exactly — and it answers only for the sizes small enough to list.

Named by 4 essays across 2 fields — each of them below, with the objects they name alongside it.

Every one of the cube's 384 nets, scored. All 384 spanning trees of the cube's face graph, unfolded and scored on the total separation their cuts leave: how far apart, in edge lengths, the two copies of each cut edge end up on the page. Every one of them is a valid net — no Platonic unfolding overlaps — and they range from 11.30 to 17.97, a factor of 1.59. The heuristic of unfolding outwards from a chosen face lands on the first of them, exactly.

What the net heuristic cannot find

Choosing a net by unfolding outwards from a face beats guessing, and the earlier measurement of that left its own limit unknown. Enumerated in full, the heuristic turns out to be exactly optimal on every solid small enough to check — rank one of 384 — and above that ceiling no sample can tell whether it still is, because eight hundred random nets never reach it.

families · Polyhedral Two pairs of ten are in every picture that can be drawn. London, New York, Tokyo, Sydney, Cape Town, drawn holding seven of their ten pairs' bearings exactly — the most any flat picture of five places can hold. Of the 15,360 ways of choosing seven pairs and a direction on each, 152 give a picture whose held arrows all point forwards, and London–Tokyo and Sydney–Cape Town, drawn heavy, appear in every one of them. The three pairs this particular choice leaves out are dashed. The price is paid by the rest: the worst, New York to Sydney, is 142.1° from the truth.

The pair that cannot be held is not the one the geometry names

Four world cities have exactly one pair of the six that no drawable picture holds, and the obvious candidate — the pair whose bearing there and bearing back disagree most — is not it. That quantity predicts something else exactly: whether a held pair's arrow has a forced direction. It separates the two groups with a clean gap and nothing in between. And at five places there is no forbidden pair at all, so the four-place case was a statement about how little room five constraints leave rather than about the map.

impossibility · Embedding Six of eighteen, where either kind alone can hold only five. London, New York, Tokyo, Sydney, drawn holding five distances exactly — the thick edges — and one bearing exactly, the arrow from London to New York. Six exact constraints, where five distances is the most any flat picture of four places can hold and five bearings is the most it can hold of those. All six are satisfied to 1.9 × 10⁻¹¹. The sixth is not free room found by luck: a picture is blind to where it sits, how it is turned and how large it is, distances see the last of those three and bearings see the second, and a picture holding both kinds is blind only to where it sits. Tokyo–Sydney, dashed, is the distance left out; the picture is wrong about it by 6.9%.

A mixed picture holds one more, and what it is holding is north

A flat picture of n places can hold 2n − 3 distances exactly, and it can hold 2n − 3 bearings exactly, and it is tempting to read one budget spent twice. It is not one budget. A distance cannot see which way the page is turned and a bearing cannot see how large it is drawn, so a picture constrained by both kinds is blind to one freedom instead of two — and holds 2n − 2. For four cities that is six of the eighteen available where either kind alone can hold five, and nothing whatever can hold seven: nought of 31,824 choices.

impossibility · Embedding A picture of four places, and the one constraint that is about the sheet. The best flat picture of four world cities, with the six distances it could be asked to hold drawn between the dots and the twelve bearings implied by them. Every one of those is a statement about how two places stand to one another, so every one is blind to where the whole picture sits on the paper. The cross is a constraint of a third kind: a stated position for one place — this dot goes here — which is two numbers and is blind to nothing. Adding both of them takes the largest independent set from 6 to 8.

Two numbers about the paper buy nothing about the places

A picture of n places has four freedoms, and a constraint holds only what its own blindness leaves: distances reach 2n − 3, bearings 2n − 3, the two together 2n − 2. Position is reached by neither, because nothing about how places stand to one another says where on the sheet to put them. A stated position does reach it, and two of them take the budget to 2n exactly. What the count conceals is that the extra two buy nothing: a largest picture holds six place constraints with the pins and six without, the same 7,596 of them, holding the same pairs at the same rates. The constraint system is a direct sum, and a pin is a different currency.

impossibility · Embedding

Named alongside it

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

BearingConstraintDegrees of freedomEmbeddingRigidityVerificationAzimuthCertificateCombinatoricsConvergence of meridiansCut locusEquidistance

All concepts